system analysis and control by harvad web

SOLUTIONS MANUAL FOR SELECTED PROBLEMS IN

PROCESS SYSTEMS ANALYSIS AND CONTROL DONALD R. COUGHANOWR

COMPILED BY

M.N. GOPINATH BTech.,(Chem) CATCH ME AT gopinathchemical@gmail.com

Disclaimer: This work is just a compilation from various sources believed to be reliable and I am not responsible for any errors.

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CONTENTS

PART 1: SOLUTIONS FOR SELECTED PROBLEMS

PART2:

LIST OF USEFUL BOOKS

PART3:

USEFUL WEBSITES

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PART 1 1.1 Draw a block diagram for the control system generated when a human being steers an automobile.

1.2 From the given figure specify the devices

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Solution:

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Inversion by partial fractions:

3.1(a)

dx 2 dx + + x = 1 x ( 0) = x ' ( 0) = 0 2 dt dt

 dx 2  L  2  = s 2 X ( s ) − sx(0) − x ' (0)  dt 

 dx  L   = s X ( s ) − x ( 0)  dt  L(x) = X(s)

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L{1} = 1/s s 2 X ( s ) − sx(0) − x ' (0) + s X ( s ) − x (0) + X ( s ) =

= ( s 2 + s + 1) X ( s ) =

X ( s) =

1 s

1 s( s + s + 1) 2

Now, applying partial fractions splitting, we get X ( s) =

X ( s) =

1 s +1 − 2 s ( s + s + 1)

3 2

s +1 1  1  2  − −    2 2 2 2 s   2  3   1   3 1  3  s +  +    s +  +  2  2  2   2   

L−1 ( X ( s )) = 1 − e

X (t ) = 1 − e

1 − t 2

Cos

3 1 −2t 3 t− e sin t 2 2 3

  3  3 1  Cos t +   Sin    2 t 2 3    

dx 2 dx + 2 + x = 1 x ( 0) = x ' ( 0) = 0 2 dt dt

when the initial conditions are zero, the transformed equation is ( s 2 + s + 1) X ( s ) =

1 s

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X ( s) =

1 s( s + s + 1) 2

1 A Bs + C = + 2 s ( s + s + 1) s s + 2 s + 1 2

1 = A(s2 + 2s +1) + Bs2 + Cs 0 = A + B(by equating the co − effecient of s 2 ) 0 = 2 A + C (by equating the co − effecients of s )

1 = A(by equating the co − effecients of const) A+ B = 0 B = −1 C = −2 A A = 1, B = −1, C = −2

1 s+2 − 2 s s + 2s + 1  1 (s + 1) + 1  L−1{ X ( s )} = L−1  − (s + 1)2  s X ( s) =

 1 1  { X (t )} = 1 − L−1  + 2   s + 1 (s + 1)  { X (t )} = 1 − e − t (1 + t ) dx 2 dx 3.1 C + 3 + x = 1 x ( 0) = x ' ( 0) = 0 2 dt dt

by Applying laplace transforms, we get

= ( s 2 + 3s + 1) X ( s ) =

X ( s) =

1 s

1 s( s + 3s + 1) 2

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X ( s) =

A Bs + C + 2 s s + 3s + 1

1 = A( s 2 + 3s + 1) + Bs 2 + Cs 0 = A + B(by equating the co − effecient of s 2 ) 0 = 3 A + C (by equating the co − effecients of s)

1 = A(by equating the co − effecients of const) A+ B = 0 B = −1 C = −3 A = −3 A = 1, B = −1, C = −3

s+3  1 L−1{ X ( s )} = L−1  − 2   s s + 3s + 1 

  −1 −1  1 L { X ( s )} = L  − s  s +      1 L−1{ X ( s )} = L−1  − s  s +   

X (t ) = 1 − e

−

3t 2

(Cos

s+3 2 3   − 2  

   2  5   2  

3 2  3   . s+ 2 5  2 − 2 2 2 3  3  5     + − s    − 2   2   2  

   2  5   2  

5 2

5t 3 5 + t sinh 2 2 5

3.2(a) dx 4 d 3 x + 3 = Cos t; x (0) = x ' (0) = x ''' (0) = 0 4 dt dt x11 (0) = 1

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Applying Laplace transforms, we get s 4 X ( s ) − s 3 x (0) − s 2 x1 (0) − sx '' (0) − x ''' (0) + s 3 X ( s ) − s 2 x (0) − sx ' (0) − x '' (0) =

X ( s ) ( s 4 + s 3 ) − ( s + 1) =

s s +1 2

 s  + 1) + ( s + 1)  s 4 + s 3 X ( s ) =  ( 2  s +1  3 s + s + s + s 2 + 1 s 3 + s 2 + 2s + 1 = 3 2 = 3 2 s ( s + 1)( s + 1) s ( s + 1)( s + 1)

s 3 + s 2 + 2s + 1 A B C D Es + F = + 2 + 3+ + 2 3 2 s ( s + 1)( s + 1) s s s s +1 s +1

s 3 + s 2 + 2 s + 1 = As 2 ( s + 1)( s 2 + 1) + Bs( s + 1)( s 2 + 1) + c( s + 1)( s 2 + 1) + Ds 3 ( s 2 + 1) + ( Es + F ) s 3 ( s + 1) A+B+E=0 equating the co-efficient of s5. A+B+E+F=0 equating the co-efficient of s4. A+B+C+D+F=0 equating the co-efficient of s3. A+B+C=0 equating the co-efficient of s2. B+C=2 equating the co-efficient of s. A+B+E=0 equating the co-efficient of s2. C=1equating the co-efficient constant. C=1 -B=-C+2=1 A=1-B-C=-1 D+F=0 E+F=0D+E=1 D-E=0 2D=1 A=-1; B=1; C=1 D=1/2; E=1/2; F =-1/2

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 − 1 1 1 1 / 2 1 / 2( s − 1)  + L−1{( s)} = L−1  + 2 + 3 +  s s +1 s2 + 1  s s 1 1 / 2 1 / 2( s − 1)  −1 1 L−1 {X ( s )} = L−1  + 2 + 3 + +  s s s +1 s2 + 1   s 2 {X (t )} = −1 + t + t + 1 e −t + 1 Cos t − 1 S int 2 2 2 2

d 2 q dq + = t 2 + 2t q(0) = 4; q1 (0) = −2 dt 2 dt applying laplace transforms,we get

s 2Q ( s ) − sq(0) − q ' (0) + sQ (( s ) − q(0) =

Q ( s )( s 2 + s ) − 4 s + 2 − 4 =

2 2 + 2 3 s s

2 1   + 1 s2  s 

2( s + 1) + ( 4 s + 2) 3 s Q( s) = ( s 2 + s)

2 s + 2 + 4 s 4 + 2s 3 s 4 ( s + 1)

2 2*3  1  + 4 Q ( s ) = 4 +  s + 1  s( s + 1) s ( s + 1)

1 L−1 (Q ( s )) = q(t ) = 4e −t + 2(1 − e −t ) + t 3 3

therefore q(t ) = 2 +

t3 + 2 e −t 3

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3s 3s  1 1  =  2 − 2 2 ( s + 1)( s + 4) 3  s + 1 s + 4 

3.3 a)

1   1 = 2 2 − 2 s + 2 2  s +1 1   1 L−1  2 2 − 2 = Cost − Cos 2t s + 2 2  s +1 b)

1 1 A B+C = = + 2 2 2 s ( s − 2 s + 5) s ( s − 1) + 2 s s − 2s + 5 2

[

]

A+B=0 -2A+C=0 5A=1 A=1/5 ;B=-1/5;C=2/5

We get

X ( s) =

1 1 2−s  + 2  5  s s − 2 s + 5 

Inverting,we get

1 1 t  t 1 2 2 + e Sin t − e Cos t  5  2

1 1  1 + e t  Sin 2t − Cos 2t   5 2 

3s 2 − s 2 − 3s + 2 A B C D = + 2+ + c) 2 2 s ( s − 1) s s s − 1 ( s − 1) 2

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As( s − 1) 2 + B( s − 1) 2 + Cs 2 ( s − 1) + Ds 2 = 3s 3 − s 2 − 3s + 2

A( s 3 − 2 s + s ) + B ( s 2 − 2 s + 1) + C ( s 3 − s 2 ) + Ds 2 = 3s 3 − s 2 − 3s + 2

A+C=3 -2A+B-C+D=-1 A-2B=-3 B=2; A=2(2)-3=1 C=3-1=2 D=2(1)-2+2-1=1 We get X ( s ) =

1 2 2 1 + 2+ + s s s + 1 ( s − 1) 2

By inverse L.T L−1 [X (t )] = 1 + 2t + 2e t + te t

L−1 [X (t )] = 1 + 2t + e t ( 2 + t )

3.4 Expand the following function by partial fraction expansion. Do not evaluate co-efficient or invert expressions X ( s) =

2 ( s + 1)( s + 1) 2 ( s + 3)

X ( s) =

A Bs + C Ds + E F + 2 + 2 + 2 s + 1 s + 1 ( s + 1) s+3

= A( s 2 + 1) 2 ( s + 3) + ( Bs + C )( s + 1)( s + 3)( s 2 + 1) + ( Ds + E )( s + 1)( s + 3) + F ( s + 1)( s 2 + 1) 2

= A( s 4 + 2 s 2 + 1)( s + 3) + ( Bs + C )( s 2 + 4 s + 3)( s 2 + 1) + ( Ds + E )( s 2 + 4 s + 3) + F ( s + 1)( s 2 + 4 s + 1)

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= s 5 ( A + B + F ) + s 4 (3 A + C + 4 B + F ) + s 3 ( 2 A + B + 4C + 3B ) + s 2 (6 A + C + 4 B + 3C ) + s( A + 4C + 3B + 4 E + F ) + 3 A + 3 AC + 3E + F = 2 A+B+F=0 -3A+C+4B+F=0 2A+B+4C+3B=0 6A+C+4B+3C=0 A+4C+3B+3D+4E+F=0 3A+3C+3E+F=2

by solving above 6 equations, we can get the values of A,B,C,D,E and 1 . X ( s) = 3 s ( s + 1)( s + 1) ( s + 3) 3 X ( s) =

A B C D E F G H + 2 + 3+ + + + + 2 s s s s +1 s + 2 s + 3 ( s + 3) ( s + 3) 3

by comparing powers of s we can evaluate A,B,C,D,E,F,G and H. 1 c) X ( s ) = s ( s + 2)( s + 3) ( s + 4)

A B C D + + + s +1 s + 2 s+3 s+4 by comparing powers of s we can evaluate A,B,C,D X ( s) =

3.5 a) X ( s ) =

Let

1 s ( s + 1)(0.5s + 1)

1 A B C = + + s ( s + 1)(0.5s + 1) s s + 1 (0.5s + 1)

 s 2 3s   s2  = A + + 1 + B  + s  + C ( s 2 + s ) = 1 2 2  2  A=1 A B B 1 + + C = 0= +C = − 2 2 2 2

3A 3 + B + C = 0= B + C = − 2 2

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B/2=1/2 *-3/2=-1; B=-2; C= -3/2+2=1/2

X ( s) =

1 2 1 1  − +   s s + 1 2  0.5s + 1 

= L − 1 ( X ( s )) = x ( t ) = 1 − 2 e − t + e − 2 t b)

dx + 2 x = 2; x (0) = 0 dt

Applying laplace trafsorms

sX ( s ) − x (0) + 2 X ( s ) = 2 / s

L−1 ( X ( s )) =

2 s ( s + 2)

 2  L−1 ( X ( s )) = 2 L−1    s ( s + 2)  1 / 2 1 / 2  = L−1 ( X ( s )) = 2 L−1  − s + 2   s

−2 t 1 − e = 3.6 a) Y ( s ) =

s +1 s + 2s + 5 2

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= Y ( s) =

s +1 s + 2s + 5 2

s +1 ( s + 1) 2 + 4

  s +1 = L−1 (Y ( s )) = L−1  2   ( s + 1) + 4  using the table,we get Y (t ) = e − t Cos2t

b) Y ( s ) =

Y ( s) =

s 2 + 2s s4

1 2 + 3 2 s s

Y(t)= L−1 (Y ( s )) = t + t 2

c) Y ( s ) =

2s ( s − 1) 3

2s − 2 + 2 ( s − 1) 3

2 2 + 2 ( s − 1) ( s − 1) 3

 2   2  Y (t ) = L−1  + L−1   2  3   ( s − 1)   ( s − 1) 

t2 t 2(te + e = e t (t 2 + 2t ) 2 t

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3.7a)

Y ( s) =

As + B Cs + D 1 = + ( s 2 + 1) ( s 2 + 1) ( s 2 + 1)

thus ( As + B ) + (Cs + D )( s 2 + 1) = 1 = Cs 3 + Ds 2 + ( A + C ) s + ( B + D ) = 1

C=0,D=0 Also A=0;B=1 1 1 A B C D Y ( s) = 2 = = + + + 2 2 2 2 ( s + 1) (s + i) (s − i) (s + i) (s + i) (s − i) (s − i)2

A( s + i )( s − i ) 2 + B( s − i ) 2 + C ( s − i )( s + i ) 2 + D( s + i ) 2 = 1

( A + C ) s 3 + ( − Ai + B + Ci + D ) s 2 + ( A − 2 Bi + C + 2 Di ) + ( − Ai − B + Ci − D ) = 1

Thus,A+C=0 -Ai+B+Ci+2Di=0 ; B=D A-2Bi+C+2Di=0 -Ai-B+Ci-D=1

Also D=-Ci;B=-Ci, A=-C,C=-i/4

A=i/4 ; B=-1/4; D=-1/4

Y ( s) =

i/4 − 1/ 4 −i/4 − 1/ 4 + + + 2 (s + i) (s + i) (s − i) (s − i)2

Y (t ) =

− 1/ 4 −i/4 −1/ 4 i/4 + + + 2 (s + i) (s + i) ( s − i) ( s − i) 2

−1/ 4 −i/4 −1/ 4 i/4 + + + 2 (s + i) (s + i) (s − i) (s − i)2

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Y (t ) = i / 4e −it − 1 / 4e −it −1 / 4e it − 1 / 4te it Y (t ) = 1 / 4(ie − it − te −it − ie it − te it ) Y (t ) = 1 / 4(i (Cost − iSin t ) − t (Cost − i Sin t ) − i(Cos t + iSin t ) − t (Cos t + i Sin t ) ) Y (t ) = 1 / 4( 2 Sin t − 2t Cos t )

Y ( t ) = 1 / 2 ( Sin

3.8

f ( s) =

= f ( s) =

t − t Cos

1 s ( s + 1) 2

A B C + + 2 s s s +1

= A( s + 1) + Bs( s + 1) + Cs 2 = 1 Let s=0 ; A=1 s=1; 2A+B+C=1 s=-1: C=1 B=-1 1 1 1 f ( s) = 2 + + s s s +1

f (t ) = (t − 1) + e − t

PROPERTIES OF TRANSFORMS

4.1 If a forcing function f(t) has the laplace transforms

f ( s) = =

1 e − s − e −2 s e −3s + − s s2 s

1 − e −3s e − s − e −2 s + s s2

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f (t ) = L−1{ f ( s )} = [u(t ) − u(t − 3)] + [(t − 1)u (t − 1) − (t − 2)u (t − 2)] = u(t ) + (t − 1) u(t − 1) − (t − 2)u(t − 2) − u (t − 3) graph the function f(t)

4.2 Solve the following equation for y(t): t

∫ y (τ ) dτ 0

dy (t ) y ( 0) = 1 dt

Taking Laplace transforms on both sides t

 dy (t )  L{∫ y (τ ) dt} = L    dt  0 1 . y ( s ) = s. y ( s ) − y (0) s

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1 . y ( s ) = s. y ( s ) − 1 s y ( s) =

s s −1 2

 s  y (t ) = L−1{ y ( s )} = L−1  2  cosh(t )  s − 1 4.3 Express the function given in figure given below the s-

t – domain and the

domain

This graph can be expressed as = {u(t − 1) − u(t − 5)} + { (t − 2)u(t − 2) − (t − 3)u (t − 3)} + {u(t − 5) − (t − 5)u (t − 5) + (t − 6)u (t − 6)} f (t ) = u (t − 1) + (t − 2)u(t − 2) − (t − 2)u(t − 3) − (t − 5)u(t − 5) + (t − 6)u(t − 6)

f ( s ) = L{ f (t )} =

e − s e −2 s e −3s e −3s e −5 s e −6 s + 2 − 2 − − 2 + 2 s s s s s s =

e − s − e −3s e −2 s + e −6 s − e −3s − e −5 s + s s2

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4.4 Sketch the following functions: f (t ) = u (t ) − 2u(t − 1) + u(t − 3)

f (t ) = 3tu(t ) − 3u(t − 1) + u(t − 2)

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4.5 The function f(t) has the Laplace transform

f ( S ) = (1 − 2e − s + e −2 s ) / s 2 obtain the function f(t) and graph f(t) 1 − 2 e − s + e −2 s f ( s) = s2

1 − e − s e − s − e −2 s − s2 s2

f (t ) = L−1{ f ( s )} = − (t − 1)u(t − 1) + tu(t ) − {(t − 1)u (t − 1) − (t − 2)u(t − 2)] = tu(t ) − 2(t − 1)u (t − 1) + (t − 2)u(t − 2) 4.6 Determine f(t) at t = 1.5 and at t = 3 for following function: f (t ) = 0.5u(t ) − 0.5u(t − 1) + (t − 3)u (t − 2)

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At t = 1.5 f (t ) = 0.5u(t ) − 0.5u (t − 1) + (t − 3)u (t − 2) f (1.5) = 0.5u(t ) − 0.5u(t − 1) f (1.5) = 0.5 − 0.5 = 0

At t = 3

f (3) = 0.5 − 0.5 + (3 − 3) = 0

RESPONSE OF A FIRST ORDER SYSTEMS 5.1 A thermometer having a time constant of 0.2 min is placed in a temperature bath and after the thermometer comes to equilibrium with the bath, the temperature of the bath is increased linearly with time at the rate of I deg C / min what is the difference between the indicated temperature and bath temperature (a) 0.1 min (b) 10. min after the change in temperature begins. © what is the maximum deviation between the indicated temperaturew and bath temperature and when does it occurs. (d) plot the forcing function and the response on the same graph. After the long enough time buy how many minutes does the response lag the input.

Consider thermometer to be in equilibrium with temperature Xs X (t ) = X S + (1° / m )t , t > 0 as it is given that the temperture varies linearly X(t)-Xs = t Let X(t) = X(t) - Xs = t

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temperature

bath at

Y(s) = G(s).X(s) Y ( s) =

1 1 A B C = + + 2 2 1 + τs s 1 + τs s s

A = τ 2 B = − τ C =1

τ2 τ 1 Y ( s) = − + 2 1 + τs s s Y ( t ) = τe − t / τ − τ + t (a) the difference between the indicated temperature and bath temperature at t = 0.1 min = X(0.1)_ Y(0.1) = 0.1 - (0.2e-0.1/0.2 - 0.2+0.1) since T = 0.2 given = 0.0787 deg C (b) t = 1.0 min X(1) - Y(1) = 1- (0.2e-1/0.2 - 0.2 +1) = 0.1986 (c) Deviation D = -Y(t) +X(t) = -τe-t/T+T =τ (-e-t/T+1) For maximum value dD/dT = τ (-e-t/T+(_-1/T) = 0 -e-t/ = 0 as t tend to infinitive D = τ (-e-t/T+(_-1/T) = τ =0.2 deg C

5.2 A mercury thermometer bulb in ½ in . long by 1/8 in diameter. The glass envelope is very thin. Calculate the time constant in water flowing at 10 ft / sec at a temperature of 100 deg F. In your solution , give a summary which includes (a) Assumptions used. (b) Source of data (c) Results

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T = mCp/hA =

( ρAL)C p h ( A + πDL)

Calculation of

NU d =

Re d =

Pr =

hD = CRem (Pr) n K

Dvρ

Cpµ K

(1 / 8 * 2.54 * 10 −2 )(10 * 0.3048)103 = 9677.4 10 −3

= 4.2 KJ / KgK

Source data: Recently, Z hukauskas has given c,m ,ξ,n values. For Re = 967704 C = 0.26 & m = 0.6 NuD = hD/K = 0.193 (9677.4)*(6.774X10-3) = 130 .h = 25380

5.3 Given a system with the transfer function Y(s)/X(s) = (T1s+1)/(T2s+1). Find Y(t) if X(t) is a unit step function. If T1/T2 = s. Sktech Y(t) Versus t/T2. Show the numerical values of minimum, maximum and ultimate values that may occur during the transient. Check these using the initial value and final value theorems of chapter 4.

Y ( s) =

T1s + 1 T2 s + 1

X(s) =unit step function = 1 X(s) = 1/s

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Y ( s) =

T1 s + 1 A B = + s (T2 s + 1) s iT2 s

A = 1 B = T1 - T2 Y ( s) =

1 T1 − T2 + s 1 + T2 s

Y (t ) = 1 +

T1 − T2 −t / T2 e T2

If T1/T2 = s then Y (t ) = 1 + 4e − t / T2 Let t/T2 = x then Y (t ) = 1 + 4e

−x

Using the initial value theorem and final value theorem Lim Y (T ) = Lim sY ( s ) S →∞

T →0

1 T s +1 s = T1 = 5 Lim 1 = Lim S →∞ T s + 1 S →∞ 1 T2 2 T2 + s T1 +

Lim Y (T ) = Lim sY ( s ) = Lim T →0

S →∞

S →0

T1 s + 1 =1 T2 s + 1

Figure:

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5.4 A thermometer having first order dynamics with a time constant of 1 min is placed in a temperature bath at 100 deg F. After the thermometer reaches steady state, it is suddenly placed in bath at 100 deg F at t = 0 and left there for 1 min after which it is immediately returned to the bath at 100 deg F. (a) draw a sketch showing the variation of the thermometer reading with time. (b) calculate the thermometer reading at t = 0.5 min and at t = 2.0 min

1 Y ( s) (τ = 1 min) = X ( s) s + 1

1 e−s  ( s ) = 10  − s  s

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 1 − e−s  Y ( s ) = 10   s   − e−s  1 Y ( s ) = 10   s ( s + 1) s ( s + 1)  Y (t ) = 10(1 − e − t ) t < 1 Y (t ) = 10( (1 − e − t ) − (1 − e − ( t −1) ) ) t ≥ 1 At t = 0.5 T = 103.93 At = 2 T =102.325

5.5 Repeat problem 5.4 if the thermometer is in 110 deg F for only 10 sec.

If thermometer is in 110 deg F bath for only 10 sec

T = 110 − 10e − t / 60

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0 < t < 10 sec & T = 60 sec T (t = 10 sec) = 101.535 T = 100 + 1.535e â&#x2C6;&#x2019; ( t â&#x2C6;&#x2019;10 ) / 60 t > 10 sec T(t=30sec) = 101.099 deg F T(t=120sec) = 100.245 deg F

5.6 A mercury thermometer which has been on a table for some time,is registering the room temperature ,758 deg F. Suddenly, it is placed in a 400 deg F oil bath. The following data are obtained for response of the thermometer

Time (sec) Temperature, Deg F 0 75 1 107 2.5 140 5 205 8 244 10 282 15 328 30 385 Give two independent estimates of the thermometer time constant.

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τ =

t  325  ln   400 − T 

From the data , average of 9.647,11.2,9.788,10.9,9.87,9.95, and 9.75 is 10.16 sec.

5.7 Rewrite the sinusoidal response of first order system (eq 5.24) in terms of a cosine wave. Re express the forcing function equation (eq 5.19) as a cosine wave and compute the phase difference between input and output cosine waves. 1   1 Aω  τ  ( s) = 2 Y ( s) = τs + 1 s +ω2 s + 1

splitting into partial fractions then converting to laplace transforms

Y (t ) =

Aαωτ −t /τ e + τ 2ω 2 + 1

τ 2ω 2 + 1

sin(ωt + φ )

where φ = tan-1 (ωτ) As t →∝ Y (t ) s =

Y (t ) = Y (t ) =

τ 2ω 2 + 1

sin(ωt + φ ) =

π  cos(ωt −  − φ  2  τ 2ω 2 + 1 A

π  A sin(ωt + φ ) = A cos − ωt  2  π  A cos ωt −  2 

The phase difference = φ −

 π − −  = φ 2  2

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)

5.8 The mercury thermometer of problem 5.6 is allowed to come to equilibrium in the room temp at 75 deg F.Then it is immersed in a oil bath for a length of time less than 1 sec and quickly removed from the bath and re exposed to 75 deg F ambient condition. It may be estimated that the heat transfer coefficient to the thermometer in air is 1/ 5th that in oil bath.If 10 sec after the thermometer is removed from the bath it reads 98 Deg F. Estimate the length of time that the thermometer was in the bath. t < 1 sec

T1 = 400 − 325e − t1 / τ

Next it is removed and kept in 75 Deg F atmosphere Heat transfer co-efficient in air = 1/5 heat transfer co-efficient in oil

hair = 1/5 hoil

τ =

mC τ oil = 10 sec hA τ air = 50 sec TF = 75 + (T1 − 75)e − t2 / 50 TF = Final Temp = 98 deg C

98 = 75 + (325 − 325e − t1 / 10 )e −10 / 50 e −t / 10 = 0.91356 t 1 = 0.904 sec.

5.9 A thermometer having a time constant of 1 min is initially at 50 deg C. it is immersed in a bath maintained at 100 deg C at t = 0 . Determine the temperature reading at 1.2 min. τ = 1 min for a thermometer initially at 50 deg C. Next it is immersed in bath maintained at 100 deg C at t = 0 At t = 1.2

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Y (t ) = A(1 − e − t / τ ) Y (1.2) = 50(1 − e −1.2 / 1 ) + 50 Y(1.2) = 84.9 deg C

5.10 In Problem No 5.9 if at, t = 1.5 min thermometer having a time constant of 1 minute is initially at 50 deg C.It is immersed in a bath maintained at 100 deg C at t = 0.Determine the temperature reading at t = 1.2 min. At t = 1.5 Y (1.5) = 88.843°C Max temperature indicated = 88.843 deg C AT t = 20 min T = 88.843 − 13.843(1 − e −18.8 / 1 )

T = 75 Deg C.

5.11 A process of unknown transfer function is subjected to a unit impulse input. The output of the process is measured accurately and is found to be represented by the function Y(t) = t e-t. Determine the unit step response in this process. X(s) = 1 Y(t) = te-t Y (s) =

G( s) =

1 ( s + 1) 2 1 Y ( s) = X ( s ) ( s + 1) 2

For determining unit step response Y ( s) =

1 ( s + 1) 2

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Y ( s) =

1 A B C = + + 2 ( s + 1) s s + 1 ( s + 1) 2

A = 1 B = -1 C = -1 Y ( s) =

1 1 1 − − s s + 1 ( s + 1) 2

Y (t ) = 1 − e − t − te − t

Response of first order system in series

7.1 Determine the transfer function H(s)/Q(s) for the liquid level shown in figure P7-7. Resistance R1 and R2 are linear. The flow rate from tank 3 is maintained constant at b by means of a pump ; the flow rate from tank 3 is independent of head h. The tanks are non interacting.

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Solution :

A balance on tank 1 gives

q â&#x2C6;&#x2019; q1 = A1

dh1 dt

where h1 = height of the liquid level in tank 1

similarly balance on the tank 2 gives q1 â&#x2C6;&#x2019; q2 = A2

dh2 dt

and balance on tank 3 gives

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q2 − q0 = A3

here q1 =

dh dt

h1 h q2 = 2 R1 R2

q0 = b

So we get

q−

dh h1 = A1 1 dt R1

dh2 h1 h2 − = A2 dt R1 R2

dh h2 − b = A3 R2 dt

writing the steady state equation

qS −

dh h1s = A1 1S = 0 dt R1

dh2 S h1S h2 S − = A2 dt R1 R2 h2 S −b = 0 R2

Subtracting and writing in terms of deviation

Q −

dH 1 H = A1 dt R1

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dH 2 H1 H 2 − = A1 dt R1 R2

dH H2 = A3 dt R2

where Q = q –qS H1= h1-h1S H1= h2-h2S H = h - hS

Taking Laplace transforms

Q( s) −

H 1 ( s) = A1 s H 1 ( s ) ---------(1) R1

H 1 (s) H 2 (s) − = A2 s H 2 ( s ) --------(2) R1 R2

H 2 (s) = A3 s H ( s ) ----------(3) R2

We have three equations and 4 unknowns(Q(s),H(s),H1(s) and H2(s). So we can express one in terms of other.

From (3)

H 2 (s) =

H 2 (s) -------------(4) R1 A3 s

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R2 H 1 ( s ) where τ2 =R2 A2------------(5) R1 (τ 2 s + 1)

H 2 (s) =

From (1)

H 1 ( s) =

R1Q( s ) , (τ 1 s + 1)

τ 1 = R1 A1 ---------(6)

Combining equation 4,5,6

H ( s) =

Q( s) ( A3 s )(τ 1 s + 1)(τ 2 s + 1)

H ( s) 1 = Q( s) ( A3 s )(τ 1 s + 1)(τ 2 s + 1)

Above equation can be written as i.e, if non interacting first order system are there in series then there overall transfer function is equal to the product of the individual transfer function in series.

7.2 The mercury thermometer in chapter 5 was considered to have all its resistance in the convective film surrounding the bulb and all its capacitance in the mercury. A more detailed analysis would consider both the convective resistance surrounding the bulb and that between the bulb and mercury. In addition , the capacitance of the glass bulb would be included. Let Ai = inside area of bulb for heat transfer to mercury. Ao = outside area of bulb, for heat transfer from surrounding fluid. .m = mass of the mercury in bulb. mb = mass of glass bulb. C = heat capacitance of mercury.

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Cb = heat capacity of glass bulb. .hi = convective co-efficient between the bulb and the surrounding fluid. .ho = convective co-efficient between bulb and surrounding fluid. T = temperature of mercury. Tb = temperature of glass bulb. Tf = temperature of surrounding fluid. Determine the transfer function resistance and capacitance on inclusion of the bulb results in overall transfer function different

between Tf and T. what is the effect of bulb the thermometer response? Note that the a pair of interacting systems, which give an from that of Eq (7.24)

Writing the energy balance for change in term of a bulb and mercury respectively

Input - output = accumulation h0 A0 (T f −Tb )−hi Ai (Tb − T ) = mb C b

hi Ai (Tb −T )−0 = m C

dTb dt

dT dt

Writing the steady state equation h0 A0 (T fs −Tbs )−hi Ai (Tbs − Ts ) = mb C b

dTbs =0 dt

hi Ai (Tbs −Ts ) = 0

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Where subscript s denoted values at steady subtracting and writing these equations in terms of deviation variables. h0 A0 (T f −Tb )−hi Ai (Tb − Tm ) = mb C b

hi Ai (Tb −Tm )−0 = m C

dTb dt

dTm dt

Here TF = Tf - TfS TB = Tb - TbS Tm = T - TS Taking laplace transforms h0 A0 (TF ( s ) −TB ( s ))−hi Ai (TB − Tm ) = mb C b TB ( s ) ----(1)

And hi Ai (TB ( s ) −Tm ( s )) = mC sTB ( s ) ------(2)

= h0 A0 (TF ( s ) −TB ( s ))−mCSTm ( s ) = mb C b sTB ( s )

From (2) we get   mC s + 1 = Tm ( s ) (τ i s + 1) TB ( s ) = Tm ( s )    hi Ai

Where τ i =

mC hi Ai

Putting it into (1)

 mC  TF ( s ) − Tm ( s ) (τ i s + 1))(τ 0 s + 1) + s = 0 h0 A0  

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 mC  = TF ( s ) = Tm ( s ) (τ i s + 1))(τ 0 s + 1) + s h0 A0   =

Tm ( s ) = TF ( s )

τ iτ 0 s 2 + (τ i + τ 0 +

Tm ( s ) = TF ( s )

mC )s + 1 h0 A0

τ iτ 0 s 2 + (τ i + τ 0 +

mC )s + 1 h0 A0

Or we can write T (s) = T f (s)

τi =

τ iτ 0 s 2 + (τ i + τ 0 +

mC )s + 1 h0 A0

mC mC and τ 0 = b b hi Ai h0 A0

We see that a loading term mC/ hoAo is appearing in the transfer function. The bulb resistance and capacitance is appear in τ 0 and it increases the delay i.e Transfer lag and response is slow down.

7.3 There are N storage tank of volume V Arranged so that when water is fed into the first tank into the second tank and so on. Each tank initially contains component A at some concentration Co and is equipped with a perfect stirrer. A time zero, a stream of zero concentration is fed into the first tank at volumetric rate q. Find the resulting concentration in each tank as a function of time. Solution:

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. ith tank balance qC i −1 − qC i = V

dC i dt

qC ( i −1) s − qC is = 0

C (i −1) − C i =

V dC i q dt

 V τ =  q 

Taking lapalce transformation C (i −1) ( s ) − C i ( s ) = τ sCi ( s ) C (i −1) ( s ) = (1 + τ s )Ci ( s )

Ci ( s) 1 = C i −1 ( s ) 1 + τ s

Similarly

Ci ( s) C1 ( s) C 2 ( s) C ( s) Ci ( s) 1 = × × − − − − − − − − − i −1 × = Co( s) C 0 ( s) C1 ( s) Ci −2 ( s) Ci ( s) (1 + τ s)i

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Or C N (s) 1 = Co( s ) (1 + τ s ) N C N ( s) =

− C0 s (1 + τ s ) N

1 τ τ τ  C N ( s) = − C0  − − −−−−−−−  N N −1 1 + τs  (1 + τ s ) (1 + τ s ) s

t t − −  t  N −1 τ τ − e t e t N −2  − N −2 . −−−−−e τ  C N (t ) = − C 0 1 − N −1 .  τ  ( N − 1)! τ ( N − 2)!  

  t −  τ   C N (t ) = − C0 1 − e .    

N −1

t   τ  +. ( N − 1)!

N −2  t    τ  − − − − − +1  ( N − 2)!  

7.4 (a) Find the transfer functions H2/Q and H3/Q for the three tank system shown in Fig P7-4 where H1,H3 and Q are deviation variables. Tank 1 and Tank 2 are interacting. 7.4(b) For a unit step change in q (i.e Q = 1/s); determine H3(0) , H3(∞) and sketch H3(t) vs t. Solution :

Writing heat balance equation for tank 1 and tank 2

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q − q1 = A1

q1 − q 2 = A2

q1 =

dh1 dt

dh2 dt

h1 − h2 h q2 = 2 R1 R2

Writing the steady state equation q s − q1s = 0 q1s − q 2 s = 0

Writing the equations in terms of deviation variables Q − Q1 = A1

dH 1 dt

Q1 − Q2 = A2

dH 2 dt

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Q1 =

H1 − H 2 H Q2 = 2 R1 R2

Taking laplace transforms Q ( s ) − Q1 ( s ) = A1 sH 1 ( s )

Q1 ( s) − Q2 ( s) = A1 sH 2 ( s) R1 Q1 ( s ) = H 1 ( s ) − H 2 ( s ) R2 Q2 ( s ) = H 2 ( s )

Solving the above equations we get

H 2 ( s) R2 = 2 Q(s) τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 )s + 1

[

Here

]

τ1 = R1 A1 τ 2 = R2 A2

Now writing the balance

q 2 − q3 = A3

for third tank

dh3 dt

Steady state equation q 2 S − q3 S = 0

q3 =

h3 R3

Q2 −

dh H3 = A3 3 dt R3

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Taking laplace transforms H 3 (s) = A3 sH ( s ) R3

Q2 ( s ) −

H 3 ( s) (τ 3 s + 1) where R3

Q2 ( s ) =

τ 3 = R3 A3

From equation 1,2,3,4 and 5 we got

Qs ( s) Q( s )

[τ τ

1 2

]

s + (τ 1 + τ 2 + A1 R2 ) s + 1 2

Putting it in equation 6

H 3 ( s) Q( s)

R3 τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 (τ 3 s + 1)

[

]

Putting the numerical values of R1,R2 and R3 and A1,A2,A3

H 3 ( s) Q( s)

H 2 (s) Q( s )

4 4 s + 6 s + 1 (2s + 1)

[

2 4s + 6s + 1

[

]

Solution (b)

Q (s) =

1 s

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H 3 (s) =

1 4 2 s 4s + 6s + 1 (2 s + 1)

[

]

From initial value theorem

H 3 (0) = Lim sH 3 ( s) S →∞

Lim S →∞

4 (2 s + 1)(4s 2 + 6s + 1)

Lim

S →∞

(2 s + 1) s 3 (4 +

6 1 + 2) s s

H3 (0) = 0

From final value theorem

H 3 (∞) = Lim sH 3 ( s ) S →0

Lim S →0

4 (2s + 1)(4 s 2 + 6 s + 1)

(∞) = 4

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7.5 Three identical tanks are operated in series in a non-interacting fashion as shown in fig P7.5 . For each

tank R=1, τ = 1. If the deviation in

flow rate to the first tank in an impulse function of magnitude 2, determine (a) an expression for H(s)

where H is the deviation in level in the third

tank. (b) sketch the response H(t) (c) obtain an expression for H(t)

solution :

writing energy balance equation for all tanks

q − q1 = A

dh1 dt

q1 − q2 = A

dh2 dt

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q2 − q3 = A

q1 =

h h1 q2 = 2 R R

dh dt

q3 =

h R

So we get

qS − q1S = 0 q1S − q2 S = 0 q 2 S − q3 S = 0

writing in terms of deviation variables and taking laplace transforms

Q( s) −

H1 ( s) = AS H 1 ( s ) R

Q1 ( s ) H 2 ( s ) − = AS H 2 ( s ) R R H 2 ( s) H ( s) − = AS H ( s ) R R

solving we get

H ( s) Q( s)

H ( s) =

R 1 = 3 (τ s + 1) ( s + 1) 3

2 Q( s) = 3 (τ s + 1) ( s + 1) 3

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H (t ) = L−1 {H ( s )} = 2

t 2 −t e 2

H (t ) = t 2 e − t dH (t ) = 2te −t − te −t = 0 dt

= 2t = t 2

at t = 2 max will occur.

7.6 In the two- tank mixing process shown in fig P7.6 , x varies from 0 lb salt/ft3 to 1 lb salt/ft3 according to step function. At what time 3

does the

salt concentration in tank 2 reach 0.6 lb/ ft ? The hold up volume of each tank is 6 ft3.

Solution

Writing heat balance equation for tank 1 and tank 2

qx − q y = V

dy dt

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q y − qc = V

dl dt

steady state equation q xs − q ys = 0

q ys − qcs = 0

writing in terms of deviation variables and taking laplace transforms

X ( s) − Y ( s) =

V s Y ( s) q

1 1 V Y ( s) ;τ = = = q X ( s)  V  τ s +1  s + 1 q 

C ( s) =

Y ( s) X ( s) = (τ s + 1) (τ s + 1) 2

C ( s) 1 = X ( s ) (τ s + 1) 2

X ( s) =

τ=

1 s

V 6 = =2 q 3

C ( s) =

X ( s) s( 2 s + 1) 2

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C ( s) =

(1 / 4) 1 s( s + ) 2 2

1 C ( s) =   4

  

1 1   s ( s + 1 )2 2

1   1 1 2 C ( s) = −   2 − 1 s  1  s +  + 1  2   2

C (t ) = 1 −

− 1 −2 te − e 2 2

C ( t ) = 0 . 61 lb salt / ft 3

t = 4.04 min

7.7 Starting

from first

and H2(s)/Q(s)

for

the

principles, derive liquid level

the transfer functions H1(s)/Q(s)

system

shown in

resistance are linear and R1= R2 = 1. Note that two

figure P7.7. The

streams are flowing

from tank 1, one of which flows into tank 2. You are expected to give numerical values

of the parameters

and

in the transfer functions and to

show clearly how you derived the transfer functions.

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Writing heat balance equation for tank 1

q − qa − q1 = A1

q1 =

dh1 dt

h1 h qa = 1 R1 Ra

=q −

dh h1 h − 1 = A1 1 dt Ra R1

writing the balance equation for tank 2

q1 − q2 = A2

dh2 dt

dh2 h1 h2 − = A2 dt R1 R2

writing steady state equations

qs −

hs hs − 1 =0 Ra R1

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h1 s h2 s − =0 R1 R2 writing the equation in terms of deviation variables

 1 dH 1 1  Q − H 1  +  = A1 dt  Ra R1 

dH 2 H1 H 2 − = A2 dt R1 R2

taking laplace transforms

 R + R2   = A1S H 1 s -----------(1) Q ( s ) − H 1 ( s ) 1  R1 Ra 

and

H1 ( s) H 2 ( s) − = A2 s H 2 ( s ) -----------(2) R1 R2

from (1) we get

H1 ( s) 1 = Q( s)  R1 + Ra   A1 s + R R  1 a  

 R1 Ra  R + R  H1 ( s) a   1 = Q( s)   R1 Ra A1  R + R s + 1 a   1

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 R1 Ra    RR A H 1 ( s )  R1 + Ra  ; τ1 = 1 a 1 = [τ 1s + 1] R1 + Ra Q( s)

and from (2 ) we get  R2 Ra  R2   R + R  R  H1 ( s) a  1  τ =R A =  1 [ τ 1s + 1](τ 2 s + 1) 2 2 2 Q( s)

putting the numerical values of parameters

2   H1 ( s) 3 =   Q( s) 4   s + 1 3  

2   H 2 (s) 3 = Q(s) 4   s + 1(s + 1)  3

8.1 A step transfer

change of magnitude 4 is introduced into a system having the

Y ( s) 10 = 2 X ( s ) s + 1.6s + 4 Determine (a) % overshoot (b)Rise time (c)Max value of Y(t) (d)Ultimate value of Y(t) (e) Period of Oscillation.

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Given X ( s ) =

4 s

Y ( s) =

The transfer function is

40 s ( s + 1.6s + 4) 2

2.5 Y ( s) 10 × 0.25 = = 2 0.25s + 0.4 s + 1) X ( s ) 0.2( s 2 ) + ( 1.6 ) s + 1) 4

τ 2 = 0.25 ; τ = 0.5 and 2ξ τ = 0.4

ξ =

0.4 = 0.4 (< 1 = system is underdamped ) 2(0.5)

we find ultimate value of Y(t) 40 s 40 = = 10 S →0 s ( s + 1.6 + 4) 4

Lt Y (t ) = Lt sY ( s ) = Lt

t →∞

S →0

thus B= 10 now, from laplace transform tables ξt   − 1 Y (t ) = 101 − e τ sin(α + φ ) 1−ξ 2  

where α =

1−ξ 2

, φ = tan −

1−ξ 2

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(a) Over shoot =

 − πξ A = exp  1−ξ 2 B 

  = exp  − π × 0.4  = 0.254  0.84     

thus % overshoot = 25.4 c)thus, max value of Y(t) = A+B = B(0.254)+B = 2.54+10 = 12.54 e) Period of oscillation =

2πτ 1−ξ 2

= 3.427

b) For rise time, we need to solve ξt   − 1 e τ sin(αt + φ ) = 10 for t = tr 101 − 1−ξ 2  

= e = e

−

ξτ r τ

0.4τ r − 0.5

sin(α t r + φ ) = 0 sin(1.833 t r + 1.1589) = 0

solving we get tr = 1.082 thus SOLUTION: % Overshoot = 25.4 Rise time = 1.0842 Max Y(t) = 12.54 U(t) Y(t) = 10 Period of oscillation = 3.427 Comment : we see that the Oscillation period is small and the decay ratio also small = system is efficiently under damped.

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8.2 The tank system operates at steady state. At t = 0, 10 ft3 of wateris added to tank 1. Determine the maximum deviation in level in both tanks from the ultimate steady state values, and the time at which each maximum occurs. A1 = A2 = 10 ft3 R1 = 0.1ft/cfm R2 = 0.35ft/cfm.

As the

tanks are non interacting the transfer functions are

H ( s) K1 0.1 = = Q ( s ) τ 1 s + 1 ( s + 1) H 2 ( s) R2 0.35 = = Q( s) (τ 1 s + 1)(τ 2 s + 1) ( s + 1)(3.5s + 1) Now, an impulse of ∂(t ) = 10 ft 3 is provided

Q ( s ) = 10 = H 1 ( s ) = and H 2 ( s ) =

1 = e −t s +1

3.5 3.5 = 2 ( s + 1)( 3.5s + 1) 3.5s + 4.5s + 1

Now τ 2 = 3.5 = τ = 1.871 2ξτ = 4.5 = ξ =

4.5 = 1.202 2τ

thus, this is an ovedamped system

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Using fig8.5, for ξ = 1.2 , we see that maximum is attained at t = 0.95, t = 1.776 min

And the maximum value is around τ 2 = 0.325 Y2 (t) = 0.174 = H2(t) = 0.174x3.5 = 0.16ft thus max deviation is H1 will be at t = 0 = H1 = 1 ft max deviation is H2 will be at t = 1.776 min = H2Max = 0.61 ft. comment : the first tank gets the impulse and hence it max deviation turns out to be higher than the deviations for the second tank. The second tank exhibits an increase response ie the deviation increases, reaches the H2Max falls off to zero.

8.3 The tank liquid level shown operates at steady state when a step change is made in the flow to tank 1.the transaient response in critically damped, and it takes 1 min for level in second tank to reach 50 % of total change. If A1/A2 = 2 ,find R1/R2 . calculate τ for each tank. How long does it take for level in first tank to reach 90% of total change?

For the first tank, transfer function

For the second tank

H 1 (s) R1 = Q (s) τ s + 11

R H 2 ( s) = Q( s) (τ 1 s + 1)(τ 2 s + 1)

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H 2 ( s) R2 = 2 Q( s) τ 1τ 2 s + (τ 1 + τ 2 ) s + 1

Q( s) =

R2 1 1 ; H 2 ( s) = 2 s s τ 1τ 2 s + (τ 1 + τ 2 ) s + 1

τ ( parameter) = τ 1τ 2 For τ ( parameter ) = τ 1τ 2

  t for ξ = 1, H 2 (t ) = R2 1 − 1 +  τ 1τ 2  

 − e  

τ 1τ 2

  

given, t = 1 for τ ( parameter) = τ 1τ 2 H 2 (t → ∞) = R2 (1 − (0) ) = R2

  1 = R2 1 − 1 +   τ 1τ 2  

 − e  

τ 1τ 2

 R  = 2 −I  2 

also 2ξτ = τ 1 + τ 2

ξ =1=

τ1 + τ 2 2

= τ = A1 R1 = A2 R2 =

R1 A2 = = 0.5 R2 A1

from I

 1 − 1 − 0.5 = 1 +  e τ  τ

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− R1 8.3) H 1 ( s ) = ; H 1 (t ) = R1 (1 − e τ 1 ) s (τ 1 s + 1)

0.94(t → ∞) = R1 (1 − e

0.9 R1 = R1 (1 − e

−

t 0.596

−

t 0.596

−

τ1

)

= 0.1; t =1.372 min

thus R1 = 0.5 R2

τ 1 =τ 2 = 0.596 min t 90% = 1.372 min

Comment : Small values of τ 1 ,τ 2 indicate the system regains the steady state quickly. Also as R1 > R2 , the sec ond tan

responds more slowly to changes than first tan k .

8.4 Assuming the flow in the manometer to be laminar function between applied pressure P1 and the manometer reading h. Calculate a) steady sate gain ,b) τ ,c) ξ . Comment on the parameters and their relation to the physical nature of this problem.

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Assumptions: Cross-sectional area =a Length of mercury in column = L Friction factor = 16/Re (laminar flow) Mass of mercury = mrg Writing a force balance on the mercury

Mass X acceleration = pressure force - drag force - gravitational force ( ALρ )

d 2h ρu 2 = Ap − Af − A( ρgh) 1 2 dt 2

p L d 2 h 8µ dh + +h= 1 2 ρgD dt ρg g dt

At Steady state,

hs =

p1s ρg

p L d 2 H 8µ dH + +H = 1 2 ρgD dt ρg g dt

p (s) 8µ L 2 s H ( s) + sH ( s ) + H ( s ) = 1 g ρgD ρg

[

]

= k1 s 2 + k 2 s + 1 H ( s ) = k 3 p1 ( s )

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H 1 (s) k3 = 2 p1 ( s ) (k1 s + k 2 s + 1)

Where

Thus

Where

k1 =

8µ 1 L ; k2 = ; k3 = ; g ρgD ρg

H 1 (s) R = 2 2 p1 ( s ) ( τ s + 2ξτ s + 1)

R =

Now b)τ =

1 8µ L ; τ 2 = ; 2ξτ = ; g ρgD ρg

L ; g

8µ 1  4 µ  L    . = c)ξ = ρgD 2τ  ρgD  g 

−1

Steady state gain Lt G ( s ) = R =

S →0

1 ; ρg

Comment : a) τ is the time period of a simple pendulum of Length L. b) ξ is inversely proportional to τ , smaller the τ ,the system will tend to move from under damped to over damped characteristics.

8.5 Design a mercury manometer that will measure pressure of upto 2 atm, and give responses that are slightly under damped with ξ = 0.7 Parameter to be decide upon :

.a) Length of column of mercury .b) diameter of tube.

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Considering hmax to be the maximum height difference to be used

p1 = ρghmax = hmax =

2 * 1.01325 * 10 5 ; 9.81 *13600

hmax =1.51 m; Assuming the separation between the tubes to be 30 cm, We get an additional length of 0.47 m; Which gives us the total length L= 1.5176.47 L = 2 M Now, ξ = 0.7 =

 4 µ  g   = 0.7     ρgD  L 

9.81 g 4 * 1.6 *10 −3 * 2 = 1.5 * 10 −7 L= D= 0.74 * 13600 * 9.81 0.7 ρg = 0.00015 4µ

As can be seen, the values yielded are not proper, with too small a diameter and too large a length. A smaller ξ value and lower measuring range of pressure might be better.

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8.6 verify that for a second order system subjected to a step response,

Y (t ) = 1 −

1− ξ 2

−

ξt τ

[

sin 1 − ξ 2

]τt + tan

−1

1− ξ 2

With ξ <1 1 1 2 2 s (τ s + 2ξτs + 1)

Y (s) =

τ 2 s 2 + 2ξτs + 1 = ( s − s1 )(s − s 2 )

where s1 =

s2 =

−ξ

ξ 2 −1 =− a + b τ

ξ 2 −1 =− a −b τ

 1   2 τ  Y (s) = s ( s − s1 )( s − s 2 ) Y (s) =

1 A B C  + +  2  τ  s ( s − s1 ) ( s − s 2 ) 

Y (s) =

1 A B C  + +  2  τ  s ( s − s1 ) ( s − s 2 ) 

A( s 2 − ( s1 + s s ) s + s1 s 2 ) + Bs( s − s 2 ) + Cs ( s − s1 ) = 1 A+ B +C = 0

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− A( s1 + s s ) − Bs 2 − Cs1 = 1 As1 s s = 1; A =

1 1 = B+C = − s1 s 2 s1 s 2

1 s1

= Bs 2 + Cs 2 = −

= Cs1 + Cs 2 =

s1 + s 2 1 1 − = s1 s 2 s1 s 2

=C=

1 1 1 1 =B=− − =− s 2 ( s 2 − s1 ) s 2 ( s 2 − s1 ) s1 s 2 s1 ( s 2 − s1 )

Y (s) =

1  1  τ 2   s1 s 2

1 1  1 1 1 . − + . .   s s1 (s 2 = s1 ) (s − s1 ) s 2 ( s 2 − s1 ) (s − s 2 ) 

 1  1 1 1 − e S1t + e s2t  2  s2 ( s 2 − s1 ) τ  s1 s2 s1 ( s2 − s1 ) 

8.6 Y (t ) =

1 = 1 τ s1 s2 2

Y (t ) = 1 −

 1 S1t 1 S2t  s e − s e  τ ( s2 − s1 )  1 2 

Y (t ) = 1 −

[

1 s2 e S1t − s1e S2t ( s2 − s1 )

Y (t ) = 1 +

τ 2 ξ −1 2

[s e 2

S1t

]

− s1e S2t

]

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Y (t ) = 1 −

τje

−

ξt τ

2 ξ −1 2

[(−a − jb)(cos bt + j sin bt) − (−a + jb)(cos bt − j sin bt]

Y (t ) = 1 −

τje

−

ξt τ

2 ξ 2 −1

Y (t ) = 1 −

− 2

[− 2 jb(b cos bt + a sin bt )]

ξt τ

[ 1−ξ −1

cos(αt ) + ξ sin(αt )

]

[ 1−ξ ] 2

α =

[

]

 1−ξ 2    ξ 

φ = tan −1  verified

8.14 From the figure in your text Y(4) for the system response is expressed

b) verify that for ξ = 1, and a step input

t  Y (t ) = 1 − 1 +  τ

 −τ e 

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Y ( s) =

Y ( s)

1 1 s τ 2 s 2 + τs + 1

A Bs + C 1 = + 2 s (τs + 1) B (τ + 1) 2 2

A(τ 2 s 2 + 2τs + 1) + Bs Cs = 1

Aτ 2 + B = 0

2 Aτ + C = 0

A=1; B = τ 2 ; C = 2τ

Y ( s) =

1 τ (τs + 1) + τ − s τ (s + 1)2

Y ( s) =

τ τ 1 − − s (τs + 1) (τs + 1) 2

Y (t ) = 1 − e

−

1 − − te τ

Y (t ) = 1 − (1 + )te

−

proved

c) for ξ > 1, prove that the step response is

Y (t ) = 1 − e

−

ξt τ

[cosh(αt ) + β sinh(αt )]

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ξ 2 −1 ξ β = 2 α = τ ξ −1

Now Y ( s ) =

1/τ 2 s ( s − B1 )( s − s2 )

Where s1 = −

ξ 2 −1 ξ + τ τ

ξ 2 −1 ξ s2 = − − τ τ from 8.6(a)

Y (t ) =

1  1 1 1 1 1 1  − +  2  τ  s1 s2 s s1 ( s2 − s1 ) ( s − s1 ) s2 ( s2 − s1 ) ( s − s2 ) 

Y (t ) = 1 −

[

1 s2 e S1t − s1e S2t ( s2 − s1 )

  − ξ − 1 − ξ 2 τ Y (t ) = 1 +  τ 2 ξ 2 − 1         e Y (t ) = 1 + − ξ e  2 ξ 2 −1   −

ξt τ

 ξ 2 −1    t τ  

 −ξt   e τ e  1

ξ 2 −1  τ

t  

]

+ ξe

−

ξt τ

 −ξ + 1−ξ 2 −  τ 

− ξ 2 − 1e

−

ξ 2 −1 t τ

 − ξt − e τ e  

− ξ 2 − 1e e

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−

ξ 2 −1 τ

ξ 2 −1 t τ

  t  

    

Y (t ) = 1 + e

−

 ξ  eαt − e −αt  − 2  ξ 2 − 1 

ξt τ

Y (t ) = 1 − e

−

ξt τ

  eαt + e −αt  −  2  

  

[cosh(αt ) + β sinh(αt )]

8.7 Verify that for a unit step-input  − πξ   (1) overshoot = exp  1−ξ 2     − 2πξ (2) Decay ratio = exp  1−ξ 2 

   

For a unit step input the response (ξ<1):  ξt  −  τ 

  1 − ξ 2  t  Sin  1 − ξ 2 + tan −1  2    τ ξ 1−ξ    (1) we have to find time t where the maxima occurs e

Y (t ) = 1 −

= dY/dt = 0  ξt  −  τ 

  1−ξ 2 dY ) ξe t Sin  1 − ξ 2 + tan −1  =   ξ dt τ 1 − ξ 2 τ   −

 ξt  −  τ 

 = tan   

  1 − ξ 2  t  = 0 Cos  1 − ξ 2 + tan −1    τ ξ      2   1−ξ 2 2 t −1  1 − ξ   = + tan 1−ξ   ξ τ ξ  

1−ξ 2t

   

= nπ

for maxima

1−ξ 2t

= 2 nπ

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πt

= t=

1−ξ 2

8.8 Verify that for X(t) =A sin ωt, for a second order system,

Y (t ) =

(1 − (ωt ) ) + (2ξτ ) 2 2

φ = − tan −1

sin(ωt + φ )

2ξωτ 1 − (ωτ ) 2

Y ( s) =

A 1 2 2 2 ( s + ω ) (τ s + 2ξτs + 1)

Y ( s) =

Aω  A1 B1 C1 D1  + + + τ 2  ( s − jω ) s + jω ( s − s1 ) ( s − s2 ) 

Now as t → ∞, Y (t ) = A11 cos ωt + B11 sin ωt

Where A11 = A1 + B1 B11 = j( A1 − B1 ) to determine A1 , B1 put s = jω ,− jω in the order

A1 =

− j j B1 = 2ω ( jω − s1 )( jω − s2 ) 2ω ( jω + s1 )( jω + s2 )

A11 =

 j  1 1 −  2ω  ( s1 + jω )( s2 + jω ) ( jω − s1 )( jω − s2 ) 

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A11 =

j  ( −ω 2 − js1ω − js2ω + s1 s2 ) − ( −ω 2 + js1ω + js2ω + s1 s2 )    2 2ω  ( s1 + ω 2 )( s22 + ω 2 ) 

    ( s1 + s2 ) ( s1 s2 − ω 2 ) 11 similarly A11 =  2 B =   2 2 2 2 2 2 2   ( s1 + ω )( s2 + ω )   ω ( s1 + ω )( s2 + ω ) 

using s1 + s2 =

= s1 + s2 =

− 2ξ

4ξ 2

τ2

−

τ2

s1 s2 =

τ2

2(2ξ 2 − 1)

τ2

  2ξ −   A ω τ A11 = 2   2 τ  1 2ω 2 4 + 2 ( 2ξ − 1) + ω τ  τ 4  − 2 Aωξ =

τ3 2

 2 1   2ξω  ω − 2  +   τ   τ  

  Aϖ  11 and B = τ   1 ω      τ 2  =

− 2 Aωξτ (1 − (ωτ ) 2 ) 2 + ( 2ξωτ ) 2

   2 2    2ξω    −ω2  +     τ     1 2  2 −ω  τ 

A(1 − (ωτ ) 2 ) (1 − (ωτ ) 2 ) 2 + ( 2ξωτ ) 2

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Thus tan φ =

And, A New =

Thus, Y (t ) =

− 2ωτξ A11 = 11 1 − (ωτ ) 2 B A

((1 − (ωτ ) 2 ) 2 + ( 2ξωυ ) 2

A ((1 − (ωτ ) 2 ) 2 + ( 2ξωυ ) 2

(using

A11 + B11 = ANew

Sin(ωt + φ )

proved

8.9) If a second- order system is over damped, it is more difficult to determine the parameters ξ & τ experimentally. One method for determining the parameters from a step response has been suggested by R.c Olderboung and H.Sartarius (The dynamics of Automatic controls,ASME,P7.8,1948),as described below. (a) Show that the unit step response for the over damped case may be written in the form. s (t ) = 1 −

r1e r2t − r2 e r1t r1 r2

Where r1 and r2 are the roots of

τ 2 s 2 + 2ξτs + 1 = 0 (b) Show that s(t) has an inflection point at ti =

ln(r2 / r1 ) (r2 − r1 )

t − ti

= s 1 (t i )

Where, s 1 (t i ) = − r1e r1t = −r2 e r2ti

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r1 ( r1 − r2 )

r  = − r1  2   r1  (d) Show that the value of step response at the inflection point is rr s (t ) 1 1 s 1 (t i ) =1 + 1 2 s 1 (t i ) and that hence 1 − 1 i = − − r1 r2 r1 r2 s (t i )

(e) on a typical sketch of a unit

step response show distances equal to

s (t i ) 1 & 1 1 s (t i ) s (t i ) (f) Relate ξ & τ to r1 & r2 1−

(a) G ( s ) =

1 τ = τ s + 2ξτs + 1)  2  2ξ  1  s +   s + 2 τ  τ 

  

τ2 ( s − r1 )( s − r2 )

1 = Y (s) =

τ2 s ( s − r1 )( s − r2 )

1 A B C = + + s ( s − r1 )( s − r2 ) s s − r1 s − r2 1 = A( s − r1 )( s − r2 ) + Bs ( s − r2 ) + cs( s − r1 ) Put s= 0 = Ar1r2 =1 ; A = τ 2 1 r1 (r1 − r2 ) 1 s = r2 = Cr2 (r2 − r1 ) = 1; C = r2 (r2 − r1 )

Put s = r1 = Br1 (r1 − r2 ) = 1; B =

Y (s) =

 1 τ 2 1 1  +  + 2  τ  s r1 (r1 − r2 )( s − r1 ) r2 (r2 − r1 )( s − r2 ) 

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Y (t ) =

 e r1t e r2t 1  2   τ + + 2  r1 (r1 − r2 ) r2 (r2 − r1 )  τ 

 1 Y (t ) =1 −  r1e r2t − r2 e r1t − r r  1 2

[

Y (t ) = 1 −

] 

φr1e r t − r2 e r t 2

(r1 − r2 )

(b) For inflection point ,

d 2s d 3s = 0 & =0 dt 2 dt 2

r r (e r2t − e r2t ) ds =− 1 2 r1 − r2 dt

r1 r2 (r2 e r2t − r1e r2t ) d 2s = − =0 r1 − r2 dt 2

= u 2 e r2t = r1e r1t =

r2 = e ( r1 −r2 ) ti r1

r  ln 2  r = ti =  1  r1 − r2

(c )

ds (t ) dt

t = ti

 r1 r2  r2  =− r1 − r2  r1 

= s ' (t i ) ]

  

r1 − r2

r −  2  r1

  

r1 r2 − r r1

   

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 r1 r2  r2  =− r1 − r2  r1 

  

 r1 (r1 − r2 )  r2  = − r1 − r2  r1 

ds (t ) dt

Also

ds (t ) dt

t = ti

=−

t = ti

  

r1 r2 − r r1

   

r1   r2 − r r1     

 r = −r1  2  r1 

ds (t ) dt

t = ti

 r2  − r1− r2  r1 2 1

r1 − r2

r1   r2 − r r1     

r1 r2 (e r2t − e r1t ) =− ( r1 − r2 )

r1 r2 − e r1t (r1 − r2 )

 r1   − 1  r2 

= −r1e r1t = − r2 e r2t

(d) s (t i ) = 1 −

r1e

− r2 e r1 − r2

r2t i

rt i1

r r  s 1 (t i )  1 − 2   r2 r1  =1+ r1 − r2

r r  s 1 (t i )  1 − 2   r2 r1  = s (t i ) = = 1 + r1 − r2

Now

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1 1  s (t i ) − 1 = = s 1 (t i )  +   r1 r2   1 1 1 − s (t i ) = = − −  ' s (t )  r1 r2  r1 + r2 =

− 2ξ

r1 + r2 =

1 r1 r2 = ;τ =

1 r1 r2

=; r1 + r2 = −2 r1r2ξ

r  1  r1 + 2  2  r2 r1   

ξ =− 

proved.

8.10 Y(0),Y(0.6),Y( ∞ ) if

Y ( s) =

1 25( s + 1) s ( s 2 + 2 s + 1)

1 1  Y ( s) =  1 +  2 2s s s  ( + + 1) 25 25

Y(s) impulse response + step response of G(s)

Where G ( s ) =

1 s 2s ( + + 1) 25 25 2

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Y (t ) =

1 1−ξ 2 τ

−

ξt τ

sin 1 − ξ 2

+ tan −1

1−ξ 2

Y(t) = 1+5.0.3e-t sin (4.899t)-1.02e-t sin(4.899t+1.369)

Y(0)= 1-1=0 Y(0.6) = 1+0.561+0.515 Y( ∞ ) =1

Comment : as we can see ,the system exhibits an inverse response by increasing from zero to more than 1 and as t tend to ∞ ,will reach the steady state value of 1.

8.11 In the system shown the dev in flow to tank 1 is an impulse of magnitude 5 . A1 = 1 ft2, A2 = A3 = 2 ft2 , R1 = 1 ft/cfm R2 = 1.5 ft/cfm . (a) Determine H1(s), H2(s), H3(s)

Transfer function for tank 1

H1 ( s) 1 = Q( s) (τ 1s + 1)

H 1 ( s) =

5 ( s + 1)

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from tank 2,

H 2 ( s) R2 1.5 = = Q( s) (τ 1 s + 1)(τ 2 s + 1) ( s + 1)(3s + 1)

for tank 3, q2 − qc = A3

dh3 dt

q3 = qc (const ) = Q2 = A3

dh3 dt

dh H2 = A3 3 dt R2

H 2 ( s ) H 3 ( s) 1 = = R2 H 2 ( s ) 3s

thus, A3 SH 3 ( s ) =

A3 SH 3 ( s ) =

H 2 ( s) H 3 ( s) 1 = = R2 H 2 ( s ) 3s

H 3 ( s) 0.5 = Q ( s ) s( s + 1)(3s + 1)

8.11© H 1 ( s ) =

5 ( s + 1)

H 1 (t ) = 5e − t H 1 (3.46) = 0.155 A H 2 ( s) 1.5 = 2 Q( s) 3s + 4 s + 1 Q (1) = 5 = H 2 ( s ) =

7.5 3s + 4 s + 1 2

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τ= 3 2ξτ = 4

ξ=

4 4 = = 1.155 2τ 2 3

from fig 8.5

ξ = 1.155 and

= ξ=

3.46 =2 3

τH 2 (t ) = 0.265 X 7.5

H 2 (t ) =

0.265 X 7.5

= 1.147

H 3 ( s) 0.5 = 2 Q( s) s(3s + 4 s + 1)

Q( s) = 5 = H 3 ( s) =

τ= 3ξ=

2.5 s (3s + 4 s + 1) 2

2 3

from fig 8.2 at

= 2, ξ = 1.155

Y(t) =0.54 H3(t) =0.54*2.5 = 1.35

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8.12 sketch the response Y(t) if

Y ( s) =

e −2 S ( s 2 + 1.2 s + 1)

Determine Y(t) for t = 0,1,5, ∞ e −2 S e −2 S 1 e −2 S (0.8) = = ( s 2 + 1.2 s + 1) ( s + 0.6) 2 + (0.8) 2 0.8 ( s + 0.6) 2 + (0.8) 2 Y (t ) = 1.25e −.6( t −2 ) sin(0.8(t − 2)) t ≥ 2

Y ( s) =

for t = 0 Y (0) = 0 t = 1, Y (1) = 0 t = 5, Y (5) = 0.14 t = ∞, Y ( ∞ ) = 0

Problem 8.13 The system shown is at steady state at t = 0, with q = 10 cfm A1 = 1ft2,A2=1.25ft2, R1= 1 ft/cfm, R2= 0.8 ft/cfm. a) If flow changes fro 10 to 11 cfm, find H2(s). b) Determine H2(1),H2(4),H2( ∞ ) c) Determine the initial levels h1(0),h2(0) in the tanks. d) obtain an expression for H1(s) for unit step change.

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Writing mass balances, q−

(h1 − h2 ) = A

dh1 ( for tan k 1) dt

At steady state q 2 −

(h1s − h2 s ) R1

= h1S − h2 S

Also for tank 2 (h1 − h2 ) − h2 = A dh2 2 R1 R2 dt At steady state

(h1s − h2 s ) 1

h2 S = h2 S = 0.8 * 10 = 8 0.8

h1S = 18 C) h1S = 18 ft h2 (0) = 8 ft

The equations in terms of deviation variables

Q − Q1 = A1

H − H2 dH 1 where Q1 = 1 dt R1

H dH 2 Q2 = 2 dt R2 H 2 ( s) R2 0.8 = = 2 2 Q( s ) τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 s + 2.8s + 1

Q1 − Q2 = A2

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H 2 ( s) =

0.8 ( Ans 8.31(a )) s ( s + 2.8s + 1) 2

Step response of a second order system

τ 2 =1=τ =1 2ξτ = 2.8; ξ =

a)t = 1 = b)t = 4 =

τ t

2.8 = 1.4 2

= 1; H 2 (t ) = 0.8(0.22) = 0.176 ft ( from fig ) = 4; H 2 (t ) = 0.8(0.78) = 0.624 ft ( from fig )

c)t → ∞ = H 2 (t → ∞) = 0.8 ft

Thus H 2 (1) = 0.176 ft H 2 (4) = 0.624 ft H 2 (∞) = 0.8 ft

8.13(d) we have

Q( s )Q1 ( s ) = A1 ( s ) H 1 ( s ) Q1 ( s ) − Q2 ( s ) = A2 ( s ) H 2 ( s ) Q( s ) − Q2 ( s ) = A1 ( s ) H 1 ( s ) + A2 ( s ) H 2 ( s )

 1  + A2 ( s ) H 2 ( s )  Q( s ) − A1 ( s ) H 1 ( s ) =   R2 

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1 + τ 2s    H 2 ( s )  =      R2  

 R (Q( s ) − A1 sH 1 ( s )   H 2 ( s ) =  2 Hτ 2 s  

We have

 H 2 ( s)  R2 R  = 2 =  2 Q( s )  τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1  Deg

R2 H 2 ( s )  R2 (Q( s ) − A1 sH 1 ( s )   =  (1 + τ 2 s ) Deg   H (s)  1 +τ 2s  = 1 − A1 s − 1  Deg Q( s)  

H ( s) Q( s)

H ( s) Q( s )

H ( s) Q( s)

1  Deg − 1 − τ 2 s  1  Hτ 2 ( s )   = 1 −  sA1  Deg  A1 s  Deg 

1  τ 1τ 2 s 2 + (τ 1 + τ 2 + A1 R2 ) s + 1 − 1 − τ 2 s    sA1  Deg 

1  τ 1τ 2 s + τ 1 + A1 R2 ) s   =  sA1  Deg 

  R2 + R1 (1 + τ 2 s )  =  2 Q( s )  τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 

H ( s)

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8.14

Y ( s) =

(2 s + 4) 2 2 s ( 4 s + 0.8s + 1) Y ( s) =

(s + 2) 4 2 s ( 4 s + 0.8s + 1)

2 1  Y ( s ) = 4 1 +  2 s  ( 4 s + 0.8s + 1) 

Y ( s) =

8 4 + 2 s ( 4 s + 0.8s + 1) ( 4 s + 0.8s + 1) 2

= (step response) + (impulse response)

Now,τ = 4 = 2 ; 2ξτ = 0.8

ξ = 0.2

also,

4 =2 2

impulse response τY(t) = 4*0.63 = 2.52 (from figure) step response = 8*1.15 = 9.2 (from figure) Y(4) = 1.26+9.2 Y(4) =10.46

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Q 9.1. Two tank heating process shown in fig. consist of two identical, well stirred tank in series. A flow of heat can enter tank2. At t = 0 , the flow rate of heat to tank2 suddenly increased according to a step function to 1000 Btu/min. and the temp of inlet Ti drops from 60oF to 52oF according to a step function. These changes in heat flow and inlet temp occurs simultaneously. (a)

Develop a block diagram that relates the outlet temp of tank2 to inlet temp of tank1 and flow rate to tank2.

(b)

Obtain an expression for T2’(s)

(c)

Determine T2(2) and T2(∞)

(d)

Sketch the response T2’(t) Vs t.

Initially Ti = T1 = T2 = 60oF and q=0 W = 250 lb/min Hold up volume of each tank = 5 ft3 Density of the fluid = 50 lb/ft3 Heat Capacity = 1 Btu/lb (oF)

q w

Solution: (a) For tank 1

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Input – output = accumulation WC(Ti – To) - WC(T1 – To) = ρ C V

dT1 -------------------------- (1) dt

At steady state WC(Tis – To) - WC(T1s – To) = 0 ------------------------------------(2) (1) – (2) gives WC(Ti – Tis) - WC(T1 – T1s) = ρ C V

dT '1 dt

dT '1 WTi - WT1 = ρ V dt ’

’

Taking Laplace transform

WTi(s) = WT1(s) + ρ V s T1(s)

T1 ( s ) 1 , where τ = ρ V / W. = Ti ( s ) 1 + τs

From tank 2

q + WC(T1 – To) - WC(T2 – To) = ρ C V

dT2 -------------------------- (3) dt

At steady state qs + WC(T1s – To) - WC(T2s – To) = 0 ------------------------------------(4) (3) – (4) gives Q ‘ + WC(T1 – T1s) - WC(T2 – T2s) = ρ C V Q ‘ + WCT1’ - WCT2’ = ρ C V

dT ' 2 dt

Taking Laplace transform

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Q (s) + WC(T1(s) - T2(s)) = ρ C V s T2(s)

T2 ( s ) =

(b)

1  Q( s)  + T1 ( s ) , where τ = ρ V / W.  1 + τs  WC 

τ = 50*5/250 = 1 min WC = 250*1 = 250 Ti(s) = -8/s and Q(s) = 1000/s

Now by using above two equations we relate T2 and Ti as below and after taking laplace transform we will get T2(t)

T2 ( s ) =

1 Q( s) 1 + Ti ( s ) 1 + τs 250 (1 + τs )2

T2 ( s ) =

4 8 − (1 + s ) (1 + s )2

1 1 1  1  1   − 8 − − T2 ( s ) = 4 − 2   s (1 + s )   s s + 1 (1 + s )  T2 (t ) = (4 + 8t )e −t − 4

(c)

T2’(2) = -1.29 T2(2) = T2’(2) + T2s = 60 – 1.29 = 58.71 oF T2’(∞) = -4 T2(∞) = T2’(∞) + T2s = 60 – 4 = 56 oF

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0.85 T2’(t)

0 0.5 t

-4

Q – 9.2. The two tank heating process shown in fig. consist of two identical , well stirred tanks in series. At steady state Ta = Tb = 60oF. At t = 0 , temp of each stream changes according to a step function Ta’(t) = 10 u(t)

Tb’(t) = 20 u(t)

(a) Develop a block diagram that relates T2’ , the deviation in the temp of tank2, to Ta’ and Tb’. (b) Obtain an expression for T2’(s) (c) Determine T2(2) W1 = W2 = 250 lb/min V1 = V2 = 10 ft3 ρ1 = ρ2 = 50 lb/ft3 C = 1 Btu/lb (oF)

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Tb W2

W3=W1+W2

Solution: (a)

For tank1

T1 ( s) 1 , where τ1= ρ V / W1. = Ta( s) 1 + τ 1 s

For tank2 W1C(T1 – To) +W2C(Tb – To) – (W1+W2)C(T2 – To)= ρ C V

dT2 ------ (1) dt

At steady state W1C(T1s – To) +W2C(Tbs – To) – (W1+W2)C(T2s – To)= 0 -----------------(2) (1) – (2) W1T1’ + W2Tb’- W3T2’ = ρ V

dT ' 2 dt

Taking L.T W1T1(s) + W2Tb(s)- W3T2(s) = ρVs T2(s)

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T2 ( s ) =

(b)

[

]

1 W1 T (S ) + W 2 T ( S ) where τ= ρ V / W3. W3 1 W3 b 1 + τs

τ1 = 50*10/250 = 2 min τ = 50*5/250 = 1 min W1/W3 = 1/2 = W2/W3 Ta(s) = 10/s and Tb(s) = 0/s

Now by using above two equations we relate T1 and Ta as below and after taking laplace transform we will get T2(t)

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1 T (s) 1 T (s) b 1 T2 ( s ) = 2 − 2 (1 + s ) (1 + s ) 1 Tb ( s ) 1 Ta ( s ) 2 T2 ( s ) = − 2 (1 + s )(1 + 2s ) (1 + s ) 5 10 T2 ( s ) = − s (1 + s )(1 + 2s ) s (1 + s ) T2 ( s ) =

15 + 20 s s (1 + s )(1 + 2s )

 15 5 20   T2 ( s ) =  − −  s s + 1 (1 + 2s )  T2 (t ) = 15 − 5e −t − 10e

(c)

−t

T2’(2) = 10.64 oF T2(2) = T2’(2) + T2s = 60 + 10.64 = 70.64 oF

Q – 9.3. Heat transfer equipment shown in fig. consist of tow tanks, one nested inside the other. Heat is transferred by convection through the wall of inner tank. 1. Hold up volume of each tank is 1 ft3 2. The cross sectional area for heat transfer is 1 ft2 3. The over all heat transfer coefficient for the flow of heat between the tanks is 10 Btu/(hr)(ft2)(oF) 4. Heat capacity of fluid in each tank is 2 Btu/(lb)(oF) 5. Density of each fluid is 50 lb/ft3 Initially the temp of feed stream to the outer tank and the contents of the outer tank are equal to 100 oF. Contents of inner tank are initially at 100 oF. the flow of heat to the inner tank (Q) changed according to a step change from 0 to 500 Btu/hr. (a) Obtain an expression for the laplace transform of the temperature of inner tank T(s). (b) Invert T(s) and obtain T for t= 0,5,10, ∞

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10 lb/hr

T1 T2

Solution: (a)

For outer tank

WC(Ti – To) + hA (T1 – T2)- WC(T2 – To) = ρ C V2

dT2 -------------------------- (1) dt

At steady state WC(Tis – To) + hA (T1s – T2s)- WC(T2s – To) = 0 ------------------------------------ (2) (1) – (2) gives

WCTi’ + hA (T1’ – T2’)- WCT2’ = ρ C V2

dT2 ' dt

Substituting numerical values 10 Ti’ + 10 ( T1’ – T2’) – 10 T2’ = 50

dT2 ' dt

Taking L.T. Ti(s) + T1(s) – 2T2(s) = 5 s T2(s) Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives

T2 ( s) 1 = T1 ( s ) 2 + 5s

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For inner tank dT1 --------------------- (3) dt

Q - hA (T1 – T2) = ρ C V1

Qs - hA (T1s – T2s) = 0 ------------------------------- (4) (3) – (4) gives Q’ - hA (T1’– T2’ ) = ρ C V1

dT1 ' dt

Taking L.T and putting numerical values Q(s) – 10 T1(s) + 10 T2(s) = 50 s T1(s) Q(s) = 500/s and

T2(s) = T1(s) / (2+ 5s)

10T1 ( s ) 500 − 10T1 ( s ) + = 50sT1 ( s ) s 2 + 5s 50 1   = T1 ( s ) 5s − + 1 s 2 + 5s   T1 ( s ) =

50(2 + 5s ) s (25s 2 + 15s + 1)

T1 ( s ) =

2( 2 + 5 s ) s s + 3.82 s + 26.18 50 50

T1 ( s ) =

100 94.71 5.29 − − 3 . 82 26 s s+ s + .18 50 50

(

)(

(

T1 (t) = 100 - 94.71 e

)

) (

− 3.82 t

- 5.29 e

− 26.18t

)

and T1 (t) = 200 - 94.71 e

− 3.82 t

− 26.18 t

For t=0,5,10 and ∞ T(0) = 100 oF T(5) = 134.975 oF T(10) = 155.856 oF T(∞) = 200 oF

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Q – 10.1. A pneumatic PI controller has an output pressure of 10 psi, when the set point and pen point are together. The set point and pen point are suddenly changed by 0.5 in (i.e. a step change in error is introduced) and the following data are obtained. Time,sec

Psig

3.5

Determine the actual gain (psig per inch displacement) and the integral time. Soln: e(s) = -0.5/s for a PI controller Y(s)/e(s) = Kc ( 1 + τI-1/s) Y(s) = -0.5Kc ( 1/s + τI-1/s2) Y(t) = -0.5Kc ( 1 + τI-1 t ) At t = 0+ y(t) = 8 Y(t) = 8 – 10 = -2 2=0.5Kc

Kc = 4 psig/in At t=20 y(t) = 7 Y(t) = 7-10 = -3 3 = 2 ( 1 + τI-1 20 )

τI = 40 sec Q-10.2. a unit-step change in error is introduced into a PID controller. If KC = 10 , τI = 1 and τD = 0.5. plot the response of the controller P(t) Soln: P(s)/e(s) = KC ( 1 + τD s+ 1/ τIs) For a step change in error

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P(s) = (10/s)(1 + 0.5 s + 1/s ) P(s) = 10/s + 5 + 10/s2 P(t) = 10 + 5 δ(t) + 10 t

10(1+t) P(t)

15 10

Q – 10.3. An ideal PD controller has the transfer function P/e = KC ( τD s + 1) An actual PD controller has the transfer function P/e = KC ( τD s + 1) / (( τD/β) s + 1) Where β is a large constant in an industrial controller If a unit-step change in error is introduced into a controller having the second transfer function, show that P(t) = KC ( 1 + A exp(-βt/ τD)) Where A is a function of β which you are to determine. For β = 5 and KC =0.5, plot P(t) Vs t/ τD. As show that β ∞, show that the unit step response approaches that for the ideal controller. Soln: P/e = KC ( τD s + 1) / (( τD/β) s + 1) For a step change, e(s) = 1/s P(s) = KC s( τD s + 1) / (( τD/β) s + 1)

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    1 τ D 1 − 1 β    = KC  +  s τ s   1+ D   β

P(t)

     τ D 1 − 1 β  − βt   e τD  = K C 1 +  τ   D   β − βt   = K C 1 + ( β − 1)e τ D   

So, A = β – 1 P(t) = 0.5 ( 1 + 4 exp(-5t/ τD))

2.5

P(t)

0.5 t/τD As β ∞ then τD/β 0 and P/e = KC ( τD s + 1) / (( τD/β) s + 1) becomes P/e = KC ( τD s + 1) that of ideal PD controller Q – 10.4. a PID controller is at steady state with an output pressure of a psig. The set point and pen point are initially together. At time t=0, the set point is moved away from

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the pen point at a rate of 0.5 in/min. the motion of the set point is in the direction of lower readings. If the knob settings are KC = 2 psig/in of pen travel τI = 1.25 min τD = 0.4 min plot output pressure Vs time

Soln: Given de/dt = -0.5 in/min s e(s) = -0.5 Y(s)/e(s) = KC ( 1 + τD s+ 1/ τIs) Y(s) = -( 1/s + 1/ τIs2 + τD ) Y(t) = -( 1 + t/1.25 + 0.4 δ(t) ) Y(t) = y(t) – 9 = - ( 1 + t/1.25 + 0.4 δ(t) ) y(t) = 8 – 0.8 t – 0.4 δ(t)

9 8 7.6

y(t)

Q – 10.5. The input (e) to a PI controller is shown in the fig. Plot the output of the controller if KC = 2 and τI = 0.5 min

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0.5

e 0

-0.5

t, min

e(t) = 0.5 ( u(t) - u(t-1) - u(t-2) + u(t-3) ) e(s) = (0.5/s) ( 1 – e-s - e-2s + e-3s ) P(s)/e(s) = KC ( 1 + (1 / τI s) ) = 2 ( 1+ 2/s ) P(s) = ( 1/s + 2/s2 ) (1 – e-s - e-2s + e-3s ) P(t)

= 1 + 2t

0≤t<1

1≤t<2

= 5 – 2t

2≤t<3

3≤t<∞

Q – 12.1. Determine the transfer function Y(s)/X(s) for the block diagrams shown. Wxpress the results in terms of Ga, Gb and Gc

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Soln. (a)

Balances at each node (1) = GaX (2) = (1) – Y = GaX – Y (3) = Gb(2) = Gb(GaX – Y) (4) = (3) + X = Gb(GaX – Y) + X Y = Gc(4) = Gc (Gb(GaX – Y) + X) = GaGbGcX – GbGcY + GcX Y Gc(GaGb + 1) = X 1 + GbGc

(b)

Balances at each node (1) = X – (4) (2) = Gb(1) = Gb( X – (4)) (5) = GcX/Ga (3) = Gc(2) = GbGc( X – (4)) (4) = (3) + (5)

--------------------------- 5

= GbGc( X – (4)) + GcX/Ga Y = Ga(4) From the fifth equation (4) = GbGcX – GbGc(4) + GcX/Ga ----------- 6 ( 4) =

(GaGbGc + Gc) X (1 + GbGc)Ga

From the sixth equation Y (GaGb + 1)Gc = X (1 + GbGc)

Q – 12.2 Find the transfer function y(s)/X(s) of the system shown

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Soln: Balance at each node (1) = X – Y

---------(a)

(2) = (1) + (3)

----------(b)

(3) = G1(2) where G1 = 1/(τ1s + 1)

----------(c)

(4) Y = G2(3) where G2 = 0.5/(τ1s/2 + 1)

----------(d)

From (d) and (c) Y = (2)G1G2 = G1G2 (X – Y + (3) )

----------(e)

Also from (b) and (c) (3) = G1((1) + (3)) (3)(1 – 1/(τ1s + 1)) = 1/(τ1s + 1) (3) τ1s = 1 (3) = 1/(τ1s ) = (X – Y) / (τ1s) Substitute this in (e) Y=

0.5

τ 1s

(τ 1 s + 1)(

 1  1 + ( X − Y ) τ s   1 + 1)

1 Y = 2 2 X τ 1 s + 2τ 1 s + 1

Q – 12.3. For the control system shown determine the transfer function C(s)/R(s)

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Soln. Balances at each node (1) = R – C

------------------(a)

(2) = 2 (1) = 2(R – C)

------------------(b)

(3) = (2) – (4) = 2(R – C) – (4)

-------------------(c)

(4) = (3)/s = (2(R – C) – (4))/s

-------------------(d)

(5) = (4) – C

-------------------(e)

C = 2(5)

-------------------(f)

Solving for (4) using (d) s (4) = 2(R – C) – (4) (4) = 2(R – C) / (s +1) Using (e) (6) = 2(R – C) / (s +1) – C  2 (R − C ) − C  C = 2  s +1 4 R = C (( s + 1) + 4 + 2( s + 1) ) C 4 = R 3s + 7

Q – 12.4. Derive the transfer function Y/X for the control system shown

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Soln. Balance at each node (1) = (5) + X

-----------------(a)

(2) = (1) – (4)

-----------------(b)

(3) = (2)/s

------------------(c)

Y = (3)/s

------------------(d)

(5) = 2 (3)

------------------(e)

(4) = 25Y

------------------(f)

From (b) (4) = (1) – (2) = (1) – s (3)

from (c)

= (1) – s2 Y

from (d)

= (5) + X - s2 Y

from (a)

= 2 (3) + X - s2 Y from (e) = 2 s Y + X - s2 Y From (f) Y = (2 s Y + X - s2 Y)/25 X = Y( 25 – 2s + s2 ) Y 1 = 2 X s − 2 s + 25

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13.1 The set point of the control system in fig P13.1 given a step change of 0.1 unit. Determine (a) The maximum value of C and the time at which it occurs. (b) the offset (c) the period of oscillation. Draw a sketch of C(t) as a function of time.

5 C ( s + 1)( 2 s + 1) = 5 R 1+ K ( s + 1)( 2 s + 1)

C 8 = 2 R 2 s + 3s + 9

0.1 s 0.8 0.8 = = 0.0889 S →0 2 s + 3s + 9 9

b) C ( ∞) = Lt

offset = 0.0111 c) K =

1 1 2 0.8 ;2ξτ = ⇒ ξ = ;τ = 3 3 9 2 2

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 πξ overshoot = exp −  1−ξ 2 

  = 0.305  

= Maximum vslue of C = 1.0305*0.0889=0.116 Maximum value of C = 0.116 0.116 =

ξt   1−ξ 2 − 1 0.8  t 1− e τ sin  1 − ξ 2 + tan −1   ξ τ 9   1−ξ 2  

τ 1−ξ 2

tan

−1

1−ξ 2

= 1.6

Time at which Cmax occurs = 1.6 (c ) Period of ociullation is T =

2πτ 1−ξ 2

= 3.166

T =3.166 Decay ratio = (overshoot)2 = 0.093

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      

13.2 The control system shown in fig P 13.2 contains three-mode controller. (a) For the closed loop, develop formulas for the natural period of oscillation τ and the damping factor ξ in terms of the parameters K, τ D , τ I and τ 1 . (b) Calculate ξ when K is 0.5 and when K is 2. (c ) Do ξ & τ approach limiting values as K increases, and if so, what are these values? (d ) Determine the offset for a unit step change in load if K is 2. (e ) Sktech the response curve (C vs t) for a unit-step change IN LOAD WHEN k is 0.5 and when K is 2. (f) In both cases of part (e) determine the max value of C and the time at which it occurs.

C = R

 1 1   k 1 + τ D s + τ 1s + 1  τ I s  1+

k  1  1 + τ D s +  τ 1s + 1  τ I s 

1 (τ 1 s + 1) C = U 1   1    1+ k 1 + τ D s +  τ 1s + 1   τ I s  

τIs =

k τ Iτ I  2  k + 1   s +  τ I τ D + τ I s + 1 k   k  

 1   k 1 + τ D s + τ I s  C  = R  1   τ 1 s + 1 + k 1 + τ D s + τ I s  

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(

)

k τ Dτ I s 2 + τ I s + 1 C = R (kτ I τ D + τ 1τ I ) s 2 + (k + 1)τ I s + k

τ2 =

τ I (kτ D + τ I )

=2×

=ξ =

=T =

;2τξ =

τ I (kτ D + τ I ) k

ξ=

(k + 1)τ I k

τI (k + 1) 2 k (kτ D + τ 1 ) 2π ×

2τπ 1− ξ

τ I (kτ D + τ 1 ) k

4k (kτ D + τ 1 ) − (k + 1) 2 τ D 2 k (kτ D + τ 1 )

4π (kτ D + τ 1 )  τ 4k  k  D  τI

  τ1  +   τ I

   − (k + 1) 2  

B) τ D = τ I =1; τ 1 .=2 For k = 0.5 ; ξ =0.75 For k = 2 ; ξ =1.5

1 = 0.671 0.5(2.5)

1 = 0.530 2×3 2

 1 1 +  τ I 2 1 (k + 1) τ I 1  k C) = ξ = = τ  2 k (kτ D + τ 1 ) 2  τ D + I  k  

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As k → ∞, ξ =

4π (kτ D + τ 1 )

 τ 4k  k  D  τI

  τ1  +   τ I

   − (k + 1) 2  

kτ I τ D + τ 1τ I k

τ=

τ = τ Iτ D + τ

1 τI = 0.3535 2 τ1

K →∞

τ Iτ 1 k

= τ I τ D =1

As k → ∞, T =

4πτ D

τ 4 D −1 τI

= 7.2552

1 τ 1s + 1 C (d) = U 1  k  1 + τ D s +  1+ τ 1s + 1  τ I s 

τIs C = U (k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k U=

1 s

τI (k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k

Lt sf ( s ) =

S →0

0 =0 k

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K=2 For a unit step change in U C (∞) = 0

Offset = 0 (e) k = 0.5, ξ =0.671 & τ =2.236 T=

2πτ 1− ξ 2

= 18.95

If k = 0.5 C 2s s ;C= 2 = 2 U 5s + 3s + 1 5s + 3s + 1 If k = 2 C 0.5s 0.5 ;C = 2 = 2 U 2 s + 1. 5 s + 1 2s + 1.5s + 1

In general C (t ) =

τ 1− ξ

The maximum occurs at t =

−

ξt τ

sin 1 − ξ 2

τ 1− ξ 2

tan −1

τ 1− ξ 2

If k = 0.5 tmax = 2.52 Cmax=0.42 If k = 2 tmax = 1.69 Cmax=0.19

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13.3 The location of a load change in a control loop may affect the system response. In the block diagram shown in fig P 13.3, a unit step chsange in load enters at either location 1 and location 2. (a) What is the frequency of the transient response when the load enters at location Z? (b) What is the offset when the load enters at 1 & when it enters at 2? (c) Sketch the transient response to a step change in U1 and to a step

change in U2.

1 U 1 = ;U 2 = 0 s  2  1  ( S ( R − C ) + U 1 ) =C   2 s + 1  2 s + 1 

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R=0

2 C 2 (2 s + 1) 2 = = 2 2 U1 4 s + 4 s + 11 5× 2 ( 2 s + 1) + 1

2 C = 2 U 1 4 s + 4 s + 11

4 1 2 2 ;2ξτ = ⇒ ξ = ;τ = 11 11 11 11

1 1 Frequency = f = = T 2π

1−ξ 2

1 2π

10 = 0.2516 2

C(∞) = 2/11 Offset = 2/11 =0.182

U1=0;U2=1/s 2   1  ⇒5× ( R − C ) + U 2  =C  2s + 1  2 s + 1 

R=0

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1 C 2s + 1 ⇒ = 10 1 U2 × +1 2s + 1 2 s + 1 C 2s + 1 = 2 U 2 4 s + 4 s + 11 C(∞) = 1/11

Offset = 1/11=0.091

1 a) if U 1 = ; frequency = 0.2516 s 1 if U 2 = ; frequency = 0.2516 s 1 b)if U 1 = ; frequency = 0.182 s 1 if U 2 = ; frequency = 0.091 s

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13.5A PD controller is used in a control system having first order process and a measurement lag as shown in Fig P13.5. (a) Find the expressions for ξ and τ for the closed –loop response. (b) If τ1 = 1 min, τm = 10 sec, find KC so that

ξ = 0.7 for the two cases: (1)

τD =0,(2) τD =3 sec,

K C (1 + τ D s ) (τ 1 s + 1) C a) = K R C (1 + τ D s ) 1+ (τ 1 s + 1)(τ m s + 1)

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K C (1 + τ D s )(1 + τ m s ) C = 2 R τ 1τ m s + (τ 1 + τ m + K Cτ D ) s + ( K C + 1)

= τ2 =

τ 1τ m KC + 1

τ 1τ m

τ=

KC + 1

ξ=

1 τ 1 + τ m + k Cτ D 2 τ 1τ m (k C + 1)

b) τ 1 = 1 min; τ m = 10s;ξ = 0.7

τD = 0 1)

1 70 ⇒ 0.7 = × 2 600(k C + 1)

kc=3.167

τ D = 3s 2) ⇒ 0.7 =

70 + 3k C 1 × 2 600(k C + 1)

k C = 5.255

k 1 c)for R = ; c(∞) = C s kC + 1

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For τ D = 0; C (∞) = 0.76; offset = 0.24 for τ D = 3s; C (∞) = 0.84; offset = 0.16 600 4.167 = 105.57 = period 1− ξ 2

2π × for τ D = 0; period =

600 6.255 = 86.17 s = period for τ D = 3s; period = 1 − ( 0.7) 2 2π

Comments:

Advantage of adding derivative mode is lesser offset lesser period 13.6The thermal system shown in fig P 13.6 is controlled by PD controller. Data ; w = 250 lb/min; ρ = 62.5 lb/ft3; V1 = 4 ft3,V2=5 ft3; V3=6ft3; C = 1 Btu/(lb)(°F) Change of 1 psi from the controller changes the flow rate of heat of by 500 Btu/min. the temperature of the inlet stream may vary. There is no lag in the measuring element. (a) Draw a block diagram of the control system with the appropriate transfer function in each block.Each transfer function should contain a numerical values of the parameters. (b) From the block diagram, determine the overall transfer function relating the temperature in tank 3 to a change in set point. (c ) Find the offset for a unit steo change in inlet temperature if the controller gain KC is 3psi/°F of temperature error and the derivative time is 0.5 min.

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WT0C + q = ρCV1 (T1 − T0 ) + WT1C WT1C = ρCV2 (T2 − T1 ) + WT2 C WT2 C = ρCV3 (T3 − T2 ) + WT3C T0 (WC + ρCV1 ) + q = T1 (W C + ρCV1 ) T1 = T0 +

q WC + ρCV1

T1= T2 = T3 T3 = T0 +

q q( s ) ⇒ T3 ( s ) = (WC + ρCV1 )s WC + ρCV1

13.6 (b) 2 T ( s) ( s + 1)(1.25s + 1)(1.5s + 1) = 2 R( s ) 1 + k C (1 + τ D s) ( s + 1)(1.25s + 1)(1.5s + 1) ' 3

k C (1 + τ D s)

T3' ( s) 2k C (1 + τ D s) = 2 R( s) ( s + 1)(1.875s + 2.75s + 1) + 2k C (1 + τ D s)

T3' ( s) 2k C (1 + τ D s) = 3 R( s) 1.875s + 4.625s 2 + (3.75 + 2k Cτ D ) s + 2k c + 1

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c) kC=3; τ D = 0.5, offset = ?,τ 0' ( s) =

1 s

T3' ( s) 1 Lt ' 3 2 s → 0 Ti ( s) 1.875s + 4.625s + (3.75 + 2k Cτ D ) s + 2k C + 1

1 2k C + 1

1 = 0.143 7

Offset =0.143 13.7 (a) For the control system shown in fig P 13.7, obtain the closed loop transfer function C/U. (b) Find the value of KC for whgich thre closed loop response has a ξ of 2.3. (c) find the offset for a unit-step change in U if KC = 4.

s +1  1 (R − C) + U  = C  KC × 0.25s + 1  s

C = = U

1 s

KC s +1 . s 0.25s + 1

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C 0.25s + 1 = 2 U 0.25s + s + K C ( s + 1) C s+4 = 2 U s + 4( K C + 1) s + 4 K C b) ξ=2.3

τ=

K +1 1 ;2ξτ = C KC 4K C K +1 1 2 × 2. 3 = C KC 4KC

KC + 1

= 2.3 KC KC=2.952 C) KC=4,U = 1/s 1 s+4 =C= × 2 s s + 20s + 16 1 4 offset = 0.25. C (∞ ) =

13.8 For control system shown in Fig 13.8 (a) C(s)/R(s) (b) C(∞) (c) Offset (d) C(0.5) (e) Whether the closed loop response is oscillatory.

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4 C s( s + 1) (a) = 4 R 1+ s( s + 1) C 4 = 2 R s +s+4 b) C(∞) =2*1=2 C(∞) =2 C) offset = 0

1 1 1 d) τ = ;2ξτ = ⇒ ξ = 2 4 4

C (t ) = 1− 2

1 1 1−   4

−

t 2

 15 t  + tan −1 15  sin   4 τ 

1  −  15  4 + tan −1 15   e 4 sin  = C (0.5) = 2 1 − 15   4   C(0.5)=0.786

.e) ξ<1, the response is oscillatory.

13.9 For the control system shown in fig P13.9,determine an expression for C(t)

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if a unit step change occurs in R. Sketch the response C(t) and compute C(2).

C = R

1 s

 1 1 + 1 +   s

C s +1 = R 2 s + 1` R=

1 s

1 s +1 −1 = + s (2 s + 1) s 2 s + 1 t

1 − C (t ) = 1 − e 2 2 C(2) = 0.816

13.10 Compare the responses to a unit-step change in a set point for the system shown in fig P13.10 for both negative feedback and positive feedback.Do this for KC of 0.5 and 1.0. compare the responses by sketching C(t).

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-ve feed back :

KC s( s + ( K C + 1))

+ve feed back

C = R

1 s +1 1 1 − KC s +1

KC s ( s + (1 − K C ))

KC ×

For KC = 0.5 , response of -ve feed back is

1 2 − 1 3 C= = 3+ s ( 2 s + 3) s 2 s + 3

− 1 1 − 1 C (t ) = − e 2 = (1 − e 2 ) 3 3 3

response of +v feed back is

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1 1 −2 = + s( 2 s + 1) s 2 s + 1

C (t ) = 1 − e

−

t 2

For KC = 1, response of -ve feed back is 1 1 − 1 C= = 2+ 2 s ( s + 2) s s + 2 1 1 C (t ) = − e −2 t 2 2 response of +ve feed back is C=

1 s2

C (t ) = t

14.1 Write the characteristics equation and construct Routh array for the control system shown . it is stable for (1) Kc= 9.5,(ii) KC =11; (iii) Kc= 12

Characteristics equation

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6 Kc =0 ( s + 1)( s + 2)( s + 3) or ( s + 1)(s + 2)(s + 3) + 6 Kc = 0

( s 2 + 6 s + 11s + (6 + 6 Kc) = 0 s 3 + 6 s 2 + 11s + (6 + 6 Kc) = 0

Routh array s3 s s

6 (6 + 6 Kc) 6(1 + Kc)

For Kc=9.5 = 10-(Kc)= 10-9.5=0.5>0 therefore stable. For Kc=11 = 10-(Kc)= 10-11=-1<0 therefore unstable For Kc=121 = 10-(Kc)= 10-12=--2<0 therefore unstable

14.2 By means of the routh test, determine the stability of the system shown when KC = 2. Characteristic equation

10   3  1 + 21 + 2 2 =0  s   2 s + 4s + 10 

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(2 s 2 + 4 s + 10) s + 2( s + 3).2.10 = 0 2s 3 + 4 s 2 + 10s + 40 s + 120 = 0 2s 3 + 4 s 2 + 50s + 120 = 0 s 3 + 2s 2 + 25s + 60 = 0 Routh Array 1 2 -10/2

25 60

The system is unstable at Kc = 2. 14.4 Prove that if one or more of the co-efficient (a0,a1,….an) of the characteristic equation are negative or zero, then there is necessarily an unstable root Characteristic equation : a 0 x n + a1 x n −1 + .................................. + a n = 0 a 0 ( x n + a1 / a 0 x n −1 + .......................a n / a 0 ) = 0

a 0 ( x − α 1 )( x − α 2 )....................( x − α n ) = 0 We have α 1 , α 2 .............................α n < 0

As we know the second co-efficient a1/a0 is sum of all the roots  n n  a1 = (−1) 2 ∑ ∑ α iα j  / 2 a0  i =1 j =1 

Therefore sum of all possible products of two roots will happen twice as α 1α 2 dividing the total by 2.

α iα j > 0 (α i < 0 α j < 0) And ∴ a 2 / a0 > 0 ⇒ a 2 > 0 Similarly

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aj a0

= (−1) j ( sum of aoll possible products of j roots)

if j = even (−1) j is 1 and the sum is > 0 so a j / a 0 > 0

if j = odd (−1) j is (−1) and the sum is < 0 so

aj a0

is again > 0

in both case a j / a 0 > 0 so a j > 0( for j = 1,.......n)

14.5 Prove that the converse statement of the problem 14.4 that an unstable root implies that one or more co-efficient will be negative or zero is untrue for all co-efficient ,n>2. Let the converse be true, always .Never if we give a counter example we can contradict. Routh array

s 3 + s 2 + 2s + 3 s3 1 2 s2 1 3 s −1 0 s0 System is unstable even when all the coefficient are greater than 0; hence a contradiction, 14.6 Deduce an expression for Routh criterion that will detect the Presence of roots with real parts greater than σ for any rectified σ >0 Characteristic equation a 0 x n + a1 x n −1 + ................................. + a n = 0

Routh criteria determines if for any root, real part > 0. Now if we replace x by X such that

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.x + σ =X Characteristic equation becomes a 0 ( X − σ ) n + a1 ( X − σ ) n −1 + ................................. + a n = 0

Hence if we apply Routh criteria, We will actually be looking for roots with real part > σ rather than >0 a 0 x n + a1 x n −1 + a 2 x n −2 ................................. + a n = 0 Routh criterion detects if any root α j is greater than zero. Is there any x = α 1 , α 2 ,...............,α j ,..........α n > 0 − − − − − (1) Now we want to detect any root

α j > −σ α j> 0 From(1)

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x = α 1 , α 2 ,............................α j ,..........................α n , > 0 implies is there any x = α1 > 0 x = α2 > 0 . . . x =αj >0 . . . x = αn > 0 add σ on both sides is there any x + σ = α1 + σ > 0 x +σ = α2 +σ > 0 . . . .x + σ = α j + σ > 0 . . . x +σ = αn +σ > 0 so, Let X = x + σ and apply Routh criteria to det ect any α j + σ > 0 or α j > −σ 14.7 Show that any complex no S1 satisfying S < 1, yields a value of

1+ s that satisfies Re(Z)>0, 1− s

Let S=x+iy,

x2 + y2 < 1

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1+ s 1â&#x2C6;&#x2019; s

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(1 + x ) + iy (1 − x) + iy (1 − x) − iy (1 − x) + iy (1 − x 2 + (1 = x + 1 − x)iy − y 2 ) 1 + x 2 − 2x + y 2 =

1 − ( x 2 + y 2 ) + 2iy 1 − 2x + (x 2 + y 2 )

Re(Z ) =

1 − (x2 + y 2 ) 1 − 2x + (x 2 + y 2 ) if Re( z ) > 0 then 1 − ( x 2 + y 2 ) > 0 and 1 − 2x + (x 2 + y 2 ) > 0 we have x 2 + y 2

x2 + y2 < 1 Ranges are − 1 < x < 1 −1 < y < 1

Po int s in the unit circle

1 − ( x 2 + y 2 ) > 0 is true therefore x 2 + y 2 < 1 Now 1 + (x 2 + y 2 ) − 2x if x = −1& y = 0 then it is 4 if x = 1 & y = 0 then it is 0 0 < (1 + ( x 2 + y 2 ) − 2 x) < 4 Re( z ) > 0 example: if s = (0.5 + i0.5) the system is unstable due to the real part   1 L−1    s − (0.5 + i 0.5) 

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  1 L−1  = e 0.5t (Cos 0.5t + Sin 0.5t )   s − (0.5 + i0.5)  14.8 For the output C to be stable, we analyze the characteristic equation of the system

1 1 × (τ 3 s + 1) = 0 τ I s (τ 1 s + 1)(τ 2 s + 1)

τ I s(τ 1 s 2τ 2 + τ 1 s + τ 2 s + 1) + τ 3 s + 1 = 0 τ I τ 1τ 2 s 3 + τ I (τ 1 + τ 2 ) s 2 + (τ I + τ 3 ) s + 1 = 0

Routh Array s3

τ I τ 1τ 2

τ I (τ 1 + τ 2 ) α

s s0

α =

τ I + τ3 1 0

τ I (τ 1 + τ 2 )(τ I + τ 3 ) − τ I τ 1τ 2 τ I (τ 1 + τ 2 )

Now (1) τ I τ 2τ 1 > 0 Since τ 1 & τ 2 τ 1 > 0;τ 2 > 0

are process time constant they are definitely +ve

(2) τ I (τ 1 + τ 2 ) > 0 (3) α > 0 ⇒ τ I (τ 1 + τ 2 )(τ I + τ 3 ) > τ I τ 1τ 2

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τ 1τ I + τ 1τ 3 + τ 2τ I + τ 2τ 3 − τ 1τ 2 > 0 τ I (τ 1 + τ 2 ) > τ 1τ 2 − τ 3 (τ 1 + τ 2 )

τI >

τ 1τ 2 −τ 3 τ1 +τ 2

also τ I > 0 14.9 In the control system shown in fig find the value of Kc for which the system is on the verge of the instability. The controller is replaced by a PD controller, for which the transfer function is Kc(1+s). if Kc = 10, determine the range for which the system is stable.

Characteristics equation

6 Kc =0 ( s + 1)( s + 2)( s + 3)

or ( s + 1)(s + 2)(s + 3) + 6 Kc = 0 ( s 2 + 3s + 2)(s + 3) + 6 Kc = 0 s 3 + 6 s 2 + 11s + (6 + 6 Kc) = 0

Routh array

1 3

3 1 + Kc

 1 + Kc  3− )  3 

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For verge of instability

 1 + Kc  ) 3=  3  Kc = 8

Characteristics equation

1 +

10(1 + kcs) =0 ( s + 1)3

s 3 + 3s 2 + s (3 + 10 Kcs) + 11 = 0 Routh Array

3 + 10τ D

s2 3 11 s 3(3 + 10τ s ) > 11 for vege 30τ S > 2

τ D > 2 / 30 14.10 (a) Write the characteristics equation for the central system shown (b) Use the routh criteria to determine if the system is stable for Kc=4 © Determine the ultimate value of Kc for which the system is unstable

(a) characteristics equation

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 s + 2  1  1 1 + kc =0    3  2s = 1  ( s + 1) ( s 2 + s )(2 s + 1) + kc( s + 2) = 0 2s 3 + 3s 2 + (1 + kc) s + 2kc = 0 s 3 + 3s 2 + 3s + (1 + kc) = 0 Kc=4Routh array

s3 2 s s

3 8 − 1/ 3

not stable 3(1 + kc) − 4kc =0 3 3 − Kc = 0; Kc = 3

For verge of instability 14.11 for the control shown, the characteristics equation is 4 3 s + 4 s + 6 s 2 + 4 s + (1 + k ) = 0 (a) determine value of k above which the system is unstable. (b) Determine the value of k for which the two of the roots are on the imaginary axis, and determine the values of these imaginary roots and remaining roots are real.

s 4 + 4 s 3 + 6 s 2 + 4s + (1 + k ) = 0

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s4 1

6(1 + k )

s3 4

s 1

1+ k 4 4 − (1 + k ) 5 1+ k

For the system to be unstable  1 + k  41 −    < 0   5  1<

1+ k 5 k>4 1+ k < 0 k < −1 k > −1

The system is stable at -1<k<4 (b) For two imaginary roots 4=

4 (1 + k ); k = 4 5

Value of complex roots 5s 2 + 5 = 0 s = ±i

s 2 + 1 s 4 + 4 s 3 + 6s 2 + 4s + 5

s 2 + 4s + 5

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s4 + 0 + s2 4s 3 + 5s 2 4s 3 + 0 + 4 s 5s 2 + 5 5s 2 + 5 0 SOLUTION:

− 4 ± 16 − 20 − 4 ± 2i = = −2 ± i 2 2

PART 2 LIST OF USEFUL BOOKS FOR PROCESS CONTROL

1. PROCESS CONTROL BY R.P VYAS, CENTRAL TECHNO PUBLICATIONS, INDIA ( WIDE VARIETY OF SOLVED PROBLEMS ARE AVAILABLE IN THIS BOOK) 2. ADVANCED CONTROL ENGINEERING BY RONALD.S .BURNS , BUTTERWORTH AND HIENEMANN. 3. PROCESS MODELLING SIMULATION AND CONTROL FOR CHEMICAL ENGINEERS, WILLIAM.L.LUYBEN, MCGRAW HILL. 4.

A MATHEMATICAL INTRODUCTION TO CONTROL THEORY BY SCHOLOMO ENGELBERG, IMPERIAL COLLEGE PRESS

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LIST OF USEFUL WEBSITES www.msubbu.com FOR BLOCK DIAGRAM REDUCTION AND OTHER CHEMICAL ENGG. LEARNING RESOURCES

Readings,Recitations,Assignments,Exams,StudyMateri als,Discussion Group,Video Lectures now study whatever u want with respect to chemical engg. http://ocw.mit.edu/OcwWeb/index.htm

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