Solution Manual for Water-Resources Engineering, 3rd Edition David A Chin by Ace Test Banks

Solution Manual for Water-Resources Engineering, 3rd Edition David A Chin

richard@qwconsultancy.com

1|Pa ge

Chapter 1

Introduction 1.1. The mean annual rainfall in Boston is approximately 1050 mm , and the mean annual evapotranspiration is in the range of 380–630 mm (USGS). On the basis of rainfall, this indicates a subhumid climate. The mean annual rainfall in Santa Fe is approximately 360 mm and the mean annual evapotranspiration is < 380 mm . On the basis of rainfall, this indicates an arid climate.

Chapter 3

Design of Water-Distribution Systems 3.1. Use Drinking Cooking Bathing Washing of cloths Washing of utensils Washing and clearing of houses and residences Flushing of water, closets etc. Lawn watering and Gardening Total

Consumption in L/d/person 5 5 55 20 10 10 30 15 150

3.2. (a) By graphical extension P2030 = 100, 000 (b) For arithmetic growth: P = kt + P0 where k=

61000 − 52000 P1990 − P1980 = = 900 10 10

Therefore P2030 = 900t + P1990 = 900(40) + 61000 = 97, 000 (c) For geometric growth: P = P0 ekt Therefore P1990 = P1970 ek(20) 61000 = 40000e20k k = 0.021 50

51 and hence P2030 = P1990 ek(40) = 61000e(0.021)(40) = 141, 298 (d) For declining growth: P = Psat − Ce−kt

(1)

where Psat = 100,000 and 1970 : t = 0, P = 40000 1990 : t = 20, P = 61000 Substituting 1970 population into Equation 1 gives 40000 = 100000 − Ce−k(0) = 100000 − C which gives C = 60, 000 Substituting 1990 population into Equation 1 gives 61000 = 100000 − 60000e−k(20) which gives k = 0.0215 Equation 1 can now be used to predict the 2030 population (t = 60) as P2030 = 100000 − 60000e−0.0215(60) = 83, 483 (e) Equations 3.11 and 3.12 give the logistic-curve parameters a and b as Psat − P0 100 − 40 = 1.5 = P0 40 1 P0 (Psat − P1 ) 1 40(100 − 52) b= ln = ln = −0.0486 Δt P1 (Psat − P0 ) 10 52(100 − 40)

The logistic curve for predicting the population is given by Equation 3.9 as P =

100, 000 Psat = bt 1 + ae 1 + 1.5e−0.0486 t

In 2030, t = 60 years and the population given by Equation 3.19 is P = 92, 487 people

(2)

52 3.3. According to Equation 3.10 – (a) 2P0 P1 P2 − P − 12 (P0 + P2 ) P0 P2 − P12 (2 × 25, 000 × 40, 000 × 70, 000) − (40, 000)2 × (25, 000 + 70, 000) 25, 000 × 70, 000 − (40, 000)2 = 80, 000

Psat =

According to Equation 3.11 a=

Psat − P0 80, 000 − 25, 000 = 2.2 = P0 25, 000

According to Equation 3.12 1 P0 (Psat − P2 ) ln t P1 (Psat − P0 ) 1 25, 000(80, 000 − 40, 000) = ln 10 40, 000(80, 000 − 25, 000) = −0.079

(b) Using Equation 3.9 Psat 1 + aebt 80, 000 = 1 + 2.2 × e−0.079t

P( ) =

Hence population after 10 years will be 66,251. 3.4. Diameter of single main pipe = 300mm = 0.3m Let the length of single pipe line be ‘L’ According to Equation 2.33– 2

Lv hf = f2gD

f Lv = 2g×0.3

53 Let v’ be the velocity of ﬂow in each parallel pipe and D’ be the diameter of each parallel pipe Again Equation 2.33 – 2

Lv hf = f2gD

Equating both the equations – 2

Lv Lv = f2gD =⇒ f2gD 2

=⇒ vv 2 = 0.3 D Total ﬂow in the single main pipe = sum of ﬂow in two parallel pipes =⇒ v × π4 × 0.32 = 2 × v × π4 × D 2 2

4Dt =⇒ vv 2 = 8.1×10 −3 t4

4D =⇒ 0.3 D t = 8.1×10−3 1 −3 5 =⇒ D = 2.43×10 = 0.2274m 4

Two pipes of 227.4 mm dia. may be used. 3.5. The NFF can be estimated by Equation 3.20 as NFFi = Ci Oi (X + P )i where the construction factor, Ci , is given by Ci = 220F

For the 5-story building, F = 1.0 (Table 3.3, Class 2 construction), and Ai = 5000 m2 , hence √ Ci = 220(1.0) 5000 = 16000 L/min where Ci has been rounded to the nearest 1000 L/min. The occupancy factor, Oi , is given by Table 3.4 as 0.85 (C-2 Limited Combustible, Office), (X + P )i can be estimated by the median value of 1.4, and hence the needed fire flow, NFF, is given by NFFi = (16000)(0.85)(1.4) = 19000 L/min This flow must be maintained for a duration of 5 hours (Table 3.6), hence the required volume, V , of water is given by V = 19000 × 5 × 60 = 5.7 × 106 L = 5700 m3 3.6. The needed fire flow for any building should not exceed 45,000 L/min for a duration of 10 hours .

54 3.7.

(i) Average daily water requirement = Average daily per capita demand × population = 150 × (4 × 5 × 4) = 12, 000 L = 12m3 An underground tank shall be provided to collect municipal water supply in the morning as well as evening, during 3 hrs each. This water shall be lifted to the overhead tank, say by pumping for 2 hrs in the morning and evening each. (ii) Capacity of underground tank Normally in shift supplies, an underground tank of 2/3rd of one day requirement & an overhead tank of 1/3rd of one day requirement may suﬃce. So let us provide an underground tank of 2/3 rd of one days requirement of 12000 L. So, Tank Capacity = 2/3 × 12000 = 8000 L = 8m3 Provide a tank of 4m × 2m × 1m(water depth) size with a free-board of 0.2 m Overall tank Size = 4m × 2m × 1.2m (iii) Capacity of Overhead Tank Let us provide overhead tank storage equal to 1/3rd of 1 day’s requirement. Capacity of overhead tank = 1/3 × 12000 = 4000 L = 4 m3 The size of the overhead tank may be decided as per site availability. (iv) Size of Rising Main Average daily water requirement to be pumped = 12000 L Pumping hrs allowing power failures = 4 hrs So, Pumping Rate = 12000/4 = 3000 L/h = 0.833 L/s Now, assume that the length of the rising main from the underground tank to the overhead tank is measured from the building plan is 20 m. Let us use 50 mm dia. rising main Let the equivalent length of all the pipe ﬁttings like elbows, gate valves etc. be 16 m (4 elbows × 3 m + 1 gate valve ×4m = 16 m) Total pipe length causing head loss = (20 + 16)m = 36m Let use 40 mm dia. Pipe for the rising main, to carry 0.833 L/s. The velocity in the pipe should be = 1000×0.833 = 0.663m/s π ×(0.04)2 4

Now, head loss HL /min (equation 3.23) 1.35

Q = 85.4 c1.35 D 4.37 H

1.35

= 85.4 1400.833 1.35 (4)4.37 = 4.88 × 10−4 m/min Total head loss in 36 m length of pipe = 36 × 4.88 × 10−4 = 0.018m (v) Design of pump capacity Rate of pumping = 0.833 L/s Head on the pump – Static head = 1.2 m (depth of underground tank)

55 Static head = 14 m (Assuming the O.H tank to be placed on roof with is 2m from the roof level, which is 12 m above the ground level) Frictional loss = 0.018 m Minimum residual head required at inlet = 1m (assumed) Total delivery head = 16.218 m B.H.P of pump,(Equation 2.131) 3 = γρhp η kW [Assuming, eﬃciency of pump and motor =0.6, γ = 9.81kN/m , Q = 0.833 × 10−3 m3 /s] −3 ×6.218

= 9.81×0.833×1 0.6 = 2.2 kW

(i) Average daily draft (Equation 3.2) = per capita average consumption × population = 150 × 1, 50, 000 l/d = 22.5 Ml/d (ii) Maximum daily draft (See Table 3.2) = (1.8 × 22.5) Ml/d = 40.5 Ml/d (iii) Maximum hourly draft of the maximum day (See Table 3.2) = (3.25 × 22.5) Ml/d = 73.125 Ml/d (iv) Fire flow = 2000 L/min (assume) (see Table 3.5) = 2.88 Ml/d 3.8. Average daily draft (Equation 3.2) = 200 × 1,00,000 = 20 Ml/d Maximum daily draft (See Table 3.2) = 1.8 × 20 = 36 Ml/d Fire flow = 2000 L/min (assume)(Table 3.5) = 2.88 Ml/d Coincident Draft = (36 + 2.88) Ml/d = 38.88 Ml/d 3.9. Let Q1 and Q2 be the discharges from tanks I and II. So, (Q1 +Q2 ) is the total incoming flow into the tank in the town. Head loss due to friction in Pipe AJ= 0.35Q1 2 Head loss due to friction in Pipe BJ= 0.46Q2 1.85 Head loss due to friction in Pipe JC= 0.6(Q1 +Q2 )1.95 =⇒ 0.35Q1 2 +0.6(Q1 +Q2 )1.95 = 100 − 85.37 = 14.63 ———— (1)

56 Again 0.46Q2 1.85 +0.6(Q1 +Q2 )1.95 = 96 − 85.37 = 10.63 ———– (2) Subtracting (2) from (1), we get 0.35Q1 2 -0.46Q2 1.85 = 4 =⇒ Q2 1.85 =0.7609 Q1 2 - 8.697——— (3) To get the value of Q1 and Q2 , solve Equation (3) by trial and substitute the values in Equation (1) & (2) to examine their satisfaction. First trial let, Q1 = 3.6 =⇒ Q2 = 1.086 Substituting the values of Q1 & Q2 in Eqn. (1) & (2) 4.536 + 12.197 = 16.733&0.536 + 12.197 = 12.733 (both on higher side) For second trial let, Q1 = 3.5 =⇒ Q2 = 0.776 Substituting the values of Q1 & Q2 in Eqn. (1) & (2) 4.288 + 10.20 = 14.488 & 0.287 + 10.20 = 10.488 (both on lower side) For third trial let, Q1 = 3.505 =⇒ Q2 = 0.793 Substituting the values of Q1 & Q2 in Eqn. (1) & (2) 4.299 + 10.306 = 14.606 & 0.299 + 10.306 = 10.606 (both equations are satisfied) So, Total inflow into the tank = (3.505 + 0.793) = 4.298 m3 /s 3.10. The required storage is the sum of three components: (1) volume to supply the demand in excess of the maximum daily demand; (2) fire storage; and (3) emergency storage. The volume to supply the peak demand can be taken as 25% of the maximum daily demand volume. Problem 3.7 gives the maximum daily flow rate as 2.50 m3 /s, hence the storage volume to supply the peak demand fluctuations over a day, Vpeak , is given by Vpeak = (0.25)(2.50 × 86400) = 54000 m3 The required fire flow, Qf , is calculated in Problem 3.7 as 0.467 m3 /s (= 28,029 L/min) and, according to Table 3.6, this fire flow must be maintained for at least 7 hours. The volume to supply the fire demand, Vfire , is therefore given by Vfire = (0.467 × 3600)(7) = 11768 m3 The emergency storage, Vemer , can be taken as the average daily demand, in which case Vemer = 200000 × 600 = 1.2 × 108 L = 120000 m3 The required volume, V , of the service reservoir is therefore given by V = Vpeak + Vfire + Vemer = 54000 + 11768 + 120000 = 185768 m3 The service reservoir should be designed to store about 185 800 m3 of water.

57 3.11. (a) For line AB: population served = 50000 + 20000 = 70000 people average demand = (70000)(0.6) = 42000 m3 /d = 0.486 m3 /s maximum daily demand = (0.486)(1.8) = 0.875 m3 /s maximum hourly demand = (0.486)(3.25) = 1.58 m3 /s fire demand = 15000 + 10000 = 25000 L/min = 0.417 m3 /s maximum daily + fire demand = 0.875 + 0.417 = 1.29 m3 /s design flow = max(1.29, 1.58) = 1.58 m3 /s For line BC: population served = 20000 people average demand = (20000)(0.6) = 12000 m3 /d = 0.139 m3 /s maximum daily demand = (0.139)(1.8) = 0.250 m3 /s maximum hourly demand = (0.139)(3.25) = 0.452 m3 /s fire demand = 10000 L/min = 0.167 m3 /s maximum daily + fire demand = 0.250 + 0.167 = 0.417 m3 /s design flow = max(0.417, 0.452) = 0.452 m3 /s (b) For the steel transmission pipe (k = 1 mm, D = 1200 mm): π π A = D2 = 1.22 = 1.131 m2 4 4 1.58 = 1.40 m/s VAB = 1.131 (1.40)(1.2) VAB D = = 1.68 × 106 ReAB = ν 10−6 1/1200 5.74 5.74 k/D 1 √ = −2 log + + = −2 log 0.9 3.7 3.7 (1.68 × 106 )0.9 fAB ReAB fAB = 0.0191 0.452 = 0.400 m/s VBC = 1.131 (0.400)(1.2) VBC D = = 4.80 × 105 ReBC = ν 10−6 1/1200 5.74 5.74 k/D 1 √ = −2 log + + = −2 log 0.9 3.7 3.7 (4.80 × 105 )0.9 fBC ReBC fBC = 0.0196 Applying the energy equation between A and B, neglecting minor losses, gives 2 V2 pB LAB VAB + hp = + AB + zB D 2g γ 2g 2 1.402 5000 1.40 550 + hp = + + 20 0 − (0.0191) 1.2 2(9.81) 9.79 2(9.81)

0 − fAB

58 which gives hp = 84.2 m. The required speciﬁc speed, ns , (where ω = 1200 rpm = 125.7 rad/s) is therefore given by 1

ns =

ωQ 2 3

(ghp ) 4

(125.7)(1.58) 2 3

(9.81 × 84.2) 4

= 1.03

This is a centrifugal pump . Applying the energy equation between B and C, neglecting minor losses, gives 2 V2 pC LBC VBC + hp = + BC + zC D 2g γ 2g 7000 0.4002 480 0.4002 0 − (0.0196) + hp = + + 20 1.2 2(9.81) 9.79 2(9.81)

0 − fBC

which gives hp = 69.97 m. The required speciﬁc speed, ns , is therefore given by 1

ns =

ωQ 2 3

(ghp ) 4

(125.7)(0.452) 2 3

(9.81 × 69.97) 4

= 0.63

This is a centrifugal pump . (c) For the storage reservoir, taking the daily service storage as 25% of the maximum daily demand volume and noting that 10000 L/min fire flow is to be maintained for 3 h, and 15000 L/min for 4 h, gives service storage = (0.25)(1.8)(42000) = 18900 m3 fire storage = (10)(60)(3) + (15)(60)(4) = 5400 m3 emergency storage = 42000 m3 required storage = 18900 + 5400 + 42000 = 66300 m3

3.12. Let us assume the following ﬁxture loads of the various ﬁxtures: (See table 3.12) Fixture Water closet Wash basin Bath taps with Kitchen Sink Total for each flat

WSFU 2.0 1.0 4.0 1.0 8.0

The maximum ﬂow rate among the above mentioned ﬁxtures pertain to both taps with showers (Table 3.10)

59 Let it be 15 L/min = 0.25 L/s Gross demand of the building = (2 flats on each floor× no. of floors × 0.25) L/s = (2 × 7 × 0.25) L/s = 3.5 L/s On fig., the main down take pipe portions between various floors are marked from (1) to (7). the branch lines from top most floor arte numbered starting from 8 onwards. Left side branch is marked first starting from extreme end, and continuing the numbers on right side branch from extreme end, as shown. The pipes are thus, numbered from 8 to 11on the left side, and from 12 to 15 on right side on 6th floor. The pipes on lower floors are similarly numbered. The probability of simultaneous use is worked out from figure 3.13 Thus, for pipe no. 1, total load of 2 × 7 × 8 WSFU=112 WSFU is used to read out the probable demand as 150 L/min i.e. 2.5 L/s. As the head on the lower floors is high, it will tend to draw all the water, and hence larger sized pipes will have to be used in upper floors, and the design shall be done for upper floors to balance the available static head up to the remotest point in the branches by accounting their frictional resistance also. The no. of upper flats for which the frictional resistance of branch lines is to be added is given by the ratio of the probable demand and gross demand. Pipe Load in Probable No. Fixture flow in Units L/s (Fig. 3.13) 1 112 2.5 2 96 2.3 3 80 2.25 4 64 2.08 5 48 2 6 32 1.5 7 16 1.3 8 1 0.13 9 2 0.12 10 6 0.3 11 8 0.3 The ratio in this case is = 2.5 3.5 = 71 The total no. of flats for which the branch frictional losses are to be added = 14 × 0.71 = 9.94 Hence the frictional loss of branch lines of both sides on top 5 floors (6th to 2nd floors) shall be computed. Friction loss of branch pipes on lower floors can be ignored. While designing the pipes, we will have to work out the head loss for each pipe as to work out the residual pressure at each of the off take points. .35

Loss of head in m/min = 85.4 (140)(Q) 1.35 D 4.37 ————————– (1)

61 Pipe

Designed Load

Probable

location Pipe No. on pipe demand in

Pipe

Assumed Assumed

length pipe dia. equivaalent

Corresponding from

In mm

Equivalent

Total

Head

Cumulative Velocity Total

Remarks Com-

length of

pipe

loss in

head

pare

in m

pipe =

avail-

with col. 14

Q/A = Q π d2 4 m/sec

able

length of

fittings col. 5 length pipe

pipe

X col. 7 in m col. 5+ in m

in m

Fixture to Loading

Fig.

fittings as

units

11.17

%age of

col.

Read from Fig. in m

col. 5

in m

11.15 in l/s(Q)

Table 11.6

units of col. 3

loss in

8 per min cal- col. 10 culated from Xcol. 9 equatio 1

Top

112

2.5

5.4

25%

1.35

6.75

0.00118

0.007954

1.27324

col.

in m

15 Head

loss

floor

branch

lines

main

on left side = 0.022 m.

The

same head loss will

occur

branches

right side, due to similarity Branch L 8

0.13

2.0

100%

2.0

4.00

0.00043

0.001722

0.009676

Branch L 9

0.17

1.5

100%

1.5

3.00

0.00071

0.002121

0.011797

Branch L 10

0.3

3.0

100%

3.0

6.00

0.00202

0.012131

0.023928

Branch L 11

0.3

1.5

100%

1.5

3.00

0.00202

0.006066

0.029994

0.02204

0.052033

Branch R 12 to 15

Same as in pipes 8 to 11

0.413803

0.05

Head

loss

almost

is bal-

anced 5th floor

1.25

3.0

25%

0.75

3.75

0.00097

0.003634

0.055667

main Branch L

Same as in pipes 8 to 11

0.02204

0.077707

Branch R

Same as in pipes 8 to 11

0.02204

0.099746

0.002405

0.102151

4th floor

3.0

25%

0.75

3.75

0.00064

0.1

0.15

main Branch L

Same as in pipes 8 to 11

0.02204

0.124191

Branch R

Same as in pipes 8 to 11

0.02204

0.14623

0.005867

0.152097

0.02204

0.174137

0.18

0.018839

0.192976

0.20

3rd floor

0.9

3.0

25%

0.75

3.75

0.00156

main Branch L 2nd

Same as in pipes 8 to 11 48

0.8

3.0

50%

1.5

4.5

0.00419

Friction losson

floor

right side pipes

main

of 3rd floor and of lower floors is ignored

1st floor

0.6

3.0

100%

3.0

6.0

0.00729

0.043733

0.236709

0.24

0.4

3.0

100%

3.0

6.0

0.00344

0.020656

0.257365

0.26

main Ground floor main

3.13. The design calculations for this problem are summarized in Table 3.1. 3.14. From the given data: Qref = 200 L/min = 0.00333 m3 /s, L1 = 20 m, L2 = 5 m, Δz1 = 2 m, Δz2 = 3 m, p0 = 380 kPa, and p2 = 240 kPa. For galvanized iron, ks = 0.15 mm = 1.5 × 10−4 . From the supply pipe to the ﬁrst ﬂoor: p1 V12 L1 V12 p0 V02 + + z0 = + + z 1 + f1 γ 2g γ 2g D1 2g

62 where π 2 D = 0.7854D2 4 2Qref 2(0.00333) 0.008487 V1 = = = A1 0.7854D2 D2 D V1 D 8487 0.008487 Re1 = = = 2 −6 ν D 10 D 0.25 0.25 f1 = 2 = 2 4.054×10−5 0.9 1.5×10−4 5.74 + 0.001671D log log + 8487 0.9 D 3.7D ( D ) A1 =

Combining the above equations and substituting known quantities yields p1 = 36.02 − γ

log

0.25 4.054×10−5 + 0.001671D0.9 D

7.342 × 10−5 D5

(1)

From the ﬁrst to the second ﬂoor, p2 V22 L2 V22 p1 V12 + + z1 = + + z 2 + f2 γ 2g γ 2g D2 2g where π 2 D = 0.7854D2 4 Qref 0.00333 0.004244 V2 = = = 2 A2 0.7854D D2 0.004244 V2 D 4244 D = Re2 = = 2 −6 ν D 10 D 0.25 0.25 f1 = 2 = 2 −5 4.054×10 0.9 1.5×10−4 5.74 + 0.003118D log log + 4244 0.9 D 3.7D ( D ) A2 =

Combining the above equations and substituting known quantities yields p1 = 27.51 + γ

log

0.25 4.054×10−5 + 0.003118D0.9 D

4.590 × 10−6 D5

(2)

Solving Equations 1 and 2 for D and taking the next larger available diameter yields D = 0.0508 m = 2 in. as the required pipe diameter. For this diameter, the actual pressure on the second floor (p2 ) is 262 kPa and the pressure on the first floor (p1 ) is 295 kPa .

Pipe AB BB’ B’C’ C’F’ C’D’ D’E’

Starting Head (m) 38.82 22.33 23.12 20.46 20.46 16.44

Flow (L/min) 409 144 144 45 45 45

Length (m) 17.0 2.7 2.5 46.0 4.0 46.0

Diam (mm) 64 51 51 25 51 25

Velocity (m/s) 2.12 1.17 1.17 2.62 0.37 1.52

Fitting Length (m) 5.12 4.66 3.05 1.52 2.13 1.52

Total Length (m) 22.12 7.36 5.55 47.52 6.13 47.52

Friction Loss (m) 1.37 0.21 0.16 5.15 0.02 5.15

Other Losses (m) 15.12 0 0 0 0 0

Elev Diﬀ (m) 0 −1.0 2.5 0 4.0 0

Table 3.1: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths Terminal Head (m) 22.33 23.12 20.46 15.31 16.44 11.29

Terminal Pressure (kPa) 216 226 200 146 161 109

Chapter 4

Fundamentals of Flow in Open Channels 1.5 4.1. n(slope)= H V = 1 = 1.5 1 Bed slope, s0 = 1000

Area of section, A= 40m2 Chezy’s constant, C= 60 For the most economical section b + 2nd = d (n2 + 1) 2 b + 2 × 1.5d =d 2

(1.52 + 1)

b = 0.606d But area of trapezoidal section, A=

2 × 0.606d + 2 × 1.5d b + (b + 2nd) ×d= × d = 2.106d2 2 2

But

A = 40 40 = 4.36m d = 2.106 & b = 2.64m Discharge for most economical sectionHydraulic Radius, R= d2 = 4.36 2 =2.18m √ 1 Discharge, Q= AC Rs0 = 40 × 60 2.18 × 1000 = 112.06m3 /s 64

65 4.2. Distance(mm) below free surface 0 50 100 150 200 250

Pressure of water(mm) 0 51 104 159 216 275

Coeﬃcient of pressure (k) value 1.02 1.04 1.06 1.08 1.1

Pressure= kρgh (N) 499.8 1019.2 1558.2 2116.8 2695

All stage coefficients are greater than 1 (unity) So flow pressure will be concave in curvature φ = 0.275×1000×9.8 Pressure coefficient k = ρgh 1000×9.8×0.250 = 1.1

Wall

O 499.8 0.050 1019 0.10

Actual pressure distribu on

1558.2 0.150

2116.8 0.20 0.250

2695 P’

Considering straight line segment instead of curvature F= 12 × 499.8 × 0.5 × (499.8 + 1019.2) × 12 × .05 + (1019.2 + 1558.2) ×

1 1 1 2 × 0.05 + (2116.8 + 1558.2) × 2 × 0.05 + 2 × .05 (2116.8 + 2695) = 327.075N/m length

of the wall Hydrostatic distributed total pressure

66 ṕ=kρgh(considering k=1) Force = 12 × P × h = 12 × 1000 × 9.8 × .250 = 306.25N/m Therefore, excess pressure=(327.08−306.25)=20.83N/m 4.3. Given U=1+2(y/h)1/2 h V=1/h 0 {1+2( hy )1/2 }dy 3

= 1/h[h + 4h 21 ] 3h 2

At h=3.5, V=8.12 m/s h α = 1/u3 h 0 u3 dy Now, h 3 0 u dy = 7h+14 = 38.5

α = (8.1231)×3.5 × 38.5 = 0.02 h β = v21h 0 u2 dy (Equation 4.8) h 2 h y 12 y 0 u dy = 0 (1 + 4( h ) + 4( h )) = 36.73 m/s β=

1 × 36.73 = 0.159 (8.12)2 ×3.5

4.4. The shear stress, τ0 , on the perimeter of the channel is given by τ0 = γRS0

(1)

From the given data b = 5 m, y = 1.8 m, m = 1.5, and the geometric properties of the channel are A = by + my 2 = 5(1.8) + 1.5(1.8)2 = 13.86 m2 P = b + 2 1 + m2 y = 5 + 2 1 + 1.52 (1.8) = 11.49 m 13.86 A = = 1.21 m R= P 11.49 From the given data, τ0 = 3.5 N/m2 , and since γ = 9790 N/m2 , Equation 1 gives the maximum allowable slope, S0 , as 3.5 τ0 = = 0.00030 S0 = γR (9790)(1.21) For the excavated channel, ks = 3 mm = 0.003 m, and ν = 1.00 × 10−6 m2 /s at 20◦ C. Substituting these data into Equation 4.38 gives the ﬂow rate, Q, as Q = −2A 8gRS0 log10 Q = −2(13.86) = 17.2 m3 /s

ks 0.625ν + 3√ 12R R 2 8gS0

8(9.81)(1.21)(0.00030) log10

0.625(1.00 × 10−6 ) 0.003 + 12(1.21) (1.21) 32 8(9.81)(0.00030)

67 Therefore, for the given ﬂow depth restrictions in the channel, the ﬂow capacity of the channel is 17.2 m3 /s . 4.5. BZ8/3 =

5/3

[(1+ Zh )h] B B √ 2/3 [1+2 Z 2 +1 h/B]

Given data Q= 60 m3 /s h=2.5 m m=2 B=3.92 m Maximum shear stress at bed (According to Equation 5.16) τbm = ρghs = 1000 × 9.81 × 2.5 × 1 × 10−3 P a = 24.525 Pa According to Equation 5.17, τ sm = ks × τbm = (0.802 × 24.525)P a = 19.66 Pa ks = 0.066×2 + 0.67 = 0.802 (if m>1.5) (Equation 5.18) 4.6. Hydraulically rough ﬂow conditions occur in open channels when u∗ ks ≥ 70 ν

(1)

gRSf

(2)

where u∗ =

Equation 4.46 can be rearranged and put in the form n 0.039

ks =

= 2.84 × 108 n6

(3)

Substituting Equations 2 and 3 into Equation 1 and noting that ν = 1.00 × 10−6 m2 /s at 20◦ C and g = 9.81 m/s2 yields √ 9.81 RSf × 2.84 × 108 n6 ≥ 70 1.00 × 10−6 which simpliﬁes to n6

RSf ≥ 7.9 × 10−14

(4)

From the given data: b = 5 m, S0 = 0.05% = 0.0005, and by deﬁnition: R=

by 5y A = = P 2y + b 2y + 5

(5)

68 Equation 4, can be combined with Equation 5 to give the following condition for fully turbulent flow, 5y 6 (0.0005) ≥ 7.9 × 10−14 (0.013) 2y + 5 This condition is satisfied when y ≥ 0.683 m . 4.7. The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as 0.625ν ks + Q = −2A 8gRS0 log10 12R R 32 √8gS0

(1)

where the following variables are known: y = 2.20 m S0 = 0.0006 ks = 2 mm = 0.002 m ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 3.6y + 2y 2 = 3.6(2.20) + 2(2.20)2 = 17.6 m2 R=

3.6(2.20) + 2(2.20)2 3.6y + 2y 2 √ = √ = 1.31 m 3.6 + 2 5y 3.6 + 2 5(2.20)

Substituting these variables into Equation 1 yields Q = 34.0 m3 /s . Since y = 2.20 m corresponds to A = 17.6 m2 , then V = 34.0/17.6 = 1.93 m/s . The Manning’s equation gives the average velocity, V , as V =

1 2 12 R 3 S0 n

Table 4.2 indicates that a mid-range roughness coeﬃcient for concrete is n = 0.015. The average velocity given by the Manning equation is V =

2 1 1 (1.31) 3 (0.0006) 2 = 1.96 m/s 0.015

and the corresponding ﬂow rate, Q, is Q = AV = (17.6)(1.96) = 34.5 m3 /s Hence, in this case, the Darcy-Weisbach and Manning equations give the similar results . 4.8. The Darcy-Weisbach uniform-ﬂow equation is given by Equation 4.38 as ks 0.625ν + Q = −2A 8gRS0 log10 12R R 32 √8gS0

(1)

69 where the following variables are either known or can be expressed in terms of the uniform-ﬂow depth, y: S0 = 0.0001 ks = 1 mm = 0.001 m Q = 18 m3 /s ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 5y + 2y 2 R=

5y + 2y 2 √ 5 + 2 5y

Substituting these variables into Equation 1 and solving for y yields y = 2.19 m . Check u∗ ks /ν and R/ks to determine the state of the ﬂow and the validity of the Manning equation. Taking y = 2.19 m gives R = 1.39 m and √ (9.81)(1.39)(0.0001)(0.001) u∗ ks gRS0 ks = = = 37 ν ν 1.00 × 10−6 1.39 R = 1390 = ks 0.001 Therefore, since 5 ≤ u∗ ks /ν ≤ 70 (i.e., 5 ≤ 37 ≤ 70) then according to Equation 5.19 the ﬂow is in transition . Since u∗ ks /ν < 70 (i.e., 37 < 70) and R/ks > 500 (i.e., 1390 > 500), then the Manning equation is not valid . 4.9. Comparing the Manning and Darcy-Weisbach equations 1 R6 8g = f n which gives

√ 1 1 1 1 1 f 2 R6 f 2 R6 fR6 = = n= √ 8.86 8g 8(9.81)

If the friction factor, f , is taken as a constant, the above relation indicates that n will be a 1 function of the depth (since R is a function of the depth). If f ∼ R− 3 , n would be a constant in the above equation. So the answer to the question is no . 4.10. A=(b+my)y= (2.5 + 2 × .9) .9 = 3.87m2 P=2 (m2 + 1) y + b=6.52m R=A/P=3.87/6.52=0.6m (i) τavg = ρgRs = 1.77Pa (ii) u∗ = (gRs) = 0.042m/s

70 Now, u∗νks = 126 > 70 Since this value is greater than 70 hence the boundary is hydrodynamically rough. (iii) Chezy’s constant=

8g f (Equation 4.40)

√1 =-2log ( KS )=-1.238 12R f

f=0.652 Therefore=10.97 1

Manning’s coeﬃcient= n1 R 6 (Equation 4.41) n=1.406 4.11. From the given information,

n = 0.039d 6

where d is in m. In this case, d = 30 mm = 0.030 m, and a 70% error in d is 0.7(0.030) = 0.021 m. Hence, d = 0.030 m ± 0.021 m. Hence, the “best estimate” of n, denoted by n̄, is given by 1

n̄ = 0.039(0.030) 6 = 0.022 The lower estimate of n, nL , is given by 1

nL = 0.039(0.030 − 0.021) 6 = 0.018 and the upper estimate of n, nU , is given by 1

nU = 0.039(0.030 + 0.021) 6 = 0.024 The maximum percentage error in estimating n is therefore given by error =

0.022 − 0.018 × 100 = 18% 0.022

4.12. According to Equation 4.45, n 1

ks6

√1 8g

R ks

1 6

2.0 log 12 kRs

Let y=

n 1

ks6 R x= ks

(1)

71 and taking g = 9.81 m/s2 , Equation 1 can be written 1

√ 1

0.1129x 6 = y= 2.0 log(12x) 2.0(log 12 + log x) 8(9.81)

(2)

0.1129x 6 0.1129x 6 = = 2.0(log 12 + 0.4343 ln x) 2.158 + 0.8686 ln x

(3)

The minimum value of n/ks2 (= y) occurs when dy/dx = 0, where 5

(2.158 + 0.8686 ln x)( 16 × 0.1129x− 6 ) − (0.1129x 6 )(0.8686x−1 ) dy = =0 dx (2.158 + 0.8686 ln x)2 which yields x = 33.63 and substituting into Equation 3 yields y = 0.0389 1

Therefore, under fully-rough ﬂow conditions, the minimum value of n/ks6 (= y) is 0.0389, or approximately 0.039 . 1

When n/ks6 diﬀers by 5% from 0.039, n 1

= 1.05(0.039) =

ks6

√1 8g

R ks

1 6

2.0 log 12 kRs

0.04095 =

0.1129x 6 2.158 + 0.8686 ln x

which yields x = 6 or

281

1 6

Therefore, n/ks is within 5% of 0.039 when 6≤

R ≤ 281 ks

It is noteworthy that this range is narrower than suggested by Yen (1991) and Hager (1999). The reason for this is that the constant value they assumed is a bit higher than 0.039. 4.13. From the given data: b =5 m, m = 3, L = 100 m, z1 = 24.01 m, z2 = 23.99 m, Z1 = 25.01 m, Z2 = 24.95 m, Q0 = 15 m3 /s, ΔQ = 2 m3 /s.

72 (a) From the given data: y1 = Z1 − z1 = 25.01 m − 24.01 m = 1.00 m y2 = Z2 − z2 = 24.95 m − 23.99 m = 0.96 m P1 = b + 2 1 + m2 y1 = 5 + 2 1 + 32 (1.00) = 11.32 m P2 = b + 2 1 + m2 y2 = 5 + 2 1 + 32 (0.96) = 11.07 m A1 = by1 + my12 = (5)(1.00) + (3)(1.00)2 = 8.000 m2 A2 = by2 + my22 = (5)(0.96) + (3)(0.96)2 = 7.565 m2 Sf = −

Δy + Δz + Δ

V2 2g

L 2

=−

Q (0.96 − 1.00) + (23.99 − 24.01) + 2(9.81)

1 − 812 7.5652

100

= 0.006 − 9.422 × 10−6 Q2

A1 8.000 = 0.7067 m = P1 11.32 A2 7.565 = 0.6834 m R2 = = P2 11.07 0.7067 + 0.6834 R 1 + R2 = = 0.6951 m R̄ = 2 2 8.000 + 7.565 A1 + A2 = = 7.783 m2 Ā = 2 2 R1 =

According to the Manning equation, 2

(7.783)(0.6951) 3 ĀR̄ 3 Sf = 0.006 − 9.422 × 10−6 Q2 n= Q Q which gives n=

6.107 0.006 − 9.422 × 10−6 Q2 Q

(1)

(b) Since Q = 15±2 m3 /s, Equation 1 gives Q (m3 /s) 13 15 17

n 0.0319 0.0254 0.0206

ks =

6 n × 1000 0.039 (mm) 299 76 22

Based on these results, Manning’s n is in the range 0.021–0.032 and the roughness height is in the range of 22–300 mm .

73 (c) From the derived data, 2/15 ΔQ/Q = = 0.52 Δn/n (0.0319 − 0.0254)/0.0254 ΔQ/Q 2/15 = = 0.71 Δn/n (0.0254 − 0.0206)/0.0254 2/15 ΔQ/Q = = 0.045 Δk/k (299 − 76)/76 2/15 ΔQ/Q = = 0.187 Δk/k (76 − 22)/76 Based on these results, the flows are much more sensitive to to specification of Manning’s n than specification of the roughness height. The relative sensitivity to Manning’s n is in the range of 0.52–0.71 , while the relative sensitivity to the roughness height is 0.05–0.19 . 4.14. For fully-turbulent flow conditions, u∗ ks > 70 ν

(1)

where u∗ is given by Equation 4.30 as u∗ =

τ0 = gRS0 ρ

(2)

Combining Equations 1 and 2 gives √

or ks

gRS0 ks > 70 ν

70ν RS0 > √ g

Taking ν = 1.00 × 10−6 m2 /s (at 20◦ C), and g = 9.81 m/s2 yields the turbulence condition ks

70(1.00 × 10−6 ) √ RS0 > 9.81

which simpliﬁes to ks

RS0 > 2.2 × 10−5

For the given trapezoidal channel, ks = 3 mm = 0.003 m, S0 = 0.1% = 0.001, b = 3 m, m = 2, and for a ﬂow depth y, R=

A by + my 2 3y + 2y 2 3y + 2y 2 √ √ = = = P 3 + 4.472y b + 2y 1 + m2 3 + 2y 1 + 22

74 For turbulent ﬂow,

0.003

ks RS0 > 2.2 × 10−5 3y + 2y 2 (0.001) > 2.2 × 10−5 3 + 4.472y

which requires that y > 0.056 m Therefore, flow conditions are fully turbulent when the depth of flow exceeds 0.056 m = 5.6 cm . At this minimum flow depth, 3(0.056) + 2(0.056)2 = 0.0536 m 3 + 4.472(0.056) 0.0536 m R = 17.9 = ks 0.003 m R=

Since R/ks is within the range for n/ks6 to be assumed constant, using the Manning equation is appropriate . 4.15. Data given, Width of the channel, b=6m Depth of flow before jump,y1 = 0.8m Discharge, Q= 20m3 /s Discharge per unit width, q= Qb = 3.33m3 /s/m Let the depth of flow after jump=y2 Depth of flow after the jump is given by (Equation 4.139), y2 = − y21 + = − 0.8 2 +

2 2×3.332 ( 0.84 + 9.81×0.8 ) = 1.328m

2 y 2q 2 ) ( 41 + gy 1

Loss of energy per kg of water due to hydraulic jump is given by3

(1.328−0.8) 2 −y1 ) = 4×0.8×1.328 =0.0346m-kg/kg Acording to Equation 4.146, E = (y4y 1 y2

4.16. From the given data: d = 15 cm = 0.15 m, Sx = 2% = 0.02, S0 = 1% = 0.01, and ks = 1 mm = 0.001 m. Assume ν = 10−6 m2 /s. The transverse slope corresponds to m = 50. (a) The following geometric characteristics of the channel can be derived from the given data: T = dm = (0.15)(50) = 7.5 m 1 1 A = dT = (0.15)(7.5) = 0.5625 m2 2 2 P = d + (d2 + T 2 ) = 0.15 + (0.152 + 7.52 ) = 7.651 m 0.5625 A = = 0.07352 m R= P 7.651

75 Using these data, the Darcy-Weisbach equation gives Q = −2A 8gRS0 log10 = −2(0.5625)

ks 0.625ν + 3√ 12R R 2 8gS0

8(9.81)(0.07352)(0.01) log10

0.625(10−6 ) 0.001 + 12(0.07352) (0.07352) 32 8(9.81)(0.01)

= 0.792 m3 /s (b) Using the Manning equation: 1

n = 0.039ks6 = 0.039(0.001) 6 = 0.0123 1 2 2 1 1 1 (0.5625)(0.07352) 3 (0.01) 2 = 0.803 m3 /s Q = AR 3 S02 = n 0.0123 (c) The following parameters are used to check the validity of the Manning equation: R 0.07352 = 74 = ks 0.001 ks RS0 = (0.001) (0.07352)(0.01) = 2.7 × 10−5 √ Since 4 < R/ks < 500 and ks RS0 > 2.2 × 10−5 , Manning’s equation is valid . (d) The discrepancy in Q calculated by the Darcy-Weisbach and Manning equations is 1

0.803 m3 /s − 0.792 m3 /s = 0.011 m3 /s. This is due to the approximation that n = 0.039ks6 . 4.17. From the given data: Q = 1.8 m3 /s, m = 2, n = 0.025, and S0 = 0.1% = 0.001. (a) Size the channel to accommodate the design flow under normal conditions. Assuming that the flow in the channel can be described by the Manning equation (i.e. fully turbulent) 1 2 1 (1) Q = AR 3 S02 n Since the lengths of the channel sides are equal to the bottom width, b, then the flow depth, y, is related to the bottom width by the relation y=√

b b =√ = 0.447b 2 1+m 1 + 22

The geometric properties of the channel are A = by + my 2 = b(0.447b) + (2)(0.447b)2 = 0.847b2 P = 3b R=

A 0.847b2 = = 0.282b P 3b

(2)

76 Substituting into the Manning equation, Equation 1, gives 1.8 =

2 1 1 (0.847b2 )(0.282b) 3 (0.001) 2 0.025

which yields b = 1.67 m According to Equation 2 the depth of ﬂow is given by y = 0.447(1.67) = 0.746 m The required channel is to have a bottom width of 1.67 m, side slopes of 2:1 (H:V), and a depth of at least 0.746 m. (b) Let y be the depth of ﬂow when the average shear stress, τ , on the channel lining is equal to the critical shear stress, τc = 4.0 Pa. The channel lining then becomes unstable and the geometric properties of the channel are A = by + my 2 = 1.67y + 2y 2 P = b + 2 1 + m2 y = 1.67 + 2 1 + 22 y1.67 + 4.47y R=

A 1.67y + 2y 2 = P 1.67 + 4.47y

The average shear stress, τ , on the perimeter of the channel is given by τ = γRS0

(3)

where γ = 9790 N/m3 . The channel lining is unstable when τ = τc = 4.0 Pa, and Equation 3 gives 1.67y + 2y 2 (0.001) 4.0 = (9790) 1.67 + 4.47y which yields y = 0.625 m Therefore, whenever the ﬂow depth exceeds 0.625 m, the channel lining becomes unstable. In terms of ﬂow, the Manning equation gives Q=

1 2 1 AR 3 S02 n

2 1 1 1.67(0.625) + 2(0.625)2 3 2 [1.67(0.625) + 2(0.625) ] = (0.001) 2 = 1.27 m3 /s 0.025 1.67 + 4.47(0.625) Therefore, whenever the ﬂow rate exceeds 1.27 m3 /s , the channel lining becomes unstable. An alternative lining should be used if the channel is to accommodate the design ﬂow of 1.8 m3 /s.

77 4.18. From the given data, b = 3 m, m = 2, S0 = 0.001, and n = 0.015. (a) For 4 < R/ks < 500, Manning’s n and the roughness height (= equivalent sand roughness), ks , are related by n 1 = 0.039 ks6 0.015 = 0.039 1 ks6 which yields ks = 0.00324 m = 3.24 mm. (b) Manning’s n can be assumed to be approximately constant for 4 < R/ks < 500 4ks <

< 500ks

4(0.00324) <

< 500(0.00324)

0.0130 m <

< 1.620 m

(1)

Fully turbulent ﬂow (an essential requirement for the validity of the Manning equation) requires that ks RS0 > 2.2 × 10−5 (0.00324) R(0.001) > 2.2 × 10−5 R > 0.0461 m

(2)

Equations 1 and 2 collectively indicate that n can be taken as a constant and the Manning equation is valid for 0.0461 m < R < 1.620 m For R = 0.0461 m, the (lower) flow depth, yL , satisfies the relation byL + myL2 √ = 0.0461 b + 2 1 + m2 yL 3yL + 2yL2 √ = 0.0461 3 + 2 1 + 22 yL which yields yL = 0.0479 m. Similarly, for R = 1.62 m, the corresponding flow depth, yU , is given by yU = 2.95 m. Therefore, the Manning equation can be applied with a constant value of n in the range 0.0479 m < y < 2.95 m . For R = 0.0461 m, A = 0.1483 m2 , and Q=

1 2 2 1 1 1 AR 3 S02 = (0.1483)(0.0461) 3 (0.001) 2 = 0.0402 m3 /s n 0.015

Similarly, for R = 1.62 m, A = 26.26 m2 , and 1 2 2 1 1 1 AR 3 S02 = (26.26)(1.62) 3 (0.001) 2 = 76.36 m3 /s n 0.015 hence the range of ﬂow for which the Manning equation can be applied with a (approximately) constant value of n is 0.0402 m3 /s < Q < 76.36 m3 /s .

78 (c) The general equation for n under fully turbulent conditions is 1

ks6 √ 8g

R ks

2 log

12R ks

(0.00324) 6 R 6 √ 0.00324 8(9.81) 12R = 2 log 0.00324 1

1 6

1 0.05644 = 3.569 + log R 63.24 + 17.72 log R

Hence, taking Q = 100 m3 /s, the Manning equation can be put in the form Q=

1 2 1 AR 3 S02 n

Q = [63.24 + 17.72 log R]

100 = 63.24 + 17.72 log

2 3

S02

P 5 1 3y + 2y 2 (3y + 2y 2 ) 3 2 √ √ 2 (0.001) 3 + 2 5y (3 + 2 5y) 3

which yields y = 3.303 m and n = 0.0148. If n is assumed to be constant and equal to 0.015, then the Manning equation requires that 5

1 1 (3y + 2y 2 ) 3 2 100 = √ 2 (0.001) 0.015 (3 + 2 5y) 3

which yields y = 3.326 m. Therefore, the error in the calculated ﬂow depth incurred by assuming a constant n (= 0.015) for Q = 100 m3 /s is 0.7% , which is a relatively small error. 4.19. From the given data: Q = 150 m3 /s, b = 10 m, m = 2.5, S0 = 0.1% = 0.001, y = 5 m, and A = by + my 2 = (10)(5) + (2.5)(5)2 = 112.5 m2 P = b + 2y 1 + m2 = 10 + 2(5) 1 + 2.52 = 36.93 m 112.5 A = = 3.046 m R= P 36.93 (a) Using the Manning equation: 1 2 1 AR 3 S02 n 2 1 1 100 = (112.5)(3.046) 3 (0.001) 2 n

which yields n = 0.0498 . Assuming that 4 < R/ks < 500, n 1

ks6 0.0498 1 6

= 0.039 = 0.039

79 which yields ks = 4.354 m and R/ks = 3.046/4.354 = 0.7. Since R/ks < 4, n/ks cannot be taken as a constant (= 0.039) that is independent of R/ks . Assuming that the ﬂow is fully turbulent,

n 1

ks6 0.0498 1 6

√1 8g

R ks

2.0 log 12 kRs √ 1

1 6

3.046 ks

8(9.81)

1 6

2.0 log 12 3.046 ks

which yields ks = 1.577 m . This result is based on the assumption of fully turbulent ﬂow, which requires that

(1.577)

RSf > 2.2 × 10−5

(3.046)(0.001) > 2.2 × 10−5 0.0870 > 2.2 × 10−5

Hence the fully-turbulent flow assumption is validated. (b) Under the flow conditions described here, the Manning equation is not valid since n is not a constant but depends on the flow depth. If n is expressed as a function of the flow depth, then the Manning equation can be used. (c) If the depth of flow increases by 50%, then y = 1.5(5.0) = 7.5 m A = by + my 2 = 10(7.5) + 2.5(7.5)2 = 215.6 m2 P = b + 2y 1 + m2 = 10 + 2(7.5) 1 + 2.52 = 50.39 m 215.6 A = = 4.279 m R= P 50.39 Taking ks = 1.577 m and R = 4.279 m, Manning’s n is given by

n 1

√1 8g

R ks

1 6

2.0 log 12 kRs 4.279 1 6 √ 1 n 8(9.81) 1.577 1 = 2.0 log 12 4.279 1.577 6 1.577 ks6

which yields n = 0.0475 .

80 4.20. The Horton (1933a) and Einstein (1934) formulae are based on the assumption that the total cross-sectional mean velocity is equal to the subarea mean velocity. According to the Manning equation 1 2 12 R 3 S0 ne 1 2 1 subarea mean velocity = Ri3 S02 ni

total cross-sectional mean velocity =

(1) (2)

Combining Equations 1 and 2, and using the Horton/Einstein assumption 1 2 12 1 2 1 R 3 S0 = Ri3 S02 ne ni 2 2 1 Ai 3 1 A 3 = ne P ni Pi 1 Ai 1 A = 3 3 ne2 P ni2 Pi 3 1 A ni2 Pi = Ai 3 P 2 ne N N 3 1 A 2 (ni Pi ) = Ai 3 ne2 P i=1 i=1

(3)

Since A=

i=1

Equation 3 simpliﬁes to 1 3 2

i=1 (ni Pi )

which yields ⎛ ne = ⎝

3 2

⎞2

i=1 Pi ni ⎠

4.21. The Lotter (1933) formula is based on the assumption that the total discharge is the sum of the subarea discharges. According to the Manning equation, 1 2 1 AR 3 S02 ne 2 1 1 subarea discharge = Ai Ri3 S02 ni

total discharge =

(1) (2)

81 Combining Equations 1 and 2 and using the Lotter (1933) assumption 1 1 2 1 2 1 AR 3 S02 = Ai Ri3 S02 ne ni N

i=1

1 A3 1 A i = ne P 23 ni P 23 5 3

i=1

5 1 PR3 = ne

N 1 i=1

Pi Ri3

which simpliﬁes to 5

ne =

PR3 N

Pi Ri3 i=1 ni

4.22. From the given shape of the ﬂoodplain (Figure 4.5), the following geometric characteristics are derived: Section, i 1 2 3 4 5 6 7

Pi (m) 20.6 100.0 6.7 15.0 6.7 150.0 20.6 319.6

Ai (m2 ) 50 500 39 120 39 750 50 1548

Ri (m) 2.42 5.00 5.81 8.00 5.81 5.00 2.42 34.46

ni 0.040 0.030 0.015 0.013 0.017 0.035 0.060

yi (m) 2.50 5.00 6.50 8.00 6.50 5.00 2.50

The total perimeter, P , of the (compound) channel is 319.6 m, the total area, A, is 1548 m2 , and hence the hydraulic radius, R, of the compound section is given by R=

1548 A = = 4.84 m P 319.6

Substituting these data into the formulae listed in Table 4.3 yields the following results: Formula Horton/Einstein Pavlovskii Einstein and Banks Lotter Krishnamurthy and Christensen Cox Yen Average

ne 0.034 0.035 0.034 0.026 0.029 0.032 0.033 0.032

82 A conservative (high) estimate of the composite roughness is 0.035, and the average composite roughness predicted by the models is 0.032 . 4.23. In the main channel: n = 0.016 and S0 = 0.005. When the main channel ﬂows full: 1 A = [30 + 30 + 3(2) + 3(3)](3) = 112.5 m2 2 P = 30 + 3( 32 + 12 + 22 + 12 ) = 46.2 m 112.5 A = = 2.44 m R= P 46.2 The Manning equation gives the capacity, Q, of the main channel as Q=

1 2 2 1 1 1 AR 3 S02 = (112.5)(2.44) 3 (0.005) 2 = 900 m3 /s n 0.016

When flow is in the floodplain, use the Horton equation to calculate the equivalent Manning’s roughness, ne . If the depth of flow in the main channel is y: i

Section

1 2 3

Left Floodplain Main Channel Right Floodplain

Pi (m) 100(y − 3) 46.2 25(y − 3)

Ai (m2 ) 50(y − 3)2 112.5 + 45(y − 3) 12.5(y − 3)2

Ri (m) 0.5(y − 3) 2.44 + 0.974(y − 3) 0.5(y − 3)

n 0.040 0.016 0.050

Using Horton equation ⎛ ne = ⎝

3 2

⎞2

i=1 Pi ni ⎠ N i=1 Pi

100(y − 3)(0.040) 2 + 46.2(0.016) 2 + 25(y − 3)(0.050) 2 = 100(y − 3) + 46.2 + 25(y − 3)

which simpliﬁes to

ne =

1.08y − 3.15 125y − 328.8

2 3

For the entire channel, including ﬂoodplains, A=

Ai = 50(y − 3)2 + 112.5 + 45(y − 3) + 12.5(y − 3)2 = 62.5(y − 3)2 + 45(y − 3) + 112.5

i=1

P =

Pi = 100(y − 3) + 46.2 + 25(y − 3) = 125y − 328.8

i=1

With Q = 1590 m3 /s, the Manning equation gives 5

1 2 1 A 3 12 1 AR 3 S02 = S Q= ne ne P 23 0 2 5 1 125y − 328.8 3 [62.5(y − 3)2 + 45(y − 3) + 112.5] 3 1590 = (0.005) 2 2 1.08y − 3.15 (125y − 328.8) 3

83 which yields

y = 5.50 m

Therefore, the encroachment on the left ﬂoodplain is (5.50 − 3.0)(100) = 250 m , and the encroachment on the right ﬂoodplain is (5.50 − 3.0)(25) = 62.5 m .

4.24. According to Equation 4.56,

v(y) = V +

1 y gdS0 1 + 2.3 log κ d

The value of y where v(y) = V occurs when

V =V +

1 y gdS0 1 + 2.3 log κ d

which can be simpliﬁed as follows:

y 1 gdS0 1 + 2.3 log κ d y 0 = 1 + 2.3 log d 1 y log = − d 2.3 y = 10−1/2.3 d y = 0.368d 0=

84 4.25. According to Equation 4.56, v(y) = V +

y 1 gdS0 1 + 2.3 log κ d

At y/d = 0.2, v(0.2d) = V +

1 1 gdS0 (1 + 2.3 log 0.2) = V + gdS0 (−0.61) 0.4 0.4

(1)

1 1 gdS0 (1 + 2.3 log 0.8) = V + gdS0 (0.78) 0.4 0.4

(2)

At y/d = 0.8, v(0.8d) = V +

Combining Equations 1 and 2 gives 2V + v(0.2d) + v(0.8d) = 2

1 √ 0.4

gdS0 (0.78 − 0.61) = V + 0.21 gdS0 2

√ Therefore, assuming that the term 0.21 gdS0 is small, then the average velocity, V , can be estimated by [v(0.2d) + v(0.8d)]/2. 4.26. Section Factor, Z1 = sQn 1/2 = z1 = 1008 B 8/3 (15) 3

50×0.02 1/2 =100 (1×10−4 )

= 0.073 = 7.3 × 10−2 5/3

[(1+ Zh )h] Z B B = √ 2/3 B 8/3 [1+2 Z 2 +1 h/B]

From Trial and error, hn = 3.044 A= h(B+Zh) = 3.044(15 + 3.044) = 54.925

√ P= B+2h Z 2 + 1 √ = 15 + 2 × 3.044 2 = 23.6 R= PA = 2.327 2

V= n1 R 3 S 2 1 = 0.02 (2.327)2/3 × (1 × 10−4 )

1/2

= 50 × 1.76 × 0.01 = 0.88 A D = BottomA width = B = 3.66 0.88 Froude no, F = √VgD = √9.81×3.66 = 0.1469 < 1

So it is sub-critical ﬂow.

85 4.27. Data given Velocity of ﬂow,V1 = 5m/s Depth of ﬂow,y1 = 0.56m Width of channel,b = 6m Discharge per unit width, q=

V 1 × y1 × b Q = = (5 × 0.56) = 2.8m3 /s/m b b

1 Froude no. of the upstream, Fr1 = √VgD

A 1 Now, D = W = b×y b = y1

(F r)1 = √

5 = 2.13 9.81 × 0.56

As the Froude no is more than one, the flow is shooting on the upstream side. Shooting flow is unstable flow and it will convert itself into streaming flow by raising its height & hence hydraulic jump will take place. Let the depth of the hydraulic jump be y2 (Equation 4.140) y1 y2 = ( 2 0.56 = 2

1 + 8(Fr )2 − 1)

2 1 + 8(2.13) − 1 = 1.429m

Height of hydraulic jump = y2 − y1 = (1.429 − 0.56) = .87 m Loss of energy per kg of water is given by3

1.434 2 −y1 ) = 4×0.56×1.429 = 0.205m − kg/kg of water Acording to Equation 4.146, E = (y4y 1 y2 3

= 45.9 hp Horse power lost= 1000×16.8×.205 75 4.28. The Darcy-Weisbach equation can be written as hf =

f¯L V̄ 2 D 2g

Deﬁning S=

hf L

and

R̄ =

D 4

and substituting into the Darcy-Weisbach equation gives S=

f¯ V̄ 2 4R 2g

86 4.29. Data given Width of the channel, b = 40m Aﬄux, (y2 − y1 ) = 1.8m Depth of channel, Depth of channel, y1 = 2.5m Therefore, y2 = 4.3m 1 Bed slope, s0 = 2000

Manning’s constant, n = 0.03 Area and perimeter of ﬂow at section 1, A1 = b × y1 = 100m2 P1 = b + 2y1 = 45m R1 =

A1 100 = = 2.22m P1 45

1 1 1/2 1 2 1/2 R 3 s0 = × (2.22)2/3 × ( ) = 1.27m/s n 0.03 2000 Speciﬁc energy at section 1, V =

E 1 = y1 +

(1.27)2 v12 = 2.5 + = 2.58m 2g 2 × 9.81

From Continuity equation, V2 =

A1 × V1 1.27 × 100 1.27 × 100 = 0.74m/s = = A2 B × y2 40 × 4.3

Area & perimeter at section 2,

A2 = 172m2 P2 = 48.6m R2 =

E 2 = y2 +

172 = 3.54m 48.6

0.742 V22 = 4.3 + = 4.33m 2g 2 × 9.81

Now, average depth,yavg = 2.5+4.3 = 3.4m 2 Vavg =

V 1 × y1 1.27 × 2.5 = 0.93m = yavg 3.4

Ravg =

2.22 + 3.54 R1 + R2 = = 2.88m 2 2

87 To ﬁnd the value of sf use, Manning’s equation, Vavg = 0.93 =

1 Ravg 2/3 sf 1/2 n

2 1 1 × (2.88) 3 × sf 2 0.03

sf = 1.9 × 10−4 The length of the backwater curve (Δ L) is (Equation 4.154), ΔL =

4.33 − 2.58 E2 −E1 = 1 = 5645.16m −4 s0 −sf 2000 − 1.9 × 10

4.30. Data given, b=2.6m So = 4×10−4 yn =1.25m i) k= n1 AR2/3 ,where k is the conveyance of the channel A =3.25m2 b×yn R=A/P= b+2y =0.64m n Therefore, k=161m3 /s per unit longitudinal slope

ii) Froude No., Fr = √VgD n D=A/T= b×y = yn b

V = n1 R2/3 SO 1/2 =0.99m/s Therefore, Fr =0.283<1 iii) Discharge, Q=(3.25 × 0.99)m3 /s=3.22m3 /s iv) τavg = ρgRs = (100 × 9.81 × 0.64 × 4 × 10−4 )Pa (Equation 4.21) = 2.511Pa 4.31. Data given Width of the channel, b = 10m Depth of the channel, y = 2m Velocity of ﬂow, V = 1.5m/s 1 Bed slope, s0 = 200

1 Slope of the energy line,sf = 5000

Rate of change of depth of water (Equation 4.123), s0 − sf dy = dx 1 − αFr2 Now, Fr = √VgD

88 A Hydraulic Depth (D) = W = b×y y =y 1.5 Therefore, Fr = √9.81×2 = 0.34 1 1 − 5000 dy −3 200 dx = 1−0.342 = 5.43 × 10

4.32. Data given Width of the channel, b= 4m Speciﬁc energy, E= 3.5m Discharge, Q= 15m3 /s We knowE =y+

V2 2g

Q 15 = 4y = 3.75 Now,V = b×y y 2

3.75 0.72 Therefore, 3.5 = y + 2×9.81×y 2 =y + y 2

y 3 − 3.5y 2 + 0.72 = 0 The alternate depths of flow are 0.489m and 3.441m 4.33. From the given data: Q = 16 m3 /s, y1 = 2 m, b1 = 10 m, and m = 3. The following preliminary calculations will be useful, A1 = b1 y1 + my12 = (10)(2) + (3)(2)2 = 32 m2 T1 = b1 + 2my1 = (10) + 2(3)(2) = 32 m A1 32 =1m = D1 = T1 32 Q 16 V1 = = 0.5 m/s = A1 32 V1 0.5 Fr1 = √ = 0.160 = gD1 (9.81)(1) (a) Denote the depth of flow and the bottom width of the contracted section as b and y, respectively. For critical flow in the contracted section: A3 Q2 = g T 2 (by + 3y 2 )3 16 = 9.81 b + 2(3)y which yields (by + 3y 2 )3 = 26.10 b + 67

(1)

89 For conservation of energy, y1 + 2+

Q2 Q2 = y + 2gA2 2gA21

162 162 = y + 2(9.81)(32)2 2(9.81)(by + 3y 2 )2

which simpliﬁes to y+

13.05 = 2.013 (by + 3y 2 )2

(2)

Both Equations 1 and 2 must be satisﬁed for choking to occur at the downstream section. (b) When b = 0, Equation 2 gives y+

13.05 = 2.013 (3y 2 )2

which yields y = 1.90 m or 1.10 m. Hence the depth in the contracted section will be 1.90 m and the ﬂow will not be choked . 4.34. From the given data: b1 = 10.0 m, y1 = 1.00 m, Q = 8 m3 /s, b2 = 6 m, and L = 7 m. (a) According to the energy equation E1 = E2 +

V12 2g

where Q 8 = 0.800 m/s = b1 y1 (10.0)(1.00) (0.800)2 V12 = = 0.0326 m 2g 2(9.81) V2 E1 = y1 + 1 = 1.00 + 0.0326 = 1.0326 m 2g Q2 82 0.0906 E 2 = y2 + = y + = y2 + 2 2g(b2 y2 )2 2(9.81)(6y2 )2 y22 V1 =

Substituting into the energy equation, Equation 1, gives 1.0326 = y2 +

0.0906 + 0.0326 y22

which simpliﬁes to 1.00 = y2 +

0.0906 y22

which yields the following positive solutions y2 = 0.383 m,

0.884 m

(1)

90 Since Fr21 =

V12 0.8002 = 0.065 = gy1 (9.81)(1.00)

the upstream flow is subcritical, and therefore the flow in the constriction must also be subcritical, and hence y2 = 0.884 m (b) To assess the effect of the energy loss, the depth of flow in the constriction must be calculated without including the energy loss. According to the energy equation E1 = E 2

(2)

where E1 = 1.0326 m 0.0906 E 2 = y2 + y22 Substituting into the energy equation, Equation 2, gives 1.0326 = y2 +

0.0906 y22

which yields the following positive solutions y2 = 0.371 m,

0.924 m

Since the upstream flow is subcritical, the flow in the constriction must also be subcritical, and hence y2 = 0.924 m Therefore, if energy losses are neglected the calculated flow depth is in error by (0.924 − 0.884)/0.884 × 100 = 4.5%. This effect is not very significant. (c) According to the energy equation E 1 = E2 +

V12 2g

where V1 = 0.800 m/s, and E1 = 1.0326 m E 2 = y2 +

Q2 82 0.1612 = y + = y2 + 2 2 2 2g(b2 y2 ) 2(9.81)(4.5y2 ) y22

Substituting into the energy equation, Equation 3, gives 1.0326 = y2 +

0.1612 0.8002 + 2(9.81) y22

(3)

91 which does not have any positive solutions. Therefore, the flow is choked and critical flow exists within the constriction. Under critical flow conditions, A3 Q2 = g T (4.5y2 )3 82 = 9.81 4.5 which yields y2 = 0.686 m (d) Since the flow is choked, the constriction influences the upstream flow depth. Under critical flow conditions, 3 3 E2 = y2 = (0.686) = 1.028 m 2 2 According to the energy equation E1 = E2 + or y1 +

V12 2g

(4)

V2 V12 = 1.028 + 1 2g 2g

which yields y1 = 1.028 m 4.35. Data given Width of the channel,b = 3m Depth of ﬂow,y = 2m Area of ﬂow,A = b × y = 6m2 1 Bed slope,s0 = 1000

C= 60 Wetted perimeter, P= b + 2y = 3 + 4 = 7m2 Hydraulic Radius,R = PA = 67 = 0.857 √ 1 = 10.54m3 /s (from Equation 4.39) The discharge,Q = AC Rs0 = 6 × 60 0.857 × 1000 For maximum discharge for a given area, slope of bed and roughness of the channel are constant. Let b’ and y’ be the new width and depth of the channel. Then, Area= b y b y =6 Also for maximum discharge, b = 2y 2y 2 = 6

92 y = 1.732m b = 3.464m New dimensions of the channel areWidth, b = 3.464m Depth, y = 1.732m 6 = 0.866m Therefore, R = PA = 6.928

Maximum discharge, Q is given by 1 Q = AC R so ) = 6 × 60 (0.866 × 1000 ) = 10.6m3 /s Therefore, increase in discharge= Q − Q = (10.6 − 10.54) = 0.06m3 /s

4.36. Flow in a rectangular open channel is choked when E1 = E2 + Δzc 3 V12 = yc + Δzc 2g 2 1 2 3 q2 3 V1 = y1 + + Δzc 2g 2 g 1 3 Q2 3 V12 = y1 + + Δzc 2g 2 gb2 1 3 (V1 by1 )2 3 V12 = y1 + + Δzc 2g 2 gb2 y1 +

V2 3 (V13 y13 ) y1 + 1 = + Δzc 2g 2 g 13 Dividing by y1 yields 3 V2 1+ 1 = 2gy1 2

V12 gy1

13 +

Δzc y1

and deﬁning V1 Fr1 = √ gy1 then Equation 1 can be written as Fr2 3 2 Δzc = 1 + 1 − Fr13 y1 2 2 From Problem 4.35: b = 3 m, Q = 4 m3 /s, y1 = 1.5 m, and Δzc = 0.15 m. Therefore Q 4 = 0.889 m/s = by1 (3)(1.5) V1 0.889 = 0.232 = Fr1 = √ gy1 (9.81)(1.5) V1 =

(1)

93 which yields Fr2 3 2 Δzc = 1 + 1 − Fr13 y1 2 2 2 (0.232)2 3 Δzc =1+ − (0.232) 3 1.5 2 2 and solving for Δzc gives Δzc = 0.69 m 4.37. Data given b= 5.5m y= 0.75m b×y 5.5×0.75 = 5.5+2×0.75 = 0.59m R= b+2y 1/2

V= n1 R2/3 s0

1 = 0.02 × (0.59) 3 (0.003)1/2 = 1.925m/s

A D= W = b×y b =y 1.925 = .71 Now, Fr = √VgD = √9.81×0.75

Therefore the ﬂow is subcritical. 2

According to Equation 4.110, yc = ( qg )

1/3

= [ (1.925×0.75) ] 9.81

1/3

= 0.60m

Since the depth of the weir is greater than yc , the ﬂow is subcritical at the weir and hence a hydraulic jump must take place. 4.38. From the given data: Q = 18 m3 /s, b = 5 m, m = 2, y1 = 2 m, and there is a 0.50-m wide bridge pier placed in the channel. A1 = (b + my1 )y1 = (5 + 2 × 2)(2) = 18 m2 A2 = [(5 − 0.5) + 2y2 ]y2 = 4.5y2 + 2y22 Q 18 = 1 m/s V1 = = A1 18 and the energy equation gives y1 +

V12 V2 = y2 + 2 2g 2g

2 18 12 1 = y2 + 2+ 2(9.81) 2(9.81) (4.5y2 + 2y22 ) 2 1 2.05 = y2 + 16.5 (4.5y2 + 2y22 ) with the solutions y2 = 0.59 m, 1.99 m

94 which correspond to supercritical and subcritical flow conditions respectively. Since the upstream flow is subcritical, choose the subcritical downstream flow y2 = 1.99 m The maximum pier width produces critical flow at the constriction such that y2 = yc or Q2 A3 = c g Tw

(1)

where Ac = (5 − wp )yc + 2yc2 Tw = (5 − wp ) + 4yc Substituting into Equation 1 gives [(5 − wp )yc + 2yc2 ]3 182 = 9.81 (5 − wp ) + 4yc

(2)

The energy equation requires that V12 V2 = yc + c 2g 2g Q2 2.05 = yc + 2gA2c

y1 +

2.05 = yc + which simpliﬁes to

182 2(9.81)[(5 − wp )yc + 2yc2 ]2

1 2 16.51 1 wp = 5 + 2yc − yc 2.05 − yc

(3)

Simultaneous solution of Equations 2 and 3 yields yc = 1.61 m

and

wp = 4.42 m

Therefore, the maximum width of the pier that will not cause a rise in the upstream water surface is 4.42 m . 4.39. From given data: Q = 15 m3 /s, b = 4.5 m, y1 = 1.9 m, Δz = 0.15 m, m = 1.5. Let b be the bottom width at the step, then b = b + 2mΔz = 4.5 + 2(1.5)(0.15) = 4.95 m A1 = (b + my1 )y1 = (4.5 + 1.5 × 1.9)(1.9) = 13.97 m2 Q 15 = 1.07 m/s V1 = = A1 13.97 Q 15 15 V2 = = = 2 A2 b y2 + my2 4.95y2 + 1.5y22

95 The energy equation gives V12 V2 = y2 + 2 + Δz 2g 2g 2 2 1 15 1.07 + 0.15 = y2 + 1.9 + 2(9.81) 2(9.81) 4.95y2 + 1.5y22 11.47 + 0.15 1.96 = y2 + (4.95y2 + 1.5y22 )2 y1 +

Solving for y2 gives y2 = 1.73 m, 0.52 m These depths correspond to subcritical and supercritical flow conditions respectively. Since the upstream flow is subcritical, the flow at the step is also subcritical and y2 = 1.73 m At the maximum step height, y2 = yc and the energy equation gives 2 Q V12 1 = yc + + Δzm y1 + 2g 2g Ac 11.47 1.96 = yc + + Δzm A2c which can be written as Δzm = 1.96 − yc − Under critical flow conditions,

11.47 [(4.5 + 3Δzm )yc + 1.5yc2 ]2

(1)

A3 Q2 = c g Tc

which, in this case, can be written as [(4.5 + 3Δzm )yc + 1.5yc2 ]3 152 = 22.94 = 9.81 4.5 + 3(yc + Δzm ) Solving Equations 1 and 2 gives yc = 0.719 m 4.40. Neglecting losses at the entry 2

H = y + v2g According to equation 4.43, v = n1 R2/3 s1/2 As the channel is wide, v = n1 y 2/3 s1/2

and

Δzm = 0.921 m

(2)

96 2

1 1 3 2 [ny s ] =⇒ y = H - 2g

2 4

1 1 =⇒ y = 3.0 - 2×9.81 [ (0.02) 2 y 3 × 0.0005]

Solving by trial and error, y = 2.755 m Bed level at the downstream end of the channel, i.e., at the section B = 102.00 – (0.0005×1200) = 101.40 m Downstream pool elevation = 101.40 +2.755 = 104.155 m

4.41. From the given data: S0 = 0.05% = 0.0005, m = 2, b = 5 m, and Q = 7 m3 /s. For a float-finished concrete channel, Table 4.2 gives n = 0.015. (a) The depth of flow, y, is given by the Manning equation as 5

1 A 3 12 S Q= n P 23 0

1 [5y + 2y 2 ] 3 1 2 7= √ 2 (0.0005) 0.015 [5 + 2 1 + 22 y] 3

which yields y = 0.897 m For the Manning equation to be valid, the ﬂow must be fully turbulent, which requires that n6 RS0 ≥ 9.6 × 10−14 (1) where R=

5(0.897) + 2(0.897)2 A √ = 0.676 m = P 5 + 2 1 + 22 (0.897)

Substituting into Equation 1 yields n6 RS0 = (0.015)6 (0.676)(0.0005) = 2.09 × 10−13 ≥ 9.6 × 10−14

97 Hence, the ﬂow is fully turbulent. For the Manning’s n to be independent of the ﬂow depth, R < 500 (2) 4< ks where ks ≈ (n/0.039)6 . In this case, 0.676 R = = 209 ks (0.015/0.039)6 which is within the range given by Equation 2. Based on the criteria given by Equations 1 and 2, the Manning equation is valid . (b) The head loss, hL , in the contraction can be estimated using the relation V22 V12 − α1 hL = C α2 2g 2g

(3)

Since the contraction is abrupt, C = 0.6, and V1 and V2 are given by Q 7 = = 1.15 m/s A1 5(0.897) + 2(0.897)2 Q 7 = V2 = A2 4y2 + 2y22 V1 =

and α1 and α2 can be taken as unity. The energy equation requires that y1 +

V12 V2 = y2 + 2 + hL 2g 2g

(4)

Combining Equations 3 and 4 gives V2 V2 y1 + (1 + C) 1 = y2 + (1 + C) 2 2g 2g

(5)

Substituting known quantities gives 1.152 = y2 + (1 + 0.6) 0.897 + (1 + 0.6) 2(9.81)

7 4y2 + 2y22

1 2(9.81)

which simpliﬁes to 1.005 = y2 +

3.996 (4y2 + 2y22 )2

which yields y2 = 0.815 m (subcritical)

0.609 m (supercritical)

At the upstream section, the Froude number is given by 1.15 V1 Fr1 = √ = 0.44 (subcritical) = gD (9.81)(0.709)

98 Therefore, the ﬂow in the contracted section is subcritical and equal to 0.815 m . If the head loss is ignored, V12 V2 = y2 + 2 2g 2g 2 2 7 1 1.15 = y2 + 0.897 + 2(9.81) 2(9.81) 4y2 + 2y22 y1 +

which simpliﬁes to 0.964 = y2 +

2.497 (4y2 + 2y22 )2

which yields y2 = 0.861 m (subcritical) Taking the head loss into account has a significant effect on the calculated flow depth in the contracted section (0.815 m vs. 0.861 m), with a depth difference of 5% when head loss is taken into account. 4.42. Let Section 1 be the upstream section, Section 2 be the “throat” section, and Section 3 be the downstream section. From the given data: Q = 100 m3 /s, n = 0.025, S0 = 0.5%, y1 = 3.000 m, b1 = b3 = 30 m, b2 = 20 m, L12 = L23 = 40 m. Neglecting Energy Losses: The energy equation applied between sections 1 and 2 is given by V22 V12 = y1 + + (z1 − z2 ) (1) y2 + 2g 2g From the given data: y1 = 3.000 m A1 = b1 y1 = (30)(3.000) = 90 m2 Q 100 = 1.111 m/s V1 = = A1 90 z1 − z2 = L12 S0 = (40)(0.005) = 0.20 m P1 = b1 + 2y1 = 30 + 2(3.000) = 36.00 m A1 90 = 2.500 m = R1 = P1 36 A2 = b2 y2 = 20y2 Q 100 5 V2 = = = A2 20y2 y2 Substituting the calculated parameters into Equation 1 gives

(1.111)2 (5/y2 )2 y2 + = 3.000 + + 0.20 2(9.81) 2(9.81)

99 which yields y2 = 3.133 m . Between sections 2 and 3, the energy equation is V2 V2 y3 + 3 = y2 + 2 + (z2 − z3 ) 2g 2g

(2)

where y2 = 3.133 m A2 = b2 y2 = (20)(3.133) = 62.66 m2 Q 100 V2 = = = 1.596 m/s A2 62.66 z2 − z3 = L23 S0 = (40)(0.005) = 0.20 m P2 = b2 + 2y2 = 20 + 2(3.133) = 26.27 m A2 62.66 = 2.385 m = R2 = P2 26.27 A3 = b3 y3 = 30y3 Q 100 3.333 V3 = = = A3 30y3 y3 Substituting the calculated parameters into Equation 2 gives 1.5962 (3.333/y3 )2 = 3.133 + + 0.20 y3 + 2(9.81) 2(9.81) which yields y3 = 3.414 m . Accounting for Energy Losses: The energy equation applied between sections 1 and 2 is given by V22 V12 (3) = y1 + + (z1 − z2 ) − he − hf y2 + 2g 2g where he and hf are the energy losses due to expansion/contraction and friction and are given by 2 V2 V12 − he = C (4) 2g 2g (5) hf = Sf L where C = 0.1 for contractions . Assuming that the friction slope, Sf , is approximately the same at all sections then Sf can be calculated at section 1 as Sf =

nQ 2

AR 3

2 =

(0.025)(100) 2

= 0.000227

(90)(2.500) 3

Between sections 1 and 2, Equations 4 and 5 give 1.1112 25 (5/y2 )2 − = 0.00510 − 1.234 he = (0.1) 2(9.81) 2(9.81) y22 hf = Sf L = (0.000227)(40) = 0.00908 m

100 Substituting the calculated parameters into the energy equation (Equation 3) gives (1.111)2 25 (5/y2 )2 = 3.000 + + 0.2 − 0.00510 − 1.234 − 0.00908 y2 + 2(9.81) 2(9.81) y22 which yields y2 = 3.116 m . Hence, A2 = (20)(3.116) = 62.32 m2 Q 100 = 1.605 m V2 = = A2 62.32 A3 = 30y3 Q 100 3.333 V3 = = = A3 30y3 y3 The energy equation applied between sections 2 and 3 is given by V2 V2 y3 + 3 = y2 + 2 + (z2 − z3 ) − he − hf 2g 2g

(6)

where he and hf are the energy losses due to expansion/contraction and friction and are given by 2 V3 V22 − he = C (7) 2g 2g (8) hf = Sf L where C = 0.3 for expansions. Between sections 2 and 3, Equations 7 and 8 give 1.6052 11.11 (3.333/y2 )2 − = 0.01529 2.576 − he = (0.3) 2(9.81) 2(9.81) y32 hf = Sf L = (0.000227)(40) = 0.00908 m Substituting the calculated parameters into the energy equation (Equation 6) gives (3.333/y3 )2 (1.605)2 1.11 y3 + − 0.00908 = 3.116 + + 0.2 − 0.01529 2.576 − 2 2(9.81) 2(9.81) y3 which yields y3 = 3.350 m . Without considering energy losses, the stage diﬀerence between the upstream and downstream sections is (3.000 m + 0.4 m) − 3.414 m = −0.014 m . Taking energy losses into account, the stage diﬀerence between the upstream and downstream sections is (3.000 m + 0.4 m) − 3.350 m = −0.050 m . Therefore an error of approximately (14-50)/50 = 72% is introduced by neglecting energy losses. 4.43. Taking Section 1 upstream of the bridge constriction, Section 2 at the bridge constriction, and Section 3 downstream of the bridge constriction (after expansion) then, from the given data, b1 = 10 m, b2 = 7 m, b3 = 10 m, Q = 20 m3 /s, and y1 = 2 m.

101 (a) The speciﬁc energy, E1 , at section 1 is given by E 1 = y1 +

V12 Q2 202 = 2 + = 2.051 m = y1 + 2g 2(9.81)(10)2 (2)2 2gb21 y12

where it is noted that V1 = 20/(10 × 2) = 1 m/s. Applying the energy equation between sections 1 and 2 gives 2 V12 V2 − E 2 = E1 − Cc (1) 2g 2g where Cc = 0.6 for an abrupt contraction, and it is assumed that V2 > V1 . Substituting the given and derived data into Equation 1 gives Q2 Q2 V12 = 2.051 − 0.6 − y2 + 2g 2gb22 y22 2gb22 y22 2 2 20 20 12 = 2.051 − 0.6 − y2 + 2(9.81)(7)2 y22 2(9.81)(7)2 y22 2(9.81) 0.6657 = 2.082 y2 + y22 which yields y2 =1.897 m , E2 = 2.013 m, and V2 = 1.506 m/s (conﬁrming V2 > V1 ). Applying the energy equation between sections 2 and 3 gives 2 V2 V32 − (2) E3 = E2 − Ce 2g 2g where Ce = 0.8 for an abrupt expansion, and it is assumed that V2 > V3 . Substituting the given and derived data into Equation 2 gives 2 Q2 Q2 V2 − = 2.013 − 0.8 y3 + 2g 2gb23 y32 2gb23 y32 202 202 1.5062 − y3 + = 2.013 − 0.8 2(9.81) 2(9.81)(10)2 y32 2(9.81)(10)2 y32 0.3670 = 1.968 y3 + y32 which yields y3 =1.862 m , E3 = 1.921 m, and V3 = 1.074 m/s (conﬁrming V2 > V3 ). (b) If energy losses are neglected, the energy equation is given by E1 = E2 Q2 = 2.051 2gb22 y22 202 = 2.051 y2 + 2(9.81)(7)2 y22 y2 +

which yields y2 = 1.941 m. When energy losses are taken into account, it was found that y2 = 1.897 m, and hence the error in the ﬂow depth at section 2 associated with

102 neglecting energy losses in the contraction is 2.3% . If energy losses are neglected in both the contraction and the expansion, then y3 = y1 = 2 m. When energy losses are taken into account, it was found that y3 = 1.862 m, and hence the error in the flow depth at section 3 associated with neglecting energy losses in the contraction and subsequent expansion is 7.4% . It is apparent that neglecting energy losses has a significant effect in this case. 4.44. From the given data: Q = 36 m3 /s, b = 10 m, n = 0.030, S0 = 0.001, and y = 3 m. Determine the normal depth, using the Manning equation 5

1 A3 Q= S0 n P 23 36 =

(10yn ) 3 √ 1 0.001 0.030 (10 + 2yn ) 23

which gives yn = 2.45 m Determine the critical depth, 2 13 2 13 q 3.6 = = 1.09 m yc = g 9.81 Since y > yn > yc , the water surface follows a M1 proﬁle . Determine Sf using the Manning equation, Sf =

nQ 2

AR 3

nQP 3 5

nQ(10 + 2y) 3

(0.030)(36)(10 + 2 × 3) 3

(10y) 3

(10 × 3) 3

= 0.00056

The slope of the water surface is given by S0 − Sf dy 0.001 − 0.00056 = = 2 2 = 0.00046 dx 1 − Fr 1.2 1 − 9.81(3) where the velocity is taken as V = Q/A = 36/(10×3) = 1.2 m/s. If y = 2 m, then yn > y > yc and the water surface follows a M2 proﬁle . Therefore the shape of the water surface would be diﬀerent than when y = 3 m. 4.45. We know E = 5.5 = Again-

(y2 − y1 )3 4y1 y2

(y2 − y1 )3 ............... 4y1 y2 q = 4.5m3 /s/m

(1)

103 We have, 2q 2 (Equation 4.38) gy 1 2 × 4.52 y22 + y1 y2 = 9.81 × y 1 4.12 y22 + y1 y2 = y1 4.12 y2 (y1 + y2 ) = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) y1 (y2 + y1 ) =

It can be proved that – (from Equation 4.140) 3 1 (−3 + 1 + 8F12 ) 16g 1/3 E = . . . . . . . . . . . . . . . . . . (3) 2/3 q 2/3 (−1 + 1 + 8F12 ) F 1

Solving equations (1) and (2) by trial and error method the sequent depths are found out to be, y1 =0.22m and y2 =3.20m 4.46. From given data: b = 6 m, m = 2, n = 0.045, S0 = 0.015, Q = 80 m3 /s, and y = 5 m. Calculate normal depth using Manning equation, 5

1 A3 Q= S0 n P 23

1 (byn + 2yn2 ) 3 1 (6yn + 2yn2 ) 3 √ 80 = S = 0.015 √ √ 0 0.045 (b + 2 5yn ) 23 0.045 (6 + 2 5yn ) 23 which gives yn = 2.21 m Calculate the critical depth, A3 Q2 = c g Tc 2 (byc + 2yc2 )3 80 = 9.81 (b + 4yc ) (6yc + 2yc2 )3 652.4 = (6 + 4yc ) which gives yc = 2.07 m Since y > yn > yc , the water surface has a M1 proﬁle and the depth increases in the downstream direction. The slope, Sf , of the energy grade line is given by the Manning equation as 2

Sf =

nQP 3 5

√ 2 nQ(b + 2 5y) 3 5

(by + 2y 2 ) 3

√ 2 (0.045)(80)(6 + 2 5 × 5) 3 5

(6 × 5 + 2 × 52 ) 3

= 0.000508

104 Other hydraulic parameters are A = by + 2y 2 = (6)(5) + 2(5)2 = 80 m2 Q 80 V = = = 1 m/s A 80 T = b + 2my = (6) + 2(2)(5) = 26 m 80 A = = 3.08 m D= T 26 12 V2 Fr2 = = = 0.0331 gD 9.81(3.08) The slope of the water surface is therefore given by S0 − Sf dy 0.015 − 0.000508 = = 0.0150 = 2 dx 1 − 0.0331 1 − Fr The depth, yu , 100 m upstream is given by yu = 5 − 100(0.0150) = 3.50 m and the depth, yd , 100 m downstream is given by yd = 5 + 100(0.0150) = 6.50 m 4.47. A= (B+Zd)d = (4 + 1.5 × 1.25)1.25 = 7.3475 √ P = 2 Z 2 + 1 d +B √ = (2 × 1.25 Z 2 + 1 + 4) = 12.125 R = PA = 7.3475 12.125 = 0.6059 U∗ = gRSf (Equation 4.30) √ = 9.81 × 0.6059 × 10−4 = 0.02 u ∗ ks 0.02×2 3 ν = 1.1×10−6 = 36.36×10 (Equation 4.31)

Q = -2A

√

ks 0.625ν √ 8gRS0 log10 ( 12R + R3/2 ) (Equation 4.38) 8gS 0

= 0.5678 m3 /s 0.5678 Average velocity, V = Q A = 7.3475 = 0.077 m/s

4.48. According to the momentum equation

Fx = ρQ(V2 − V1 )

105 In this case, Q Q − γA1 ȳ1 − γA2 ȳ2 + γV sin θ = ρQ A2 A1 Q Q A1 ȳ1 − A2 ȳ2 + V sin θ = Q − gA2 gA1 For a rectangular channel of width b, A1 = by1 A2 = by2 y1 ȳ1 = 2 y2 ȳ2 = 2 y1 + y 2 5y2 b V= 2 sin θ = S0 Q = qb and substituting into the momentum equation gives y 2 y1 + y2 y2 q 2 b2 q 2 b2 5y2 bS0 = b 1 −b 2 + − 2 2 2 gby2 gby1 2 2 y y1 + y2 q2 q2 y1 − 2 + 5y2 S0 = − 2 2 2 gy2 gy1 or q2 q2 y2 y 1 + y2 y12 + 5y2 S0 = 2 + − 2 gy1 2 gy2 2 4.49. Let ‘A’ be the section of the Vena Contracta ya = depth of the vena contracta = 0.6 × 0.7 = 0.42m 20 = 14.9m/s Va = (3.2×0.42) 14.9 a (since D = ya ) = √9.81×0.42 = 7.33 Fa = F roude N o.at vena contracta = √VgD

If y2 = sequent depth required f or a jump at vena contracta 1 y2 = (1 + 8Fa2 ) − 1 ya 2 Which yield y2 = 4.15 m Here the tail water depth (yt ) is 3.5 m, so yt <y2 , a free repelled jump will form.

106 4.50. The head loss, hL , is deﬁned by y1 + Dividing by y1 gives 1+

V12 V2 = y2 + 2 + hL 2g 2g

V12 y2 V2 hL = + 2 + 2gy1 y1 2gy1 y1

which can be put in the form y2 V2 V2 hL =1− + 1 − 2 y1 y1 2gy1 2gy1 Deﬁne Fr21 =

V12 gy1

(2)

Combining Equations 1 and 2 gives V22 y2 Fr21 hL 1− =1− + y1 y1 2 gy1 Fr21 ⎤ ⎡ V22 y2 Fr21 ⎣ ⎦ 1− + =1− V12 y1 2 gy =1−

y2 Fr1 + y1 2

V2 1 − 22 V1

gy1

Since V1 = q/y1 and V2 = q/y2 , then

y2 Fr21 (q/y2 )2 hL =1− + 1− y1 y1 2 (q/y1 )2

which simpliﬁes to hL y2 Fr21 1− =1− + y1 y1 2

(1)

y1 y2

4.51. (a) From the given data: Q = 20 m3 /s, y1 = 1 m, b1 = 1 m, and m = 2, which gives A1 = b1 y1 + my12 = (1)(1) + (2)(1)2 = 3 m2 T1 = b1 + 2my1 = (1) + 2(2)(1) = 5 m A1 3 = = 0.6 m D1 = T1 5 Q 20 V1 = = 6.67 m/s = A1 3 V2 6.672 = 7.56 Fr21 = 1 = gD1 (9.81)(0.6) Since Fr1 > 1 a hydraulic jump will occur in the channel.

107 (b) The upstream and downstream depths in a hydraulic jump are related by Q2 Q2 + A1 ȳ1 = + A2 ȳ2 gA1 gA2 where 1 y1 y1 1 y1 + by12 · + my12 · A1 ȳ1 = my12 · 2 3 2 2 3 3 3 3 3 my1 by 2(1) 1(1) = + 1 = + = 1.167 m3 3 2 3 2 Substituting the given and derived data: y2 202 202 + 1.167 = + (5y2 ) (9.81)(3) (9.81)(5y2 ) 2 8.15 + 2.5y22 14.76 = y2 which yields y2 = 2.08 m, 0.586 m, and −2.67 m. Hence the only feasible (subcritical) depth in the rectangular channel is 2.08 m . The energy loss in the hydraulic jump is equal to the change in speciﬁc energy, E1 − E2 , where 6.672 V12 =1+ = 3.268 m 2g 2(9.81) Q 20 V2 = = 1.92 m/s = A2 2.08 × 5 1.922 V2 = 2.27 m E2 = y2 + 2 = 2.08 + 2g 2(9.81) E 1 = y1 +

Therefore, the energy loss is 3.628 m − 2.27 m = 1.358 m, and the power loss is given by Power loss = γQΔE = (9.79)(20)(1.358) = 266 kW 4.52. The general hydraulic jump equation is given by Q2 + Aȳ = constant gA

(1)

For a trapezoidal channel, ȳ =

by 2 2

my 3 3 2 by + my

(2)

Combining Equations 1 and 2 yields Q2 + (by + my 2 ) g(by + my 2 )

by 2 2

my 3 3 2 by + my

= constant

108 which simpliﬁes to

Q2 + gy(b + my)

by 2 2

my 3 3

= constant

which demonstrates that Q2 + gy1 (b + my1 )

by12 2

my13 3

Q2 = + gy2 (b + my2 )

by22 2

my23 3

4.53. From the given data: Q = 21 m3 /s, b = 2 m, m = 1, and y1 = 1 m. The momentum equations requires that Q2 Q2 + A1 ȳ1 = + A2 ȳ2 (1) gA1 gA2 where A1 = cby1 + y12 = 2(1) + 12 = 3 m2 Q 21 = 7 m/s V1 = c = A1 3 A1 = cby2 + y22 = 2y2 + y22 Q 21 V1 = c = A2 2y2 + y22 and (1)1 1 1 1(2)(0.5) + (1)1 Aȳ 2 3 + 2 3 = = 0.44 m ȳ1 = A 3 y2 1 2 y2 2y2 2 + 2 2 y2 3 y2 + y22 /3 = ȳ2 = 2 2 + y2 2y2 + y2 Substituting into Equation 1 gives 212 212 y23 2 + y + 3(0.44) = + 2 9.81(3) 3 9.81(2y2 + y22 ) which yields y2 = 2.59 m The energy equation gives the energy loss, ΔE, as 72 V2 V2 − 2.59 − ΔE = y1 + 1 − y2 − 2 = 1 + 2g 2g 2(9.81) 4.54. Data given V1 = 6.5m/s y1 = 1.2m

21 2×2.59+2.592

2(9.81)

= 0.748 m

109 6.5 1 Froude Number, Fr1 = √VgD = √9.81×1.2 = 1.89

Hence the ﬂow is supercritical. We know- Equation 4.140 y2 1 1 2 −1 2 = 1 + 8F − 1 = 1 + 8×1.89 r y1 2 2 1 hL = 2.22m 3

(2.22−1.2) 2 −y1 ) Loss of energy, hL = (y4y = 4×1.2×2.22 = 0.1m 1 y2

4.55. From given data: Q = 10 m3 /s, b = 5.5 m, S0 = 0.0015, n = 0.038, y2 = 2.2 m. (a) Using the direct-integration method, y 1 = y2 −

S0 −

nQP̄ 3

Ā 3

(x2 − x1 ) 2 1 − V̄gȳ 2 2 nQ(b+2ȳ) 3 S0 − 5 (bȳ) 3 (x2 − x1 ) = y2 − 2 1 − gbQ2 ȳ3 2 2 (0.038)(10)(5.5+2ȳ) 3 0.0015 − 5 (5.5ȳ) 3 (100 − 0) = 2.2 − 102 1 − (9.81)(5.5) 2 ȳ 3

where ȳ =

(1)

y1 + 2.2 y1 + y2 = 2 2

(2)

Solving Equations 1 and 2 gives y1 = 2.12 m (b) Using the standard-step equation 2

ΔL =

y + V2g

S̄f − S0

(3)

This equation is solved iteratively until ΔL = 100 m, and the iterations are summarized in the following table: y2 2.2 2.2 2.2 2.2

A2 12.1 12.1 12.1 12.1

P2 9.4 9.4 9.4 9.4

R2 1.29 1.29 1.29 1.29

V2 0.826 0.826 0.826 0.826

Therefore y1 = 2.12 m .

S2 0.00070 0.00070 0.00070 0.00070

y1 2.20 2.10 2.11 2.12

A1 12.1 11.6 11.6 11.7

P1 9.4 9.2 9.22 9.24

R1 1.29 1.26 1.26 1.26

V1 0.826 0.866 0.862 0.857

S1 0.00070 0.00080 0.00079 0.00078

S̄f 0.00070 0.00075 0.00075 0.00074

ΔL

0 129 115 100

110 Find the uniform ﬂow depth, yn , using the Manning equation 5

1 A3 Q= S0 n P 23 which can be written as 5

1 (5.5yn ) 3 10 = S0 n (5.5 + 2yn ) 23 5

(5.5yn ) 3 √ 1 10 = 0.0015 0.038 (5.5 + 2yn ) 23 which gives yn = 1.719 m Plugging this value of y into the direct-step equation, Equation 3 gives y2 2.2

A2 12.1

P2 9.4

R2 1.29

V2 0.826

S2 0.000704

y1 1.719

A1 9.455

P1 8.94

R1 1.00

V1 1.06

S1 0.00149

S̄f 0.001

ΔL 1230

Therefore ΔL = 1230 m . 4.56. From the given data: Q = 5 m3 /s, b = 4 m, S0 = 0.04, n = 0.05, and y2 = 1.5 m. Let the given section be Section 2, then A2 = by2 = (4)(1.5) = 6 m2 P2 = b + 2y2 = 4 + 2(1.5) = 7 m A2 6 R2 = = = 0.857 m P2 7 Q 5 V2 = = = 0.833 m/s A2 6 ⎤2 ⎡ 2 (0.05)(5) nQ ⎦ ⎣ S2 = = = 0.00213 2 2 (6)(0.857) 3 A2 R23 Taking y1 = 1 m, then A1 = 4 m2 ,

P1 = 6 m,

and S̄f =

R1 = 0.667 m,

V1 = 1.25 m/s,

S1 = 0.00670

0.00213 + 0.00670 S1 + S2 = = 0.00442 2 2

and ΔL = =

[y1 + V12 /2g] − [y2 + V22 /2g] S̄f − S0 [1 + 1.252 /2(9.81)] − [1.5 + 0.8332 /2(9.81)] = 12.9 m 0.00442 − 0.04

Therefore the depth is equal to 1 m at a location 12.9 m upstream .

111 4.57. Here y1 = 0.75m A1 = b × y1 = 5.5 × 0.75 = 4.125m2 P1 = b + 2y1 = (5.5 + 2 × 0.75) = 7m V1 = 1.926m/s 4.125 1 R1 = A P1 = 7 = 0.59m

Q= 4.125 × 1.926 = 7.94m3 /s 2

0.02×7.94 −3 ] = [ 4.125×0.59 sf1 = [ AnQ 2/3 ] = 2.99 × 10 R2 1

Again, y2 = 2.2m A2 = 5.5 × 2.2 = 12.1m2 P2 = (5.5 + 2 × 2.2) = 9.9m 12.1 2 R2 = A P2 = 9.9 = 1.22 7.94 2 V2 = Q A2 = 12.1 = 0.66m/s 2

0.02×7.94 −4 ] = [ 12.1×1.22 sf2 = [ AnQ 2/3 ] = 1.32 × 10 R2 2

Avg. friction slope, sf =

sf1 +sf2 = 1.561 × 10−3 2

1.926 E1 = y1 + 2g1 = 0.75 + 2×9.81 = 0.94m V2

0.66 E2 = y2 + 2g2 = 2.2 + 2×9.81 = 2.22m −E1 2.22−0.94 = 0.003−1.561×10 ΔL = Es02−s −3 = 889.5m f

Hence the depth of the channel increases to 2.20m at a location that is approximately 889.5m upstream of the section where the depth is 0.75m. ! ! yL 1 2 −1 2) − 1 = 1 (1 + 8F = 1 + 8 × 7.33 a ya 2 2 yL = 4.15m Where the tail water depth, yt = 3.5m Since yt < yL a free repelled jump will form. 4.58. From the given data: S0 = 0.01, m = 3, b = 3.00 m, n = 0.015, Q = 20 m3 /s, and y = 1.00 m. (a) The normal depth of ﬂow is calculated using the Manning equation: 5

1 2 1 A 3 12 1 S Q = AR 3 S02 = n n P 23 0

20 =

1 1 [3yn + 3yn2 ] 3 2 √ 2 (0.01) 0.015 [3 + 2 1 + 32 yn ] 3

which gives yn = 0.82 m

112 For the Manning equation to be valid, n6

RS0 ≥ 9.6 × 10−14

In this case, R=

3(0.82) + 3(0.82)2 A √ = = 0.55 m P 3 + 2 1 + 32 (0.82)

which gives n6

RS0 = (0.015)6

(0.55)(0.01) = 8.45 × 10−13 ≥ 9.6 × 10−14

Hence, the Manning equation is valid, and yn = 0.82 m . (b) When flow conditions are critical, A3 Q2 = g T which requires that 202 [3yc + 3yc2 ]3 = 9.81 3 + 2(3)yc which gives yc = 1.15 m (c) Since yc > yn the slope is a steep slope . Since yn < y < yc , the water surface has a S2 profile . (d) In a S2 water surface profile, the depth decreases in the downstream section, so the depth of 1.1 m must occur upstream of the section where the depth is 1 m. At the location where the depth is 1.1 m: y1 = 1.1 m A1 = [3 + 3y1 ]y1 = [3 + 3(1.1)](1.1) = 6.93 m2 √ √ P1 = 3 + 2 10y1 = 3 + 2 10(1.1) = 9.96 m A1 6.93 = 0.696 m = R1 = P1 9.96 Q 20 = 2.89 m/s V1 = = A1 6.93 ⎤2 ⎡ 2 nQ ⎦ (0.015)(20) ⎣ Sf 1 = = = 0.00304 2 2 (6.93)(0.696) 3 A1 R13

113 and where the depth is 1.00 m: y2 = 1.00 m A2 = [3 + 3y2 ]y2 = [3 + 3(1.00)](1.00) = 6.00 m2 √ √ P2 = 3 + 2 10y2 = 3 + 2 10(1.00) = 9.32 m A2 6.00 = 0.644 m = R2 = P2 9.32 Q 20 V2 = = 3.33 m/s = A2 6.00 ⎡ ⎤2 2 nQ (0.015)(20) ⎣ ⎦ Sf 2 = = = 0.00450 2 2 (6.00)(0.644) 3 A2 R 3 2

Substituting the hydraulic parameters at sections 1 and 2 into the direct-step equation and taking α = 1 gives ! 2 1 y + α V2g 2 ΔL = S̄f − S0 2

2.89 3.33 1.1 + 2×9.81 − 1.00 + 2×9.81 0.00304+0.00450 = − 0.01 2

= 6.26 m Hence, the depth in the channel increases to 1.1 m at a location that is approximately 6.26 m upstream of the section where the depth is 1.0 m. (e) The speciﬁc energy, E1 , at the gaging station is given by E 1 = y1 +

3.332 V12 = 1.0 + = 1.565 m 2g 2(9.81)

Just downstream of the gaging station (at the hump), the speciﬁc energy, E2 , is given by E2 = E1 − 0.2 = 1.565 − 0.2 = 1.365 m Therefore, y2 + y2 +

Q2 = 1.365 2gA2

202 = 1.365 2(9.81)(3y2 + 3y22 )2 20.4 y2 + = 1.365 (3y2 + 3y22 )2

There is no solution to this equation. Therefore, critical depth will occur over the hump, the ﬂow will be choked, and y2 = yc = 1.15 m

114

1.0m

4m 1

3 m

6.0m

4.59. Region 1 A1 =(b + my) y = (6 + 1 × 3) 3 =27m2 Region 2 Bottom width =(6 + 2 × 3 + 2 × 3) = 18m A2 =(18 + 1 × 1) = 19m2 Total Area=46m2 √ √ Wetted Perimeter= 2 × 2 + 2 × 3 + 2 × 3 × 2 + 6 = 23.31m 46 = 1.97m Hydraulic Radius, R=A/P= 23.31

Q= n1 AR2/3 S 1/2 = 32.72m3 /s

4.60. From the given data: ΔL = 100 m, b = 5 m, y1 = 1 m, y2 = 0.9 m, Q = 2.5 m3 /s, and S0 = 0.5%. The energy equation requires that:

ΔL =

y + V2g

S − S0

(1)

115 The following parameters can be calculated from the given data: A1 = by1 = (5)(1) = 5 m2 P1 = b + 2y1 = 5 + 2(1) = 7 m A1 5 = = 0.7143 m R1 = P1 7 Q 2.5 V1 = = 0.5 m/s = A1 5 A2 = by2 = (5)(0.9) = 4.5 m2 P2 = b + 2y2 = 5 + 2(0.9) = 6.8 m A2 4.5 = 0.6618 m = R2 = P2 6.8 Q 2.5 V2 = = 0.5556 m/s = A2 4.5 S0 = 0.005 Substituting the calculated parameters into Equation 1 yields 1 1 V2 S = S0 + y+ ΔL 2g 2 0.52 0.55562 1 1+ − 0.9 + = 0.005970 S = 0.005 + 100 2(9.81) 2(9.81) ⎤2 nQ ⎦ ⎣ ⎡ Sf 1 =

2 3

⎡ Sf 2 =

A1 R1

A2 R2

2.5 (5)(0.7143)

= 0.3915n2

2 3

⎤2

⎣ nQ ⎦ 2 3

(2)

2.5 (4.5)(0.6618)

2 3

= 0.5352n2

1 1 (Sf 1 + Sf 2 ) = (0.3915 + 0.5352)n2 = 0.4634n2 2 2 Combining Equations 8.111 and 8.112 gives S=

(3)

0.005970 = 0.4634n2 which yields n = 0.114 . 4.61. From the given data: Q = 11 m3 /s, b = 5 m, S0 = 0.001, n = 0.035, and y1 = 2 m. Calculate the uniform depth, yn , using the Manning equation 5

1 A3 S0 Q= n P 23 which gives

1 (5yn ) 3 √ 11 = 0.001 0.035 (5 + 2yn ) 23

116 and leads to yn = 2.19 m and therefore 95% of the normal depth is equal to 0.95 × 2.19 = 2.08 m. Using the standardstep method yields the following results: x (m) 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360

y2 (m) 2.00 2.01 2.01 2.02 2.02 2.03 2.03 2.04 2.04 2.05 2.05 2.06 2.06 2.06 2.07 2.07 2.08 2.08

A2 (m2 ) 10.0 10.0 10.1 10.1 10.1 10.1 10.2 10.2 10.2 10.2 10.3 10.3 10.3 10.3 10.3 10.4 10.4 10.4

P2 (m) 9.00 9.01 9.02 9.04 9.05 9.06 9.07 9.08 9.09 9.10 9.10 9.11 9.12 9.13 9.14 9.14 9.15 9.16

R2 (m) 1.11 1.11 1.11 1.12 1.12 1.12 1.12 1.12 1.12 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.14

V2 (m/s) 1.10 1.10 1.09 1.09 1.09 1.08 1.08 1.08 1.08 1.07 1.07 1.07 1.07 1.07 1.06 1.06 1.06 1.06

y1 (m) 2.01 2.01 2.02 2.02 2.03 2.03 2.04 2.04 2.05 2.05 2.06 2.06 2.06 2.07 2.07 2.08 2.08 2.08

0.0013 0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

A1 (m2 ) 10.0 10.1 10.1 10.1 10.1 10.2 10.2 10.2 10.2 10.3 10.3 10.3 10.3 10.3 10.4 10.4 10.4 10.4

P1 (m) 9.01 9.02 9.04 9.05 9.06 9.07 9.08 9.09 9.10 9.10 9.11 9.12 9.13 9.14 9.14 9.15 9.16 9.16

R1 (m) 1.11 1.11 1.12 1.12 1.12 1.12 1.12 1.12 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.13 1.14 1.14

V1 (m/s) 1.10 1.09 1.09 1.09 1.08 1.08 1.08 1.08 1.07 1.07 1.07 1.07 1.07 1.06 1.06 1.06 1.06 1.06

S̄

0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

0.0013 0.0013 0.0013 0.0013 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012 0.0012

The upstream water surface proﬁle is described by x and y2 . 4.62. From the given data: S0 = 0.0005, n = 0.040, Q = 250 m3 /s, y2 = 8 m, b2 = 12 m, m2 = 2, b1 = 16 m, and m1 = 3. Using these data A2 = b2 y2 + m2 y22 = 12(8) + 2(8)2 = 224 m2 P2 = b2 + 2y2 1 + m22 = 12 + 2(8) 1 + 22 = 47.8 m A2 224 = 4.69 m = P2 47.8 Q 250 = 1.12 m/s V2 = = A2 224 ⎡ ⎤2 nQ ⎦ (0.040)(250) Sf 2 = ⎣ = 2 2 (224)(4.69) 3 A2 R23 R2 =

= 0.000254

Use the standard-step equation

ΔL =

!1 V2 y + 2g 2 S̄f − S0

100 = 100 =

y1 + 2g1 − y2 − 2g2 Sf 1 +Sf 2 − S0 2 2

1.12 1 y1 + 2(9.81) − 8 − 2(9.81) Sf 1 +0.000254 − 0.0005 2 y1 + 0.05097V12 − 8.064

0.5Sf 1 − 0.000373

(1)

117 where Q Q 250 = = 2 A1 b1 y1 + m1 y1 16y1 + 3y12 ⎤2 ⎡ ⎤2 ⎡ nQ nQ ⎦ =⎣ 5 ⎦ Sf 1 = ⎣ 2 2 3 3 3 A1 R1 A1 /P1 V1 =

(0.040)(250)

(2)

√ 2 5 (16y1 + 3y12 ) 3 /(16 + 2y1 10) 3

(3)

Substituting Equations 2 and 3 into Equation 1 and solving for y1 gives y1 = 8.00 m Therefore the depth 100 m upstream is 8.00 m . 4.63. Let d be the depth of ﬂow in the ﬂoodplain, and using the Horton equation, 3

2(d + 45)(0.1) 2 + (16)(0.05) 2 ne = 2d + 90 + 6 + 10

2 3

3.023 + 0.0632d = 106 + 2d

2 3

A = 30 + 100d P = 106 + 2d Substituting into the Manning equation gives 5

1 A 3 12 S Q= n P 23 0 2 5 3 (30 + 100d) 3 1 106 + 2d 2 110 = 2 (0.0075) 3.023 + 0.0632d (106 + 2d) 3 which gives d = 0.842 m. Verify that the ﬂow is non constrained in the main channel: 5

100 =

1 1 (10d) 3 (0.0075) 2 0.05 (10 + 2d) 23

which gives d = 3.80 m. Therefore the flow is in the floodplain as calculated previously (d = 0.842 m). (a) If 15 m of the floodway is filled in, 3

(d + 30)(0.1) 2 + (16)(0.05) 2 + (d + 45)(0.1) 2 ne = 2d + 75 + 6 + 10 A = 30 + 85d P = 91 + 2d

2 3

2.551 + 0.0632d = 91 + 2d

2 3

118 Substituting into the Manning equation gives

91 + 2d 110 = 2.551 + 0.0632d

2 3

(30 + 85d) 3 (91 + 2d)

2 3

(0.0075) 2

which gives d = 0.904 m. Therefore, ﬁlling in 15 m of the ﬂoodplain causes the water-level to rise 0.904 m − 0.842 m = 0.062 m = 6.2 cm . (b) At the upstream section (Section 1), y1 = 3 + 0.904 = 3.904 m 2 3.023 + 0.0632(0.904) 3 = 0.0935 n1 = 106 + 2(0.904) A1 = 30 + 100(0.904) = 120.4 m2 V1 = Q/A1 = 0.914 m/s V12 = 0.0425 m 2g P1 = 106 + 2(0.904) = 107.8 m 2

S1 =

(110)(0.0935)(107.8) 3

= 0.006297

(120.4) 3

At the downstream section (Section 2), y2 = 54.50 − [50 − (0.0075)(150)] = 5.625 m 2 3.023 + 0.0632(2.625) 3 = 0.0937 n2 = 106 + 2(2.625) A2 = 30 + 100(2.625) = 292.5 m2 V2 = Q/A1 = 0.376 m/s V22 = 0.00721 m 2g P2 = 106 + 2(2.625) = 111.3 m 2

S2 =

(110)(0.0937)(111.3) 3 5

= 0.0003424

(292.5) 3

Combining the results, 0.006297 + 0.0003424 S1 + S2 = = 0.003320 2 2 ! 2 1 y + V2g [3.904 + 0.0425] − [5.625 + .00721] 2 ΔL = = 0.003320 − 0.0075 S̄ − S0 S̄ =

which yields ΔL = 403 m .

119 4.64. From the given data: z1 = 102.05 m, Z1 = 105.27 m, m = 3, b1 = 20 m, S0 = 2%, n = 0.07, L = 100 m, b2 = 10 m, Q = 12 m3 /s, and y1 = 1.60 m. (a) From the given data: A1 = b1 y1 + my12 = 20(1.60) + 3(1.60)2 = 39.68 m2 P1 = b1 + 2y1 1 + m2 = 20 + 2(1.60) 1 + 32 = 30.1 m ⎞2 ⎛ ⎞2 ⎛ 2 2 2 3 3 (0.07)(12)(30.1) nQ nQP 1 ⎠ ⎠ =⎝ = = 0.0003100 Sf 1 = ⎝ 2 5 5 3 3 39.68 3 A1 R1 A1 ⎛ ⎞2 2 4 4 3 3 P23 nQP 2 P2 2 ⎠ ⎝ = (0.07 × 12) 10 = 0.07056 10 Sf 2 = 5 A23 A23 A23 4

1 P3 S̄f = (Sf 1 + Sf 2 ) = 0.0001550 + 0.3528 210 2 A3 2

The energy equation is 2 1.6 + 0.3024 2(9.81)

100 =

1 − y2 + 2×9.81

12 A2

S̄f − 0.02

Solving gives y2 = 3.586 m . Checking the Froude numbers of the upstream and downstream flows verifies that both are subcritical. (b) From the given data, the elevation of the floodplain at the bridge section is 105.27 − (0.02)(100) = 103.27 m. The elevation of the water surface at the bridge section is 102.05 − (0.02)(100) + 3.586 = 103.64 m. Therefore, since the water surface elevation is higher than the elevation of the floodplain, the floodplain will be flooded . 4.65. The usual energy equation that is used in the standard-step method is ! 2 B y + α V2g A ΔL = S̄f − S0 which can be written as S̄f ΔL − S0 ΔL = yB + αB

VB2 V2 − yA + α A A 2g 2g

(1)

where

zB − zA (2) ΔL and zA and zB are the elevations of the bottom of the channel at stations A and B respectively. Denoting the water-surface elevations at A and B by ZA and ZB respectively, then S0 =

Z A = z A + yA

(3)

Z B = z B + yB

(4)

120 Combining Equations 1 to 4 gives VB2 V2 − (zA + yA ) − αA A 2g 2g 2 2 V V = ZB + αB B − ZA − αA A 2g 2g 2 VB VA2 = ZB + αB − ZA + αA 2g 2g

S̄f ΔL = (zB + yB ) + αB

which can be written as

B V2 Z +α = S̄f ΔL 2g A

(5)

From the given data: ΔL = 140 m, Q = 280 m3 /s, n = 0.040, and ZA = 517.4 m. From the given cross-section information, AA = 144.98 m2 PA = 48.09 m AA 144.98 = 3.01 m = RA = PA 48.09 Q 280 = = 1.93 m/s VA = AA 144.98 ⎡ ⎤2 (0.040)(280) nQ ⎦ = SA = ⎣ 2 2 (144.98)(3.01) 3 AA R 3

= 0.001373

Taking S̄f =

SA + SB ΔL

and αA = αB = 1, the energy equation between sections A and B, Equation 5, can be written as V 2 SB VA2 SA + ΔL − B + ΔL ZB = ZA + 2g 2 2g 2 VB2 1.932 0.001373 SB = 517.4 + + (140) − + 140 2(9.81) 2 2(9.81) 2 which simpliﬁes to

where

ZB = 517.69 − 0.05097VB2 + 70SB

(6)

⎤2 ⎡ ⎤2 nQ (0.040)(280) ⎦ =⎣ ⎦ = 125.44 SB = ⎣ 2 2 AB RB3 AB RB3 A2B RB3

(7)

⎡

121 Combining Equations 6 and 7 gives the following form of the energy equation that is most useful for backwater computations, ZB = 517.69 − 0.05097VB2 +

8778 4

A2B RB3

(8)

Iterative calculations to determine ZB are indicated in the following table, where the initial estimate of ZB is 517.60 m. (1) ZB (m) 517.60 517.48 517.47

(2) AB (m2 ) 86.96 84.33 84.11

(3) PB (m) 32.77 32.45 32.42

(4) RB (m) 2.65 2.60 2.59

(5) VB (m/s) 3.22 3.32 3.33

(6) ZB (m) 517.48 517.47 517.47

The calculations begin with an assumption of ZB in Column 1, and the corresponding area, AB , in Column 2 and wetted perimeter, PB , in Column 3 are obtained from the given (tabular) data. The hydraulic radius, RB in Column 4 is obtained using RB = AB /PB , and the average velocity, VB , in Column 6 is obtained using VB = Q/AB . The values of AB , RB , and VB corresponding to the assumed value of ZB are substituted into the energy equation, Equation 8, to yield the calculated value of ZB shown in Column 6. If the calculated value of ZB in Column 6 is not equal to the assumed value of ZB in Column 1, then the calculations are repeated with assumed value of ZB equal to the calculated value of ZB . Based on the above calculations, the water-surface elevation at station B is 517.47 m . At Station B, the channel invert elevation is 515.10 m and hence the depth of ﬂow, yB is given by yB = 517.47 − 515.10 = 2.37 m Since AB = 84.11 m2 , and the channel is approximately rectangular, the the width of the channel at section B is 84.11/2.37 = 35.49 m. Also, since the piers present an obstruction 2.50 m wide, the width of the channel adjacent to the piers is 35.49 m − 2.50 m = 32.99 m. The speciﬁc energy at section B, EB , is given by E B = yB +

VB2 3.332 = 2.37 + = 2.94 m 2g 2(9.81)

Let yP be the depth of ﬂow adjacent to the bridge pier, then neglecting energy losses requires that VP2 2g Q2 2.94 = yP + 2gA2P E B = yP +

2.94 = yP +

2802 2g(32.99yP )2

122 which simpliﬁes to

2.94 = yP +

3.672 yP2

(9)

which yields yP = 2.13 m

1.78 m

Since the flow at station B is subcritical (Fr < 1), the flow adjacent to the pier is probably subcritical also, and therefore the depth of flow at the pier is 2.13 m. The water surface elevation is 2.13 m + 515.10 m = 517.23 m .

4.66. From the given data: Q = 220 m3 /s, Z1 = 13.5 m, y1 = 1.00 m, Z2 = 21.5 m, b = 10 m, and n = 0.01. Based on these data, the downstream depth in the stilling basin is given by

y2 = Z2 − (Z1 − y1 ) = 21.5 − (13.5 − 1.0) = 9 m

Let L be the length of the stilling basin, so the hydraulic jump occurs at L/2. For y2 = 9 m, the conjugate depth equation gives

1 y2 = y1 2 9 1 = y1 2

−1 +

Q2 1+8 2 3 gb y1

−1 +

2202 1+8 (9.81)(10)2 y13

1 = 2

−1 +

394.7 1+8 3 y1

which gives y1 = 1.087 m. The next step is to ﬁnd the distance, ΔL, over which the depth increases from 1.0 m to 1.087 m. This distance is given by

ΔL =

y + V2g

S̄f − S0

(1)

123 and the variables to be substituted into this equation are:

y1 = 1 m V1 = 22 m/s A1 = 1 × 10 = 10 m2 P1 = 2 × 1 + 10 = 12 m A1 10 = 0.8333 m = R1 = P1 12 2 nQ 2 (0.01)(220) Sf 1 = = = 0.06172 2 2 AR 3 (10)(0.8333) 3 y2 = 1.087 m A2 = 1.087 × 10 = 10.87 m2 220 m/s = 20.24 m/s V2 = 10.87 P2 = 2 × 1.087 + 10 = 12.17 m A2 10.87 = 0.8606 m = R2 = P2 12.17 2 nQ 2 (0.01)(220) Sf 2 = = = 0.05004 2 2 AR 3 (10.87)(0.8606) 3 Sf 1 + Sf 2 0.06172 + 0.05004 = = 0.05588 S̄f = 2 2 S0 = 0

Substituting into Equation 1 gives

ΔL =

! ! 222 20.242 − 1.087 + 2(9.81) 1 + 2(9.81) 0.05588 − 0

= 66.3 m

Therefore, the length of the stilling basin should be 2 × 66.3 m = 133 m .

4.67. From the given data: b = 5 m, m = 2, n = 0.018, S0 = 0.001, T W = 1.00 m, and Q = 20 m3 /s. Use the standard-step method, where 2

ΔL =

Q Q [y1 + α1 2gA 2 ] − [y2 + α2 2gA2 ] 1

S̄f − S0

124 where ΔL = 100 m, α1 = α2 = 1, and the objective is to ﬁnd the depth y2 at the gate, given the depth y1 at a location 100 m upstream of the gate. In this case, y1 = 2.20 m − S0 ΔL = 2.20 − (0.001)(100) = 2.10 m A1 = by1 + = my12 = 5(2.10) + 2(2.10)2 = 19.32 m2 P1 = b + 2 1 + m2 y1 = 5 + 2 1 + 22 (2.10) = 14.39 m A2 = by2 + = my22 = 5y2 + 2y22 P2 = b + 2 1 + m2 y2 = 5 + 2 1 + 22 y2 = 5 + 4.472y2 ⎡ ⎤2 ⎡ ⎤2 2 3 nQ ⎦ nQP1 ⎦ Sf 1 = ⎣ =⎣ 2 5 A1 R13 A13 2

(0.018)(20)(14.39) 3

= 0.000234

(19.32) 3 ⎤2 2

⎡ Sf 2 = ⎣

nQP23 ⎦ 5

A23

(0.018)(20)(5 + 4.472y2 ) 3

= 0.1296

(5y2 + 2y22 ) 3

(5 + 4.472y2 ) 3 10

(5y2 + 2y22 ) 3

S̄f = 0.5(Sf 1 + Sf 2 ) = 0.0001172 + 0.0648

(5 + 4.472y2 ) 3 10

(5y2 + 2y22 ) 3

Substituting into the standard-step equation yields 2

100 =

20 20 [2.10 + 2(9.81)(19.32) 2 ] − [y2 + 2(9.81)(5y +2y 2 )2 ] 2

0.0001172 + 0.0648

4 (5+4.472y2 ) 3 10 (5y2 +2y22 ) 3

− 0.001

which rearranges to 4

20.39 (5 + 4.472y2 ) 3 + 6.48 y2 + 10 − 2.243 = 0 2 2 (5y2 + 2y2 ) (5y2 + 2y 2 ) 3 2

Viable solutions to this equation are y2 = 2.18 m and y2 = 0.73 m, which correspond to subcritical and supercritical flow conditions respectively (F2 = 0.26 and F2 = 1.75). Since the upstream flow is subcritical (y1 = 2.10 m, F1 = 0.28), the flow at the gate must also be subcritical, in which case y2 = 2.18 m. Under these conditions, Q = 20 m3 /s, HW = 2.18 m, T W = 1.00 m, and the gate discharge relationship requires that √ Q = 13.3h HW − T W √ 20 = 13.3h 2.18 − 1.00 which yields h = 1.38 m. Therefore, the gate must be opened at least 1.38 m to prevent the water elevation 100 m upstream of the gate from exceeding an elevation of 2.20 m.

125 4.68. (a) y2 = 1.25 m: In this case the ﬂow is contained entirely in the main channel and the ﬂow variables are as follows: A = 23(1.25) = 28.75 m2 P = 180 + 2(1.25) = 28.04 m 5

1 28.75 3 1 A3 = 1169 m3 /s K= 2 = n2 P 3 0.025 28.04 23 α= 1 Fr∗ =

αQ2 T (1)(250)2 (23) = = 6.17 gA3 (9.81)(28.75)3

(b) y2 = 3.25 m: In this case the ﬂow is contained in all three parts of the compound channel and the ﬂow variables are as follows: A1 = 180(3.25 − 2.52) = 131.4 m2 P1 = 180 + (3.25 − 2.52) = 180.7 m 5

1 A13 1 131.4 3 K1 = = 1180 m3 /s 2 = n1 P 3 0.090 180.7 23 1

A2 = 23(3.25) = 74.75 m2 P2 = 23 + 2(2.52) = 28.04 m 5

1 A23 1 74.75 3 K2 = = 5749 m3 /s 2 = n2 P 3 0.025 28.04 23 2

A3 = 130(3.25 − 2.52) = 94.9 P3 = 130 + (3.25 − 2.52) = 130.7 m 5

1 A33 1 94.9 3 = 1095 m3 /s K3 = 2 = n3 P 3 0.070 130.7 23 3

A = A1 + A2 + A3 = 131.4 + 74.75 + 94.9 = 301.1 m2 K = K1 + K2 + K3 = 1180 + 5749 + 1095 = 8024 m3 /s A2 K13 K23 K33 α= 3 + 2 + 2 = 6.01 K A21 A2 A3 Δα 6.006 − 5.999 dα ≈ = 0.750 = dy Δy2 3.25 − 3.24 T = w1 + w2 + w3 = 180 + 23 + 130 = 333 m Fr∗ =

αQ2 T Q2 dα = 0.441 − gA3 2gA2 dy

126 An increment of Δy2 = 0.01 m was used in estimating dα/dy. At y2 = 1.25 m the flow is supercritical , and at y2 = 3.25 m the flow is subcritical . 4.69. The calculated flow variables at section BD are as follows: yD1 = 2.29 m AD1 = 11.4 m2 PD1 = 7.29 m KD1 = 309.1 m3 /s yD2 = 4.79 m AD2 = 239.4 m2 PD2 = 102.9 m KD2 = 14015 m3 /s yD3 = 2.29 m AD3 = 11.4 m2 PD3 = 7.29 m KD3 = 309.1 m3 /s AD = 262.3 m2 KD = 14633 m3 /s αD = 1.06 Sf D = 0.00042 VD = 1.14 m/s This gives the stage at BD as 4.79 m + 83.21 m = 88.00 m .

Chapter 5

Design of Drainage Channels

WPSinθ WP

τS

WPτb

5.1. Assumptions 1. A & B have the same physical properties. Let ‘α’ be the same internal friction angle. Naturally the bank inclination ‘θ’ should be less than ‘Φ’, for the particle ‘B’ to remain stable, even under a dry canal condition. When there is a flow of water, there is a tendency for the particle ‘A’ to be dragged along the direction of canal bed slope. Whereas the particle ‘B’ tries to get disloged in an inclined direction due to shear stress of the flowing water. Consider a particle of a noncohesive material with a(submerged) weight, WP on the bottom of a channel. This particle resists the shear force of the flowing fluid by the friction force between the particle & surrounding physically on the bottom of the channel. The friction 127

128 force is given by μP WP where μP is the co-eﬃcient of friction between particles on the bottom of the channel. When particle motion on the bottom of the channel is incipient, the shear stress on the bottom of the channel is equal to the permissible shear stress, τP and AP τP = μP W

(equation 5.23)

Where, AP is the eﬀective surface area of a particle on the bottom. The co-eﬃcient of friction between particles on the bottom of the channel is related to the angle of repose (α) of the particle material by μP = tan α

(equation 5.24)

Combining equation 5.23 and 5.24 lead to the following expression. P τP = W AP tan α

(equation 5.25)

Again for particle B on the side of the channel, The total force, FP tending to move a particle on the side of the channel is given by equation 5.26. FP = (τs AP )2 + (WP Sinθ)2 Where τS is the shear stress exerted by the flowing fluid of the side of the channel. Again equation 5.27 Ff = WP cos θ tan α When motion is incipient, FP = Ff and the shear stress on the side of the channel is equal to permissible shear stress on the side of the channel τP s WP cos θ tan α (τP s AP )2 + (WP sin θ)2 tan2 θ P ⇒ τP s = W cos θ tan α 1 − tan (equation 5.28) 2α AP tan2 θ ⇒ ττPPs = cosθ 1 − tan 2α Hence, the shear stress required for moving a sand grain on the side slope is less than that required to move it on the bed. 5.2. From the given data: Q= 15 m3 /s, S0 =1×10−4 , m = 1.5 and d50 = 25 mm Use the most efficient trapezoidal section with m = 1.5, in which case equation 5.11 gives the bottom width, b as b = 2( (1 + m2 ) − m)y =2( (1 + 1.52 ) − 1.5) y=0.606 y The corresponding flow area, A & wetted perimeter, P are given by A = by + my 2 = (0.606y)y + 1.5y 2 = 2.106y 2 P = b + 2ys (1 + m2 ) = (0.606) + 2y (1 + 1.52 ) = 4.212 y Step 1: consider the case where lining is used. In this case, the perimeter of the channel consists of gravel mulch linings.

129 Step 2: Manning equation (as given in table 5.6) as n = 0.031

(equation 5.36)

1/2

Q = n1 AR2/3 S0 5/3

1/2

⇒ Q = n1 PA2/3 S0

2 5/3

1 (2.106 y ) ⇒ 15 = 0.031 (1 × 10−4 )1/2 (4.212)2/3

Which yield y=3.80 m b= 2.3 m step 3: the maximum shear stress on the bottom of the channel τb = γy S0 = 9790 × 3.8 × (1 × 10−4 ) = 3.72 Pa From equation 5.18 Ks = 0.066 m + 0.67 = 0.066 × 1.5 + 0.67 = 0.76 And so the maximum shear stress exerted on the side of the channel, τs is given by τs = Ks τb

(equation 5.38)

= 2.83 Pa Step 4: The permissible shear stress, τp on the gravel mulch lining can be estimated from table 5.7 for d50 = 25 mm as 19 Pa. The side slope angle θ is given by 1 ) = 33.7◦ θ = tan−1 ( m

The tractive-force ration, K is given by equation 5.29 as 2 (33.7) = 0.253 K = 1 − sin sin2 (35) & hence the permissible shear stress on the side of the channel, τP s is given by, τP s = Kτp = 0.253 × 19 = 4.81 Pa Step 5: Since the maximum shear stress on bottom of the channel (3.72 Pa) is less than the permissible shear stress on the bottom of the channel (1.9 Pa), and the maximum shear stress on the side of the channel (2.83 Pa) is less permissible shear stress on the side of the channel (4.81 Pa), the lining is adequate. For a depth of flow of 3.8 m & bottom width of 2.3 m, the flow area is 31.41 m2 & the flow velocity (for Q= 15 M3 /s) is 0.5 m/s. Acoording to equation 5.34 F= 0.3 m The height of the lining above the bottom of the channel should be at least (3.8+0.3)=4.1 m

130 Size of the channel, b=2.3 m & the lining should extend at least 4.1 m above the bottom of the channel. 5.3. According to equation 4.152 K = n1 AR2/3

(i)

For a constant value of n it is thus possible to express K as a unique function of y for a given channel. In gradually varied ﬂow computation it is convenient to use a relation of the following form K 2 = C.y N

(ii)

Where C is a co-eﬃcient & N is the second hydraulic exponent for uniform ﬂow computation ⇒ 2 ln K = 0 + N ln y d ⇒ 2 dy (ln k) = Ny

(iii)

From equation (i) ⇒ ln K = ln A + 23 ln R − ln n d 2 dR ⇒ dy (ln k) = A1 dA dy + 3R dy

(iv)

Equating (iii) & (iv) N 2 dR ⇒ 2y = A1 dA dy + 3R dy T 2 d T 2T 2 A + 3R dy (A/P ) = A + 3A − 3P (dP/dy)

(as P=b+2y So, dP dy = 2)

2y [5T − 2R dP ⇒ N = 3A dy ]

For a deep, narrow rectangular channel b/y → 0 2y b×y Therefore, N = 3×(b×y) [5b − 2( b+2y ) × 2] 2 4b [5b − b/y+2 ] = 3b 4 ] = 23 [5 − b/y+2

= 23 × [5 − 42 ]

[as b/y→ 0]

Therefore, N=2 5.4. Lining on the sides only

√ Here for the bed, n1 =0.012 . P2 =2×1.6× 1 + 1.52 =5.77 m

P=P1 +P2 =5+5.77=10.77 m Equivalent roughness by formula in table 4.3 3 2

3 2 2

+5.77×0.012 ] /3 n= [5×0.025 10.77 = 0.019 2/3

5.5. From the given data: b = 3 m, m = 2, ymax = 1 m, S0 = 0.005, n = 0.020, and τp = 25 Pa. It can be assumed that γ = 9790 N/m3 (at 20◦ C). Under limiting conditions, the bottom shear stress, τb , is equal to τp , in which case 25 = γRS0 = 9790

(3)y + (2)y 2 by + my 2 √ √ (0.005) = 9790 (0.005) b + 2 1 + m2 y 3 + 2 1 + 22 y

131 which yields y = 0.714 m. At this depth of ﬂow, A = by + my 2 = (3)(0.714) + (2)(0.714)2 = 3.162 m2 P = b + 2 1 + m2 y = 3 + 2 1 + 22 (0.714) = 6.193 m 5

1 1 (3.162) 3 1 A 3 12 2 = 7.14 m3 /s Q= 2 S0 = 2 (0.005) nP3 0.020 (6.193) 3

Therefore, the maximum ﬂow rate for which the lining will be stable is 7.14 m3 /s . 5.6. The maximum shear stress on the channel boundary, τmax , and the average ﬂow velocity, vave , in the channel are given by τmax = γyS0 1 2 1 vave = R 3 S02 n

(1) (2)

Since vave = vp when τmax = τp , and eliminating S0 from Equations 1 and 2 yields 1 2 vp = R 3 n

τp γy

1 2

nγ

1 2

R4 y3

τp2

and taking γ = 9790 N/m2 gives vp =

1 n(9790)

which simpliﬁes to 0.010 vp = n

1 2

R4 y3

τp2

(3)

It is better to design a channel based on maximum permissible shear stress rather than maximum permissible velocity since the maximum permissible shear stress remains constant under all flow conditions, while the maximum permissible velocity varies with flow conditions, as evidenced by Equation 3. 5.7. From the given data: ds = 2.5 mm = 0.0025 m and SG = 2.65. At 20◦ C, γ = 9790 kN/m3 and ν = 1.00 × 10−6 m2 /s. Hence, ds γs 0.0025 − 1 gds = 0.1 0.1 [(2.65) − 1] (9.81)(0.0025) = 159 ν γ 1.00 × 10−6 Using this value (159) in the the Sheilds diagram (Figure 5.5) yields an intersection point of around τ∗ = 0.050 and Re∗ = 120. Using the definition of τ∗ given by Equation 5.22, the critical shear stress, τc , is given by τc = τ∗ (γs − γ)ds = τ∗ (SG − 1)γds = (0.050)(2.65 − 1)(9790)(0.0025) = 2.0 N/m2 = 2.0 Pa Therefore, the (maximum) permissible stress on the bottom of the channel is 2.0 Pa .

132 5.8. Dividing Equation 5.28 by Equation 5.25 yields K=

τps = cos θ τp

1−

tan2 θ = cos θ tan2 α

tan2 α − tan2 θ tan2 α

Taking the cos θ under the square root yields cos2 θ tan2 α − sin2 θ K= tan2 α Multiplying the numerator and denominator by cos2 α yields cos2 θ sin2 α − sin2 θ cos2 α K= sin2 α Using the identities cos2 θ = 1 − sin2 θ and sin2 α = 1 − cos2 α in the numerator yields (1 − sin2 θ)(1 − cos2 α) − sin2 θ cos2 α K= sin2 α which simpliﬁes to

1 − sin2 θ − cos2 α sin2 α

Using the identity cos2 α = 1 − sin2 α yields 1 − sin2 θ − (1 − sin2 α) sin2 α − sin2 θ = K= sin2 α sin2 α which simpliﬁes to

1−

sin2 θ sin2 α

5.9. The tractive-force ratio, K, is given by Equation 5.29 as τps sin2 θ = 1− K= τp sin2 α Equations 5.16 and 5.17 indicate that the actual shear stress on the side of the channel, τs , is Ks times of the concurrent shear stress on the bottom of the channel. Therefore, as long as the tractive force ratio is greater than Ks , the side slope will be stable when the bottom is unstable. Hence, the condition for side-slope failure is sin2 θ 1− < Ks sin2 α

133 which gives sin2 θ > 1 − Ks2 2 sin α or sin θ > sin α 1 − Ks2

(1)

where it is noted that Ks is a function of θ as given by Equation 5.18 which can be expressed in the form ⎧ ⎪ 0.77 m ≤ 1.5 ⎪ ⎪ ⎨ Ks = 0.066 + 0.67 1.5 < m < 5 ⎪ ⎪ tan θ ⎪ ⎩ 1.0 m≥5 and m = 1/ tan θ. When θ = 25◦ and α = 30◦ , then m=

1 = 2.14 tan θ

Ks = 0.066m + 0.67 = 0.811 sin 25◦ sin θ =√ = 0.722 1 − 0.8112 1 − Ks2 sin α = sin 30◦ = 0.5 Therefore, Equation 1 is satisﬁed and so yes the side slope will become unstable before the bottom. 5.10. From the given data: S0 = 0.01, b = 4 m, m = 2.5, d50 = 250 mm, d85 = 300 mm, and τp = 200 Pa. The side slope angle, θ, is given by 1 1 −1 −1 = tan = 21.8◦ θ = tan m 2.5 Since the stone lining is derived from an open gradation, the angle of repose, α, can be estimated from Table 5.2 (Equation 1) as d85 0.125 300 0.125 = (37.1) = 38.0◦ α = α0 d50 250 Therefore, the tractive force ratio, K, is given by Equation 5.29 as sin2 (21.8) = 0.798 K = 1− sin2 (38.0) The permissible shear stress on bottom of the channel, τp , is 200 Pa, and the permissible shear stress on the side of the channel, τps , is given by τps = Kτp = (0.798)(200) = 160 Pa

134 The limiting condition for stability of the bottom lining is 200 = γyS0 = (9790)y(0.01) which yields y = 2.04 m. The tractive force ratio, K, is given by Equation 5.18 as K = 0.066m + 0.67 = 0.066(2.5) + 0.67 = 0.835 Therefore, the limiting condition for stability on the side lining is 160 = KγyS0 = (0.835)(9790)y(0.01) which yields y = 1.96 m. Therefore, the side lining controls the channel and the depth of ﬂow should not be greater than 1.96 m . 5.11. From the given data: τp = 60 Pa, nb = 0.023, b = 2 m, m = 4, S0 = 0.005, and y = 0.80 m. Taking γ = 9790 N/m3 , the maximum shear stress in the straight segment of the channel, τb , is given by τb = γyS0 = (9790)(0.80)(0.005) = 39.2 Pa Therefore, the allowable bend factor, Kr , is Kr =

τr 60 = 1.53 = τb 39.2

According to Equation 5.31, Kr = 2.38 − 0.206

r c

+ 0.0073

r 2 c

T T r r 2 c c 1.53 = 2.38 − 0.206 + 0.0073 T T which yields rc /T = 23.2 or 5.0. Since Equation 5.31 is valid for 2 < rc /T < 10, the only valid solution is rc /T = 5.0. Therefore, rc = 5.0T = 5.0[b + 2my] = 5.0[2 + 2(4)(0.80)] = 42 m So the radius of curvature of the channel should not be less than 42 m for the lining to remain stable. 5.12. Let us consider here the trapezoidal section of side slope m:1 (H:V) The ﬂow area, A & the wetted perimeter ‘P’ are given by A = my 2 + by √ p = 2y m2 + 1 + b

(equation 5.3) √

√ Thus P = Ay − my + 2y m2 + 1 = P = Ay + y(2 m2 + 1 − m)

Considering A & m to be constants dP dy = 0

⇒ (b + my) = (2 (m2 + 1) − m)y

(equation 5.4)

135 ⇒ b = (2 (m2 + 1) − m)y Top surface width √ T=b+2my=2 m2 + 1y Again ΔOPQ, OQ= OP sin θ = T2 √m12 +1 √

= 2 m2 +1 × √m12 +1 = y Thus semi circle with ‘0’ as centre & ‘y’ as radius is tangential to the bed & sides of the most eﬃcient trapezoidal section. (b+my)y √ R = (b+2 = y2 m2 +1y

If the side slope is assumed to be available & A and y are constant, the condition for the most eﬃcient section dP dm = 0

⇒ y 2√m22 +1 2m − y = 0 √ ⇒ 2m = m2 + 1 ⇒ m = √13 Therefore, θ=60◦ Thus the most eﬃcient section in a trapezoidal section is one half of a hexagon. 5.13. The recommended freeboard given by Equation 5.33 is F = 0.15 +

V2 2g

and so for F > 0.30 m the required velocity is given by 0.15 +

V2 > 0.30 2g

→

V >

2g(0.15)

→

V > 1.72 m/s

Also for channels on steep gradients with ﬂow depths greater that 30 cm will require freeboards greater than 30 cm. 5.14. From the given data: b = 10 m, m = 2, S0 = 0.00053, n = 0.030, Q = 28 m3 /s, and rc = 100 m. Determine the normal depth using the Manning equation 5

1 A3 S0 Q= n P 23 which gives

1 (10y + 2y 2 ) 3 √ 28 = 0.00053 √ 0.030 (10 + 2y 5) 23

136 and solving for y gives y = 2.00 m and the corresponding velocity is given by V =

Q 28 = = 1.00 m/s A 10(2) + 2(2)2

Since y ≥ 0.30 m and V ≤ 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is 0.30 m. The superelevation, hs , around a bend is given by hs =

V 2T V 2 [b + 4y] 1.002 [10 + 4(2)] = 0.018 m = = grc grc (9.81)(100)

The required additional freeboard around the bend is hs /2 = 0.009 m = 0.9 cm. Since this value is small and a conservative freeboard of 30 cm is already being used in the straight section of the channel, then maintaining a freeboard of 30 cm in the channel bend is appropriate. Hence the design flow depth is 2.00 m and the minimum required freeboard is 30 cm . 5.15. From the given data: Q = 30 m3 /s, S0 = 0.002. For float-finished concrete, take n = 0.015. Use the best hydraulic section: b = 1.15y, m = 0.58, A = 1.73y 2 , P = 3.46y, and R = 0.5y. According to the Manning equation 5

1 A3 S0 Q= n P 23

1 (1.73y 2 ) 3 √ 30 = 0.002 0.015 (3.46y) 23 which gives y = 2.30 m The ﬂow velocity, V , is given by V =

30 30 Q = = = 3.27 m/s A 1.73y 2 1.73(2.30)2

Since y ≥ 0.30 m and V > 1.72 m/s, the freeboard, F , estimated by Equation 5.34 is F = 0.15 +

V2 3.272 = 0.15 + = 0.70 m 2g 2(9.81)

Therefore, for the most eﬃcient channel, the minimum required channel depth is 2.30 m + 0.70 m = 3.00 m , the bottom width is 1.15(2.30) = 2.65 m and the side slope is 0.58:1 (H:V). 5.16. According to Table 4.1 1

6 n=0.047d50

=0.047×(3.5 × 10−6 )1/6

137 =0.018 From Trapezoidal channel A=by+my 2 = 15.8 × 1.5 + 2 × (1.5)2 = 28.2 m2 √ P= b+2y 1 + m2 15.8 + 2 × 1.5 (1 + 22 ) = 22.51 m 28.2 = 1.25 m R = PA = 22.51

The side share stress factor Ks , is given by Equation 5.18 as:Ks = 0.066m+0.67 = 0.06 × 2 + 0.67 =0.802 & so that maximum shear stress exerted on the side of the channel, τs = Ks × τb = Ks × vyS0 The side slope angle θ is given by: 1 ) = tan−1 12 = 26.6◦ = tan−1 ( m Again, the tractive-force ratio, K is given by Equation 5.29 as: 2 26.6◦ K = 1 − sin = 0.642 sin2 36◦ It is worth nothing that since K<Ks it is guaranted that the side shear stress controls the stability of the channel. Equation 5.47 gives the permissible shear stress on the soil. τps = 0.5d50 = 0.5 × (3.5) = 1.75P a & so the permissible shear stress on the side of the channel, τps is given by τps = k × τp = 0.642 × 1.75 = 1.12Pa The maximum shear stress exerted by ﬂowing water on the sides of the channel, τp s is given by τp s = Ks × vyS0 At the limit of stability τs = τP s 1.12 = 0.802 × 9790 × 1.5 × S0 S0 = 9.51 × 10−5 The permissible longitudinal slope is 9.51×10−5 5.17. Discharge, Q= 2 m3 /s Velocity, V= 1.2 m/s N= 0.03

138 According to table 5.1 A=y2 , P= 4y, T= 2y, R=y/2 Hence the Manning’s equation gives 1/2

Q= n1 AR2/3 S0

1/2

1 ⇒ 2 = 0.03 ( y2 )2/3 S0

× y2

(1)

Again 1/2

1 ⇒ 1.2 = 0.03 ( y2 )2/3 S0

(2)

Equation (1)÷(2) 2 = y2 ⇒ 1.2

Therefore, y=1.29≈ 1.3 m. and b=2.6 m. The best dimension of the channel is, (2.6 m× 1.3 m) Again 1/2

1 × (1.3/2)2/3 × S0 ⇒ 1.2 = 0.03

Therefore, S0 = 2.3 × 10−3 m/m The slope required for the channel is 2.3×10−3 m/m. 5.18. For most eﬃcient section in case of rectangular channel (table 5.1) b=2y R=y/2 P=b+2y ⇒ 3.5 = b + 2y = 4y Therefore, y= 3.5 4 = 0.875 m and b= 1.75 m 1/2

Q = n1 AR2/3 S0

1 = 0.015 (1.75 × 0.875) × (1/2000)1/2 m3 /s = 2.283 m3 /s

Again equation 4.46 gives n 6 −3 m )6 = ( 0.015 Ks = ( 0.039 0.039 ) m = 3.24 × 10

√ Since Ks Rs > 2.2×10−5 , the flow is fully turbulent, however since R/Ks < 500, so the Manning’s equation is valid. 5.19. From the given data: Q = 0.42 m3 /s, b = 0.4 m, m = 3, and S0 = 0.008. The flow area, A, and the wetted perimeter, P , can be expressed in terms of the flow depth, y, by the relations A = by + my 2 = 0.4y + 3y 2 P = b + 2y 1 + m2 = 0.4 + 2y 1 + 32 = 0.4 + 6.325y Step 1: Try a gravel mulch lining with d50 = 25 mm.

139 Step 2: For typical gravel-mulch linings, the Manning’s n depends on the ﬂow depth as shown Table 5.6. This functional relationship can be expressed as n(y) and the Manning equation gives 5

1 A 3 12 Q= S n P 23 0

1 1 (0.4y + 3y 2 ) 3 2 0.42 = 2 (0.008) n(y) (0.4 + 6.325y) 3

(1)

Solving Equation 1 simultaneously with the n versus y relationship in Table 5.6 yields y = 0.349 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 ) τb = γyS0 = (9790)(0.349)(0.008) = 27.3 Pa The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(27.3) = 23.7 Pa Step 4: The permissible shear stress, τp , on the gravel-mulch lining is estimated from Table 5.7 for d50 = 25 mm as 19 Pa. Since the lining is non-cohesive, the permissible shear stress on the side of the channel, τps (=Kτp ), will be less than 19 Pa. Step 5: The maximum shear stress on bottom of the channel (27.3 Pa) is greater than the permissible shear stress on the gravel-mulch lining with d50 = 25 mm (19 Pa), and so the gravel-mulch lining with d50 = 25 mm is inadequate. Try another a larger stone size. Step 1: Try gravel mulch lining with d50 = 50 mm. Step 2: Solving the Manning equation (Equation 1) simultaneously with the n versus y relationship in Table 5.6 yields y = 0.389 m. Step 3: The maximum shear stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.389)(0.008) = 30.5 Pa and the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks τb = (0.868)(30.5) = 26.5 Pa Step 4: The permissible shear stress, τp , on the gravel-mulch lining can be estimated from Table 5.7 for d50 = 50 mm as 38 Pa. The angle of repose, α, of the gravel mulch can be estimated from Table 5.2 (Equation 2) for subangular-shaped stones as α = α0 (d50 )0.00778 = (34.3)(50)0.00778 = 35.4◦

140 The side-slope angle θ is given by θ = tan

−1

1 m

= tan

−1

1 = 18.4◦ 3

The tractive-force ratio, K, is given by Equation 5.29 as K=

sin2 θ 1− = sin2 α

1−

sin2 (18.4) = 0.839 sin2 (35.4)

and hence the permissible shear stress on the side of the channel, τps is given by τps = Kτp = (0.839)(38) = 31.9 Pa Step 5: Since the maximum shear stress on bottom of the channel (30.5 Pa) is less than the permissible shear stress on the bottom of the channel (38 Pa), and the maximum shear stress on the side of the channel (26.5 Pa) is less permissible shear stress on the side of the channel (31.9 Pa), the lining is adequate. Final speciﬁcation: The recommended channel design has a gravel mulch lining with d50 = 50 mm. 5.20. From the given data: S0 = 0.05%, Q = 0.2 m3 /s. Concrete Channel, √ Best Hydraulic Section: For the best hydraulic section, m = 1 , A 2 = y , P = 2 2y, and for unﬁnished concrete n = 0.017. Using the Manning equation gives 5

1 A3 1 S0 Q= n P 23 2

1 (y 2 ) 3 1 0.2 = √ 2 (0.0005) 0.017 (2 2) 3 2 which yields y = 0.640 m. For this depth of ﬂow, A = y 2 = (0.640)2 = 0.4096 m2 0.2 Q = = 0.49 m/s V = A 0.4096 Using a freeboard of 0.30 m gives a total depth of 0.640 m + 0.30 m = 0.940 m and hence the excavated volume per km length is Vexcav = (0.940)2 (1000) = 884 m3 Since the charge is $100/m3 , the total cost per km is 884 m3 × $100/m3 = $88,400/km .

141 Gravel Mulch, d50 = 50 mm: For serviceability, take m = 3 . The best hydraulic section (m = 1) cannot be used because of likely slope-stability problems. For a triangular section, A = my 2 = 3y 2 P = 2 1 + m2 y = 2 1 + 32 y = 6.325y Using these geometric relationships in the Manning equation gives 5

1 A3 1 Q= S0 n P 23 2

1.2 =

1 1 (3y 2 ) 3 2 (0.0005) n (6.325y) 3 2

which yields

y = 1.815n 8

(1)

Assuming that 0.5 m < y < 1.0 m, the n-function is given by n = 0.046 − 0.008y

(2)

Solving Equations 1 and 2 simultaneously yields y = 0.551 m and n = 0.0416. The assumption associated with Equation 2 is validated. The maximum shear stress on the side of the channel, τs , is given by τs = Ks γRS0

(3)

From the given data: Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 R=

A 3(0.551)2 = = 0.261 m P 6.325(0.551)

Substituting these calculated parameters into Equation 3, with γ = 9790 N/m3 , gives τs = (0.868)(9790)(0.261)(0.0005) = 1.1 Pa The permissible shear stress on the side of the channel, τps , is given by τps = Kτb where τb = 38 Pa. For round stone, α = 31.5◦ (50)0.00778 = 32.5◦ 1 θ = tan−1 = 18.4◦ 3 2 sin θ sin2 18.4◦ = 1− K = 1− = 0.809 2 sin α sin2 32.5◦ A 3(0.551)2 R= = = 0.261 m P 6.325(0.551)

(4)

142 Substituting the calculated parameters into Equation 4 gives τps = (0.809)(38) = 31 Pa Since τs τps (i.e., 1.1 Pa 38 Pa), the channel is more than adequate. For y = 0.551 m and m = 3: A = my 2 = 3(0.551)2 = 0.911 m2 0.2 Q = = 0.22 m2 V = A 0.911 Using a freeboard of 0.30 m gives a total depth of 0.551 m + 0.30 m = 0.851 m and hence the excavated volume per km length is Vexcav = 3(0.851)2 (1000) = 2170 m3 Since the charge is $70/m3 , the total cost per km is 2170 m3 × $70/m3 = $151,900/km . Comparison: For the concrete-lined best section the cost is $88,400/km, and for the gravel mulch section the cost is $151,900/km. The concrete lined channel is the better alternative in terms of cost. The diﬀerence is $151,900/km − $88,400/km = $63,500/km . 5.21. From the given data: b = 0.90 m, m = 3, S0 = 0.03, soil classiﬁcation SC, PI = 16, e = 0.5, Q = 0.5 m3 /s, sod-grass lining in good condition, and h = 0.075 m. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 P = b + 2y 1 + m2 = (0.9) + 2y 1 + 32 = 0.9 + 6.325y R=

A 0.9y + 3y 2 = P 0.9 + 6.325y

Step 1: Determine the ﬂow depth in the channel. Equation 5.41 gives Cs = 106, and Equation 5.42 gives Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.075)0.528 = 0.142 The average shear stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 ) τ0 = γRS0 = (9790)

0.9y + 3y 2 0.9 + 6.325y

(0.03) =

293.7(0.9y + 3y 2 ) 0.9 + 6.325y

and the Manning’s n is given by Equation 5.43 as n = Cn τ0−0.4 = (0.142)

293.7(0.9y + 3y 2 ) 0.9 + 6.325y

−0.4

= 0.0146

0.9 + 6.325y 0.9y + 3y 2

0.4 (1)

143 The Manning equation gives 5

1 A 3 12 Q= S n P 23 0

1 1 (0.9y + 3y 2 ) 3 2 0.5 = 2 (0.03) n (0.9 + 6.325y) 3

(2)

Solving Equations 1 and 2 simultaneously yields y = 0.213 m and n = 0.032. Step 2: Determine the eﬀective stress on the underlying soil. The maximum stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.213)(0.03) = 62.6 Pa For sod grass in good condition, Table 5.9 gives Cf = 0.90, and assuming that d75 < 1.3 mm Equation 5.45 gives ns = 0.016, and the eﬀective shear stress on the soil underlying the grass lining is given by Equation 5.46 as τe = τb (1 − Cf )

n 2 s

= (62.6)(1 − 0.9)

0.016 0.032

2 = 1.6 Pa

Step 3: Determine the the permissible shear stress on the soil underlying the vegetative lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (1.6 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is adequate . 5.22. From the given data: Q = 35 m3 /s, S0 = 0.1%, d75 = 30 mm (moderately rounded), and T = 17 m. For moderately angular gravel, Figure 5.7 gives α = 38.5◦ . Assume that the tractive force on the side is limiting. The tractive force ratio, K, is given by 2 sin θ sin2 θ 2.58 1 − 2.58 sin2 θ = 1 − 1 − = (1) = K = 1− 2 2 1 + m2 sin α sin 38.5◦ where the side slopes are m:1 (H:V). Equation 5.47 gives the permissible shear stress on the soil (= permissible shear stress on the bottom of the channel) as τp = 0.75d75 = 0.75(30) = 22.5 Pa

144 The permissible shear stress on the side of the channel is therefore given by τps = Kτp = 22.5 1 −

2.58 1 + m2

(2)

Assuming that 1.5 < m < 5, the side-shear stress factor, Ks is estimated by Equation 5.18 as Ks = 0.066m + 0.67 and so the maximum shear stress exerted on the side of the channel, τs , is given by τs = Ks γyS0 = (0.066m + 0.67)(9790)y(0.001) = (0.646m + 6.56)y

(3)

The maximum depth of ﬂow for stability of the channel boundary, y, occurs when τs = τps and so combining Equations 2 and 3 gives (0.646m + 6.56)y = 22.5 1 −

2.58 1 + m2

or 2.58 1 + m2

(4)

b + 2my = 17 → b = 17 − 2my

(5)

22.5 0.646m + 6.56

1−

The top-width constraint requires that

For d75 = 30 mm, Equation 5.45 estimates Manning’s n as 1

6 n = ns = 0.015d75 = 0.015(10) 6 = 0.022

and the Manning equation requires that 5

1 A 3 12 S Q= n P 23 0 35 =

1 1 (by + my 2 ) 3 2 √ 2 (0.001) 0.022 (b + 2y 1 + m2 ) 3

(6)

Equations 4 to 6 are three equations in three unknowns which yield m = 3.34 , b = 1.70 m , and y = 2.29 m. The channel is practical since m ≥ 3 as would be required for practicality, and the side-slope angle (θ = tan−1 (1/3.34) = 16.7◦ ) is much less than the angle of repose (38.5◦ ). 5.23. From the given data: Q = 0.5 m3 /s, S0 = 0.002, PI = 25, e = 0.4, and the soil is classiﬁed as CH. For practicality and functionality, take the side slope as 3:1 (H:V), so m = 3. Step 1: Determine the maximum flow depth for stability. For a soil classiﬁcation of CH and PI = 25, Table 5.10 gives c3 = 0.097, c4 = 1.38, c5 = −0.373, and c6 = 48.

145 Therefore, the permissible shear stress on the channel boundary, τp , is given by Equation 5.48 as τp = τp,c = [c1 PI2 + c2 PI + c3 ][c4 + c5 e]2 c6 = [0.097][1.38 − 0.373(0.4)]2 (48) = 7.1 Pa The maximum shear stress on the bottom of the channel, τb , is given by (for γ = 9790 N/m3 ) τb = γyS0 = (9790)(y)(0.002) = 19.6y The maximum allowable flow depth causes τb = τp , which requires that 19.6y = 7.1 Pa which yields y = 0.362 m. Step 2: Determine the bottom width of the channel. The flow must be of sufficient size to accommodate the design flow of 0.5 m3 /s. Manning’s n can be estimated by Equation 5.45 as n = 0.016, and the Manning equation requires that 5

1 [by + my 2 ] 3 1 1 A 3 12 2 S = Q= √ 2 2 S0 0 nP3 n [b + 2y 1 + m2 ] 3 5

1 1 [b(0.362) + (3)(0.362)2 ] 3 2 0.5 = √ 2 (0.002) 0.016 [b + 2(0.362) 1 + 32 ] 3

which yields b = 0.368 m. The corresponding flow area and average velocity are 0.53 m2 and 0.94 m/s respectively. The required freeboard, F , according to Equation 5.34 is 0.30 m. Hence, the designed channel should have a bottom width of 0.368 m , side slopes of 3:1 , and a total (minimum) depth of 0.362 m + 0.30 m = 0.66 m . Step 3: Compare the channel with the most efficient section. The bottom width of the most efficient section is given by Equation 5.11 as b = 2y( 1 + m2 − m) = 2(0.291)( 1 + 32 − 3) = 0.094 m Since the minimum required channel width (0.362 m) is greater than the bottom width of the most efficient section, then the most efficient section cannot be used . 5.24. From the given data: S0 = 0.003, d75 = 30 mm (moderately angular), d = 3 m, b = 6 m, and m = 2.5 Step 1: Calculate the stable flow depth. The maximum allowable shear stress on the bottom of the channel, τp , can be estimated using Equation 5.47 as τp = 0.75d75 = 0.75(30) = 22.5 Pa Assuming a freeboard of F = 0.30 m, the maximum allowable flow depth is d − F = 3 m − 0.30 m = 2.70 m. Check that flows with this depth will not erode the channel boundary. The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(2.5) + 0.67 = 0.835

146 For d75 = 30 mm and moderately rounded, Figure 5.7 gives the angle of repose as α = 38.1◦ , and for m = 2.5 the side slope angle is θ = 21.8◦ . The tractive force ratio, K, is therefore given by Equation 5.29 as sin2 θ sin2 21.8◦ 1 − = 0.798 K = 1− = sin2 α sin2 38.1◦ Since K < Ks , the side slope is limiting for channel erosion. The permissible shear stress on the side, τps , and the shear stress exerted by the ﬂowing water on the side, τs , are given by τps = Kτp = (0.798)(22.5) = 18.0 Pa τs = Ks τb = Ks γyS0 = (0.835)(9790)y(0.003) = 24.5y At the limit of stability, τs = τps , which requires that 24.5y = 18.0 Pa which yields y = 0.735 m. This result shows that the physical maximum ﬂow depth in the channel (2.70 m) cannot be safely achieved without erosion. The capacity of the channel must be calculated using y = 0.735 m. Step 2: Determine the capacity of the channel. Manning’s n in the channel can be estimated using Equation 5.45 which yields 1

6 = 0.015(30) 6 = 0.0264 n = 0.015d75

and the capacity of the channel is given by the Manning equation as 5

1 A 3 12 Q= S n P 23 0

1 [by + my 2 ] 3 1 2 Q= S √ n [b + 2y 1 + m2 ] 32 0 5

1 1 [(6)(0.735) + (2.5)(0.735)2 ] 3 2 = 8.30 m3 /s Q= √ 2 (0.003) 0.0264 [(6) + 2(0.735) 1 + 2.52 ] 3

The safe capacity of the channel is 8.30 m3 /s . 5.33. From the given data: b = 0.90 m, m = 3, S0 = 0.03, soil classiﬁcation SC, PI = 16, e = 0.5, Q = 0.5 m3 /s, and vegetation lining with class D retardance. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 P = b + 2y 1 + m2 = (0.9) + 2y 1 + 32 = 0.9 + 6.325y R=

A 0.9y + 3y 2 = P 0.9 + 6.325y

147 Step 1: Determine the ﬂow depth in the channel. Table 5.11 gives Cn = 0.147. The average shear stress on the channel boundary, τ0 , is given by Equation 5.44 as (assuming γ = 9790 N/m3 ) 293.7(0.9y + 3y 2 ) 0.9y + 3y 2 (0.03) = τ0 = γRS0 = (9790) 0.9 + 6.325y 0.9 + 6.325y and the Manning’s n is given by Equation 5.43 as n = Cn τ0−0.4 = (0.147)

293.7(0.9y + 3y 2 ) 0.9 + 6.325y

−0.4 = 0.0151

0.9 + 6.325y 0.9y + 3y 2

0.4 (1)

The Manning equation gives 5

1 A 3 12 S Q= n P 23 0

1 1 (0.9y + 3y 2 ) 3 2 0.5 = 2 (0.03) n (0.9 + 6.325y) 3

(2)

Solving Equations 1 and 2 simultaneously yields y = 0.216 m and n = 0.032. Step 2: Determine the eﬀective stress on the underlying soil. The maximum stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.216)(0.03) = 63.4 Pa For mixed vegetation in good condition, Table 5.9 gives Cf = 0.75, and assuming that d75 < 1.3 mm Equation 5.45 gives ns = 0.016, and the eﬀective shear stress on the soil underlying the grass lining is given by Equation 5.46 as n 2 0.016 2 s τe = τb (1 − Cf ) = (63.4)(1 − 0.75) = 4.0 Pa n 0.032 Step 3: Determine the the permissible shear stress on the soil underlying the vegetative lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (4.0 Pa) is greater than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is inadequate . 5.26. From the given data, the channel properties are: b = 5 m, m = 3, d = 1 m, and S0 = 0.001.

148 (a) Find capacity of grass channel. For grass with retardance C, Table 5.11 gives Cn = 0.220, and for sodded grass in good condition Table 5.9 gives Cf = 0.90. For sandyclay soil assume d75 < 1.3 mm so Manning’s n of the bare soil is given by Equation 5.45 as ns = 0.016. Assuming a required freeboard of F = 0.30 m, the maximum ﬂow depth in the channel is y = d− F = 1 m − 0.30 m = 0.70 m. Using these variables, the relevant channel and ﬂow properties are calculated as follows: A = by + my 2 = (5)(0.70) + (3)(0.70)2 = 4.97 m2 P = b + 2y 1 + m2 = 5 + 2(0.7) 1 + 32 = 9.43 m 4.97 A = = 0.527 m R= P 9.43 τ0 = γRS0 = (9790)(.527)(0.001) = 5.2 Pa n = Cn τ0−0.4 = (0.220)(5.2)−0.4 = 0.114 5

1 1 (4.97) 3 1 A 3 12 (0.001) 2 = 0.90 m3 /s Q= 2 S0 = nP3 0.114 (9.43) 23

τb = γyS0 = (9790)(0.70)(0.001) = 6.9 Pa n 2 0.016 2 s τe = τb (1 − Cf ) = (6.9)(1 − 0.90) = 0.01 Pa n 0.114 Hence the eﬀective stress on the underlying soil is 0.01 Pa. The permissible stress on the underlying soil can be calculated using the parameters in Table 5.10 for CL soil with PI = 14 and e = 0.4, which correspond to the parameters: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.48, c5 = −0.57, and c6 = 4.8 × 10−3 . Substituting these parameters into Equation 5.48 gives the permissible shear stress on the underlying soil, τp,c , as τp,c = [c1 PI2 + c2 PI + c3 ][c4 + c5 e]2 c6 = [(1.07)(14)2 + (14.3)(14) + (47.7)][(1.48) + (−0.57)(0.4)]2 (4.8 × 10−3 ) = 2.8 Pa Since τe τp,c (i.e., 0.01 Pa 2.8 Pa), the lining is adequate when y = 0.70 m, which corresponds to Q = 0.90 m3 /s . (b) Find capacity of bare-soil channel. Since the maximum bottom stress (5.2 Pa) is greater than the allowable soil shear stress (2.8 Pa), the maximum allowable depth under bare-soil conditions is less than 0.70 m. The limiting depth occurs when τ0 = τp,c , which requires that γRS0 = 2.8 (9790)R(0.001) = 2.8 which yields R = 0.286 m. Using the deﬁnition of the R requires that by + my 2 √ = 0.286 m b + 2y 1 + m2 5y + 3y 2 √ = 0.286 m 5 + 2y 1 + 32

149 which yields y = 0.340 m. For this ﬂow depth, P = 7.15 m, A = 2.05 m2 , and Manning’s equation gives 5

1 1 A 3 12 1 (2.05) 3 2 = 1.76 m3 /s Q= 2 S0 = 2 (0.001) nP3 0.016 (7.15) 3

Therefore the maximum ﬂow rate under bare-soil conditions is 1.76 m3 /s . (Almost double the capacity when there is no lining. The is caused by the much lower value of n for the case of no lining.) 5.27. From the given data: m = 3, S0 = 1.4% = 0.014, and Q = 1.1 m3 /s. For a Bermuda grass of height 4 cm, Table 5.8 indicates a retardance classiﬁcation of E. From the given channel dimensions, the area, A, and wetted perimeter, P , and hydraulic radius, R, are given by A = 3y 2 P = 2 1 + 32 y = 6.325y A = 0.4743y R= P

(1)

Step 1: Determine the ﬂow depth in the channel. For a retardance of E, Equation 5.56 gives a = 52.1, and Equation 5.55 gives Manning’s n as 1

1.22R 6 1.22R 6 = n= 52.1 + 19.97 log[R1.4 (0.014)0.4 ] 52.1 + 19.97 log[R1.4 S00.4 ]

(2)

The Manning equation requires that 5

1 A 3 12 S n P 23 0

1 1 (3y 2 ) 3 0.8 = (0.014) 2 n (6.325y) 23

(3)

Solving Equations 1, 2, and 3 simultaneously yields y = 0.578 m . Step 2: Determine the maximum shear stress on the bottom of the channel. The maximum shear stress on the bottom of the channel, τb , is given by (assuming γ = 9790 N/m3 ) τb = γyS0 = (9790)(0.578)(0.014) = 79.2 Pa Step 3: Determine the the permissible shear stress on the channel lining. The permissible shear stress on a (retardance) class D lining, τp , is given in Table 5.12 as 16.8 Pa. Since the maximum shear stress exerted on the bottom of the channel (79.2 Pa) is greater than the permissible shear stress on the grass lining (16.8 Pa) the proposed lining is inadequate .

150 5.28. From the given data: Q = 10 m3 /s, S0 = 0.001, b = 3 m, m = 2, and the total depth of the channel is 2 m. The channel is lined with Kentucky bluegrass, which has an estimated retardance of C (Table 5.8). For any ﬂow depth, y, the ﬂow area, A, wetter perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = 3y + 2y 2 √ P = b + 2 1 + m2 y = 3 + 2 5y = 3 + 4.472y R=

3y + 2y 2 A = P 3 + 4.472y

(1)

Step 1: Determine the ﬂow depth in the channel. For a retardance of C, Equation 5.56 gives a = 44.6, and Equation 5.55 gives Manning n as 1

1.22R 6 1.22R 6 = n= 0.4 44.6 + 19.97 log[R1.4 (0.001)0.4 ] 44.6 + 19.97 log[R1.4 S0 ]

(2)

and the Manning equation gives 5

1 A 3 12 S Q= n P 23 0 10 =

1 1 (3y + 2y 2 ) 3 2 2 (0.001) n (3 + 4.472y) 3

(3)

Simultaneous solution of Equations (1) to (2) yields y = 2.31 m. Since this ﬂow depth exceeds the depth of the channel (= 2 m), the channel is inadequate . 5.29. The required freeboard, F , and Manning’s n (for retardance C) are given by F = 0.152 +

Q2 2gA2

(1) 1

1.22R 6 n= 44.6 + 19.97 log(R1.4 S00.4 )

(2)

√ For a depth of ﬂow, y, the geometric properties of the channel are: A = 4y 2 , P = 2 17y = 8.246y, R = A/P = 0.485y, T = 8y, and D = A/T = 0.5y. Substituting into Equation 2 gives 1

1.081y 6 1.22(0.485y) 6 = 44.6 + 19.97 log((0.485y)1.4 (0.010.4 )) 44.6 + 19.97 log(0.0575y 1.4 )

The maximum permissible shear stress is 98.9 Pa, therefore, at the limit of stability, γyS0 = 98.9 Pa → (9790)(y)(0.01) = 98.9 Pa → y = 1.01 m

151 At this ﬂow depth (y = 1.01 m), 1

1.081(1.01) 6 = 0.0543 n= 44.6 + 19.97 log(0.0575 × 1.011.4 ) R = 0.485(1.01) = 0.490 m A = 4(1.01)2 = 4.08 m2 1 2 2 1 1 1 (4.08)(0.490) 3 (0.01) 2 = 4.67 m3 /s Q = AR 3 S02 = n 0.0543 Q 4.67 V = = = 1.14 m/s A 4.08 D = 0.5(1.01) = 0.505 V2 1.142 = = 0.262 (subcritical, OK) gD (9.81)(0.505) 1.142 = 0.218 m → Use F = 0.30 m F = 0.152 + 2(9.91)

Fr2 =

Therefore the maximum channel depth without lining is 1.01 + 0.30 = 1.31 m, and the corresponding ﬂow capacity is 4.67 m3 /s . 5.30. From the given data: Q = 4 m3 /s, S0 = 0.8% = 0.008, y = 1 m, h = 15 cm, e = 0.45, d75 = 0.8 mm, and PI = 8. Take side slope as m = 3 . (a) Following the conventional design process: Cs = 106 (good condition) Cn = 0.35(106)0.10 (0.15)0.528 = 0.205 A = by + my 2 = b(1) + 3(1)2 = b + 3 P = b + 2y 1 + m2 = b + 2(1) 1 + 32 = b + 6.325 b+3 A = R= P b + 6.325 b+3 b+3 τ0 = γRS0 = (9790) (0.008) = 78.32 b + 6.325 b + 6.325 −0.4 −0.4 b+3 b+3 −0.4 = 0.205 78.32 = 0.0358 n = C n τ0 b + 6.325 b + 6.325 Substituting into the Manning equation yields 5

1 A 3 12 S Q= n P 23 0 0.4 5 1 b+3 (b + 3) 3 2 4 = 27.91 2 (0.008) b + 6.325 (b + 6.325) 3 1.602 =

(b + 3)2.067 (b + 6.325)1.067

which yields b = 0.345 m and n = 0.0472.

152 (b) The maximum stress on the bottom of the channel is τb = γyS0 = (9790)(1)(0.008) = 78.32 Pa For sod grass in good condition, Cf = 0.90, and for the underlying soil ns = 0.016. The eﬀective stress on the underlying soil is given by τe = τb (1 − Cf )

n 2 s

= (78.32)(1 − 0.90)

0.016 0.0472

2 = 0.900 Pa

Since the soil is cohesionless, the permissible stress is τp = 1 Pa. Since τe < τp the lining is adequate . (c) The maximum shear stress on the bottom of the channel is given by τb = γyS0 = (9790)(1)(0.008) = 78.32 Pa In the alternative retardance-based approach, for Class C lining the permissible shear stress, τp , is 47.9 Pa. Since τb > τp the lining is inadequate . 5.31. From the given data: b = 0.9 m, m = 3, S0 = 0.03, Q = 0.3 m3 /s, soil classiﬁcation SC, PI = 16, and e = 0.5. For the proposed RECP lining, τI = 100 Pa. The ﬂow area, A, wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = 0.9y + 3y 2 P = b + 2y 1 + m2 = 0.9 + 2y 1 + 32 = 0.9 + 6.325y R=

0.9y + 3y 2 A = P 0.9 + 6.325y

Step 1: Determine the ﬂow depth in the channel. For the RECP lining, the parameters a and b are given by Equations 5.62 and 5.63 as nupper 0.036 0.033 nmid ln = −1.44 ln ln = −0.138 b = −1.44 ln nlower nmid 0.040 0.036 −b a = nmid · τmid = (0.036) · (100)0.138 = 0.0680

The average shear stress on the channel boundary, τ0 , is given by (assuming γ = 9790 N/m3 ) τ0 = γRS0 = (9790)

0.9y + 3y 2 0.9 + 6.325y

(0.03) = 293.7

0.9y + 3y 2 0.9 + 6.325y

The Manning’s n is given by Equation 5.61 as n = aτ0b = (0.0680)

293.7

0.9y + 3y 2 0.9 + 6.325y

−0.138

= 0.0310

0.9y + 3y 2 0.9 + 6.325y

−0.138 (1)

153 The Manning equation requires that 5

1 A 3 12 Q= S n P 23 0 0.3 =

1 1 (0.9y + 3y 2 ) 3 2 2 (0.03) n (0.9 + 6.325y) 3

(2)

Solving Equations 1 and 2 simultaneously gives y = 0.188 m and n = 0.041. Step 2: Determine the eﬀective stress on the underlying soil. The shear stress on the bottom of the channel, τb , is given by τb = γyS0 = (9790)(0.188)(0.03) = 55.2 Pa and the eﬀective shear stress on the underlying soil, τe , is given by Equation 5.64 as 100 6.5 τI 6.5 = 55.2 − = 2.1 Pa τe = τb − 4.3 τI 4.3 100 Step 3: Determine the the permissible shear stress on the soil underlying the RECP lining and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (2.1 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed lining is adequate .

5.32. From the given data: d50 = 0.15 m, γs = 25.9 kN/m3 , b = 0.6 m, m = 3, S0 = 2%, and Q = 1.13 m3 /s. For any ﬂow depth, y, the wetted perimeter, P , the ﬂow area, A, the top width, T , and the average depth, ȳ, are given by P = b + 2y 1 + m2 = 0.60 + 2y 1 + 32 = 0.6 + 6.325y A = by + my 2 = (0.60)y + 3y 2 = 0.6y + 3y 2 T = b + 2my = (0.60) + 2(3)y = 0.6 + 6y ȳ =

A 0.6y + 3y 2 = T 0.6 + 6y

(1)

Step 1: Determine the ﬂow depth in the channel. Assume that 1.5 ≤ ȳ/d50 ≤ 185, then Manning’s n is given by Equation 5.67 as 1

0.319ȳ 6 2.25 + 5.23 log

ȳ d50

(2)

154 and the Manning Equation requires that 5

1 A 3 12 Q= S n P 23 0

1 1 (0.6y + 3y 2 ) 3 2 1.13 = 2 (0.02) n (0.6 + 6.325y) 3

(3)

Solving Equations 1 to 3 simultaneously yields y = 0.501 m, ȳ = 0.292 m, and ȳ/d50 = 1.95. Since 1.5 ≤ ȳ/d50 ≤ 185, the assumed expression for n (Equation 2) is validated. Step 2: Determine the maximum shear stresses exerted on the bottom and sides of the channel. The maximum shear stress on the bottom of the channel, τp , is given by Equation 5.73 as (assuming γ = 9790 N/m3 ) τb = SF · γyS0 = SF · (9790)(0.501)(0.02) = 98.1 · SF Pa

(4)

The side-shear-stress factor, Ks , is given by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 and so the maximum shear stress on the sides of the channel, τs , is given by τs = Ks τp = (0.868)(98.1 · SF) = 85.2 · SF Pa

(5)

Step 3: Determine the permissible shear stress on the bottom and sides of the channel and assess the adequacy of the lining. The shear-velocity Reynolds number, Re, under design ﬂow conditions is given by Equation 5.77 as (assuming ν = 10−6 m2 /s) √ (9.81)(0.292)(0.02)(0.15) g ȳS0 d50 = = 3.6 × 10−4 Re = ν 10−6 Using this value of Re, the corresponding values of τ∗ and SF given by Equation 5.76 as τ∗ = 0.047 and

SF = 1.0

and hence the permissible shear stress on the bottom of the channel, τp , is given by Equation 5.75 as τp = τ∗ (γs − γ)d50 = (0.047)(25900 − 9790)(0.15) = 114 Pa The side-slope angle, θ is given by θ = tan

−1

1 = 18.4◦ 3

and the angle of repose, α can be estimated from Figure 5.6, which yields 41.3◦ , very angular α= 38.5◦ , very rounded

155 Since the riprap is subangular, taking the average of α for “very angular” and “very rounded” stones gives α = 39.9◦ and hence the tractive force ratio, K, is given by Equation 5.82 as sin2 θ sin2 (18.4◦ ) K = 1− = = 0.871 1 − sin2 α sin2 (39.9◦ ) The permissible shear stress on the side of the channel, τps is given by 5.81 as τps = Kτp = (0.871)(114) = 99.3 Pa Using the calculated safety factor of SF = 1.0, the maximum shear stress on the bottom of the channel is given by Equation 4 as τb = 98.1(1.0) = 98.1 Pa and the maximum shear stress on the side of the channel is given by Equation 5 as τs = 85.2(1.0) = 85.2 Pa. Since the maximum shear stress on the bottom of the channel (98.1 Pa) is less than the permissible shear stress on the bottom of the channel (114 Pa), and the maximum shear stress on the sides of the channel (85.2 Pa) is less than the permissible shear stress on the sides of the channel (99.3 Pa), the proposed lining is adequate . 5.33. From the given data: S0 = 0.002, d50 = 25 mm (moderately rounded), and Q = 0.2 m3 /s. The lining will function as a gravel-mulch lining. Assume γs = 25.9 kN/m3 . Specify the typical recommended value of 3:1 (H:V) side slope (m = 3), and using Equation 5.18 the side-shear-stress factor, Ks , can be estimated as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868 For d50 = 25 mm Figure 5.6 gives

α=

38.5◦ , 31.5◦ ,

very angular very rounded

Since the lining is moderately, taking the average of α for “very angular” and “very rounded” stones gives α = 35◦ and hence the tractive force ratio, K, is given by Equation 5.82 as sin2 θ sin2 (18.4◦ ) = = 0.835 1 − K = 1− sin2 α sin2 (35◦ ) Assume a value for the bottom width, b, substitute the given parameters into the following equations, and solve for the ﬂow depth, y: A = by + 3y 2 P = b + 2y 1 + 32 T = b + 2(3)y A ȳ = T 1 0.319ȳ 6 n= 2.25 + 5.23 log(ȳ/d50 ) 5

1 A 3 12 S n P 23 0

(1)

156 For the solved value of y, verify that 1.5 ≤ ȳ/d50 ≤ 185 to validate the assumed functional form of n. Determine Re using Equation 5.77 as (with ν = 10−6 m2 /s) √ Re =

g ȳS0 d50 ν

Based on the calculated value of Re, determine τ∗ and SF from Equation 5.76. The calculate the following shear stresses: τb = SF · γyS0 τs = Ks τb τp = τ∗ (γs − γ)d50 τps = Kτp = 0.835τp If τb < τp and τs < τps the assigned value of b is adequate. For an optimal design it is recommended that b be changed incrementally in intervals of 0.1 m until these criteria are just satisfied. In the present case this occurs at b = 3.1 m, which has a flow depth of 9.1 cm. In accordance with the freeboard guidelines us a freeboard of 15 cm, so the minimum depth of the lined channel is 9.1 cm + 15 cm ≈ 24 cm. So the designed channel should have a bottom width of 3.1 m , side slopes of 3:1 , and a minimum lined depth of 24 cm . 5.34. From the given data: MT = 0.23 m, d50 = 0.15 m, γs = 25.9 kN/m3 , b = 0.60 m, m = 3, S0 = 9%, and Q = 0.28 m3 /s. For any flow depth, y, the wetted perimeter, P , the flow area, A, the top width, T , and the average depth, ȳ, are given by P = b + 2y

1 + m2 = 0.60 + 2y

1 + 32 = 0.60 + 6.325y

A = by + my 2 = (0.60)y + 3y 2 = 0.6y + 3y 2 T = b + 2my = (0.60) + 2(3)y = 0.6 + 6y ȳ =

0.6y + 3y 2 A = T 0.6 + 6y

(1)

Step 1: Determine the ﬂow depth in the channel. Assume that 1.5 ≤ ȳ/d50 ≤ 185, then Manning n is given by Equation 5.67 as 1

0.319ȳ 6

2.25 + 5.23 log

ȳ d50

(2)

and the Manning Equation requires that 5

1 A 3 12 S n P 23 0

1 1 (0.6y + 3y 2 ) 3 2 0.28 = 2 (0.09) n (0.6 + 6.325y) 3

(3)

157 Solving Equations 1 to 3 simultaneously yields y = 0.244 m, ȳ = 0.158 m, and ȳ/d50 = 1.05. Since ȳ/d50 < 1.5, the assumed expression for n (Equation 2) is not validated. Assume that 0.3 < ȳ/d50 < 1.5, then Manning n is given by Equation 5.67 as 1

ȳ 6 n= √ gf1 f2 f3

(4)

where 0.28 (0.6y + 3y 2 ) (9.81)ȳ 0.453 ȳ 0.814 0.15 d50 0.453 ȳ 0.814 = 1.14 β = 1.14 T d50 0.6 + 6y 0.15 log(0.755/β) 0.28Fr f1 = β 0.6 + 6y 0.492 1.025((0.6+6y)/0.15)0.118 f2 = 13.434 β 0.15 0.6 + 6y −β f3 = ȳ

Fr =

(5) (6) (7) (8) (9)

Solving Equations 3 to 9 simultaneously yields y = 0.182 m, ȳ = 0.123 m, and ȳ/d50 = 0.82. Since 0.3 < ȳ/d50 < 1.5 the assumed expression for n (Equation 4) is validated. It is also noted that Equation 5 gives Fr = 1.22, indicating that the ﬂow is supercritical. Step 2: Determine the maximum shear stress exerted on the bottom of the channel. The default safety factor, SF, for gabions is 1.25, which is based on an assumed τ∗ = 0.10. The safety factor can also be calculated based on the friction-velocity Reynolds number, Re, where (assuming ν = 10−6 m2 /s) √ (9.81)(0.123)(0.09)(0.15) g ȳS0 d50 = = 4.9 × 104 Re = ν 10−6 and interpolating with Equation 5.76 gives SF = 1.03. Use the more conservative SF = 1.25. The maximum shear stress on the bottom of the channel, τb is therefore given by (assuming γ = 9790 N/m3 ) τb = SF · γyS0 = (1.25) · (9790)(0.182)(0.09) = 200 Pa Step 3: Determine the permissible shear stress on the bottom of the channel and assess the adequacy of the lining. Taking τ∗ = 0.10, Equation 5.88 gives τp = τ∗ (γs − γ)d50 = (0.10)(25900 − 9790)(0.15) = 242 Pa and taking Equation 5.89 gives τp = 0.0091(γs − γ)(MT + 1.24) = 0.0091(25900 − 9790)(0.23 + 1.24) = 216 Pa Therefore, taking the maximum of the two estimations gives τp = 242 pa. Since the maximum shear stress on the bottom of the channel (200 Pa) is less than the permissible shear stress (242 Pa), the proposed lining is adequate .

158 5.35. From the given data: m = 3, b = 0.9 m, S0 = 2%, h = 0.20 m, soil classification SC, PI = 16, e = 0.5, Q = 0.28 m3 /s. For a concrete lining, use nL = 0.013. For the grass lining, Equation 5.41 gives Cs = 106 and Equation 5.42 gives Cn = 0.35Cs0.10 h0.528 = 0.35(106)0.10 (0.20)0.528 = 0.239 For a depth of flow, y, expressions for the flow area, A, and wetted perimeter, P , and low-flow wetted perimeter, PL , are given by A = by + my 2 = (0.9)y + (3)y 2 = 0.9y + 3y 2 P = b + 2y 1 + m2 = 0.9 + 2y 1 + 32 = 0.9 + 6.325y PL = b = 0.9 m Step 1: Determine the flow depth in the channel. The Manning’s n is given by Equation 5.99 as

3 23 PL PL ns 2 n= nL + 1− P P nL 2 0.9 ns 32 3 0.9 + 1− = (0.013) 0.9 + 6.325y 0.9 + 6.325y 0.013

(1)

For grassed sides ns is given by Equation 5.43 as ns = Cn τ0−0.4 = Cn [γRS0 ]−0.4 −0.4 0.9y + 3y 2 = (0.239) 9790 (0.02) 0.9 + 6.325y −0.4 0.9 + 3y 2 = 0.0290 0.9y + 6.325y

(2)

From the Manning equation 5

1 A 3 12 S n P 23 0

1 1 (0.9y + 3y 2 ) 3 2 0.28 = 2 (0.02) n (0.9 + 6.325y) 3

(3)

Solving Equations 1 to 3 simultaneously yields the ﬂow depth, y = 0.214 m and n = 0.046. Step 2: Determine the eﬀective stress on the soil underlying the side of the channel. There is no need to calculate the maximum shear stress on the bottom of the channel, since with a concrete bottom lining erosion will not be an issue. For a side slope of 3:1 (H:V), the side-shear-stress-factor, Ks , can be estimated by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(3) + 0.67 = 0.868

159 and so the maximum shear stress on the sides of the channel, τs , can be estimated as (assuming γ = 9790 N/m3 ) τs = Ks τp = Ks [γyS0 ] = Ks [γyS0 ] = (0.868)[(9790)(0.214)(0.02)] = 36.4 Pa For mixed grass in good condition, Table 5.9 gives Cf = 0.75, for d75 < 1.3 mm (assumed) Equation 5.45 gives ns = 0.016, and the maximum effective shear stress, τe , on the soil underlying the grass lining on the side of the channel is given by Equation 5.46 as n 2 0.016 2 s τe = τs (1 − Cf ) = (36.4)(1 − 0.75) = 1.1 Pa n 0.046 Step 3: Determine the permissible shear stress on the side of the channel and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 (for SC soil) as: c1 = 1.07, c2 = 14.3, c3 = 47.7, c4 = 1.42, c5 = −0.61, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(16)2 + 14.3(16) + 47.7][1.42 + (−0.61)(0.5)]2 (4.8 × 10−3 ) = 3.3 Pa Since the shear stress on the underlying soil (1.1 Pa) is less than the permissible shear stress on the underlying soil (3.3 Pa), the proposed grass lining is adequate . 5.36. From the given data: m = 4, b = 1.2 m, S0 = 1.5%, soil classification ML, PI = 12, e = 0.3, and Q = 0.7 m3 /s. For a concrete lining, use nL = 0.013. For class B vegetation, Table 5.11 gives Cn = 0.418. For a depth of flow, y, expressions for the flow area, A, and wetted perimeter, P , and low-flow wetted perimeter, PL , are given by P =

by + my 2 = (1.2)y + (4)y 2 = 1.2y + 4y 2 b + 2y 1 + m2 = 1.2 + 2y 1 + 42 = 1.2 + 8.246y

PL =

b = 1.2 m

Step 1: Determine the ﬂow depth in the channel. The Manning’s n is given by Equation 5.99 as 3 23 PL ns 2 PL + 1− nL n= P P nL 2 1.2 ns 32 3 1.2 + 1− = (0.013) (1) 1.2 + 8.246y 1.2 + 8.246y 0.013 For grassed sides ns is given by Equation 5.43 as ns = Cn τ0−0.4 = Cn [γRS0 ]−0.4 −0.4 1.2y + 4y 2 (0.015) = (0.418) 9790 1.2 + 8.246y −0.4 1.2y + 4y 2 = 0.0568 1.2 + 8.246y

(2)

160 From the Manning equation 5

1 A 3 12 Q= S n P 23 0

1 1 (1.2 + 4y 2 ) 3 2 0.7 = 2 (0.015) n (1.2 + 8.246y) 3

(3)

Solving Equations 1 to 3 simultaneously yields the flow depth, y = 0.409 m and n = 0.081. Step 2: Determine the effective stress on the soil underlying the side of the channel. For a side slope of 4:1 (H:V), the side-shear-stress-factor, Ks , can be estimated by Equation 5.18 as Ks = 0.066m + 0.67 = 0.066(4) + 0.67 = 0.934 and so the maximum shear stress on the sides of the channel, τs , can be estimated as (assuming γ = 9790 N/m3 ) τs = Ks τp = Ks [γyS0 ] = Ks [γyS0 ] = (0.934)[(9790)(0.409)(0.015)] = 56.1 Pa For mixed grass in good condition, Table 5.9 gives Cf = 0.75, for d75 < 1.3 mm (assumed) Equation 5.45 gives ns = 0.016, and the maximum effective shear stress, τe , on the soil underlying the grass lining on the side of the channel is given by Equation 5.46 as n 2 0.016 2 s τe = τs (1 − Cf ) = (56.1)(1 − 0.75) = 0.5 Pa n 0.081 Step 3: Determine the permissible shear stress on the side of the channel and assess the adequacy of the lining. Since PI > 10, the soil is cohesive and the parameters to estimate the permissible shear stress are given in Table 5.10 (for ML soil) as: c1 = 1.07, c2 = 7.15, c3 = 11.9, c4 = 1.48, c5 = −0.57, and c6 = 4.8 × 10−3 . The permissible shear stress on the underlying soil, τp,c , is then given by Equation 5.48 as τp,c = (c1 PI2 + c2 PI + c3 )(c4 + c5 e)2 c6 = [1.07(12)2 + 7.15(12) + 11.9][1.48 + (−0.57)(0.3)]2 (4.8 × 10−3 ) = 2.1 Pa Since the effective shear stress on the underlying soil (0.5 Pa) is less than the permissible shear stress on the underlying soil (2.1 Pa), the proposed grass lining is adequate . 5.37. The area of the of the triangular bottom section is Tb Tb A1 = 2 2m1 and the area of the trapezoidal section above the triangular bottom section is Tb Tb A2 = Tb + y − m2 y − 2m1 2m1

(1)

(2)

161 Adding Equations (1) and (2) gives the total area of the channel as Tb A= 2

Tb 2m1

Tb Tb + Tb + y − m2 y − 2m1 2m1

The perimeter of the triangular bottom section is Tb 2 P1 = 2 1 + m1 2m1 and the perimeter of the trapezoidal section above the triangular bottom section is Tb 2 P2 = 2 1 + m2 y − 2m1 Adding Equations (3) and (4) gives the total wetted perimeter of the channel as Tb Tb 2 2 + 2 1 + m2 y − P = 2 1 + m1 2m1 2m1

(3)

(4)

Chapter 6

Design of Sanitary Sewers 6.1. From the given data: A = 65 ha, 10% large lots, 75% small single-family lots, and 15% small two-family lots. The population densities are given as 6/ha for large lots, 75/ha for single-family lots, and 125/ha for two-family lots. Hence, the estimated population, P , of the residential development is given by P = 65[0.1(6) + 0.75(75) + 0.15(125)] = 4914 people Taking the average per capita flow rate as 350 L/d = 0.35 m3 /d, then the average total flow rate at the end of the design period is 4914(0.35) = 1720 m3 /d = 0.0199 m3 /s. Since the average flow at the beginning of the design period is 30% of the flow at the end of the design period, then the average flow when the sewers are first installed is 0.30(0.0199 m3 /s) = 0.00597 m3 /s. Using Equation 6.5, the peaking factors can be estimated as = 1.88(0.00597)−0.095 = 3.06 PFmin = 1.88Q−0.095 avg1 = 1.88(0.0199)−0.095 = 2.73 PFmax = 1.88Q−0.095 avg2 The maximum and minimum flows are estimated by multiplying the corresponding average wastewater flows by these factors, thus Maximum flow = 2.73(0.0199) = 0.0543 m3 /s = 54.3 L/s Minimum flow = 3.06(0.00597) = 0.0183 m3 /s = 18.3 L/s 6.2. From the given data, the total area of the city is 45 km2 = 4500 ha, and the residential area is 65% of 4500 ha = 2925 ha. Taking the per-capita flow rate as 500 L/d/person (= 5.79 × 10−6 m3 /s/person) gives the wastewater flows in the following table Type Large lots Small single-family lots Multi-story apartments Total

Area (ha) 0.15(2925) = 438.8 0.75(2925) = 2193 0.10(2925) = 292.5

162

Density (persons/ha) 6 75 2500

Population 2633 164531 731250 898414

Flow (m3 /s) 0.015 0.953 4.234 5.20

163 The commercial sector of the city covers 25% of 4500 ha = 1125 ha, with a flow rate per unit area of 50,000 L/d/ha = 5.79 × 10−4 m3 /s/ha. Hence the average flow from the commercial sector is (5.79 × 10−4 )(1125) = 0.65 m3 /s. The industrial sector of the city covers 10% of 4500 ha = 450 ha, with a flow rate per unit area of 90,000 L/d/ha = 1.04 × 10−3 m3 /s/ha. Hence the average flow from the commercial sector is (1.04 × 10−3 )(450) = 0.47 m3 /s. The infiltration and inflow from the entire area is 1500 L/d/ha × 4500 ha = 6.75 × 106 L/d = 0.08 m3 /s. On the basis of these calculations, the average daily wastewater flow (excluding I/I) is 5.20 + 0.65 + 0.47 = 6.32 m3 /s. Since the average flow at the beginning of the design period is 35% of the flow at the end of the design period, then the average flow when the main sewer is first installed is 0.35(6.32 m3 /s) = 2.21 m3 /s. Using Equation 6.5, the peaking factors can be estimated as = 1.88(2.21)−0.095 = 1.74 PFmin = 1.88Q−0.095 avg1 = 1.88(6.32)−0.095 = 1.58 PFmax = 1.88Q−0.095 avg2 The maximum and minimum flows are estimated by multiplying the corresponding average wastewater flows by these factors and adding the I/I. Thus Maximum flow = 1.58(6.32) + 0.08 = 10.1 m3 /s Minimum flow = 1.74(2.21) + 0.08 = 3.9 m3 /s 6.3. From the given data: D = 760 mm, h = (3/4)×760 mm = 570 mm, and Q = 260 L/s = 0.260 m3 /s. Use Equation 6.13 to calculate θ, such that D θ h= 1 − cos 2 2 760 θ 570 = 1 − cos 2 2 which yields θ = 4.189 radians. Substitute for θ in Equation 6.14 to determine A yields 4.189 − sin(4.189) θ − sin θ D2 = (760)2 = 3.65 × 105 mm2 = 0.365 m2 A= 8 8 The average velocity, V , is given by V =

0.260 Q = = 0.712 m/s A 0.365

Therefore, the average velocity is estimated as 0.71 m/s .

164 6.4. For circular pipe:

θ − sin θ 8

(1)

1 P = Dθ 2 A (θ − sin θ)D R= = P 4θ Q 8Q Q = θ−sin θ = V = A (θ − sin θ)D2 D2 8

(2) (3) (4)

The Manning equation is given by

1 2 12 R 3 S0 n and substituting Equations 1 to 4 into Equation 5 gives

(5)

V =

2 1 (θ − sin θ)D 3 12 8Q = S0 (θ − sin θ)D2 n 4θ which simpliﬁes to

1 (θ − sin θ) 3 8 12 D 3 S0 20.16Q = 2 n θ3 and can be re-arranged to give 2

−1

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 = 0 6.5. From the given data: Q = 7 m3 /s, S0 = 0.01, D = 1.6 m, and assume n = 0.015. For uniform ﬂow, 2

−1

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 = 0 2

θ− 3 (θ − sin θ) 3 − 20.16(0.015)(7)(1.6)− 3 (0.01)− 2 = 0 which gives θ = 4.35 radians The flow depth, h, is therefore given by θ 1.6 4.35 D 1 − cos = 1 − cos = 1.25 m h= 2 2 2 2 The flow area, A, is given by 4.35 − sin 4.35 θ − sin θ 2 D = (1.6)2 = 1.69 m2 A= 8 8 and hence the flow velocity, V , is given by V =

7 Q = = 4.14 m/s A 1.69

165 When the pipe ﬂow three-quarters full, D 3 h= D= 4 2

θ 1 − cos 2

which gives θ = −0.5 2 θ = 4.19 radians cos

Substituting into the uniform-ﬂow equation gives 2

(4.19)− 3 (4.19 − sin 4.19) 3 − 20.16(0.015)(7)D− 3 (0.01)− 2 = 0 which simplifies to D = 1.63 m 6.6. At Maximum flow, hmax D = 0.7 And vmax = 0.9 m/s h = 0.7 From fig. 6.2, for D v V = 1.12 q Q = 0.838

v = vmax = 0.9 m/s 0.9 = 0.8 m/s V = 1.12

qmax =0.838Q At average ﬂow max = 0.8838Q = 0.335Q qav = q2.5 2.5

From Fig. 6.2 q Q = 0.335 h D = 0.41 v V = 0.9

Vav = 0.9V = 0.9 × 0.8 = 0.72 m/s. Hence while carrying the average discharge, the sewer remains 41% full and the velocity then generated is 0.72 m/s. 6.7. From the given data: Q = 3.5 m3 /s, D = 1.4 m, n = 0.015, h/D = 0.5. Equation 6.13 gives 1 θ h = 1 − cos D 2 2 or

θ 1 1 − cos 0.5 = 2 2

166 which leads to θ=π From Equation 6.17, the value of S0 is given by 2 8 8 20.16nQD− 3 20.16(0.015)(3.5)(1.4)− 3 = S0 = 2 5 2 5 θ− 3 (θ − sin θ) 3 π − 3 (π − sin π) 3

= 0.019

6.8. When the pipe in running 23 × D h 2 1 α = = × (1 − cos ) D 3 2 2 α 4 ⇒ = 1 − cos 3 2 4 1 α ⇒ cos = 1 − = − 2 3 3 ∴ α = 219◦ 2

A = D4 × 2

= D4 ×

πα sinα 360◦ − 2 π×219◦ sin219◦ 360◦ − 2

= D4 [1.9102 + 0.315] 2

= 2.23 × D4

α P = πD × 360 = πD × 219 360 = 1.91 D 2.23 × D = 0.292D R = 4×1.91

Using Manning’s equation, 2

1 × (0.292 × 0.45) 3 × V = 0.013

1 1 350

= 1.06 > 0.6m/s(Checked for self cleansing) 2

× 1.06m3 /s = 0.12m3 /s Q = 2.23 × (0.45) 4 6.9. According to the Manning equation, the average velocity, V , is given by 1 2 12 R 3 S0 n

V = For full-pipe ﬂow, A=

πD2 , 4

P = πD,

A D = P 4

πD2 , 8

P =

πD , 2

A D = P 4

and for half-pipe ﬂow,

Hence, for both full-pipe and half-pipe ﬂow, 1 V = n

D 4

2 3

S02

167 6.10. For a pipe ﬂowing half full, A = πD2 /8, P = πD/2, R = D/4, and the Manning’s equation gives Q=

1 2 1 AR 3 S02 n

1 πD2 0.030 = 0.015 8

D 4

2 3

(0.005) 2

which gives D = 0.301 m = 301 mm. The nearest commercial size is a diameter of 305 mm . For the Darcy-Weisbach (DW) equation, ks = (n/0.039)6 = 0.00324 m, μ = 0.001 N·s/m2 , and ρ = 998.2 kg/m3 . The DW equation gives

ks 0.625μ Q = −2A 8gRS0 log + 3√ 12R ρR 2 8gS0 0.625(0.001) D πD2 0.00324 8(9.81) (0.005) log + 0.030 = −2 3 8 4 12(D/4) (998.2)(D/4) 2 8(9.81)(0.005)

which yields D = 0.302 m = 146 mm. The nearest commercial size in 305 mm . √ The same Manning’s n should be used if 3.6 < R/ks < 360 and ks RS0 > 2.2 × 10−5 . When the pipe is half full, R (0.305/4) = 24 = ks 0.00324 0.305 × 0.005 = 6.326 × 10−5 ks RS0 = 0.00324 4 √ When the pipe flows full, ks , R, and S0 are the same. Since ks RS0 > 2.2 × 10−5 , the flow is fully turbulent and Manning’s n will be the same for both half-full and full flow conditions. 6.11. According to Equation 6.17, for any circular pipe flowing under partially full conditions, 8

−1

20.16nQD− 3 S0 2 = θ− 3 (θ − sin θ) 3

(1)

When this same pipe is ﬂowing full with the same ﬂow rate, Q, then θ = 2π and Equation 6.17 gives −1

20.16nQD− 3 S0 2 = (2π)− 3 [2π − sin(2π)] 3 = 2π

(2)

Combining Equations 1 and 2 yields 2

θ− 3 (θ − sin θ) 3 = 2π which gives θ = 4.529 radians and 6.283 (= 2π) radians. Hence the flow rate under partially full conditions is the same as the flow rate under full-flow conditions when 1 θ 1 4.529 h = 1 − cos = 1 − cos = 0.82 D 2 2 2 2 Hence, when h/D = 0.82 the flow rate is the same as the full-flow flow rate.

168 0.72 6.12. By applying Kirpich equation, tc=0.019 SL0.333 o 1 1 + 1 Here, L = 1000 m; S0 = 350 2 1000 = 519 0.77

= 0.019 × ((1000) 1 0.335 ) 519

= 43.06 min. Again Eqn (10.13) 2/5 nL √ tc = 6.99 2/5 So ie

2/5

6.99 = 43.06

0.05×1000 √ 1

2 5

519

ie = 12.09 mm/hr The storm runoﬀ, Qp =CiA =ie A 12.09 Assume catchment area=1km2 = 106 m2 = 1000×3600 × 106 m3 /s

= 3.36 m3 /s Assume Avg. per-capita wastewater Residential ﬂow rate= 250 L/d/person (see Table 6.2) Average wastewater discharge=(250×50000) L/d =1250000 L/d = 1250 m3 /d = 0.014 m3 /s Total combined discharge= (0.014 + 3.36) m3 /s = 3.37 m3 /s Here considered full ﬂow pipe 2

1/2

→ n1 AR s S0

= 3.37 2

1 → 0.05 × π4 × D2 × (D/4) s × (1/519) 2 = 3.37

→ D = 2.56 m → Diameter of sewer = 2.56 m 6.13. Per capita water supply = 250 L/d/person Population per ha = 250 person/ha So, total population = 250 × 50 = 12, 500 3125000 = 0.036 m3 /s Average water supply daily = 12500 × 250L/d = 1000×24×3600

Average sewage discharge=80% of water supply=0.8 × 0.036 = 0.029 m3 /s Maximum discharge for which sewer should be designed running half = 3 × 0.029 = 0.087 m3 /s

169 When pipe is running half full A=

D π π × D2 , P = × D, R = 8 2 4

We have, 2

Q = n1 AR 3 S02 1 ⇒ 0.087 = 0.013 × π8 D2 ×

D 23 4

1 1 1000

D=0.576 m 0.087 0.087 V= Q A = π ×D 2 = π ×(0.576)2 = 0.334 < 0.6 m/s 4

Hence, the condition of self cleansing capacity is not satisfied. 6.14. Total population of the area = 200 persons/ha × (40 × 100)ha = 8 × 105 persons Average sewage flow = 270 × 8 × 105 L/d = 2.5 m3 /s Maximum sewage flow =(1.5 × 2.5) = 3.25 m3 /s 20 1 m3 /s × 24×3600 Storm water flow z= 40 × 106 × 1000 = 9.259 m3 /s Total flow of the considered sewer =Sewage flow+Storm flow =(2.5+9.259) m3 /s =11.759 m3 /s Hence the capacity of sewer=11.759 m3 /s. 6.15. From the given data: D = 915 mm = 0.915 m, Qmax = 0.60 m3 /s, Qmin = 0.030 m3 /s, = 1.5 mm = 0.0015 m, τc = 2.0 Pa, T = 20◦ C, (h/D)max = 0.75, and Vlim = 4.0 m/s. Step 1: From the given information, any pipe slope greater than 0.1% meets the minimumcover requirement. Therefore, Sref1 = 0.001. Step 2: The pipe diameter under consideration is D = 915 mm.

170 Step 3: When Q = Qmax = 0.60 m3 /s and h/D = (h/D)max = 0.75, the following calculations are made: h −1 θmax = 2 cos 1−2 = 2 cos−1 [1 − 2(0.75)] = 4.189 radians D max 0.915 sin θmax sin(4.189) D = 1− 1− = 0.276 m Rmax = 4 θmax 4 4.189 θmax − sin θmax 4.189 − sin(4.189) 2 D = (0.915)2 = 0.529 m2 Amax = 8 8 (Qmax /Amax )Rmax (0.60/0.529)(0.276) Remax = = 3.13 × 105 = ν 10−6 0.0015 = 0.00543 = Rmax 0.276 −2 1.65 + fmax = 0.25 log 14.8Rmax Re0.9 max −2 1.65 0.00543 + = 0.25 log = 0.0215 14.8 (3.13 × 105 )0.9 1

6 2 nmax = 0.1129Rmax fmax = 0.1129(0.276) 6 (0.0215) 2 = 0.0133 ⎤2 ⎡ − 38 20.16nmax Qmax D ⎦ Smax = ⎣ 2 5 −3 θmax [θmax − sin θmax ] 3 2 8 20.16(0.0133)(0.60)(0.915)− 3 = = 0.00127 2 5 (4.189)− 3 [4.189 − sin(4.189)] 3

Based on these results, the minimum slope required for the sewer to ﬂow no more than 75% full is 0.00127. Hence, Sref2 = 0.00127. Step 4: When Q = Qmin = 0.030 m3 /s, for any given value of the pipe slope, Sref3 , the corresponding central ﬂow angle, θmin , is determined by simultaneous solution of the following relations: sin θmin sin θmin 0.915 D 1− 1− = Rmin = 4 θmin 4 θmin θmin − sin θmin θmin − sin θmin 2 Amin = D = (0.915)2 8 8 (Qmin /Amin )Rmin (Qmin /Amin )Rmin = Remin = ν 10−6 0.0015 = Rmin Rmin −2 1.65 fmin = 0.25 log + 14.8Rmin Re0.9 min

171 1

6 2 fmin nmin = 0.1129Rmin ⎡ ⎤2 ⎡ ⎤2 − 38 − 83 20.16nmin Qmin D min Qmin (0.915) ⎦ = ⎣ 20.16n ⎦ Sref3 = ⎣ 2 5 5 −3 − 32 3 3 θmin [θmin − sin θmin ] θmin [θmin − sin θmin ]

Based on the value of θmin calculated from these equations, the value of Rmin is calculated and and the boundary shear stress, τmin is calculated using the relation τmin = γRmin Sref3 The calculations in Step 4 are repeated for incremental values of Sref3 until τmin ≈ τc = 2.0 Pa. In the present case, these calculations yield Sref3 = 0.00306. Step 5: To meet the physical constraints, the maximum ﬂow depth limitation, and the minimum boundary shear stress requirements, the minimum required pipe slope, S0 , is given by S0 = max(Sref1 , Sref2 , Sref3 ) = max(0.00100, 0.00127, 0.00306) = 0.00306 Step 6: Determine the average velocity when Q = Qmax and S0 = 0.00306. Application of the Manning equation yields Vmax = 1.62 m/s. Since the maximum velocity (1.62 m/s) is less than the given limit (4.0 m/s) the calculated slope of 0.00306 is acceptable. 6.16. From the given data: A = 1700 ha, Pmin = 10000, Pmax = 50000, q = 250 L/d/person, and I/I = 2 m3 /d/ha. First determine the minimum and maximum ﬂows: −3 10 = 0.0289 m3 /s Qavg1 = Pmin q = (10000)(250) 86400 = 1.88(0.0289)−0.095 = 2.63 PFmin = 1.88Q−0.095 min 2 × 1700 = 0.0394 m3 /s QI/I1 = 86400 Qmin = PFmin × Qavg1 + QI/I1 = 2.63 × 0.0289 + 0.0394 = 0.115 m3 /s −3 10 Qavg2 = Pmax q = (50000)(250) = 0.145 m3 /s 86400 = 1.88(0.145)−0.095 = 2.26 PFmax = 1.88Q−0.095 min QI/I2 = QI/I1 = 0.0394 m3 /s Qmax = PFmax × Qavg2 + QI/I2 = 2.26 × 0.145 + 0.0394 = 0.367 m3 /s It is required that S0 = 1% = 0.01. For D = 150 mm, the Manning roughness is 0.0106 for concrete in typical condition. For Q = Qmax = 0.367 m3 /s and h/D = 0.75, 1 θ h = 1 − cos D 2 2 θ 1 1 − cos 0.75 = 2 2

172 which yields θ = 4.189 radians. Using the Manning equation (with D as the subject of the formula) ⎡

⎤3

D=⎣ θ =

8 −1 20.16nQS0 2 ⎦ 5 − 23 3

(θ − sin θ)

20.16(0.0106)(0.367)(0.01)− 2 2

3 8

= 0.474 m

(4.189)− 3 (4.189 sin 4.189) 3

Tentatively select the closest commercial size pipe of 525 mm, readjust the Manning roughness to n = 0.0114 which corresponds to D = 525 mm and recalculate as follows: D=

20.16(0.0114)(0.367)(0.01)− 2 2

(4.189)− 3 (4.189 sin 4.189) 3

3 8

= 0.487 m

Therefore tentative pipe diameter of 525 mm is conﬁrmed. Check that the pipe is selfcleansing at the minimum ﬂow when Q = Qmin = 0.115 m3 /s. Under this condition, the Manning equation gives 2

−1

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 = 0 2

θ− 3 (θ − sin θ) 3 − 20.16(0.0114)(0.115)(0.525)− 3 (0.01)− 2 = 0 which yields θ = 2.445 radians and hence D sin θ 0.525 sin 2.445 R= 1− = 1− = 0.0968 m 4 θ 4 2.445 τ = γRS0 = (9790)(0.0968)(0.01) = 9.5 Pa Since the minimum shear stress (= 9.5 Pa) is much greater than the minimum required shear stress (= 0.97 Pa) corresponding to a particle size of 1.5 mm, then the sewer is self cleansing. Check the maximum velocity, θ = 4.189 2 θ − sin θ 2 4.189 − sin 4.189 A=D = (0.525) = 0.174 m2 8 8 0.367 Qmax = = 2.11 m/s Vmax = A 0.174 Since the maximum velocity is less than the maximum allowable velocity of 3 m/s, a 525 mm diameter pipe meets all of the design requirements. 6.17. Let us ﬁrst assume that the town is provided a planned water supply at an average per capita rate equal to 270 L/person/day. Also assume that 80% of this water supply will be reaching the sewer or sanitary sewage. 80 × 270 × 5000 L/d ∴ Quantity of sanitary sewage produced per day= 100

173 0.8×270×5000 = 1000×24×3600 m3 /s

Average sewage discharge= 0.0125 m3 /s Assume maximum sewage discharge= 3 × Average sewage discharge = 3 × 0.0125 m3 /s = 0.0375 m3 /s The peak runoﬀ, Qp = CiA Time of concentration, tc=(10+20) min = 30 min From equation 9.3, a 100 = 30+20 = 2 cm/h = 20mm/h i = tc+b 20 1 × 60×60 × 50 × 104 m3 /s Qp = 0.68 × 1000

∴ Considered discharge= 1.89 m3 /s. 6.18. From the given data: Q = 0.5 m3 /s, T = 23◦ C, S0 = 0.009, D = 1220 mm = 1.22 m, and n = 0.013. Substituting the given data into the Manning equation (Equation 6.17) gives 2

−1

θ− 3 (θ − sin θ) 3 − 20.16nQD− 3 S0 2 = 0 2

θ− 3 (θ − sin θ) 3 − 20.16(0.013)(0.5)(1.22)− 3 (0.009)− 2 = 0 which gives θ = 2.061 radians. The wetted perimeter, P , and the top width, B, are given by (1.22)(2.061) Dθ = = 1.257 m 2 2 θ 2.061 B = D sin = 1.22 sin = 1.046 m 2 2 P =

Hydrogen sulﬁde is not a problem when Z ≤ 5000, hence Z = 0.308

EBOD P 1 1 S 2 Q3 B 0

5000 = 0.308

1.257 (0.009) (0.5) 1.046 EBOD 1 2

1 3

which gives EBOD = 1017 mg/L Since EBOD = BOD5 · 1.07T −20 then 1017 = BOD5 · 1.0723−20 which gives BOD5 = 830 mg/L

174 6.19. W cos θ t

Unit length

Consider a layer of sediment of unit width & unit length & of thickness‘t’ deposited at the invert of a sewer of gradient ‘θ’. Net γsub is the submerged unit wt of sediment. Then the wt of sediment considered W= γsub(1)(1)t But γsub = γw S−1 1+e Where, γw => unit wt. of water; S=> speciﬁc gravity of the sediment e=> void ratio. But the porosity of sediment n=e/(1+e) 1-n=1/(1+e) γsub = γw (S − 1)(1 − n) w= γw (S − 1)(1 − n)t τ0 = wsin θ [therefore θ is small tanθ = sin θ] = γw (S − 1)(1 − n)t sin θ ————————————- (i) Again from equation 6.27 τ0 = γw RS0 —————————————————- (ii) using K=(1–n)sinθ, as an important characteristics of the sediment, we have (S-1)kt= RS0 S0 = k/R(S–1)d For single grains, the vol. per unit area (i.e. t) below f of the diameter of the grain ‘d ’ as an inverse measure of the surface area of the individual grain expected to drag or friction S0 = K R (S − 1)d ——————————————(iii)

Now from Chezy’s equation √ V=C RS0 √ ∴ c = 8g f k Vs = 8g R× R (s − 1)d f

175

8k f (s − 1)gd

Here, f=0.03 K=0.04 d = 0.2 mm S=2.65 Vs =

8 × 0.04/0.03(2.65 − 1) × 0.2/1000

=0.059 m/s 2

1/2

Now, Manning’s equation Vs = n1 R 3 S0 Therefore, R=D/4=0.2/4=0.05 2

1 × (0.05) 3 × S02 Vs = 0.03 1 S0 = 8.756

∴ The minimum valce of gradient is 1 in 8.756. 6.20. (a) From the given data: Qmin = 0.15 m3 /min = 150 L/s, D = 915 mm, and the design particle size is 1.5 mm. From Table 6.8, = 0.00942(150)−0.5791 = 0.000517 Smin = 0.00942Q−0.5791 min (b) From the given data, Qmax = 0.29 m3 /s, and Table 6.4 gives n = 0.0118. Substituting these values into the Manning equation (Equation 6.17) and varying S0 until h/D = 0.75 yields S0 = 0.000233. This corresponds to a maximum velocity of 0.55 m/s, so the depth of ﬂow controls the derived slope. Since this slope (= 0.000233) is less than the slope required for self cleansing (= 0.000517), then the design slope is the larger of the two, which is 0.000517 . (c) From the given data: Qavg = 0.25 m3 /s, T = 25◦ C, and BOD5 = 250 mg/L. Substituting into the Manning equation (with D = 915 mm, n = 0.0118, and S0 = 0.000517) yields θ = 3.21 radians and hence: P θ 3.21 = = = 1.60 B 2 sin(θ/2) 2 sin(3.21/2) EBOD = BOD5 × 1.07T −20 = 250 × 1.0725−20 = 351 Z = 0.308

EBOD 1 2

S0 Q

1 3

351 P = 0.308 1 1 × (1.60) = 12093 B (0.000517) 2 (0.25) 3

Based on this value of Z, Table 6.7 indicates that sulﬁde generation is likely to occur . 6.21. From the given data: D = 535 mm, Qmax = 0.30 m3 /s, Qmin = 0.07 m3 /s = 70 L/s, and the design particle size is 1.5 mm.

176 (a) From Table 6.8, the minimum slope is Smin = 0.00830Q−0.5767 = 0.00830(70)−0.5767 = 0.000716 min For a 535-mm diameter pipe in Typical condition, from Table 6.4 it can be estimated that n = 0.0114. Substituting these values along with Qmax = 0.30 m3 /s into the Manning equation (Equation 6.17) and varying S0 until h/D = 0.75 yields S0 = 0.000402. This corresponds to a maximum velocity of 0.48 m/s, so the depth of ﬂow controls the derived slope. Since this slope (= 0.000402) is less than the slope required for self cleansing (= 0.000716), then the design slope is the larger of the two, which is 0.000716. Using this slope, the cover at the downstream manhole is 4.50 m − (2.50 m −150 × 0.000716 m) = 2.11 m. Since this is greater than the minimum cover of 2.00 m, use a pipe slope of 0.000716 . Maintain a diameter of 535 mm since both self-cleansing and ﬂow capacity requirements are met. (b) At the upstream end, the crown elevation is 2.50 m and the invert elevation is 2.50 m − 0.535 m = 1.965 m . At the downstream end, the crown elevation is 2.50 m − 150 × 0.000716 m = 2.393 m and the invert elevation is 2.393 m − 0.535 m = 1.858 m . These elevations assume that no invert drop is necessary at the manhole. 6.22. The maximum one hour rainfall having 2 year frequency=40 mm= ρ24 According to equation 10.18, i=

5.70 × ρ24 t0.62

5.70 × ×40 = 35.59mm/hr = 9.88 × 10−6 m/s 200.62

Storm sewer, QP = (0.55 × 20 × 104 × 9.88 × 10−6 )m3 /s = 1.0868m3 /s Average sewage flow= (0.8 × 10000 × 200)L/d = 16, 00, 000L/d Maximum sewage flow=3×Average sewage flow = 3 × 16, 00, 000 = 48, 00, 000L/d = 0.056 m3 /s 6.23. Preliminary calculations and specifications: From the given data, the average wastewater flow at the beginning and end of design period, Qavg2 and Qavg2 , respectively, and the I/I flow can be expressed in convenient units as follows: Qavg2 = 250 L/d/person × 600 persons/ha = 150000 L/d/ha = 1.736 L/s/ha Qavg1 = 0.20Qavg2 = 0.20(1.736) = 0.347 L/s/ha QI/I2 = 100 m3 /d/km = 0.00116 L/s/m QI/I1 = QI/I2 = 0.00116 L/s/m

177 Since the design particle diameter is 1.0 mm, the design boundary shear stress (i.e., tractive force), τc , is given by Equation 6.28 as τc = 0.867d0.277 = 0.867(1.0)0.277 = 0.87 Pa Therefore, sewers in which the boundary shear stress is greater than 0.87 Pa under minimum-flow conditions will be taken as self cleansing. In accordance with conventional practice, the sewer system will be designed to meet the following additional constraints: Vmax = 3.5 m/s h ≤ 0.75 D Hydraulics of existing sewer: The results of the design computations are shown in Figure 6.1. The computations begin with Line 0, which is the existing sewer main that must be extended to accommodate the sewer lines in the proposed residential development. The average flow in the sewer main at the end of the design period is 0.400 m3 /s = 400 L/s, and the average flow at the beginning of the design period is 0.2(0.400) = 0.080 m3 /s = 80 L/s. Using Equation 6.5, the peaking factors and corresponding flows are given by = 1.88(0.400)−0.095 = 2.05 PFmax = 1.88Q−0.095 avg2 PFmin = 1.88Q−0.095 = 1.88(0.08)−0.095 = 2.39 avg1 Qmax = PFmax Qavg2 = (2.05)(0.400) = 0.820 m3 /s = 820 L/s Qmin = PFmin Qavg1 = (2.39)(0.08) = 0.191 m3 /s = 191 L/s Hence, the maximum flow is 820 L/s (column 10) and the minimum flow is 191 L/s (column 13). The n value for a 1220-mm concrete pipe in typical condition is 0.0121. With a slope of 0.008 (column 14) and a diameter of 1220 mm (column 15), the depth of flow at the maximum flow rate is 379 mm (column 17) and the corresponding maximum velocity is 2.65 m/s (column 18). Under minimum-flow conditions, the depth of flow is 184 mm with a corresponding boundary shear stress of 8.91 Pa (column 16). The invert elevation of the main sewer at MH 5 is 55.00 m (column 22) and the ground-surface elevation at MH 5 is 60.04 m (column 24). Sewer Line 1: The design of the sewer system begins with Line 1 on A Street, which goes from MH 1 to MH 2 and is 55 m long. The area contributing wastewater flow is 0.47 ha (column 7). The maximum and minimum wastewater flows are calculated as follows: QI/I2 = 0.00116 L/s/m × 55 m = 0.0638 L/s = 0.000064 m3 /s QI/I1 = QI/I2 = 0.000064 m3 /s Qavg2 = 1.736 L/s/ha × 0.47 ha = 0.816 L/s = 0.000816 m3 /s Qavg1 = 0.2Qavg2 = 0.2(0.000816) = 0.000245 m3 /s −0.44 = 6.42 PFmax = 0.281Q−0.44 avg2 = 0.281(0.000816) −0.44 PFmin = 0.281Q−0.44 = 10.9 avg1 = 0.281(0.000245)

178 Qmax,sewage = PFmax Qavg2 = (6.42)(0.000816) = 0.00524 m3 /s = 5.24 L/s Qmin,sewage = PFmin Qavg1 = (10.9)(0.000245) = 0.00178 m3 /s = 1.78 L/s Qmax = Qmax,sewage + QI/I2 = 5.24 + 0.0638 = 5.30 L/s Qmin = Qmin,sewage + QI/I1 = 1.78 + 0.0638 = 1.84 L/s Hence, the maximum ﬂow is 5.30 L/s (column 10) and the minimum ﬂow is 1.84 L/s (column 13). The n value for a 150-mm concrete pipe in typical condition is 0.0106. The minimum slope for self-cleansing, Smin calculated from the appropriate equation in Table 6.8 along with the ground slope, Sground as follows: Smin = 0.00539Q−0.5692 = 0.00539(1.84)−0.5692 = 0.00381 min Sground =

65.00 − 65.80 = 0.0218 55

Hence, to maintain a minimum cover of 1 m, specify the pipe slope, S0 , to equal the ground slope (= 0.0218). With a pipe slope of 0.0218 (column 14) and a diameter of 150 mm (column 15), the depth of flow at the maximum flow rate is 34 mm (column 17) and the corresponding maximum velocity is 1.77 m/s (column 18). Under minimum-flow conditions, the depth of flow is 26 mm with a corresponding boundary shear stress of 3.42 Pa (column 16). The pipe has adequate capacity, is self cleansing, and meets all design constraints. The drop in the sewer, Δz, is given by Δz = LS0 = (55)(0.0218) = 1.20 m The sewer invert at the upper end to have a 1.00 m cover is 65.00 m − 1.00 m − 0.150 m = 63.85 m (column 21), and the sewer invert at the lower end is 63.85 m − 1.20 m = 62.65 m. Sewer Lines 2 to 4: The design of Lines 2 and 3 follows the same sequence as for Line 1, with the exception that the wastewater flows in each pipe are derived from the sum of the contributing areas of all upstream pipes plus the pipe being designed and that the I/I flow in each pipe is the sum of all upstream I/I flows plus the I/I contribution to the pipe being designed. Using this approach, the invert elevation at the end of Lines 3 and 4 is 58.89 m (column 22), where the sewer laterals join the main sewer. The invert elevation of the sewer main is 55.00 m, which is 58.89 m − 55.00 m = 3.89 m below the invert of the laterals. A special drop-manhole structure will be required at this intersection. Sewer Line 5: The main sewer leaving MH 5 (Line 5) is designed next. The tributary area to Line 5 is the sum of the contributing areas of all contributing sewer laterals (Lines 1 to 4, 1.41 + 0.9 = 2.31 ha) plus the equivalent area of the average flow in the main sewer upstream of MH 5 (0.40 m3 /s ÷ 1.736 L/s/ha = 230.3 ha) plus the area that contributes directly to Line 5 (0.17 ha). Hence the total contributing area is 2.31 + 230.3 + 0.17 = 232.78 ha (column 7). The I/I contribution to Line 5 is the sum of the I/I contributions to all upstream laterals (0.285 + 0.105 = 0.390 L/s) plus the I/I contribution directly to Line 5 (70 m × 0.00116 L/s/m = 0.0812 L/s) for a total I/I

1 2 3 4 5

A Street A Street A Street A Street Main Street

Line no. (1) (2) 0 Main Street

1 2 3 4 5

(3) -

2 3 5 5 12

(4) 5

Manhole no.

55 90 100 90 70

0.47 0.50 0.44 0.90 0.17

0.47 0.97 1.41 0.90 232.78

Length Increment Total (m) (ha) (ha) (5) (6) (7) -

Area

0.064 0.168 0.285 0.105 0.471

I/I (L/s) (8) 5.30 8.03 9.98 7.65 828

Total (L/s) (10) 820 0.064 0.168 0.285 0.105 0.471

I/I (L/s) (11) 1.78 2.67 3.30 2.57 185

Sewage (L/s) (12) -

Minimum flow

1.84 2.83 3.58 2.67 185

Total (L/s) (13) 191

Sewer invert Ground surface

0.0218 150 0.0156 150 0.0236 150 0.0204 150 0.0008 1220

3.42 3.18 4.87 3.83 1.46

34 61 61 55 731

1.77 1.20 1.48 1.30 1.14

0.03

1.20 1.40 2.36 1.84 0.06

63.85 62.65 61.25 60.73 54.97

62.65 61.25 58.89 58.89 54.91

65.00 63.80 62.40 61.88 60.04

63.80 62.40 60.04 60.04 60.04

Manhole Min Max Max invert Fall in Upper Lower Upper Lower Slope of Diam force depth velocity drop sewer end end end end sewer (mm) (Pa) (mm) (m/s) (m) (m) (m) (m) (m) (m) (14) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24) 0.008 1220 8.90 379 2.65 55.00 60.04

Figure 6.1: Sewer Design Calculations

5.24 7.87 9.70 7.53 828

Sewage (L/s) (9) -

Maximum flow

179

180 contribution of 0.390 + 0.0812 = 0.471 L/s (column 8). The maximum and minimum ﬂows are calculated using the peaking ﬂow factors as previously described. Since the the required minimum slope is less than the practical slope of 0.08%, a slope of 0.08% is used. A manhole drop at the end of the pipe of 0.03 m is used to account for energy losses at the manhole, where laterals intersect.

Chapter 7

Design of Hydraulic Structures 7.1. From the given data: D = 1.220 m, Q = 2.5 m3 /s, and S0 = 0.005. Assume that n = 0.013 for concrete pipe. Substituting in Equation 7.1 gives 5

(θ − sin θ) 3 θ

2 3

= 20.16

nQ (0.013)(2.5) = 5.453 √ = 20.16 8√ D S0 (1.220) 3 0.005 8 3

which yields θ = 4.052 radians. Substituting θ into Equation 7.2 gives D 1.220 θ 4.052 y= 1 − cos = 1 − cos = 0.878 m 2 2 2 2 Therefore, the normal depth of flow is 0.878 m . Substituting the given data into Equation 7.3 gives (θ − sin θ)3 (2.5)2 Q2 θ = 512 5 = 512 = 120.7 gD (9.81)(1.220)5 sin 2 which yields θ = 4.0153 radians. Substituting θ into Equation 7.2 gives θ 1.220 4.0153 D 1 − cos = 1 − cos = 0.8681 m y= 2 2 2 2 Therefore, the critical depth of flow is 0.868 m . An approximation of the critical depth can be obtained by using Equation 7.4, which in this case gives 2 0.25 0.25 Q (2.5)2 1.01 1.01 = = 0.857 m yc = D0.26 g (1.220)0.26 9.81 7.2. From the given data: D = 0.5 m, n = 0.013, ke = 0.05, Cd = 0.95, S0 = 0.02, and L = 20 m. The elevation difference, Δh, between the headwater and the top of the culvert at the exit is given by Δh = 0.2 + S0 L = 0.2 + (0.02)(20) = 0.6 m Assume that the exit is not submerged, flow is either Type 2 or Type 3. Assume the flow is Type 2: 2gΔh 2(9.81)(0.6) 2 Q=A = π(0.25) = 0.464 m3 /s L 2(9.81)(0.013)2 20 4 + 0.05 + 1 2gn2 4 + ke + 1 (0.5/4) 3

181

182 Determine if the culvert ﬂows full. The Manning equation gives 5

1 A3 Qfull = n P 23

1 (π0.252 ) 3 √ S0 = 0.02 = 0.534 m3 /s 0.013 (π0.5) 23

and the culvert flows full when the discharge exceeds 1.07(0.534) = 0.571 m3 /s. Since Q < 0.571 m3 /s, the culvert does not flow full, and the assumption of Type 2 flow is not supported. Assuming Type 3 flow, Q = Cd A 2gh = 0.95π(0.25)2

2(9.81)(0.25 + 0.2) = 0.55 m3 /s

Since Q < 0.571 m3 /s, the culvert does not flow full, Type 3 flow is confirmed, and the discharge through the culvert is 0.55 m3 /s . 7.3. From the given data: b = 3.5 m, m = 2, S0 = 0.005, n = 0.025, Q = 3 m3 /s, L = 10 m, and Hp = 3 m. Find the normal depth of flow in the channel using the Manning equation: 5

1 A 3 12 S Q= n P 23 0

1 [by + my 2 ] 3 1 2 Q= S √ n [b + 2y 1 + m2 ] 32 0 5

1 [3.5yn + 2yn2 ] 3 1 2 3= √ 2 (0.005) 0.025 [3.5 + 2y 1 + m2 ] 3

which gives yn = 0.465 m. This can be taken as the tailwater depth for the culvert. Assume Type II ﬂow through the culvert, the ﬂow is given by 2gΔh Q = A 2gn2 L + ke + 1 4

(1)

Taking the diameter to be D, ke = 1, and Δh = 3 − D, then Equation 1 gives 2(9.81)(3 − D) πD2 3= 2 (10) 2(9.81)(0.013) 4 + 0.1 + 1 4 (0.25D) 3

Which gives D = 2.12 m and confirms Type II flow. For a reinforced concrete pipe, the next-larger commercial size is D = 2135 mm . 7.4. From the given data: Q = 1 m3 /s and H = 1 m . Assume Type 3 flow to use NIST equation. For the submerged condition, c = 0.0398, Y = 0.67, and S = 0. The NIST equation gives H = c · gFr2 + Y − 0.5S D 1 = (0.0398) · (9.81) D

1 √ π 2 9.81D 4D

+ 0.67 − 0.5(0)

183 which gives two solutions D = 1.47 m or 0.568 m. Since only the latter solution is admissible take D = 0.568 m = 568 mm. This corresponds to Fr = 1.67 which is within the range of applicability of the NIST equation which requires Fr ≥ 0.7. The next higher commercial size is D = 610 mm . If a grooved end is selected, c = 0.0292, Y = 0.74, and the NIST equation gives H = c · gFr2 + Y − 0.5S D 1 = (0.0292) · (9.81) D

1 √ π 2 9.81D 4D

+ 0.74 − 0.5(0)

which gives two solutions D = 1.33 m or 0.528 m. Since only the latter solution is admissible take D = 0.528 m = 528 mm. This corresponds to Fr = 2.01 which is within the range of applicability of the NIST equation which requires Fr ≥ 0.7. The next higher commercial size is D = 535 mm . The corresponding appropriate equation is the oriﬁce equation given by Q = Cd A gh where conventional values for Cd are 0.62 for square-edged entrances and 1.0 for rounded entrances. For a square-edged entrance, π D 2 D (9.81) 1 − 1 = (0.62) 4 2 which gives two solutions D = 1.94 m or 0.952 m. Since only the latter solution is admissible take D = 0.952 m = 952 mm. The next higher commercial size is D = 1065 mm . For a rounded entrance, 1 = (1.0)

π 4

D (9.81) 1 − 2

which gives two solutions D = 1.98 m or 0.715 m. Since only the latter solution is admissible take D = 0.715 m = 715 mm. The next higher commercial size is D = 760 mm . Based on these results, using the non-NIST equation will lead to an overdesign of the culvert. 7.5. From the given data: T W = 10.00 m, D = 380 mm = 0.38 m, and L = 8 m. Since the water depth at the outlet is 10.00 m − 9.50 m = 0.50 m, and the culvert diameter is 0.38 m, then Type 1 ﬂow is expected. Accordingly, the discharge, Q, is given by Equation 7.12 as 2gΔh Q=A (1) 4 2 2gn L/R 3 + ke + 1

184 For a corrugated metal pipe, Table 7.3 gives n = 0.028 (conservative value) and Table 7.4 gives ke = 0.7 for a mitered entrance. Substituting into Equation 1 gives 2(9.81)(HW − 10.00) π Q = (0.38)2 4 4 2(9.81)(0.028)2 (8)/(0.38/4) 3 + 0.7 + 1 which simpliﬁes to the following culvert performance curve Q = 0.236 (HW − 10)

(2)

If the culvert is required to pass 0.30 m3 /s, then the required headwater elevation satisfies the relation 0.3 = 0.236 (HW − 10) which gives HW = 11.62 m . When the tailwater elevation is 9.75 m, the flow through the culvert is Type 2, and the flow rate is based on the crest elevation of the culvert exit, which is equal to 9.50 m + 0.38 m = 9.88 m. Hence the discharge relation is given by Q = 0.236 (HW − 9.88) For HW = 11.62 m, Q = 0.236 (11.62 − 9.88) = 0.31 m3 /s Therefore the culvert capacity increases by 0.01 m3 /s or 3% . Type 3 flow is not a possibility in this case since the culvert is hydraulically long (i.e. L > 10D). 7.6. From the given data: culvert dimensions = 1.5 m × 1.5 m, n = 0.013, Cd = 0.95, ke = 0.05, S0 = 0.007, and L = 40 m. (a) Free outlet conditions. Assume Type 2 flow, where Δh = 0.5 m + 0.007(40) = 0.78 m and therefore Q=A

2gΔh = (1.5)2 2 2gn L4 + ke + 1

2(9.81)(0.78) = 7.08 m3 /s 2(9.81)(0.013)2 40 4 + 0.05 + 1 (1.52 /6) 3

Determine if the culvert ﬂows full. Calculate the normal depth, yn , using the Manning equation, 5

1 An3 12 Q= S n P 23 0 n

1 (1.5yn ) 3 1 7.08 = (0.007) 2 0.013 (1.5 + 2yn ) 23

(1.5yn )5 (1.5 + 2yn )2 yn = 1.22 m

1.33 =

185 Therefore, the assumption that the culvert flows full (Type 2 flow) is not supported. Assuming Type 3 flow, Q = Cd A 2gh = 0.95(2.25)

2(9.81)(0.75 + 0.5) = 10.6 m3 /s

Determine if the culvert ﬂows full. Calculate the normal depth, yn , using the Manning equation, 5

1 An3 12 S Q= n P 23 0 n

1 (1.5yn ) 3 1 2 10.6 = 2 (0.007) 0.013 (1.5 + 2yn ) 3

(1.5yn )5 (1.5 + 2yn )2 yn = 1.7 m

4.468 =

Therefore, the assumption that the culvert does not flow full (Type 3 flow) is not supported. The flow is somewhere between Type 2 and Type 3 flow, and the discharge is in the range 7.08 to 10.6 m3 /s . (b) For a submerged outlet with tailwater 0.5 m above the crown of the culvert at the exit, Δh = S0 L = 0.007(40) = 0.28 m and

Q=A

2gΔh = (1.5)2 2gn2 L4 + ke + 1 R3

2(9.81)(0.28) = 4.25 m3 /s 2(9.81)(0.013)2 40 4 + 0.05 + 1 (1.52 /6) 3

Find Δh for Q = 10.6 m2 /s, where Δh =

2gn2 L 4

Q2 2(9.81)(0.013)2 (40) 10.62 + ke + 1 = + 0.05 + 1 = 1.74 m 4 2gA2 2(9.81)(2.25)2 (2.25/6) 3

So the headwater must be 1.5 + 0.5 + 1.74 − 0.28 = 3.46 m above the inﬂow-channel invert. For Q = 7.08 m3 /s, Δh = 0.78 m. Therefore the range of headwater elevations is from 2.50 to 3.46 m above the channel invert. H = 0.6 0.9 = 1.5 Hw According to equation 7.70

7.7. 1st case

Cd = 0.611 + 0.075 HHw = 0.611 + 0.075 = 0.724

0.6 0.9

186 Equation 7.67 2 2 √ Q = Cd 2gb(H) 3 3 √ 3 2 = × 0.724 × 2 × 9.81 × 2 × (0.9) 2 3 = 3.46 m3 /s This solution assumes that the length of the weir is such that HHw < 5. 2nd case Contracted weir, L=1.5 m Equation 7.76 3

Q = Cw (b − 0.1nH)H 2 Here, b=1.5, n=2 From Equation 7.74 2 √ Cw = Cd 2g = 2.14 3 3

3.46 = 2.14(1.5 − 0.1 × 2 × H)H 2 3

1.62 = (1.5 − 0.2H)H 2 H=1.18 Change in the upstream water level = (1.18 − 0.9) = 0.28 m. 7.8. From the given data: H = 2 m, TW = 1 m, Q = 1 m3 /s, L = 15 m, and S0 = 0.015. Assuming Type 1 ﬂow, then Equation 7.12 gives Q=A

2gΔh

(1)

2gn2 L/R 3 + ke + 1

where A = πD2 /4 = 0.785D2 , Δh = H − TW + S0 L = 2 − 1 + (15)(0.015) = 1.225 m, n = 0.013, R = D/4 = 0.25D, and taking ke = 0.50, Equation 1 gives 1 = 0.785D

1 = 3.85D

2(9.81)(1.225)

2(9.81)(0.013)2 (15)/(0.25D) 3 + 0.50 + 1 1

0.316D

− 34

+ 1.50

Solving gives D = 0.61 m. This indicates that the exit is submerged, conﬁrming Type 1 ﬂow. Use a culvert with D = 61 cm .

187 7.9. From the given data: b = 2 m, m = 3, d = 3 m, and Q = 8 m3 /s. The minimum culvert size causes ponding to within 30 cm of the top of the approach channel. For a culvert diameter D with Type 3 ﬂow, Q = Cd A 2gh π

D 8 = 0.62 D 2(9.81) 3 − 0.3 − 4 2 D 2.7 − 8 = 2.157D2 2 2

which yields D = 1.645 m. The next step is to confirm the Type 3 flow assumption by demonstrating that the pipe does not flow full under design conditions. Taking n = 0.013 and assuming full-flow conditions, 2 1 1 1 2 8 1 πD2 π D 3 2 1 2 S0 = D 3 S02 Q = AR 3 S0 = n n 4 4 16n 8 1 π (1.645) 3 (0.02) 2 = 8.054 m3 /s = 16(0.013) Since the actual flow rate is 8 m3 /s (< 8.054 m3 /s), the culvert does not flow full under design conditions and Type 3 flow is confirmed. The next larger commercial size is 1675 mm and this is the size that is recommended. The culvert should be constructed of reinforced concrete pipe . 7.10. From the given data: b = 2 m, Q = 4 m3 /s, S0 = 0.1, L = 25 m, ke = 0.1, and for a concrete culvert n = 0.013. Find the normal depth of flow in the culvert using the Manning equation, 5

1 A 3 12 S Q= n P 23 0

1 (2yn ) 3 1 2 4= 2 (0.1) 0.013 (2 + 2yn ) 3

which gives yn = 0.24 m Find the critical depth, yc , where Q2 A3 = c g Tc 2 (2yc )3 4 = 9.81 2 which gives yc = 0.74 m

188 Therefore the culvert slope is steep. Assume Type 5 ﬂow. Applying Equation 7.23 (neglecting the upstream velocity) gives 2g(Δh + V12 /2g − hi ) 4 = (0.74 × 2) 2(9.81) Δh − 0.1 ×

Q = Ac

42 2(9.81)(2 × 0.74)2 )

which gives Δh = 0.41 m Since yc occurs at about 1.4yc downstream of the culvert entrance, the headwater depth is yc + Δh − S0 (1.4yc ) = 0.74 + 0.41 − 0.1(1.4 × 0.74) = 1.05 m . Since the entrance is unsubmerged, Type 5 flow is confirmed. 7.11. From the given data: D = 1.83 m, L = 6.1 m, h1 = 1.170 m, and h2 = 0.920 m. For RCP take n = 0.013 and for vertical headwall take ke = 0.5. Assume Type 6 flow with negligible upstream velocity, hence the flow is described by Δh −

Q2 V2 = hi + hf = 0.5 + S̄f L 2g 2gA21

where

Sf =

(1)

2 (2)

AR 3

From the culvert geometry and using Equation 8.113 gives

h (m) 0.920 1.170

h/D 0.5027 0.6393

θ (rad) 3.154 3.706

A (m2 ) 1.325 1.776

P (m) 2.886 3.391

R (m) 0.4591 0.5237

Sf 2.718Q2 × 10−4 1.269Q2 × 10−4

Substituting into Equation 8.112 gives Q2 1 Q2 = 0.5 + Q2 (1.170 − 0.920) − 2 2 2(9.81)(1.325) 2(9.81)(1.776) 2

2.78 + 1.269 2

10−4 × 6.1

which gives Q = 2.574 m3 /s. Since there are two barrels, the total ﬂow through the culvert is 2 × 2.574 = 5.15 m3 /s .

189 To verify Type 6 ﬂow, show that y > yc within the culvert. Under critical ﬂow conditions, with D = 1.83 m, A3 Q2 = g T 2 2.574 A3 = 9.81 T θ−sin θ

D2 θ

0.6754 =

16.85 =

(θ − sin θ)3 sin(θ/2)

D sin 2 θ−sin θ 2 3 D 8 θ 0.6754 = D sin 2

which gives θ = 2.846 radians, and

θ D yc = 1 − cos = 0.780 m 2 2

Since y > yc , at both the entrance and the exit, Type 6 flow in confirmed. 7.12. From the given data: D = 0.915 m, Q = 1.5 m3 /s, L = 15 m, S0 = 0, ke = 0.2, and TW = 0.40 m. For a concrete culvert it can be assumed that n = 0.013. (a) For Type 2 flow, the difference between the headwater elevation and the crown of the culvert exit, Δh, is given by Equation 7.11. From the given data, π π A = D2 = (0.915)2 = 0.6576 m2 4 4 1.5 Q = = 2.281 m/s V = A 0.6576 0.915 D = = 0.2288 m R= 4 4 and substituting into Equation 7.11 gives V2 V2 + 2g 2g R 2 2 (0.013) (2.281) (15) (2.281)2 (2.281)2 + = 0.412 m H −D = + 0.2 4 2(9.81) 2(9.81) (0.2288) 3 Δh =

n2 V 2 L 4 3

+ ke

The calculated result that H − D = 0.412 m gives H = D + 0.295 m = 0.915 m + 0.412 m = 1.327 m (b) For the Type 2 flow variation shown in Figure 7.4, calculate the critical flow depth, yc . Under critical flow conditions, Equation 7.3 gives (θ − sin θ)3 (1.5)2 Q2 θ = 512 5 = 512 = 183.1 gD (9.81)(0.915)5 sin 2

190 which yields θ = 4.3686 radians and hence D θ 0.915 4.3685 yc = 1 − cos = 1 − cos = 0.721 m 2 2 2 2 Therefore, assume that the HGL at the exit is (yc + D)/2 = (0.721 + 0.915)/2 = 0.818 m. Using the previously calculated value of Δh = 0.412 m, the headwater depth is H = 0.818 m + 0.412 m = 1.23 m 7.13. The relationship between the depth of ﬂow, y1 , at the culvert entrance and the depth of ﬂow, y2 at the culvert exit can be estimated using the standard-step method, where 1 Q2 y + 2gA 2 2 L= S̄f − S0 which can be put in the form L Q2 Q2 (Sf 1 + Sf 2 − 2S0 ) = y1 + − y − 2 2 2gA21 2gA22

(1)

where ⎡

⎤2

Sf 1 = ⎣

nQ ⎦

⎡ Sf 2 = ⎣

(2)

A1 R13

⎤2

nQ ⎦

(3)

A2 R23

In this case, 5

2 3

A1 R1 = 2 3

A13

(2y1 ) 3

A2 R2 =

A23

(4)

(5)

(2 + 2y1 ) 3

P13

(2y2 ) 3

(2 + 2y2 ) 3

P23

Combining Equations 2 and 4, and taking n = 0.013 and Q = 10 m3 /s gives 2

Sf 1 = (0.013)(10)

(2 + 2y1 ) 3 5

= 0.00168

(2y1 ) 3

(2 + 2y1 ) 3 10

(6)

y13

and similarly

Sf 2 = 0.00168

(2 + 2y2 ) 3 10

y23

(7)

191 The culvert tailwater depth, y2 , satisﬁes the Manning equation in the downstream channel, hence 5 1 2 1 1 A 3 12 2 Q = AR 3 S0 = S (8) n n P 23 0 where Q = 10 m3 /s, n = 0.022, S0 = 0.005, b = 5 m, m = 2, and A = by2 + my22 = 5y2 + 2y22 P =b+2

1 + m2 y2 = 5 + 2 1 + 22 y2 = 5 + 4.47y2

Substituting into Equation 8 gives 5

1 1 (5y2 + 2y22 ) 3 2 10 = 2 (0.005) 0.022 (5 + 4.47y 3

which yields y2 = 0.713 m

(9)

From the given data, L = 10 m, and combining Equations 1, 6, 7, and 9 gives ⎡ ⎤ 4 4 10 ⎣ (2 + 2y1 ) 3 (2 + 2 × 0.713) 3 0.00168 + 0.00168 − 2(0.005)⎦ 10 10 2 3 3 0.713 y 1

= y1 +

102 102 − 0.713 − 2(9.81)(2y1 )2 2(9.81)(2 × 0.713)2

which yields y1 = 0.677 m Therefore, the depth of flow at the culvert entrance is 0.68 m and the depth of flow at the culvert exit is 0.71 m . Since the culvert is 2 m × 2 m, then the culvert has adequate capacity when the flow is 10 m3 /s. It is interesting to note that the critical flow depth in the culvert is 1.36 m, the normal flow depth is 1.33 m, and hence there is a S3 water surface profile in the culvert. 7.14. From the given data: Q = 0.20 m3 /s, Er = 50.17 m, E0 = 48.01 m, b = 2 m, m = 2, S0 = 0.1% = 0.001, and for riprap n = 0.028 (typical). For normal flow in the exit channel. 5

1 A 3 12 Q= S n P 23 0

1 1 (2y + 2y 2 ) 3 0.20 = √ 2 (0.001) 2 0.028 (2 + 2y 5) 3

which yields y = 0.224 m. Allow upstream ponding to within 46 cm (= 18 in.) of roadway gives a headwater depth of HW = (50.17 − 0.46) − 48.01 = 1.70 m

192 Assume Type 2 ﬂow and take ke = 0.05 gives Q=A

2gΔh 2gn2 L 4

π 0.20 = D2 4

+ ke + 1 2(9.81)(1.70 − D)

2(9.81)(0.013)2 (15) 4

(D/4) 3

+ 0.05 + 1

which yields D = 0.283 m. Verify the assumption that the culvert ﬂows full. The culvert capacity, Q0 , is given by 5

1 A03 12 S Q0 = 1.07 n P 23 0 0

where n = 0.013, A0 = πD2 /4 = 0.0629 m2 , P0 = πD = 0.8891 m, and hence 5

1 1 (0.0629) 3 (0.001) 2 = 0.028 m3 /s Q0 = 1.07 0.013 (0.8891) 23

Since Q > Q0 the culvert flows full and Type 2 flow is confirmed. Also since the minimum required diameter (283 mm) is smaller than the minimum regulatory size (455 mm), use a culvert diameter of 455 mm . 7.15. For the drainage channel upstream and downstream of the culvert: b = 2 m, m = 3, d = 2 m, S0 = 0.5% = 0.005, and n = 0.018. For the culvert: D = 455 mm = 0.455 m, and S0 = 0.005. For both the channel and the culvert: Q = 1.5 m3 /s. The normal flow depth in the channel, yn is given by the Manning equation, 1 2 1 AR 3 S02 n 5 1 [2yn + 3yn2 ] 3 1 1.5 = (0.005) 2 √ 0.018 [2 + 2 10yn ] 23

which yields yn = 0.330 m. The culvert outlet is therefore not submerged. (a) Taking the required freeboard in the channel as 30 cm, the ﬂow is either Type 2 or Type 3, with Type 2 more likely since the culvert is hydraulically long. Assuming Type 2 ﬂow for a N-barreled culvert, Q 2gΔh (1) = A 2gn2 L N + ke + 1 4 R3

193 where Δh = (h1 − h2 ) + S0 L = (1.70 − 0.455) + (0.005)(15) = 1.32 m n = 0.013 L = 15 m π π A = D2 = (0.455)2 = 0.163 m2 4 4 0.455 D = = 0.114 m R= 4 4 ke = 0.5 (headwall with square-edged entrance) Q = 1.5 m3 /s Substituting into Equation 1 gives 1.5 = (0.163) N

2(9.81)(1.32) 2(9.81)(0.013)2 (15) 4

(0.114) 3

+ 0.5 + 1

which yields N = 2.80. Therefore consider using a 3-barrel culvert. To confirm the Type-2 flow assumption, verify that each culvert flows full when Q = 1.5/3 = 0.5 m3 /s. The capacity of the culvert, Qref , is given by 1 2 2 1 1 1 (0.163)(0.114) 3 (0.005) 2 = 0.22 m3 /s Qref = 1.07 AR 3 S02 = 1.07 n 0.013

Since 0.5 m3 /s > Qref , the culvert flows full, Type 2 flow is confirmed, and a 3-barrel culvert is selected. (b) To find the headwater elevation, take N = 3 in Equation 1, 1.5 2(9.81)Δh = (0.163) 2(9.81)(0.013)2 (15) 3 + 0.5 + 1 4 (0.114) 3

which yield Δh = 1.15 m, and hence the headwater depth, h1 , is given by h1 = Δh + h2 − S0 L = 1.15 + 0.455 − (0.005)(15) = 1.530 m Use the direct step method to estimate the distance upstream to where the depth is equal to 50% of the depth at the culvert: y1 = 0.5(1.530) = 0.765 m A1 = (2)(0.765) + 3(0.765)2 = 3.29 m2 √ P1 = 2 + 2 10(0.765) = 6.84 m A1 3.29 = 0.481 m = R1 = P1 6.84 Q 1.5 V1 = = 0.456 m/s = A1 3.29

194 ⎤2 nQ ⎦ ⎣ ⎡ Sf 1 =

2 3

A1 R1

(0.018)(1.5) (3.29)(0.481)

2 3

= 1.787 × 10−4

y2 = 1.530 m A2 = (2)(1.530) + 3(1.530)2 = 10.08 m2 √ P2 = 2 + 2 10(1.530) = 11.68 m A2 10.08 = 0.863 m = R2 = P2 11.68 Q 1.5 V2 = = 0.149 m/s = A2 10.08 ⎡ ⎤2 2 nQ ⎦ (0.018)(1.5) ⎣ Sf 2 = = = 0.8732 × 10−5 2 2 3 3 (10.08)(0.863) A2 R2 S̄f = 0.5(Sf 1 + Sf 2 ) = 0.5(1.787 × 10−4 + 0.8732 × 10−5 ) = 0.9372 × 10−4 2 1 0.4562 0.1492 y + V2g − 1.530 + 2(9.81) 0.765 + 2(9.81) 2 = 154 m ΔL = = 0.9372 × 10−4 − 0.005 S̄f − S0 Therefore the distance upstream to where the water depth is 50% of the water depth at the culvert entrance is 154 m . 7.16. A ﬂow scenario in which the culvert entrance is not submerged and the exit is submerged is illustrated below:

Figure 7.1: Culvert Flow The flow capacity of the culvert could be calculated as follows: (1) assume a flow, Q; (2) use backwater calculations to find where the water surface intersects the top of the pipe; (3) using this intersection point and the tailwater elevation, calculate the flow through the pressurized portion of the pipe; (4) repeat steps 1 to 3 until the flow in step 1 equals the flow in step 3. This would yield the capacity of the culvert under the given headwater and tailwater conditions.

195 7.17. From the given data, D = 450 mm = 0.45 m, L = 4 m, and S = 3% = 0.03. For an unsubmerged inlet (Type 5 ﬂow), the appropriate USFHWA equation is Equation 7.26, which is given by M H Ec = + Kg 2 FrM − 0.5S (1) D D which is applicable for Fr ≤ 0.6. Under critical-ﬂow conditions (rectangular section), 2 13 1 q 3 Q2 3 = g 2 gD2

3 3 E c = yc = 2 2

(2)

For a box culvert with 45◦ wingwalls, Table 7.1 gives K = 0.026 and M = 1. Combining Equations 1 and 2 and substituting known parameters gives H = D

3 2

2 H = 1.48

Q2 gD 2

1 3

+ Kg 2 FrM − 0.5S 1

Q2 9.81×0.452

0.45

+ 0.026(9.81)

Q/0.452

1 2

(9.81)(0.45)

− 0.5(0.03)

which simpliﬁes to 2

H = 1.19Q 3 + 0.155Q − 0.00457

(3)

where H is in m and Q is in m3 /s. Equation 3 deﬁnes the culvert performance curve provided that Fr ≤ 0.6 Q/0.452 (9.81)(0.45)

≤ 0.6

which yields Q ≤ 0.264 m3 /s Therefore, for unsubmerged conditions, when Q ≤ 0.264 m3 /s the culvert performance curve is given by Equation 3. In the case where the culvert entrance is submerged (Type 3 ﬂow), the appropriate USFHWA equation is Equation 7.15, which is given by H = cgFr2 + Y − 0.5S D

(4)

which is applicable for Fr ≥ 0.7. For a box culvert with 45◦ wingwalls, Table 7.1 gives c = 0.0347 and Y = 0.86. Substituting known parameters into Equation 4 gives H = 0.0347(9.81) 0.45

Q/0.452 (9.81)(0.45)

+ 0.86 − 0.5(0.03)

196 which yields H = 2.75Q2 + 0.258

(5)

where H is in m and Q is in m3 /s. Equation 5 deﬁnes the culvert performance curve provided that Fr ≥ 0.7 Q/0.452 (9.81)(0.45)

≥ 0.7

which yields Q ≥ 0.302 m3 /s Therefore, for submerged conditions, when Q ≥ 0.302 m3 /s the culvert performance curve is given by Equation 5. Using the analytic expressions given by Equations 3 and 5, the culvert performance curve is tabulated as follows: Q (m3 /s) 0.10 0.20 0.25 0.30 0.40 0.50 0.60

H (m) 0.27 0.43 0.50 0.51 0.70 0.95 1.25

The assumptions made for Type 3 and Type 5 flow are: low tailwater, hydraulically short culvert (for Type 3 flow), and a steep culvert slope (for Type 5 flow). For D = 0.45 m = 1.48 ft, and Q = 0.6 m3 /s = 21.2 ft3 /s, B = 0.45 m = 1.48 ft, Q/B = 21.2/1.48 = 14.3 (ft3 /s)/ft, and the USFHWA nomogram gives (for 30◦ to 75◦ wingwall flare) H = 3.1 D which yields H = 3.1D = 3.1(0.45 m) = 1.4 m This estimate of H obtained from the USFHWA nomogram (1.4 m) differs by approximately 10% from the calculated value of 1.25 m. This difference could be attributed to nomogram distortion (after repeated copying) and the uncertainty in reading the nomogram.

197 h1 0.6 = = 0.3 L 2 Using Equation 7.102, 0.65 Cd = 1 (1 + HHw ) 2

7.18. Here

Here, H L = Upstream of the weir, h1 = 0.6 m V2

Q −3 Q2 H = h1 + 2g1 = 0.6 + 2×9.81×(1.6×3) 2 = 0.6 + 2.21 × 10

Cd =

0.65 −3Q2

1 + 0.6+2.21×10 1

1 2

According to Equation 7.10, 3 √ Q = Cd gb 23 H 2 =

× 2 1

0.65

(1.6+2.21×10−3Q ) 2

√

9.81 × 3 × 2 3

(0.6 + 2.21 × 10−3Q ) ) = 3.32 (1.6 + 2.21 × 10−3Q2 )

2 −3Q2 ) 3 (0.6 + 2.21 × 10

3 2

1 2

Which yield Q = 1.23 m3 /s This solution assumes that the length of the weir is such that 0.08< hL1 <0.33. Again, Second case, L=3.0 m Cd =

0.65

H 2 (1+ 1.0 )

0.65

(1+H) 2

3 √ =⇒ Q = Cd gb 23 H 2 √ 3 =⇒ 1.23 = 0.65 1 × 9.81 × 3.0 × 23 H 2 (1+H) 2

=⇒ 1.23 = 3.32 ×

H3 (1+H)

1 2

=⇒ h = 0.01 =⇒ 0.01 = h1 +

Q2 2 × 9.81 × [(h1+1 ) × 3]2

(1.23)2 2 × 9.81 × [(h1+1 ) × 3]2 ∴ h1 = 0.005m ∼ = 0.01 m = h1 +

Water surface elevation upstream of this second weir= (1+ h1 )m = 1.01 m. This solution assumes that the length of the weir is such that 0.08 < hL1 < 0.33. 7.19. The possible ﬂow regimes are Types 2, 3, and 6, and the calculated headwater depths for these cases are as follows:

198

Type

Headwater depth (m)

2 3 6

1.515 1.044 1.143

Based on these results, the minimum-performance method would require a headwater depth of 1.515 m corresponding to Type 2 flow. This is the same as found previously. 7.20. For the given design flow rate, the tailwater elevation can be derived from the normal-flow condition in the downstream channel. Characteristics of the downstream trapezoidal channel are given as: b = 1 m, m = 2, So = 0.02, and n = 0.040. Taking Q = 3.00 m3 /s, the Manning equation gives Q=

1 2 1 AR 3 S02 n

1 (yn + 2yn2 ) 3.00 = 0.040

yn + 2yn2 √ 1 + 2 5yn

(0.02) 2

which yields a normal-flow depth, yn = 0.673 m. Since the invert elevation of the downstream channel at the culvert outlet is 11.71 m, the tailwater elevation, TW, under the design condition is given by TW = 11.71 m + 0.673 m = 12.38 m Since the diameter of the culvert is 0.76 m and the tailwater depth is 0.673 m, the culvert outlet is not submerged; and since the roadway elevation is 13.50 m and the tailwater elevation is 12.38 m, the tailwater is below the roadway. Assuming that roadway overtopping (by the headwater) occurs under the design condition, the design flow rate is equal to the sum of the flow rate through the culvert and the flow rate over the roadway such that 3 2gΔh Q = A 2gn2 L + Cd LR Hr2 (1) + ke + 1 4 R3

where Type 2 ﬂow through the culvert is assumed. From the given data: Q = 3.00 m3 /s, D = 0.76 m, A = πD2 /4 = 0.454 m2 , n = 0.012 (Table 7.3 for concrete pipe, good joints, smooth walls), L = 20 m, R = D/4 = 0.19 m, ke = 0.7 (Table 7.4), LR = 15.0 m, and Δh = HW + S0 L − D = (13.50 − 12.11 + Hr ) + 0.02 × 20 − 0.76 = 1.03 + Hr Combining Equations 1 and 2 with the given data yields 3 2(9.81)(1.03 + Hr ) + Cd (15)Hr2 3 = 0.454 2(9.81)(0.012)2 (20) + 0.7 + 1 4 0.19 3

3 = 1.35 1.03 + Hr + 15Cd Hr2

(2)

199 The simultaneous solution this equation with the graphical relations in Figure 7.6 yields Cd = 1.50 Hr = 0.17 m The ﬂow rate over the roadway, Qr , is given by 3

Qr = Cd LR Hr2 = (1.50)(15)(0.17) 2 = 1.58 m3 /s The corresponding ﬂow rate through the culvert is equal to 3.00 m3 /s − 1.58 m3 /s = 1.42 m3 /s. The ﬂow capacity of the culvert is given by 1 2 1 Qcapacity = 1.07 AR 3 S02 = 1.9 m3 /s n

Since this capacity exceeds the computed flow rate in the culvert, the culvert does not flow full and Type 3 flow is indicated. For Type 3 flow, 3

Q = Cd A 2gh + Cd LR Hr2 3 = Cd (0.454) 3 = Cd (0.454)

2(9.81)(road elevation + Hr − inlet invert −

0.76 2(9.81) 13.5 + Hr − 12.11 − 2

3 D ) + 15Cd Hr2 2 3

+ 15Cd Hr2

3 = 2.01Cd

1.01 + Hr + 15Cd Hr2

which yields Cd = 1.385 Hr = 0.037 m The ﬂow rate over the roadway, Qr , is given by 3

Qr = Cd LR Hr2 = (1.385)(15)(0.037) 2 = 0.15 m3 /s and the corresponding flow rate through the culvert is equal to 3.00 m3 /s − 0.15 m3 /s = 2.85 m3 /s. Since the flow through the culvert (2.85 m3 /s) exceeds the full-flow capacity of the culvert (1.9 m3 /s), Type 3 flow is not confirmed. Since neither Type 2 nor Type 3 flow can be confirmed, the flow conditions are intermediate between these types of flow. Type 2 conditions are more severe and should be taken as design conditions, in which case the depth of flow over the roadway is 0.17 m , the flow rate over the roadway is 1.58 m3 /s , and the flow rate through the culvert is 1.42 m3 /s . 7.21. From the given data: D = 0.450 m, H = 0.60 m, S0 = 0.01, and L = 3.6 m. From the given culvert characteristics it can be assumed that n = 0.013 and ke = 0.5.

200 (a) Assuming Type 2 ﬂow yields the following results: π π A = D2 = (0.45)2 = 0.159 m2 4 4 Δh = 0.6 − 0.45 + 3.6(0.01) = 0.186 m 0.45 D = = 0.1125 m R= 4 4 2gΔh Q = A 2gn2 L + ke + 1 4 R3 2(9.81)(0.186) = 0.232 m3 /s = (0.159) 2(9.81)(0.013)2 (3.6) + 0.5 + 1 4 (0.1125) 3

The next step is to confirm that the culvert flows full when Q = 0.232 m3 /s. According to the Manning equation, the full-flow capacity of the culvert is given by 2 1 2 1 1 1 AR 3 S 2 = (0.159)(0.1125) 3 (0.01) 2 = 0.285 m3 /s n 0.013 Since the calculated flow rate assuming Type 2 is less than Qfull , the culvert doest not flow full and the assumption of Type 2 flow in not validated. Assume Type 3 flow and that the flow is given by the orifice equation,

Qfull =

Cd = 0.62 0.45 D = 0.6 − = 0.375 m 2 2 Q = Cd A 2gh = (0.62)(0.159) 2(9.81)(0.375) = 0.267 m3 /s h=H−

Since Q < Qfull , Type 3 flow is confirmed and the capacity according to the orifice equation is 0.267 m2 /s. Assume Type 3 flow and that the flow is given by the NIST equation, where c = 0.0398, Y = 0.67, and H = 32cFr2 + Y − 0.5S D 0.6 = 32(0.0398)Fr2 + (0.67) − 0.5(0.01) 0.45 which yields Fr = 0.722, which indicates that the NIST equation is valid (Fr > 0.7), and Q = Fr · A gD = (0.722) · (0.159)

(9.81)(0.45) = 0.241 m3 /s

Since Q < Qfull , Type 3 flow is confirmed and the capacity according to the NIST equation is 0.241 m2 /s. Since the NIST equation gives the least capacity, the capacity of the culvert should be taken as 0.241 m3 /s . (b) For the outlet design, first design the stone size of the riprap. Taking Q = 0.241 m3 /s and D = 0.45 m, calculate the depth of flow in the culvert using the Manning equation, 5

(θ − sin θ) 3 θ

2 3

= 20.16

(0.013)(0.241) nQ = 20.16 √ 8√ D 0.01 (0.45) 3 0.01 8 3

201 which yields θ = 3.988 radians. Hence, h 1 θ 1 3.988 = 1 − cos = 1 − cos = 0.705 D 2 2 2 2 h TW = D = (0.705)(0.45) = 0.317 m D 4 4 0.044 Q 3 0.044 0.241 3 = = 0.0604 m = 60.4 mm d50 = TW D 0.317 0.45 Hence the median stone size should be 60 mm and the thickness should be 3 × d50 = 180 mm and length =

5.43Q D

3 2

5.43(0.241) 3

= 4.34 m

(0.45) 2

width = 3D + 0.4La = 3(0.45) + 0.4(4.34) = 3.09 m Since the calculated width is greater than three times the culvert diameter, the calculated width is conﬁrmed as the required width. 7.22. From the given data: b = 7 m, y1 = 3.5 m, yg = 0.5 m. The gate discharge, Q, is given by Q = Cd byg

2gy1 = Cd (3)(0.5)

2(9.81)(3.5) = 12.43Cd

For vertical gates, Cc = 0.61 and Cc 0.61 = = 0.585 Cd = y 1 + Cc yg1 1 + 0.61 0.5 3.5 Therefore Q = 12.43(0.585) = 7.27 m3 /s 7.23. A vertical gate with a rounded edge directs the flow under the gate in a downward direction and then gradually turns the flow into a horizontal direction. This produces a greater contraction than a sharp-edged gate which rapidly turns the flow into a horizontal direction. Consequently, the contraction coefficient of a vertical gate with a rounded edge is greater than the contraction coefficient of a vertical gate with a sharp edge. 7.24. The variables η and Cc are defined as follows yg y1

(1)

y2 yg

(2)

η = Cc and Cc =

202 Combining Equations 7.44 and 7.45 gives Cc Q= byg y 1 + Cc yg1

2gy1

Cc2 2 2 yg b yg (2gy1 ) 1 + C c y1 C c yg 2 16 y1 y1 3 16η 2 y1 3 16 8Q2 = = = yg y 1 + η y η(1 + η) gb2 y23 2 2 1+ C Q2 =

(3)

c y1

Equation 7.48 is given by y2 y3 = 2

−1 +

8Q2 1+ 2 3 gb y2

(4)

Substituting Equation 3 into Equation 4 and noting that y2 = Cc yg yields C c yg y3 = 2

16 −1 η(1 + η)

which is the same as Equation 7.49. 7.25. From the given data: b = 5 m, Q = 16 m3 /s, yg = 0.7 m. The gate discharge is given by Q = Cd byg

2gy1

where for vertical gates, Cc = 0.61 and Cc Cd = y 1 + Cc yg1 Contracted depth, y2 , is given by y2 = Cc yg = 0.61(0.7) = 0.427 m Find the downstream depth, y3 , using the hydraulic jump equation 8q 2 y2 −1 + 1 + 3 y3 = 2 gy2 where q= and therefore 0.427 y3 = 2

16 Q = = 3.2 m2 /s b 5

−1 +

8(3.2)2 1+ (9.81)(0.427)3

= 2.01 m

203 7.26. From the given data: Q = 4 m3 /s, y2 = 1.1 m, b = 3 m, and q = Q/b = 4/3 = 1.33 m2 /s. The hydraulic jump equation gives y1 8q 2 y2 = −1 + 1 + 3 2 gy1 y1 8(1.33)2 −1 + 1 + 1.1 = 2 (9.81)y13 which simpliﬁes to

−1 +

1.44 1+ 3 y1

= 2.2

and solving iteratively gives y1 = 0.24 m. For a vertical sluice gate, the coeﬃcient of contraction is 0.61, and therefore the minimum gate opening to prevent the formation of a hydraulic jump is 0.24/0.61 = 0.39 m . The energy loss in the hydraulic jump is given by ΔE =

(1.1 − 0.24)3 (y2 − y1 )3 = 0.60 m = 4y1 y2 4(1.1)(0.24)

and the corresponding power loss, ΔP , is ΔP = γQΔE = (9.79)(4)(0.60) = 23.5 kW 7.27. Using equation 7.103 ⎡ 5

13 ⎤0.1 + 1500 hL1 ⎥ 3 ⎦ 1 + 1000 hL1

h1 L

⎢ Cd = 0.5 + 0.1 ⎣

Here, h1 =0.85 m, L=2.5 m, Hw = 1.5 m, b=3.5 m 0.85 5

13 + 1500 0.85 2.5 3 1 + 1000 0.85 2.5

2.5

Cd = 0.5 + 0.1 = 0.541

Here, H = h1 + 2g1 (Equation 7.99) Since there is Since there is no inﬂow into the tank, So, H = h1 = 0.85 m Equation 7.101 √

=⇒ Q = Cd gb

2 H 3

3 2

0.1

204 = 0.541 ×

√

9.81 × 3.5

3 × 0.85

= 2.53 m3 /s Again, if the level in the tank falls from 0.85 to 0.6m, then the upstream depth be ‘h2 ’ When time,‘t’ seconds after commencement of fall in reservoir level. 3 2 2 √ H Q = Cd gb 3 √ 3 = 0.541 × 9.81 × 3.5 23 h 2 3

= 3.23 h 2 Therefore, during a time interval, ∂t, the outﬂow will be, 3

Q∂t = 3.23h 2 ∂t And the corresponding reduction in tank water level will be -∂h. Thus outﬂow will also be given, =−5000 ∂h 3

∴ 5000 ∂h = (−)3.23h 2 ∂t ∂h ⇒ ∂t = (−) 5000 3.23 × 3 h2

5000 0.6 ∂h

⇒ t = (−) 3.23

0.85 h 32

= (−) 5000 3.23

(0.85) 2

−

(0.6) 2

= 638.83 s 7.28. According to the equation 7.42 Q= by1 y2 y12g +y2 =⇒ 8.0 = 2.5 × 1.5 × y2 2×9.81 1.5+y2 =⇒ 19.62y22 − 4.55y2 − 6.825 = 0 ∴ y2 = 4.55±23.59 2×19.62 = 0.717 ∴ Downstream depth is 0.717 m Now, we know from continuity equation Q= Ai Vi Q 8 =⇒ V2 = AQ2 = b×y = 2.5×0.717 = 4.463 m/s 2

∴ V1 = 2.13m/s

205 Froude no. at the downstream 4.463 (Fr ) − √Vgy2 2 = √9.81×0.717 = 1.68 > 1

So, the ﬂow is supercritical. 7.29. The energy equation (Equation 7.51) can be put in the form y = y1 +

Q2 Q2 − 2yb2 y12 2yb2 y22

(1)

and the momentum equation (Equation 7.52) can be put in the form y 2 = y32 +

2Q2 2Q2 − gb2 y3 gb2 y2

(2)

Combining Equations 1 and 2 gives 2 1 2Q2 1 1 Q2 1 2 = y1 + y3 + 2 − − gb y3 y2 2gb2 y12 y22 which simpliﬁes to Q4 4g 2 b4

1 1 − y12 y22

Q2 gb2

Deﬁne the following relations

ξ=

1 y1 2 2 − − + y1 y22 y3 y2

+ (y12 − y32 ) = 0

2 1 − 1 + 2(λ − 1) η

(3)

(4)

y1 y3 y 2 = C c yg yg y2 η = Cc = y1 y1

(5)

λ=

(6) (7)

Combining Equations 3 to 7 yields 2 2 2 2 Cc2 yg2 1 1 Q Q η2 + 1 − =0 − 1 − ξ 4Cc2 yg2 η 2 gb2 y1 gb2 y1 η2 λ2

(8)

Equation 8 is a quadratic equation in Q2 /(gb2 y1 ), and applying the quadratic formula to Equation 8 gives 2 1 − λ12 ξ ± ξ 2 − η12 − 1 2 Q = 2 gb2 y1 η2 1 − 1 2C 2 y 2 η 2 c g

which simpliﬁes to ξ− Q = Cc

2 2 1 − λ12 ξ 2 − η12 − 1 1 η −η

byg

2gy1

206 7.30. From the given data Yg = 0.65 m, b = 1.0 m, Q = 6.5 m3 /s Using Cc = 0.61(Equation 7.46) Equation 7.43 Y2 = Cc × yg = (0.61 × 0.65) = 0.397 m Using equation 7.42 Q = by1 y2 y12g +y2 =⇒ 6.5 = 1 × y1 × 0.397 ×

2×9.81 y2 +0.397

19.62y12 − 268y1 − 106 = 0 Solving the equation y1 = 14.04 m The upstream water level is 14.04 m. According to equation 7.45 Cd = Cc yg 1+Cc y

0.61 0.65 1+0.61 1.404

= 0.60 Equation 7.44 √ Q = Cd byg 2gy1

√ = 0.60 × 1 × 0.65 2 × 9.81 × 14.04

= 6.47 m3 /s/m Discharge per unit width=6.47 m3 /s/m 7.31. From the given data: y1 = 5 m, b = 8 m, and Q = 50 m3 /s under free-ﬂow conditions. Taking Cc = 0.61, Equation 7.50 gives η = Cc

yg yg = 0.61 = 0.122yg y1 5

and Equation 7.45 gives Cd = √

Cc = 1+η

0.61 1 + 1.22yg

and Equation 7.44 gives the (free) ﬂow through the gate as Q = Cd byg 2gy1 0.61 50 = (8)(yg ) 1 + 1.22yg which yields yg = 1.88 m

2(9.81)(5)

207 For yg = 1.88 m, and η = 0.122(1.88) = 0.229, the distinguishing condition is given by Equation 7.49 as C c yg 0.61(1.88) 16 16 y3 = −1 = − 1 = 3.79 m 1+ 1+ 2 η(1 + η) 2 0.229(1 + 0.229) Since the tailwater depth of 4 m exceeds the distinguishing condition (y3 = 3.79 m), then submerged ﬂow conditions exist and the ﬂow through the gate is given by Equation 7.53. In this case, Equations 7.54 and 7.55 give y1 5 = = 1.25 y3 4 2 2 1 1 − 1 + 2(λ − 1) = − 1 + 2(1.25 − 1) = 11.8 ξ= η 0.229

λ=

and Equation 7.53 gives ξ− Q = Cc

ξ2 −

2 2 1 − λ12

1 −1 η2

byg

1 η −η

2gy1

1 2 1 2 1 2 1 − 1.252 11.8 − 11.8 − 0.2292 − 1

= 0.61

1 0.229 − 0.229

(8)(1.88)

2(9.81)(5)

= 58.7 m3 /s Therefore, when the tailwater depth is 4 m, the ﬂow through the gate is submerged and equal to 58.7 m3 /s . 7.32. From the given data: y1 = 5 m, y3 = 2 m, b = 3 m, and for a vertical gate with a sharp edge Cc = 0.61. At the distinguishing condition, C c yg 16 y3 = −1 (1) 1+ 2 η(1 + η) where η = Cc

yg yg → η = 0.61 = 0.122yg y1 5

Substituting into Equation 1 gives 0.61yg 2= 2

16 −1 0.122yg (1 + 0.122yg )

which yields yg = 0.383 m. So for gate openings < 0.383 m a hydraulic jump will occur downstream of the gate.

208 For gate openings ≤ 0.383 m, Q = Cd byg where

2gy1

(2)

Cc 0.61 = = Cd = yg y 1 + C c y1 1 + 0.61 5g

0.61 1 + 0.122yg

Substituting into Equation 2 gives the following discharge equation for yg ≤ 0.383 m: Q=

0.61 (3)yg 1 + 0.122yg

18.1yg , 1 + 0.122yg

2(9.81)(5)

which simpliﬁes to for yg ≤ 0.383 m

For gate openings > 0.383 m, ξ−

2 2 1 − λ12 ξ 2 − η12 − 1

Q = Cc

byg

1 η −η

2gy1

where 5 y1 = = 2.5 y3 2 yg yg η = Cc = 0.61 = 0.122yg y1 5 2 2 2 1 1 8.2 − 1 + 2(λ − 1) = ξ= − 1 + 2(2.5 − 1) = −1 +3 η 0.122yg yg

λ=

Substituting into Equation 3 gives the following discharge equation for yg > 0.383 m: ξ− Q = 18.1

2 2 ξ 2 − η12 − 1 (0.84)

yg ,

1 η −η

for yg > 0.383 m

where ξ=

2 8.20 −1 +3 yg

η = 0.122yg 7.33. Since

Cc = 1 − 0.75

θ 90

+ 0.36

θ 90

(3)

209 To ﬁnd θ that minimizes Cc , 0.75 dCc =− + 0.72 dθ 90

θ 90

1 =0 90

which simpliﬁes to −0.00833 + 0.0000889θ = 0 or

θ = 93.7◦

Since

d2 Cc >0 dθ2 then θ = 93.7◦ is a minimum. Since θ must be between 0◦ and 90◦ , then, within this range, the minimum value of Cc occurs at θ = 90◦ . 7.34. From Equation 7.73,

Q = Cw bH 2

(1)

If Hw is the height of the weir, then H + Hw = 2 m or H = 2 − Hw

(2)

Combining Equations 1 and 2 gives 3

Q = Cw b(2 − Hw ) 2 and re-arranging gives

Hw = 2 −

Q Cw b

2 3

(3)

Equation 7.74 gives 2 Cw = Cd 3

(4)

and Equation 7.70 gives Cd = 0.611 + 0.075

H Hw

(5)

Combining Equations 4 and 5 gives

H 0.611 + 0.075 Hw H = 1.80 + 0.221 Hw

2 Cw = 3

2g (6)

Combining Equations 3 and 6 with Q = 10 m3 /s and b = 5 m, gives

2 Hw = 2 − 1.80 + 0.221H/Hw

2 3

210 Since H = 2 − Hw , this equation can be written as

2Hw Hw = 2 − 1.58Hw + 0.442

2 3

(7)

Solving Equation 7 by trial and error gives Hw = 1.01 m and H = 2 − 1.01 = 0.99 m. Verify the validity of Equation 5: 0.99 H = 0.98 < 5 = Hw 1.01 Therefore, Equation 5 is valid. 7.35. A sill differs from a weir in that the height of the crest (Hw ) is much less than the height of water above the crest (H) such that H/Hw > 15 . The discharge coefficient for a weir (H/Hw ≤ 10) can be estimated using Equation 7.70, and the discharge coefficient for a sill (H/Hw ≥ 15) can be estimated using Equation 7.71. The efficacy of using Equation 7.72 to estimate the discharge coefficients for both sills and weirs can be measured by the percentage difference between Equations 7.70 and 7.72 for weirs and Equations 7.70 and 7.72 for sills. These results are presented in the following table: H/Hw 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Cd (Eq.7.72) 1.01 1.03 1.04 1.04 1.05 1.06 1.06 1.07 1.07 1.07 – – – – 1.07 1.07 1.07 1.07 1.07 1.07

Eq.7.70 −32% −26% −19% −13% −6% 0% 7% 13% 20% 27% – – – – – – – – – –

Eq.7.71 – – – – – – – – – – – – – – 9% 9% 8% 8% 7% 7%

211 These results indicate Equation 7.72 is most accurate in estimating the discharge coeﬃcient of weirs when 5 ≤ H/Hw ≤ 7 and Equation 7.72 appears satisfactory for estimating the discharge coeﬃcient of sills. 7.36. From the given data: ymax = 3.0 m, Qmax = 3 m3 /s, and b = 5 m. The height, H, of water above the weir is given by H = 3 − Hw and the discharge equation is 3

Q = 1.83bH 2 3

3 = 1.83(5)H 2 which gives H = 0.475 m and therefore the height of the weir, Hw , is given by Hw = 3 − H = 3 − 0.475 = 2.52 m Verify H/Hw = 0.475/2.52 = 0.19 < 0.4, therefore the formulation is okay. 7.37. According to the equation 7.70 H Cd = 0.611 + 0.075 Hw Taking Cd = 0.62 From equation 7.75 Q = 1.83bH3/2 Here, Q1 = 72 m3 /hr = 0.02m3 /s Q2 = 50m3 /hr = 0.014 m3 /s 3

0.02 =⇒ H12 = 1.83×b 2

=⇒ H1 =

(0.02) 3

(1)

=⇒ H2 = (0.014) 2

(2)

(1.83×b) 3

Similarly, 2 3

(1.83×b) 3

∴ Change in level, Δz= 35 mm = 0.035 m 2 3

∴ 0.035 = (0.02) −(0.014) 2

2 3

(1.83×b) 3

=⇒ b = 0.166m ∴ Width of the weir = 166 mm.

212 7.38. From the given data: y1 = 3 m, b = 4 m, and yg = 0.5 m. When the weir is at its minimum height the distinguishing condition exists, in which case the discharge is free and a hydraulic jump occurs just downstream of the gate. A free discharge, with Cc = 0.61, gives Cc 0.61 = = 0.581 Cd = y 1 + Cc yg1 1 + 0.61 0.5 3 Q = Cd byg

2(9.81)(3) = 8.915 m3 /s

2gy1 = (0.581)(4)(0.5)

The hydraulic jump equation is given by 1 y3 −1 + 1 + 8Fr22 = y2 2 Using Q = 8.915 m3 /s in the hydraulic jump equation yields y2 = Cc yg = (0.61)(0.5) = 0.305 m A2 = by2 = (4)(0.305) = 1.22 m2 Q 8.915 = 7.307 m/s V2 = = A2 1.22 V2 7.307 Fr2 = √ = 4.224 = gy2 (9.81)(0.305) 1 y3 −1 + 1 + 8Fr22 = y2 2 1 y3 = −1 + 1 + 8(4.224)2 0.305 2 which yields y3 = 1.676 m. This is the depth upstream of the weir, so H + Hw = 1.676 m

(1)

The weir discharge is governed by the equation 3 2 Q = Cd 2gbH 2 3 H 2 0.611 + 0.075 8.915 = 3 Hw

Combining Equations 1 and 2 gives 0.611 + 0.075

H 1.676 − H

2(9.81)(4)H 2

(2)

H 2 = 0.7549

which yields H = 1.024 m and hence Hw = 1.676 − 1.024 = 0.652 m. Therefore, the minimum weir height to ensure submergence of the gate discharge is approximately 0.65 m . 7.39. From the given data: b = 5 m, Q = 1 m3 /s, H ≥ 0.5 m, and y ≤ 1.7 m. The discharge coeﬃcient, Cd , is given by H Cd = 0.611 + 0.075 Hw

213 Requiring H = 0.5 m and y = 1.7 m when Q = 1 m3 /s, Hw = y − H = 1.7 − 0.5 = 1.2 m 0.5 H = 0.417 = Hw 1.2 H Cd = 0.611 + 0.075 = 0.611 + 0.075(0.417) = 0.642 Hw 2 2 Cw = Cd 2g = (0.642) 2(9.81) = 1.90 3 3 and the discharge equation requires 3

Q = Cw [b − 0.1nH]H 2 3

1 = 1.90[b − 0.1(2)(0.5)](0.5) 2 which gives b = 1.59 m and since Hw = 1.2 m, then the weir crest is 2 − 1.2 = 0.8 m below the top of the channel. 7.40. From the given data: Q = 55 m3 /s and b = 10 m. (a) The weir equations require that 3

Q = Cw (b − 0.2H)H 2 H Cd = 0.611 + 0.075 Hw 2 Cw = Cd 2g 3 H + Hw = 5 Combining the above equations and substituting the given data yields 3 5 − Hw 2 0.611 + 0.075 2(9.81)[10 − 0.2(5 − Hw )](5 − Hw ) 2 55 = 3 Hw 3 5 − Hw [10 − 0.2(5 − Hw )](5 − Hw ) 2 18.63 = 0.611 + 0.075 Hw which yields Hw = 2.95 m . Verify the validity of the assumed relation for Cd : H/Hw = (5-2.95)/2.95 = 0.69, which is less than 5, therefore, the assumed expression for Cd is applicable.

214 (b) At the downstream end of the lake (Section 2), y2 = 5 m A2 = b2 y2 = (15)(5) = 75 m2 P2 = b2 + 2y2 = 15 + 2(5) = 25 m A2 75 =3m = R2 = P2 25 ⎤2 ⎡ 2 nQ ⎦ (0.018)(55) ⎣ Sf 2 = = = 4.027 × 10−5 2 2 3 3 (75)(3) A2 R 2

Q 55 = 0.733 m/s v2 = = A2 75 At the upstream end of the lake, the depth is y1 and A1 = b1 y1 = 15y1 P1 = b1 + 2y1 = 15 + 2y1 A1 15y1 R1 = = P1 15 + 2y1 ⎡ ⎤2 ⎢ Sf 1 = ⎣

v1 =

3 (0.018)(55) ⎥ −4 (15 + 2y1 ) 10 2 ⎦ = 1.178 × 10 3 15y1 y13 (15y1 ) 15+2y 1

Q 55 3.667 = = A1 15y1 y1

Substituting into the energy equation,

ΔL =

2 1 y + v2g 2

S̄f − S0

100 =

(3.667)2 0.7332 y1 + 2(9.81)y − 5 + 2 2(9.81) 1

4.027×10−5 +1.178×10−4 2

100 =

4 (15+2y1 ) 3 10 y13

− 0.005

y1 + 0.6853 − 5.027 y2 1

4 3

1) 5.888 × 10−5 (15+2y − 0.004980 10

y13

which yields y1 = 4.50 m. Therefore, the depth at the upstream end of the lake is 4.50 m .

215 7.41. For the upstream location, y1 = 3.50 m A1 = 6(3.5) + 2(3.5)2 = 45.5 m2 V1 = Q/A1 = 0.549 m/s V12 = 0.01536 m 2g √ P1 = 6 + 2(3.5) 5 = 21.65 m 2

S1 =

(0.05)(25)(21.65) 3

= 0.0002803

(45.5) 3

Applying the energy equation between the upstream location and the weir gives 252 [3.50 + 0.01536] − y2 + 2(9.81)(6y +2y2 )2 2 2 100 = √ 2 2 (0.05)(25)(6+2y2 5) 3 5 2) 3 (6y2 +2y2

+0.0002803

− 0.001

which gives y2 = 3.57 m. For the weir, the Rouse equation gives 3 H 2 0.611 + 0.075 2gbH 2 Q= 3 Hw This equation is valid as long as H/Hw < 5. The weir will be taken as a Cipolletti weir. Taking Z as the elevation of the weir crest, b as the length of the weir crest, and noting that the elevation of the bottom of the river at the weir location is 2.00 m − 0.1 m = 1.90 m, then H = 3.57 − Hw Z = Hw + 1.90 Combining the Rouse equation with the expression for H in terms of Hw gives the following design equation, 3 0.268 8.466 = 0.536 + b(3.57 − Hw ) 2 Hw This equation can be solved for alternative values of b. A possible design is: b = 2 m, Z = 2.65 m , Hw = 0.75 m, H = 2.82 m, H/Hw = 3.77 (which validates using the Rouse equation for Cd ). 7.42. From the given data: Q = 5 m3 /s, b = 10 m, and yn = 5 m. For a 5-m high suppressed sharp-crested weir, 3

Q = 1.83bH 2 3

5 = 1.83(10)H 2 which gives H = 0.421 m

216 Check H/Hw = 0.421/5 = 0.084 < 0.4, hence the weir discharge formula is validated, and the new ﬂow depth is 5 + 0.421 = 5.42 m . Consider a rectangular weir with b = 7 m, and a height of 5 m. 1st iteration, H = 0.5 m: Cd = 0.611 + 0.075 2 Cw = Cd 3

H 0.5 = 0.62 = 0.611 + 0.075 Hw 5

2 2g = (0.62) 3

2(9.81) = 1.83

and the discharge equation gives 3

Q = Cw [b − 0.1nH]H 2 3

5 = 1.83[7 − 0.1(2)H]H 2 which yields H = 0.54 m 2nd iteration, H = 0.54 m: Cd = 0.611 + 0.075 2 Cw = Cd 3

H 0.54 = 0.619 = 0.611 + 0.075 Hw 5

2 2g = (0.619) 3

2(9.81) = 1.83

and the discharge equation gives 3

Q = Cw [b − 0.1nH]H 2 3

5 = 1.83[7 − 0.1(2)H]H 2 which yields H = 0.54 m Hence, if the weir is contracted, the ﬂow depth is 5 + 0.54 = 5.54 m . For a 7-m wide Cipolletti weir, 3

Q = Cw bH 2 and taking Cw = 1.83 gives 3

5 = 1.83(7)H 2 which yields H = 0.53 m, and hence a ﬂow depth of 5 + 0.53 = 5.53 m . Since H/Hw = 0.53/5 = 0.106 < 0.4, the applied formula is correct.

217 7.43. From the given data: H + Hw = 2 m, and Q = 5 m3 /s. The discharge formula is given by 3

Q = Cw bH 2

(1)

where H 2 0.611 + 0.075 2g = 3 Hw H = 1.80 + 0.221 Hw

2 Cw = Cd 3

2g (2)

If the top-width of the weir is 10 m, then b+2

H 4

or b = 10 −

= 10

H 2

(3)

Combining Equations 1 to 3, and substituting the given data yields 3 H H 5 = 1.80 + 0.221 10 − H2 2−H 2 which gives H = 0.42 m and hence Hw = 2 − 0.42 = 1.58 m and b = 10 −

0.42 H = 10 − = 9.79 m 2 2

7.44. From the given data: Hw = 0.8 m, b = 1 m, and H = 0.2 m. Check H/Hw = 0.2/0.8 = 0.25 < 0.4, therefore 3

Q = 1.83bH 2 = 1.83(1)(0.2) 2 = 0.164 m3 /s When the weir is submerged,

y 3 0.385 Qs d 2 = 1− Q H

which gives Qs = Q 1 −

y 3 0.385 d

= (0.164) 1 −

0.05 0.2

3 2

0.385

= 0.156 m3 /s

218 7.45. The flow over the weir is submerged, and hence either Equation 7.78 or Equation 7.79 can be used to estimate the ratio of submerged flow, Qs , to free flow, Q. From the given data: H = 2.5 m − 2.0 m = 0.5 m, and yd = 2.3 m − 2.0 m = 0.3 m. Equation 7.78 gives 3 y 3 0.385 0.3 2 Qs d 2 = 1− = 1− Q H 0.5 and Equation 7.79 gives yd Qs = 1+ Q 2H

0.385

= 0.786

0.3 yd 0.3 = 1+ = 0.822 1− 1− H 2 × 0.5 0.5

These results indicate that Qs /Q is in the range 0.786–0.822. From the given data: Hw = 2 m, and H/Hw = 0.5/2 = 0.25. Since H/Hw < 5, then either Equation 7.70 or Equation 7.72 can be used to estimate the discharge coeﬃcient, Cd . According to Equation 7.70, Cd = 0.611 + 0.075

H = 0.611 + 0.075(0.25) = 0.630 Hw

and Equation 7.72 gives 10 15 −0.01 14.14 H/Hw + Cd = 1.06 8.15 + H/Hw H/Hw + 1 10 15 −0.01 0.25 14.14 = 1.06 + 8.15 + 0.25 0.25 + 1

= 1.01 These results indicate that Cd is in the range 0.630–1.01. From the given data, b = 5 m and the free-ﬂow discharge over the weir is given by Equation 7.67 2 Q = Cd 3

3 2 2gbH 2 = Cd 3

2(9.81)(5)(0.5) 2 = 5.22Cd

For Cd = 0.630, Q = 5.22(0.630) = 3.29 m3 /s and for Cd = 1.01, Q = 5.22(1.01) = 5.27 m3 /s. The minimum submerged flow, Qs , corresponds to Q/Qs = 0.786, which yields Qs = 0.786(3.29) = 2.59 m3 /s, and the maximum estimate of Qs corresponds to Q/Qs = 0.822, which yields Qs = 0.822(5.27) = 4.33 m3 /s. Available empirical formulae yield flow rates over the weir in the range 2.59– 4.33 m3 /s . For a more precise estimate of the flow rate, field calibration of this particular weir would be necessary. 7.46. From the given data: z1 = 101.40 m, z2 = 96.00 m, Q = 4.00 m3 /s, and b = 6 m. Also, H + Hw = 101.40 − 96.00 = 5.40 m. The weir equation gives: 3 2 Q = Cd 2gbH 2 3 H 2 0.611 + 0.075 4= 3 5.40 + H

2(9.81)(6)H 2

219 which yields H = 0.51 m. Therefore, Hw = 5.40 − H = 4.89 m, and H/Hw = 0.014. Since H/Hw < 5, the equation used for Cd is valid. Based on these derived data, the crest elevation of the weir is 101.40 m − 0.51 m = 100.89 m . Using Villemonte’s formula, for a 90% flow reduction y 0.385 Qs d = 1− Q H y 0.385 Qs d = 1− Q H which yields yd /H = 0.386, and hence yd = 0.386(0.51) = 0.20 m. The limiting tailwater elevation is 100.89 m + 0.20 m = 101.09 m . 7.47. From the given data: S0 = 0.1% = 0.001, b = 2 m, m = 2, yT = 2 m, Q = 3 m3 /s, and y = 1.5 m. (a) For a contracted rectangular weir with the Rouse equation (Equation 7.70) used to estimate the discharge coefficient, the flow equation is given by 3

Q = Cw (b − 0.1nH)H 2 3 2 = Cd 2g (b − 0.1nH)H 2 3 3 H 2 0.611 + 0.075 = 2g (b − 0.1nH)H 2 3 Hw Substituting the given parameters yields H 2 0.611 + 0.075 3= 3 Hw

2g (2 − 0.2H)H 2

(1)

(2)

Also, since the maximum ﬂow depth is 1.5 m, then H + Hw = 1.5 m

(3)

Solving Equations 2 and 3 gives H = 0.85 m Hw = 0.65 m These results were derived using the Rouse equation (Equation 7.70) to estimate the discharge coeﬃcient, and the Rouse equation is strictly valid only when H/Hw < 5. In this case, H/Hw = 1.31 so the Rouse approximation is valid. However, the contraction formula given by Equation 1 is only valid when b > 3H, and in this case b = 2.35H and so the results do not validate using the contraction discharge formula given by Equation 1. Alternative estimates of Cd can be derived from either the Chaudhry equation (Equation 7.71) or the Swamee equation (Equation 7.72). The Chaudhry equation is for “sill-like” conditions which are unlikely to exist here, and so the general equation for estimating the discharge coeﬃcient proposed by Swamee will also be considered.

220 For a contracted rectangular weir with the Chaudhry equation (Equation 7.71) used to estimate the discharge coeﬃcient, the ﬂow equation is given by ⎡ ⎤ 15 −0.01 10 3 2 14.14 H/Hw Q = ⎣ 1.06 + 2g ⎦ (b − 0.1nH)H 2 (4) 3 8.15 + H/Hw H/Hw + 1 Substituting the given parameters into Equation 4 and solving simultaneously with Equation 3 yields H = 0.66 m Hw = 0.84 m In this case b = 3.03H and so the results validate using the contraction discharge formula given by Equation 1. Analysis: It is apparent that the uncertainty in Cd (Rouse vs. Swamee) is such that the contraction discharge formula (Equation 1) might be valid in some cases. The results presented here collectively indicate that the required height of the weir is in the range of 0.65–0.84 m and that Cd estimated by the Chaudhry equation is higher than Cd estimated by the Rouse equation. To be conservative, the lower weir height should be selected along with the lower Cd (i.e., Rouse solution) so as to ensure that the total depth of 1.5 m is not exceeded. Therefore, the required weir can be taken as 0.65 m high , and 2 m wide . (b) When the weir is submerged, Villemonte’s formula gives y 3 0.385 Qs d 2 = 1− Q H where yd = (1.5 − 0.20) − 0.65 = 0.65 m Therefore, Qs = 1− 3

0.65 0.85

3 2

0.385

which yields Qs = 1.96 m3 /s Since the free-ﬂow weir capacity is 3 m3 /s, then submergence reduces the weir capacity by 3 − 1.96 = 1.04 m3 /s . (c) For Qs /Q = 0.9, the submerged weir-ﬂow equation gives 0.9 = 1 −

y 3 0.385 d

0.85

which yields yd = 0.33 m

221 The corresponding tailwater elevation is 0.65 m + 0.33 m = 0.98 m, and hence the difference between the headwater and tailwater elevations is 1.5 m − 0.98 m = 0.52 m. As long as the elevations of the headwater and tailwater differ by less than 52 cm the flow will deviate by less than 10% from the free-flow weir discharge. 7.48. The relative (percentage) difference, E, in using Equation 7.79 instead of Equation 7.78 to estimate Qs /Q is given by yd 3 0.385 yd yd 2 1− H − 1− H 1 + 2H × 100% E= 3 0.385 1 − yHd 2 Which yields the following tabular relation: yd /H 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

E (%) 0.0 0.9 2.0 3.1 4.0 4.6 4.6 3.8 1.6

The maximum diﬀerence between Equations 7.78 and 7.79 is approximately 4.6% , which is within the uncertainty of Equations 7.78 and 7.79 in predicting the the submerged ﬂow rates. On the basis of these results, neither equation is preferable . 7.49. From the given data: Q = 12 L/s = 0.012 m3 /s, and θ = 30◦ . The discharge equation requires that 5 8 θ Q = Cd 2g tan H2 15 2 and assuming that Cd = 0.58 gives 8 0.012 = (0.58) 15

2(9.81) tan

30◦ 2

which simpliﬁes to H = 0.25 m Since H > 50 mm, Cd = 0.58 is validated. 7.50. From the given data: H = 0.5 m, and θ = 50◦ . The discharge equation gives ◦ 5 5 θ 8 50 8 H 2 = (0.58) 2(9.81) tan (0.5) 2 = 0.11 m3 /s Q = Cd 2g tan 15 2 15 2

222 7.51. From the given data: θ = 90◦ , and Q = 1 L/min = 1.667 × 10−5 m3 /s. The properties of water at 20◦ C are: σ = 0.0734 N/m, ρ = 998.2 kg/m3 , and μ = 0.001 N·s/m2 . Therefore 1

1 (998.2)(9.81) 2 H 2 ρg 2 H 2 = = 3.13 × 106 H 2 Re = μ 0.001 (998.2)(9.81)H 2 ρgH 2 = = 1.33 × 105 H 2 We = σ 0.0734 1.19 1.19 0.0138 Cd = 0.583 + 1 = 0.583 + 1 1 = 0.583 + 5 (ReWe) 6 [(3.13 × 106 H 2 )(1.33 × 105 H 2 )] 6 H 12

Substituting into the discharge equation gives 5 θ 8 H2 Q = Cd 2g tan 15 2 ◦ 5 0.0138 8 90 −5 0.583 + H2 1.67 × 10 = 2(9.81) tan 5 15 2 H 12 which gives H = 0.010 m 7.52. From the given data: H = 40 mm, and θ = 70◦ , therefore 5 θ 8 H2 Q = Cd 2g tan 15 2

(1)

Since H < 50 mm, Cd = 0.583 + 1

1.19 1

(ReWe) 6

(9.81) 2 (0.04) 2 g2H 2 = Re = = 6.26 × 105 ν 1.00 × 10−6 ρgH 2 (998.2)(9.81)(0.04)2 We = = 215.2 = σ 72.8 × 10−3 where a temperature of 20◦ C has been assumed. Substituting into Equation 1 gives ◦ 5 1.19 70 8 0.583 + (0.04) 2 2(9.81) tan Q= 1 15 2 (6.26 × 105 × 215.2) 6 = 3.36 × 10−4 m3 /s = 0.336 L/s 7.53. From the given data: Q = 0.30 m3 /s, H + Hw = 1 m, and T = 1 m. By geometry, θ T = 2H tan 2 and the discharge relation is given by 8 Q = Cd 15

5 θ H2 2g tan 2

(1)

(2)

223 Combining Equations 1 and 2, and taking Cd = 0.58 gives 3 T 8 Q = (0.58) 2(9.81) H2 15 2 3 8 1 H2 0.30 = (0.58) 2(9.81) 15 2 which gives H = 0.577 m T 1 −1 −1 θ = 2 tan = 2 tan = 82◦ 2H 2 × 0.577 Hw = 1 − H = 1 − 0.577 = 0.423 m 7.54. (a) The Kindsvater and Shen equation can be expressed in the form 5 2 5 θ 8 (H + k) 2g tan Cd H2 Q= 5 15 2 H2 or

5 k 2 8 Cd 1 + Q= 15 H

5 θ H2 2g tan 2

Using the Kindsvater and Shen expression for Cd and k yields 4.42 − 0.1035θ + 1.005 × 10−3 θ2 − 3.24 × 10−6 θ3 Cd = (0.6072 − 0.000874θ + 6.1 × 10−6 θ2 ) 1 + 1000H

where θ is in degrees and H is in meters. (b) For θ = 90◦ , the Kindsvater and Shen parameters are Cd = 0.578 and

k = 0.8835 mm

and so the Kindsvater and Shen equation can be put in the form 5 θ 8 (H + k) 2 Q = Cd 2g tan 15 2 π H + 0.8835 52 8 = (0.5780) 2(9.81) tan 15 4 1000 where H is in millimeters. The above equation simpliﬁes to 2

H = 882.9Q 5 − 0.8835

(1)

To release 20,000 m3 in 24 hours requires an average ﬂow rate of 20000/(24 × 60 × 60) = 0.2315 m3 /s. Using this ﬂow rate in Equation 1 yields H = 490 mm = 0.49 m Therefore using a weir height of 0.49 m will release the stored water from the detention pond in no less than 24 hours.

224 7.55. From the given data: θ = 90◦ , H2 = 20 cm = 0.20 m, and H1 = 1.20 − H2 = 1.20 − 0.20 = 1.00 m. The discharge, Q, over the weir is given by Equation 7.92 as 2 Q = Cd 3

3 2

H2 2gbH1 + Cvd A 2g H1 + 3

(1)

The discharge coeﬃcient, Cd , is given by Equation 7.70 as Cd = 0.611 + 0.075

H1 H2

(2)

which yields Cd = 0.611 + 0.075

1.00 = 0.986 0.20

Since H1 /H2 = 5, Equation 2 is valid, and Cd can be taken as 0.986 in the weir design. The area, A, of the V-notch is given by ◦ θ 90 2 2 = 0.20 tan = 0.04 m2 A = H2 tan 2 2 Taking Cvd = 0.6 and Q = 1.5 m3 /s, substituting into Equation 1 gives 2 1.5 = (0.986) 3

3 2

2(9.81)b(1.00) + 0.60(0.04)

0.20 2(9.81) 1.00 + 3

which yields b = 0.48 m Therefore, a compound weir with a crest length of 48 cm will yield a discharge of 3 m3 /s when the water depth in the detention area is 1 m. 7.56. From the given data: Hw = 0.25 m, b = 1.5 m, y = 0.75 m, and the height of water, h1 , above the weir is h1 = y − Hw = 0.75 − 0.25 = 0.50 m. Q2 Q2 Q2 = h + = 0.50 + = 0.50 + 0.0403Q2 1 2gA2 2g(by)2 2(9.81)(1.5 × 0.75)2 0.65 0.65 0.65 Cd = 1 = 1 1 = 2 2 2 (3 + 0.161Q2 ) 2 1 + HHw 1 + 0.50+0.0403Q 0.25 3 3 2 2 1 1 2 2 0.65 2 2 2 H = (0.50 + 0.0403Q ) Q = Cd bg 1 (1.5)(9.81) 3 3 (3 + 0.161Q2 ) 2 H = h1 +

which can be solved iteratively for Q to give Q = 0.343 m3 /s

225 7.57. From the given data: L = 1 m, b = 1 m, and Hw = 0.3 m. The valid operating range is 0.08 < h1 /L < 0.33, so 0.08 m < h1 < 0.33 m. V2 Q2 = h1 + 2g 2gb2 (h1 + Hw )2 Q2 Hmin = 0.08 + = 0.08 + 0.353Q2 2(9.81)(1)2 (0.08 + 0.3)2 Q2 Hmax = 0.33 + = 0.33 + 0.0796Q2 2(9.81)(1)2 (0.5 + 0.3)2 0.65 Cd = 1 2 1 + HHw H = h1 +

Cd,min = Cd,max =

0.65 1 + 0.08+0.353Q 0.3

1 = 2

0.65

1 + 0.33+0.0796Q 0.3 3 2 1 2 H Q = Cd bg 2 3

Qmin =

0.65

1 = 2 2

0.65 1

(1.267 + 1.177Q2min ) 2

(1.267 + 1.177Q2 ) 2 0.65 1

(2.10 + 0.265Q2 ) 2

(1)(9.81)

1 2

3 2 2 2 (0.08 + 0.353Qmin ) 3

which yields Qmin = 0.0223 m3 /s and Qmax =

0.65 1

(2.10 + 0.265Q2max ) 2

(1)(9.81)

1 2

3 2 2 2 (0.33 + 0.0796Qmax ) 3

which yields Qmax = 0.146 m3 /s Therefore, the operating range for the weir is given by 0.0223 m3 /s < Q < 0.146 m3 /s 7.58. From the given data: b = 3 m, h1 + Hw = 4 m, Q = 5 m3 /s, and (Q/A)2 [5/(3 × 4)]2 V12 = = = 0.00885 m 2g 2g 2(9.81) which gives H = h1 +

V12 = h1 + 0.00885 2g

(1)

226 The discharge rate is given by √ Q = Cd gb

2 H 3

3 2

(2)

and 0.65

Cd =

(3)

(1 + H/Hw ) 2

Combining Equations 1 to 3 and substituting the given data yields √

0.65

(1 + H/Hw ) 5=

1 2

0.65 1 + h1 +0.00885 4−h1

√

2 H 3

3 2

3 2 2 9.81(3) (h1 + 0.00885) 3

which gives h1 = 1.53 m

3.84 m

Hw = 4 − h1 = 0.16 m

2.47 m

This indicates two possible ﬂow conditions:

Condition 1 2

h1 (m) 1.53 3.84

H (m) 1.54 3.85

Hw (m) 2.47 0.16

H/Hw

0.62 24.06

0.51 0.13

√

3 gb 23 H 2 (m3 /s) 9.77 38.64

Only Condition 1 gives a Cd that is typical of properly-operated broad-crested weirs (typically 0.527 < Cd < 0.540), so take Hw = 2.47 m . For proper operation of the weir, take h1 /L = 0.25, so L=

1.53 h1 = = 6.12 m 0.25 0.25

Therefore the weir should be 2.47 m high and 6.12 m long. 7.59. From the given data: b = 3 m, Q = 1.5 m3 /s, y = 2 m, and therefore h1 = 2 − Hw

227 and Q2 1.52 = (2 − H ) + = 2.003 − Hw w 2gb2 (h1 + Hw )2 2(9.81)(3)2 (2 − Hw + Hw )2 1 0.65 0.65 2 Cd = 1 = 1 = 0.459Hw 2 2.003−Hw 2 H 1 + Hw 1+ Hw 3 2 1 2 H Q = Cd bg 2 3 3 1 2 1 2 2 2 (2.003 − Hw ) 1.5 = 0.459Hw (3)(9.81) 3 H = h1 +

which gives Hw = 0.056 m, 1.33 m To achieve critical ﬂow conditions over the weir use Hw = 1.33 m , which corresponds to h1 = 2 − Hw = 2 − 1.33 = 0.67 m The weir length, L, is then given by the relation 0.08 < h1 /L < 0.33 12.5 > L/h1 > 3.03 12.5h1 >

> 3.03h1

12.5(0.67) >

> 3.03(0.67)

8.38 m >

> 2.03 m

Therefore a weir length in the range 2.0 m < L < 8.4 m would be satisfactory. 7.60. From the given data: H = 11.70 m − 11.20 m = 0.50 m, Q = 0.75 m3 /s, and Hw = 11.20 − 9.10 m = 2.10 m. Using the Chow (1959) equation, Cd =

0.65 (1 + H/Hw )

1 2

0.65

= 0.584

(1 + 0.50/2.10) 2

Applying the weir equation, √

Q = Cd gb

2 H 3

√ 0.75 = 0.584 9.81b

3 2

3 2 2 0.50 3

which yields b = 2.13 m. For broad-crested weirs to function properly requires that 0.08 < h1 /L < 0.33. Using the Swamee (1988) equation with h1 /L = 0.20 yields

(0.20)5 + 1500(0.20)1 3 Cd = 0.5 + 0.1 1 + 1000(0.20)3

0.1 = 0.536

228 Applying the weir equation, √ 0.75 = 0.536 9.81b

3 2 2 0.50 3

which gives b = 2.32 m. Therefore, the Swamee (1988) equation gives the more conservative design with b = 2.32 m and L = h1 /0.20 = 0.50/0.20 = 2.50 m . 7.61. When the depth is 0.50 m: Q = 0.08 m3 /s, h1 = 0.25 m, and Hw = 0.25 m. The discharge is given by 3 2 2 √ (1) Q = Cd gb h1 3 where h1 + 0.89 − 0.38 h1 + Hw 2 0.25 0.25 = 0.95 + 0.89 = 0.938 − 0.38 0.25 + 0.25 0.25 + 0.25

Cd = 0.95

h1 h1 + Hw

Substituting known quantities into Equation 1 yields 3 2 2 h1 3 3 √ 2 2 0.08 = (0.938) 9.81b (0.25) 3 √ Q = Cd gb

which gives b = 0.400 m . When the depth is 0.75 m: Q = 0.50 m3 /s. The total flow can be taken as the flow in the center section, Qc , plus the flow in the side sections, Qs . Take the side sections to each have the same width, bs . For flow in the center section: h1c = 0.50 m, Hwc = 0.25 m, and the discharge coefficient, Cdc , is given by

h1c − 0.38 + 0.89 Cdc = 0.95 h1c + Hwc 2 0.50 0.50 + 0.89 = 1.06 − 0.38 = 0.95 0.50 + 0.25 0.50 + 0.25 h1c h1c + Hwc

For ﬂow in the side section: h1s = 0.25 m, Hws = 0.50 m, and the discharge coeﬃcient, Cds , is given by h1s Cds = 0.95 + 0.89 − 0.38 h1s + Hws 2 0.25 0.25 = 0.95 + 0.89 = 0.869 − 0.38 0.25 + 0.50 0.25 + 0.50

h1s h1s + Hws

229 The total ﬂow, Q, is therefore given by 3 3 2 2 2 2 B−b √ h1s Q = Cdc gb h1c + 2Cds g 3 2 3 3 3 √ √ 2 2 2 2 B − 0.400 (0.25) 0.50 = (1.06) 9.81(0.400) (0.50) + 2(0.869) 9.81 3 2 3 √

which yields B = 1.720 m . For proper operation of the weir it is required that: h1 0.25 h1 > 0.08 → L < →L< → L < 3.125 m L 0.08 0.08 h1 0.50 h1 < 0.33 → L > →L> → L > 1.515 m L 0.33 0.33 Based on these results, it would be appropriate to take L in the range of 1.52 m < L < 3.13 m . 7.62. The surface of the spillway in the quadrant downstream of the crest is given by Equation 7.116 as n x 1 y = Hd K Hd The slope of the spillway is given by dy/dx where n−1 n−1 Hd x x n 1 dy = n = dx K Hd Hd K Hd If x = XDT when dy/dx = α (= downstream slope) then n−1 x n α= K Hd which yields

Taking n = 1.85,

1 n−1 1 XDT (Kα) = Hd n 1 1.85−1 1 XDT (Kα) = Hd 1.85

which simplifies to XDT = 0.485(Kα)1.176 Hd 7.63. Assuming that the spillway is sufficiently high that the ratio of the spillway height, P , to the design head, Hd , is greater than 3, Figure 7.28(a) gives the basic discharge coefficient as C0 = 2.18. Assuming that the effective head, He , is less than 9.1 m, then for the limiting case where the water pressure on the spillway is equal to −4.6 m, Equation 7.115 requires that 1 He = 1.43 = Hd 0.7

230 and Figure 7.28(b) gives Cincl /Cvert = 0.99, and Figure 7.28(c) gives C/C0 = 1.05. The discharge coeﬃcient, C, under maximum headwater conditions can now be calculated using the relation Cincl C C= C0 = (0.99)(1.05)(2.18) = 2.27 Cvert C0 From the given data, Q = 1600 m3 /s, Le = 40 m, and therefore the eﬀective head, He , on the spillway is given by Equation 7.114 as He =

Q CLe

2 3

1600 (2.27)(40)

2 3

= 6.78 m

Since this calculated value of He is less than 9.1 m, the initial assumption that He < 9.1 m used in estimating the discharge coeﬃcient is validated. The design head, Hd , corresponding to He = 6.78 m is given by Equation 7.115 as Hd = 0.7He = 0.7(6.78) = 4.75 m The depth of water, d, upstream of the spillway is given by d = 200 m − 170 m = 30 m and the approach velocity, vo , can be estimated by v0 =

1600 Q = = 1.33 m/s Le d (40)(30)

with a velocity head, hv , given by hv =

1.332 v02 = = 0.09 m 2g 2(9.81)

Therefore, at the maximum pool elevation, the height of water above the spillway is He − hv = 6.78 m − 0.09 m = 6.69 m, and the required crest elevation of the spillway is 200 m − 6.69 m = 193.31 m . Having determined the required crest elevation of the spillway (= 193.31 m) to pass the design ﬂow rate (= 1600 m3 /s), the next step is to determine the required shape of the spillway in the vicinity of the crest. Since the maximum pool elevation is 200 m and the bottom elevation behind the spillway is 170 m, the height of the spillway crest, P , is given by P = 200 m − 170 m − 6.69 m = 23.31 m and

23.31 P = 4.91 = Hd 4.75

which indicates a “high spillway” with negligible approach velocity. Figure 7.27(a) gives K = 2.0 for the spillway coordinate coeﬃcient, and the proﬁle of the downstream quadrant of

231 the spillway, taking n = 1.85, is given by n x y 1 = Hd K Hd y 1 x 1.85 = 4.75 2.0 4.75 which simplifies to y = 0.133x1.85 Since the downstream slope of the spillway is 1:1.5 (H:V), then α = 1.5 and the horizontal distance from the apex to the downstream tangent point, XDT , is given by Equation 7.117 as XDT = 0.485(Kα)1.176 Hd XDT = 0.485(2.0 × 1.5)1.176 4.75 which gives XDT = 8.39 m Therefore, at a distance of 8.39 m downstream of the crest, the curved spillway profile merges into the linear profile of the spillway chute, which has a slope of 1:1.5 (H:V). The correspond1.85 = 0.133(8.39)1.85 = 6.80 m. ing y value is y = 0.133XDT Taking P/Hd = 4.91 in Figure 7.27(b) and (c) gives A/Hd = 0.28 and B/Hd = 0.165, which yields A = 0.28(4.75) = 1.33 m, B = 0.165(4.75) = 0.784 m, and the profile of the upstream quadrant of the spillway is given by Equation 7.118 as x2 (B − y)2 + =1 A2 B2 x2 (0.784 − y)2 + =1 1.332 0.7842 which simplifies to x2 + 2.878(0.784 − y)2 = 1.769

(1)

Since the upstream face of the spillway has a 1:1 slope, then Fs = 1 and the horizontal distance from the apex to the upstream tangent point is given by XU T =

A2 Fs

1 =

[A2 Fs2 + B 2 ] 2

(1.33)2 (1) 1

[(1.33)2 (1)2 + (0.784)2 ] 2

= 1.146 m

Substituting x =1.146 m into Equation 1 gives two solutions: y = 0.386 m and y = 1.182 m. Referring to Figure 7.2, it is apparent why we have two solutions (y1 and y2 ), and that the value of y to be selected is the lower value of y = 0.386 m. Therefore, the shape of the spillway upstream of the crest is given by Equation 1 up to the point (1.146 m, 0.386 m) where it merges with the planar upstream face.

232

Spillway Surface (0,0) (X UT ,y 1)

B (0,B)

A (X UT ,y 2)

x 2/A2 + (B-y) 2/B 2 = 1 y

Figure 7.2: Upstream Spillway Shape [As an aside, Equation 1 can be diﬀerentiated to give x dy = dx 2.878(0.784 − y)

(2)

At (−1.146 m, 0.386 m) Equation 2 gives dy/dy = 1, and at (−1.146 m, 1.182 m) Equation 2 gives dy/dy = −1. Since dy/dy = 1 at the upstream face, our result is conﬁrmed.] 7.64. The discharge over the spillway is given by Equation 7.120 as 3

Q = CLe He2

(1)

Let X be the required crest elevation, then taking the pool elevation as 200 m, the eﬀective head, He , is given by He = 200 − X (2) Combining Equations 1 and 2 yields 3

Q = CLe (200 − X) 2

(3)

The eﬀective length of the spillway, Le , is given by Equation 7.121 as Le = L − wN − 2(N Kp + Ka )He

(4)

From the given data: w = 1 m, N = 2, for round-nosed piers Kp = 0.01, and for square abutments Ka = 0.2. Substituting these values, along with Equation 2 into Equation 4 yields Le = L − (1)(2) − 2(2 × 0.01 + 0.2)(200 − X) = L + 0.44X − 90

(5)

233 Combining Equations 3 and 5 and putting Q = 500 m3 /s gives the following relationship between the crest length and the crest elevation, 3

500 = C(L + 0.44X − 90)(200 − X) 2 which can be put in the convenient form L=

500 3

C(200 − X) 2

− 0.44X + 90

(6)

Determination of the discharge coeﬃcient requires speciﬁcation of the crest height, P , which is given by P = X − 150 (7) where the bottom elevation behind the spillway is taken as 150 m. Also needed is the design head, Hd , given by Equation 7.114 for He > 9.1 m, and Equation 7.115 for He < 9.1 m. Using the above-derived equations, the relationship between the crest elevation, X, and the spillway length, L, is given in the following table for selected crest elevations between 195 m and 180 m. X (m) 195 190 185 180

P (m) 45 40 35 30

He (m) 5 10 15 20

Hd (m) 3.7 7.5 12.4 17.9

P/Hd

He /Hd

Ci /Cv

C/C0

12.1 5.3 2.8 1.7

1.4 1.3 1.2 1.1

2.18 2.18 2.18 2.18

0.99 0.99 0.99 0.99

1.05 1.04 1.02 1.00

2.27 2.24 2.20 2.16

L (m) 23.9 13.5 12.5 13.4

7.65. From the given data: He = 0.50 m, Q = 0.75 m3 /s, Ka = 0 (abutments will be well-rounded), and P = 2.10 m. The design head, Hd , is given by Hd = 0.7He = 0.7(0.50) = 0.35 m and hence P/Hd = 2.10/0.35 = 6.00, and He /Hd = 0.50/0.35 = 1.43. The discharge coeﬃcient is determined as: C0 = 2.18, C/C0 = 1.06 (using a vertical upstream face) and hence C = C0 ×

C = (2.18)(1.06) = 2.31 C0

Using the spillway equation, 3

Q = CLe He2 3

0.75 = (2.31)Le (0.50) 2 which yields Le = 0.918 m . Therefore, by using a spillway instead of a broad-crested weir a smaller crest length is required, hence a smaller “footprint” associated with the structure.

234 For P/Hd = 6.00, K = 2.00, A/Hd = 0.282, B/Hd = 0.167, and n can be taken as 1,85. Therefore, A = 0.282(0.35) = 0.0987 m, B = 0.167(0.35) = 0.0585 m, and the elevation of the upstream spillway surface is given by n y x 1 = Hd K Hd y 1 x 1.85 = 0.35 2.00 0.35 which yields y = 1.22x1.85 For the downstream surface of the spillway, (B − y)2 x2 + =1 A2 B2 x2 (0.0585 − y)2 + =1 0.09872 0.05852 which yields (0.0585 − y)2 x2 + = 10−3 9.74 3.42 7.66. From the given data: Le = 5.0 m, P = 4.2 m, Hd = 2.0 m, and hd = 2.0 m − 0.7 m = 1.3 m. From these data, P/Hd = 4.2/2.0 = 2.1, and Figure 7.28(a) gives the basic discharge coefficient, C0 , as 2.17. Since the upstream face of the spillway is vertical, and assuming the upstream velocity head is negligible, then He ≈ Hd and Figure 7.28(b) and (c) indicate that the free-flow discharge coefficient, C, is equal to C0 , and hence C = 2.17. The free-flow discharge, Q0 , is given by Equation 7.120 as

Q20 Q0 = CLe He = (2.17)(5.0) 2.0 + 2gA2 3 2

Q20 = (2.17)(5.0) 2.0 + 2(9.81)(6.2 × 5)2

which yields Q0 = 31.9 m3 /s. This ﬂow corresponds to an upstream velocity, va , and velocity head, va2 /2g, given by va =

31.9 v2 Q0 = = 1.03 m/s → a = 0.05 m A 6.2 × 5 2g

Therefore He = 2.0 m + 0.05 m = 2.05 m, and hd 1.3 = 0.63 = He 2.05 hd + d 1.3 + 4.8 = 2.98 = He 2.05 Using these parameter values in Figure 7.30 gives a discharge-coeﬃcient reduction of approximately 0.75% caused by submergence. Hence, the adjusted discharge coeﬃcient is C = (1 − 0.0075)(2.17) = 2.15

235 The corresponding discharge, Q, under the given submerged condition is

Q2 Q = CLe He = (2.15)(5.0) 2.0 + 2gA2 3 2

Q2 = (2.15)(5.0) 2.0 + 2(9.81)(6.2 × 5)2

which yields Q = 31.6 m3 /s. This ﬂow corresponds to an upstream velocity, va , and velocity head, va2 /2g, given by va =

31.6 v2 Q = = 1.02 m/s → a = 0.05 m A 6.2 × 5 2g

Since this approach velocity is approximately the same as calculated for the free-ﬂow condition, no further iteration is necessary. The submerged discharge is 31.6 m3 /s . 7.67. From the given data: z0 = 0.50 m, z1 = 2.5 m, L = 10 m, and Hd = 0.70 m. The maximum allowable head on the spillway, He , is given by He =

0.7 Hd = = 1.0 m 0.7 0.7

and

P 2.5 − 2 = 2.85 = Hd 0.7

(a) From Figure 7.28, the discharge coeﬃcient is given by C0 = 2.18. Under design conditions, H = 0.7 m, and the spillway discharge, Q, is given by 3

Q = C0 LH 2 = 2.18(10)(0.7) 2 = 12.8 m3 /s (b) Under maximum-ﬂow conditions, H = He = 1.0 m, which corresponds to a maximum stage upstream of the spillway of 2.5 m + 1.0 m = 3.5 m , and using Figure 7.28 gives the corresponding discharge coeﬃcient: C 1.0 H = 1.43 → = = 1.05 → C = (1.05)(2.18) = 2.29 Hd 0.7 C0 and hence the discharge capacity of the spillway under this condition is given by 3

Q = CLHe2 = (2.29)(10)(1.0) 2 = 22.9 m3 /s (c) If the maximum stage is exceeded, then the pressure on the downstream side of the spillway will likely lead to cavitation and possible pitting of the concrete structure. 7.68. For free discharge (when the gate is completely out of the water) the discharge, Q0 , is given by the relation 3 (1) Q0 = Cd bH 2 where Cd is the discharge coeﬃcient, b is the width of the spillway, and H is the height of the water above the spillway crest. Under critical conditions (when the gate is above the water), the critical ﬂow depth, yc , is calculated using the relation A3 Q2 Q20 = → 0 = (byc )3 b = b2 yc3 → yc = g T g

Q20 b2

13 1 g

(2)

236 When the gate is lowered such that the gate opening, yg , is less than yc , then the discharge is given by Q = Cd byg 2gH (3) From the given data, Cd = 2.2, b = 10 m, and H = 2 m, Equations 1 and 2 give 3

Q0 = (2.2)(10)(2) 2 = 62.2 m3 /s 1 1 3 62.22 yc = = 1.58 m 102 9.81 Therefore, when the gate opening, yg , is greater than 1.58 m, the discharge is equal to 62.2 m3 /s. When the gate opening is less than yc , we have free orifice flow. For free orifice flow, take Cc = 0.61, which gives Cc 0.61 Cd = = = y yg 1 + Cc H 1 + 0.61 2g

0.61 1 + 0.305yg

and hence Equation 3 gives Q= √

0.61 (10)yg 1 + 0.305

2(9.81)(2) =

38.2yg m3 /s 1 + 0.305yg

Using Equation 4 gives the following results: Gate Opening, yg (m) > 1.58 1.58 1.40 1.20 1.00 0.80 0.60 0.40 0.20 0.00

Flow, Q (m3 /s) 62.2 49.6 44.8 39.2 33.4 27.4 21.1 14.4 7.4 0.0

7.69. Taking y1 as the depth of ﬂow at the entrance to the stilling basin & ignore energy loss. Ec + 4 = y1 +

Q2 2g(by1 )2

where the speciﬁc energy at critical ﬂow, 1 1 3 q2 3 3 (11.5)2 3 3 Ec = 2 g = 2 9.81 (As q= Qb = 115 10 = 11.5 m /s/m)

(4)

237 = 3.56 m 1 =⇒ 7.56 = y1 + (11.5)2 × 2×9.81 × y12 1

Which yield, y1 = 1.01 m Using Manning’s equation, 1

Q = n1 AR 3 S02 1 =⇒ 115 = 0.025 (10 × y3 )

10y3 10+2y3

2 3

(0.015) 2

=⇒ y35 − 0.517y32 − 5.17y3 − 12.92 = 0 Solving by trial & error we get, y3 = 1.9 m Again from equation 7.48, y3 8Q2 1 = 1 + −1 + y2 2 gb2 y 3 2

=⇒ y23 + 7.6y22 + 14.44y2 = 1+8Q = 1+8×(115) = 107.85 gb2 9.81×(10)2 Solving by trial & error we get, y2 = 2.62 m Again equation 4.154 V2

1 )−(y +α 2 ) (y1 +α 2g 2 2g ΔL = Sf −S0

q 11.5 Here, α = 1, V1 = yq1 = 11.5 1.01 =11.39 m/s, V2 = y2 = 2.62 = 4.39 m/s 2 ∴ Vavg = V1 +V = 11.39+4.39 = 7.89 2 2

Again equation 4.151 2 n2 V 2 = 4avg Sf = nQ2 AR 3

3 Ravg

Assuming the stilling basin is of the same width as the spillway crest, the hydraulic radius Ravg is, 10×y1 10×y2 1 10×1.01 10×2.62 m = + + Ravg = 12 10+2y 10+2y2 2 10+2(1.01) 10+2(2.62) = 1.28 m 1 2

∴ Sf = (0.025) ×(7.89) = 0.028 4 (1.28) 3

Again equation 4.154

ΔL =

1 y1 +α 2g

V22 − y2 +α 2g

Sf −S0

= (7.62 − 3.60)/(0.028 − 0.015) = 309 m To compute the length of the jump, the Froude no, Fr1 is,=2.27 From ﬁg 7.39 (c), for Fr1 =2.27 L1 /y2 = 4.2 L1 = 11.0 m The length of stilling basin should be at least (ΔL + L1 )= 311 m to prevent the jump from leaving the basin.

238 11.39 Fr1 = √Vgy1 1 = √9.81×1.01 = 3.62

So, Type –IV basin. 7.70. From the given data: X = 2.00 m, Hd = 7.00 m, a = 2.10 m, h = 0.50 m, R = 2.75 m, and L = 5.00 m. The gate angle, θ, is given by Equation 7.127 as −1 a − h −1 2.10 − 0.50 = cos = 54◦ θ = cos R 2.75 From the given data, X/Hd = 2.00/7.00 = 0.29 and θ = 54◦ , and Figure 7.34 gives the discharge coeﬃcient as Cd = 0.67. When the ponded water is 1.00 m above the crest, the head on the gate opening, H, can be approximated by H = 0.87 m + 1.00 m −

0.50 h cos(54◦ ) = 0.87 m + 1.00 m − cos(54◦ ) = 1.72 m 2 2

and the corresponding ﬂow rate under gate is given by Equation 7.126 as Q = Cd Lh

2gH = (0.67)(5.00)(0.50)

2(9.81)(1.72) = 9.73 m3 /s

This discharge is approximately 58% of the ﬂow of 16.9 m3 /s when the gate opening is 1.00 m. 7.71. From the given data: Q0 = 10 m3 /s, H1 = 0.90 m, and h = 1.00 m. Under the given operating conditions the reservoir head on the gate seat, H2 , is H2 = H1 + h = 0.90 m + 1.00 m = 1.90 m and the head on the crest of the spillway is 1.00 m + 0.90 m − 0.25 m = 1.65 m. Assuming that the pool elevation remains as shown in Figure 7.37 when the gate is fully open, then H0 = 1.65 m. Substituting the given and derived parameters into Equation 7.128 gives 3

H2 −H2 Q = 2 3 1 Q0 H2 0

(1.90) 2 − (0.90) 2 Q = 3 10 (1.65) 2 which yields Q = 8.3 m3 /s . This discharge is approximately 84% higher than the ﬂow of 4.5 m3 /s when the gate opening is 0.50 m. 7.72. The equation of the area-elevation curve, will generally be of the form: A = α + β · h + γ · h2 + · · · η · hn−1

(1)

Where, A represent the area at any elevation h; and α, β γ . . . . . . η are all constants, This equation can be determined and then integrated to obtain storage (capacity), as explained below:

239 Let y be the height of the water surface in the reservoir above any assumed datum, over which the storage /capacity is to be worked out. Let Ay represents symbolically the area of the contour at this height. Then, assume that the equation of area-elevation curve is given by: Ay = α + β · y + γ · y 2 + ·η · y n−1

(2)

Where,α, β, γ · · · · · · are all constants. From the actual survey, or from the points falling on the area-elevation curve, the area of any required no. of contours (n) are known. Let them be A0 , A1 , A2 . . . . . . . . . . . . corresponding to the known level values, i.e., heights 0, y1, y2,. . . . . . . above the datum. Substituting these equations in equation (2) we get, A0 = α A1 = α + β · y1 + γ · y12 + · · · η · y1n−1 A2 = α + β · y2 + γ · y22 + · · · η · y2n−1 A3 = α + β · y3 + γ · y32 + · · · η · y3n−1 Thus, we get n simultaneous equations to determine n number of constants (α, β, γ · · · · · · η). Hence, the equation of the area-elevation curve, i.e., Ay = α+β ·y +γ ·y 2 +· · · η ·y n−1 becomes deﬁned, with α, β, γ · · · · · · η all known. This can now be integrated between the limits 0 to y, y−y

y−y

Ay · dy =

y=0

(α + β · y + γ · y 2 + · · · η · y n−1 )dy

y=0

Or Sy = Storage (capacity) between 0 to y y−y

(α + β · y + γ · y 2 + · · · η · y n−1 )dy

y=0

2 3 n = α · y + β · y2 γ · y3 + · · · η · yn + K Where K is a constant, which obviously is the reservoir capacity at datum. Use equation (1) wherein y is the height above 100 m, as Ay = α + βy + γy 2 Now substituting A0 = 5 ha (as y = 0) Also, 20.5 = 5 + β(20) + γ(20)2 =⇒ 15.5 = 20β + 400γ

(ii)

Again,A2 = 42.6 ha, at y2 = 40 m =⇒ 42.6 = 5 + 40β + 1600γ =⇒ 37.6 = 40β + 1600γ Equating (ii) & (iii)

(iii)

240 β = 0.61 γ = 8.25 ×10−3 Hence the equation of the area elevation can be Ay = 5 + 0.61y + (8.25 × 10−3 )y2 7.73. The ﬁrst step is to calculate the ﬂow rate over the spillway under maximum-pool conditions. This is given by Equation 7.120 as 3

Q = CLe He2

(1)

When the pool elevation behind the spillway is 6.00 m, the eﬀective head, He , is given by He = 6.00 m − 5.00 m = 1.00 m and Equation 7.115 gives the design head, Hd , (without piers) Hd = 0.7He = 0.7(1.00) = 0.70 m The eﬀective length of the spillway crest, Le , is given by Equation 7.121 as Le = L − wN − 2(N Kp + Ka )He

(2)

Since there are no piers, N = 0, Ka = 0 (well rounded abutments), and L = 10 m, therefore Equation 2 gives Le = L = 10 m The height of the spillway crest, P , is 5 m, hence 5 P = =5 Hd 1 and

He 1 = 1.43 = Hd 0.7

Using these values in Figure 7.28 gives C0 = 3.95, C/C0 = 1.05, and hence C (C0 ) = (1.05)(3.95) = 4.15 C= C0 Converting from U.S. Customary units to SI units yields C = 0.552(4.15) = 2.29 Substituting the calculated parameters into Equation 1 yields 3

Q = CLe He2 = (2.29)(10)(1.00) 2 = 22.9 m3 /s Hence the ﬂow rate down the spillway is 22.9 m3 /s.

241 Neglecting energy loss down the spillway, at the entrance to the stilling basin the energy equation gives Q2 2gA2 22.92 6 = y2 + 2(9.81)(10y2 )2

E 1 = y2 +

which yields y2 = 5.99 m

0.21 m

Since the ﬂow at the base of the spillway will be supercritical, y2 = 0.21 m, and the ﬂow velocity, V , and the Froude number, Fr, are given by 22.9 Q = = 10.9 m/s by (10)(0.21) V 10.9 Fr = √ = = 7.6 gy (9.81)(0.21) V =

Since Fr > 4.5 and V < 18 m/s, use a Type III stilling basin . 7.74. From the given data: Q = 100 m3 /s, wg = 3 m, w = 1 m, N = 3, ztr = 80.00 m, zbr = 67.00 m, and ztw = 70.00 m. The ﬂow, Q, over the spillway crest is given by 3

Q = CLe He2

(1)

The equivalent length of the spillway crest, Le , is given by Le = L − wN − 2(N Kp + Ka )He Specify pointed-nose piers and rounded abutments, so Kp =0 and Ka = 0. Three gates will require two piers, so the spillway length, L, is given by L = 3 × gate width + 2 × pier width = 3(3) + 2(1) = 11 m The equivalent crest length is therefore given by Le = 11 m − (2)(1 m) − 0 = 9 m Assuming P/Hd > 3 and He < 9.1 m, then the discharge coeﬃcient is C = 2.18. With piers, He /Hd = 1/0.74 = 1.35 which gives C/C0 = 1.04 and hence C = (1.04)(2.18) = 2.27. Substituting known quantities into Equation 1 gives 3

100 = (2.27)(9)He2 which yields He = 2.88 m. This validates the assumption that He < 9.1 m, and also gives Hd = 0.74He = 2.13 m. Estimate the velocity head upstream of the reservoir as follows: d = 80.00 m − 67.00 m = 13.00 m 100 Q = = 0.855 m/s v0 = Le d (9)(13) v02 0.8552 = = 0.04 m 2g 2(9.81)

242 If h is the water level above the crest, then v02 = He 2g h + 0.04 = 2.88 h+

which yields h = 2.84 m and elevation of spillway crest = 80.00 m − 2.84 m = 77.16 m height of spillway, P = 77.16 m − 67.00 m = 10.16 m Since P/Hd = 10.17/2.13 = 4.77, the initial assumption that P/Hd > 3 is validated. The width of the spillway is the same as the crest length of the spillway and is therefore equal to 11 m. Since there is a 10% energy loss down the spillway, the speciﬁc energy at the entrance to the stilling basin, E1 , is estimated by (neglecting velocity heads) E1 = (80 − X) − 0.1(80 − 70) = 79 − X which requires that 79 − X = y1 +

1002 2(9.81)(11y1 )2

which simpliﬁes to X = 79 − y1 −

4.21 y12

(2)

The hydraulic jump equation requires that y2 8(100/11)2 −1 + 1 + y2 = 2 (9.81)y13 which simpliﬁes to y2 y2 = 2

−1 +

67.40 1+ y13

(3)

Consistency of y2 with the given tailwater elevation requires that y2 = 70 − X

(4)

Combining Equations 2 to 4 yields y1 = 0.555 m X = 64.76 m y2 = 5.24 m v1 = 16.4 m/s Fr1 = 7.02 Therefore the invert elevation of the stilling basin is 64.76 m . Since Fr1 > 4.5 and v1 < 18 m/s, a Type III stilling basin is required.

243 7.75. From the given data, the average ﬂow rate into the reservoir is 2500 m3 /s = 7.88 × 1010 m3 /year, and the average suspended-sediment concentration is 250 mg/L = 0.250 kg/m3 , therefore the average sediment load entering the reservoir is given by sediment load = inﬂow rate × suspended-sediment concentration = 7.83 × 1010

kg m3 × 0.250 3 = 1.96 × 1010 kg/year year m

The area of the reservoir is 850 km2 = 8.5 × 108 m2 , and the average depth of the reservoir is 18.7 m, therefore the reservoir storage capacity is given by storage capacity = area of reservoir × average depth = 8.5 × 108 m2 × 18.7 m = 1.59 × 1010 m3 and storage capacity 1.59 × 1010 m3 = = 0.20 year annual inﬂow 7.88 × 1010 m3 /year Based on this ratio of storage capacity to annual inﬂow (= 0.20 year), the percent of sediment trapped in the reservoir is estimated from Figure 7.48 as 93%. Since the average sediment load delivered by the river to the reservoir is 1.96 × 1010 kg/year, the rate at which sediment is accumulating in the reservoir is 0.93 × 1.96 × 1010 kg/year = 1.78 × 1010 kg/year. Since the bulk density of the sediment accumulating at the bottom of the reservoir is 1,600 kg/m3 , and the area of the reservoir is 850 km2 = 8.5 × 108 m2 , the rate at which sediment volume is accumulating is given by sediment trap rate sediment bulk density 1.78 × 1010 kg/year = = 1.11 × 107 m3 /year 1600 kg/m3

sediment volume accumulation rate =

Since the plan area of the reservoir is 850 km2 = 8.5 × 108 m2 , the rate of sediment accumulation on the bottom of the reservoir is given by sediment volume accumulation rate reservoir area 7 1.11 × 10 m3 /year = = 0.013 m/year = 1.3 cm/year 8.5 × 108 m2

rate of sediment accumulation =

At this rate, it will take approximately 1400 years for the reservoir capacity to decrease by 10% due to sediment accumulation. 7.76. From the given data, the cumulative inflow, demand, and difference between inflow and demand is tabulated below:

244 Month

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Cumulative Inﬂow (×1010 m3 ) 0.23 0.50 0.85 1.11 1.42 1.75 1.99 2.24 2.64 3.07 3.42 3.74 4.00 4.20 4.47 4.81 5.17 5.50 5.80 6.13 6.47 6.81 7.13 7.42

Cumulative Demand (×1010 m3 ) 0.27 0.53 0.86 1.19 1.57 1.93 2.31 2.68 3.02 3.34 3.62 3.89 4.16 4.43 4.75 5.08 5.46 5.82 6.20 6.57 6.91 7.23 7.52 7.78

Cumulative Inﬂow − Demand (×1010 m3 ) −0.04 −0.03 −0.01 −0.08 −0.15 −0.18 −0.32 −0.44 −0.38 −0.27 −0.20 −0.15 −0.16 −0.23 −0.28 −0.27 −0.29 −0.32 −0.40 −0.44 −0.44 −0.42 −0.39 −0.36

These data indicate that the reservoir storage must be sufficient accommodate the cumulative deficit between demand and reservoir inflow that occurs between Month 3 and Month 21, and this deficit is equal to (−0.01 + 0.44) × 1010 m3 = 0.43 × 1010 m3 . Assuming that the reservoir is full at the beginning of this critical interval, then an active reservoir storage of 0.43 × 1010 m3 will be sufficient. 7.77. From the given data: H = 85 m, D = 0.60 m, L = 300 m, ks = 8 mm, Dj = 50 mm, kj = 0.8, k = 0.5, v2 = 2 m/s, and V2 = 6 m/s. The flow rate, Q, is determined by application of the energy equation between the upstream reservoir and the exit of the nozzle, which requires that H−

Q2 Q2 f L Q2 − k = j D 2gA2 2gA2j 2gA2j

Using the Swamee-Jain equation yields Q = 0.0597 m3 /s. The corresponding velocities in the delivery pipeline and nozzle jet are V = 0.21 m/s and Vj = 30.4 m/s, and the friction factor

245 is f = 0.0425. Using these derived data yields the following results: he = H − hL = Y −

(0.0425)(300) 0.212 fL V 2 = 85 − = 84.95 m D 2g (0.60) 2(9.81)

Vj2 v2 V2 30.42 22 62 − k 2 − 2 = 84.95 − 0.8 − 0.5 − = 45.3 m 2g 2g 2g 2(9.81) 2(9.81) 2(9.81) PT = γQhT = (9.79)(0.0597)(45.3) = 26.5 kW hT 45.3 = 0.53 = ηT = he 84.95 hT = he − kj

For the given conﬁguration, the expected power from the system is 26.5 kW with a hydraulic eﬃciency of 53% . 7.78. From the given data: H = 100 m, D = 2.0 m, L = 500 m, ks = 15 mm, Q = 20 m3 /s, ΔhDT = 5.0 m, and V = 0.8 m/s. The velocity (= Q/A) in the penstock can be calculated as Vp = 6.37 m/s. Using the given values of Q, D, and ks , and assuming that the kinematic viscosity of water, ν, is 10−6 m2 /s, (at 20◦ C), the friction factor, f , of the penstock can be calculated using the Swamee-Jain equation which yields f = 0.0345. The head, hT , extracted by the turbine is given by Equation 7.140 as hT = H −

(0.0345)(500) 6.372 0.82 f L Vp2 V2 − ΔhDT − = 100 − − 5.0 − = 77.16 m D 2g 2g 2.0 2(9.81) 2(9.81)

which gives PT = γQhT = (9.79)(20)(77.16) = 15100 kW = 15.1 MW Therefore the system will extract 15.1 MW of power from the water ﬂowing through the turbine. 7.79. From solution 7.72 Ay = 5 + 0.61y + (8.25 × 10−3 )y2 Integrating this equation we get, 2

=⇒ Sy = 5y + 0.61 y2 (8.25 × 10−3 ) y3 + k The constant ‘k’ is obviously the reservoir capacity upto 100 m, which is given to be 12.4 ha. =⇒ Sy = 5y + 0.305y2 + (2.75 × 10−3 )y3 + 12.9 This is a equation of capacity elevation curve. Reservoir capacity at RL 145 m is Sy = 5 × 45 + 0.305 × (45)2 + (2.75 × 10−3 )(45)3 + 12.9 ha-m = 1106.12 ha-m 7.80. According to equation 7.120 3

Q = CLe He2 Value of P = 25.0 m, Hd = 3.5 m

246 P 25 = 7.14 = Hd 3.5 From Fig. 7.28(a) Basic discharge co-eﬃcient P = 7.14, C0 = 2.18 Hd Since the minimum pressure head of 5.0 m. From equation 7.115, 5.0 He P V = 1.43 Where, H = Hd + ρg = , He = H + 2g Hd 3.5 And Fig. 7.28(c) gives, C = 1.05 C0 Under min. head water condition can now be calculated using the relation, c C0 = 1.05 × 2.18 = 2.29 C= c0 3 Q Now, = 2.29 × (5) 2 = 25.60 m3 /s/m. Le The discharge intensity over the spillway is 25.60 m3 /s/m.

Chapter 10

Fundamentals of Surface-Water Hydrology II: Runoff 10.1. The discharge per unit width y, q is given by, q= vy = (i − f )L0 cos θ From data given, i = 25mm/h, f = 0, θ = tan−1 s0 = tan−1 0.05 = 2.860 Therefore, cos θ = 0.999, L0 = 30m Thus q = 10−3 × 30 × 0.999 m3 /h/m = 749.25×10−3 m3 /h/m Reynolds no is, Re = 4VΥR For a unit width sheet flow,R = y, the flow remains laminar Therefore, Re = 4VΥy = 4q Υ = 756 < 2000. So the flow is laminar The resistance coefficient ‘CL ’ is given by, C = 96+108 i0.4 , where i = 25mm/hr = 1in/hr

Therefore, C = 204 The Friction factor is, f = The depth is, y = Velocity, V=

f q02 8gs0

204 C = 0.27 (Equation 10.15) = Re 756

1/3

= 1.44×10−3 m (Equation 10.14 and 10.16)

2.08 × 10−4 q = = 0.145m/s y 1.44 × 10−3

10.2. Data given Flow length, L = 1km. = 1000 m Average slope, S0 = 0.007 312

313 The time of concentration is (by using Kirpich Equation 10.27) L0.77 s0 0.385 = 0.019 × (1000)0.77 × (0.007)−0.385

tc = 0.019

= 26.21 min By interpolation Maximum depth of rainfall for 26.21 min duration × (26.21 − 20) + 40 = (50−40) 10 = 46.21 mm Average intensity, i = 46.21 26.21 × 60 = 105.8 mm/hr. Peak flow rate, QP (m3 /s) = CiA 3.6 Given C = 0.35 A = 0.90 km2 Therefore, QP = 0.35×105.8×0.9 m3 /s 3.6 = 9.26 m3 /s 10.3. The maximum flow distance that should be described by sheet flow is 100 m (as per NRCS guidelines). 10.4. Discharge per unit width, q = 2.5×10−4 m3 /s/s Υ = 1.1 × 10−6 m2 /s Reynolds no, Re = 4q Υ = 909 < 2000 So the flow is laminar For laminar flow, y = αq m 1

f Again, α = ( 8gs )3 0

Given, f = 60, s0 = 0.05, α =

60 8×9.81×0.05

1 3

= 2.48,

m = 23 2

Y = 2.48×(2.5 × 10−4 ) 3 = 9.84 × 10−3 m = 9.84 mm −4

= 0.03m/s Velocity, V = yq = 2.54×10 9.84×10−3 10.5. Curve Number, CN = 69 for group B & 79 for group C = 74 Average Curve No for the watershed is, CN = 69+79 2 Potential maximum Retention (cm) (equation 10.76), S = 2.54 The excess rainfall/runoﬀ (before urbanization),

1000 74

− 10 = 89.09 mm

314 2

According to Equation 9.80, Q = (PP−0.2S) +0.8S = 87mm (since P = 160mm) After urbanization, Q = 100mm So, the impact of urbanization is to cause (100-87) = 13mm in additional runoff from this storm, i.e. 13% increase. 10.6. From the given data: the area of the watershed is 121 ha. The first step is to separate the base flow using the concave base-flow separation method. The inflection point occurs at 1000 h (t = 63 min) and a base flow of 0.009 m3 /s is assumed up to the time of the peak discharge. Using these points, the base flow is shown in Column 6 of Figure 10.1, and the corresponding runoff is shown in Column 7, where the resulting runoff volume is 5362 m3 . The rainfall

Figure 10.1: Hydrograph Separation characteristics are shown in Figure 10.2, where the cumulative rainfall volume of 34 465 m3 is shown at the bottom of Column 4.

315

Figure 10.2: Rainfall/Runoff Analysis (a) Since the rainfall duration is 38 min, this indicates an average infiltration capacity, f0 , given by (34465 − 5362) 1000 × 60 = 38 mm/h f0 = × 121 × 104 38 The infiltration capacity must be increased from 38 mm/h to account for the times when the actual infiltration is equal to the rainfall rate (in which cases the rainfall rate is less than the infiltration capacity). As shown in Figure 10.2, an adjusted infiltration capacity of 75 mm/h will yield the the calculated direct runoff (5362 m3 ). This indicates that the watershed has predominantly Class A soils. (b) From the above analysis, P =

34465 × 1000 = 25.4 mm 121 × 104

and

5362 × 1000 = 4.43 mm 121 × 104 Applying the NRCS curve number equation, Q=

(P − 0.2S)2 P + 0.8S (28.5 − 0.2S)2 4.43 = 28.5 + 0.8S Q=

316 which yields S = 53.6 mm = 2.11 inches. This corresponds to a curve number, CN, given by 1000 1000 = = 83 CN = 10 + S 10 + 2.11 Hence CN = 83 . (c) The time of concentration, tc , can be estimated by the NRCS relation tl = 0.6tc

(1)

where tl is the time lag between the centroid of the rainfall excess and the time to peak of the runoﬀ hydrograph. From the rainfall analysis in Figure 10.2, the centroid of the rainfall excess, t̄r , is given by t̄r =

7.5(1609) + 21(3759) = 17.0 min 1609 + 3759

and since the peak runoﬀ occurs at t = 49 min, Equation 1 gives (49 − 17.0) = 0.6tc which yields tc = 53 min . (d) Practical uses of these data include calculating the watershed runoﬀ for a given design rainfall. 10.7. The time of concentration, tc , is estimated by the kinematic wave equation (Equation 10.13) as 3 6.99 nL 5 √ tc = 2 S0 i5 e

where n = 0.25, L = 60 m, and S0 = 0.005, and therefore 3 6.99 0.25 × 60 5 174 √ = 0.4 min tc = 2 ie 0.005 ie5 where ie is in mm/h. Equating the storm duration to the time of concentration, then the eﬀective rainfall rate, ie , for a 10-year storm is given by the IDF relation as ie = Ci =

1500C mm/h (tc + 6.19)0.78

Combining the latter two equations, with C = 0.5, yields ie =

1500(0.5) 750 0.78 = 0.78 174 174 + 8.96 + 8.96 i0.4 i0.4 e

Solving by trial and error yields ie = 35 mm/h, and a corresponding time of concentration of 42 min, which exceeds the minimum allowable time of concentration of 5 minutes. The peak runoﬀ, Qp , from the residential development is given by the rational formula as Qp = CiA = ie A

317 where ie = 35 mm/h = 9.72 ×10−6 m/s, and A = 2 ha = 2 × 104 m2 , therefore Qp = (9.72 × 10−6 )(2 × 104 ) = 0.19 m3 /s 10.8. From the given data: A = 20 ha = 20 × 104 m2 , C = 0.7, n = 0.25, L = 100 m, and S0 = 0.6% = 0.006. The (maximum) eﬀective rainfall intensity, ie , and corresponding time of concentration, tc , are given by the simultaneous solution of the following equations, 1020C (tc + 8.7)0.75 3 6.99 nL 5 √ tc = 2 S0 ie5

ie =

(1) (2)

Substituting the given data into Equations 1 and 2 yields 1020(0.7) 714 = (tc + 8.7)0.75 (tc + 8.7)0.75 3 6.99 0.25 × 100 5 223.8 √ = 0.4 tc = 2 ie 0.006 ie5

ie =

(3) (4)

Solving Equations 3 and 4 gives ie = 31.1 mm/h = 8.64 × 10−6 m/s and tc = 56.6 minutes. Using the rational formula to calculate the peak runoﬀ, Qp , yields Qp = ie A = (8.64 × 10−6 )(20 × 104 ) = 1.73 m3 /s Repeating the previous calculations and changing each parameter by 10% to increase the peak discharge yields the following results Parameter

Value

A C n L So

22 ha 0.77 0.225 90 m 0.0066

ie (mm/h) 31.1 35.3 32.8 32.8 31.9

Qp (m3 /s) 1.90 1.96 1.82 1.82 1.77

On the basis of these results, it is clear that the highest increase in the peak runoff is associated with a 10% error in the runoff coefficient, C, and lowest increase in the peak runoff is associated with a 10% error in the slope, S0 . Assuming a worst-case 10% error in all parameters, then A = 22 ha = 22 × 104 m2 , C = 0.77, n = 0.225, L = 90 m, and S0 = 0.66% = 0.0066. The (maximum) effective rainfall intensity,

318 ie , and corresponding time of concentration, tc , are given by the simultaneous solution of the following equations, 1020(0.77) 785.4 = 0.75 (tc + 8.7) (tc + 8.7)0.75 3 6.99 0.225 × 90 5 191.6 √ = 0.4 tc = 2 ie 0.0066 ie5

ie =

(5) (6)

Solving Equations 5 and 6 gives ie = 40.4 mm/h = 1.12 × 10−5 m/s and tc = 43 minutes. Using the rational formula to calculate the peak runoff, Qp , yields Qp = ie A = (1.12 × 10−5 )(22 × 104 ) = 2.46 m3 /s 10.9. For a given duration, storms with longer return periods generally have higher average intensities. Since infiltration losses are relatively insensitive to the rainfall intensity (particularly when the rainfall rate exceeds the infiltration capacity of the soil), then storms with higher average intensities will result is a greater fraction of runoff (i.e. a higher runoff coefficient). Therefore, storms with longer return periods will have higher runoff coefficients. 10.10. From the given data: A = 2 ha = 2 × 104 m2 , 40% impervious, L = 100 m, P2 = 12 cm, and S0 = 0.7% = 0.007. The runoff coefficient of the area can be taken as 0.4. The first step is to calculate the time of concentration, tc , using all available methods. • The kinematic wave equation with n = 0.012 (for asphalt) gives tc =

6.99

2 5

nL √ S0

which gives

3 5

6.99 (0.4i)

2 5

0.012 × 100 √ 0.007

3 5

308 i0.4

i = 1.66 × 106 tc−2.5

(1)

Combining Equation 1 with the given IDF curve yields 1.66 × 106 tc−2.5 =

2029 (tc + 7.24)0.73

which gives tc = 4.9 min • The NRCS method with only overland ﬂow gives 5.5 tc = √ P24

nL √ S0

4 5

5.5 =√ 120

0.12 × 100 √ 0.007

4 5

= 4.2 min

• The Kirpich equation gives tc = 0.019

L0.77 (100)0.77 = 0.019 = 4.45 min (0.007)0.385 S00.385

319 • The Izzard equation with cr = 0.0070 (for smooth asphalt) gives 1 −6 1 2.8×10−6 ×0.4i+0.0070 530 2.8×10 1 ie +cr L 3 (100) 3 530 1 3 S0 (0.007) 3 tc = = 2 2 (0.4i) 3 ie3 1.12 × 10−6 i + 0.0070 = 23700 2 i3

(2)

Combining Equation 2 with the IDF curve gives

i= 23700

2029 1.12×10−6 i+0.0070 2 i3

0.73

+ 7.24

which gives i = 359 mm/h

and

tc = 3.5 min

• The Kerby equation with r = 0.10 (for asphalt) gives tc = 1.44

Lr √ S0

0.467

= 1.44

100 × 0.10 √ 0.007

0.467 = 13.4 min

Based on the above results, the minimum expected time of concentration is 3.5 min with a corresponding rainfall intensity of 359 mm/h = 9.97 × 10−5 m/s. Using the rational method, the peak runoﬀ, Qp , is given by Qp = CiA = (0.4)(9.97 × 10−5 )(2 × 104 ) = 0.798 m3 /s Since the design rainfall return period is 10 years, there is a probability of 0.1 or 10% of ﬂooding in any given year. 10.11. The time of concentration, tc , of the impervious area is given by tc =

6.99 2 5

nL √ S0

3 5

where n = 0.035, L = 30 m, and S0 = 0.005, and therefore tc =

6.99 2 5

0.035 × 30 √ 0.005

3 5

35.3 min ie0.4

where ie is in mm/h. Taking the storm duration as tc , the 10-year eﬀective rainfall rate, ie , is given by the IDF curve as ie = Ci =

1500C mm/h (tc + 8.96)0.78

320 Combining the latter two equations, with C = 0.9, yields ie =

1500(0.9) 1350 0.78 = 0.78 35.3 35.3 + 8.96 + 8.96 0.4 0.4 i i e

Solving by trial and error yields ie = 178 mm/h, and a corresponding time of concentration of 4.4 min, which is less than the minimum allowable time of concentration of 5 minutes. Taking tc = 5 min, the IDF relation yields ie =

1500(0.9) = 173 mm/h (5 + 8.96)0.78

The peak runoff, Qp , from the impervious area is given by the rational formula as Qp = CiA = ie A where ie = 173 mm/h = 4.81 ×10−5 m/s, and A = 0.5 ha = 5000 m2 , therefore Qp = (4.81 × 10−5 )(5000) = 0.24 m3 /s Problem 10.7 indicated that the peak runoff from the entire composite catchment is 0.19 m3 /s, and the present example shows that the peak runoff from the directly-connected impervious area is 0.24 m3 /s. The peak runoff from this development is therefore 0.24 m3 /s . 10.12. (a) For a single-family residential area, assume C = 0.40, and from the given data: A = 0.5 ha = 5000 m2 , tc = 10 min, and i=

1209 1209 = = 119 mm/h = 3.30 × 10−5 m/s 0.79 (t + 8.86) (10 + 8.86)0.79

The peak runoﬀ rate is given by the rational formula as Qp = CiA = (0.40)(3.30 × 10−5 )(5000) = 0.066 m3 /s (b) For t = 24 h = 1440 min the total rainfall amount is given by P = it =

1209 24 · = 0.0923 m = 9.23 cm 0.79 (1440 + 8.86) 1000

From the given data, S = 8 cm and the runoﬀ, Q, is given by Q=

(9.23 − 0.2 × 8)2 (P − 0.2S)2 = = 3.72 cm = 0.0372 m P + 0.8S P + 0.8(8)

For the 10-ha site, the runoﬀ volume, V , is given by V = QA = (0.0372)(10 × 104 ) = 3720 m3 Since S = 8 cm = 80 mm, the curve number for the site is given by CN =

1000 1000 = = 76 10 + 0.0394S 10 + 0.0394(80)

321 10.13. Runoﬀ Coeﬃcient, C = 0.3 +1 ) Intensity of Rainfall, i = ic = tPR ( ttRc +1

Maximum precipitation, P = 40mm = 4cm Storm Period, tR = 45min Time of concentration, tc = 32min. 4 45+1 ic = 45 32+1 = 7.43cm/hr Q = Cic A = 0.3×7.43 × 150 36 = 9.288cumec 10.14. For t = tc = 20 min T = 25 yr. i = ic = 15 cm/hr. A = 25 ha Q = CiA ) cumec = ( 0.56×15×25 36 = 5.833 cumec 10.15. (a) Take n = 0.25 (lawns). Using kinematic-wave equation, (nL)0.6 (0.25 × 100)0.6 236.3 = 0.4 tc = 6.99 0.4 0.3 = 6.99 0.4 0.3 ie (0.005) ie ie S0

(1)

The 10-year IDF curve for Atlanta (from Table 9.2) is given by i=

64.1 (t + 8.16)0.76

in./h =

1628 (t + 8.16)0.76

mm/h

For a suburban area, take C = 0.33 and therefore ie =

1628(0.33) 537.2 = 0.76 (t + 8.16) (t + 8.16)0.76

mm/h

(2)

Combining Equations 1 and 2 gives tc = 19.12(tc + 8.16)0.304 which gives tc = 72.7 min

For the NRCS equation, the 2-year IDF curve for Atlanta is required. Using the Chen 10 = 5.5 in., R100 = 3.6 in., T = 2 years → T = 1.443 years, method: R110 = 2.6 in., R24 p 1 10 10 x = 1.385, R1 /R24 = 47.3%, a1 = 26, b1 = 9.0, c1 = 0.80, and a = 45.72. Therefore the 2-y IDF curve is given by i=

45.2 (t + 9.0)0.80

in./h

322 For a 2-year 24-h storm, 45.2 × 24 × 25.4 = 81.5 mm = 8.15 cm (24 × 60 + 9.0)0.80

P2 =

Therefore the overland travel time ( = time of concentration) is given by tf = 0.0288

(nL)0.8 (0.25 × 100)0.8 = 0.0288 = 1.103 h = 66.18 min (8.15)0.5 (0.005)0.4 P20.5 S00.4

For the Kirpich equation, tc = 0.019

L0.77 1000.77 = 0.019 = 5.07 min (0.005)0.385 S00.385

For the Izzard equation (taking cr = 0.036), 530

2.8×10−6 ×537.2(tc +8.16)−0.76 +0.036 1 0.005 3

tc =

(100) 3

[537.2(tc + 8.16)−0.76 ] 3

which gives tc = 72.9 minutes. Check ie L, 537.2 = 19.0 mm/h (72.9 + 8.16)0.76 1 (100) = 1.9 m2 /h 19.0 × ie L = 1000 ie =

So the Izzard equation is valid. For the Kerby equation (taking r = 0.40), 100 × 0.40 Lr √ = 19.30 min = 1.44 tc = 1.44 √ S0 0.005 From the above analyses, the expected range of tc is 19 min < tc <73 min . The Kirpich tc ( = 5 min) appears to be an outlier and unreasonably low. (b) For tc = 19 minutes, the peak runoﬀ using the rational method is given by Qp = i e A =

537.2 1 × (1 × 104 ) = 0.121 m3 /s × 0.76 (19 + 8.16) (1000)(3600)

and for tc = 73 minutes, the peak runoﬀ using the rational method is given by Qp = i e A =

537.2 1 × (1 × 104 ) = 0.0528 m3 /s × (73 + 8.16)0.76 (1000)(3600)

Therefore the peak runoﬀ by the rational method is 0.0528 m3 /s < Qp < 0.121 m3 /s . Consider now the TR-55 method, for the 24-h storm: P =

1628 × 24 = 155 mm (24 × 60 + 8.16)0.76

323 For a residential area (lot sizes of 0.4 ha, sandy loam, Type B soil), CN = 68 and hence 68 =

1000 → S = 120 mm → Ia = 0.2S = 24 mm → Ia /P = 24/155 = 0.155 10 + S/25.4

The runoﬀ, Q, is given by Q=

(155 − 0.2(120))2 (P − 0.2S)2 = = 68.4 mm = 6.84 cm (P + 0.8S) (155 + 0.8 × 120)

For Ia /P = 0.155 and Type II rainfall → C0 = 2.267, C1 = −0.5086, and C2 = −0.1020. An from the NRCS method for tc , tc 1.103 h, log(tc ) = 0.04258, and log qu = C0 + C1 log tc + C2 (log tc )2 − 2.366 = 2.267 − 0.5086(0.04258) − 0.1020(0.04258)2 − 2.366 = −0.1209 which gives qu = 0.7571 m3 /s/cm/km2 . Taking A = 1 ha = 0.01 km2 , Q = 6.84 cm, and Fp = 1 gives Qp = qu AQFp = (0.7571)(10−2 (6.84)(1) = 0.0518 m3 /s Comparing this result with the rational method, it is apparent that for approximately the same time of concentration (66 min for TR-55 and 73 min for rational method) both approaches give approximately the same result, 0.0518 m3 /s versus 0.0528 m3 /s. Based on these results, it appears reasonable to take 66 min < tc < 73 min and design the drainage system for a peak runoff of 0.0528 m3 /s. 10.16. From the given data: CN = 70, A = 10 ha = 0.1 km2 , and tc = 15 min = 0.25 h. The peak runoff can be calculated using the TR-55 method, where the peak runoff is given by qp = qu AQFp and it can be assumed that Fp = 1. The 24-hour precipitation, P , is given by the IDF curve with t = 24 h = 1440 min, where P = iΔt =

2029 × 24 = 240 mm (1440 + 7.24)0.73

The storage, S, can be derived from the curve number, CN ( = 70), by 1 1 1000 1000 − 10 = − 10 = 109 mm S= 0.0394 CN 0.0394 70 and therefore

0.2S 0.2 × 109 Ia = = = 0.091 P P 240 The runoﬀ, Q, from the site is given by Q=

[240 − 0.2 × 109]2 [P − 0.2S]2 = = 146 mm = 14.6 cm P + 0.8S 240 + 0.8 × 109

324 Atlanta is characterized by Type II rainfall and, since Ia /P < 0.1, Table 6.20 gives C0 = 2.55323, C1 = −0.61512, and C2 = −0.16403. Substituting into Equation 6.102 gives log(qu ) = C0 + C1 log tc + C2 (log tc )2 − 2.366 = 2.55323 − 0.61512 log 0.25 − 0.16403(log 0.25)2 − 2.366 = 0.498 which yields

qu = 3.15 (m3 /s)/(cm·km2 )

Therefore, the peak runoﬀ rate by the TR-55 method is given by qp = qu AQFp = (3.15)(0.1)(14.6)(1) = 4.60 m3 /s The rational method gives the peak discharge, qp , as qp = CiA

(1)

where C is the runoﬀ coeﬃcient, i is the average rainfall intensity over a duration equal to the time of concentration ( = 15 min), and A is the catchment area. In this case, i=

2029 = 210 mm/h = 5.83 × 10−5 m/s (15 + 7.24)0.73

Substituting into Equation 1 gives 4.60 = C(5.83 × 10−5 )(10 × 104 ) which yields C = 0.79 10.17. From the given data: T = 10 y, A = 1 ha = 0.01 km2 , L = 100 m, S0 = 1%, n = 0.25, Type B soil, and C = 0.4. The 10-y IDF curve for Atlanta is given by i=

64.1 1628 in./h = mm/h 0.76 (t + 8.16) (t + 8.16)0.76

Using the kinematic wave equation, (nL)0.6 (0.25 × 100)0.6 192.0 3.2 tc = 6.99 0.4 0.3 = 6.99 0.4 = 0.4 min = 0.4 h 0.3 ie (0.01) ie ie ie S0

(1)

The IDF curve with a storm duration equal to tc requires that ie = Ci =

1628(0.4) mm/h (tc + 8.16)0.76

(2)

and the peak runoﬀ, Qp is given by Qp = ie A = 104 ie mm2 ·mm/h = 0.002778ie m3 /s

(3)

325 Simultaneous solution of Equations 1 and 2 yields ie = 29.7 mm/h and tc = 49 min, and substituting into Equation 3 yields Qp = 0.0825 m3 /s. According to the TR-55 method, the peak runoﬀ is given by (with Fp = 1), qp = qu AQFp = qu (0.01)Q(1) = 0.01qu Q Using the IDF curve, the 24-h ( = 1440-min) rainfall is given by P =

1628 × 24 = 154.8 mm (1440 + 8.16)0.76

For the rational method and the TR-55 method to give the same result, the following relationships must hold, 0.0825 = 0.01qu Q

(4)

(P − 0.2S)2

(154.8 − 0.2S)2

= P+ 0.8S 154.8 + 0.8S 1000 S = 25.4 − 10 mm CN Ia = 0.2S 0.2S Ia = = 0.001292S P 154.8 log qu = C0 + C1 log tc + C2 [log tc ]2 − 2.366 Q=

(5) (6) (7) (8)

= C0 + C1 log(49/60) + C2 [log(49/60)]2 − 2.366 = C0 − 0.08796C1 + 0.007736C2 − 2.366

(9)

where C1 , C2 , and C3 are known functions of Ia /P . Solving Equations 4 to 9 yields CN = 61 . 10.18. Direct runoﬀ Peak(DRO) = Flood Peak-Base Flow = 160–6 = 152cumec Pnet =

DROpeak 152 = 3.8cm = U H peak 40

6 Depth of storm rainfall, P = Pnet +losses = 3.8+( 10 × 6) = 7.4cm

Hence multiplying the given UH by 3.8 cm & adding 6cumec, the stream ﬂow ordinates at successive 3 hr intervals are6, 63, 158, 120, 72.5, 38.3, 17.4 cumec respectively 10.19. (a) The area, Ah , under the unit hydrograph can be estimated by numerical integration as Ah = (1800)(1.4 + 3.2 + 1.5 + 1.3 + 1.1 + 1.0 + 0.66 + 0.49 + 0.36 + 0.28 + 0.25 + 0.17) = 21, 078 m3 where the time increment between the hydrograph ordinates is 30 min = 1800 s. Since the area of the catchment is 2.1 km2 = 2.1 × 106 m2 , then the depth, h, of rainfall excess is given by 21078 = 0.01 m = 1 cm h= 2.1 × 106

326 Since the depth of rainfall excess is 1 cm, the given hydrograph qualifies as a unit hydrograph. (b) For a 15-min rainfall excess of 2.8 cm, the runoff hydrograph is estimated by multiplying the ordinates of the unit hydrograph by 2.8. This yields the following runoff hydrograph: Time (min) Runoff (m3 /s)

0 0

30 3.9

60 9.0

90 4.2

120 3.6

150 3.1

180 2.8

210 1.8

240 1.4

270 1.0

300 0.78

330 0.70

360 0.48

390 0

(c) For a 30-min rainfall excess of 10.3 cm, the runoff hydrograph is that resulting from two consecutive 15-min rainfall excesses of 5.15 cm. The runoff hydrograph is calculated by adding the runoff hydrographs from two consecutive events, separated by 15 minutes. To facilitate this computation, the unit hydrograph must first be interpolated for 15-minute intervals. The computations are summarized in Table 10.1, where Runoff-1 and Runoff-2 are the runoff hydrographs of the two 15-min 5.15-cm events respectively. Table 10.1: Hydrograph Computation Time (min) 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360 375 390 405

Runoﬀ-1 (m3 /s) 0 3.61 7.21 11.85 16.48 12.10 7.73 7.21 6.70 6.18 5.67 5.41 5.15 4.27 3.40 2.99 2.52 2.21 1.85 1.65 1.44 1.39 1.29 1.08 0.88 0.46 0 0

Runoﬀ-2 (m3 /s) 0 0 3.61 7.21 11.85 16.48 12.10 7.73 7.21 6.70 6.18 5.67 5.41 5.15 4.27 3.40 2.99 2.52 2.21 1.85 1.65 1.44 1.39 1.29 1.08 0.88 0.46 0

Total Runoﬀ (m3 /s) 0 3.61 10.82 19.06 28.33 28.58 19.83 14.94 13.91 12.88 11.85 11.08 10.56 9.42 7.67 6.39 5.51 4.73 4.06 3.50 3.09 2.83 2.68 2.37 1.96 1.34 0.46 0

327 10.20. Width of the channel, b0 = 2.5m Let L be the length of the channel Q Q = 2.5d Q = 0.001L, V = b×d

Distance along channel,L(m) Flow rate(m3 /s) Depth,d(m) Velocity,V(m/s)

0.03

0.06

0.09

0.12

0.15

0 0

0.016 0.75

0.025 0.96

0.032 1.125

0.038 1.26

0.043 1.40

1/2

Again, Q = n1 AR2/3 s0

120

150

1 A5/3 1/2 n P2/3 s0 , [A = b × d, P = b + 2d]

1 (2.5 × d)5/3 (0.030)1/2 0.015 (2.5 + 2d)2/3 Q = 53.174 ×

=⇒

Q 53.174

d5/3 (2.5 + 2d)2/3 =

d5 (2.5 + 2d)2

Therefore, Q = 0.03 10.21. The ﬁrst step is to estimate the 30-min unit hydrograph from the 15-min unit hydrograph. These computations are given in the following table: Time (min) 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240

UH1 (m3 /s) 0 0.7 1.4 2.3 3.2 2.4 1.5 1.4 1.3 1.2 1.1 1.1 1.0 0.83 0.66 0.58 0.49

UH2 (m3 /s) 0 0 0.7 1.4 2.3 3.2 2.4 1.5 1.4 1.3 1.2 1.1 1.1 1.0 0.83 0.66 0.58

UH1 + UH2 (m3 /s) 0 0.7 2.1 3.7 5.5 5.6 3.9 2.9 2.7 2.5 2.3 2.2 2.1 1.8 1.49 1.24 1.07

30-min UH (m3 /s) 0 0.4 1.1 1.9 2.8 2.8 2.0 1.5 1.4 1.3 1.2 1.1 1.1 0.9 0.75 0.62 0.54

328 255 270 285 300 315 330 345 360 375 390 405

0.43 0.36 0.32 0.28 0.27 0.25 0.21 0.17 0.09 0 0

0.49 0.43 0.36 0.32 0.28 0.27 0.25 0.21 0.17 0.09 0

0.92 0.79 0.68 0.60 0.55 0.52 0.46 0.38 0.26 0.09 0

0.46 0.40 0.34 0.30 0.28 0.26 0.23 0.19 0.13 0.05 0

The given 120-min storm can be viewed as 4 consecutive 30-min storms with rainfall amounts of 2.4 cm, 4.5 cm, 2.1 cm, and 0.8 cm. The runoﬀ from each storm is estimated by multiplying the 30-min unit hydrograph by 2.4, 4.5, 2.1, and 0.8 respectively. The hydrographs are then lagged by 30 minutes and summed to estimate the total runoﬀ from the storm (see Equation 10.53). These computations are summarized in the following table: Time (min) 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360

UH × 2.4 (m3 /s) 0 1.0 2.6 4.6 6.7 6.7 4.8 3.6 3.4 3.1 2.9 2.6 2.6 2.2 1.8 1.5 1.3 1.1 1.0 0.8 0.7 0.7 0.6 0.6 0.5

UH × 4.5 (m3 /s) 0 0 0 1.8 5.0 8.6 12.6 12.6 9.0 6.8 6.3 5.9 5.4 5.0 5.0 4.1 3.4 2.8 2.4 2.1 1.8 1.5 1.4 1.3 1.2

UH × 2.1 (m3 /s) 0 0 0 0 0 0.8 2.3 4.0 5.9 5.9 4.2 3.2 2.9 2.7 2.5 2.3 2.3 1.9 1.6 1.3 1.1 1.0 0.8 0.7 0.6

UH × 0.8 (m3 /s) 0 0 0 0 0 0 0 0.3 0.9 1.5 2.2 2.2 1.6 1.2 1.1 1.0 1.0 0.9 0.9 0.7 0.6 0.5 0.4 0.4 0.3

Total Runoﬀ (m3 /s) 0 1.0 2.6 6.4 11.7 16.1 19.7 20.5 19.1 17.3 15.6 13.9 12.6 11.0 10.4 8.9 7.9 6.7 5.8 4.9 4.3 3.7 3.2 2.9 2.6

329 375 390 405 420 435 450 465 480

0.3 0.1 0 0 0 0 0 0

1.0 0.9 0.6 0.2 0 0 0 00

0.6 0.5 0.5 0.4 0.3 0.1 0 0

0.3 0.2 0.2 0.2 0.2 0.2 0.1 0

2.2 1.8 1.3 0.8 0.5 0.3 0.1

10.22. When there is no rain, flow in the channel is derived primarily from the inflow of ground water . The name given to this no-rain channel flow is baseflow . Computations for this problem are tabulated in Table 10.2, where the measured rainfall is given in column 2 and the flow hydrograph is given in column 3. From the given hydrograph, the baseflow in the channel is 100 m3 /s and the direct runoff can be determined by subtracting the baseflow (i.e. 100 m3 /s) from the total hydrograph, which gives direct-runoff hydrograph in column 4. The area under the direct-runoff hydrograph is summed in column 5, and the total area under the direct-runoff hydrograph is 12.6 m·km2 . Since the area of the catchment is 315 km2 , then the total runoff is 12.6/315 m = 4 cm. The unit hydrograph is computed by dividing the direct runoff hydrograph (column 4) by 4 which yields the unit hydrograph in column 6. Table 10.2: Hydrograph Separation into Baseflow and Direct Runoff (1) Time (h) 0

(2) Rainfall (cm)

(3) River Flow (m3 /s) 100

(4) Direct Runoﬀ (m3 /s) 0

100

300

200

0.5 1

50 1.44

700

600

0.5 4

0 0.36

2.5 3

(6) Unit Hydrograph (m3 /s) 0

2.5 2

(5) Area (m·km2 )

150 2.70

1000

900

225 2.88

800

700

600

500

400

300

175 2.16 125 1.44 75 0.90

300

200

50 0.54

200

100

25 0.18

330 Table 10.2: Hydrograph Separation into Baseﬂow and Direct Runoﬀ (cont’d) (1) Time (h) 10

(2) Rainfall (cm)

(3) River Flow (m3 /s) 100

(4) Direct Runoﬀ (m3 /s) 0

100

(5) Area (m·km2 )

(6) Unit Hydrograph (m3 /s) 0

0 11 6.0

0 12.60

The total rainfall depth is 6 cm and the total runoff depth is 4 cm, hence the total losses during the storm is 6 cm − 4 cm = 2 cm . If this loss is distributed uniformly over the duration of the storm event (4 h), then the average loss rate is 2 cm/4 h = 0.5 cm/h. Subtracting this loss rate from the rainfall rate given in Table 10.2 gives an effective rainfall of 2 h, lasting from the first to the third hour. Therefore, the duration of the effective rainfall associated with the unit hydrograph is 2 h . The given rainfall excess consists of four 2-hour sequences of 1 cm, 3 cm, 4 cm, and 2 cm. Using the unit hydrograph, the runoff from each of these rainfall sequences are given in columns 2 to 5 in Table 10.3, and the total runoff is given in column 6. Table 10.3: Direct Runoff Estimated From Unit Hydrograph (1) Time (h) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 10.23. tR = 15h tL = 40h

(2) R1 (m3 /s) 0 50 150 225 175 125 75 50 25 0 0 0 0 0 0 0

(3) R2 (m3 /s) 0 0 0 150 450 675 525 375 225 150 75 0 0 0 0 0

(4) R3 (m3 /s) 0 0 0 0 0 200 600 900 700 500 300 200 100 0 0 0

(5) R4 (m3 /s) 0 0 0 0 0 0 0 100 300 450 350 250 150 100 50 0

(6) Total (m3 /s) 0 50 150 375 625 1000 1200 1425 1250 1100 725 450 250 100 50 0

331 We know, tL = 5.5tr (Equation 10.66) 40 = 7.27h tr = 5.5

Again, Equation 10.68, tLR = tL +0.25(tR –tr ) = 40+0.25(15–7.27) = 41.93h p (QPR = 160m3 /s, A = 4000km2 ) So, Equation 10.65, QPR = 2.75 tLR

C A

CP = 0.61 Again,Equation 10.67, tL = 0.75Ct (LLc )0.3 40 = 0.75 × Ct (160 × 80)0.3 Therefore, Ct = 3.12 10.24. Velocity of ﬂow, Vf = 1m/s (in the range 15m–150m) 15 The travel time over the 15m of pasture is t = l Vf = 1 = 15sec.

For the rectangular channel the velocity at 30m intervals was calculated in Problem 10.20 30 = 80 sec. First 30m, Δt = 0.375

The total travel time for the channel is = (80+35.09+28.85+25.21+22.56) = 191.71sec.

Distance along the channel,l(m) Δl Calculated velocity,V(m/s) Avg. Velocity,Vavg (m/s) Travel time,Δt l σ

120

150

30 0.75

30 0.96

30 1.125

30 1.26

30 1.40

0.375

0.855

1.04

1.19

1.33

35.09

28.85

25.21

22.56

Time of conc.,tc = (15+191.71)s = 206.71sec 10.25. Use the NRCS dimensionless unit hydrograph given in Table 10.11. From the given data: A = 16 km2 and tc = 14.4 h. The time to peak, Tp , is given by Equation 10.73 as 1 Tp = t r + t l 2

(1)

where tr = 30 min = 0.5 h is the duration of rainfall excess, and tl is the time lag that can be estimated by Equation 10.74 as tl = 0.6tc = 0.6(14.4) = 8.64 h

332 Substituting tr = 0.5 h and tl = 8.64 h in Equation 1 gives 1 1 Tp = tr + tl = (0.5) + 8.64 = 8.89 h 2 2 which can be rounded to the next whole number multiple of tr as Tp = 9.00 h. The peak runoﬀ rate, Qp , is given by Equation 10.80 as Qp = 2.08

A 16 = 3.70 (m3 /s)/cm = 2.08 Tp 9.00

Using Tp = 9.00 h and Qp = 3.70 (m3 /s)/cm with the dimensionless unit hydrograph in Table 10.11 gives the 30-min unit hydrograph in Columns 1 and 2 of the following table: (1) Time (h) 0 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 18.0 19.8 21.6 23.4 25.2 27.0 30.6 37.8 41.4 45.0

(2) Q (m3 /s) 0 0.37 1.14 2.45 3.44 3.70 3.44 2.89 2.07 1.44 1.04 0.77 0.54 0.39 0.29 0.21 0.11 0.04 0.01 0

(3) Q1 (m3 /s) 0 1.09 3.32 7.09 9.99 10.73 9.99 8.39 6.10 4.17 3.01 2.23 1.58 1.14 0.83 0.60 0.32 0.12 0.03 0

(4) Q2 (m3 /s) 0 0.40 1.40 3.13 4.75 5.45 5.27 4.57 3.45 2.43 1.73 1.18 0.91 0.65 0.47 0.35 0.19 0.07 0.02 0

(5) Q1 + Q2 (m3 /s) 0 1.49 4.72 10.22 14.24 16.18 15.26 12.96 9.55 6.60 4.74 3.41 2.49 1.79 1.30 0.95 0.51 0.19 0.05 0

The hydrograph in Column 3 results from the ﬁrst 30 min of rainfall and is obtained by multiplying the unit hydrograph (Column 2) by 2.9 cm, the hydrograph in Column 4 results from the second 30 min of rainfall and is obtained by lagging the unit hydrograph in Column 2 by 30 min ( = 0.5 h) and multiplying by 1.5 cm. The runoﬀ hydrograph in Column 5 results from the rainfall excess in the entire storm and is the sum of the hydrographs in Columns 3 and 4.

333 10.26. Data obtained from problem 10.23, Ct = 3.12 Cp = 0.61 We know, tL = 0.75Ct (LLC )0.3 = 0.75 × 3.12 × (120 × 60)0.3 = 33.61 h tL = 33.61 Also, tr = 5.5 5.5 = 6.11h

For a 6-h unit hydrograph, tR = 6h Now, tLR = tL +0.25(tR –tr ) = 33.61+0.25(6–6.11) = 33.58h pA QPR = 2.75 tLR (since tL >tLR ) = 2.75 × 0.61×3000 = 149.7m3 /s 33.61

At 75% of peak discharge, (Equation 10.70 and 10.71) W50 = W75 =

2.14

(

QP R 1.08 ) A

(

QP R 1.08 ) A

1.22

= 54.51h = 31.08h

Tp = 12 tR + tLR =

2 × 6 + 33.61

= 36.61h (from Equation 10.72)

TB = 3×36.61 = 109.83h t LR

Excess rainfall

Time(hr) tR = 6h 149.7 Flow rate (m3/s)

160 140

112.3 120

W75= 31.08

74.9

100

W50=54.51h

60 40

TB=109.83h 20

Time(hr) 0

100

120

334 10.27. From the given equation Qp = 2.08

A Tp

(1)

where Qp is in (m3 /s)/cm, A is in km2 , and Tp is in hours. If TB is the time base of the unit hydrograph (in hours), and the unit hydrograph corresponds to a runoﬀ of 1 cm, then 1 (3600TB )Qp = 10−2 (106 A) 2

(2)

where the factor of 3600 converts TB (in hours) to seconds, 10−2 converts the runoﬀ depth of 1 cm to meters, and 106 converts A (in km2 ) to m2 . Combining Equations 1 and 2 yields A 1 = 104 A (3600TB ) 2.08 2 Tp and solving for TB gives TB = 2.67Tp Hence, the runoﬀ duration is 2.67Tp . 10.28. For cat (33.61+3) i.e. 36.61 h tR = 2.0

6-h Snyder units hydrograph

tLR = tL +0.25(tR –tr ) tL Also, tr = 5.5 tL = tL + 0.25(tR − 5.5 )

= (tL − 0.05tL ) + 0.25tR = 0.95tL + 0.25tR Time to peak from beginning of ER(Eﬀective rainfall) TP = t2R + tLR tLR = (8 − 22 ) = 7hr ⇒ 7 = 0.95tL + 0.25 × 2 ⇒ tL = 6.84h We know tL = 0.75Ct (LLc )0.3 ⇒ 6.84 = 0.75 × Ct (15 × 7)0.3 ⇒ Ct = 2.26 Again PA QP R = 2.75 CtLR

335 ⇒ 55 = 2.75 × CP ×200 7

⇒ CP = 0.7

For catchment N tL = 0.75Ct (LLc )0.3 = 0.75 × 2.26 × (40 × 25)0.3 = 13.46hr tL tR = 5.5 = 2.45 hr.

tLR = 0.95tL + 0.25tR = 0.95 × 13.46 + 0.25 × 2 = 13.29 hr. QP R = 2.75×0.7×350 13.46 = 50 m3 /s W50 = =

2.14

Q R 1.08 ( P ) A

2.14 50 1.08 ( 350 )

= 17.5 ∼ = 18 hr. &W75 =

1.22

(

QP R 1.08 ) A

= 9.98 ∼ = 10 hr. 10.29. First, tabulate the contributing area for each of the given time intervals:

Time Interval (min) 0–5 5–10 10–15 15–20 20–25 25–30

Contributing Area (ha) 2 5 12 19 30 7

The runoﬀ hydrograph is calculated using Equation 10.81, and the computations are given in the following table (a factor of 0.002778 is used to convert ha·mm/h to m3 /s):

336 Time (min) 0 5 10 15 20 25 30 35 40 45 50 60 70

Runoﬀ (m3 /s) [120(2)](0.002778) = [70(2) + 120(5)](0.002778) = [50(2) + 70(5) + 120(12)](0.002778) = [30(2) + 50(5) + 70(12) + 120(19)](0.002778) = [20(2) + 30(5) + 50(12) + 70(19) + 120(30)](0.002778) = [10(2) + 20(5) + 30(12) + 50(19) + 70(30) + 120(7)](0.002778) = [ 10(5) + 20(12) + 30(19) + 50(30) + 70(7)](0.002778) = [ 10(12) + 20(19) + 30(30) + 50(7)](0.002778) = [ 10(19) + 20(30) + 30(7)](0.002778) = [ 10(30) + 20(7)](0.002778) = [ 10(7)](0.002778) =

0 0.7 2.1 5.3 9.5 15.9 12.0 7.9 4.9 2.8 1.2 0.2 0

10.30. Time 80 (hr.) Flow 1070 3 (m /s)

680

390

240

150

We know, t

Qt = Q0 e− k When , K =

t Q ln( Q0 ) t

‘Q vs t’ is plotted on the semi log paper K value is the slope of the recession ﬂood = lnΔt ΔQ 31−59 = ln 1000 100

= 12 hr 10.31. The non-linear reservoir model is given by Equation 10.97, where C = 1, W = 250 m, S0 = 0.005, A = 2 ha = 2 ×104 m2 , n = 0.35 (from Table 10.1), and yd = 10 mm = 0.010 m. Taking Δt = 5 min = 300 s, and substituting the given parameters into Equation 10.97 gives 1

y 2 − y1 (1)(250)(0.005) 2 = īe − 300 (2 × 104 )(0.35)

y 1 + y2 − 0.010 2

5 3

which simpliﬁes to y2 = y1 + 300 2.78 × 10

−7

īe − 0.00253

y 1 + y2 − 0.010 2

5 3

337 where the factor 2.78 × 10−7 is introduced to convert īe in mm/h to m/s. This equation is solved iteratively for y as a function of time, and the corresponding runoﬀ, Q, is given by Equation 10.96 as 1 5 5 1 5 (1)(250) CW (y − yd ) 3 S02 = (y − 0.010) 3 (0.005) 2 = 50.5(y − 0.010) 3 n 0.35 Starting with y1 = 0 m at t = 0, the runoﬀ computations are tabulated for t = 0 to 60 min as follows:

t (min) 0

īe (mm/h)

y (m) 0

Q (m3 /s) 0

0.0100

0.0000

0.0141

0.0053

0.0173

0.0139

0.0179

0.0158

0.0177

0.0152

0.0175

0.0145

0.0173

0.0139

0.0171

0.0132

0.0169

0.0126

0.0167

0.0120

0.0165

0.0114

0.0163

0.0109

110 5 50 10 40 15 10 20 0 25 0 30 0 35 0 40 0 45 0 50 0 55 0 60

Beyond t = 60 min, the runoﬀ, Q, decreases gradually to zero. 10.32. From the given data, Δt = 10 min and tc = 25 min, therefore Equation 10.103 gives Kr =

10 Δt = = 0.17 2tc + Δt 2(25) + 10

Also, from the given data, x = 0.45, A = 1 km2 , and Equation 10.98 gives the instantaneous runoﬀ, I, as I = [ix + ie (1.0 − x)]A = [i(0.45) + ie (1.0 − 0.45)](1)(0.278) = 0.278[0.45i + 0.55ie ]

(1)

338 where the factor 0.278 is required to give I in m3 /s. The catchment runoﬀ is given by Equation 10.102 as Qj = Qj−1 + Kr (Ij−1 + Ij − 2Qj−1 ) = Qj−1 + (0.17)(Ij−1 + Ij − 2Qj−1 )

(2)

Beginning with I1 = Q1 = 0, Equation 1 is applied to calculate I at each time step, and Equation 2 is applied to calculate the runoﬀ hydrograph. These calculations are summarized in the following table: t (min) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

i (mm/h) 0 50 200 103 52 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

ie (mm/h) 0 0 130 91 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

I (m3 /s) 0 6.3 44.9 26.8 8.2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Q (m3 /s) 0 1.1 9.4 18.4 18.1 13.3 8.8 5.8 3.8 2.5 1.7 1.1 0.7 0.5 0.3 0.2 0.1 0.1 0.1 0.0 0.0

10.33. From the given data: A = 3 km2 and tc = 50 minutes. (a) For a rainfall-excess duration, tr , of 10 min, 1 1 Tp = tr + tl = tr + 0.6tc 2 2 1 = (10) + 0.6(50) = 35 min = 0.583 h 2 and

A 3 = 10.70 (m3 /s)/cm = 2.08 Tp 0.583 Therefore, the 10-min NRCS triangular unit hydrograph has a time base of 2.67Tp = Qp = 2.08

2.67(0.583) = 1.57 h = 94 min and a peak of 10.70 (m3 /s)/cm which occurs at Tp = 0.583 h = 35 min .

339 A 1-h rainfall excess corresponds to 6 10-min rainfall excesses in sequence. To construct the 1-h unit hydrograph we need the following points: t (min) Q (m3 /s)

0 0

10 3.06

20 6.11

30 9.17

40 9.79

50 7.98

60 6.17

70 4.35

80 2.54

90 0.73

100 0

The 1-h hydrograph is calculated by 10-min lagged superposition of the above triangular unit hydrograph and this superposition is given in Columns 1 and 2 of Table 10.4. The 1-h unit hydrograph is obtained by dividing Column 2 by 6 cm and the result is given in Column 3 of Table 10.4. Table 10.4: Calculation of 1-h Unit Hydrograph and Runoﬀ Hydrograph (1) t (min) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

(2) Q (m3 /s) 0 3.06 3.06 + 6.11 = 9.17 9.17 + 9.17 = 18.34 18.34 + 9.79 = 28.13 28.13 + 7.98 = 36.11 36.11 + 6.17 = 42.28 42.28 − 3.06 + 4.35 = 43.57 43.57 − 6.11 + 2.54 = 40.00 40.00 − 9.17 + 0.73 = 31.56 31.56 − 9.79 = 21.77 21.77 − 7.98 = 13.79 13.79 − 6.17 = 7.62 7.62 − 4.35 = 3.27 3.27 − 2.54 = 0.73 0.73 − 0.73 = 0.00

(3) QUH (m3 /s) 0.00 0.51 1.53 3.06 4.69 6.02 7.05 7.26 6.67 5.26 3.63 2.30 1.27 0.55 0.12 0.00

(4) Q4.96cm (m3 /s) 0.00 2.53 7.59 15.18 23.26 29.86 34.97 36.01 33.08 26.09 18.00 11.41 6.30 2.73 0.60 0.00

(b) From the given data i = 6 cm/h, dp = 4 mm, and di = 2 mm. (i) For the pervious area, the runoff is 6 cm − 1 cm − 0.4 cm = 4.6 cm. For the impervious area, the runoff is 6 cm − 0.2 cm = 5.8 cm. Since the catchment area is 30% impervious, the runoff depth from the catchment area is 4.6 (0.7) + 5.8(0.3) = 4.96 cm . (ii) Multiplying the 1-h unit hydrograph by 4.96 cm gives the actual runoff hydrograph shown in Column 4 of Table 10.4. (iii) For the SBUH method, the distribution of rainfall intensity and effective rainfall intensity are shown in Columns 2 and 3 of Table 10.5. For the SBUH method, x = 0.30, ie = 6 cm/h − 1 cm/h = 5 cm/h, A = 3 km2 , and I is given by I = [ix + ie (1 − x)]A

340 The calculated values of I are given in Column 4 of Table 10.5. Taking Δt = 10 min and tc = 50 min gives

Kr =

Δt = 0.167 tc + Δt

and hence the recursion formula for the SBUH method is Qj = Qj−1 + 0.167(Ij−1 + Ij − 2Qj−1 ) The results of applying this recursion formula are shown in Column 5 of Table 10.5.

Table 10.5: Calculation of Runoﬀ Hydrograph Using SBUH Method (1) t (min) 0 10 30 40 50 60 70 80 90 100 110 120 130 140 150

(2) i (cm/h) 0 6 6 6 6 6 0 0 0 0 0 0 0 0 0

(3) ie (cm/h) 0 5 5 5 5 5 0 0 0 0 0 0 0 0 0

(4) I (m3 /s) 0 44.20 44.20 44.20 44.20 44.20 0 0 0 0 0 0 0 0 0

(5) Q (m3 /s) 0.00 7.39 19.68 33.32 36.95 39.37 33.60 22.38 14.91 9.93 6.61 4.40 2.93 1.95 1.30

(iv) Comparing the runoff hydrograph calculated using the 1-h unit hydrograph with the runoff hydrograph calculated using the SBUH method, the SBUH runoff hydrograph gives higher inflows to the detention pond leading up a (higher) peak runoff and will likely lead to a larger and more conservative pond design. 10.34. From the given data: A = 5 km2 , I = 35%, tc = 1 h, and fp = 200 mm/h. Using the alternating block method, the runoff at 10-min intervals is calculated as follows:

341 t (min) 0–10 10–20 20–30 30–40 40–50 50–60

i (mm/h) 281 201 159 133 114 101

it (mm) 46.8 67.0 79.5 88.7 95.0 101.0

Amt (mm) 46.8 20.2 12.5 9.2 6.3 6.0

Rate (mm/h) 281 124 75 55 38 36

Intensity (mm/h) 38 75 281 124 55 36

Since the area is 35% impervious, the average infiltration capacity is given by average infiltration capacity = 0.65(200) + 0.35(0) = 130 mm/h Comparing this with the 10-min rainfall intensities derived from the alternating block method, it is apparent that the duration of rainfall excess, tr is 10 minutes, occurring in the 20–30 min time interval. The time lag, tl , time to peak, Tp , and the peak runoff, Qp , in the NRCS unit hydrograph can be taken as tl = 0.6tc = 0.6(60) = 36 min 1 1 Tp = tr + tl = (10) + 36 = 46 min 2 2 A 5 = 15.2 (m3 /s)/cm = 2.08 Qp = 2.08 Tp 41/60 Subtracting the rainfall intensity from the infiltration rate gives the depth of runoff as depth of runoff = (281 − 130)

10 = 25.2 mm = 2.52 cm 60

So multiply the unit hydrograph by 15.2 (m3 /s)/cm × 2.52 = 38.3 m3 /s. So the NRCS runoﬀ hydrograph has a peak of 38.3 m3 /s and a time base of 5Tp = 5(41 min) = 205 min = 3.42 h. Using the SBUH method with Δt = 10 min and x = 0.35 gives 10 Δt = = 0.0769 2tc + Δt 2(60) + 10 I = [ix + ie (1 − x)]A = [i(0.35) + ie (1 − 0.35)](5)(0.278) = 1.39[0.35i + 0.65ie ] m3 /s

Kr =

Qj = Qj−1 + (0.10)(Ij−1 + Ij − 2Qj−1 ) The above SBUH relationships yield the following SBUH runoﬀ hydrograph,

342 t (min) 0 10 20 30 40 50 60 70 80

i (mm/h) 0 38 75 281 124 55 36 0

ie (mm/h) 0 0 0 151 0 0 0 0

I (m3 /s) 0 19 37 273 60 27 18 0

Q (m3 /s) 0 1.85 6.98 36.5 62.5 58.7 51.4 42.9 ( = 0.8Qj−1 )

Therefore the maximum runoff rate from the SBUH is 62.5 m3 /s . 10.35. This problem requires that the runoff hydrograph be routed through the detention basin. Using Δt = 30 min, the storage and outflow characteristics can be put in the convenient tabular form: Stage (m) 8.0 8.5 9.0 9.5 10.0 10.5 11.0

S (m3 ) 0 1041 2288 3761 5478 7460 9726

O (m3 /s) 0 1.16 3.29 6.04 9.31 13.0 17.1

2S/Δt + O (m3 /s) 0 2.32 5.83 10.2 15.4 21.3 27.9

and the computations in the routing procedure are summarized in the following table: (1) Time (min) 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420

(2) I (m3 /s) 0 3.6 8.4 5.1 4.2 3.6 3.3 2.7 2.3 1.8 1.5 0.84 0.51 0 0

(3) 2S/Δt − O (m3 /s) 0 −0.28 −2.26 −2.14 −1.10 −0.98 −0.78 −0.62 −0.44 −0.28 −0.14 0.0 −0.01 0.0 0.0

(4) 2S/Δt + O (m3 /s) 0 3.6 11.7 11.2 7.16 6.70 5.92 5.22 4.38 3.66 3.02 2.20 1.35 0.50 0.0

(5) O (m3 /s) 0 1.94 6.98 6.67 4.13 3.84 3.35 2.92 2.41 1.97 1.58 1.10 0.68 0.25 0.0

343 10.36. In accordance with Equation 10.126, select Δt such that 2KX ≤ Δt ≤ 2K(1 − X) =⇒ 2(35)(0.3) ≤ Δt ≤ 2(35)(1 − 0.3) =⇒ 21 min ≤ Δt ≤ 49 min and according to Viessman and Lewis (2003) it is recommended that K ≤ Δt ≤ K 3 or

35 min ≤ Δt ≤ 35 min 3 Taking Δt = 30 min, the Muskingum constants are given by Equations 10.122 to 10.124 as 30 − 2(35)(0.3) Δt − 2KX = = 0.114 2K(1 − X) + Δt 2(35)(1 − 0.3) + 30 30 + 2(35)(0.3) Δt + 2KX = = 0.645 C2 = 2K(1 − X) + Δt 2(35)(1 − 0.3) + 30 2(35)(1 − 0.3) − 30 2K(1 − X) − Δt = = 0.241 C3 = 2K(1 − X) + Δt 2(35)(1 − 0.3) + 30 C1 =

These results can be veriﬁed by taking C1 + C2 + C3 = 0.114 + 0.645 + 0.241 = 1. The Muskingum routing equation, Equation 10.121, is therefore given by Oj+1 = C1 Ij+1 + C2 Ij + C3 Oj = 0.114Ij+1 + 0.645Ij + 0.241Oj This routing equation is applied repeatedly to the given inﬂow hydrograph, and the results are as follows: (1) Time (min) 0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480

(2) I (m3 /s) 0 12.5 22.1 15.4 13.6 12.4 11.7 10.8 9.9 8.4 8.1 7.5 4.2 0 0 0 0

(3) O (m3 /s) 0 1.43 10.9 18.6 16.0 14.0 12.7 11.8 10.9 9.98 8.75 8.19 7.29 4.47 1.08 0.26 0.06

344 (1) Time (min) 510 540 570 600

(2) I (m3 /s) 0 0 0 0

(3) O (m3 /s) 0.02 0.00 0.00 0.00

10.37. From the given data: Δt = 0.5 h, and the Muskingum routing equation is given by the combination of Equations 10.121 to 10.124 as Δt − 2KX Δt + 2KX 2K(1 − X) − Δt Ij+1 + Ij + Oj Oj+1 = 2K(1 − X) + Δt 2K(1 − X)Δt 2K(1 − X) + Δt

(1)

For given values of Δt, X, and K, Equation 1 is used to calculate the outﬂow hydrograph corresponding to the given inﬂow hydrograph, and the sum of absolute errors between the calculated and measured hydrograph is given by Equation 10.127. For various assumed values of X and K, the sum of absolute errors are given in the following table:

[1]

[2]

[3]

0.5 0.4 0.3 0.2 0.1 0.0 0.21 0.19

(min) 0.5 37.5 39.7 40.1 39.9 39.9 40.0 40.1

Sum of Absolute Errors (m6 /s2 ) 22.8 12.5 5.9 0.3 4.9 9.3 0.7 0.6

For each given value of X in Column 1, the value of K that minimizes the sum of absolute errors is given in Column 2, and the corresponding sum of absolute errors is given in Column 3. Based on the tabulated results, the values of X and K that minimize the sum of absolute errors are X = 0.20 and K = 40.1 min. The sum of squared residuals between a calculated and measured hydrograph is given by Equation 10.128, and the sum of squared residuals for various values of X and K are given in the following table:

345 [1]

[2]

[3]

0.5 0.4 0.3 0.2 0.1 0.0 0.21 0.19

(min) 30.0 37.5 39.8 40.0 39.9 39.5 40.0 40.0

Sum of Squared Residuals (m6 /s2 ) 113.4 36.8 7.6 0.01 5.0 17.4 0.09 0.06

For each value of X in Column 1, the value of K that minimizes the sum of squared residuals is given in Column 2, and the corresponding sum of squared residuals is given in Column 3. Based on the tabulated results, the values of X and K that minimize the sum of squared residuals are X = 0.2 and K = 40.0 min. The optimal values of X and K based on the sum of absolute errors and the sum of squared residuals are approximately the same and equal to X = 0.2 and K = 40 min . A P B×y = (in case of wide rectangular channel) B + 2y R=y

10.38. R =

dA = Bdy We know, Ck = dQ dA = B1 dQ dy 1 AR2/3 S 1/2 n 1 d 1 2/3 1/2 (B × y)y s Ck = B dy n Q=

1 d 1 × B × × S 1/2 × (y 5/3 ) B n dy 5 1 1/2 5 = S × × y 3 −1 n 3 5 1 1/2 2/3 S y = 3 n 5 Ck = V (Where, V = n1 S 1/2 y 2/3 ) 3 =

346 10.39. Data given, A= 2000 km2 tc = t×N = 3 × 8 = 24h K = 12h Number of isochrones = N−1 = 8−1 = 7 tc Computation interval t = tc between successive isochrones = 3h = 24/8 = N

According to Equation 10.87 and 10.88, Clark’s approach, QL = C I + C2 Q2 C =

t 3 = 0.22 t = K+2 12 + 32

C2 =

K − 2t 13 − 32 = = 0.78 K + 2t 13 + 32

Check C + C2 = (0.22 + 0.78) = 1.0; ; T heref ore OK Again, I = 2.78 Atr = 2.78 A3r = 0.927Ar Clark’s Q2 = C I + C2 Q1 C2 Q1 = 0.78Q, Q2 = IU HO C I = 0.22 × 0.927Ar = 0.203Ar Time(hr.)

Ar (km2 )

C I = 0.203 Ar

0 3 6 9 12 15 18 21 24 27 30 33

0 30 65 90 115 140 230 580 750

0 6.1 13.2 18.3 23.3 28 46.7 117.7 152.3 0 0 0

C2 Q1 = 0.78Q1

0 4.8 14 24.4 37.2 50.9 76.1 151.2 236.7 184.6 144

IUHO Q2 = C I + C 2 Q1 0 6.1 18 31.3 47.7 65.2 97.6 193.8 303.5 236.7 184.6 144

3-h UH (cumec) 0 3.1 12.1 24.7 39.5 56.5 81.4 145.7 248.7 270.1 210.7 164.3

10.40. From the given data: N = 84 storms, BCF = 1.521 (Table 10.15 for SS), DA = 70 ha, IA = 50%, M AR = 95 cm, M JT = 8.1◦ C, and X2 = 0 (since commercial plus industrial use is less than 75%). These variables are within the ranges given in Table 10.17, and therefore the USGS regression equation, Equation 10.146 can be used. The regression constants taken

347 from Table 10.15 (for SS) are a = 0.5926, b = 0.0988, d = 0.0104, e = −0.0535, and c = f = 0. Substituting data into Equation 10.146 gives √ Y = 0.454(N )(BCF )10[a+b (DA)+c(IA)+d(M AR)+e(M JT )+f (X2)] √ = 0.454(84)(1.521)10[0.5926+0.0988 (70)+0.0104(95)−0.0535(8.1)] = 5460 kg 10.41. From the given data: N = 84 storms, BCF = 1.365 (Table 10.15 for PB), DA = 70 ha, IA = 50%, M AR = 95 cm, M JT = 8.1◦ C, and X2 = 0 (since commercial plus industrial use is less than 75%). These variables are within the ranges given in Table 10.17, and therefore the USGS regression equation, Equation 10.146 can be used. The regression constants taken from Table 10.15 (for PB) are a = −1.9679, b = 0.1183, c = 0.0070, d = 0.00504, and e = f = 0. Substituting data into Equation 10.146 gives √ Y = 0.454(N )(BCF )10[a+b (DA)+c(IA)+d(M AR)+e(M JT )+f (X2)] √ = 0.454(84)(1.365)10[−1.9679+0.1183 (70)+0.0070(50)+0.00504(95)] = 37 kg 10.42. From the given data: α = 0.0336 (Table 10.18: PO4 , Residential), P = 98 cm, D = 15 persons/ha, Ns = 14 days, Equation 10.148 gives f = 0.142 + 0.134D0.54 = 0.142 + 0.134(15)0.54 = 0.72 and Equation 10.149 gives 14 Ns = = 0.7 20 20 Substituting these data into the EPA model, Equation 10.147, gives s=

Ms = 0.0442αP f s = 0.0442(0.0336)(98)(0.72)(0.7) = 0.073 kg/ha For a 80-ha development, the annual phosphate load is 80 ha × 0.073 kg/ha = 6 kg . 10.43. (a) Using the USGS method for suspended solids: N = 62/year, BCF = 1.521, DA = 10 ha, IA = 65%, M AR = 98.5 cm, M JT = 9.6◦ C, X2 = 0, a = 0.5926, b = 0.0988, d = 0.0104, e = −0.0535. The annual load, Y , is given by Equation 10.146 as √ Y = 0.454(N )(BCF )10[a+b (DA)+c(IA)+d(M AR)+e(M JT )+f (X2)] √ = 0.454(62)(1.521)10[0.5926+0.0988 (10)+0.0104(98.5)−0.0535(9.6)] = 1115 kg Use Equation 10.150 to calculate the volume of runoﬀ, R, where I P − 3.004d0.5957 R = 0.15 + 0.75 100 = [0.15 + 0.75(0.65)](98.5) − 3.004(0.15)0.5957 = 61.8 cm

348 The average concentration, c, is therefore given by c=

1115 × 106 = 18 mg/L (0.618)(10 × 104 )(103 )

(b) Parameters of the EPA method are: α = 16.3 (Table 10.18), P = 98.5 cm, D = 25 persons/ha, Equation 10.148 gives f = 0.142 + 0.134D0.54 = 0.142 + 0.134(25)0.54 = 0.904 and s = 1.0. Hence Equation 10.147 gives Ms = 0.0442αP f s = 0.0442(16.3)(98.5)(0.904)(1.0) = 64.2 kg/ha So for 10 ha, mass = 10 × 64.2 = 642 kg. Since the runoﬀ, R, is given by R = 61.8 cm, then the average concentration, c, is given by c=

642 × 106 = 10 mg/L (0.618)(10 × 104 )(103 )

Chapter 11

Design of Stormwater Collection Systems 11.1 The time of conc. Is given by Kirpich’s eqn, tc = 0.019 ×

L0.77 20000.77 = 0.019 × = 50.87 min 0.0050.385 S00.385

Run oﬀ coeﬃcient, C = (3×0.3)+(3×0.7) = 0.5 6 all depth 62 = 50.87 × 60 mm/h = 70.77 mm/h The average rainfall intensity, i − maximum rainf tc

= 1.96 × 10−5 m/s Therefore the peak runoff rate, Qp = CiA = 0.5 × 1.96 × 10−5 × 6 × 106 = 58.8 m3 /s 11.2. From the given data: Q = 0.09 m3 /s, Sx = 2.5% = 0.025, and S0 = 1.5% = 0.015. Assuming that a significant portion of the gutter flow extends onto the rough-asphalt pavement, Table 11.1 gives n = 0.015. Using the Manning equation (Equation 11.6) to find the depth, d, at the curb gives 1 8 1 d 3 S02 Q = 0.375 nSx 8 1 1 d 3 (0.015) 2 0.09 = 0.375 0.015 × 0.025 and solving for d leads to d = 0.067 m = 6.7 cm This flow depth is less than the curb height of 15 cm, and therefore the flow is constrained within the gutter. The top-width, T , of the gutter flow, is given by T =

0.067 d = 2.68 m = Sx 0.025

The spread of 2.68 m is much larger than the gutter width of 90 cm, verifying the assumption that a signiﬁcant portion of the gutter ﬂow extends onto the rough-asphalt pavement, and 356

357 hence justifying the assumed n value of 0.015. For an inlet width, Wo , of 0.7 m, the ratio, Rw , of frontal flow to the total gutter flow is given by Equation 11.14 as 8 8 W0 3 0.7 3 Rw = 1 − 1 − =1− 1− = 0.55 T 2.68 (a) For a gutter depression, a, of 30 mm (= 0.030 m), the cross slope, Sw , of the gutter relative to the pavement slope is given by Equation 11.13 as a 0.030 = 0.043 = Sw = W0 0.70 Hence the equivalent cross slope, Se , of the depressed inlet is given by Equation 11.12 as Se = Sx + Sw Rw = 0.025 + (0.043)(0.55) = 0.049 The length, LT , of the curb inlet required to intercept all of the gutter flow is given by Equation 11.10, with Se replacing Sx , as 0.6 1 0.6 1 = 0.817(0.09)0.42 (0.015)0.3 = 6.40 m LT = 0.817Q0.42 S00.3 nSe 0.015 × 0.049 (b) If there is no inlet depression, then Se = Sx = 0.025 and the length, LT , of the curb inlet required to intercept all of the gutter flow is given by Equation 11.10 as 0.6 1 0.6 1 = 0.817(0.09)0.42 (0.015)0.3 = 9.58 m LT = 0.817Q0.42 S00.3 nSx 0.015 × 0.025 The fraction, R, of gutter flow removed by an inlet of length L is given by Equation 11.11. When R = 70% = 0.70, Equation 11.11 gives L 1.8 0.70 = 1 − 1 − LT Solving for L/LT gives L = 0.487 LT Therefore, the length of the curb inlet required to intercept 70% of the flow is 0.487(6.40 m) = 3.12 m for a depressed inlet, and 0.487(9.58 m) = 4.67 m for an undepressed inlet. 11.3 (a) Since the flow depth (9 cm) is less than the height of the inlet (15 cm), the curb inlet acts as a weir. In this case, Qi = 0.1 m3 /s, W0 = 0.3 m, d = 0.09 m, and the weir equation for a depressed inlet (Equation 11.15) can be put in the form Qi − 1.8W0 1.25d1.5 The required weir length, L, is therefore given by 0.1 L= − 1.8(0.3) = 2.42 m 1.25(0.09)1.5 L=

With an inlet depression, the length of the curb opening should therefore be at least 2.42 m .

358 (b) In the case of no inlet depression, the inlet equation for a nondepressed inlet (Equation 11.15) can be put in the form Qi L= 1.60d1.5 and the required weir length is given by L=

0.1 = 2.31 m 1.60(0.09)1.5

These calculations show that a longer inlet is required with a depression than without a depression, which is unlikely to be the case. Consequently, a curb opening of at least 2.42 m should be used with or without a depression. 11.4 (a) Guo equation. From the given data: L = 1.5 m, h = 0.15 m, and d = 0.05 m. It is estimated that Cw = 0.44, Co = 0.67, and Cm = 0.93. Equation 11.17 gives ⎧ √ 3 3 2 = (0.44) C 2gLd 2(9.81)(1.5)(0.05) 2 = 0.0327 m3 /s, ⎪ w ⎪ ⎨ √ Qi = Cm Qw Qo = (0.93) (0.0327)(0.149) = 0.0649 m3 /s, ⎪ ⎪ 1 1 ⎩ √ Co 2gLhd 2 = (0.67) 2(9.81)(1.5)(0.15)(0.05) 2 = 0.149 m3 /s,

weir flow mixed flow orifice flow

Therefore the estimated capacity of the sump inlet is min(0.0327 m3 /s, 0.0649 m3 /s, 0.149 m3 /s) = 0.0327 m3 /s . (b) USFHWA equation. Since the curb inlet is undepressed, Equation 11.15 is the appropriate USFHWA relationship, which gives Qi = 1.60Ld1.5 = 1.60(1.5)(0.05)1.5 = 0.0268 m3 /s 11.5. According to the Manning equation, Equation 11.6, the ﬂow rate, Q, in the gutter is given by 1 8 1 d 3 S02 Q = 0.375 nSx Since d = T Sx , the Manning equation in terms of the top width, T , is Q=

0.375 53 12 8 Sx S0 T 3 n

(1)

The ﬂow, Qw , over a top width, W0 , measured from the curb is therefore given by Qw =

8 0.375 53 12 8 0.375 53 12 Sx S0 T 3 − Sx S0 (T − W0 ) 3 n n

Combining Equations 1 and 2 gives Qw =1− Rw = Q

T − W0 T

8 3

(2)

359 which yields

8 W0 3 Rw = 1 − 1 − T

This equation is important in the design of grate inlets since it gives the fraction of the gutter flow that flows directly over the grate. If the flow velocity in the gutter is less than the splash-over velocity, the grate will intercept 100% of the frontal flow. 11.6. From the given data: Q = 0.07 m3 /s, Sx = 2% = 0.02, and S0 = 2.5% = 0.025. Assuming that a significant portion of the gutter flow extends onto the smooth-asphalt pavement, Table 11.1 gives n = 0.015. Using the Manning equation (Equation 11.6) to find the depth, d, at the curb gives 1 8 1 d 3 S02 Q = 0.375 nSx 8 1 1 d 3 (0.025) 2 0.07 = 0.375 0.015 × 0.02 and solving for d leads to d = 0.051 m = 5.1 cm This flow depth is less than the curb height of 8 cm, and therefore the flow is constrained within the gutter. The top-width, T , of the gutter flow, the flow area, A, and the flow velocity, V , are given by 0.051 d = 2.55 m = Sx 0.02 1 1 A = dT = (0.051)(2.55) = 0.065 m2 2 2 0.07 Q = = 1.08 m/s V = A 0.065 T =

The spread of 2.55 m is much wider than the gutter width of 0.90 m, verifying the assumption that a significant portion of the gutter flow extends onto the rough-asphalt pavement, and hence justifying the assumed n value of 0.015. Since the spread (T = 2.55 m) exceeds the maximum transverse dimension of typical grates (= 914 mm), use a grate with a transverse dimension, W0 , of 914 mm. According to Figure 11.7, the length of a reticuline grate corresponding to a splash-over velocity of 1.08 m/s is approximately 43 cm. Therefore, any grate longer than 43 cm will intercept 100% of the frontal flow, and have a frontal flow efficiency, Rf , equal to 1.0. To maximize side flow interception, use a (typical) maximum available grate length of 1220 mm. The selected grate therefore has dimensions of 914 mm × 1220 mm . The ratio, Rw , of frontal flow to total gutter flow is given by Equation 11.20 as 8 8 W0 3 0.914 3 =1− 1− = 0.69 Rw = 1 − 1 − T 2.55

360 The ratio, Rs , of the side ﬂow intercepted to the total side ﬂow is given by Equation 11.19 as Rs = 1 +

0.0828V 1.8 Sx L2.3

−1

= 1+

0.0828(1.08)1.8 (0.02)(1.22)2.3

−1

= 0.25

Therefore, the ratio, R, of the intercepted flow to the total gutter flow is given by Equation 11.21 as R = Rf Rw + Rs (1 − Rw ) = (1.0)(0.69) + (0.25)(1 − 0.69) = 0.77 Based on this result, the grate intercepts 0.77(0.07 m3 /s) = 0.054 m3 /s . 11.7 Since the depth of flow is less than 12 cm, the inflow to the inlet is given by the weir equation (Equation 11.23), which can be put in the form P =

Qi 1.66d1.5

where P is the grate-inlet perimeter, not including the side adjacent to the curb, Qi = 0.1 m3 /s, and d = 0.09 m, and 0.1 = 2.23 m P = 1.66(0.09)1.5 The minimum length of the grate can be derived from Figure 11.7 based on the type of grate and the ﬂow velocity in the gutter, V0 . Assume a reticuline grate (worst-case scenario), and V0 =

Qi A

where A is the ﬂow area in the gutter, given by 1 d 0.09 1 = (0.09) = 0.27 m2 A= d 2 Sx 2 0.015 The ﬂow velocity, V0 , in the gutter is therefore given by V0 =

0.1 Qi = = 0.37 m/s A 0.27

and the minimum length of the grate inlet to intercept all frontal ﬂow is given by Figure 11.7 as L = 0.15 m. Therefore, the grate inlet must have a minimum length along the gutter of 15 cm , and a minimum perimeter of 223 cm . 11.8 (a) Guo equation. From the given data: L = 0.914 m, W = 0.457 m, and d = 0.05 m. For a vane grate, it is estimated that Co = 0.67, Cm = 0.93, Cw = 0.30, Nw = 0.62, and No = 0.32. Substituting these data into Equation 11.24 gives ⎧ 3 3 ⎨Nw Cw √2g[2W + L]d 2 = (0.62)(0.30) 2(9.81)[2(0.457) + (0.914)](0.05) 2 = 0.0168 m3 /s, weir flow √ 3 mixed flow Qi = Cm Qw Qo = (0.93) (0.0168)(0.0887) = 0.0359 m /s, ⎩ 1 1 √ 3 No Co

2gW Ld 2 = (0.32)(0.67)

2(9.81)(0.457)(0.914)(0.05) 2 = 0.0887 m /s,

orifice flow

The minimum estimated ﬂow into the grate inlet is 0.0168 m3 /s , and hence this can be taken as in inlet ﬂow capacity for a ponded depth of 5 cm.

361 (b) USFHWA equation. Since d ≤ 12 cm the weir ﬂow is appropriate and Equation 11.23 gives 3 3 Qi = 1.66P d 2 = 1.66[2(0.457) + (0.914)](0.05) 2 = 0.0339 m3 /s Hence the USFHWA-estimated capacity of the inlet for a ponded depth of 5 cm is 0.0339 m3 /s . Hence the Guo et al. (2009) equations provide a more conservative estimate of the inlet capacity. 11.9. Flow Capacity, Q = 0.3 m3 /s Using equation 11.6, 8 1 1 d 3 s02 Q = 0.375 nSx Using typical valuesSX = cross slope = 2% = 0.02 S0 = longitudinal slope = 0.5% = 0.005 n=0.016 0.3 = 0.375

1 0.016 × 0.02

× d 3 × (0.005) 2

d = 0.12 m Maximum depth in gutter,‘d’, is 0.12m. Again in equation 11.7, d = T Sx T =

d =6 Sx

d From Fig. 11.2, 2A d = Sx

Area of gutter, A = 2S1x d2 = 0.36m2 Maximum velocity in gutter V0 =

0.3 Q = = 0.83 m/s A 0.36

Length of curb opening (eqn. 11.15), Qi = 1.25(L + 1.8W0 )d1.5 0.3 = 1.25(L + 1.8 × 0.5)(0.12)1.5 L=4.814 m 11.10. The Manning equation for ﬂow in a triangular gutter is given by Q=

0.375 53 12 8 Sx S0 T 3 n

where Q is the flow in the gutter, n is the Manning roughness coefficient, Sx is the cross slope, S0 is the longitudinal slope, and T is the top width of the gutter flow. The gutter is

362

Figure 11.1: Gutter Flow depressed over a width W0 as shown in Figure 11.1. The ﬂow, Q , in the undepressed portion of the gutter is given by 8 0.375 53 12 Sx S0 (T − W0 ) 3 Q = (1) n and the ﬂow, Q in the depressed portion of the gutter is given by Q =

8 0.375 53 12 83 0.375 53 12 Sw S0 T − Sw S0 (T − W0 ) 3 n n

(2)

where the top width T is shown in Figure 11.1. The ratio, Rw , of the ﬂow in the depressed portion of the gutter to the total gutter ﬂow is given by Q Q + Q

Rw = which can be put in the form Rw =

Q Q

−1

Equations 1 and 2 can be combined to yield 5

Sx3 (T − W0 ) 3 Q = 5 5 8 8 Q S 2 T 3 − S 3 (T − W ) 3 w

Dividing the numerator and denominator by Sx3 W03 yields Q = Q

Sw Sx

5 3

( WT0 − 1) 3 8 5 8 Sw 3 T T 3 − Sx ( W0 − 1) 3 W0

363 The following geometric relations are apparent from Figure 11.1, d = Sw , T − W0

d = Sx T − W0

Eliminating d gives T − W0 Sw = T − W0 Sx which simpliﬁes to T =1+ W0

T Sx −1 W0 Sw

(3)

Combining Equations 11 and 3 and simplifying gives Q = Q

Sw /Sx x 1 + STw /S −1 W

8 3

(4) −1

Substituting Equation 4 into Equation 11 yields ⎧ ⎪ ⎪ ⎪ ⎨ Rw =

⎪ ⎪ ⎪ ⎩

⎫−1 ⎪ ⎪ ⎪ ⎬

Sw /Sx x 1 + STw /S −1 W

8 3

⎪ ⎪ ⎭ − 1⎪

This equation is important in the design of combination inlets since it gives the fraction of the gutter flow that flows over the grate in the depressed gutter. If the gutter velocity is less than the splash-over velocity, then all of this (frontal) flow is intercepted. 11.11. Step 1: Calculate the interception capacity of the curb opening upstream of the grate inlet. From the given data, Q = 0.09 m3 /s, S0 = 1.5% = 0.015, n = 0.015 (Table 11.1), Sx = 3.5% = 0.035, a = 30 mm = 0.03 m, and W0 = 0.6 m. The fraction, Rw , of gutter flow over the depressed section in front of the curb opening is given by Equation 11.25, where ⎧ ⎪ ⎪ ⎪ ⎨ Rw =

⎪ ⎪ ⎪ ⎩

⎫−1 ⎪ ⎪ ⎪ ⎬

Sw /Sx x 1 + STw /S −1

8 3

⎪ ⎪ ⎭ − 1⎪

In this case, Sw = Sx +

a 0.03 = 0.085 = 0.035 + W0 0.6

(1)

364 and Equation 1 can be written as ⎫−1 ⎪ ⎪ ⎪ ⎬

⎧ ⎪ ⎪ ⎪ ⎨

0.085/0.035 Qw = 1+ 8 ⎪ ⎪ Q 3 ⎪ ⎪ 0.085/0.035 ⎪ ⎪ ⎭ ⎩ 1+ − 1 T −1 0.6 ⎫−1 ⎧ ⎪ ⎪ ⎬ ⎨ 2.43 = 1+ 8 ⎪ ⎪ 3 ⎩ 2.43 1 + 1.67T − 1⎭ −1

(2)

where Qw is the ﬂow over width W0 , and Q is the total gutter ﬂow given as 0.09 m3 /s. Hence, Equation 2 can be written as ⎫−1 ⎪ ⎬

⎧ ⎪ ⎨ Qw = 0.09

2.43 1+ 8 ⎪ ⎪ 3 ⎩ 2.43 1 + 1.67T −1 − 1 ⎭

(3)

It is convenient for subsequent analyses to work with the ﬂow, Qs , over the section outside of the depressed section, in which case

Qs = Q − Qw = 0.09 − 0.09

⎧ ⎪ ⎨ ⎪ ⎩

2.43 2.43 1 + 1.67T −1

⎫−1 ⎪ ⎬ 8 3

⎪ − 1⎭

(4)

This equation must be solved simultaneously with the Manning equation, Equation 11.6, which can be written as 8 0.375 53 12 Sx S0 (T − W0 ) 3 (5) Qs = n which, in this case gives Qs =

5 1 8 0.375 (0.035) 3 (0.015) 2 (T − 0.6) 3 0.015

which simpliﬁes to 8

Qs = 0.0115(T − 0.6) 3

(6)

Simultaneous solution of Equations 4 and 6 gives Qs = 0.0261 m3 /s and T = 1.96 m. The ﬂow ratio, Rw , over width W0 is therefore given by Rw =

0.09 − 0.0261 Q − Qs = = 0.71 Q 0.09

The equivalent cross slope, Se , in the gutter depression is given by a 0.03 0.71 = 0.071 Rw = 0.035 + Se = Sx + Sw Rw = Sx + W0 0.6

365 Equation 11.10 gives the length, LT , of curb opening for 100% interception as LT = 0.817Q0.42 S00.3

1 nSe

0.6 = 0.817(0.09)

0.42

(0.015)

0.3

1 0.015 × 0.071

0.6 = 5.12 m

Since the length, L, of curb opening upstream of the grate is 3 m − 1.2 m = 1.8 m, the efficiency of the curb opening is given by Equation 11.11 as L 1.8 1.8 1.8 R=1− 1− =1− 1− = 0.541 LT 5.12 The flow, Qic , intercepted by the curb opening is therefore given by Qic = 0.541(0.09 m3 /s) = 0.049 m3 /s Step 2: Calculate the interception capacity of the grate inlet. The gutter flow immediately upstream of the grate, Qg , is given by Qg = Q − Qic = 0.09 − 0.049 = 0.041 m3 /s The ratio of flow over the grate, Qw to the gutter flow upstream of the grate, Qg , is given by Equation 2 (with Qg replacing Q) and, similar to Equation 4, the side flow, Qs , is given by

Qs = 0.041 − 0.041

⎧ ⎪ ⎨ ⎪ ⎩

2.43 2.43 1 + 1.67T −1

⎫−1 ⎪ ⎬ 8 3

⎪ − 1⎭

(7)

The Manning equation is given by Equation 5 , which leads to Equation 6. Solving Equations 6 and 7 simultaneously gives Qs = 0.00515 m3 /s and T = 1.34 m. The ratio, Rw , of ﬂow over W0 to the total gutter ﬂow is therefore given by Rw =

Qg − Qs 0.041 − 0.00515 = 0.87 = Qg 0.041

Next, calculate the frontal flow interception efficiency, Rf . The total flow area, A in the gutter is given by 1 1 A = [T 2 Sx + aW0 ] = [(1.34)2 (0.035) + (0.03)(0.6)] = 0.0404 m2 2 2 and hence the average velocity, V , in the gutter is given by V =

Qg 0.041 = = 1.01 m/s A 0.0404

For a 1.2-m long reticuline grate, the splash-over velocity is greater than 1.01 m/s (see Figure 11.7), and therefore the frontal ﬂow interception eﬃciency is 100% and hence Rf = 1.0

366 The side ﬂow interception eﬃciency, Rs , is given by Equation 11.19 as Rs = 1 +

0.0828V 1.8 Sx L2.3

−1

= 1+

0.0828(1.01)1.8 (0.035)(1.2)2.3

−1

= 0.387

The ﬂow intercepted by the grate, Qig , is given by Equation 11.21 as Qig = Qg [Rf Rw + Rs (1 − Rw )] = 0.041[(1.0)(0.87) + (0.387)(1 − 0.87)] = 0.038 m3 /s

(8)

Step 3: Calculate the total interception capacity of the combination inlet. The interception capacity of the combination inlet, Qi , is the sum of the curb opening capacity, Qic , and the grate capacity, Qig , hence Qi = Qic + Qig = 0.049 + 0.038 = 0.087 m3 /s 11.12. A = 100 × 100 m2 = 104 m2 C=0.95 From problem 11.23, using equation, 2261 mm/hr I= (t + 8.5)0.754 Before development, 2261 mm/hr = 3.36 × 10−5 m/s Ibd = (40 + 8.5)0.754 After development, 2261 mm/h = 4 × 10−5 m/s Iad = (30 + 8.5)0.754 Runoff from the land before development, = 0.95 × 104 × Ibd = 0.3192 m3 /sec Runoff from the land after development=0.95 × 104 × Iad = 0.38 m3 /sec 11.13 The weighted value of C, 20 × 0.4 20 × 0.5 20 × 0.36 + + = 0.41 C= 60 60 60 The velocity of flow, Vf = 1.5m s The time of concentration, 1000 = 11 min tc = lengthVfof run = 1.5×60

When the rainfall intensity for one hour duration is 62.5mm,then for 11min duration,the rainfall intensity is 150 m2 /h. Using the rational formula, 150 × 3600 × 104 = 13.284 × 107 m3 /s Qp = CiA = 0.41 × 1000

367 11.14 From the given data, Qi = 0.2 m3 /s, S0 = 0.005, n = 0.017, Sx = 0.015, and Equation 11.37 gives the length of the drain as L = 0.817Q0.42 S00.3 i

0.6 1 0.6 1 = 0.817(0.2)0.42 (0.005)0.3 = 12.1 m nSx (0.017)(0.015)

Hence the slotted drain must be at least 12.1 m long to remove the gutter ﬂow. 11.15 From the given data: A = 4.96 ha = 49600 m2 , and S0 = 0.1% = 0.001. Assume C = 0.2 and n = 0.2 for the pervious area and C = 0.9 and n = 0.1 for the impervious area. Take x as the impervious fraction. For the pervious area, the time of concentration is given by (0.2 (1 − x)49600)0.6 (1 − x)0.3 = 1031 (1) tcp = 6.99 (0.2i)0.4 (0.001)0.3 i0.4 and for the impervious area

(0.1 x(49600))0.6 x0.3 = 373 tci = 6.99 (0.9i)0.4 (0.001)0.3 i0.4

(2)

From the given IDF curve: i=

8426 7836 ⇒ tc = − 43.8 40.7 + 0.930t i

(3)

When the entire area is contributing, the peak runoﬀ, Q1 , is given by Q1 = C1 i1 A1 = (0.7x + 0.2)i1 (49600)

(4)

and the peak runoﬀ generated by the DCIA, Q2 , is given by Q2 = C2 i2 A2 = 0.9i2 (49600x)

(5)

where Equations 4 and 5 combine to give 0.7x + 0.2 i1 Q1 = Q2 0.9x i2

(6)

DCIA controls the runoﬀ when Q1 /Q2 < 1. Standard calculations for Q1 /Q2 are given in Table 11.1. Based on the calculations DCIA will control the design of the drainage sysTable 11.1: Peak Flow Computations x 0.10 0.05 0.06 0.055

i1 (mm/h) 27.1 26.6 26.7 26.6

tcp (min) 267 263 272 273

i2 (mm/h) 118 129 126 127

tci (min) 27.6 21.5 23.1 22.5

Q1 /Q2 0.69 1.08 0.95 1.01

tems when DCIA exceeds 5.5% . These calculations indicate that roadway and parking-lot drainage systems are likely to control the design of the drainage system, since this is the primary DCIA and could likely exceed 5.5%.

368 11.16 Using the given IDF curve, the eﬀective rainfall rate, ie , is given by the rational formula as ie = Ci = C

8000 t + 40

(1)

where C is the runoﬀ coeﬃcient. The storm duration, t, is taken to be equal to the time of concentration, tc , given by Equation 10.13 as tc =

6.99 2 5

nL √ S0

3 5

(2)

and simultaneous solution of Equations 1 and 2 using the given catchment characteristics leads to the following times of concentration, tc : Catchment

Surface

pervious impervious pervious impervious

tc (min) 45 11 70 12

Consider now the flows at specific locations. Inlet 1, and Pipe 1: When the entire catchment A is contributing, the time of concentration is 45 min, the average rainfall rate, i, from the IDF curve is 94.1 mm/h (= 2.61 × 10−5 m/s), the weighted average runoff coefficient, C̄ is given by C̄ = 0.6(0.9) + 0.4(0.2) = 0.62 Since the area of the catchment is 0.5 ha (= 5000 m2 ), then the peak runoff rate, Qp , from the catchment is given by the rational formula as Qp = C̄iA = (0.62)(2.61 × 10−5 )(5000) = 0.0809 m3 /s Considering only the directly-connected impervious portion of the catchment, the time of concentration is 11 min, the average rainfall rate, i, from the IDF curve is 157 mm/h (= 4.36 × 10−5 m/s), the runoff coefficient, C, is 0.9, the contributing area is 0.3 ha (= 3000 m2 ), and the peak runoff rate, Qp , is given by Qp = CiA = (0.9)(4.36 × 10−5 )(3000) = 0.118 m3 /s The calculated peak runoff from the directly-connected impervious area is greater than the calculated runoff from the entire area, and therefore the design discharge for Inlet 1 and Pipe 1 is controlled by the directly-connected impervious area and is equal to 0.118 m3 /s . Inlet 2: When the entire catchment B is contributing, the time of concentration is 70 min, the average rainfall rate, i, from the IDF curve is 72.7 mm/h (= 2.02 × 10−5 m/s), the weighted average runoff coefficient, C̄ is given by C̄ = 0.15(0.9) + 0.85(0.2) = 0.31

369 Since the area of the catchment is 1 ha (= 10000 m2 ), then the peak runoff rate, Qp , from the catchment is given by the rational formula as Qp = C̄iA = (0.31)(2.02 × 10−5 )(10000) = 0.0626 m3 /s Considering only the directly-connected impervious portion of the catchment, the time of concentration is 12 min, the average rainfall rate, i, from the IDF curve is 154 mm/h (= 4.27 × 10−5 m/s), the runoff coefficient, C, is 0.9, the contributing area is 0.15 ha (= 1500 m2 ), and the peak runoff rate, Qp , is given by Qp = CiA = (0.9)(4.27 × 10−5 )(1500) = 0.0576 m3 /s The calculated peak runoff from the directly-connected impervious area is less than the calculated runoff from the entire area, and therefore the design discharge for Inlet 2 is controlled by the entire catchment and is equal to 0.0626 m3 /s . Pipe 2: First consider the case were the entire tributary area of 1.5 ha (= 15000 m2 ) is contributing runoff to Pipe 2. The time of concentration of catchment A is equal to 45 min plus the time of flow in pipe 1 which, in lieu of hydraulic calculations, can be taken as 2 min. Therefore, the time of concentration of catchment A is 47 min. The time of concentration of catchment B is 70 min, and therefore the time of concentration of the entire tributary area to pipe 2 (including both catchments A and B) is equal to 70 min. The average rainfall intensity corresponding to this duration (from the IDF curve) is 72.7 mm/h (= 2.02 × 10−5 m/s), area-weighted runoff coefficient, C̄, is given by C̄ =

1 [(0.30 + 0.15)(0.9) + (0.20 + 0.85)(0.2)] = 0.41 1.5

and the rational formula gives the peak runoff rate, Qp , as Qp = CiA = (0.41)(2.02 × 10−5 )(15000) = 0.124 m3 /s Consider now the case where only the directly-connected impervious portions of catchments A and B are contributing. In this case, the contributing area is 0.45 ha (= 4500 m2 ), the time of concentration is 13 min (equal to the time of concentration for Inlet 1 plus travel time of 2 min in pipe), the corresponding average rainfall intensity from the IDF curve is 151 mm/h (= 4.19 × 10−5 ), the runoff coefficient is 0.9, and the rational formula gives a peak runoff, Qp , of Qp = CiA = (0.9)(4.19 × 10−5 )(4500) = 0.170 m3 /s Therefore the peak runoff rate calculated by using the entire catchment is less than the peak runoff rate calculated by considering only the directly-connected impervious portion of the tributary area. The design flow for pipe 2 is therefore controlled by the impervious area and is equal to 0.170 m3 /s . 11.17 Pipe I: For all the area, tc = 25 min, C = 0.65(0.9) + 0.35(0.3) = 0.69

370 and eﬀective rainfall, ie , given by ie = Ci = 0.69

6000 6000 = 0.69 = 92 mm/h = 2.56 × 10−5 m/s t + 20 25 + 20

Therefore, peak runoﬀ, Qp , given by Qp = ie A = (2.56 × 10−5 )(104 ) = 0.256 m3 /s For DCIA, tc = 12 min, C = 0.9, and eﬀective rainfall, ie , given by ie = Ci = 0.90

6000 6000 = 0.90 = 169 mm/h = 4.69 × 10−5 m/s t + 20 12 + 20

Therefore, peak runoff, Qp , given by Qp = ie A = (4.69 × 10−5 )(0.65 × 104 ) = 0.305 m3 /s Hence the design flow for pipe I is 0.305 m3 /s . Pipe II: For all the area, tc = 30 min, C = 0.69, and effective rainfall, ie , given by ie = Ci = 0.69

6000 6000 = 0.69 = 82.8 mm/h = 2.3 × 10−5 m/s t + 20 30 + 20

Therefore, peak runoff, Qp , given by Qp = ie A = (2.3 × 10−5 )(104 ) = 0.23 m3 /s Hence the design flow for pipe II is 0.23 m3 /s . Pipe III: For all the area, tc = 30 + 3 = 33 min, A = 2 ha = 2 × 104 m2 , C = 0.69, and effective rainfall, ie , given by ie = Ci = 0.69

6000 6000 = 0.69 = 78.1 mm/h = 2.17 × 10−5 m/s t + 20 33 + 20

Therefore, peak runoﬀ, Qp , given by Qp = ie A = (2.17 × 10−5 )(2 × 104 ) = 0.434 m3 /s For DCIA, tc = 12 + 3 = 15 min, C = 0.9, A = 0.65 ha = 0.65 × 104 m2 , and eﬀective rainfall, ie , given by ie = Ci = 0.90

6000 6000 = 0.90 = 154 mm/h = 4.29 × 10−5 m/s t + 20 15 + 20

Therefore, peak runoﬀ, Qp , given by Qp = ie A = (4.29 × 10−5 )(0.65 × 104 ) = 0.279 m3 /s

371 Hence the design ﬂow for pipe III is 0.434 m3 /s . For a concrete pipe, n = 0.013, and Manning’s equation gives 3.21Qn √ D= S0

3 8

3.21(0.434)(0.013) √ = 0.02

3 8

= 0.462 m

Verify the validity of the Manning equation, n6 RS0 = (0.013)6 (0.462/4)(0.02) = 2.32 × 10−13 > 9.6 × 10−14 Therefore, Manning’s equation is valid. Check the ﬂow velocity: V =

0.434 Q = π = 2.58 m/s 2 A 4 (0.462)

This velocity is adequate to prevent sedimentation and scour. 11.18 For PVC pipe, n=0.013 Discharge, Q=5m3 /s S0 =4%=0.04 For typical value, SX =2%=0.02 By equation 11.6, 8 1 Q = 0.375 nS1 x d 3 × (0.04) 2 Therefore, d = 0.218 m Depth in the pipe is 0.218 m Let ‘d’ be the size of concrete pipe. Again, Top width, T = sdx = 0.218 0.02 , m = 10.9 m Flow area, A = 12 × d × T = 12 × 0.218 × 10.9 = 1.881 V =

5 Q = m/s = 2.658 m/s A 1.881 1

Now, V = n1 R s × s02 23 1 1 2.658 = 0.012 × D × (0.04) 2 4 2

Therefore, D=0.2554 m 11.19 The Darcy-Weisbach equation (Equation 11.44) gives 0.811f Q2 D= gS0 which can be put in the form

1 5

0.811f (0.50)2 = (9.81)(0.009) f = 0.435D5

1 5

= 1.181f 5

(1)

372 The friction factor, f , also depends on D via the Colebrook equation (Equation 2.35) which is given by 1 2.51 ks /D √ √ = −2 log (2) + 3.7 f Re f The equivalent sand roughness, ks , of concrete is in the range 0.3–3.0 mm (Table 2.1) and can be taken as ks = 1.7 mm. Assuming that the temperature of the water is 20◦ C, the kinematic viscosity, ν, is equal to 1.00 × 10−6 m/s2 , and the Reynolds number, Re, is given by Re =

4Q 4(0.50) 6.37 × 105 VD = = = ν πDν πD(1.00 × 10−6 ) D

(3)

Combining Equations 1 to 3 gives √

1 0.435D5

= −2 log

2.51 0.0017/D + 6.37×105 √ 3.7 0.435D5

which simpliﬁes to 1.52D− 2 = −2 log(4.59 × 10−4 D−1 + 5.97 × 10−6 D− 2 ) 5

and yields D = 0.57 m = 57 cm Therefore the Darcy-Weisbach equation requires that the sewer pipe be at least 57 cm in diameter. 8 1 11.20 Q = 0.375 nS1 x d 3 s02 Given n=0.016 S0 =0.01 Sx =1/20 d=0.25m Therefore, Q=1.172m3 /s Area of the ﬂow, A = 12 × T × d Where T = sdx = 0.25 = 5m 1 20

Therefore, A = 12 × 5 × 0.25 = 0.625 m2 1.172 Velocity in the gutter, V = Q A = 0.25 = 1.8752 m/s Again by equation 11.10, 0.6 1 LT = 0.817Q0.42 s0.3 0 nsx Here, n=0.015 S0 =0.03 Sx =1/20

373 Q=0.3m3 /s Therefore, LT =12.89 m From equation 11.15, Qi = 1.25(L + 1.8W0 )d1.5 Here L=LT =12.89m W0 =0.5m d=0.25m Qi =2.155m3/s 11.21 The service manholes placed along the pipe will each cause a head loss, hL , where hL = K

V2 2g

For inﬂow and outﬂow pipes aligned opposite to each other, K is between 0.12 and 0.32, and can be assigned an average value of K = 0.22. Since V = 1.96 m/s, the head loss, hL , is therefore given by (1.96)2 = 0.043 m hL = 0.22 2(9.81) 1

11.22 V = n1 R s S02 2

In case of full ﬂow, D 0.5 R= = = 0.14 m 4 4 Velocity, V = 1.5 m/s Typical value, n = 0.016 1 2 1 1.5 = × (0.14) 3 × s02 0.016 s0 = 0.008 Again from equation 11.48, 1.52 V2 = 0.3 × = 0.037 m 2g 2 × 9.81 Again, for solving the normal depth of ﬂow, (eqn 11.2), hL = k

V = n1 y 3 s02 2

y 3 = 1.5×0.011 2

(0.008) 2

3 0.015 2 y= = 0.069 m 0.089 Required depth from equation 11.49, Δz = y +

v2 + hL 2g

374 = 0.069 + 0.1146 + 0.034 = 0.2176 m However since the pipe diameters at the incoming and outgoing is same, there is no drop. Therefore,invert elevation= (1.55 − Δz) = (15.5 − −0.2178) = 15.2824 m 11.23 From table 10.1, n=0.2 (assume) From equation 10.13, 3 6.99 nL 5 tc = 2 √ s0 ie5 Here n=0.2, L=50 m, s0 =1%=0.01 6.99 0.2×50 35 √ tc = 2 min = 110.78 min 2 0.01 5 ie5 te Again the eqution is given by, 2261 2261 mm/h = i= 0.754 (tc + 8.5)0.754 110.78 2 ie5

+8.5

The eﬀective rainfall rate, ie = Ci = 0.35 ×

2261 0.754 110.78 2 + ie5

Solving this equation we get, ie = 61.1 mm/h = 1.697 × 10−5 m/sec And a corresponding time of concentration of tc =21.38min The peak run-oﬀ rate, QP , is given by the rational formula asQP =CiA=ie A=16.97m3 /s

Chapter 12

Design of Stormwater Management Systems 12.1. From the given data: A = 7.5 ha = 7.5 × 104 m2 , L = 100 m, W = 25 m, and d = 2.0 m. The water-quality volume, WQV, area of the lake, Alake , and ponded height, H0 , are given by WQV = 0.020A = 0.020(7.5 × 104 ) = 1500 m3 Alake = L × W = (100)(25) = 2500 m2 WQV 1500 = 0.600 m H0 = = Alake 2500 Assuming that the WQV is added instantaneously to the lake, the outﬂow hydrograph, Q(t), satisﬁes the continuity equation: dH (1) Q = Alake dt where H is the height of the water surface above the control elevation. Combining Equation 1 with the given weir discharge equation, and substituting Alake = 2500 m2 , gives the following equation describing the stage hydrograph, 5

0.080H 2 = 2500

dH dt

Solving this equation with the initial condition that H = H0 = 0.600 m when t = 0 sec yields H=

1667 3586 − 0.080t

2 3

and combining this result with the given weir discharge equation gives the outﬂow hydrograph as 18749 (2) Q(t) = 5 (3586 − 0.080t) 3 375

376 where Q is in m3 /s and t is in seconds. The evacuation time is calculated using Equation 12.1 which requires that Te O(t)dt = WQV 0

18749

(3586 − 0.080t) 3

dt = 1500 m3

which yields Te = 28975 s = 8.0 h. Therefore, the evacuation time of the WQV in the detention pond is 8.0 h. From the given data, the average annual rainfall, d¯rain , is 1.20 m and the runoff coefficient, C, is 0.75; the average runoff, Q, and the volume of the pond, Vpond , are given by (1.20)(7.5 × 104 )(0.75) d¯rain AC = = 185 m3 /d 365 365 Vpond = LW d = (100)(25)(2.0) = 5000 m3 Q=

Therefore the detention time, Td , is given by Equation 12.2 as Td =

Vpond 5000 = 27 days = 185 Q

In summary, the detention pond will have an evacuation time of approximately 8.0 h and a detention time of approximately 27 days . 12.2. From the given data: D = 1.50 m, Z0 = 1.00 m, Z1 = 2.00 m, Lw = 0.40 m (oriﬁce) and Lw = πD = π(1.50 m) = 4.71 m (top of riser), A0 = 0.40 m × 0.40 m = 0.16 m2 , and it can be assumed that Cw = 1.83 and Cd = 0.6. Substituting these values into Equation 12.9 yields the following results: Z (m) 1.00 1.20 1.40 1.60 1.80 2.00 2.20 2.40

Q Formula – Cw Lw (Z − Z0 )1.5 Cw Lw (Z − Z0 )1.5 Cd A0 2g(Z − Z0 ) Cd A0 2g(Z − Z0 ) Cd A0 2g(Z − Z0 ) Cw Lw (Z − Z1 )1.5 + Cd A0 2g(Z − Z1 ) Cw Lw (Z − Z1 )1.5 + Cd A0 2g(Z − Z1 )

(m3 /s) 0.0 0.065 0.185 0.329 0.380 0.425 0.961 2.45

12.3 The storage volume, S, is given by S=

A0 + Ah 2

(1)

377 where A0 is the area of the base of the reservoir given by A0 = LW and Ah is the area at a height h above the base of the reservoir, given by 2h L Ah = W + tan α

(2)

(3)

Combining Equations 1 to 3 gives S=

L 2 h + (LW )h tan α

12.4 From the given data: A = 5 ha = 50000 m2 , and Q0 = 1 m3 /s. For a development with 0.4-ha (= 1-ac) lots and sandy loam (Type B) soil: CN = 68. For 25 mm of runoff, the runoff volume, V25 , is V25 = (0.025)(50000) = 1250 m3 Determine the stage/storage curve for the retention area. If the base of the retention area is L × L and the ponded depth is h, then the stage/storage curve is given by S = (L + 3h)2 h The maximum allowable ponding depth is 0.3 m and the minimum infiltration rate in Type B soil is 3.8 mm/h. So the time to infiltrate 0.3 m (= 300 mm) is 300/3.8 = 79 h. Since this is greater than the maximum of 72 h, then the maximum height in the retention area has to be reduced to 72 h × 3.8 mm/h = 274 mm = 0.274 m For a ponding depth of 0.274 m, h = 0.274 m, S = 1250 m3 , and the stage/storage curve gives 1250 = (L + 3 × 0.274)2 (0.274) which gives L = 66.72 m. The area at the top of the retention basin is therefore given by A = (L + 6h)2 = (66.72 + 6 × 0.274)2 = 4674 m2 Therefore, reserve 4674/50000 = 9.3% of the site area for retention. For a 25-y 1-day storm, the rainfall depth, P , is given by P = it =

7836(24) 7836t = = 179.2 mm 34.11 + 0.7052t 34.11 + 0.7052(24 × 60)

and the available site storage, S0 , is given by 1000 10 + 0.0394S0 1000 68 = 10 + 0.0394S0

CN =

378 which gives S0 = 119 mm. The runoﬀ depth, Q, is given by Q=

(179.2 − 0.2 × 119)2 (P − 0.2S0 )2 = 88.00 mm = P + 0.8S0 179.2 + 0.8 × 119

and this yields a runoﬀ volume, V0 , given by V0 = (0.088)(50000) = 4400 m3 The height of ponded water corresponding to this runoﬀ volume is derived from the stage/storage curve, V0 = (L + 3h)2 h 4400 = (66.72 + 3h)2 h which yields h = 0.912 m. The elevation of the weir crest is to be 0.274 m above the bottom of the retention area, and the crest width, b, can be initially estimated using the weir equation, 3

Q0 = 1.83bhw2 3

1 = 1.83b(0.912 − 0.274) 2 which gives b = 1.07 m . The final crest length will be determined by routing the runoff hydrograph through the detention basin. The discharge structure will consist of a weir box in which the water from the detention basin enters the box by flowing over the weir and exits the box through a culvert pipe. 12.5 The required detention basin volume is first estimated by subtracting the pre-development runoff volume, V1 , from the post-development runoff volume, V2 . From the given hydrographs: V1 = (30)(60)[2.0 + 7.5 + 1.7 + 0.90 + 0.75 + 0.62 + 0.49 + 0.30 + 0.18 + 0.50] = 26892 m3 and V2 = (30)(60)[3.5+10.6+7.5+5.1+3.0+1.5+0.98+0.75+0.62+0.51+0.25+0.12] = 61974 m3 A preliminary estimate of the required volume, V , of the drainage basin is V = V2 − V1 = 61974 − 26892 = 35082 m3 From the storage-elevation function, the head h corresponding to a storage volume of 35,082 m3 is 1.31 m. The maximum pre-development runoff, Q, is 7.5 m3 /s, and the weir equation gives 3 Q = 1.83bh 2 or b=

Q 1.83h

3 2

7.5 3

1.83(1.31) 2

= 2.73 m

379 Based on this preliminary estimate of the crest length, use a trial length of 2.5 m, and the corresponding weir discharge equation is 3

Q = 1.83bh 2 = 1.83(2.5)h 2 = 4.58h 2 The post-development hydrograph can be routed through the detention basin using Δt = 30 min, and the storage and outﬂow characteristics can be put in in the following form Elevation (m) 0 0.5 1.0 1.5

Storage, S (m3 ) 0 11022 24683 41522

Outﬂow, O (m3 /s) 0 1.62 4.58 8.41

2S/Δt + O (m3 /s) 0 13.87 32.01 54.55

The routing computations (using the modiﬁed Puls method) are summarized in the following table: Time (min) 0 30 60 90 120 150

Inﬂow, I (m3 /s) 0 3.5 10.6 7.5 5.1 3.0

2S/Δt − O (m3 /s) 0 2.68 12.60 21.96 24.54 23.26

2S/Δt + O (m3 /s) 0 3.5 16.78 30.70 34.56 32.64

O (m3 /s) 0 0.41 2.09 4.37 5.01 4.69

From these results, the maximum outflow is less than the pre-development discharge, and therefore a crest length of 2.5 m is adequate. From an economic viewpoint, it is not reasonable to increase the size of the outlet structure in order to make the post-development peak runoff equal to the pre-development peak runoff. In fact, it makes more (economic) sense to reduce the crest length of the outlet structure until the maximum elevation in the storage reservoir is equal to 1.5 m, making maximum use of the reservoir and using the shortest crest length. This condition occurs when b = 0.645 m , in which case the storage-outflow characteristics are given by Elevation (m) 0 0.5 1.0 1.5

Storage, S (m3 ) 0 11022 24683 41522

Outﬂow, O (m3 /s) 0 0.42 1.18 2.17

2S/Δt + O (m3 /s) 0 12.66 28.61 48.30

The routing computations (using the modiﬁed Puls method) are summarized in the following table:

380 Time (min) 0 30 60 90 120 150 180 210 240

Inﬂow, I (m3 /s) 0 3.5 10.6 7.5 5.1 3.0 1.5 1.0 0.8

2S/Δt − O (m3 /s) 0 3.27 16.08 31.26 39.97 43.76 43.93 42.26 40.16

2S/Δt + O (m3 /s) 0 3.5 17.37 34.18 43.86 48.07 48.26 46.41 43.99

O (m3 /s) 0 0.12 0.64 1.46 1.95 2.16 2.17 2.07 1.92

The maximum water elevation in the reservoir is therefore equal to 1.5 m . 12.6 From the given data: A = 5 km2 = 5 × 106 m2 , WQV = 0.025 × 5 × 106 = 125000 m3 , and retention area = 0.5 × 106 m2 . Therefore, height of ponding of WQV =

125000 = 0.25 m 0.5 × 106

So the height of the crest of the discharge (weir) structure is 0.25 m . Need to route the discharge through a weir structure. The discharge equation of a sharpcrested weir is O = 1.83bH 1.5 Taking Δt = 20 min = 1200 s, the storage table is given by Elev (m) 0 0.25 z

S (m3 ) 0 125000 5z × 105

O (m3 /s) 0 0 1.83b(z − 0.25)1.5

2S/Δt + O (m3 /s) 0 208.3 833.3z + 1.83b(z − 0.25)

Try b = 1 m. Routing the runoﬀ through the retention area gives: t (min) 0 20 40 60 80 100 120

I (m3 /s) 0 7.4 14.8 11.1 7.4 3.7 0

2S/Δt − O (m3 /s) 0 7.4 29.6 55.5 74.0 85.1 88.8

O (m3 /s) 0 0 0 0 0 0 0

381 Therefore, retaining the water-quality volume will retain all the runoff on site. The maximum volume of water in the retention area is equal to the runoff volume, which is given by 1 runoff volume, V = (120 × 60)(14.8) = 53280 m3 2 The maximum depth of ponding is therefore given by ponding depth =

53280 V = = 0.107 m Ar 0.5 × 106

12.7 Under pre-development conditions: I = 10%, d = 0.25 cm, P = 98.5 cm, and the (annual) runoff, R, is given by Equation 10.150 as I P − 3.004d0.5957 R = 0.15 + 0.75 100 = [0.15 + 0.75(0.10)](98.5) − 3.004(0.25)0.5957 = 20.8 cm Since there are 62 storms/year, the runoff per storm is 0.208/62 = 0.00335 m/storm. Under post-development conditions: I = 65%, d = 0.15 cm, P = 98.5 cm, and the (annual) runoff, R, is given by Equation 10.150 as I P − 3.004d0.5957 R = 0.15 + 0.75 100 = [0.15 + 0.75(0.65)](98.5) − 3.004(0.15)0.5957 = 61.8 cm Since there are 62 storms/year, the runoff per storm is 0.618/62 = 0.00997 m/storm. Therefore, the required volume, V , of the storage area is given by V = (0.00997 − 0.00335)(10 × 104 ) = 662 m3 From the storage-elevation function, h = 1.10 m. Taking Q = 3 m3 /s, the orifice equation gives Q 3 √ A= = 0.993 m2 = 0.65 2gh 0.65 2(9.81)(1.10) Therefore, the orifice diameter, D, is given by D=

0.993 ×

4 = 1.12 m π

12.8. From the given data: A0 = 5 ha = 5 × 104 m2 , A1 = 2.55 ha = 2.55 × 104 m2 , e0 = 6.200 m, e1 = 6.300 m, P = 25.4 cm, and S = 7 cm. (a) The runoﬀ depth is given by the NRCS equation as Q=

[25.4 − 0.2(7)]2 [P − 0.2S]2 = = 18.6 cm = 0.186 m P + 0.8S 25.4 + 0.8(7)

382 Therefore, the runoﬀ volume, V0 , is given by V0 = QA0 = (0.186)(5 × 104 ) = 9300 m3 Between elevations e0 = 6.200 m and e1 = 6.300 m, the site storage, S01 is given by 2.5 × 104 A1 = (6.300 − 6.200) = 1250 m3 2 2 and the additional above-ground storage required is 9300 − 1250 = 8050 m3 . If h is the height of the berm above elevation 6.300 m, then 8050 8050 h= = = 0.322 m A1 2.5 × 104 S01 = (e1 − e0 )

Therefore the elevation of the top of the berm is 6.300 m + 0.322 m = 6.622 m . (b) For 1.3 cm of runoff, the storage volume required is 0.013 × 5 × 104 = 650 m3 . If e is the top elevation of the V notch, then A1 (e − e0 ) = 650 2 2.5 × 104 (e − 6.200) = 650 2 which yields e= 6.252 m and the height of the V-notch weir is 0.052 = 52 mm. The maximum discharge rate from the weir is 650/(24 × 60 × 60) = 0.00752 m3 /s. According to the Kindsvater and Shen equation 5 θ 8 Q = Cd 2g tan (H + k) 2 15 2 8 θ −6 2 0.00752 = (0.6072 − 0.000874θ + 6.1 × 10 θ ) 2(9.81) tan 15 2 5 52 + 4.42 − 0.1035θ + 1.005 × 10−3 θ2 − 3.24 × 10−6 θ3 2 1000 which yields a notch angle of θ = 166◦ . 12.9. Since the infiltration capacity of the soil is 50 mm/h, the runoff versus time is given in the following table: Time (min) 0–5 5–10 10–15 15–20 20–25 25–30 30–35 35–40 40–45 Total

Rainfall Rate (mm/h) 24.1 30.4 41.6 67.4 169. 97.3 51.6 36.2 26.8 544.4

Runoﬀ Rate (mm/h) 0.0 0.0 0.0 17.4 119. 47.3 1.6 0.0 0.0 185.3

383 Since Δt = 5 min = 0.0833 h, and the catchment area, A, is 5 ha = 50 000 m2 , the volume of runoﬀ, V , is given by V = (185.3 × 10−3 Δt)(A) = (185.3 × 10−3 × 0.0833)(50000) = 772 m3 Therefore the storage volume required in the retention area is 772 m3 . Since the retention area covers 700 m2 , the depth, d, of ponding expected in the retention area during the design storm is 772 = 1.10 m d= 700 12.10. From the given data: A = 4 ha = 4 × 104 m2 , L = 200 m, S0 = 0.5%, ADCIA = 0.8 ha = 8000 m2 , LDCIA = 80 m, and SDCIA = 0.7%. The IDF curve for Miami for T = 25 years is given by 7836

48.6T −0.11 + t(0.5895 + T −0.67 )

48.6(25)−0.11 + t(0.5895 + 25−0.67 )

mm/h

7836

mm/h

7836 mm/h 34.1 + 0.705t

(1)

First, consider the entire contributing area. According to Table 10.7, for single-family residential areas, C = 0.4 is a mid-range value. This value should be increased by 10% to account for a 25-year return period storm, which yields C = 0.44. Using the kinematic wave equation (Equation 10.13) with n = 0.25 (typical of lawns in Table 10.1) gives (nL)0.6 (0.25 × 200)0.6 358 = 0.4 min tc = 6.99 0.4 0.3 = 6.99 0.4 0.3 ie (0.005) ie ie S0

(2)

Taking t = tc in Equation 1 and substituting Equation 2 gives ie = Ci = 0.44

7836 34.1 + 0.705 i358 0.4

mm/h

which yields ie = 36.8 mm/h = 1.02 × 10−5 m/s, and tc = 89 min. The peak runoﬀ from the entire catchment is therefore given by Qp = CiA = ie A = (1.02 × 10−5 )(4 × 104 ) = 0.41 m3 /s Considering the DCIA consisting of asphalt pavement, Table 10.1 gives a mid-range value of C = 0.83, which should be increased by 10% to account for the 25-year return period of the storm, which yields C = 1.1(0.83) = 0.91. Using the kinematic wave equation with nDCIA = 0.011 (Table 10.1 for asphalt) gives tc = 6.99

(nDCIA LDCIA )0.6 (0.011 × 80)0.6 28.7 = 6.99 = 0.4 min 0.3 ie0.4 (0.007)0.3 ie ie0.4 SDCIA

(3)

384 Taking t = tc in Equation 1 and substituting Equation 3 gives ie = Ci = 0.91

7836 34.1 + 0.705 28.7 i0.4

mm/h

= 195 mm/h = 5.28 × 10−5 m/s, and t

which yields ie c = 3.5 min. Since tc < 5 min, take tc = 5 min, which gives an effective rainfall of 7836 = 190 mm/h = 5.28 × 10−5 m/s ie = 0.91 34.1 + 0.705(5) The peak runoff from the entire catchment is therefore given by Qp = CiADCIA = ie ADCIA = (5.28 × 10−5 )(8000) = 0.42 m3 /s Therefore the peak runoff is determined by the DCIA and is estimated as 0.42 m3 /s . For a 25-year 3-day storm, the average intensity can be determined by taking t = 3 days = 4320 min in the equation for the IDF curve, which gives i=

7836 = 2.54 mm/h 34.1 + 0.705(4320)

which gives a 3-day rainfall amount, P , of P = 2.54(24)(3) = 183 mm For a residential area with 1/2-acre lots on Group B soil, the curve number, CN, can be estimated from Table 9.14 as CN = 70, where CN =

1000 10 + S

which yields S = 4.29 in. = 109 mm, and hence the runoﬀ, Q, is given by Q=

(P − 0.2S)2 (183 − 0.2 × 109)2 = = 96 mm P + 0.8S 183 + 0.8 × 109

The volume of runoﬀ, V , to be stored in the lake can be estimated by multiplying the runoﬀ depth, Q (= 96 mm = 0.096 m) by the site area, A (= 4 × 104 m2 ), hence V = QA = (0.096)(4 × 104 ) = 3840 m3 At the control elevation, the receiving lake is 15 m × 15 m and has 6:1 side slopes above the control elevation. For a height, h, above the control elevation, the volume stored in the lake, Vs , can be estimated by Vs =

1 [15 × 15 + (15 + 12h)(15 + 12h)] h = 9h(25 + 20h + 8h2 ) 2

When Vs = 3840 m2 , then

3840 = 9h(25 + 20h + 8h2 )

which yields a (real) solution of h = 2.87 m, which corresponds to a setback distance of 2.87 m × 6 = 17.2 m .

385 12.11 The volume, V of runoff corresponding to a depth of 2.5 cm = 0.025 m on an area of 20 ha = 2 × 105 m2 is given by V = (0.025)(2 × 105 ) = 5000 m3 At a minimum infiltration rate of 100 mm/h = 0.10 m/h, the maximum area, A of infiltration basin required to infiltrate 5000 m3 in 36 h is given by A=

5000 = 1390 m2 (0.10)(36)

12.12. For T = 10 years, the average storm intensity for a storm of duration t (in minutes) is given by 7836 7836 = i= −0.11 −0.67 48.6(10) + t(0.5895 + 10 ) 37.72 + 0.8034t The rainfall depth, D, for a storm of duration t (in minutes) is given by 130.6t 7836 t it = = D= 60 37.72 + 0.8034t 60 37.72 + 0.8034t

(1)

For the maximum value of D, (37.72 + 0.8034t)(130.6) − (130.6)(0.8034) dD = dt (37.72 + 0.8034t)2

(2)

When dD/dt = 0, Equation 2 gives 4926 + 104.9t − 104.9t = 0 which yields 492.6 = 0 which is clearly impossible. Therefore, there is no maximum rainfall amount, and the rainfall amount monotonically increases with the duration of the storm. As t → ∞, then dD/dt → 0 and Equation 1 gives 130.6t = 163 mm D→ 0.8034t For a site with a runoff coefficient of 0.4, the maximum amount of runoff is 0.4 (163 mm) = 65 mm. If the site is capable of retaining 50 mm of runoff, then 65 mm − 50 mm = 15 mm of runoff will be discharged from the site. Therefore, yes a portion of the 10-year runoff will be discharged from the site. 12.13 The length of the swale is determined by Q fP

(1)

2 1 1 AR 3 S 2 n

(2)

L= and the Manning equation requires Q=

386 For a triangular channel, A = my 2 P = 2 1 + m2 y

(3) (4)

where y is the depth of ﬂow in the channel. Combining Equations 2 to 4 and re-arranging yields 3 3 2 1 1 n 8 Q 8 (1 + m ) 8 (5) y = 24 5 3 m 8 S 16 and combining Equations 1 and 4 yields L=

Q 2f 1 + m2 y √

(6)

Finally, combining Equations 5 and 6 to eliminate y gives 5

L = 151400

Q 8 m 8 S 16 3

n 8 (1 + m2 ) 8 f

where the conversion factor has been included for f in cm/h. 12.14 From the Manning equation: Q=

2 1 1 AR 3 S 2 n

(1)

and for a trapezoidal section A = (b + my)y P = b + 2y 1 + m2

(2) (3)

Combining Equations 1 to 3 and simplifying gives 5

1 y 3 (b + my) 3 Q= √ n (b + 2y 1 + m2 ) 23

(4)

For the most eﬃcient trapezoidal section b = 2( 1 + m2 − m) y

(5)

b 2( 1 + m2 − m)

(6)

or y=

√

Combining Equations 4 and 6 gives 5

√ bm 1 y 3 b + 2( 1+m2 −m)

√

√ 1+m b + 2(2b 1+m2 −z)

2 3

5 3 1

(7)

387 Solving for y gives Q y=

√ 1+m2 b + (√b1+m 2 −m)

3 5

2 5

(8)

bm S 10 b + 2(√1+m 2 −m)

Combining Equations 8 and 5 to eliminate b gives 3 8

√ 1 (2 1 + m2 − m)S 2

y = 1.19

(9)

Combining Equations 3 and 9 gives the following expression for P , P = b + 2.38

3 8

√ 1 (2 1 + m2 − m)S 2

1 + m2

(10)

The swale length, L, is given by Q fP 3 If L is in m, Q is in m /s, and f in cm/h, then L=

L = 360000

Q fP

(11)

Combining Equations 10 and 11 gives L=

360000Q

b + 2.38

3 8

Qn √ 1 (2 1+m2 −m)S 2

√

1 + m2

(12) f

12.15 From the given data: Q = 0.01 m3 /s, m = 5:1 (H:V) = 5, S = 0.015, n = 0.030, f = 200 mm/h = 20 cm/h, and Equation 12.11 gives the required swale length, L, as 5

L = 151400

n 8 (1 + m2 ) 8 f 5

= 151400

Q 8 m 8 S 16 5

(0.01) 8 (5) 8 (0.015) 16 3

(0.030) 8 (1 + 52 ) 8 20

= 257 m 12.16 With b = 1 m, Equation 12.12 gives L=

Qn √ 1 (2 1+m2 −m)S 2

b + 2.38 =

1 + 2.38

= 107 m

360000Q

3 8

360000(0.01) (0.01)(0.03) √ 1 (2 1+52 −5)(0.015) 2

√

1 + m2

3 8

√

1 + 52

388 12.17 From the given data: Q = 0.002 m3 /s, So = 0.02, and the average height of the vegetation is 100 mm. This given data covers the specifications in steps 1 to 3 of the design procedure. Step 4: The design depth in the swale should be at least 50 mm below the height of the vegetation (100 mm − 50 mm = 50 mm), with a maximum height of 75 mm. Therefore, in this case, the design flow depth is taken as 50 mm. Step 5: Use a trapezoidal section for the swale. Step 6: Use side slopes of 4:1 (m = 4), a bottom width b, and a depth y = 50 mm (= 0.050 m). The flow area, A, wetted perimeter, P , and hydraulic radius, R, are given by A = by + my 2 = b(0.050) + (4)(0.050)2 = 0.050b + 0.01 P = b + 2 1 + m2 y = b + 2 1 + 42 (0.050) = b + 0.412 0.050b + 0.01 A = R= P b + 0.412

(1)

where 0.6 m < b < 2.5 m. Step 7: The Manning equation requires that 5

1 2 1 A 3 12 1 AR 3 S02 = S n n P 23 0

In this case,

0.002 = or

1 1 (0.050b + 0.01) 3 (0.01) 2 n (b + 0.412) 23

1 (0.050b + 0.01)5 = 8 × 10−6 n3 (b + 0.412)2

(2)

For Class E retardance, Manning’s n is given by Equation 5.55 as 1

1.22R 6 n= 52.1 + 19.97 log(R1.4 S00.4 )

(3)

Simultaneous solution of Equations 1, 2 and 3 gives b = 2.0 m and the corresponding ﬂow velocity (= Q/A) is 0.0182 m/s. Since the ﬂow velocity is less than the maximum velocity of 0.3 m/s, a swale with a bottom width of 2.0 m should be used. Step 8: Using a detention time of 5 minutes with the design velocity of 0.0182 m/s gives the length, L, of the swale as L = V t = (0.0182)(5 × 60) = 5.46 m which is shorter than the minimum length of 30 m. Therefore use a length of 30 m.

389 In summary, a swale with a a trapezoidal cross-section, a bottom width of 2.00 m , side slopes of 4:1, and 30 m long should be used. 12.18 From the given data: Q = 0.03175 m3 /s and Kt = 10 m/d = 1.157 × 10−4 m/s. The length,L, of the exﬁltration trench is given by L=

Qt CiAt = [nW + Kt t]H [nW + Kt t]H

Taking W = 1 m, H = 2 m, and n = 0.4 gives L=

0.03175t (0.03175)t = −4 [(0.4)(1) + (1.157 × 10 )(2)]H 0.8 + 2.314 × 10−4 t

In this case L increases monotonically with t, so take t → ∞, which yields L = 137 m. With a factor of safety of 2, the trench length should be 274 m. So the dimensions of the exﬁltration trench are 1 m × 2 m × 274 m . 12.19 From the given data: C = 0.5, A = 3 ha = 30000 m2 , and Kt = 35/3 = 11.7 m/d = 0.486 m/h (using a safety factor of 3). According to ASCE (1996) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform eﬃciently. Substituting these values into Equation 12.24 gives (0.5)i(30000)t 15000it CiAt = = (1) L= (nW + Kt t)H (0.4 × 1 + 0.486t)(2) 0.8 + 0.972t where i in m/h, and t in hours are related by i=

0.403 (60t + 8.16)0.69

m/h

(2)

Combining Equations 1 and 2 gives the required trench length, L, as a function of the storm duration, t, as 6045t L= (3) (0.8 + 0.972t)(60t + 8.16)0.69 Taking the derivative with respect to t gives dL dt

= =

[(0.8 + 0.972t)(60t + 8.16)0.69 ]6045 − 6045t[(0.8 + 0.972t)0.69(60t + 8.16)−0.31 60 + (60t + 8.16)0.69 0.972] [(0.8 + 0.972t)(60t + 8.16)0.69 ]2 6045(0.8 + 0.972t)(60t + 8.16)0.69 − 250260t(0.8 + 0.972t)(60t + 8.16)−0.31 − 5876t(60t + 8.16)0.69 [(0.8 + 0.972t)(60t + 8.16)0.69 ]2

and the maximum-value criterion, dL/dt = 0, yields 6045(0.8 + 0.972t)(60t + 8.16)

0.69

− 250260t(0.8 + 0.972t)(60t + 8.16)

−0.31

− 5876t(60t + 8.16)

0.69

which gives t = 0.628 h, and substituting into Equation 3 yields L = 192 m. On the basis of these results, a trench length of 192 m gives the trench volume required to handle the design storm without causing ponding. Since the seasonal high water table is 4.6 m below the ground surface, and the minimum allowable spacing between the bottom of the trench and the water table is 1.2 m, a (maximum) trench height of 4.6 m − 1.2 m = 3.4 m would still be adequate. The trench is designed with dimensions 1 m × 2 m × 192 m .

390 12.20. The required length of trench is given by L=

CiAt (nW + Kt t)H

(1)

From the given data: W = 1 m, Kt = 5/2 = 2.5 m/d = 0.104 m/h (assuming a factor of safety of 2), H = 3 m, and for the gravel pack it can be assumed that n = 0.4. Taking the entire contributing area, then the average runoﬀ coeﬃcient, C, is given by C = 0.35(0.9) + 0.65(0.4) = 0.58 where the contributing area, A, is 0.03 ha = 300 m2 . Substituting known values into Equation 1 gives 58it (0.58)i(300)t = (2) L= (0.4 × 1 + 0.104 × t)(3) 0.4 + 0.104t In this case, i is in m/h and t is in hours. Using these same units in the IDF equation gives i=

7.836 7836(10−3 ) = 37.7 + 0.803(60)t 37.7 + 48.2t

(3)

Combining Equations 2 and 3 yields 454.5t 7.836 58t = L= 0.4 + 0.104t 37.7 + 48.2t 15.1 + 23.2t + 5.01t2 The maximum-value criterion requires that (15.1 + 23.2t + 5.01t2 )(454.5) − 454.5t(23.2 + 10.02t) dL = =0 dt (15.1 + 23.2t + 5.01t2 )2 which yields t = 1.74 h and hence L = 11.2 m. Considering only the DCIA, C = 0.9, A = 0.35(300) = 105 m2 and Equation 1 gives L=

(0.9)i(105)t 31.5t = 0.4 × 1 + 0.104t 0.4 + 0.104t

Since this value of L is equal to a fraction of the value of L given previously by Equation 2, it is clear that the minimum value of L will still occur at t = 1.74 h, and the corresponding value of L will be less than 11.2 m. Therefore, the required trench length is 11.2 m . For t = 1.74 h (= 6264 s) the average intensity is given by Equation 3 which yields i=

7.836 7.836 = = 0.0645 m/h 37.7 + 48.2t 37.7 + 48.2(1.74)

Therefore the depth of runoff retained by the trench is equal to Cit = (0.58)(0.0645)(1.74) = 0.065 m = 6.5 cm . 12.21. The first step is to determine whether the exfiltration trench should be designed for a rainfall return period of 10 years or a return period of a storm in which 25 mm of runoff occurs in 1

391 hour. Since the runoff coefficient is 0.6, a rainfall of 25/0.6 = 42 mm is expected to cause a runoff of 25 mm. Taking i = 42 mm/h and t = 60 min in the IDF curves yields 42 =

7836 48.6T −0.11 + 60(0.5895 + T −0.67 )

which gives T = 0.48 years. Therefore, designing the trench for a rainfall with a 10-year return period (T = 10 years) will result in an exfiltration trench capable of retaining/exfiltrating more than 25 mm of runoff in 1 hour. For T = 10 years, the design IDF curve is given by i=

7836 48.6(10)−0.11 + t(0.5895 + 10−0.67 )

7836 37.73 + 0.8033t

(1)

From the given data: C = 0.6, A = 0.5 ha = 5000 m2 , and Kt = 8/2 = 4 m/d (using a safety factor of 2) = 0.167 m/h. According to ASCE (1998) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform eﬃciently. Since the depth to the water table is 3.5 m, there is suﬃcient space to accommodate a trench depth of 2 m and a vertical separation of at least 1.2 m between the bottom of the trench and the water table. Substituting these values into Equation 12.24 gives L=

(0.6)i(5000)t 3000it CiAt = = (nW + Kt t)H (0.4 × 1 + 0.167t)(2) 0.8 + 0.333t

(2)

where Equation 1 gives i=

7.836 7.836 = m/h 37.73 + 0.8033(60t) 37.73 + 48.20t

(3)

Combining Equations 2 and 3 gives the required trench length, L, as a function of the storm duration, t, as L=

23510t 1465t = 2 (0.8 + 0.333t)(37.73 + 48.20t) t + 3.176t + 1.880

(4)

Taking the derivative with respect to t gives (t2 + 3.176t + 1.880)(1465) − (1465t)(2t + 3.176) 1465(t2 − 1.880) dL = = − dt (t2 + 3.176t + 1.880)2 (t2 + 3.176t + 1.880)2 and the maximum-value criterion, dL/dt = 0, yields t2 − 1.880 = 0 which gives t = 1.37 h, and substituting in Equation 4 yields L = 248 m. On the basis of these results, a 248 m × 1 m × 2 m (L × W × H) trench is capable of handling all 10-year storms without causing surface ponding. Alternative Solution: The above solution applies the factor of safety to the trench hydraulic conductivity. In many

392 cases, the factor of safety is applied to the trench length instead of the trench hydraulic conductivity. In this case, the solution is as follows: The first step is to determine whether the exfiltration trench should be designed for a rainfall return period of 10 years or a return period of a storm in which 25 mm of runoff occurs in 1 hour. Since the runoff coefficient is 0.6, a rainfall of 25/0.6 = 42 mm is expected to cause a runoff of 25 mm. Taking i = 42 mm/h and t = 60 min in the IDF curves yields 42 =

7836 48.6T −0.11 + 60(0.5895 + T −0.67 )

7836 7836 = 48.6(10)−0.11 + t(0.5895 + 10−0.67 ) 37.73 + 0.8033t

(5)

From the given data: C = 0.6, A = 0.5 ha = 5000 m2 , and Kt = 8 m/d = 0.333 m/h. According to ASCE (1998) guidelines, the porosity, n, of the gravel pack can be taken as 40% (n = 0.4), and a trench width, W , of 1 m and height, H, of 2 m can be expected to perform eﬃciently. Since the depth to the water table is 3.5 m, there is suﬃcient space to accommodate a trench depth of 2 m and a vertical separation of at least 1.2 m between the bottom of the trench and the water table. Substituting these values into Equation 12.24 gives L=

(0.6)i(5000)t 3000it CiAt = = (nW + Kt t)H (0.4 × 1 + 0.333t)(2) 0.8 + 0.0.667t

(6)

where Equation 5 gives i=

7.836 7.836 = m/h 37.73 + 0.8033(60t) 37.73 + 48.20t

(7)

Combining Equations 6 and 7 gives the required trench length, L, as a function of the storm duration, t, as L=

731.3t 23510t = 2 (0.8 + 0.0.667t)(37.73 + 48.20t) t + 1.982t + 0.939

(8)

Taking the derivative with respect to t gives (t2 + 1.982t + 0.939)(731.3) − (731.3t)(2t + 1.982) 731.3(t2 − 0.939) dL = = − dt (t2 + 1.982t + 0.939)2 (t2 + 1.982t + 0.939)2 and the maximum-value criterion, dL/dt = 0, yields t2 − 0.939 = 0 which gives t = 0.939 h, and substituting in Equation 8 yields L = 187 m. Applying a factor of safety of 2 yields a required trench length of 2 × 187 m = 374 m. On the basis of these results, a 374 m × 1 m × 2 m (L × W × H) trench is capable of handling all 10-year storms without causing surface ponding.

393 12.22. (a) In accordance with the definition of the trench hydraulic conductivity, the maximum flow rate out of the side of the trench and above the water table is given by Qu = 2KLDu (H2 − 0.5Du ) and the maximum flow rate out of the side of the trench and below the water table is given by Qs = 2KLDs H2 Hence the maximum total flow is given by Q = Qs + Qu = 2KL[Du (H2 − 0.5Du ) + Ds H2 ] which gives L=

Q K(2H2 Du − Du2 + 2H2 Ds )

(1)

(b) For tc = 10 minutes and T = 5 years, the average rainfall intensity is 308.5

48.6T −0.11 + t(0.5895 + T −0.67 )

48.6(5)−0.11 + (10)(0.5895 + 5−0.67 )

308.5

= 6.17 in./h = 4.35 × 10−5 m/s

From the given data, H2 = 2.088 m − 1.250 m = 0.838 m Du = 2.088 m − 0.305 m − 1.250 m = 0.533 m Ds = 1.250 m − (−2.484 m) = 3.734 m K = 1.74 × 10−3 s−1 Q = CiA = (0.6)(4.35 × 10−5 )(0.8 × 10−4 ) = 0.209 m3 /s Substituting into Equation 1 gives Q K(2H2 Du − Du2 + 2H2 Ds ) 0.209 = 17.5 m = 1.74 × 10−3 [2(0.838)(0.533) − (0.533)2 + 2(0.838)(3.734)]

a b+t

where i is the rainfall intensity in m/s, t is the duration in seconds, and a and b are constants given by 0.130 0.5895 + T −0.67 2916T −0.11 b= 0.5895 + T −0.67

a =

394 The runoff volume, V , resulting from a storm event of duration t is given by V = CiAt The storage volume, Vs , in the trench is given by V0 = 0.5W Du L and the exfiltrated volume, VS , out of the sides of the trench in time t is given by VS = 2KL[Du (2 −0.5Du ) + Ds H2 ]t Equating the runoff volume to the stored and exfiltrated volume gives V = V 0 + VS CAa t = 0.5W Du L + 2KL[Du (2 −0.5Du ) + Ds H2 ]t b+t which can be rearranged and put in the form L=

at (b + t)(c + dt)

(2)

where 0.130CA 0.5895 + T −0.67 2916T −0.11 b= 0.5895 + T −0.67 c = 0.5W Du

d = 2K[Du (H2 − 0.5Du ) + Ds H2 ] The maximum length occurs at t = t∗ , where dL =0 dt t=t∗ Combining Equations 2 and 3 yields ∗

t =

bc d

and the (maximum) required trench length is given by L=

at∗ (b + t∗ )(c + dt∗ )

(3)

395 (d) From the given data, 0.130(0.6)(0.8 × 104 ) 0.130CA = = 671 0.5895 + T −0.67 0.5895 + 5−0.67 2916(5)−0.11 2916T −0.11 = = 2628 b= 0.5895 + T −0.67 0.5895 + 5−0.67 c = 0.5W Du = 0.5(1)(0.533) = 0.267

d = 2K[Du (H2 − 0.5Du ) + Ds H2 ] = 2(1.74 × 10−3 )[(0.533)(0.838 − 0.5 × 0.533) + (3.734)(0.838)] = 0.01195 bc (2628)(0.267) = = 242 s = 4.04 min t∗ = d 0.01195 (671)(242) = 17.9 m L= (2628 + 242)(0.267 + 0.01195 × 242) Choose a trench length of 17.9 m (instead of 17.5 m) to be more conservative. The diﬀerence is about 2%. (e) For T = 5 years and t0 = 10 minutes, i0 = 4.35 × 10−5 m/s 0.130CA 0.130CA = = 0.141CA a= −0.67 0.5895 + T 0.5895 + 5−0.67 b = 2628 c bc 2628c = → = 3.8 × 10−4 t∗ t∗ = d d d a 4.35 × 10−5 0.141 Ci0 A = → a = 3230dL0 L0 = d d (3230dL0 )t∗ at∗ = L= (b + t∗ )(c + dt∗ ) (2628 + t∗ )(c + dt∗ ) 3230t∗ L = L0 (2628 + t∗ ) dc + t∗ Since c/d << t∗ , 3230 L = L0 2628 + t∗ which demonstrates that L > L0 when t∗ < 600 s, which means that c/d < 3.8 × 10−4 (600) = 0.228, which, in terms of trench parameters gives 0.5W Du < 0.228 2K[Du (H2 − 0.5Du ) + Ds H2 ] 12.23. From the given data: A = 0.5 ha = 5000 m3 , dWQV = 2.5 cm = 0.025 m, Kt = 10 m/d, W = 1 m, H = 1.5 m, and n = 0.4. The water-quality volume, WQV, is WQV = A × dWQV = (5000)(0.025) = 125 m3

396 If L is the length of the trench required to store the WQV, then nLW H = WQV (0.4)L(1)(1.5) = 125 m3 which yields L = 208 m. The evacuation time, Te , of the runoﬀ stored in the trench is given by nW nLW H (0.4)(1) = = 0.04 days = 1 hour = Te = Kt HL Kt (10) A trench length of 208 m will provide adequate retention storage. The evacuation time of 1 h is much less than the regulatory limit of 3 days.

Chapter 13

Estimation of Evapotranspiration 13.1. The mean daily evaporation from free surface= 0.7 × 40 mm/d/unit area = 28 mm/d/unit area Surface area of the canal stream = 30 × (30 × 1000) m2 = 9× 105 m2 = 90 ha 28 ) m/d = 2.52 m/d Losses due to evaporation = 90×( 1000

13.2. According to equation no 13.1 ET = ρw1 λ [

(Rn −G)+ρa Cp esr−ea a ] m/d +γ(1+ rrs ) a

Data given, P = 760 mm Hg = 1 atm = 101.32 kPa Tavg = 30◦ C According to equation no 13.49 P Depth of air, ρa = 3.450 T +273 101.32 = 3.450 × (30+273) = 1.154 kg/m3

Taking density of water, ρw = 998.2 kg/m3 Latent heat of vaporization, λ = 2.45 MJ/kg. Speciﬁc heat of moist air, Cp = 1.013 kJ/(Kg◦ C) = 1.013×10−3 MJ/(kg◦ C) According to Equation (13.36) Psychometric constant, γ = 0.0016286 Pλ kPa/◦ C ◦ = 0.0016286 × 101.32 2.45 = 0.0674 kPa/ C

According to equation no 13.38

17.27T es (T ) = 0.6108 exp T + 237.3 17.27 × 30 = 0.6108 exp 30 + 237.3 = 4.24 kPa 403

404 Vapour pressure of air, ea =13 mm Hg = 1.73 kPa The slope of the vapour pressure vs. temperature curve, is According to equation no 13.42 =

4098[0.6108 exp

17.27 T T +237.3 2

(T + 237.3) 4098 × 4.24 ◦ ◦ = 2 KPa/ C = 0.243 KPa/ C (30 + 237.3)

Soil heat ﬂux, G = 0 MJ/m2 . d (assumed) The aerodynamic resistance (ra ) of open water can be estimated by According to equation no 13.8 ra =

4.72[ln ZZm0 ]

1 + 0.536U2 Where, Zm = measurement height above water surface = 2 m Z0 = aerodynamic roughness of the surface =1.37 mm (assumed) Wind speed, U2 = 2.5 m/s ra =

4.72[ln 2×1000 ] 1.3 1+0.536×2.5 s/m

= 14.8 s/m = 1.713×10−4 d/m Bulk surface resistance, rs = 70 s/m = 8.101×10−4 d/m (For a standard grass reference surface with height of 0.12 m & astomatal resistance of 100 s/m) According to equation no 13.13 Net radiation, Rn = Sn + Ln According to Equation 13.14, Sn = (1 − α)Rs From Table 13.1, for open water, α = 0.08 n )S0 According to equation no 13.17 Rs = (as + bs N

Where, N = no. of bright sunshine hrs/day = 7.5 hr Use as = 0.25 & bs = 0.50 N = total no. of day light hrs. in the day = 7.5 (for clear days i.e. n = N) Use φ = 23.63◦ = 0.412 rad. According to Equation 13.18 S0 = [ 1440 π Gsc dr (Ws Sinφ Sinδ + Cosφ Cosδ SinWs )] Use, GSC = 0.0820 MJ/(m2 -min)

405 J = Julian day for mid February = 4 According to Equation 13.19 2π dr = 1 + 0.033 cos ( 365 J) = 1.024

& According to Equation 13.20,

2π δ = 0.4093 sin ( 365 J − 1.405) = −0.241 rad.

According to Equation 13.21 Ws = Cos−1 [−tan φ tanδ] = 1.463 rad. 1440 Sn = (1 − 0.08) [0.25 + 0.50 × 1] (0.0820)(1.024) [1.463 Sin (0.412) [Sin (−0.241) π +Cos (0.412) Cos (−0.241) Sin(1.463)]} = 19.78 MJ/(m2 .d) Again The net long wave radiation, Ln is √ Rs − 0.35) Ln = −σT 4 (0.34 − 0.14 ea ) (1.35 Rs0 Where, σ = 4.903 × 10−9 MJ/m2 /K4 /d Rs = (as + bs

n )S0 = (0.25 + 0.5 × 1)S0 = 0.75S0 N

Where S0 is the extraterrestrial radiation & the clear sky radiation RS0 (taking z = 0) Rs0 = [0.75 + 2 × 10−5 × 0]S0 = 0.75S 0 √ Ln = − 4.903 × 10−9 × (303)4 0.34 − 0.14 1.73 (1.35 × 1 − 0.35) = − 6.44 MJ/(m2 .d) Rn = Sn + Ln = 19.78 − 6.44 = 13.34 MJ/m2 · · · d (4.24−1.73) 0.243 (13.34 − 0) + 1.154(1.013 × 10−3 ) 1.713×10 −4 1 [ ET = ] −4 8.10×10 998.2 × 2.45 0.243 + 0.0674(1 + −4 ) 1.713×10 −4 20.37 = 4.09 × 10 0.629 = 0.0132 m/d = 13.2 mm/d

13.3 The evapotranspiration rate from any vegetated surface is given by the Penman-Monteith (PM) equation (Equation 13.1) as ⎤ ⎡ es −ea − G) + ρ c Δ(R 1 ⎣ n a p ra ⎦ ETc = ρw λ Δ + γ 1 + rs ra

406 The calculations to determine the variables in the PM equation are as follows • Adjust wind speed to 2 m above the ground. For grass, u2 =

4.87 4.87 uz = u3 = 0.921u3 ln[67.8z − 5.42] ln[67.8(3) − 5.42]

For alfalfa, h = 0.50 m 2 2 d = h = (0.50) = 0.333 m 3 3 zom = 0.123h = 0.123(0.50) = 0.615 m ln 2−d ln 2−0.333 zom 0.0615 uz = u3 = 0.875u3 u2 = u2 = ln 3−0.333 ln z−d 0.0615 zom • Calculate resistance factors. For grass, 208 0.00241 s/m = d/m u2 u2 rs = 70 s/m = 8.10 × 10−4 d/m

ra =

and for alfalfa, 110 0.00127 s/m = d/m u2 u2 rs = 45 s/m = 5.21 × 10−4 d/m

ra =

• The net radiation, Rn , is given by Rn = (1 − α)Rs + Ln where α can be taken as 0.23 for grass and 0.17 for alfalfa. The extraterrestrial solar radiation, S0 , is given by S0 =

24(60) Gsc dr (ωs sin φ sin δ + cos φ cos δ sin ωs ) π

where Gsc = 0.0820 MJ/(m2 ·min 2π J = 1 + 0.033 cos(0.0172J) dr = 1 + 0.033 cos 365 2π J − 1.405 = 0.4093 sin(0.0172J − 1.405) δ = 0.4093 sin 365 ωs = cos−1 (− tan φ tan δ)

407 The elevation of Greeley, Colorado, is 1462.4 m (z = 1462.4 m), and hence the clear-sky radiation, Rso , can be estimated by Rso = [0.75 + 2 × 10−5 z] = [0.75 + 2 × 10−5 (1462.4)]S0 = 0.78S0 The net longwave radiation, Ln , can be estimated by 4 4 Tmax,K + Tmin,K √ Rs (0.34 − 0.14 ea ) 1.35 − 0.35 Ln = −σ 2 Rs0 For daily time intervals, it can be assumed that G = 0 MJ/(m2 ·d • Take λ = 2.45 MJ/kg • The pressure, p, at elevation z = 1462.4 m is given by 293 − 0.0065(1462.4) 5.26 293 − 0.0065z 5.26 = 101.3 = 85.2 kPa p = 101.3 293 293 • The psychrometric constant, γ, is given by γ = 0.0016286

85.2 p = 0.0016286 = 0.0566 kPa/◦ C λ 2.45

• The saturation vapor pressure, es , can be estimated by 17.27Tmin 17.27Tmax + exp es = 0.3054 exp Tmax + 237.3 Tmin + 237.3 • The vapor pressure gradient, Δ, can be estimated by Δ=

4098es (T + 237.3)2

where T is the average of the maximum and minimum temperatures. • The air density, ρa , can be estimated using the realtion ρa = 3.450

p kg/m3 T + 273

• The speciﬁc heat of moist air, cp , can be taken as 1.013 × 10−3 MJ/(kg·◦ C). • The density of water, ρw , can be taken as 998 kg/m3 . The above-itemized relations are used in the Penman-Monteith equation to calculate the reference grass and alfalfa ET, and these calculations are tabulated in Figure 13.1. These results indicate that, over the 7-day interval, the grass-reference ET varies in the range of 5.8–7.8 mm/d , and the alfalfa-reference ET varies in the range 7.7–10.1 mm/d . These results show that over a 7-day interval there can be signiﬁcant variation in the reference ET (up to around 50%).

Figure 13.1: Daily ET Spreadsheet

408

409

T mean ( oC) o

T max ( C) o

T min ( C) RH min(%) RH max (%) u10 (m/s) -2

-1

R s (MJ m d ) u2 (m/s)

Jan

Feb

Mar

Ap r

Jun

Jul

17.3

19.1

20.9

22.5

25.1

May

26.4

27.2

A ug

26.2

Sep

24.8

Oct

21.4

Nov 18.8

Dec

23.2

23.4

25.0

25.8

27.3

28.3

28.8

28.9

28.4

27.4

25.3

23.4

8.5 70.9 96.1 3.52

12.1 71.9 94.8 3.60

15.1 71.3 96.0 3.84

17.8 67.2 94.0 3.45

22.0 74.0 94.1 2.91

24.0 79.5 95.0 2.66

24.9 80.6 93.9 2.49

24.8 81.1 95.8 2.44

24.2 80.8 96.9 2.56

21.0 75.8 95.2 3.11

16.1 76.0 96.9 3.39

10.3 74.3 97.1 3.35

12.51 2.63

14.60 2.69

18.18 2.87

20.03 2.58

21.70 2.18

19.54 1.99

19.95 1.86

18.33 1.83

15.97 1.91

16.11 2.33

13.69 2.54

11.59 2.51

ra (d m -1 )

9.14E-04 8.94E-04 8.38E-04 9.33E-04 1.11E-03 1.21E-03

rs (d m -1 )

8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 8.10E-04 0 0 0 0.03 0.11 0.12 0.14 0.13 0.11 0.08 0 0 12.51 14.60 18.18 19.35 19.25 17.13 17.16 15.95 14.21 14.90 13.69 11.59 2.84 2.88 3.17 3.32 3.63 3.85 3.96 3.98 3.87 3.65 3.22 2.88 1.11 1.41 1.72 2.04 2.64 2.98 3.15 3.13 3.02 2.49 1.83 1.25 1.76 1.87 2.13 2.25 2.69 3.01 3.13 3.18 3.09 2.68 2.26 1.86 0.15 0.15 0.14 0.13 0.11 0.10 0.09 0.09 0.09 0.11 0.13 0.15 0.69 0.68 0.77 0.78 0.71 0.74 0.76 0.75 0.72 0.72 0.67 0.65 0.72 0.71 0.79 0.80 0.74 0.77 0.78 0.78 0.75 0.75 0.70 0.69 -3.82 -3.72 -3.90 -3.88 -3.15 -2.93 -2.88 -2.79 -2.77 -3.18 -3.34 -3.55

-2 -1 S n (MJ m d ) e s (T max ) (kPa) e s (T min) (kPa) e a (kPa) e’ n/N f -2 -1 L n (MJ m d )

R n (MJ m

-2

-1

d )

1.29E-03 1.32E-03 1.26E-03 1.03E-03 9.49E-04

8.69

10.88

14.28

15.47

16.09

14.20

14.28

13.16

(MJ kg ) (kPa oC -1 ) e s (kPa)

2.45 0.06752 1.98

2.45 0.06752 2.14

2.45 0.06752 2.44

2.45 0.06752 2.68

2.45 0.06752 3.14

2.45 0.06752 3.42

2.45 0.06752 3.55

2.45 0.06752 3.56

15.85 0.12 1.21

17.75 0.13 1.20

20.05 0.15 1.19

21.80 0.16 ) 1.19

24.65 0.19 1.17

26.15 0.20 1.17

26.85 0.21 1.17

26.30 0.20 1.17

24.20 0.18 1.18

20.70 0.15 1.19

998

-1

T ( C) o -1 (kPa C ) 3 (kg/m ) a 3 w (kg/m ) -1 o -1

c p (MJ kg

C ) -1

1.01E-03 1.01E-03 1.01E-03 1.01E-03 1.01E-03 1.01E-03

998 1.01E-03

11.45

11.72

9.61E-04

10.35

8.04

2.45 2.45 2.45 0.06752 0.06752 0.06752 3.44 3.07 2.53

2.45 0.06752 2.07

1.01E-03 1.01E-03 1.01E-03 1.01E-03

16.85 0.12 1.21 998

1.01E-03

ET PM (mm d )

2.18

2.81

3.71

4.33

4.69

4.23

4.32

3.97

3.45

3.49

2.82

ET Abtew (mm d -1 )

2.71

3.16

3.93

4.33

4.69

4.23

4.32

3.97

3.45

3.49

2.96

2.51

Diff (mm d -1

0.53

0.35

0.23

0.00

0.14

0.45

ET PM1 (mm d -1 ) Error-PM1 (%)

3.23 48

3.91 39

5.07 37

5.56 28

5.77 23

5.24 24

5.29 23

4.92 24

4.36 26

4.44 27

3.86 37

3.05 48

2.05

Figure 13.2: ET Calculations I 13.4. (a) The PM calculations and the monthly albedos (given in the row after Rs ) that give the best agreement between the PM and Abtew equations are shown in Figure 13.2. Optimal values of α vary between 0 (not realistic) and 0.14. It appears that albedos are relatively small in the fall and winter when the sky is more cloudy, compared to the summer when there is relatively more sunshine and albedos are higher. The corresponding agreement between the PM and Abtew equations are shown in Figure 13.3. (b) If the wet season is deﬁned as when rainfall exceeds ET, then the wet season in June to October and the dry season is November to May . These results are shown in Figure 13.4. (c) Comparing the ET using the full PM equation (ETPM ) with the ET using only the solar radiation (ETPM1 ) shows that the greatest discrepancies occur during the wet season (∼ 50%) and the least discrepancies occur during the dry season (∼ 25%). These results indicate that wind speed and humidity have a greater eﬀect on dry-season ET than wet-season ET. One would suspect that the Abtew equation is most accurate during the wet season and least accurate during the dry season.

410 8.00 PM Abtew

7.00

6.00

ET (mm/d)

5.00

4.00

3.00

2.00

1.00

0.00 1

Month

Figure 13.3: Final ET Values Jan 31 7

Feb 28 8

Mar 31 11

Apr 30 13

May 31 15

Jun 30 13

Jul 31 13

Aug 31 12

Sep 30 10

Oct 31 11

Nov 30 8

Dec 31 6

Total

days ET PM (cm) ET PM1 (cm)

166

ET Abtew (cm) Rain (cm)

8 3

9 5

12 9

13 6

15 8

13 19

13 16

12 18

10 18

11 14

9 8

8 6

133 130

128

Figure 13.4: ET Calculations II 13.5 At the beginning of week water in the class A pan was 195 mm. In week there was a rainfall of 45 mm & 10 mm of water removed from the pan to keep the water level in speciﬁc range. Pan evaporation EP = rainfall + water added or water removed To maintain water level at 195 mm EP = 45 − 10 + 5 = 40 Assuming pan coeﬃcient CP = 0.70 evaporation(EL ) Therefore we know CP = Lake P an evaporation(Ep )

=⇒ 0.7 =

EL 40

So, EL = 40× 0.70 = 28 mm Therefore pan evaporation for the period is 28 mm. 13.6 Note to determine the potential evapotranspiration (PET), Here we consider, rs = 0 ra = 14.8 s/m = 1.713×10−4 d/m (used data from example 13.2) α = 0.25 (given) Sn = 19.78 × (1−0.25) (1−0.08)

= 16.125 MJ/(m2 . d) (used data from example 13.2)

411 Rn = 16.125 + (−6.44) = 9.685 MJ/m2 .d (used data from example 13.2) According to equation no 13.1 (4.24−1.73) 0.243 (9.685 − 0) + 1.154(1.013 × 10−3 ) 1.713×10 −4 1 ] ET = [ 0 998.2 × 2.45 0.243 + 0.0674(1 + 1.713×10−4 )

= 4.09 × 10−4

19.48 = 0.02567 m/d = 25.67 mm/d 0.3104

13.7. The latent heat of vaporization at 32◦ C is L = 2500 − 2.37 × T kJ/kg. = 2500 − 2.37 × 32 = 2425.76×103 J/kg. Water density, ρw = 996 kg/m3 400 Evaporation rate = 2425.76×10 3 ×996

= 1.656 × 10−7 × 1000 × 86400 mm/d = 14.3 mm/d

Chapter 14

Fundamentals of Ground-Water Hydrology I: Governing Equations 14.1. The hydraulic conductivity of an aquifer is a better measure of the productivity of an aquifer than the porosity. The hydraulic conductivity measures the ﬂow for a given head gradient, while the porosity is simply a measure of the amount of void space. 14.2. The hydraulic conductivity of the aquifer (K) for a temperature of 20◦ C is T = K×b = 83.7× 10.5 = 879 m2 /d The discharge per unit width of the aquifer (q) is, 2 q = −T dφ L = −879 (38.3–42.1) = 3340.2m /d

The speciﬁc discharge, q of the aquifer is, dΦ q=Q A = −k L = −83.7 (−0.0038) = 0.32 m/d [ From Equation no 14.5]

The pore velocity, Va = nqe = 0.32 0.2 = 1.6 m/d [ From Equation no 14.10] Water travel time from piezometer A to B is Time =

L 1000 Va = 1.6 = 625 days

14.3. The equivalent vertical permeability, (Kz ) of the aquifer of length n

Kz = n1 Lii

1 Li Ki

300

225 75 + 10−5 1.92×1.92−12

= 7.7 × 10−12 m/s The equivalent horizontal parameters of the aquifer of length Kz = = 10

n 1 K i ×Li n 1 Li

−5 ×255+1.9210−5 ×75

300

= 1.23 × 10−5 m/s The equivalent Hydraulic conductivity, Ke = 412

√

Kx × K2 = 7.49×10−6 m/s.

413 14.4. Cross sectional area, A = 10 ha Length(L) = 6116 m Hydraulic head diﬀerence, φ1 − φ2 = (306 − 294) = 12 m 2 [ From Equation no 14.3] The speciﬁc discharge (Q/A) = k φ1 −φ L

0.286 −3 ) ⇒ 10×10 4 = k × (1.962 × 10

⇒k = 1.46 m/d The linear velocity, Va = nqe = 2.86×10 0.3

−6

= 9.53×10−6 m/d [ From Equation no 14.10]

14.5. Let us consider here area of aquifer = 1 km2

(i) Change in groundwater storage = area of the aquifer×ﬂuctuation in ground water table × speciﬁc yield = (1×106 )×1 × 0.2 m3 = 0.2 Mm3 (ii) Again

Change in groundwater storage = co-eﬃcient

area of the aquifer × piezometric surface × storage

= (1×106 )× (1) × (0.0005) m3 = 0.0005 Mm3 14.6. From the given data n = 0.20 and the grain size distribution gives: d10 = 0.124 mm, d17 = 0.167 mm, d20 = 0.180 mm, d60 = 0.711 mm, and Uc = d60 /d10 = 0.711/0.124 = 5.73. For water at 20◦ C, ν = 1.004 × 10−6 m2 /s, and the hydraulic conductivity, K, is given by Equation 14.18 as K = α · φ(n) · d2e ·

g 9.81 = α · φ(n) · d2e · = 9.77 × 106 · α · φ(n) · d2e ν 1.004 × 10−6

(1)

where K is in m/s if de is expressed in meters. If de is expressed in mm and K in m/d, then Equation 14.23 is given by K = 8.44 × 105 · α · φ(n) · d2e

(2)

To facilitate calculation of de by the empirical formulae in Table 14.3, the given grain size distribution is more conveniently expressed in the following tabular form

414 Size (mm)

Fraction ﬁner

4.760

0.98

2.000

0.86

0.840

0.64

0.420

0.51

0.250

Interval, i

Fraction retained, Δgi

Upper Size, dgi (mm)

Lower Size, ddi (mm)

0.12

4.760

2.000

0.22

2.000

0.840

0.13

0.840

0.420

0.15

0.420

0.250

0.23

0.250

0.149

0.09

0.149

0.074

0.36

0.149

0.13

0.074

0.04

From the above-tabulated grain-size distribution: Δg1 = 0.04 and d1 = 0.074 mm. Using the formulas in Table 14.3 with the grain-size distribution data to calculate the parameters in Equation 14.24 gives the following results Name Hazen

α (-) 6 × 10−4

φ(n) (-) 0.4

de (mm) 0.124

K (m/d) 3.1

Restriction met N

Slichter

1 × 10−2

0.00504

0.124

0.7

Terzaghi

8.4 × 10−3

0.00569

0.124

0.6

Beyer

1.16 × 10−3

0.124

0.1

Sauerbrei

3.75 × 10−3

0.0125

0.167

1.1

Krüger

4.35 × 10−5

0.313

0.357

1.5

Kozeny

8.3 × 10−3

0.0125

0.244

5.2

Zunker

1.24 × 10−3

0.0625

0.250

4.1

Zamarin

8.3 × 10−3

0.0119

0.629

33.0

USBR

2.87 × 10−4

0.180

7.8

Therefore the hydraulic conductivity is in the range of 0.1–33 m/d .

415 14.7. The piezometric head h = 25 cm. Area of soil sample, A = Π/4 × (10.2)2 cm2 = 81.72 cm2 Length of soil specimen , L = 14 cm 162 Discharge, Q = 1.70×60 cm3 /s = 1.6 cm3 /s

We have, 1.60×14 −4 m/s Co eﬃcient permeability, k = QL Ah = 81.72×25 cm/s = 1.1×10

14.8.

Figure 14.1: Problem 14.8. Considering Figure 14.1, continuity requires that the ﬂow into a cylinder of radius r be given by Qw = 2πrHne v which re-arranges to v= 14.9. The given relationship is

Qw 2πrHne

Kij = K11 αi1 αj1 + K22 αi2 αj2

(1)

From Figure 14.2, α11 = cos θ α22 = cos θ α21 = cos(90 + θ) = − sin θ α12 = cos(90 − θ) = sin θ and substituting into Equation 1 gives cos2 θ + K22 sin2 θ K11 = K11

(2)

416

Figure 14.2: Problem 14.9. Noting the identities 1 cos2 θ = (1 + cos 2θ) 2 sin2 θ = 1 − cos2 θ

(4)

sin 2θ = 2 sin θ cos θ

(5)

(3)

and substituting Equations 3 and 4 into Equation 2 gives K11 = K11 cos2 θ + K22 (1 − cos2 θ) = K22 + (K11 − K22 ) cos2 θ + = K22

− K K11 K − K22 22 + 11 cos 2θ 2 2

which gives K11 =

+ K K − K22 K11 22 + 11 cos 2θ 2 2

K22 =

+ K K − K22 K11 22 − 11 cos 2θ 2 2

Similarly,

K12 = K21 = −

− K K11 22 sin 2θ 2

14.10. The given equations are: + K K − K22 K11 22 + 11 cos 2θ 2 2 K − K22 K + K22 − 11 cos 2θ K22 = 11 2 2 K − K22 sin 2θ K12 = K21 = − 11 2

K11 =

(1) (2) (3)

417 Equations 1 and 2 combine to give K11 + K22 = K11 + K22

(4)

K11 − K22 = (K11 − K22 ) cos 2θ

which means

K11 − K22 2

from Equation 3 2 K12 =

− K K11 22 2

(5)

cos2 2θ

(6)

sin2 2θ

(7)

Combining Equations 6 and 7 gives ±

K11 − K22 2

1 2 + K12

=±

− K K11 22 2

(8)

where the identity cos2 2θ + sin2 2θ = 1 has been used. From Equation 4: − K K11 K11 + K22 22 = − K22 2 2 K11 + K22 K − K22 = − K11 − 11 2 2

(9) (10)

Combining Equations 10 and 8 gives K11 + K22 + K11 = 2

K11 − K22 2

1 2 + K12

and combining Equations 9 and 8 gives K11 + K22 − K22 = 2

K11 − K22 2

1 2 + K12

14.11. Adding Equations 14.30 and 14.31 gives K11 + K22 = K11 + K22

(1)

and Equation 14.32 can be put in the form − K22 =− K11

K12 sin 2θ

Substituting Equation 1 and Equation 2 into Equation 14.30 gives 2K11 = K11 + K22 −

2 K12 cos 2θ sin 2θ

(2)

418 which can be rearranged to give tan 2θ =

−2K12 K11 − K22

and solving for θ yields 2K12 1 θ = − tan−1 2 K22 − K11 According to the deﬁnition of θ, this is the angle of rotation a principal axis to the corresponding coordinate axis, counterclockwise positive. Therefore, the angle of rotation needed to reach a principal axis from the corresponding coordinate axis is θ = −θ, and is given by θ =

1 2K12 tan−1 2 K22 − K11

14.12. Trace record the actual velocity of water Va = 42×100 6×3600 = 0.194 cm/s Discharge velocity, V = nVa = 0.2×0.194 = 0.0388 cm/s −2 Hydraulic gradient, i = 0.42 42 = 1 × 10

0.0388 Co-eﬃcient of permeability, k = 1×10 −2 = 3.88 cm/s

Intrinsic permeability 3.88×0.01 = 3.95×10−5 cm2 k0 = kv g = 981

Since, 9.87×10−9 cm2 = 1 darcy Therefore k0 = 4002 darcy 14.13. Seepage discharge, Q = 0.8 m3 /d 1 Hydraulic gradient, J = Φ2 −Φ L

Area, A = Π4 × (1.2)2 = 1.13 m2 The vertical n hydraulic conductivity of an aquifer is L (k z )V = ni Lii 1 ( ki )

where, L = L1 + L2 + L3 = 150 + L3 , (where, L3 = length of the pipe ﬁlled up by silty sand) 150+L3 100+50+L3 ⇒ (k z )V = 100 L3 = 11+10L3 50 10

+ 50 + 0.1

We have, Q = kJA 150+L3 20 ) × (150+L × 1.13 ⇒ 0.8 = ( 11+10L 3 3)

⇒L3 = 1.73 m 14.14. Given,

h0 =10 m, h1 = 8 m R = 0.002 m3 /day/m2 L = 1200 m And k = 3 m/day

419 (i) The water table proﬁle 2

(h0 2 −h1 2 − RL ) k x + h0 2 [ From Equation no 15.11] L 2 (102 −82 − 0.002×1200 ) 0.002x2 2 3 x + 102 ⇒h =− 3 − 1200 2 −4 2 ⇒ h = 6.66 × 10 x + 0.45x + 100 2

h2 = − Rx k −

(ii) Location of water table divides h0 2 −h1 2 a = L2 − K ] R[ 2L 3 102 −82 1200 ⇒ a = 2 − 0.002 ( 2×1200 ) = 577.5 m At x = a = 577.5 m, h = hm = height of water table divide hm 2 = −6.66 × 10−4 x2 + 0.77x + 100 ⇒ hm 2 = −6.66 × 10−4 (577.5)2 + 0.77 × 577.5 + 100 Therefore, hm = 17.95 m Discharge per unit width of aquifer K (h0 2 − h1 2 ) [ From Equation no 15.11] q x = R x − 12 + 2L At x = 0, K 3 (h0 2 − h1 2 ) = −( 0.002×1200 ) + 2×1200 (102 − 82 ) q 0 = −R L2 + 2L 2

Therefore, q 0 = – 1.155 m3 /day per meter width The negative sign indicates that discharge is in (–x) direction i.e. into water body. At x = L, q1 = qL and from qL = RL+q0 Hence q1 = discharge into water body B = 0.002×1200 + (−1.155) = 1.245 m3 /day/m width (vi) When distance of water table divide a = 0 L k h0 2 −h1 2 =0 a= 2 −R 2 2L 2 k h0 −h1 ⇒ L2 = R 2L 2 2 k 3.0 2 = 7.5×10−5 m3 /day/m2 ⇒ R = L2 h0 − h1 2 = 1200 2 10 − 8 Since, a = 0, q0 = 0 q1 = qL = RL = 1.244 m3 /day/m width 2

×2×60×24 Φ1 −4 m/d 14.15. The hydraulic conductivity, K = A(taL ln Φ = Π/4×0.06 ln 30 25 = 3.3×10 100×(60−15) 2 2 −t1 )

14.16. The vertical permeability (kz ) of the confined aquifer if the flow is normal to the stratification = 37.03 m/d kz = 4 4+5+6 +5+6 25

The horizontal permeability of the aquifer, if the ﬂow is parallel to the stratiﬁcation = 40 m/d kx = 25×4+40×5+50×6 15 The equivalent hydraulic conductivity, ke =

√

kx × kz = 38.49 m/d

420 r

Qlog 10 2

r1 14.17. K = 1.36(h 2 −h 2 [ According to equation no-15.41] ) 2

Q = 1500 lit/min = 0.025 m3 /s Now given, r1 = 16 m r2 = 34 m h1 = 14.5 − 2.2 − 2.45 = 9.85 m h2 = 14.5 − 2.2 − 1.2 = 11.1 m 0.025×log

( 34 )

10 16 −4 m/s K = 1.36[11.12 −9.85 2 ] = 2.29 × 10

14.18. From the deﬁnition of the Laplacian operator ∇2 f =

∂2f ∂2f ∂2f + 2 + 2 2 ∂x ∂y ∂z

and x = r cos θ,

y = r sin θ −→ r = (x2 + y 2 ) 2 ,

θ = tan−1 (y/x)

By the chain rule, ∂f ∂r ∂f ∂θ ∂f = · + · ∂x ∂r ∂x ∂θ ∂x

(1)

Noting that 1 du d tan−1 u = dx 1 + u2 dx then 1 ∂r 1 = (x2 + y 2 )− 2 (2x) = ∂x 2

r cos θ x = = cos θ r r

x2 + y 2 ∂ 1 −y y r sin θ sin θ ∂θ = tan−1 (y/x) = · 2 =− 2 =− 2 =− 2 2 ∂x ∂x 1 + (y/x) x x +y r r

(2) (3)

Combining Equations 1 to 3 gives ∂f sin θ ∂f ∂f = cos θ − ∂x ∂r r ∂θ 2 ∂ ∂f ∂ f = ∂x2 ∂x ∂x ∂f sin θ ∂f sin θ ∂ ∂f sin θ ∂f ∂ cos θ − − cos θ − = cos θ ∂r ∂r r ∂θ r ∂θ ∂r r ∂θ 2 2 2 2 sin θ cos θ ∂f sin θ ∂f sin2 θ ∂ 2 f ∂ f 2 sin θ cos θ ∂ f = cos2 θ 2 − + + + ∂r r ∂r∂θ r2 ∂θ r ∂r r2 ∂θ2 By the chain rule, ∂f ∂r ∂f ∂θ ∂f = · + · ∂y ∂r ∂y ∂θ ∂y

(4)

421 and ∂r 1 = (x2 + y 2 )−1/2 (2y) = sin θ ∂y 2 ∂ x ∂θ 1 1 r cos θ cos θ = tan−1 (y/x) = · = 2 = = ∂y ∂y 1 + (y/x)2 x x + y2 r2 r therefore ∂f ∂f cos θ ∂f = sin θ + ∂y ∂r r ∂θ ∂2f ∂ ∂f = ∂y 2 ∂y ∂y ∂f cos θ ∂f cos θ ∂ ∂f cos θ ∂f ∂ sin θ + + sin θ − = sin θ ∂r ∂r r ∂θ r ∂θ ∂r r ∂θ 2 2 2 2 sin θ cos θ ∂f cos θ ∂f cos2 θ ∂ 2 f ∂ f 2 sin θ cos θ ∂ f − + + = sin2 θ 2 + ∂r r ∂r∂θ r2 ∂θ r ∂r r2 ∂θ2 Combining Equations 4 and 5 (and adding ∂ 2 f /∂z 2 ) to both sides gives ∇2 f =

1 ∂2f ∂2f 1 ∂f ∂2f + + + ∂r2 r ∂r r2 ∂θ2 ∂z 2

Cylindrical coordinates are useful in cases where there is cylindrical symmetry . 14.19.

Kz =

6 + 6−8 10−7 10

m/s = 1.81× 10−8 m/s

Taking the top of the gravel as datum, Head of water due to artesian pressure = 15.5 m Head of water due to groundwater = (2×6)+1 = 13 m Therefore excess head causing ﬂow = (15.5 – 13) = 2.5 m The quantity of ﬂow per unit area, 3 2 −9 m3 /s/m2 q = Ki = 1.81× 10−8 × 2.5 12 m /s/m = 3.77×10

(5)

422 14.20. (a) Volume pumped out = area × drop in water table× speciﬁc yield (Sy ) 1.5 × 106 = 3 × 106 × (103.2 − 102.0) × Sy ⇒ Sy = 0.25 (b) Recharge volume = 0.25× (103.2-102.0)×3×106 = 0.9 Mm3 14.21. The eﬀective hydraulic conductivity components, K̄xx and K̄yy , are given by Equations 14.98 and 14.99 as n n 1 1 i i Kxx Δzi and K̄yy = Kyy Δzi K̄xx = h h i=1

i=1

When the water table is 2 m below the ground surface, h = 20 − 2 = 18 m, and therefore 1 (7 × 3 + 9 × 2 + 14 × 4 + 11 × 4 + 6 × 4 + 2 × 1) = 9.2 m/d 18 1 K̄yy = (20 × 3 + 21 × 2 + 12 × 4 + 17 × 4 + 9 × 4 + 5 × 1) = 14.4 m/d 18

K̄xx =

When the water table is 3 m below the ground surface, h = 20 − 3 = 17 m, and therefore 1 (7 × 2 + 9 × 2 + 14 × 4 + 11 × 4 + 6 × 4 + 2 × 1) = 9.3 m/d 17 1 K̄yy = (20 × 2 + 21 × 2 + 12 × 4 + 17 × 4 + 9 × 4 + 5 × 1) = 14.1 m/d 17

K̄xx =

Hence, K̄xx increases slightly and K̄yy decreases slightly when the water table falls from 2 m to 3 m below the ground surface. 14.22. Speciﬁc yield of the unconﬁned aquifer = 0.2 The recharge rate to the aquifer = 76 mm/yr. = 0.076 m/yr The amount of water pumped from the aquifer for irrigation = 254 mm/yr. = 0.254 m/yr. The saturated thickness of the aquifer = 15.2 m The amount of water that has to be taken from the saturated thickness of the aquifer = (0.254 – 0.076) m/yr. = 0.178 m/yr. The aquifer is giving 0.178 m of water per year for irrigation. 3.04 = The total no. of years that the aquifer can supply water to the irrigation load is = 0.178 17.08 years

14.23. The speciﬁc discharge (q) is (30−10) q=Q A = KJ = 1.5×103 × 20 = 0.266 m/d

Actual velocity, Va = nqe = 0.266 0.3 = 0.887 m/d 3

Time to contaminate = 1.5×10 0.887 d = 4.62 yrs.

423 14.24. The storage coeﬃcient, S, can be expressed in terms of the elastic properties of water and the elastic properties of a porous medium using Equation 14.103 (Bear, 1979) where α S = γnb Ew + n

(1)

From the given data: n = 0.3, b = 20 m, and α = 8 × 10−9 m2 /N. At 15◦ C, γ = ρg = (999.1)(9.81) = 9801 N/m3 , and Ew = (2.14 × 109 )−1 = 4.67 × 10−10 m2 /N. Substituting into Equation 1 yields S = (9801)(0.3)(20) 4.67 × 10

−10

8 × 10−9 + 0.3

= 0.0016

The value of S estimated using the Lohman (1972) equation is given by S = 3 × 10−6 (20) = 0.00006 Hence the calculated storage coefficient using the Lohman (1972) equation is much different from that calculated using the Bear (1979) equation, Equation 1. The reason for this is that Lohman (1972) used 3 × 10−6 instead of γn(Ew + α/n) as the specific storage, Ss . As shown in Table 14.5, the specific storage, Ss , can vary from 10−3 to 10−7 , and 3 × 10−6 as assumed by Lohman (1972) is not representative of all aquifer materials, and certainly not the aquifer material in the present case. 14.25. The storage coefficient measures the volume of water released from storage due to the elasticity of the water and solid matrix. Since the elasticities are quite small, the volume of water released from storage in a confined aquifer due to a 1 m fall in the pressure head is also quite small. In contrast, the specific yield measures the volume of water released from storage in the pores of an unconfined aquifer as the porous medium is drained. Therefore a 1 m drop in the water table yields much more water from an unconfined aquifer than a 1 m drop of the piezometric head in a confined aquifer. 14.26. Let, H = elevation of water table HA = 210.00–11.00 = 199.00 m HB = 207.00–7.00 = 200.00 m HC = 212.00–15.00 = 197.00 m Along, ΔHY = HB – HA = 200.00 – 199.00 = 1.00 m Hy 1 1.00 = = LAB 1200 1200 k Vy = k.iy = m/s 1200 iy =

424

199

1200 m

C 197

B 200

1000 m k Vy = k × iy = 1200 m/s

Along BC (x direction) − Hx = HB − HC = 200 − 197.00 = 3 H x 3.00 = LBC 1000 3k m/s Vx = k × i x = 1000 ix = −

V = (Vx 2 + Vy 2 )

1/2

= 3.11 × 10−3 km/s tanθ =

1 9 1/2 k [ + ] 100 144 100

Vy 5 = Vx 18

⇒ θ = 15.52◦ Where θ = Inclination of V to x axis (west direction). The ground water flows will be in a direction which makes 15.52◦ with line BC and 74.48◦ with line BA. Thus the direction of ground water flow is 74◦ 28’48”. 14.27. In terms of Cartesian coordinates (x , y ), the head distribution in the isotropic aquifer is given by R R 2 Qw Qw φ(x , y ) = φ0 − = φ0 − ln ln , r < R 2 + y 2 2 2 2πT 4πT x x +y where r 2 = x 2 + y 2 and R 2 = X 2 + Y 2 . The coordinates (X , Y ) define the boundary of the zero-drawdown circle surrounding the well. Application of the head distribution in an

425 isotropic medium to an anisotropic medium requires a scaling of the spatial coordinates and the transmissivity in accordance with Equations 14.118 and 14.124. These equations yield Tyy T T Txx = , = T = Txx Tyy ⇒ Txx Txx Tyy Tyy Tyy 2 T 2 x 2 = x = x Txx T xx T 2 Txx 2 y 2 = y = y Tyy Tyy Tyy 2 T 2 2 X = X = X Txx T xx T 2 Txx 2 Y 2 = Y = Y Tyy Tyy Substituting these relationships into Equation 14 gives the head distribution in a homogeneous anisotropic aquifer as Qw φ(x, y) = φ0 − ln 4π Txx Tyy

Tyy /Txx X 2 +

Txx /Tyy Y 2

Tyy /Txx x2 +

Txx /Tyy y 2

r<R

where r2 = x2 + y 2 , and R2 = X 2 + Y 2 is the distance (squared) to where the drawdown is zero in the anisotropic formation. 14.28. (a) The partial diﬀerential equation that describes the piezometric head distribution in a two-dimensional homogeneous anisotropic conﬁned aquifer with principal hydraulic conductivities Kxx and Kyy is given by Kxx

S ∂φ ∂2φ ∂2φ + K = yy 2 2 ∂x ∂y b ∂t

Application of the given coordinate transformations is given in the textbook, and yields the following equation if the transformed domain, K

∂2φ ∂2φ S ∂φ + K = ∂x 2 ∂y 2 b ∂t

(b) If a φ contour has an elliptical distribution in the (x, y) domain, then the contour is described by x2 y 2 + 2 =1 a2 b Using the given coordinate transformations, this equation in the (x , y ) domain becomes Ky y 2 Kx x 2 Kx x 2 Ky y 2 + = 1 → + =1 K a2 K b2 Ky a2 Kx b2

426 which yields the following circular equation in the transformed domain, x 2 y 2 + 2 =1 r2 r which requires that

r 2 = a2

Ky = b2 Kx

a Kx → = Ky b

Kx Ky

This proves that a circular contour in the (x , y ) domain corresponds to an elliptical contour in the (x, y) domain, where the ratio of principal axes in the (x, y) domain is equal to Kx /Ky . 14.29. The retention curve is aﬀected by the path of wetting, leading to a hysteresis eﬀect. 14.30. (a) According to Darcy’s law Q = – kJA = 30× −2.75 × 10−3 × (2000 × 25) = 4125 m3 /d −3 1 = 39.5−45 Where, J = hydraulic gradient = φ2 −φ L 2000 = –2.75×10

A = (2000×25) m2

(b) The piezometer head at an observation well located 300 m from the upstream = φ2 Now, 4125 = 30 × (

45 − φ2 ) × 2000 × 25 300

Therefore φ2 = 44.175 m 14.31. From the measured retention curve: θs = 0.46, and the van Genuchten empirical relation can be expressed as 1 1+ n θ − θr 1 = 0.46 − θr 1 + (αψ)n or 1 1+ n 1 θ = θr + (0.46 − θr ) 1 + (αψ)n From the measured retention curve, the points in Columns 1 and 2 in Table 14.1 can be obtained. Using a least-squares analysis, a good match is obtained when θr = 0.135, n = 0.61, and α = 0.042. Therefore the van Genuchten retention function for Krome soil is given by θ = 0.135 + 0.325

2.64 1 1 + (0.042ψ)0.61

The moisture content estimated from the van Genuchten equation is given in Column 3 of Table 14.1, and the squared errors and percentage errors are shown in Columns 4 and 5. The van Genuchten equation is superimposed on the measured retention curve in Figure 14.3. The maximum percentage error is approximately 2.3% .

427

Table 14.1: Estimation of van Genuchten Retention Curve (1) ψ (cb) 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 25.0 30.0 40.0 50.0 60.0 70.0 80.0 Sum

(2) θ (-) 0.460 0.317 0.270 0.252 0.235 0.217 0.209 0.200 0.196 0.183 0.178 0.170 0.161 0.157 0.152 0.148

(3) θvG (-) 0.460 0.314 0.272 0.248 0.231 0.218 0.209 0.201 0.195 0.185 0.178 0.168 0.162 0.157 0.154 0.151

(4) (θ − θvG )2 (-) 0 7.56 × 10−6 5.40 × 10−6 1.94 × 10−5 1.84 × 10−5 1.69 × 10−6 8.20 × 10−8 1.13 × 10−6 1.44 × 10−6 4.58 × 10−6 6.11 × 10−10 3.09 × 10−6 7.15 × 10−7 1.23 × 10−7 4.12 × 10−6 1.21 × 10−5 7.99 × 10−5

(5) Error (%) 0 −0.9 +0.9 −1.7 −1.8 +0.6 −0.1 +0.5 −0.6 +1.2 +0.0 −1.0 +0.5 +0.2 +1.3 +2.4

Figure 14.3: Measured retention curve compared with van Genuchten function for Krome Soil.

Chapter 15

Fundamentals of Ground-Water Hydrology II: Applications 200 15.1. R= 1000×365 = 5.47 × 10−4 m/d

Location of water table divide 2 2 ha −hb 1.5 112 −82 a = L2 − K = 4000 R 2×L 2 − 5.47×10−4 × 2×4000 =1998 m Groundwater table divided at a distance of 1998 m from A river Now K 5.57×10 2 2 qa = – RL 2 + 2L (ha − hb ) = – 2

−3

3 × 4000+ 1.5×57 8000 = −1.0833 m /d/m

qb = RL+ qa = 1.1047 m3 /d/m Avarager groundwater recharge to river A is −1.0833 m3 /d/m and For River B is 1.1047 m3 /d/m 15.2. Equation 15.8 gives the equation of the phreatic surface as K1 2 h + (K2 − K1 )b2 h = C1 x + C2 2 Taking the derivative of this equation yields K1 h

dh dh + (K2 − K1 )b2 = C1 dx dx

which simpliﬁes to dh dh − K2 b2 dx dx Equation 15.11 gives the ﬂow, Q, between aquifers as −C1 = −K1 (h − b2 )

Q = −K1 (h − b2 )

dh dh − K2 b2 dx dx

(1)

(2)

Comparing Equations 1 and 2 indicates that Q = −C1 428

(3)

429 where C1 is given by Equation 15.9 as C1 =

K1 2 b2 (h − h2L ) + (K2 − K1 ) (hR − hL ) 2L R L

(4)

Combining Equations 3 and 4 gives Q=

K1 2 b2 (h − h2R ) + (K1 − K2 ) (hR − hL ) 2L L L

15.3. For each side of the canal, the governing ﬂow equation is given by dh d Kh =0 dx dx which can be put in the form d2 h2 =0 dx2 Integrating this equation twice yields h2 = C1 x + C2 Deﬁning the drawdown, s, as

(1)

s=H −h

then Equation 1 can be put in the form (H − s)2 = C1 x + C2

(2)

On the left-hand side of the canal, the boundary conditions are s(0) = 0

(3)

s(L) = sL

(4)

Combining Equations 2 to 4 gives s2 − 2HsL (H − sL )2 − H 2 = L L L C2 = H 2

C1 =

(5) (6)

The flow from the left-hand side of the canal (i.e. leakage) per unit length of canal, QL , is given by K dh2 K dh K =− = − C1 = − (s2L − 2HsL ) QL = −Kh dx 2 dx 2 2L which simplifies to KsL (2H − sL ) (7) QL = 2L Similarly, the flow (i.e. leakage) from the right-hand side of the canal per unit length of canal, QR , is given by KsR (2H − sR ) (8) QR = 2L

430 and the total ﬂow (i.e. leakage) out of the canal per unit length of canal, QT , is given by QT = QL + QR

(9)

Combining Equations 7 to 9 yields the following expression for leakage out of the canal per unit length of canal K [sL (2H − sL ) + sR (2H − sR )] QT = (10) 2L From the given data: K = 30 m/d, H = 20 m, L = 70 m, and sL = sR = 0.05 m. Substituting into Equation 10 gives QT =

30 [(0.05)(2 × 20 − 0.05) + (0.05)(2 × 20 − 0.05)] = 0.856 m2 /d 2(70)

15.4. The governing equation (with recharge) is given by d dx

dh [K1 (h − b2 ) + K2 b2 ] +N =0 dx

Integrating gives [K1 (h − b2 ) + K2 b2 ]

dh + N x = C1 dx

Separating terms gives dh dh + (K2 − K1 )b2 + N x = C1 dx dx K1 d 2 dh (h ) + (K2 − K1 )b2 + N x = C1 2 dx dx K1 h

which integrates to K1 2 N x2 h + (K2 − K1 )b2 h + = C1 x + C2 2 2 The boundary conditions are h = hL at x = 0, and h = hR at x = L. Applying the boundary conditions at x = 0 yields K1 2 h + (K2 − K1 )b2 hL C2 = 2 L at x = L, N L2 K1 2 hR + (K2 − K1 )b2 hR + = C1 L + C2 2 2 which gives C1 =

K1 2 (K2 − K1 )b2 NL (hR − h2L ) + (hR − hL ) + 2L L 2

431 15.5. The given relation is K1 2 N x2 h + (K2 − K1 )b2 h + = C1 x + C2 2 2 Diﬀerentiating with respect to x gives dh dh K1 + (K2 − K1 )b2 + N x = C1 2h dx 2 dx

(1)

At the mound location, dh =0 dx

→

N x 0 = C1

→ x0 =

C1 N

(2)

From Problem 15.4, C1 =

K1 2 (K2 − K1 )b2 NL (hR − h2L ) + (hR − hL ) + 2L L 2

(3)

Combining Equations 2 and 3 yields x0 =

K1 (K2 − K1 )b2 (hR − hL ) L (h2 − h2L ) + + 2N L R NL 2

The mound does not occur between reservoirs when x0 ≤ 0 or L (K2 − K1 )b2 (hR − hL ) K1 (h2R − h2L ) + ≤− 2N L NL 2 which can be put in the form N≤

2(K2 − K1 )b2 (hL − hR ) K1 2 (hL − h2R ) + 2 L L2

Since the right-hand side of this inequality can clearly be positive, then the statement is false . A mound will not always be located between the reservoirs. 15.6. For a 3-layer aquifer, d dx

dh [K1 (h − b2 − b3 ) + K2 b2 + K3 b3 ] =0 dx

Integrating gives dh = C1 dx dh dh + (K2 b2 + K3 b3 − K1 b2 − K1 b3 ) = C1 K1 h dx dx [K1 (h − b2 − b3 ) + K2 b2 + K3 b3 ]

and integrating again K1 2 h + [(K2 − K1 )b2 + (K3 − K1 )b3 ]h = C1 x + C2 2

432 Apply the boundary conditions: h = hL @ x = 0, and h = hR @ x = L: x=0: x=L:

C2 =

K1 2 h + [(K2 − K1 )b2 + (K3 − K1 )b3 ]hL 2 L

K1 2 h + [(K2 − K1 )b2 + (K3 − K1 )b3 ]hR = C1 L + C2 2 R

which gives C1 =

K1 2 (K2 − K1 )b2 + (K3 − K1 )b2 (h − h2L ) + (hL − hR ) 2L R L

15.7. From the given data: Qw = 400 L/s = 34560 m3 /d, b = 24 m, s = 1 m @ r = 50 m, and s = 0.5 m @ r = 100 m. The transmissivity, T , is given by r2 100 34560 Qw ln ln = 7630 m2 /d = T = 2π(s1 − s2 ) r1 2π(1 − 0.5) 50 and the hydraulic conductivity, K, is given by K=

T 7630 = = 318 m/d b 24

There is no steady-state solution for the drawdown. If rw = 0.5 m, sw = 4 m, and T = 7630 m2 /d, Qw s(r) = sw − ln 2πT when s = 0: 0=4−

r rw

r 34560 ln 2π(7630) 0.5

which gives r = 128 m The steady-state drawdown is not valid beyond this distance because the drawdown becomes negative, which is not realistic. 15.8. (a) When the Thiem equation is derived using zero drawdown at a radius of inﬂuence, R, as a boundary condition, then the drawdown at the well is not used as a boundary condition. Consequently, the derived form of the Thiem equation does not accurately describe the drawdown at the well. When the Thiem equation is derived using the radius of inﬂuence boundary condition, the distribution of piezometric head, φ, surrounding the well is given by R Qw ln (1) φ(r) = H − 2πKb r

433 and when the Thiem equation is derived using the saturated thickness of the aquifer at the well as a boundary condition, the distribution of piezometric head surrounding the well is given by r Qw (2) ln φ(r) = hw + 2πKb rw Equating the head distributions in Equations 8.62 and 8.110 yields R r Qw Qw ln = hw + ln H− 2πKb r 2πKb rw which can be put in the form 2πKb (H − hw ) = ln Qw

R rw

(3)

Deﬁning the drawdown, sw , at the well by sw = H − hw then Equation 8.111 can be written as 2πKb sw = ln Qw which rearranges to

R = rw exp

R rw

2πT sw Qw

Using this value of R will give the same piezometric head distribution no matter which boundary condition is used in deriving the Thiem equation. The primary limitation of the Thiem equation is that it assumes a steady state, where a steady state is in fact not possible. The Thiem equation works reasonably well within the radius of inﬂuence when the changes in head with time have stabilized and are very small. This corresponds to pseudo-steady state conditions. (b) From the given data: Qw = 30 L/s = 0.030 m3 /s, rw = 0.1/2 = 0.05 m, and L = 1 m. Therefore, the speciﬁc discharge, q, at the well is given by q=

0.03 Qw = = 0.0955 m/s 2πrw L 2π(0.05)(1)

Darcy’s law is valid up to qd = 10 (4) ν where d is the pore scale, and ν is the kinematic viscosity of ground water. At 20◦ C, ν = 1.00 × 10−6 m2 /s. Substituting into Equation 8.112 gives 0.0955d = 10 1.00 × 10−6

434 which yields d = 0.000105 m Using d as d10 in the Hazen equation for intrinsic permeability, k, gives k = 1.02 × 10−3 d210 = 1.02 × 10−3 (0.000105)2 = 1.12 × 10−11 m2 which corresponds to a hydraulic conductivity, K, of K=k

γ μ

At 20◦ C, γ = 9790 N/m3 and μ = 1.00 × 10−3 N·s/m2 . Hence, K = 1.12 × 10−11

9790 = 1.096 × 10−4 m/s = 9.47 m/d 1.00 × 10−3

Therefore, as long as the aquifer hydraulic conductivity exceeds 9.47 m/d the Thiem equation can be applied without violating Darcy’s law. 15.9. From the given data: b = 20 m, rw = 0.70/2 m = 0.35 m, Q0 = 40.5 L/s = 3500 m3 /d, Δz1 = 3 m, n = 0.18, K1 = 5K2 , Q1 = 4.05 L/s = 350 m3 /d, φ0 = 42.301 m, φ1 = 42.015, φ2 = 42.030, φw = 41.103 m, r1 = 30 m, and r2 = 60 m. (a) From the Thiem equation,

350 60 = ln = 2574 m2 /d 2π(42.030 − 42.015) 30 K1 (17) = 2574 m2 /d T = K1 Δz1 + K2 Δz2 = K1 (3) + 5 Q1 ln T = 2π(s1 − s2 )

r2 r1

which yields K1 = 402 m/d and K2 = K1 /5 = 80.4 m/d . (b) The radius of inﬂuence, R, satisﬁes the following form of the Thiem equation, R Q0 ln φw = H − 2πT rw R 3500 ln 41.103 = 42.301 − 2π(2574) 0.35 which yields R = 89 m . (c) At any distance r from the well, the Thiem equation gives −1 Q dφ R Q R =− − 2 = dr 2πT r r 2πrT Therefore at 100 m from the well in the high-hydraulic conductivity layer, the seepage velocity, v, is given by v=

K1 dφ K1 Q (402)(3500) = = = 4.8 m/d n dr 2πnrT 2π(0.18)(100)(2574)

435 and in the low-hydraulic conductivity layer, the seepage velocity, v, is given by v=

K2 Q (80.4)(3500) K2 dφ = = = 0.97 m/d n dr 2πnrT 2π(0.18)(100)(2574)

15.10. For an unconﬁned aquifer Qw ln T = 2π(s 1 − s 2 )

r2 r1

where Qw = 400 L/s = 34560 m3 /d, H = 24 m, s1 = 1 m, r1 = 50 m, s2 = 0.5 m, r2 = 100 m. 12 s21 =1− = 0.979 m 2H 2(24) 0.52 s2 = 0.495 m s 2 = s2 − 2 = 0.5 − 2H 2(24)

s 1 = s1 −

and therefore 34560 ln T = 2π(0.979 − 0.495)

100 50

= 7880 m2 /d

The corresponding hydraulic conductivity, K, is given by K=

7880 T = = 328 m/d b 24

There is no steady state for the drawdown. The drawdown distribution is Qw (H − s) = (H − sw ) + ln πK 2

r rw

and when sw = 4 m, rw = 0.5 m, and s = 0 m, (24 − 0)2 = (24 − 4)2 +

34560 r ln π(328) 0.5

which gives r = 95 m If the actual drawdowns used to calculate the transmissivity are s1 = 1 m, and s2 = 0.5 m, then 100 34560 ln = 7625 m2 /d T = 2π(1 − 0.5) 50 and percent diﬀerence =

7625 − 7880 × 100 = −3.2% 7880

436 15.11. In this case the governing equation is dh2 r =0 dr

(1)

dh = Qw 2πrhK dr r=rw

(2)

h(R) = H

(3)

d dr with boundary conditions

Integrating Equation 1 gives r

dh2 =A dr

or 2rh

(4)

dh =A dr

Comparing this with Equation 2 gives A=

Qw πK

(5)

Integrating Equation 4 gives h2 = A ln r + B =

Qw ln r + B πK

(6)

Comparing Equations 3 and 6 gives H2 =

Qw ln R + B πK

Qw ln R πK Finally, combining Equations 5, 6, and 7 gives the equation of the water table as B = H2 −

(7)

h2 = H 2 +

r Qw ln πK R

(8)

h2w = H 2 +

rw Qw ln πK R

(9)

At the well, r = rw and

From the given data: R = 1 km = 1000 m, rw = 0.5 m, Qw = 333 L/s = 28800 m3 /d, K = 40 m/d, and H = 45 m. Substituting into Equation 9 gives h2w = 452 +

0.5 28800 ln π(40) 1000

which yields hw = 16.8 m

437 Hence, the water table elevation at the well is equal to 0 − (45 − 16.8) = −28.2 m . The seepage velocity half-way between the well and the coastline is given by v=

28800 45.8 Qw = = m/d 2π(R/2)hn 2π(1000/2)h(0.2) h

(10)

Substituting known parameters into Equation 8 to ﬁnd h gives h2 = 452 +

500 28800 ln π(40) 1000

which yields h = 43.2 m

(11)

Combining Equations 10 and 11 gives v=

45.8 = 1.06 m/d 43.2

Therefore, the estimated time, tp , to travel from the perimeter of the island to the well is tp =

1000 = 943 days = 2.6 years 1.06

15.12. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, b = 24 m, r1 = 50 m, s1 = 1 m, r2 = 100 m, and s2 = 0.5 m. s1 K0 (r1 /λ) = K0 (r2 /λ) s2 K0 (50/λ) 1 = =2 K0 (100/λ) 0.5 which (by trial and error) gives λ = 121 m The transmissivity, T , of the aquifer is given by T =

34560 K0 (50/121) K0 (r1 /λ) Qw = = 5980 m2 /d 2πs1 (rw /λ)K1 (rw /λ) 2π(1) (0.5/121)K1 (0.5/121)

and the hydraulic conductivity, K, is given by K=

5980 T = = 249 m/d b 24

If leakage is neglected, the problem is the same as Problem 15.7, which gave T = 7630 m2 /d

438 15.13. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, b = 24 m, r1 = 50 m, s1 = 1 m, r2 = 100 m, and s2 = 0.5 m. Applying Equation 15.74 gives s1 K0 (r1 /λ) = K0 (r2 /λ) s2 1 K0 (50/λ) = =2 K0 (100/λ) 0.5 which (by trial and error) gives λ = 121 m Since rw /λ = 0.5/121 = 0.0041 1, using Equation 15.74 is validated, and λ = 121 m . The transmissivity, T , of the aquifer is given by T =

Qw 34560 K0 (50/121) = 5980 m2 /d K0 (r1 /λ) = 2πs1 2π(1)

and the hydraulic conductivity, K, is given by K=

T 5980 = = 249 m/d b 24

If leakage is neglected, the problem is the same as Problem 15.7, which gave T = 7630 m2 /d 15.14. Leakage, q, through the semi-conﬁning layer is q = −K Recall that λ2 =

φ(r) − φ0 b

(1)

Kbb K

and from Problem 15.12, λ = 121 m, K = 249 m/d, b = 24 m, T = 5980 m2 /d, and therefore Kb (249)(24) K = 2 = = 0.408 d−1 b λ 1212

(2)

Also, φ0 − φ(r) =

K0 (r/λ) 34560 K0 (r/121) Qw = 2πT (rw /λ)K1 (rw /λ) 2π(5980) (0.5/121)K1 (0.5/121)

= 0.920K0 (r/121) m

(3)

Combining Equations 1 to 3 gives the leakage, q, as a function of the distance, r, from the well, hence q = 0.408[0.920K0 (r/121)] = 0.376K0 (r/121) m/d

439 S 15.15. q = 2πT = 2πkbS [ According to equation no-15.50] R R ln

ln r

Given, h2 = (20 + 30) − 3 = 47 m r1 = 10 m h1 = (50 − 4) = 46 m

r2 = 60 m

Q{ln R } K = 2πkbsr

= 1.43×10−3 m/sec Now hw is the draw down at the production well. Hence, hw = S0 –Sw 2πT S Q = log r2

e r1 r

(h2 –hw ) = (h2 –hw ) = (h2 –hw ) =

2 Qlog e r1 2πT r2 Qlog 10 r1 2.72∗20∗1.43∗10−3 60 .1log 10 0.25 2.72∗20∗1.43∗10−3

(h2 –hw ) = 3.0588 hw = 43.94 m Sw = 50–43.94 = 6.06 m Draw down in main well = 6.06 m 2

15.16. Q = 1.36k(h2 r−h1 ) [ According to equation no-15.41] log10 r2

Given-

Q = 545L/min = 9.083 ∗ 10−3 Type equation here. h1 = 12.5 − 2.3 = 10.2 h2 = 12.5 − 2.1 = 10.4 Q log10 rr21 k= 1.36(h2 2 − h1 2 ) = 3.89 × 10−4 m/sec 15.17. The theoretical equation is given by s(r) =

K0 (r/λ) Qw 2πT (rw /λ)K1 (rw /λ)

By adjusting T and λ to ﬁt the data with least-squares error gives: T = 1110 m2 /d and λ = 680 m/d .

440 15.18. We know q=

πk(H 2 −h2 )

[ According to equation no-15.41]

loge r2

1.36k(h2 2 −h1 2 ) r

log10 r2

where r1 and r2 are radial distance from main production well. Now in above given problem q = 1500 L/min = 1.5 m3 /min = .025 m3 /sec And draw down in two observation wells are 3.5 and 2.5. h1 = 10–3.5 = 6.5 m h2 = 10–2.5 = 7.5 m Putting these value in above mentioned equation we get, 2

−6.5 ) 0.025 = 1.36k(7.5 log 75 10 25

0.025log

10 25 K = 1.36(7.52 −6.5 2)

= 6.26 ∗ 10−4 T = 6.26 ∗ 10−4 ∗ 10 = 6.26 ∗ 10−3 m2 /sec Also at the well face 2

k ∗ (3.5 −hw ) q = 1.36 ∗ log r1 10 rw

−3

∗ 10 (3.5 −hw ) 0.025 = 1.36 ∗ 6.26log 75 10 0.3

log

∗ 0.025

0.3 (3.52 − hw 2 ) = 1.3610∗ 6.26 ∗ 10−3

hw 2 = 12.25–5.63 hw = 2.57 Sw = 10–2.57=7.43 m Hence, Drawdown at Production well = 7.43 m 15.19. q = vne [ According to equation no-14.10] ne =0.30 20 H1 = 32 m, H2 = 12 m − dh dl = 2000 = 0.01 = i

We Know

441 q = −ki = 20×.01 = 0.2 m3 /d[ According to equation no-14.6] 3 v = nqe = 0.2 0.3 = 0.666 m /d

Time required for pollutant to cover 2 km = 2000/0.666 = 3003 = 8.21 year 15.20. Drawdown, sp , is given by sp = s + Δs For a conﬁned aquifer, top penetration, Δs =

Qw 1 − p (1.2 − p)hs ln 2πT p rw

In this case, Qw = 400 L/s = 34560 m3 /d, T = 7630 m2 /d, b = 24 m, and rw = 0.5 m. If the top half of the aquifer is penetrated: hs = 12 m, p = 0.5 = hs /b, and Δs =

34560 1 − 0.5 (1.2 − 0.5)(12) ln = 2.03 m 2π(7630) 0.5 0.5

For full penetration, the drawdown, s, is 4 m. Therefore, with 1/2 penetration from the top sp = s + Δs = 4 + 2.03 = 6.03 m Using the general expression for drawdown resulting from partial penetration, s = 4 + 0.721

1−p 1 − p (1.2 − p)pb ln ln 48(1.2 − p)p = 4 + 0.721 p rw p

The general guidelines for distance from the well is 2 aquifer depths (48 m in this case). The drawdowns at 50 m and 100 m should be little affected . In this case, 10(rw /b) = 10(0.5/24) = 0.21 and p = 0.5, hence the requirement that 10(rw /b) ≤ p ≤ 0.8 is satisfied and Equations 15.87 and 15.91 are applicable. 15.21. From Problem 15.7: Qw = 400 L/s = 34560 m3 /d, T = 7630 m/d, and rw = 0.5 m. When the well penetrates the top 40% of the (confined) aquifer, the additional drawdown, Δs, at the well is given by Δs =

34560 1 − 0.4 (1.2 − 0.4)(0.4 × 24) Qw 1 − p (1.2 − p)hs ln ln = 2.95 m = 2πT p rw 2π(7630) 0.4 0.5

If the well penetrates the middle 40% of the aquifer, the additional drawdown, Δs, at the well is given by Δs =

Qw 1 − p (1.2 − p)hs 34560 1 − 0.4 (1.2 − 0.4)(0.4 × 24) ln ln = 2.20 m = 2πT p 2rw 2π(7630) 0.4 2(0.5)

15.22. Discharge of each well with interference given by w) Q = 2πkb(H−h R3

ln r

w ∗2a∗2a

442 = 278.76 m3 /d Discharge of each well without interference is w) Q = 2πkb(H−h ln R rw

∗ 2.5 = 2π ∗ 20 ∗ 15 = 512.21 m3 /d 500 2.3 log .05

Reduction in discharge because of Interference = 512.21−278.76 × 100 = 45.57% 512.21 15.23. From the given information, r2 S 4T t 2rS 2u ∂u = = ∂r 4T t r r2 S ∂u u =− =− 2 ∂t 4T t t u=

Hence, ∂s ∂u 2u ∂s ∂s = = ∂r ∂u ∂r r ∂u and 2u ∂ ∂2s = 2 ∂r r ∂u

2u ∂s r ∂u

∂r 2u 2u ∂ 2 s 2r − 2u ∂u 2u 2u ∂ 2 s 2r − 2u(r/2u) ∂s ∂s = + + r r ∂u2 r2 ∂u r r ∂u2 r2 ∂u

which gives 2u 2u ∂ 2 s 1 ∂s ∂2s = + ∂r2 r r ∂u2 r ∂u and ∂s ∂u u ∂s ∂s = =− ∂t ∂u ∂t t ∂u 15.24. From the given data: Qw = 400 L/s = 34560 m3 /d, rw = 0.5 m, H = 24 m, S = 0.0012, K = 300 m/d, and T = KH = 300 × 24 = 7200 m2 /d. The drawdown, s(r, t), is given by s(r, t) =

34560 Qw W (u) = W (u) 4πT 4π(7200)

which simpliﬁes to s = 0.3820W (u) and u= where

r2 (0.0012) r2 S r2 = = 4.17 × 10−8 4T t 4(7200)t t

443 r (m) 0.5 50 100

u 1.04 × 10−8 /t 1.04 × 10−4 /t 4.17 × 10−4 /t

Use these relations to calculate the drawdown as a function of time (in days) at r = 0.5, 50, and 100 m. 15.25. Q = 2πTR sd [ According to equation no-15.50] ln

Qln R So, T= 2πsrd

In ﬁrst case Sd = 3m Q = .6 m3 /min R = 500 m and r = .250 m So, T =

500 .6∗ln .250 2π∗3

= .24 m2 /min

If sd = 6 m sd Then Q = 2πT = 2π∗.24∗6 = 1.2 m3 /min = 1200 lit/m ln 500 ln R r

0.250

In second case rw = 30/2 = 15 cm = 150mm Sd = 3 m ∗6 So, Q = 2ln∗ π.24 500 0.150

= .557 m3 /min = 557.69 Lit /min = 558 Lpm 15.26. From the given data: Qw = 200 L/s = 17280 m3 /d, Sy = 0.15, K = 100 m/d, H = 28 m, r1 = 50 m, and r2 = 100 m. The transmissivity, T , of the aquifer is therefore given by T = 100(28) = 2800 m2 /d The drawdown, s, is given by s=

Qw W (u) 4πT

and

r2 Sy 4T t

444 Hence, 0.03348 (50)2 (0.15) = 4(2800)t t 2 0.1339 (100) (0.15) = u2 = 4(2800)t t 17280 W (u) = 0.4911W (u) s= 4π(2800) u1 =

Using the Thiem equation, the transmissivity is estimated from the drawdown measurements by 17280 Qw r2 100 1906 = T = ln ln = m2 /d 2π(s1 − s2 ) r1 2π(s1 − s2 ) 50 s1 − s 2 where

s s2 =s 1− 2H 2H The transmissivities are computed in the following table: s = s −

Time

1s 1 min 1h 1 day 1 month 1 year

2893 48.2 0.80 0.033 0.00112 0.000186

W (u1 ) 0 0 0.3106 2.88 6.248 8.037

W (u2 )

1570 192.8 3.21 0.134 0.00446 0.000742

0 0 0 1.64 4.846 6.632

s 1 (m) 0 0 0.153 1.378 2.900 3.669

s 2 (m) 0 0 0 0.795 2.279 3.068

T (m2 /d) ∞ ∞ 12460 3269 3069 3171

On the basis of these results, it would not be recommended to use the steady-state Thiem equation to calculate the transmissivity. 15.27. From the given data: b = 24 m, Qw = 400 L/s = 34560 m3 /d, and r = 50 m. Plotting the given data as ln s versus ln r2 /t and comparing to ln W (u) versus ln u yields α = 17.1 and β = −0.998. Equation 15.132 yields T =

34560 0.998 Qw −β = 7460 m2 /d e = e 4π 4π

and Equation 15.133 yields S = 4T e−α = 4(7460)e−17.1 = 0.00112 The hydraulic conductivity, K, of the aquifer is given by K=

7460 T = = 311 m/d b 24

15.28. Anisotropic formations can be mapped into isotropic formations using a coordinate transformations. For example, if a formation is anisotropic in the x, y plane, then the following coordinate transformations map the anisotropic x, y domain into an isotropic x , y domain, T T x, y = y (1) x = Txx Tyy

445 where T =

Txx Tyy

(2)

Isotropic solutions can be applied in the transformed domain and mapped back into the (real) anisotropic domain. Many aquifers are isotropic in the horizontal plain because the flow paths are created preferentially in the direction of ground water flow. This is particularly true in limestone formations where the flow tends to be primarily through dissolution cavities that are created in the flow direction. From the given data: Kxx = 30 m/d, Kyy = 15 m/d, Kxy = 10 m/d, b = 20 m, S = 0.005, and n = 0.15. Since the principal hydraulic conductivities are used in the coordinate transformation, and axis rotation is required. The principal components of the hydraulic conductivities, Kxx Kyy are given by (see Problem 7.9) Kxx + Kyy + = Kxx 2 Kxx + Kyy − Kyy = 2 which yield 30 + 15 + Kxx = 2 30 + 15 − = Kyy 2

Kxx − Kyy 2 Kxx − Kyy 2

30 − 15 2 30 − 15 2

1 2

2 + Kxy

1 2

2 + Kxy

1 2

+ 102 2

= 35 m/d 1 2

+ 102

= 10 m/d

and the angle of rotation, θ, between the original and principal axes is given by Equation 7.25 as 2Kxy 1 2(10) 1 = −26.6◦ = sin−1 θ = sin−1 2 Kyy − Kxx 2 10 − 35 Therefore, the principal axes correspond to the original axes rotated clockwise by 26.6◦ . The principal transmissivities in the anisotropic aquifer are Txx = Kxx b = (35)(20) = 700 m2 /d = Kyy b = (10)(20) = 200 m2 /d Tyy

The isotropic transmissivity, T , in the mapped isotropic domain is given by Equation 8.2 as T = (700)(200) = 374 m2 /d T = Txx yy

446 and the relationships between the anisotropic (principal) coordinates (x, y) and the coordinates in the equivalent isotropic domain (x , y ) are given by Equation 8.113 as x =

y =

T x= Txx

374 x = 0.731x 700

T y= Tyy

374 y = 1.367y 200

According to the Theis equation, the drawdown, s, as a function of time in the equivalent isotropic domain is given by Qw W (u) s= 4πT Where s = 1m and Qw = 40 L/s = 3456 m3 /d, then the Theis equation gives 1=

3456 W (u) 4π(374)

which yields W (u) = 1.36 which corresponds to u = 0.158 =

r2 S 4T t

For S = 0.005, and t = 1 day, then r=

0.158(4T t) = S

0.158(4)(374)(1) = 217 m 0.005

Therefore, any x , y coordinate with r = 217 m will meet the drawdown criterion for locating the monitoring well. One possible location is x = 217 m, y = 0 m, which corresponds to the following location in the (real) anisotropic formation: 217 x = = 297 m 0.731 0.731 0 y = =0m y= 1.367 1.367

Since these coordinates are in the principal coordinate system, which is rotated 26.6◦ from the actual coordinate system, the the coordinates of the monitoring well (xm , ym ) are given by xm = x cos θ + y sin θ = 297 cos 26.6◦ + 0 sin 26.6◦ = 266 m ym = −x sin θ + y cos θ = −297 sin 26.6◦ + 0 cos 26.6◦ = −133 m A monitoring well located at (266 m,−133 m) is expected to yield a drawdown of 1 m after 1 day of pumping. There are an inﬁnite number of locations for the monitoring well, each meeting the criterion that r = 217 m in the equivalent isotropic formation.

447 15.29. (a) The drawdown given by the Theis equation in an equivalent homogeneous porous medium is 2 r S Qw W (1) s= 4πT 4T t The mapping between the heterogeneous medium and the equivalent homogeneous medium is given by T T x, y = y, T = Txx Tyy (2) x = Txx Tyy where the primed coordinates are in the equivalent homogeneous domain. Combining Equations 1 and 2 and noting that r2 = (x )2 + (y )2 yields the following expression for the drawdown in the (real) anisotropic domain, ⎞ ⎛ Tyy 2 Txx 2 S⎟ Txx x + Tyy y ⎜ Qw ⎟ ⎜ s= W (3) ⎠ 4π Txx Tyy ⎝ 4 Txx Tyy t (b) From the given data Qw = 3.47 L/s = 300 m3 /d, x1 = 86.60 m, y1 = 50.00 m, x2 = −75.00, y2 = 129.90 m. Using the Cooper-Jacob approximation, Q Qw w W (u1 ) = (−0.5772 − ln u1 ) 4π Txx Tyy 4π Txx Tyy Qw Qw s2 = W (u2 ) = (−0.5772 − ln u2 ) 4π Txx Tyy 4π Txx Tyy u2 Qw Qw ln [ln u2 − ln u1 ] = s1 − s2 = u1 4π Txx Tyy 4π Txx Tyy s1 =

which can be written as

⎡

⎤

x22 y22 Qw Txx + Tyy ⎦ ln ⎣ x2 s1 − s2 = y12 1 4π Txx Tyy + Txx Tyy

(4)

Comparing the drawdown differences given by Equation 4 with the measured drawdown differences (0.12–0.13 cm) for various values of Txx and Tyy yields a best fit (in terms of mean square error) for Txx = 1000 m2 /d and Tyy = 5000 m2 /d . The theoretical drawdown at monitoring well 1 is given by ⎡ ⎛ 2 ⎞⎤ Tyy 2 Txx 2 S ⎟⎥ ⎢ ⎜ Txx x1 + Tyy y1 Qw ⎢ ⎜ ⎟⎥ s1 = ⎢−0.5772 − ln ⎜ ⎟⎥ ⎝ ⎠⎦ 4π Txx Tyy ⎣ 4 Txx Tyy t

(5)

Comparing the drawdown predicted by Equation 5 with the measured drawdown at monitoring well 1 yields (with minimum square error) S = 0.001 . The adequacy of using the Cooper-Jacob approximation for all calculations in this problem is demonstrated by showing that u1 < 0.004 for all drawdown measurements.

448 15.30. From the given data: rw = 0.5 m, b = 24 m, Qw = 400 L/s = 34560 m3 /d, and r = 50 m. Plotting the given data as ln s versus ln r2 /t and comparing to ln W (u) versus ln u yields α = 13.4 and β = −2.1. Equation 15.132 yields T =

Qw −β 34560 2.1 e = e = 22500 m2 /d 4π 4π

and Equation 15.133 yields S = 4T e−α = 4(22500)e−13.4 = 0.14 This result indicates that the aquifer is probably unconfined. The hydraulic conductivity, K, of the aquifer is given by 22500 T = 938 m/d K= = b 24 15.31. From the given data: Qw = 270 L/s = 0.27 m3 /s, H = 24 m, and r = 50 m. The Cooper-Jacob equation gives 2 2 r S 0.27 50 S Qw −0.5772 − ln = −0.5772 − ln s= 4πT 4T t 4πT 4T t 0.02149 S = −0.5772 − 6.438 − ln T Tt which simpliﬁes to s=

0.02149 T

0.02149 ln t + T

S −7.015 − ln T

(1)

The Cooper-Jacob approximation requires that u ≤ 0.004 or 502 S S r2 S ≤ 0.004 → ≤ 0.004 → ≤ 6.4 × 10−6 t s/m−2 4T t 4T t T From the given drawdown measurements, t (s) 1 10 100 1000 10000 100000

ln t 0 2.30 4.61 6.91 9.21 11.51

s (m) 0 0.0639 0.5118 1.0052 1.7209 2.1421

S/T (s/m−2 ) − 6.4 × 10−5 6.4 × 10−4 0.0064 0.064 0.64

The slope of the line in the s vs. ln t space that goes through the points at t = 1000 s, and t = 100000 s is 2.1421 − 1.0052 slope = = 0.2472 11.51 − 6.91

449 Therefore, based on the Cooper-Jacob equation (Equation 1), 0.02149 = 0.2472 T which leads to T = 0.0869 m2 /s = 7510 m2 /d The corresponding hydraulic conductivity, K, is K=

T 7510 = = 313 m/d H 24

Using the calculated slope (= 0.2472), the Cooper-Jacob equation can be written as S 0.02149 −7.015 − ln s = 0.2472 ln t + T T Since s = 1.0052 m when ln t = 6.91, then 1.0052 = 0.2472(6.91) +

0.02149 0.0869

−7.015 − ln

S 0.0869

which simpliﬁes to S = 0.0013 Verify that the Cooper-Jacob approximation is valid: 0.0013 S = = 0.01496 s/m2 T 0.0869 From the above table, the Cooper-Jacob approximation is reasonable after about 2300 seconds . 15.32. From the given data: Qw = 22 L/s = 1900 m3 /d, and r = 70 m. Step 1: Matching the early-time drawdown data to the Type A curve yields the following values Γ = 0.60 α = 20.2 β = −2.60 and hence the transmissivity, T , and storage coeﬃcient, S, are given by 1900 −(−2.60) Qw −β e = e = 2035 m2 /d 4π 4π S = 4T e−α = 4(2035)e−20.2 = 0.0000137

T =

450 Step 2: Taking Γ = 0.60 and matching the late-time drawdown data to the Type B curve yields the following values α = 11.00 β = −2.60 and hence the transmissivity, T , and speciﬁc yield, Sy , are given by 1900 −(−2.60) Qw −β e = e = 2035 m2 /d 4π 4π Sy = 4T e−α = 4(2035)e−11.00 = 0.136 T =

Step 3: For a transmissivity, T , equal to 2035 m2 /d and a saturated thickness, H, equal to 30 m, the horizontal hydraulic conductivity, Kh is given by Kh =

2035 T = = 67.8 m/d H 30

Taking Γ = 0.60 and using the deﬁnition of Γ given by Equation 15.158 yields r2 Kv H 2 Kh (70)2 Kv 0.60 = (30)2 (67.8) Γ=

which gives Kv = 7.47 m/d In summary, the test data indicate that the aquifer has a horizontal hydraulic conductivity of 68 m/d , a vertical hydraulic conductivity of 7 m/d , a specific yield of 0.14 , and a storage coefficient of 0.0000137 . The curves used to solve this problem are shown in Figure 15.1. 15.33. From the given data: H = 25 m, Qw = 40 L/s = 3456 m3 /d, and r = 40 m. Matching the early-time drawdown data to the well function given in Table 15.5 (type A curve) gives Γ = 0.2, and the origin shifts are α = 14.9 and β = −1.00. The transmissivity, T , and storage coefficient, S, are therefore given by Qw −β 3456 −(−1.00) e = e = 748 m2 /d 4π 4π S = 4T e−α = 4(748)e−14.9 = 0.001

T =

451

Figure 15.1: Neuman Curves

452 Matching the late-time drawdown data to the well function given in Table 15.6 (type B curve) with Γ = 0.2 gives α = 9.90 and β = −1.00. which yields 3456 −(−1.00) Qw −β e = e = 748 m2 /d 4π 4π Sy = 4T e−α = 4(748)e−9.90 = 0.15 T =

The horizontal hydraulic conductivity, Kh , can be estimated as Kh =

T 748 = = 29.9 m/d H 25

Since Γ is deﬁned by the relation r2 Kv H 2 Kh then the vertical hydraulic conductivity, Kv , is given by 2 2 H 25 ΓKh = (0.2)(29.9) = 2.3 m/d Kv = r 40 Γ=

In summary, the aquifer properties derived from the pump-drawdown data are: S = 0.001 , Sy = 0.15 , Kh = 29.9 m/d , and Kv = 2.3 m/d . The hydraulic conductivities can reasonably be approximated by Kh = 30 m/d and Kv = 2 m/d. 15.34. From the given data: Qw = 68.14 L/s = 5887 m3 /d, and r = 22.25 m. Comparing the ln W (u, Γ) vs. ln u curve with the ln s vs. ln(r2 /t) curve to determine the origin shift (α, β) and curve parameter Γ yields the following values: Γ = 0.2 α = 14.9 β = −1.27 and hence the transmissivity, T , and storage coeﬃcient, S, are given by Qw −β 5887 −(−1.27) = 1668 m2 /d e = e 4π 4π S = 4T e−α = 4(1668)e−14.9 = 0.00226

T =

Taking Γ = 0.2 and comparing the ln W (uy , Γ) vs. ln uy curve with the ln s vs. ln(r2 /t) curve to determine the origin shift (α, β) yields α = 9.65 β = −1.24 and hence the transmissivity, T , and speciﬁc yield, Sy , are given by Qw −β 5887 −(−1.24) e = e = 1619 m2 /d 4π 4π Sy = 4T e−α = 4(1619)e−9.65 = 0.42 T =

453 The average transmissivity, T , is equal to (1668 + 1619)/2 = 1644 m/d. For a transmissivity, T , equal to 1644 m2 /d and a saturated thickness, H, equal to 23.77 m, the horizontal hydraulic conductivity, Kh , is given by Kh =

1644 T = = 69 m/d H 23.77

Taking Γ = 0.2 and using the deﬁnition of Γ yields r2 Kv H 2 Kh (22.25)2 Kv 0.2 = (23.77)2 (69) Γ=

which gives Kv = 16 m/d In summary, the test data indicate that the aquifer has a horizontal hydraulic conductivity of 69 m/d , a vertical hydraulic conductivity of 16 m/d , a speciﬁc yield of 0.42 , and a storage coeﬃcient of 0.00226 . 15.35. From the given data: Kxx = 45 m/d, Kyy = 15 m/d, Kxy = 0 m/d, b = 14 m, S = 10−4 , Qw = 16.7 L/s = 1440 m3 /d, and t = 1 week = 7 days. Using an equivalent isotropic medium T = Txx Tyy = Kxx Kyy b2 = (45)(15)(14)2 = 364 m2 /d and the (primed) coordinates in the equivalent isotropic medium are given by T 364 x= x = 0.760x x = Txx (45)(14) T 364 y = 1.32y y= y = Tyy (15)(14) Point x = 100 m and y = 100 m corresponds to x = 0.760(100) = 76 m y = 1.32(100) = 132 m r 2 = x 2 + y 2 = (76)2 + (132)2 = 23200 m2 Drawdown in the equivalent isotropic medium is given by the Theis equation as 2 Qw r S 1440 23200 × 10−4 s= W = W = 0.315W (2.28 × 10−4 ) 4πT 4T t 4π(364) 4 × 364 × 7 = 0.315(7.83) = 2.47 m

454 With a leakage factor, λ, of 2000 m, √ 2 r S r 23200 Qw W , = 0.315W 2.28 × 10−4 , = 0.315W 2.28 × 10−4 , 0.0761 s= 4πT 4T t λ 2000 = 0.315(5.404) = 1.70 m 15.36. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, b = 24 m, K = 300 m/d, and S = 0.0012. The drawdown is given by Equation 15.177 as s(r, t) =

r Qw W u, 4πT λ

(1)

In this case, T = Kb = (300)(24) = 7200 m2 /d, r = 50 m, and u is given by u=

(50)2 (0.0012) 1.04 × 10−4 r2 S = = 4T t 4(7200)t t

(2)

Combining Equations 1 and 2 gives s=

34560 W 4π(7200)

which simpliﬁes to

s = 0.382W

1.04 × 10−4 50 , t λ

This equation indicates that at any given location and at any given time, increased leakage factors result in increased drawdowns. The leakage factor, λ, is deﬁned by λ2 =

Kbb K

K =

Kbb λ2

which gives

For b = 5 m and λ = 10 m, K =

(300)(24)(5) = 360 m/d 102

and for b = 5 m and λ = 100 m, K =

(300)(24)(5) = 3.6 m/d 1002

It is interesting to note that when λ = 10 m, the hydraulic conductivity of the semiconﬁning layer exceeds the hydraulic conductivity of the main aquifer. In this case, the aquifer is not semiconﬁned.

455 15.37. Solution summary: 1. Plot the data as s vs. r2 /t. 2. Match the W (u, Γ) function, which gives: Γ = 0.01, S = 1.5 ×10−4 , and T = 50000 m2 /d . 3. Match the W (u, r/λ) function, which gives: r/λ = 0.08, λ = 625 m , K = 0.64 m/d , and Kv = 10 m/d . 15.38. From the given data: b = 80 m, b = 28 m, and Qw = 7.23 L/s = 625 m3 /d. Using the leaky well function gives: K = 20 m/d , K = 0.05 m/d , and S = 0.0018 . 15.39. From the given data: b = 4.5 m, Qw = 1.58 L/s = 136.8 m3 /d, and r = 20 m. From the curve match, α = 12.76, β = −0.90, r/λ = 0.1, and Equation 7.256 gives the transmissivity, T , of the aquifer as 136.8 −(−0.90) Qw −β e e = = 26.8 m2 /d T = 4π 4π Equation 7.257 gives the storativity, S, of the aquifer as

S = 4T e−α = 4(26.8)e−12.76 = 0.00031 Since the leaky well function that matches the drawdown data has the parameter r/λ = 0.1, and the distance, r, of the observation well from the pumping well is 20 m, then λ=

20 r = = 200 m 0.1 0.1

The leakage factor, λ, is deﬁned by Equation 7.168 as λ=

Kbb = K

T b K

which can be put in the form

T b λ2 where K and b are the hydraulic conductivity and thickness of the semi-pervious layer respectively. In this case, T = 26.8 m2 /d, b = 4.5 m, λ = 200 m, and therefore K =

K = 15.40. Q =

πk(h2 2 −h1 2 ) r

loge r2

(26.8)(4.5) = 0.0030 m/d (200)2

[ According to equation no-15.41]

2 2 2 −h1 ) = 1.36k(h r2 log10 r 1

h1 = 30–5 = 25, h2 = 30–4.2 = 25.8, and r1 = 15, r2 = 30 k = 22 m/day

456 Q=

1.36 ∗ 22 ∗ (25.82 −252 ) log10 30 15

= 4039 m3 /day = 168 m3 /h 1.36 ∗ k(h1 2 −hw 2 ) Q= r1 log e10 rw

22 ∗ (252 −hw 2 ) 4039 = 1.36 ∗ log 25 10 .25

h2w = 625 − 245.92 hw = 19.46 sw = 30 − 19.46 = 10.53m 15.41. From the given data: H = 20 m, K = 40 m/d, (x, y) = (100 m, 100 m), T = KH = (40)(20) = 800 m2 /d, Sy = 0.2, and Qw = 200 L/s = 17280 m3 /d. Point A B C

(x, y) (m,m) (0,0) (200,200) (200,-200)

r (m) 141.4 141.4 316.2

r2 S

u = 4T ty 1.243/t 1.243/t 6.249/t

where t is in days. Using the Theis equation s=

17280 Qw [W (u1 ) + W (u2 ) + W (u3 )] = [2W (u1 ) + W (u2 )] = 1.719[2W (u1 ) + W (u2 )] 4πT 4π(800)

This equation gives the drawdown as a function of time. To illustrate the use of this equation, start with t = 1 day: t (days) 1

W (u1 )

W (u2 )

1.243

0.1780

6.249

0.00033

s (m) 0.613

15.42. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, H = 24 m, K = 300 m/d, and S = 0.012. The equation describing the drawdown adjacent to a constant-head boundary is Qw R r2 r2 R Qw 34560 r2 s= ln ln ln − ln = = = 0.7639 ln 2πT r1 r2 2πT r2 2π(300 × 24) r1 r1 Applying this equation at (100 m, 0 m) and (−100 m, 0 m) gives Point (100,0) (−100,0)

r1 (m) 100 100

r2 (m) 900 1100

s (m) 1.678 1.832

457 15.43. For Well No.2 at (0 m,200 m), and Qw = 400 L/s = 34560 m3 /d, induced drawdown, s2 , is given by r r2 Qw ln = 0.7639 ln 2 s2 = 2πT r1 r1 Point (100,0) (−100,0)

r1 (m) 223.6 223.6

r2 (m) 922.0 1118.0

s2 (m) 1.082 1.229

s1 (m) 1.678 1.832

s 2.760 3.061

If Well No.2 is to be located at (0,y), then @ (100 m,0 m), 1 9002 + y 2 (9002 + y 2 ) 2 = 0.3820 ln s2 = 0.7639 ln 1 1002 + y 2 (1002 + y 2 ) 2 and the percent drawdown contributed by Well No.2 is % of total =

s2 100s2 × 100 = s1 + s2 s1 + s2

The value of y that gives a 1% error is y = 4917 m, and hence the coordinates of Well No.2 are (0 m, 4917 m) . 15.44. From the given data: rw = 0.8 m, Qw = 500 L/s = 43200 m3 /d, b = 24 m, K = 250 m/d, and Sy = 0.2. Assuming that the drawdown is small relative to the saturated thickness of the aquifer, the drawdown distribution surrounding the well is given by s=

Qw (x − x0 − 2L)2 + (y − y0 )2 ln 4πT (x − x0 )2 + (y − y0 )2

where (x0 , y0 ) are the coordinates of the well, and L is the distance from the well to the constant-head boundary. Taking (x0 , y0 ) = (0,0) and L = 500 m, the drawdown distribution is given by s=

(x − 1000)2 + y 2 43200 (x − 1000)2 + y 2 ln = 0.5730 ln 4π(250 × 24) x2 + y 2 x2 + y 2

The seepage out of the canal, q, is given by ! +∞ ∂s dy −Kb q= ∂x x=500 m −∞

(1)

where ∂s x2 + y 2 2(x2 + y 2 )(x − 1000) − [(x − 1000)2 + y 2 ]2x = 0.5730 ∂x x=500 m (x − 1000)2 + y 2 (x2 + y 2 )2 x=500 m 1146 =− 5002 + y 2

458 Substituting into Equation 1 gives ! +∞ ! +∞ 1146 1 6 (250)(24) dy = 6.876 × 10 dy q= 2 2 2 2 500 + y −∞ −∞ 500 + y y +∞ 1 tan−1 = 6.876 × 106 500 500 −∞ " π π # − − = 13752[tan−1 (+∞) − tan−1 (−∞)] = 13752 2 2 3 = 43200 m /d Since the withdrawal rate from the canal (= 43200 m3 /d) is equal to the pumping rate, then 100% of the pumped water originates from the canal. 15.45. The drawdown distribution, s(x, y), is the sum of the drawdowns induced by two pumping wells placed symmetrically about an the impermeable boundary. Hence, 2 R R R Qw Qw ln + ln ln = s(x, y) = sp + si = 2πT r r 2πT rr where R is the radius of inﬂuence, r is the radial distance of (x, y) from the pumping well, and r is the radial distance of (x, y) from the image well. From the given data, Qw = 400 L/s = 34560 m3 /d, T = Kb = (300)(24) = 7200 m2 /d, and R = 1200 m. Substituting the given data into the drawdown equation yields 10002 12002 34560 ln = 0.764 ln s(x, y) = 2π(7200) rr rr The drawdowns at (100 m, 0 m) and (−100 m, 0 m) are given in the following table: Location (100 m,0 m) (−100 m,0 m)

r (m) 100 100

r (m) 900 1100

s (m) 2.12 1.96

15.46. Using the method of images, the drawdown distribution is the sum of the drawdowns induced by four pumping wells placed symmetrically about the impermeable boundary. The real wells are at (0 m, 0 m) and (0 m, 200 m), the image wells are at (1000 m, 0 m) and (1000 m, 200 m), and the impermeable boundary is along the line x = 500 m. The drawdown distribution is therefore given by 2 2 Qw Qw Qw R R R4 + = s= ln ln ln 2πT r1 r1 2πT r2 r2 2πT r1 r1 r2 r2 where r1 is the distance from the well at (0 m, 0 m), r2 is the distance from the well at (0 m, 200 m), and r1 and r2 are the distances from the corresponding image wells. From the given data: Qw = 400 L/s = 34560 m3 /d, T = Kb = (300)(24) = 7200 m/d, and R = 1200 m, therefore 34560 12004 12004 s= = 0.764 ln ln 2π(7200) r1 r1 r2 r2 r1 r1 r2 r2

459 The drawdown calculations are summarized in the following table: Point

r1 (m) 100 100

(100 m, 0 m) (−100 m, 0 m)

r1 (m) 900 1100

r2 (m) 224 224

r2 (m) 922 1118

s (m) 3.60 3.30

The drawdown, s1 , at any location caused by the well at (0 m, 0 m) is given by 2 R Qw ln s1 = 2πT r1 r1 and the drawdown, s2 , at any location caused by the well at (0 m, 200 m) is given by 2 R Qw s2 = ln 2πT r2 r2

(1)

(2)

If the second well contributes 1% to the drawdown, then s2 = 0.01(s1 + s2 )

(3)

Combining Equations 1 to 3 gives ln(R2 /r2 r2 ) = 0.01 ln(R4 /r1 r1 r2 r2 ) which can be written as

ln(12002 /r2 r2 ) = 0.01 ln(12004 /r1 r1 r2 r2 )

This equation is satisfied at location (100 m, 0 m) when the second well is at (0 m, 1023 m), and is satisfied at (−100 m, 0 m) when the second well is at (0 m, 957 m). Hence, when the second well is further than (0 m, 1023 m) from the first well, the drawdown contributed by the second well is less than 1% of the total drawdown. 15.47. From the given data: rw = 0.5 m, Qw = 400 L/s = 34560 m3 /d, H = 24 m, K = 300 m/d, S = 0.012, and R = 1200 m. Since the radius of influence is 1200 m, and the required image wells would be at least 1800 m from the locations of interest, then there is no need to consider any of the image wells. The drawdown, s, 200 m from the well is given by s=

R 34560 1200 Qw ln = ln = 1.37 m 2πT r 2π(300 × 24) 200

Hence the drawdown 200 m east of the well is 1.37 m and the drawdown 200 m west of the well is 1.37 m . 15.48. The drawdown distribution caused by a single well in an inﬁnite aquifer is given by 2 R Qw R R2 Qw Qw ln = ln ln = s= 2πT r 4πT r2 4πT (x − x0 )2 + y 2

460 Taking the coordinates of the pumping well as (x0 ,0), then to account for the streams on both sides of the well, image injection wells must be placed at (−x0 ,0) and (2d − x0 ,0). These image wells create imbalances at the opposite boundaries that must be balanced by image pumping wells at (2d + x0 ,0) and (−2d + x0 ,0) respectively. In turn, these image pumping wells produce imbalances that must be countered by image injection wells at (−2d − x0 ,0) and (4d − x0 ,0). This sequence of corrections continues indeﬁnitely, giving the total drawdown as R2 R2 R2 Qw ln − ln + ln − s= 4πT (x − x0 )2 + y 2 (x + x0 )2 + y 2 (x − x0 + 2d)2 + y 2 R2 R2 R2 ln + ln − ln + (x + x0 − 2d)2 + y 2 (x − x0 − 2d)2 + y 2 (x + x0 + 2d)2 + y 2 R2 R2 − ln + ··· ln (x − x0 − 4d)2 + y 2 (x + x0 + 4d)2 + y 2

which can be compactly written as s=

∞ Qw $ (x + x0 − 2nd)2 + y 2 ln 4πT n=−∞ (x − x0 − 2nd)2 + y 2

15.49. See Figure 15.2.

Figure 15.2: Problem 15.49. 15.50. This can be shown by deriving the drawdown equation for an injection well in an unconﬁned aquifer. The governing ﬂow equation is given by d2 (h2 ) 1 d(h2 ) =0 + dr2 r dr which can also be written as d dr

dh2 r =0 dr

(1)

461 The boundary conditions for an injection well are dh = −Qw 2πrw hK dr r=rw h(rw ) = hw Integrating Equation 1 twice with respect to r yields h2 = A ln r + B

(2)

Applying the boundary conditions yield A=−

Qw , πK

B = h2w − A ln rw

(3)

Substituting Equation 3 into Equation 2 yields the following expression for the piezometric head distribution in an unconﬁned aquifer surrounding an injection well r Qw 2 2 ln (4) h = hw − πK rw The drawdown, s, is deﬁned by s(r) = H − h(r) in which case Equation 4 can be written as Qw ln (H − s) = (H − sw ) − πK 2

r rw

where sw is the drawdown at the well. The drawdown equation for an injection well can be put in the form r Qw ln s = sw − (5) 2πKH rw where s = s −

s2 , 2H

s w = sw −

s2w 2H

The drawdown equation for a pumping well is s

= s w +

Qw ln 2πKH

r rw

(6)

where s w at a pumping well is equal and opposite to s w at an injection well for the same flowrate, Qw . Therefore, comparing Equations 5 and 6 demonstrates that the buildup distribution caused by injecting water at a rate Qw into a fully penetrating well in an unconfined infinite aquifer is exactly the same as the drawdown distribution caused by withdrawing water at a rate Qw from the same aquifer.

462 15.51. T = 600 m2 /d = 6.94×10−3 m2 /s 2

U = r4TSt =

0.152 ×0.005 = 2.34×10−8 4×6.94×10−3 ×48×60×60

Using the thies method W(u) to four signiﬁcant digit W(u) = –0.5772-ln(2.34×10−8 )+(2.34×10−8 )2 –

(2.34×10−8)2 (2.34×10−8)3 + = 18.14 2×2! 3×3!

We know Q w(u) = 10×4×π×6.94×10 Sn = 4πT 18.14

−3

= .048 m3 /s = 2884 lpm

The pumping rate should be 2884 lpm 15.52. From the given data: x = 3 km = 3000 m, K = 40 m/d, and z = 50 m. Using the Glover solution, Equation 15.256 requires that Q2 2 2 2 z= (Qx − C) = (Qx + QW ) = Qx + (1) K K K 2 K For the given information and assuming = 0.025, Equation 15.257 is a quadratic equation in Q in which the positive root of Q is given by Q= K x2 + z 2 − x = (0.025)(40) 30002 + 502 − 3000 = 0.417 m2 /d Therefore, the fresh-water discharge per kilometer of shoreline is 0.417×1000 = 417 (m3 /d)/km. 15.53. (a) The Ghyben-Herzberg equation (Equation 15.248) is used to calculate the depth of the saltwater interface corresponding to the given water-table contours. The intersection of the saltwater interface with the bottom of the surﬁcial aquifer is equated to the extent of saltwater intrusion, and the extent of saltwater intrusion corresponding to both high and low water table conditions should be shown on an area map (not included in this solution). (b) The freshwater inﬂow is calculated using Equation 15.252, which is given by Q=

K K 2 hL ≈ (40)h2L 2L 2L

Estimating the hydraulic conductivity, K, from the transmissivity contours and using the water-table elevations, hL , and distances L from the coast yields a freshwater inﬂow to Florida Bay of approximately 1–2 m2 /d in the wet season and 0.1–1 m2 /d in the dry season. 15.54. Let the water surface in the canal upstream of the gate be a distance h above the water surface downstream of the gate, which can be taken as 30 cm above mean sea level. The water surface in the canal upstream of the gate should be suﬃcient to keep the toe of the salt water wedge at the bottom of the aquifer, which is a distance z below the elevation of the sea water downstream of the gate. Since the elevation of the bottom of the canal is 3 m below sea level, then z = 24 + 3 + 0.3 = 27.3 m

463 and the Ghyben-Herzberg equation gives h≈

1 1 = = 0.68 m 40z 40(27.3)

Hence the elevation of the water surface upstream of the gate should be at least 0.68 m + 0.30 m = 0.98 m above mean sea level to prevent salt water intrusion. 15.55. The maximum pumpage, Qm , is related to the distance, d, below the well by Qm = 0.6πd2 K where (at 20◦ C) =

(1)

1.025 − 0.998 Δρ = 0.0271 = ρf 0.998

From the given data: Qm = 40 L/s = 3456 m3 /d, and K = 500 m/d, therefore Equation 1 gives 3456 Qm = = 135 m2 d2 = 0.6πK 0.6π(500)(0.0271) which gives d = 11.6 m Since the well is 15 m above the bottom of the aquifer, then the well will become unusable when the thickness of the saltwater wedge is 15 m − 11.6 m = 3.4 m . w 15.56. T = 2π(sQ1− s

ln ( rr21 ) [According to equation 15.34]

1.5 ln ( 93 ) = 179.84 m2 /d = 2π(6.8−4.7)

Chapter 16

Design of Ground-Water Systems 16.1. Bottom of the well, d = 30 m at a distance (r1 ) of 6 m draw down, s1 = 6m, at a distance (r2 ) of 15 m draw down, s2 = 1.5m, Maximum drawdown in the well, SW = 2.2 m Therefore, hW = (30–2.2) = 27.8m Radius of well, rW = 150 mm = 0.15 m We have, Discharge of well, QW = 50 m3 /hr. = 0.014 m3 /s Co-eﬃcient of permeability, K = 0.00014 m/s We have, 2

−h w ) QW = πK(d R ln

π × 0.00014(302 − 27.82 ) ln rRW R = 3.995 =⇒ ln 0.15 =⇒ R = 8 m =⇒ 0.014 =

−29 ) = 6.53 × 10−3 m3 /s Speciﬁc capacity of the well (for unit drawdown) = πK(30 8 ln( 0.15 )

16.2. So far 5 MGD capacity with 20 hrs pumping per day = 5 M20GD = 25 × 104 GPH Also discharge in 1m length of lateral = 10750.33 GPD = 538 GPH 4

Required length lateral = 25×10 m = 465 m 538 According to equation (16.8) QW vs = Cπds Ls P Discharge of the well, Qw = 10750.33 GPD/m = 0.0113 m3 /s 464

465 Clogging Co-eﬃcient, C = 0.5 Diameter of the screen, dS = 0.3 m Entrance velocity, vs = 0.5 cm/s = 0.005 m/s Percentage of open area, P = 0.2 Screen length of each lateral;, LS = pipe No. of laterals =

0.0113 = 24 m with 4 m blind 0.5 × 3.14 × 0.3 × 0.005 × 0.2

465 = 20 m 24

Provide 10 nos. in two layers to be placed Total length lateral = (465+24×4) = 561 m

16.3. Discharge of tube well, QW = 50 m3 /h = 0.014 m3 /s Depression head, S = 3.5 m. Maximum depth of water table = 5 m Velocity of ﬂow in the tube well, V = 2.5 m/s −3 2 Area of the tube well = 0.014 2.5 = 5.6 × 10 m Diameter of the tube well (dW ) = π4 × 5.6 × 10−3 = 0.08 m = 80 mm

The actual velocity of ﬂow in 80 mm diameter pipe, Va = Velocity head =

0.014 = 2.79 m/s π/4 × 0.082

2.792 = 0.4 m 2 × 9.81

Loss of head due to friction in pipe, hf =

f LV 2 (f = 0.025) 2gdw

The length of pipe line, L = vertical length of pipe including blind pipe & strainer + horizontal length of delivery pipe = (60+5) = 65 m hf =

0.025 × 65 × (2.79)2 = 8.06 m 2 × 9.881 × 0.08

Total head against which the motor has to work, H = maximum depth of water table + depression head + velocity head + losses = 5+3.5+0.4+8.06 = 16.96 m Hence power required for the motor =

1000 × 9.81 × 0.014 × 16.96 W QH = = 47.9 hp. 75η 7.5 × 0.65

466 16.4. From the given data: L = 100 m, W = 75 m, Qw = 467 L/s = 40,320 m3 /d, Sy = 0.2, and s = 2 m. Under existing conditions, 2 r Sy Qw W s=4 4πT 4T t r = 502 + 37.52 = 62.5 m 62.52 × 0.2 40320 W 2=4 4πT 4T (365) 12834 0.5351 2= W T T 0.5351 12834 −0.5772 − ln 2= T T which yields T = 72,100 m2 /d. With a new well (which is 137.5 m from the center of the rectangular area), Qw Qw W (u1 ) + W (u2 ) 4πT 4πT 0.5351 137.52 × 0.2 Qw 2= 4 −0.5772 − ln + −0.5772 − ln 4π(72100) 72100 4(72100)(365) 2=4

2 = 1.104 × 10−6 Qw [44.94 + 9.657] which yields Qw = 33181 m3 /d = 384 L/s. Therefore the total wellfield capacity is 5× 384 = 1920 L/s. The previous wellfield capacity is 4 × 467 L/s = 1868 L/s. The increase in capacity is 1920 L/s − 1868 L/s = 52 L/s . 16.5. The least drawdown at the site should be greater than 1.5 m. The potential points of interest that might have the least drawdown are at the center of the site and at the midpoint on each side of the site. Consider first the midpoint of the site, r

402 + 402 = 56.6 m

r2 S (56.6)2 (0.16) = = 0.00267 4T t 4(1600)(30) Q Q (−0.5772 − ln u) = 4 × (−0.5772 − ln 0.00267) = 0.001064Q s=4× 4πT 4π(1600)

which gives Q = 1409 m3 /day when s = 1.5 m.

467 At the midpoint of each side of the site, r = 802 + 402 = 89.4 m (40)2 (0.16) r12 S = = 0.001333 4T t 4(1600)(30) (89.4)2 (0.16) r2 S = 0.006666 u2 = 2 = 4T t 4(1600)(30) Q s=2× [(−0.5772 − ln u1 ) + (−0.5772 − ln u1 )] = 0.001041Q 4πT u1 =

which gives Q = 1441 m3 /day when s = 1.5 m. Therefore the critical drawdown will be on the side of the site, and the required pumping rate for dewatering is 1441 m3 /day = 16.7 L/s . 16.6. If the concentration decays as a ﬁrst-order process, c = e−λt c0 Deﬁning T50 as the time when c/c0 = 0.5, then 0.5 = e−λT50 ln 0.5 = −λT50 which gives T50 = −

ln 0.5 λ

or T50 =

0.693 λ

16.7. From the given data: R = 10−4 , and R is related to the viral concentration, c (in #/L), by R = 1 − (1 + 4.76c)−94.9 Hence, the allowable viral concentration is given by 1

(1 − 10−4 )− 94.9 − 1 (1 − R)− 94.9 − 1 = = 2.21 × 10−7 /L 4.76 4.76

The travel time, t, from the school to the well is given by t=

πr2 bn Qw

where r = 100 m, b = 20 m, n = 0.17, and Qw = 400 L/s = 34560 m3 /d, hence t=

π(100)2 (20)(0.17) = 3.09 days 34560

468 If the decay constant, λ for viruses is 0.3 d−1 , then the maximum allowable concentration, c0 , at the school is related to the allowable concentration, c, at the well (2.21 × 10−7 /L) by c = e−λt c0 or c0 = ceλt which gives c0 = (2.21 × 10−7 )e0.3×3.09 = 5.58 × 10−7 /L 16.8. From the given data: Qw = 50 L/s = 4320 m3 /d, H = 25 m, and n = 0.2. At a distance r from the pumping well, the induced seepage velocity, v, is given by v=

Qw 4320 137.5 = = m/d 2πrHn 2πr(25)(0.2) r

An identical image recharge well located on the other side of the river will account for the eﬀect of the river, and induce a seepage velocity such that the combined seepage velocity towards the (real) pumping well along a line joining the two wells is given by 137.5 41250 137.5 + = m/d r 300 − r r(300 − r)

The velocity towards the pumping well is equal to −dr/dt, hence 41250 41250 dr =− = dt r(300 − r) r(r − 300) If r = 150 m when t = 0, then r 150

r (r − 300)dr = r

41250dt

r 2 r 3 t − 300 = 41250t 0 3 2 150

3 r 2 − 150r − 1.125 × 106 − 3.375 × 106 = 41250(t − 0) 3 which gives t=

3 r 1 − 150r2 + 2.25 × 106 41250 3

The contaminant reaches the well when r = rw = 0.15 m, hence the minimum travel time, tr , from the river is given by 0.153 1 − 150(0.15)2 + 2.25 × 106 = 54.5 days tr = 41250 3 Assuming ﬁrst-order decay, then c = e−λtr = e−0.01×54.5 = 0.580 c0

469 From the given data, c = 1 μg/L, hence c0 =

1 = 1.72 μg/L 0.580

Therefore, the maximum allowable concentration in the river is 1.72 μg/L . This is a conservative result since: (1) inflow to the well from other parts of the river will have longer travel times, and (2) not all of the water entering the well comes from the river. 16.9. Required flow rate from wellfield, Q, is given by Q = (50000)(0.58) = 29000 m3 /d Neglecting interference, the allowable drawdown is 6/2 = 3 m, and the average transmissivity, T , is given by T = K H̄ = 85(35 − 6/2) = 2720 m2 /d For a single well, uw =

2S rw (0.5)2 (0.15) y = = 9.442 × 10−9 4T t 4(2720)(365)

Therefore, with less than 0.1% error (since u < 0.004), W (u) = −0.5772 − ln uw = −0.5772 − ln(9.442 × 10−9 ) = 17.90 The required pumpage for sw = 3 m, is Qw =

4π(2720)(3) 4πT sw = = 5729 m3 /d W (uw ) 17.90

Since the required ﬂow is 29000 m3 /d, then the number of wells required is given by No. of wells =

29000 = 5.06 5729

Therefore, try using 6 wells . This requires 29000/6 = 4833 m3 /d = 55.9 L/s per well. For maximum spacing, place 4 wells on the corners of the square parcel of land. The area of the land is 50 ha = 500000 m2 , and so the dimensions are 707.1 m × 707.1 m. See Figure 16.1, which requires that 707.1 − x 2 707.1 2 + = x2 2 2 and gives x = 388 m At center well, u = u1 =

2S rw (0.50)2 (0.15) y = = 17.90 4T t 4(2720)(365)

470

Figure 16.1: Problem 16.9. Assuming other 5 wells are 388 m away (a conservative estimate), then for each of these wells, u=

R2 Sy (388)2 (0.15) = = 0.00569 4T t 4(2720)(365)

and from Table 15.1, W (u) = 4.601 Hence, the drawdown, sb , induced by the 5 boundary wells is sb = 5 ×

4833 Qw W (u) = 5 × 4.601 = 3.25 m 4πT 4π2720

and the drawdown at the central well, sc , is sc =

Qw 4833 W (uw ) = 17.90 = 2.53 m 4πT 4π2720

The maximum drawdown induced by the wells is therefore less than 3.25 m + 2.53 m = 5.78 m. Hence the 6 wells will meet the drawdown limitation. Calculated the screen entrance velocity, vs =

Qw cπds Ls P

Assume c = 0.5, then vs =

4833 = 616 m/d = 0.41 m/min (0.5)π(0.5)(20)(0.5)

Since the maximum allowable entrance velocity is within the range 1.8–2.1 m/min, the screen is okay .

471 16.10. Maximum discharge, QW = 50 m3 /hr. = 0.014 m3 /s. Area of opening per m. length of screen = (π × ds × P ) × 0.15 = (π × 0.2) × 0.15 = 0.094m2 /m length of screen. Let us consider here the safe entrance velocity be vS = 1.8 m/min. = 0.03 m/sec. QW 0.014 Maximum length of the screen LS = 0.5×0.094×0.03 cπds P vs = 9.93 m ≈ 10 m. Since the aquifer thickness is 30 m & the minimum required screen length is 10 m. 16.11. From the given data: b = 40 m, dmin = 0.25 mm, dmax = 0.5 mm, K = 1803 m/d, and Qw = 384 L/s = 23.04 m3 /min. Since the aquifer particles are distributed uniformly between dmin and dmax , then d10 = 0.275 mm d60 = 0.400 mm d60 0.400 = 1.45 Uc = = d10 0.275 Without gravel pack: Since d10 > 0.25 mm and Uc < 3, the required slot size is in the range of d40 –d60 , which is 0.35–0.40 mm = 0.0138–0.0157 in. Therefore use a 15-slot screen . The entrance velocity, vs , is given by vs =

Qw cπds Ls P

Taking vs = 3.7 m/min, c = 0.5, and P = 0.15, gives 3.7 =

23.04 (0.5)πds Ls (0.15)

which yields ds Ls = 26.43 m2 . For Ls = 30 m, ds = 26.43/30 = 0.881 m. Therefore, use a screen length of 30 m and a screen diameter of 900 mm . With gravel pack: Since Uc of aquifer is less than 2.5, then for the gravel pack use Uc = 1–2.5 and d50 ≤ 6 × d50 = 6 × 0.375 which gives d50 ≤ 2.25 mm. For the gravel pack, specify d60 = 2.25 mm and Uc = 2 , then d60 =2 d10 2.25 = 1.13 mm d10 = 2 The required slot size is in the range of d5 –d10 of the gravel pack, which is 1.02–1.13 mm = 0.040–0.045 in. Therefore use 40-slot screen . (Note that d5 is estimated by extrapolation using d60 and d10 . The entrance velocity, vs , is given by vs =

Qw cπds Ls P

472 Taking vs = 3.7 m/min, c = 0.5, and P = 0.15, gives 3.7 =

23.04 (0.5)πds Ls (0.15)

which yields ds Ls = 26.43 m2 . For Ls = 30 m, ds = 26.43/30 = 0.881 m. Therefore, use a screen length of 30 m and a screen diameter of 900 mm . This is the same screen length and diameter as without the gravel pack 16.12. Size of the sieve (mm) >2.54 1.8 0.3 0.25 0.21 0.160 0.12 <0.12

Weight of material retained (gm.) 0 6 15 320 5 50 34 70 = 500 gm.

% of material retained 0 1.2 3.0 64 1.0 10 6.8 14.0 100%

Cumulative % retained 0 1.2 4.2 68.2 69.2 79.2 86.0 100

% ﬁner 100 98.8 95.8 31.8 30.8 20.8 14.0 0.0

From the drawn curve, the following characteristics of the aquifer material are read out as D60 = 0.27, D50 = 0.265, D10 0.102, 60 Cu = D D10

473 Since Cu ≥ 2.0, we should use Pack Aquifer (PA) ratio for designing the gravel pack lying between 12 and 15.5. Hence PA ratio = (D50 of gravel pack/D50 of aquifer) Using PA ratio as 12, we have → 12 = (D50 of gravel pack/0.265) D50 of gravel pack = 3.18 mm Again using PA ratio as 15.5, we have → 15.5 = (D50 of gravel pack/0.265) D50 of gravel pack = 4.11 mm D50 of the gravel pack should lie between these two limiting values i.e. 3.18 mm and 4.11 mm. These values are marked on 50% horizontal line as A and B. The dotted lines are now drawn through these two points parallel to the grain size of the aquifer material. The minimum size of the gravel between these two curves is represented by point C as 2.7 mm and the maximum size is represented by point D as 4.7 mm. Hence the gravel pack size should vary between 2.7 mm and 4.11 mm. The slot size of D10 size of this curve is 3 mm. 16.13. Aquifer discharge, Q = 30 m3 /hr. = 8.33 × 10−3 m3 /s Radial distance, r1 = 3.5 m and r2 = 1.5 m Steady state drawdowns, S1 = 3 m, S2 = 10 m, Saturated thickness, H = 20 m Therefore, h1 = (20–3) = 17 m and h2 = (20-10) = 10 m We have Q=

πK(h22 −h21 ) ln(r2 /r1 ) 2

−17 ) =⇒ 8.33 × 10−3 = πK(10 ln(1.5/3.5)

Therefore, hydraulic conductivity, K = 1.19× 10−5 m/s Transmissivity, T = 2.88×10−4 m2 /s 16.14. According to VonHofe and Helweg (1998), for optimum performance the pump diameter should be approximately 60% of the screen diameter. To provide ﬂexibility in placing the pump within the screened intake, the same diameter will be used for the casing and the screen. Therefore, using the 60% rule, casing diameter = screen diameter =

300 mm pump diameter = = 500 mm 0.60 0.60

The recommended borehole diameter is given by borehole diameter = casing diameter + 2(100 mm) = 500 mm + 200 mm = 700 mm

474 The screen length must be such that the screen entrance velocity, vs , is 1.8 m/min, where vs =

Qw cπds Ls P

From the given data: Qw = 41.7 L/s = 2.5 m3 /min, and P = 0.15. Taking c = 0.5 and ds = 500 mm = 0.5 m gives 2.5 1.8 = (0.5)π(0.5)Ls (0.15) which yields Ls = 11.8 m This puts the screen within the lower one-half to one-third of the aquifer and is therefore acceptable. Since Uc = d60 /d10 = 2.1/0.5 = 4.2 > 3.0 and d10 > 0.25 mm, a gravel pack is not required . Since 3 < Uc < 5 and d10 > 0.25 mm, Table 16.5 indicates that a slot size in the range of d40 − −d70 is desirable. Taking a midrange size of d55 , since d10 = 0.5 mm and d60 = 2.1 mm, interpolation gives d55 = 1.9 mm. Since a unit slot is equal to 0.025 mm, required slot size =

1.9 d55 = = 76 0.025 0.025

Using the next-lower slot size indicates a 75-slot screen . It is recommended that the pump intake be set in the middle of the screen . 16.15. From the given data: b = 20 m, K = 30 m/d, T = Kb = (30)(20) = 600 m2 /d, rw = 0.15 m, S = 10−4 , and t = 1 year = 365 days. Substituting these data into Equation 16.11 gives speciﬁc capacity =

4π(600) 4πT = = 289 m2 /d 2 2 W (rw S/4T t) W (0.15 × 10−4 /4 × 600 × 365)

= 3.35 (L/s)/m Since the allowable drawdown of the potentiometric surface is 5 m, the well yield is given by well yield = 3.35 (L/s)/m × 5 m = 16.8 L/s and therefore the maximum allowable pumping rate is 16.8 L/s . According to Table 16.8, a well with a speciﬁc capacity of 3.35 (L/s)/m has a moderate productivity . 16.16. From the given data: T = 270Sc , where T is the transmissivity in ft2 /d and Sc is the speciﬁc capacity in gpm/ft. Since 1 gpm = 5.451 m3 /d and 1 ft = 0.3048 m, the Fish and Stewart (1991) relation can be expressed in the form 0.3048 T = 270Sc (0.3048)2 5.451 or T = 1.40Sc

475 where T is the transmissivity in m2 /d and Sc is the specific capacity in m2 /d. In theory, the relationship between the transmissivity and specific capacity is given by the Theis equation, which can be expressed in the form ⎡ r2 S ⎤ w y W 4T tp ⎦ Sc ⎣ T = 4π where tp is the pumping time corresponding to the specific capacity estimation. Using the Cooper-Jacob approximation, the Theis equation can be further approximated by 2 r Sy 1 T = −0.5772 − ln w + ln T Sc (1) 4π 4tp 2 S /4T t ≤ 0.004. Using the given data for r , t , S = Q/s , and which is valid when uw = rw y p w p c w taking Sy = 0.23, the value of T is calculated (iteratively) using Equation 1. A summary of the input and calculated results are shown in Table 16.1. Using the calculated value of T shown in Column 8 of Table 16.1, the value of uw is calculated to verify the validity of the CooperJacob approximation. In all cases, uw ≤ 0.004 confirming the validity of the approximation. For the given specific capacities (Sc ), the transmissivities are calculated using the Fish and Stewart (1991) equation and these calculated results are shown in Column 10 of Table 16.1. A plot of the transmissivity versus specific capacity using the Theis equation is compared with

Table 16.1: Estimation of Transmissivity from Speciﬁc Capacity in Surﬁcial Aquifer, Miami-Dade County, Florida (1) Site

(2) Well

S-3011 S-3012 S-3013 S-3014 S-3005 S-3006 S-3007 S-3008 S-3009 S-3010

5 6 7

S-981 S-983 S-3065 S-3066 S-3045

(3) rw (m) 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.550 0.100 0.100 0.100 0.255 0.255 0.255 0.150 0.150 0.150 0.150 0.100

(4) Q (m3 /d) 38189 38189 38189 38189 15120 15120 15120 15120 15120 15120 2851 2851 3715 21168 21168 21168 6394 5962 6566 6566 5184

(5) sw (m) 4.600 5.270 4.150 4.300 0.518 1.067 0.610 1.219 0.610 0.610 0.305 0.305 0.457 1.280 1.219 0.975 1.280 1.280 1.219 1.219 0.671

(6) tp (d) 0.333 0.333 0.333 0.333 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.083 0.010 0.021 0.021 0.094 0.083 0.333 0.333 0.333

(7) Sc (m2 /d) 8302 7246 9202 8881 29189 14171 24787 12404 24787 24787 9348 9348 8130 16538 17365 21711 4995 4658 5387 5387 7726

(8) T (m2 /d) 7960 6426 8352 8034 25903 11677 21642 10076 21642 21642 10133 10133 8716 13062 14850 18991 4846 4443 5851 5851 9166

(9) uw (-) 6.555 × 10−6 8.120 × 10−6 6.248 × 10−6 6.495 × 10−6 8.058 × 10−6 1.787 × 10−5 9.644 × 10−6 2.072 × 10−5 9.644 × 10−6 9.644 × 10−6 6.809 × 10−7 6.809 × 10−7 7.916 × 10−7 2.748 × 10−5 1.209 × 10−5 9.450 × 10−6 2.848 × 10−6 3.494 × 10−6 6.633 × 10−7 6.633 × 10−7 1.882 × 10−7

(10) TFS (m2 /d) 11623 10145 12883 12434 40865 19839 34702 17365 34702 34702 13087 13087 11381 23153 24311 30395 6993 6521 7541 7541 10816

476

Transmissivity (m 2 /d)

45000 40000

Fish and Stewart (1991)

35000 30000 25000 20000

Best fit: T = 0.886 S

15000 10000

Measurements

5000 0 0

5000

10000

15000

20000

25000 30000 Specific capacity (m 2 /d)

35000

Figure 16.2: Specific capacity function for surficial aquifer system, Miami-Dade county, Florida . the Fish and Stewart (1991) equation in Figure 16.2. This comparison demonstrates that the Fish and Stewart (1991) approximation significantly overestimates the transmissivity in the surficial aquifer system of Miami-Dade county. A better approximation is T = 0.886Sc 16.17. The flow rates used in the step drawdown test are 800 m3 /d and 1200 m3 /d, and the given data can be put in the following tabular form QW (m3 /d) 800 1200

SW (m) 2.75 4.37

SW / QW (d/m2 ) 3.44 ×10−3 3.64 ×10−3

According to equation 16.14, Sw Qw = β + αQw

& considering the equation with the step drawdown data gives ⇒ 3.44 × 10−3 = β + 800α ⇒ 3.64 × 10−3 = β + 1200α Solving the equations simultaneously yields, α = 2 × 10−7 d2 /m5 , β = 3.4 × 10−3 d/m2 Hence the formation and well loss co-eﬃcient are 2× 10−7 d/m2 and 3.4 × 10−3 d/m2 respectively, and the drawdown, SW is given by, Sw = βQw + αQ2w The well eﬃciency, eW is equal to the percentage of total drawdown caused by formation losses and cases expressed in the form w ew = βQwβQ × 100 +αQ2 w

β × 100 = e β+αQ w

At, Qw = 1500 m3 /d, ew = 91.89 %

477 16.18. Conﬁned aquifer thickness = B Radius of penetrating well = r0 Pumping discharge = Q Distance of observation well from pumping well = R Let ‘t’ is the travel time for water to travel from observation well to pumping well. Total volume of water is coming from the observation well in time ‘t’ Area between the observation well & pumping well = π(R2 − r02 ) The volume of water supplied from observation well to pumping well = π(R2 − r02 )B × η Equating the two values in steady state, Q×t = π(R2 − r02 )B × η ⇒t=

π(R2 −r02 )B×η = f(B, η, Q, R, r0 ) Q

So, the travel time for water to travel from well M to the pumping well is a function of B,η, Q, R, and r0 16.19. The Rorabaugh method assumes that the drawdown, sw , and pumping rate, Qw , are related by sw = βQw + αQnw Plotting log(sw /Qw ) versus log Qw yields α = 0.0129, β = 0.288, and n = 1.5, which gives 1.5 sw = 0.288Qw + 0.0129Qw

where sw is in cm and Qw is in L/min. When Qw = 5.33 L/s = 500 L/min, the eﬃciency, ew , is given by ew =

βQw 0.288(500) × 100 = × 100 = 50% βQw + αQnw 0.288(500) + 0.0129(500)1.5

and the speciﬁc capacity is given by Qw Qw 1 = = sw βQw + αQnw β + αQn−1 w 1 = = 1.73 (L/min)/cm = 2.88 (L/s)/m 0.288 + 0.0129(500)0.5

speciﬁc capacity =

16.20. The empirical relation between the pumping rate, Q, and the drawdown, sw , can be expressed in the form sw = BQ + CQn which can be put in the form log

sw −B Q

= (n − 1) log Q + log C

478 Plotting log(sw /Q − B) versus log Q gives a linear relation for B = 0.0045 d/m2 , the slope of this line gives n−1 = 1.6 or n = 2.6, and the intercept gives log C = −7.854 or C = 1.40×10−8 . The empirical relation between sw and Q is therefore given by sw = 0.0045Q + 1.40 × 10−8 Q2.6 The well loss as a percentage of total losses is given by well loss =

CQn 1.40 × 10−8 Q2.6 × 100 = × 100 BQ + CQn 0.0045Q + 1.40 × 10−8 Q2.6

For the pumping rates in the pump-drawdown test Pumping Rate, Q (m3 /d) 500 1000 1500 2000 2500 3000 3500

Well Loss (%) 6 16 27 37 46 53 59

16.21. Assuming that estimates of the hydraulic conductivity and storage coeﬃcient are available, then estimates of the drawdowns at proposed monitoring well locations as a function of time can be made. This calculation is essential to the design of the pump test, since the design pumping rate must be suﬃcient to produce drawdowns at the monitoring wells (during the pump test) that are measurable with a reasonable degree of accuracy. 16.22. The drawdown, s0 , induced by a well that begins pumping at t = 0 is given by the Theis equation as Qw W (u) s0 = 4πT where r2 S (1) u= 4T t If this well were to begin pumping at time t1 , then the induced drawdown, s1 , would be given by Qw W (u1 ) s1 = 4πT where r2 S (2) u1 = 4T (t − t1 ) Applying the principle of superposition (in time), then the drawdown, s, induced by a well that begins pumping at t = 0 and stops at t = t1 is given by s = s0 − s1

479 which yields s=

Qw [W (u) − W (u1 )] 4πT

where u and u1 are given by Equations 1 and 2 respectively. 16.23. From Equation 16.21 s =

Qw W (u) − W (u ) 4πT

(1)

and the Cooper-Jacob approximations are given by W (u) = −0.5772 − ln u

W (u ) = −0.5772 − ln u

(2)

(3)

Combining Equations 1 to 3 gives Qw Qw (−0.5772 − ln u) − (−0.5772 − ln u ) = ln u − ln u 4πT 4πT u Qw ln = 4πT u

s =

(4)

Since r2 S 4T t

and

u =

r2 S 4T t

then S t u = u St

(5)

and combining Equations 4 and 5 gives s = or Qw s = 4πT

S t Qw ln 4πT St

t S ln + ln t S

16.24. The recovery measurements should match the theoretical relation given by Equation 16.24, which can be written as Qw Qw t S ln + ln (1) s = 4πT t 4πT S The recovery measurements expressed as s versus ln t/t are:

480 s (m) 1.01 0.9 0.83 0.75 0.7 0.61 0.55 0.6 0.42 0.37 0.31 0.26 0.23 0.19 0.15

t (min) 1 2 3 5 7 10 15 20 30 40 60 80 100 140 180

t (min) 241 242 243 245 247 250 255 260 270 280 300 320 340 380 420

ln(t/t ) 5.48 4.80 4.39 3.89 3.56 3.22 2.83 2.56 2.20 1.95 1.61 1.39 1.22 1.00 0.85

From the given data: Qw = 33.3 L/s = 2880 m3 /d, t = 4 h = 240 min, and S = 0.0001. The best-ﬁt line to the s versus ln t/t measurements is t s = 0.1872 ln + 0.0143 t and matching this equation with Equation 1 gives Qw = 0.1872 4πT S Qw ln = 0.0143 4πT S Solving Equation 2 for T gives T =

2880 Qw = = 1220 m2 /d 4π(0.1872) 4π(0.1872)

and solving Equation 3 for S gives 0.0143 0.0143 = 0.0001 exp = 0.00011 S = S exp Qw /4πT 0.1872 where S is not signiﬁcantly diﬀerent from S. 16.25. We have, 35 × 20 = 8.10 × 10−3 m2 /s Transmissivity, T = KH = 86400

According to equation 16.22 2

100 ×0.0003 −4 U = r4TSt = 4×8.10×10 −3 ×(48×3600) = 5.36 × 10

(2) (3)

481 Using Theis method & calculate W(u) (Equation 15.124) −4

W(u)= −0.5772 − ln(5.36 × 10−4 ) + (5.36 × 10−4 ) − 5.36×10 2.2! QW

= 6.95

= 1000 m3 /d = 0.0116 m3 /s

According to equation 15.123 Qw S(u) = 4πT W(u)

S100 = 0.79 m The drawdown at a distance of 100 m from the well is 0.79 m. 16.26. From the given data: rw = rc = 0.076 m, and Le = Lw = b = 98 m. Determine the following quantities: 98 Le = 1290 = rw 0.076 C = 12.8 1.1 12.8 −1 Re = = 6.12 + ln rw ln 1290 1290 Plot the drawdown, s, versus time, t, and this gives the following points (t, s) on the straightline portion of the relation: (0 s, 0.46 m) and ( 33 s, 0.149 m). Note that the s vs. t line becomes nonlinear after t = 33 s. Using the points on the straight-line portion of the curve: K=

0.0762 (6.12) 1 0.49 ln = 6.5 × 10−6 m/s = 0.56 m/d 2(98) 33 0.149

The support volume for this hydraulic conductivity is about 1 m × 98 m surrounding the hole used for the slug test. 16.27. The measured data of ln yt versus t indicates two straight line segments as shown in Figure 16.3. Fitting a straight line to the second linear segment yields ln y = −0.0805t + 3.069 which has a slope, m, of −0.0805. From the given dimensions of the well and aquifer, Lw = 5.5 m, H = 80 m, rc = 0.076 m, Le = 4.56 m, rw = 0.12 m, and Le /rw = 4.56/0.12 = 38.0. Equation 16.28 is appropriate for calculating ln(Re /rw ) (since the well is partially penetrating), and the dimensionless parameters A and B are given by Figure 16.8 as A = 2.6,

B = 0.42

Analyses by Bouwer and Rice (1976) indicated that if ln[(H − Lw )/rw ] > 6 then a value of 6 should be used in Equation 16.28. In this case, ln[(H − Lw )/rw ] = ln[(80 − 5.5)/0.12] = 6.43. Therefore, using ln[(H − Lw )/rw ] = 6 in Equation 16.28 yields A + B ln[(H − Lw )/rw ] −1 1.1 Re + = ln rw ln(Lw /rw ) (Le /rw ) 2.6 + 0.42(6) −1 1.1 + = = 2.37 ln(5.5/0.12) (4.56/0.12)

482

Figure 16.3: Plot of slug test measurements. and putting this result into Equation 16.27 gives r2 ln(Re /rw ) K=− c m 2Le (0.076)2 (2.37) (−0.0805) = 1.21 × 10−4 m/s = 10 m/d =− 2(4.56) 16.28. The ﬁrst step is to plot ln yt versus t, where the given measurements can be put in the form: t (s) 0 3 6 9 12 15 18 21 24

yt (mm) 700 392 260 137 91 47 31 16 11

ln yt 6.55 5.97 5.56 4.92 4.51 3.85 3.43 2.77 2.40

The plotted relation between ln yt and t is very much linear (r2 = 0.998), and is matched by the regression equation ln yt = −0.1752t + 6.5431 which has a slope, m, of −0.1752. From the given dimensions of the well and aquifer, Lw = 4 m, H = 12 m, rc = 75 mm, Le = 2 m, rw = rc + 110 mm = 185 mm, and

483 Le /rw = 2/0.185 = 10.8. Equation 16.28 is appropriate for calculating ln(Re /rw ) (since the well is partially penetrating), and the dimensionless parameters A and B are given by Figure 16.8 as A = 1.8, B = 0.25 Substituting these data into Equation 16.28 yields Re = ln rw =

A + B ln[(H − Lw )/rw ] 1.1 + ln(Lw /rw ) (Le /rw )

−1

1.8 + 0.25 ln[(12 − 4)/0.185] 1.1 + ln(4/0.185) (2/0.185)

−1 = 1.63

and putting this result into Equation 16.32 gives r2 ln(Re /rw ) (0.075)2 (1.63) K=− c m=− (−0.1752) = 4.02 × 10−4 m/s = 34.7 m/d 2Le 2(2) This result indicates an average (horizontal) hydraulic conductivity of 34.7 m/d in the immediate vicinity of the well. The transmissivity, T , of the aquifer can be estimated by T = KH = (34.7)(12) = 416 m2 /d 16.29. The relation between ln yt and t is still matched by the regression equation ln yt = −0.1752t + 6.5431 which has a slope, m, of −0.1752. From the given dimensions of the well and aquifer, Lw = H = 12 m, rc = 75 mm, Le = 2 m, rw = rc + 110 mm = 185 mm, and Le /rw = 2/0.185 = 10.8. Equation 16.29 is appropriate for calculating ln(Re /rw ) (since the well is fully penetrating), and the dimensionless parameter C is given by Figure 16.8 as C = 1.25 Substituting this value of C into Equation 16.29 yields Re ln = rw

C 1.1 + ln(Lw /rw ) (Le /rw )

−1

1.25 1.1 + ln(12/0.185) (2/0.185)

−1 = 2.64

and putting this result into Equation 16.32 gives (0.075)2 (2.64) r2 ln(Re /rw ) (−0.1752) = 6.50 × 10−4 m/s = 56 m/d m=− K=− c 2Le 2(2) This result indicates an average (horizontal) hydraulic conductivity of 56 m/d in the immediate vicinity of the well. The transmissivity, T , of the aquifer can be estimated by T = KH = (56)(12) = 672 m2 /d

484 16.30. The length, L, of the trench is given by Equation 16.38 as L=

Q Kt (W + H)

(1)

where the dimensions H and W are illustrated in Figure 16.10. Following ASCE (1996) guidelines, specify W = 1 m (a typical backhoe dimension), and H = 2 m (a typical depth for vertical slope stability). The design injection rate, Q, is 5.79 L/s = 500 m3 /d, and the design trench hydraulic conductivity, Kt , is 35/2.5 = 14 m/d. Substituting these values for Q, Kt , W , and H into Equation 1 yields L=

500 Q = = 11.9 m Kt (W + H) 14(1 + 2)

A trench dimension of 12 m long by 1 m wide by 2 m deep , when filled with gravel, will be capable of transferring water into the aquifer at a rate of 5.79 L/s ( = 500 m3 /d). Since the seasonal high water table is 5.22 m below the ground surface, there is sufficient room to install a 2-m deep trench and still maintain a distance of at least 1.2 m between the bottom of the trench and the seasonal high water table. The next question is whether the aquifer will be able to transport the effluent away from the trench as fast as it is supplied, without causing the water table to rise to within 1.2 m of the bottom of the trench. The height of the water table above the base of the aquifer as a function of time is given by W 2N L 2 2 ∗ √ , √ νtS (2) hm (t) = hi + K 4 νt 4 νt where hi = 15 m, N is given by N=

500 Q = = 41.7 m/d LW (12)(1)

The hydraulic conductivity of the aquifer, K, is 70 m/d, and the parameter ν in Equation 2 is given by Kb K(hi + hm )/2 70(15 + hm )/2 ν= = 159.1(15 + hm ) = = Sy Sy 0.22 Substituting the trench dimensions (L = 12 m, W = 1 m), and aquifer properties into Equation 2 yields 1 12 2(41.7) 2 2 ∗ [159.1(15 + hm )]tS hm = 15 + , 70 4 [159.1(15 + hm )]t 4 [159.1(15 + hm )]t 0.0198 0.238 ∗ , (3) = 225 + 189.6(15 + hm )tS (15 + hm )t (15 + hm )t This equation relates the thickness of the saturated zone below the trench to the time since the trench began operation. Values of hm for several values of t are shown in the following table:

485 t (days) 1 10 100 1000 10000

hm (m) 15.28 15.37 15.46 15.54 15.62

where S ∗ (α, β) has been estimated using Equation 16.45. The calculated values of hm indicate that the saturated thickness under the trench steadily increases, and after 10,000 days (27.4 years) the saturated thickness will be 15.62 m. This indicates that the water table below the exfiltration trench will rise (mound) by about 0.62 m ( = 62 cm) and will remain at an acceptable depth (> 1.2 m) below the trench. 16.31. The trench design in Problem 16.30 yielded a trench that is 2-m deep, and is 5.22 − 2 − 0.62 = 2.60 m above the mounded water table. If a backfill of 1 m is placed above the trench, the design is unaltered, and the trench will be 2.60 − 1 = 1.60 m above the mounded water table. This exceeds the minimum distance of 1.2 m, and therefore the trench dimensions are the same as in Problem 16.30 , with 1 m of backfill above the trench. 16.32. Amount of water that may be harvested through rainwater roof catchment system = 1.6×(3.141 × 104 ) × 0.65 m3 = 32666.4 m3 According to equation 16.8 vs = πCdQsWLs P (Given Entrance velocity, vS = 200 L/h/m2 = 0.2 m/hr) Quantity of water, QW = (0.2× 3.14× 0.15× 30× 0.6) m3 /hr. = 0.8478 m3 /hr Recharging rate through the bore well is 0.8478 m3 /hr 16.33. The trench length, L, is given by L=

Q Kt (W + H)

(1)

Following ASCE guidelines, use W = 1 m and H = 2 m. From the given data, Q = 5.79 L/s = 500 m3 /d, Kt = 20/2 m/d = 10 m/d, and therefore Equation 1 gives L=

500 = 16.7 m 10(1 + 2)

For a trench depth, H, of 2 m and 30 cm of backﬁll, there is 4 m − (2 m + 0.3 m) = 1.7 m between the bottom of the trench and the water table. The height, hm , of the water table after 20 years is given by Equation 16.39 as 2N νtS ∗ h2m (t) = h2i + K

W L √ , √ 4 νt 4 νt

(2)

486 where hi = 17 m, K = 7 m/d, W = 1 m, L = 16.7 m, and t = 20 years = 20(365) d = 7300 d. Hence, with b = 17 m, and Sy = 0.14, 500 Q = = 29.9 m/d LW (16.7)(1) (7)(17 + hm )/2 Kb = 25.0(17 + hm ) = ν= Sy 0.14

(3)

(4)

Combining Equations (2) to (4) with the given data yields 2(29.9) [25.0(17 + hm )](7300)S ∗ h2m (t) = 172 + 7

= 289 + 1.559 × 10 (17 + hm )S

∗

16.7

, 4 [25.0(17 + hm )](7300) 4 [25.0(17 + hm )](7300)

0.000585 0.00977 √ ,√ 17 + hm 17 + hm

(5)

which gives hm = 21.22 m This indicates that the mound will rise 21.22 m − 17 m = 4.22 m above the water table. Since the bottom of the trench is only 1.7 m above the water table, then the mounding is unacceptable and a longer trench is required. The minimum trench length required is such that h2m = (17 + 1.7)2 = 350 m2 and

7(17 + 1.7/2) = 892.5 m2 /d 0.14 in which case W L 2N ∗ √ , √ = 350 − 172 νtS K 4 νt 4 νt which can be put in the form 500 2 L(1) (892.5)(7300)S ∗ 9.794 × 10−5 , 9.794L × 10−5 = 61 7 ν=

9.308 × 108 S ∗ (9.794 × 10−5 , 9.794L × 10−5 ) = 61 L

(6)

which yields L = 1420 m Therefore, the required trench dimensions are 1 m × 2 m × 1420 m . A exﬁltration trench with these dimensions is clearly not practical, and an alternative means of disposal should be considered. 16.34. From the given data: L = W = 100 m, N = Q/(W L) = 5 m/d, K = 90 m/d, Sy = 0.2, hi = 35 m, and t = 20 years = 7300 days. Taking b=

35 + hm hi + hm = 2 2

487 then ν=

m (90) 35+h Kb 2 = Sy 0.2

= 7875 + 225hm

(1)

Mounding beneath the pond is given by Equation 16.39 as W L 2N 2 2 ∗ √ , √ νtS hm (t) = hi + K 4 νt 4 νt

(2)

Combining Equations 1 and 2 and substituting the given data yields

2(5) h2m = 352 + (7875 + 225hm )(7300)S ∗

100 100 , 90 4 (7875 + 225hm )(7300) 4 (7875 + 225hm )(7300) 0.2926 0.2926 h2m = 1225 + 811.1(7855 + 225hm )S ∗ √ ,√ 7875 + 225hm 7875 + 225hm

(3)

which yields hm = 48.2 m Therefore, after 20 years hm = 48.2 m and the water table beneath the pond rises 48.2 m − 35 m = 13.2 m. Since the bottom of the pond is 20 m above the water table at the beginning of recharge operations, this means that the recharge pond will still be adequate in 20 years. 16.35. From the given data: W = 5 m, L = 100 m, N = 0.2 m/d, K = 2 m/d, Sy = 0.15, hi = 5.333 m, and t = 6 days. (a) Using the Hantush function

W L √ , √ hm = K 4 νt 4 νt 5 100 (2)(0.2) ν(6)S ∗ , hm = 5.3332 + 2 4 ν(6) 4 ν(6) 0.5103 10.21 ∗ √ √ , hm = 28.44 + 1.2νS ν ν 2N νtS ∗ h2i +

(1)

where ν=

Kb K(hi + hm )/2 2(5.333 + hm )/2 = = Sy Sy 0.15 (2)

ν = 35.55 + 6.667hm

Equations 1 and 2 can be solved iteratively as indicated in the following tabulated calculations: hm (m) 8.00 6.32 6.27

√ ) α(= 0.5103 ν

√ ) β(= 0.5103 ν

S ∗ (α, β)

88.89 77.69 77.35

0.0540 0.0574 0.0580

1.080 1.158 1.161

0.1080 0.1169 0.1172

hm (m) 6.32 6.27 6.27

488 Therefore the mounding height is 6.27 m . (b) Using the Swamee and Ojha (1997) approximation results in the following tabulated calculations: √ ) √ ) hm ν α(= 0.5103 β(= 0.5103 S ∗ (α, β) hm ν ν (m) (m) 6.27 77.35 0.0580 1.161 0.1240 6.32 6.32 77.69 0.0579 1.158 0.1138 6.32 Therefore the Swamee and Ojha (1997) mounding height is 6.32 m , which is 0.8% diﬀerent than calculated using the Hantush function. 16.35. From the given data: N = 45.7 cm/d = 0.457 m/d, hi = 30.5 m, hm = 45.7 m, T = 929 m2 /d, K = 929/30.5 = 30.5 m/d, n ≈ Sy = 0.2, ν = T /Sy = 4645 m2 /d, W = 30.5 m, and L = 3220 m. Using the Hantush equation, W 2N L 2 2 ∗ √ √ νtS , hm = hi + K 4 νt 4 νt 3220 305 2(0.457) 2 2 ∗ √ 45.7 = 30.5 + , √ (4645)tS 30.5 4 4645t 4 4645t which gives tS ∗ Assume β ≥ 3.0 and

1.119 11.81 √ , √ t t

= 8.321

(1)

S ∗ (α, β) = 1 − 4i2 erfc(α) = 1 + 4erfc(α)

Therefore, Equation 1 becomes

t 1 + 4erfc

1.119 √ t

= 8.321

√ which gives t = 3.3 days and β = 11.81/ 3.3 = 6.51, which conﬁrms β ≥ 3.0 as originally assumed. If t = 1 year = 365 days and N = 0.457 × 305/W = 139.4/W , then 139.4 2 3220 W W 45.72 = 30.52 + , (4645)(365)S ∗ 30.5 4 (4645)(365) 4 (4645)(365) which gives

W 1 ∗ S , 0.6182 = 7.478 × 10−5 W 5208 None of the approximate solutions appear applicable, so use tabulated values with the iteration formula W ∗ W = 13380S , 0.6182 (2) 5208 which converges to W = 10680 m. Therefore, for waterlogging in one year, the lake would have to be approximately 10.7 km wide .

Chapter 17

Water-Resources Planning 17.1. If A is invested at the end of year j, then its value, Fj , at the end of year n is given by Fj = A(1 + i)n−j Therefore, if A is invested at the end of each year, the total future value, F , is given by F =

Fj =

j=1

A(1 + i)

n−j

j=1

(1 + i)n−j

(1)

j=1

Note the following, n

(1 + i)n−j = (1 + i)n−1 + (1 + i)n−2 + . . . + 1

j=1

then

j=1 (1 + i)

(1 + i)n−1 + (1 + i)n−1 + and solving for

n−j − 1

(1 + i) n n−j − 1 j=1 (1 + i) (1 + i) n n−j − 1 j=1 (1 + i)

j=1 (1 + i)

(1 + i)

= (1 + i)n−2 + (1 + i)n−3 + . . . + 1 = (1 + i)n−1 + (1 + i)n−2 + . . . + 1 =

(1 + i)n−j

j=1

n−j yields n

(1 + i)n−j =

j=1

(1 + i)n − 1 i

Combining Equations 1 and 2 yields F =A

(1 + i)n − 1 i 488

i A = F (1 + i)n − 1

(2)

489 17.2. The sinking-fund factor (Equation 17.4) is given by A i = F (1 + i)n − 1

(1)

and the present-worth factor (Equation 17.3) is given by F = (1 + i)n P

(2)

Combining Equations 1 and 2 gives the capital-recovery factor as A F i A = × = × (1 + i)n P F P (1 + i)n − 1 which yields A i(1 + i)n = P (1 + i)n − 1 17.3. The present worth at year 0 of only the gradient is equal to the sum of the present worths of the individual values, where each value is considered a future amount. Hence, P P P P , i, 2 + 2G , i, 3 + . . . + (n − 2)G , i, n − 1 + (n − 1)G , i, n P =G F F F F Factor out G and use the P/F formula, 2 3 n−2 n−1 1 + + + ... + + P =G (1 + i)2 (1 + i)3 (1 + i)4 (1 + i)n−1 (1 + i)n Multiplying both sides by (1 + i) yields 1 2 3 n−2 n−1 P (1 + i) = G + + + . . . + + (1 + i)1 (1 + i)2 (1 + i)3 (1 + i)n−2 (1 + i)n−1

(1)

(2)

Subtract Equation 1 from Equation 2 and simplify, 1 n 1 1 1 iP = G −G + + ... + + (1 + i)1 (1 + i)2 (1 + i)n−1 (1 + i)n (1 + i)n The left-bracketed expression can be evaluated using the following identity, (1 + i)n − 1 1 1 1 1 = + + ... + + i(1 + i)n (1 + i)1 (1 + i)2 (1 + i)n−1 (1 + i)n Combining Equations 2 and 3 yields

G (1 + i)n − 1 n P = − i i(1 + i)n (1 + i)n

which simpliﬁes to (1 + i)n − in − 1 P = G i2 (1 + i)n

(3)

490 17.4. The series starts in year 1 at an initial amount A0 . The relation to determine the total present worth, P for the entire cash ﬂow series can be achieved by multiplying each geometricallyincreasing annual return by the P/F factor 1/(1 + i)n , which yields A0 (1 + g) A0 (1 + g)2 A0 (1 + g)n−1 A0 + + + . . . + (1 + i)1 (1 + i)2 (1 + i)3 (1 + i)n 2 1 (1 + g) (1 + g) (1 + g)n−1 + P = A0 + + ... + (1 + i) (1 + i)2 (1 + i)3 (1 + i)n

P =

(1)

Multiply both sides by (1 + g)/(1 + i), subtract Equation 1 from the result, factor out P , and obtain (1 + g)n 1+g 1 − 1 = A0 P − 1+i (1 + i)n+1 1 + i Solve for P and simplify ⎡ P = A0 ⎣

1−

1+g 1+i

n⎤

i−g

⎦ , quadg = i

(2)

The term in brackets is the geometric-gradient series present worth factor for g = i. When g = i, substitute g for i in Equation 1 to obtain 1 1 1 1 + + + ... + P = A0 1+i 1+i 1+i 1+i which simpliﬁes to P =

nA0 (1 + i)

(3)

Combining Equations 2 and 3 yields ⎧ n 1−( 1+g 1+i ) ⎪ ⎨ i−g

P = A0 ⎪ ⎩ n

1+i

g = i g=i

17.5. The cost diagram of the projects A & B are shown in Fig. Below

491 It this situation, it can not directly compare present worth of project A and project B because they are based on diﬀerent time periods(i.e. 2 years and 3 years) one-way to handle the problem would be to use a common 6yrs period, in which it would replace project A three times & replace project B twice. But a more direct approach is to convert the present worth based on a period n1 to an equivalent present worth based on n2 byPn2 =

Pn1 F(A/P, i,n1 ) F(A/P,i,n2 )

In this problem, we convert present worth of project B from a 3 year time period to a 2 year period. At ﬁrst we calculate the present worth of project B for 3 yrs PB3 = $300, 000 + $50, 000FAP,10%,3 − $20, 000FF P,10%,3 n

−1 −1 We know F(P/A, i, n) = F( P , 10%, 3) = [ (1+i) ] = [ (1+0.1) ] = 2.487 [according to Equation 17.7] i(1+i)n 0.1(1+0.1)3 A

(P/A, i, n) =

1 1 = 0.751 n = (1 + i) (1 + 0.1)3

PB3 = $300, 000+$50, 000 × 2.478 − $20, 000 × 0.751 = $408,880 Now, The present worth of project B for 2 years, PB2 = PB3

F(A/p, 10%, 3) 0.402 = $285, 364.17 = $408, 880 × F(A/P, 10%, 2) 0.576

i(1 + i)n 0.1(1.1)3 ]=[ ] = 0.402 [According to Equation 17.6] n (1 + i) − 1 (1.1)3 − 1 F( A , 10%, 2) = 0.576

Now F(A/P, 10%, 2) = [ P

Now present worth of project A for 2 yrs PA2 = $200, 000 + $60, 000FAP,10%,2 − $10, 000FF P,10%,2 = $200, 000 + $60, 000 × 1.74 − $10, 000 × 0.826 = $296, 140 Now, when compared on the basis of equal time periods (i.e. for 2 yrs). Project B is the most economical. 17.6. Both the plans A & B are compared on a 30 year basis. Now, Plan A: Present work of initial cost = $550, 000 1 = $1576.41[ according to Equation 17.3] Present work of salvage value= $12, 000 (1+0.07) 30

492 30

1.07 −1 Present worth of maintenance cost= $12, 000[ 0.07×(1.07) 30 ] = $148, 908.50 [according to Equation 17.7]

Present worth of Plan A = $ (550, 000 − 1576.41 + 148, 908.50) = $697,332.08 Plan B: Present worth of initial cost = $440, 000 1 Present worth of $440, 000 at 20yrs= $440, 000 (1+0.07) 20 = $113,704.36

Present worth of salvage value for 20 yrs = $10, 000 Present worth of salvage value for 30 yrs= $10, 000

1 (1+0.07)20

1 (1+0.07)30

= $2584.19

= $1313.67

(1+0.07) −1 Present worth of maintenance costs = $50, 000[ 0.07(1+0.07) 30 ] = $620, 452.06

Present worth of Plan B = $ (440, 000 + 113, 704.36 − 258419 − 1313.67 + 620, 452.06) = $1170, 258.56 Therefore, Plan A is preferred over Plan B 17.7. All proposed alternatives are described by an initial cost, I, first-year revenue, R0 , revenue growth rate, gr , first-year cost, C0 , cost growth rate, gc , and design life, n, equal to 15 years. The return rate, i∗ , for each alternative satisfies the relation −I +

P P ∗ ∗ , gr , i , n R0 − , g c , i , n C0 = 0 R0 C0

(1)

where (P/A0 , g, i, n) is the geometric-gradient present-worth factor given by Equation 17.11, and (P/F, i, n) is the single-payment present-worth factor given by Equation 17.3. Combining Equations 17.3, 1, and 17.12 gives ⎡ −I + ⎣

1−

1+gr 1+i∗

i∗ − gr

n⎤

⎡

⎦ R0 − ⎣

1−

1+gc 1+i∗

i∗ − gc

n⎤

⎦ C0 = 0

(2)

Substituting given values of I, R0 , gr , C0 , and gc for each of the proposed alternatives yields the following results:

Alternative 1 2

I $130,000 $75,000

R0 $16,000 $12,000

gr 4.5% 3.5%

C0 $6,000 $3,200

gc 5% 3.5%

i∗ 4.5%, 5% 3.5%

Since the minimum attractive rate of return is 6%, neither alternative is economically feasible.

493 17.8. We know-

Annual cost = i

(1+i)n ×capital cost−salvage value (1+i)n −1

×$500,000−$7,000 ] = $155, 847.87 = 0.07[ (1+0.07)(1+0.07) 50 −1

Annual maintenance cost = $15,000 Total annual cost = $170,847.87

Flood Frequency(yrs) Annual Damages($)

0 0

5 20,000

15 40,000

30 60,000

50 80,000

Flood frequency Vs. Annual Damage Curve 80000 70000 60000 50000 40000 30000 20000 10000 0 0

Frequency Ordinate of Damages from fig 0 0 5 20, 000 10 35, 000 15 40, 000 20 48, 000 25 55, 000 30 60, 000 35 68, 000 40 70, 000 45 75, 000 50 80, 000

494 20000 + 35000 35000 + 40000 0 + 20, 000 ×5+ ×5+ ×5 2 2 2 48000 + 55000 55000 + 60000 40000 + 48000 ×5+ ×5+ ×5 + 2 2 2 68000 + 70000 70000 + 75000 60000 + 68000 ×5+ ×5+ ×5 + 2 2 2 75000 + 80000 × 5] = $2555, 000 + 2

Area under the curve = [

= 51, 100 Annual average beneﬁt = 2555,000 50 Benef its 51,100 Beneﬁt cost(B/C) ratio = Annual Annual Costs 170847.87 = 0.3 < 1

The project is not economically viable.