methanol-production-plant-design by Carlos Ele Ge

1.INTRODUCTION •

Methanol, CH3 OH , M= 32.042 , also termed methyl alcohol or

carbinol , is one

of the most important chemical raw materials .

From its

discovery in late 1600s , methanol has grown to become 21st largest commodity chemical with over 12× 10 6 metric tonnes annually produced in the world . Methanol has been called wood alcohol ( or wood spirit) because it was abtained commercially from destructive distillation of wood for over a century [1] It is the ninth largest organic chemical produced in U .S .

& one of the largest

volume chemical produced in the world . Today it is produced mainly from the steam reforming of natural gas via a synthesis gas intermediate. ethanol can & is , however being produced from such alternative feedstocks as coal & residual fuel oil [2]

Table 1. Specifications for Pure Methanol

Property

Grade A

Grade A A

methanol content,wt%,min

99.85

acetone & aldehydes,ppm,max

acetone,ppm,max

ethanol,ppm,max

acid ( as acetic acid),ppm,max

water content,ppm,max

1500

1000

specific gravity,20/20°C

0.7928

permanganate time,min

odor

characteristic

distillation range at 101 Kpa

1°C,must

charactristic

include color,platinum-cobalt scale,max

64.6°C

appearance

residual on evaporation,g/100 ml

clear-

carbonizable impurities

1°C,must include 64.4°C 5

colorless

clear-

0.001

colorless

0.001 30

. PROPERTIES & USES Physical properties:Methanol is a clear ,colorless,highly polar liquid with a mild odor.It is miscible with water , alcohol,ether & most common organic solvents[2].General physical properties of methanol are presented in table 2.

Table 2.Physical properties of methanol[1]

Property

value

Freezing pt. °C

-97.68

Boiling pt. °C

64.7

Critical temp.°C

239.43

Critical pressure,kPa

8096

Critical volume,ml/mol

118

Critical compressibility factor z in PV=znRT

0.224

Heat of formation (liquid)at 25°C,kJ/mol -239.045 Free energy of formation of liquid at 25oC,J’g Heat of fusion,J/g

103

Heat of vaporization at b.p..J/g

1129

Heat of combustion at 25°C,J/g

22662

Flammable limits in air Lower,vol%

6.0

Upper,vol

Autoignition temperatureoC

470

Flash point oC

Surface tension , dyn/cm

22.6

Specific heat Of vapor at 25oC,J/gK

1.370

Of liquid at 25oC.J/gK

2.533

-166.81

Vapor pressure at 25oC, Kpa

16.96

Solubility in water

miscible

Density at 25oC ,g’cm3

0.78663

Refractive index,n

1.3284

Viscosity of liquid,cP

0.541

Dielectric constant at 25oC

32.7

Thermal conductivity at 25oC ,W/mK

0.202

Chemical Properties: Methanol is the simplest aliphatic alcohol. As a typical representative of this

class of substances,

its reactivity is determined

functional hydroxyl groups.Reactions of methanol take place via cleavage of the C-O or O-H bond & are characterized by the substitution of the –H or –OH group. In contrast to higher aliphatic

alcohols however,

β-elimination with

formation of a multiple bond cannot occur. Important industrial reactions of methanol include following a. Dehydrogenation & Oxidative dehydrogenation b. Carbonylation c. Esterification with organic or inorganic acids & acid derivatives. d. Etherification e. Addition to unsaturated bonds. f. Replacement of hydroxyl groups Uses: Fig. 1 gives methanol derivatives & uses.[2] Other uses: Potentially methanol can be used as a replacement for diesel fuel & gasoline or as a gasoline extender.It can also be used as a clean-burning boiler or turbine fuel to generate electricity .Methanol can also be used to make gasoline in Mobil’s MTG(methanol to gasoline) process. Another large market for methanol is its use in the production of single-cell protein(SCP).Methanol is used as o feedstock to produce olefins,as a reducing-gas source for steel mills,to remove nitrogen from sewage sludge & use in fuel cells[1]

VARIOUS COMMERCIAL PROCESSES The oldest process for the industrial production of methanol is the dry distillation of wood,but this no longer has practical

importance. Other

processes such as oxidation of hydrocarbons & production as a byproduct othe Fisher-Tropsh synthesis according to Synthol process,have no importance today. Methanol is currently produced on an industrial scale exclusively by catalytic conversion of synthesis gas. Processes are classified according to the pressure used: a. High pressure process

25-30Mpa

b. Medium pressure process

10-25Mpa

c. Low pressure process

5-10mpa

The main advantages

og low pressure process are are lower

investment & production cost,improved operational reliability,& greater flexibility in the choice of plant size. Industrial methanol production can be subdivided into 3 steps. 1. Production of synthesis gas 2. Synthesis of methanol 3. Processing of crude methanol[3]

4. PRODUCTION

TECHNOLOGY

1. Reference flow sheet Fig.2 2. Chemical reactions : CO + 2H2 → CH3OH

∆H=-26.4 Kcal

Methanation reactions: CO + 3H2 → CH4 + H2O

∆H= -50 Kcal

2CO + 2H2 →CH4 + CO2

∆H= -60.3Kcal

Side reactions to higher mol. Wt. Compounds [ignored] Methanol decomposition: 2CH3OH → CH3-O-CH3 + H2O 3. Process description: Feed gas comprising of hydrogen & carbon monoxide is compressed to 3000-5000 psi, mixed with recycle gas, 7 fed to a high pressure converter. Internal preheat is usually employed. The reactor is

copper

lined

steel

contains

mixed

catalyst

zinc,chromium,manganese, or aluminium oxides.The temperature is maintained at 300-375 °C by

proper space velocity & heat exchanger

design. The ratio of hydrogen & CO in the feed gas is 1:4 to 1:8[1,5].Excess hydrogen improves catalyst effectiveness. Exit gases from the reactor are cooled by heat exchange with reactants, then with water. Methanol condenses under full operating pressure to maximize yields-50% [3,4].The liquid methanol is depressurized, sent to a fractionator to separate ether, 7 then to another tower to minimize the water content. Reactor design: Thick walled pressure vessels are constructed of relatively cheap steel but must be lined with Cu to avoid the formation of iron carbonyl Fe(CO)5, a volatile compound which poisons the catalyst in addition to causing severe corrosion problems.

Heat exchange to control the highly exothermic reaction is accomplished by a simple design which looks like the Montecatini-Fauser design in which the combination of a circulating high pressure water & waste heat boiler acts as the principle heat control-as shown in fig.2 Catalyst fouling: Excess hydrogen over the minimum theoritical 2:1 H2/CO ratio is used to avoid fouling of catalyst with higher molecular wt. Materials which adsorbs on the catalyst under high CO partial pressures. Earlier , synthesis gas is produced by steam reforming of natural gas which contains mainly methane. CH4

+ H2O ↔ CO + 3H2

Inert gas accumalation: In carrying

high recycle loads , possibility of

accumulating inert gas is avoided by maintaining a side stream purge of about 5% on recycle gas.

MATERIAL

BALANCE

Reactor : Recycle

purge

CO/H2

Basis: 100 tonnes of product---methanol Production per hour=130.045 kmoles/hr Accounting for the 5% fraction of methanol that decomposes to dimethyl ether , total methanol that has to be produced is = 136.9 kmoles/hr.(rxn. C)

CO Balance: CO required to produce methanol is same as the methanol qty. produced.(rxn. A) But selectivity of this rxn.=87.5%[6] Selectivity = amt. of desired product / amt. of undesired product Therefore, total CO required = 136.9/.875 = 156.681 kmoles/hr out of which (156.681- 136.9) = 19.591 kmoles/hr is utilized for side rxn.s b (methanation) Now, 25% of CO is consumed in first rxn, & rest in second. CO consumed here = (4.897 + 14.693) = 19.591 kmoles/hr

Ether Balance: Ether produced by decomposition of methanol = (136.9130.045)/2 = 3.4225 kmoles/hr.

Methane Balance: Methane produced in 2 side rxn.s = (4.897 + 14.693)= 12.2435 kmoles/hr

Water Balance: Water produced = (4.897 + 6.845/2) = 8.3195 kmoles/hr

Carbon dioxide: CO2 produced = 7.3465 kmoles/hr

Hydrogen Balance: Hydrogen required = 136.92 x 2 + 4.817 x 3 + 14.693 = 303.564 kmoles/hr But synthesis gas contains CO & H2 in the ratio 1:4 [5]. Hence hydrogen present in feed gas = 626.724 kmoles/hr. Excess hydrogen in rxn. = 323.16 kmoles/hr

Conversion per pass: In one pass only about 50% of synthesis gas is converted because thermodynamic equilibrium is reached[3,4] A part of the product steam is purged & rest is recycled .Since CO is the limiting reactant , we find recycle gas amount w.r.t. CO. Unreacted CO = Purged CO + Recycled CO 0.5(CO + X) = 0.05 x 0.5 x (CO + X) + X CO converted to methanol is found to be 130.045 kmoles/hr From here we find the recycled amount of CO to be = 125.0855 kmoles/hr

Distillation column 1[Light Ends Coloumn]: Since we have ignored all other side rxn.s,only ether is distilled in the coloumn.

Since we have ignored all other side reactions,only ether is distilled in column. We assume that xD = 0.998 & xW = 0.005 Feed F = methanol + water + ether = 130.045 + 8.315 + 3.4225 = 141.7865 kmoles/hr xF = 0.02414 D = 2.256 kmoles/hr

W =138.369 kmoles/hr

Distillation column 2[Purification column] : Since we have ignored higher alcohol formation reactions, only methanol & water enter this column.We also ignore the very small amount of ether that may be present.Hence, feed to this tower is just the binary mixture of methanol & water. G

Enriching Section

Stripping Section

Reflux L

D, XD

W,XW

Feed to the tower is saturated liquid. A total condensor is used & reflux returns to the column at its bubble point. Total moles of feed = 138.36 kmoles/hr.This contains 130.045 kmoles/hr of methanol. Hence xF = 130.045/138.36 = 0.939 Specific grade A methanol is 99.85 wt%[1,3] Then xD = (99.85/32.04)/(1-0.9985)/18 + 0.9985/32) We assume a xW of 0.01 Using relations F = D + F & FxF = DxD + WxW, We obtain D,distillate(methanol) = 130.189 kmoles/hr W,residue(water) = 8.171 kmoles/hr\

HEAT BALANCE Reactor:CP data is obtained from Smith & Van Ness Assumptions:The reactant stream is heated to a temperature of 100oC by heat exchange with product stream.Reaction takes place at temperature of 340C.The temp. of product Gases reduces 200o C due to coolant & reactant stream.

Heat i/p + Heat generated = Heat o/p + Heat removed by coolant + Heat to waste heat boiler Heat i/p Qi = [mCP∆T]CO + [mCP∆T]H2 CO,CP = 30.142 kJ/kmolesK H2 , CP=29.98 kJ/kmolesK ∆T = 100 –25 =75 oC ; 25oC = reference temperature then Qi = 1702472.244 kJ/hr Heat generated = ∑(∆H)n = 20101 x 103 kJ/kmolesK Heat o/p, Qo = ∑[mCP∆T] CP, kJ/kmolesK CH3OH

69.86

30.83

29.346

CH4

53.67

H2 O

36.507

CO2

48.135

∆T = 340 – 200 = 140 oC

then Qo = 4239286.35 kJ/hr

Heat removed by coolant water,Qw = 3391429.08 kJ/hr Then heat sent to waste heat boilerQb = 14172756.8 kJ/hr Condenser of Distillation column-2

64.571°C

CH3 OH Vapour

40°C

64.571°C Water

Heat Q = [m λ]CH3OH = [mCP∆T]H2O Mass flow rate of methanol = 1.157 kg/s λ = 1737.522 kJ/kg K then Q = 2010.72 kW mass flow rate of coolant water = 32.069 kg/s log mean temperature difference = 31.4775 oC

25°C

PROCESS DESIGN-MAJOR EQUIPMENT Distillation Tower – 2. This tower has a feed, binary mixture of methanol and water.

Assumptions : 1. Feed is the saturated liquid entering at b.p of methanol. 2. All assumptions for McCabe – Thiele method hold good here. 3. No heat loss from column.

From the material balance of this tower, we obtain, D, Distillate (methanol)

= 130.189 kmoles/hr

W, Residue (water)

= 8.171 kmoles/hr

Since feed is saturated liquid, slope of q-line is ∞.Using equilibrium data for methanol from Perry. XD/ (Rm + 1) = 0.655; Rm = 0.5226

Assume reflux ratio 1.5 times the minimum, then R = 0.7839; XD/ (R +1) = 0.559 The plot gives

No. of ideal stages = 17 No. of ideal stages in tower = 16 No. of enriching section stages = 11 No. of stripping section stages = 6 Feed enters at 11th tray. L = R.D = 102.055 kmoles /hr G = L + D = 232.244 kmoles /hr L = L + qF = 240.415 kmoles /hr G = G = 232.244 kmoles /hr Table 3 :

Enriching section

Stripping section

Top

Bottom

Liq. Kmoles/hr

102.055

240.415

Vap. Kmoles/hr

232.244

(M)liq kg/kmol

31.962

31.146

18.1404

(M)vap kg/kmol

31.962

31.538

18.1404

0.9973

0.939

0.01

0.9973

0.967

0.01

Tliq

64.512

64.981

92.501

Tvap

64.571

67.201

98.418

Liq kg/hr

3261.88

3178.6

7487.9

4361.13

Vap kg/hr

7422.98

7324.5

4212.9

PL kg/m3

453.56

484.58

978.25

Pv kg/m3

1.1536

1.1295

0.5928

(L/G (PG/PL)0.5

0.02216

0.02095

0.04936

0.02548

Surface tension for mixture is evaluated as follows (Perry) σmix ¼ = ψw σw ¼ + ψo σo ¼ σw, surface tension of pure water = 71.4 dynes/cm σo, surface tension of pure methanol = 22.6 dyn/cm ψw is defined by relation, log (ψw)q =

log

1 - ψw

(XwVw)q (XwVw +XoVo) 1-q XoVo

Vw, Molar volume of water = 18.09 cm3 /mol Vo, molar volume of methanol = 40.98 cm3/mol For methanol, q = 1 Xo = 0.9675; Xw = 0.03285 Solving, Ψw = 7.5 x 10-3; Ψo = 0.9925

+ 0.441 q T

σo Vo 2/3 - σw Vw 2/3 q

Then σmix = 22.827 dyn/cm ENRICHING SECTION Tray spacing ts

Plate hydraulics = 18” = 500 mm

Hole diameter dh

= 5.0 mm

Hole pitch lp

= 15 mm ∆

Tray thickness tT

= 3 mm

Total Hole Area / Perforated area, Ah/Ap = 0.1 From table 3 it is seen – (L/G)(ρG/ρL) 0.5

= 0.02216, is maximum at top.

Using Perry : For ts = 18”, Csb, flood = 0.295 ft/s Unf

= (Csb, flood) ( σ )0.2

(ρL – ρv) 0.5 ρr

20 = (0.295) (22.827)0.2

(484.58 – 1.1295)0.5

1.1295

= 6.267 ft/s = 1.91 m/s Un

= 0.8 Unf

= 1.52 m/s

Vol flow rate of vapour at top

= 1.801 m3/s

7324.5 3600 × 1.1295

Net area, An = 1.801 / 1.52 = 1.1786 m2

Æ Weir length D Æ Plate diameter

Assume Lw/Dc = 0.75

Lw c

θc = 2 sin-1 (Lw/Dc) = 97.2° Ac = 0.785 Dc2

Downcomer area, Ad

= 0.785 Dc2 ×

θc 360

= 0.08795 Dc2

0.75

Dc2 cos 97.2

= Ac - Ad

1.1786= 0.69705 Dc2 Dc

= 1.3 m

= 0.975 m ≈ 1m

Then Lw/Dc = 0.77 Ac

= 1.3266 m2

= 0.1486 m2

Active area, Aa = Ac – 2 Ad = 1.029 m2 Perforated area, Ap = Aa – Acz – A2z = 0.8039 m2 Calming zone, Acz = 0.134 Ac = 0.1459 m2 Waste peripheral zone, Awz = 0.06 Ac = 0.07959 m2 Then Ah = 0.08039 m2 No. of holes, nh = 0.08039 / π (0.005)2 = 4094.23 4 Weir height, h2 = 50 mm (assumption)

Weeping check : Head loss through holes, hd = k1 + k2

PG Uh2 PL

k1 = 0; k2 = 50.8 /Cv2 For Ah / Aa = 0.07812 and tt / dL = 0.6, Cv = 0.74; K2 = 92.77 (Sieve trays) Vol. flow of vap at top = 1.7874 m3/s Uh (at top)

= 1.7874 / 0.08039 = 22.34 m/s

Uh (at bottom)

= 1.801 / 0.08039 = 22.4 m/s

Then hd (at top)

= 92.768 × (1.1536 / 453.56) × 22.342 = 146.79 mm clear liquid

hd (at bot.)

= 108.49 mm (min)

hσ = 409 (σ/dLρL) = 4.117 mm clear liq how = Fw (664) (q/Lw) 2/3 q = 18.22 × 10-4 m3/s; q1 = 28.882 GPM q1 / Lw2.5 = 28.882 / (3.2802)2.5 = 1.4814 Lw / Dc = 0.77

Æ For these values, F = 1.005 w

how = 1.005 × 664 (18.22 × 10-4 / 1.0)2/3 = 9.955 mm Now, (hd + hσ) (min) = 112.60 mm (hw + how ) = 59.955 These are design values For An / Aa = 0.07812, (hd + hσ) = 17 mm which is less than design value. Hence no weeping.

how hσ hds hhg ht

Æ Æ Æ Æ Æ

Segmental weir hit Head loss due to bubble formation Dynamic seal Hydraulic gradient Total pressure drop across plate

D.C. Flooding Check hds = hw + how + (hhg / 2) (hhg neglected) q (max. at top) = 19.97 × 10-4 m3/s = 31.667 GPM For q1 / Lw2.5 = 1.624 and Lw/Dc = 0.77, Fw = 1.01 how = 10.64 mm (max. at top) hds = 60.64 mm Ua (bot.) = 1.75 m/s = 5.7414 ft/s Ua (top) = 1.737 m/s = 5.699 ft/s Fg a = Uaρg ½ = 1.86

β = 0.59 ; φt = 0.2 he1 = β.hds = 35.78 mm hf = he1 / φt = 178.9 mm had = 165.2 (q/Ada)2 C

Æ 13 to 38 mm

Ada - Min. flow area under down comer.

Let C = 1 – in = 25 mm Lap = hds - C = 35.64 mm Ada = Lw × hap = 0.03564 m2 had = 0.5187 mm ht = hd (top) + he1 = 146.79 + 35.78 = 165.75 mm D.C. Back-up hdc = ht + hw + how + hhg + had = 24.36 mm Taking φdc = 0.5 ; h1dc S = 243.6 / 0.5 = 487 mm hdc < ts. NO FLOODING.

STRIPPING SECTION ts = 500 mm dh

= 5.0 mm

= 15 mm ∆ pitch

= 3 mm

Ah / Ap = 0.1 (L/G) (fG / PL)0.5

= 0.04936 (max. at top)

Csb , flood = 0.28 ft/s Surface tension is calculated to be 38.289 dyne / cm = σ mix Unf

= 6.596 ft/s

= 0.8 Unf = 5.277 ft/s = 1.6084 m/s

Vol. flow rate of vapour at top (max) An

= 1.1199 m2

= 1.8013 m2 /s

Assuming a Lw / Dc = 0.75; θc = 97.2°C. Dc is found to be 1.267 m. We assume Dc = 1.3 m for simplification of design and to bring down tower cost. Lw

= 0.955 ≈ 1m

Lw / Dc = 0.77

= 1.3266 m2

Acz

= 0.1459 m2

= 0.1486 m2

= 0.08039 m2

= 4094.23

= 0.8039 m2

= 1.029 m2

Awz

= 0.07959 m2

Weeping Check Ah / Aa = 0.07812 ; tT / dh = 0.6 Then Cv = 0.74 ; K2 = 92.77 Vol. flow of vapour at top = 1.8013 m3 /s Vol. flow of vapour at bot. = 1.9741 m3/s Ah = 0.08039 m2 Un (at top) = 22.4 m/s Uh (at bot.) = 24.56 m/s hd (at top) = 108.499 mm clear liq. (max) hd (bottom) = 33.909 mm clear liq. (min) hσ = 3.2 mm liq. q (bottom) = 12.3836 x 10-4 m3/s ; q1 = 19.63 GPM For q1 / Lw2.5 = 1.0069 and Lw / dc = 0.77, Fw = 1/02 how = 7.793 mm clear liquid (hd + hσ) min = 37.109 mm; hw + how = 57.793 mm For Ah / Aa

= 0.07812

(hd + hσ) plot = 15.5 < design value NO WEEPING

D.C. Flooding Check q (max. at top) = 42.92 × 10-4 m3/s

q1 = 680.4 GPM For q1 / (Lw)2.5 = 45.42 and Lw / Dc = 0.77 Fw = 1.75 how

= 20.57 mm liq

hds

= hw + how = 70.57 mm

Ua (top)

= 1.7507 m/s = 5.743 ft/s

Fga = 1.104

Æ φ = 0.21 ; β = 0.61

he1

= 43.048 mm

= 215.34 mm

= 25 mm, hap = 45.57 mm

Ada

= 0.04557 m2

hda

= 1.4655 mm

= 151. 547 mm

Actual D.C. Back-up = 223.58 mm hdc hdc1

= hdc / φdc

φdc = 0.5

= 447.165 < ts

NO FLOODING

Column Efficiency a. Enriching Section Calculation of EOG : hw = 50 mm Avg. vapour rate = 7373.74 kg/hr Avg. density = 1.14155 kg/m3 Aa = 1.029 m2 ; Ua (top) = 1.737 m/s Df = (Dc + Lw)/2 = 1.15 m Avg. liquid rate = 3220.24 kg/hr Avg. liquid density = 469.07 kg/m3 q = 19.97 × 10-4 m3/sec

w = q/Df = 17.36 × 10-4 m3/sec. m Ng = (0.776 + 0.00457 hw – 0.238 UaPg0.5 + 105 w) (Nscg)-0.5 Nscg = (µg/ρgDg) = 0.578, Schimdt No. Gas viscosity and diffusivity in mixture are evaluated as follows : Tvap = 65.886°C (µg)CH3OH = 0.015 × 10-3 p. ; y1 = 0.98215

(µg)H2O = 1200 x 10-8 p.

y2 = 0.01785

ρg = 1.153 kg/m3 10-3 T1.75 [ 1/µA + 1/µB ] 0.5

Dg = DAB =

P [ (Σ γi A)0.33 + (ΣγiB)0.33 ]2+ Σ γiA = 29.9 Σ γiB = 9.44 DAB

= 29.38 × 10-6 m2/s

= 0.977

= KL. a. θL

KL a

= (3.875 × 108 × Dl) 0.5 (0.4 UaPg 0.5 + 0.17)

Liquid diffusivity Calculation Dl = ( 7.4 × 10-8 (θ MB) 0.5 T ) (nB γA) –6 φ CH3OH

= 1.9

= 32.04

= 337.5 K

= 0.34 cp

γA

= 14.8 cm3 / gmol, molar volume of water

De = DAB = 113.7 × 10-10 m2/s KL . a = 1.259 (sec)-1 θL

= hL . Aa / (1000 q)

hL = hL1 = 35.78 mm Aa

= 1.029 m2

= 19.95 × 10-4 m3 / sec

θL

= 18.43 sec

∴ NL = 23.2115 mtop

= 0.241

mbot

= 0.35

Gm / Lm 2.27 λt

= 0.547

λb

= 0.7945

Nog

= [ (1/Ng) + (λ / NL) ]-1

λ = 0.7877 = 0.9436

EOG = 1 – e-NOG = 0.6115

Murphee Plate Efficiency, EMV = DcCos (θc/2) = 0.8597 m ZL θL

= 18.43 sec

= 6.675 × 10-3 × Ua1.44 + 0.92 × 10-4 x hL – 0.00562 = 0.01246 m2/s

NPe, Peclet No. = Ze2 / DE θL = 3.218 For λ EOG = 0.4816

and

NPe = 3218,

Emv / EOG = 1.1 ∴ Emv = 0.67265 Overall Column Efficiency, EOC EOC = Nth / Nact EOC

= log (1 + Ea (λ - 1)) / log λ

Ea / Emv = 1/ [ 1+Emv ( ψ / 1-4) ] For 80% flooding and (L/G) (PG / PL)0.5 = 0.0215, ψ

= 0.14

= 0.6063

Eoc

= 0.5773

Nact

= Nth / Eoc

∴ Tower height

= 19.05 = 19 trays = ts × Nact = 9.5 m

Stripping Section Cal. of EOG : hw = 50 mm Avg. vapour rate = 5768.7 kg/hr Avg. density = 0.7964 kg / m3 Ua

= 1.7507 m/s

= 1.15 m

Avg. liquid rate

= 5924.15 kg/hr

Avg. density = 731.415 kg/m3 q

= 27.6518 × 10-4 m3/s

= 24.045 × 10-5 m3/sec m

Nsg

= 0.429

Properties of gas mixtures after evaluation are as follows : Mg

= 1.29 x 10-5 Pas

= 0.9358 kg/m3

= 32.14 × 10-6 m2/s

∴ Ng = 0.9563 Liquid diffusivity, De = 39.6 × 10-10 m2/s θL

= 16.03 sec

∴ NL = 16.832 mtop = 0.35 mbot

= 4.44

Gm / Lm = 0.978

λt

= 0.342

λb

= 4.32

Nog

= 0.844

λ = 2.3425

EOG = 0.5701 EMV : ZL

= 0.8597

= 0.0133 m2/s

NPe

= 3.47

For λ EOG = 1.335 and NPe = 3.47 Emv / EOG = 1.5 ∴ Emv = 0.855 EOC : For 80% flooding and (L/G) (PG / PL)0.5 = 0.04936, ψ = 0.07 Ea

= 0.8033

Eoc

= 0.859

= 6.98 = 7 trays

Tower height = 3.5 m

8. PROCESS DESIGN -MINOR EQUIPMENT

Total condensor (horizontal) Methanol vapours leaving top of the distillation tower are condensed using cooling water on tube side. From energy balance and mass balance, we have – mass m1 of methanol vapors = 1.1586 kg/s mass m2 of cooling water

= 32.069 kg/s.

Log Mean Temperature difference is found to be 31.4775°C Assume overall h.t. coeff. Ud = 500 W/m2k H.T. area A = Q/U (LMTD) Q from heat balance, = 2010.72 kJ/s Choosing 3/4 – in O.D. 16 BWG tubes with

O.D. = 19.05 mm I.D. = 15.748 mm a

And with shell length No. of tubes Nt

= 0.05987 m2/m length

L = 16 ft = 4.88 m,

= 127.75 / (4.88 – 0.05) (0.05987) = 441.77 tubes

Let us choose TEMA L or M-type with 480 tubes of 3/4 –in O.D. on 1 - in ∆ pitch 1 –4 passes Shell dia = 27” - 686 mm Corrected H.T. area = 138.768 m2 Corrected Uod

= 460.43 W/m2 . K

Shell side mean temp.

= 64.571°C

Tube side mean temp.

= 32.5°C

Assuming a condensing film transfer co-efficient of 1500 W/m2K, (64.571 – Tw) A (1500) = (64.571 – 32.5) A (460.4) Tw

= 54.726°C, Wall temp.

= (64.571 + 54.726) / 2) = 59.6485°C, film temp.

Tube side velocity :

Flow area at = π/4 × [15.748 × 10-3]2 × 480 4 = 0.02379 m Vt

= 32.069 / (994 × 0.02379) = 1.35 m/s

Let Nv be avg. number of tubes in vertical row. ρL = 746 kg/ m3 ; µ = 0.59 cp; K = 0.163 W/mk. Tubes in central row = Db / PT Db

= Do [ Nt / K1 ] 1/n = 586 mm

K1 = 0.249 and n1 = 2.207 → Table 12.4 of Coulson and Richardson Then tubes in central row = 23 (PT = 25. mm)

and

Nv = (2/3) × (23) = 16

Film transfer coefficients : ho

ρL (ρL –ρv)

= 0.95 KL

1/3

[ Nv -1/6 ]

µL. Γh

Γh → the tube loading, condensate flow per unit length of the tube. Γh

= 1.157 / (4.88 × 48) = 4.93 × 10-6 kg/ms

ρv

= 1.87 kg/m3

= 1296 W/m2K

Tube side : T = 32.5°C P

= 994 kg / m3

= 0.724 cp

= 0.62 w/mk

(NRe) = DiVP / µ = 2.9228 (NPo) = µ Cp / K NNu

= 4.87

= 0.023 (NRe)0.8 (NPr)0.4 = 161.949

hi Do / K = 161.949 1/Uod = 1/1295.5 + 19.05 / (15.746 × 6375) + 19.03 × 10-3 × ln (19.05 / 15.748) × 1 / (2 × 45) + 5 × 10-4 Dirt factor is assumed. Uod

= 597.65

W/m2K > Uad assumed

Pressure drop : Bell’s method, using Perry Tube side – cooling water. (NRe) = 29228 (NPr) = 4.87 hi

= 6375 W/m2K

∆PL

= 2fLVt2 ρt / D1

= 0.079 (NRe) –0.25 = 0.006

∴ ∆ PL = 6739.85 N/m2 ∆ PE = 2.5 [Pt Vt2 / 2]

= 2264.45 N/m2

∆ Ptotal = Np [ ∆PL + ∆PE ] = 36.012 KN/m2 Shell Side : Methanol vapours ρvap (64.571 °C) → 1.93 kg/m3 µvap → 0.0135 cp Let

Nb, No. of baffle = 5 Ls, baffle spacing = L / (Nb + 1) = 0.81

= [ (P1 = Do) Ls] (Ds/P1) = 0.1136 m2

= m/ρsm = 7.24 m/s

NRe

= 91703

Let us consider 30% baffle cut, Lc = 0.3 × 0.686 = 0.205 m PP, pitch parallel to flow = (√ 3 / 2) × 25.4 = 22m Nc, No. of tube rows crossed in each cross – flow region,

= Ds (1 – 2 (lc/Ds) / PP = 12.47 ∆P in cross flow section ∆Pc

= [ b. fK . W2 × Nc / ρt. Sm2 ] [µw / µb] 0.14

= 2 × 10-3 fK

∆Pc

= 0.469 KPa

= 0.1

(∆P)t = 2∆P# + (Nb – 1) + Nb ∆ Pw ∆P in end zones, ∆PE = ∆Pc [ 1 + Ncw / Nc ] New number of effective tubes,

= 0.8 Lc/PP = 7.45 = 8

∆PE

= 0.796 KPa

Window Zone : ∆Pw = bw2 (2 + 0.6 Ncw) / Sm. Sw. P. B = 5 × 10-4 Sw, area of flow through window = 150 – inch2 Swt, area occupied by tubes = (Nt/8) (1-Fc) π Do2 Ls / Ds = 0.3, Fc = 0.55 Swt

= 0.009798 m2

∴ ∆Pw = 0.831 KN/m2 (∆P)t = 2∆PE + (Nb – 1) ∆Pc + Nb. ∆ Pw = 8.923 KN/m2

9. MECHANICAL DESIGN Major : Distillation Tower – 2, Shell diameter

= 1300 mm = 51.18” = 4.2651

Working pressure

= 1 atm = 14.7 lb/sq- in gauge

Design pressure

= 16.17 lb / sq- in gauge

Shell length

= 13 m = 42.6504 - in

Shell material

= SA – 283, Grade C

Shell, double welded butt joints stress relieved but not radiographed. Tray spacing

= 18” – in

Skirt height

= 3m

Corrosion allowance, C = 1/8 – in Tray support rings = 2.5 – in × 2.5 – in 3/8 – in angles Insulation

= 3 – in on column

Accessories → one caged ladder Overhead vapor line - 12 – in, outside dia Making use of Brownell and Young, calculation of minimum shell thickness. tsh

= P/(SXE – 0.4 P) +C = 0.1605 – in

But min. shell thickness must be 8 mm = 5/16 – in SA – 283, Grade C, has allowable stress. f

= 12m650 psi

E, Joint Efficiency

Selection of head – torispherical Thickness + = 0.885 pr / (fE – 0.1 p) + C = 0.16 - in Min. thickness should be 15/16 – in Black Dia = O.D. + O.D / 24 + 2 Sf + 2/3 icr

Sf → Standard st. flange = 3 – in icr → Inside corner radius = 15/16” then B.D. = 70.6 – in Weight of head

= (π/4) d2 t (ρ/1728) = 346.89 lb = 157 kg

Calculation of axial stress Assume di = d0 = 5.18 – in Axial stress, fap = Pd/4 (ts – c) = 1003.13 psi

Calculation of dead weight = 3.4x ƒ dead wt. shell ƒ dead wt. insulation

= ρins × tins / 144 (ts – c) = 4.44x

insulation – asbestos, ρins = 40 lb/cuft Weight of top head

= 346.89 lb

Weight of ladder

= 25.00 lb per ft

Weight of 12 – in schedule 30 pipe (Appendix K) = 43.8 lb per ft Weight of pipe insulation = π/4 (1.52 – 1.02) 40 = 39.3 lb per ft. Adding up all these weights, W = 346.89 + 108.1x f dead wt. attachments (not including trays) = ∑ W/πd (ts – c) = 11.506 + 3.586 x The wt. of trays plus liquid (below × = 4) is calculated as follows – n = (x/2) – 1 f dead wt. (liq + trays)

= (x/2 – 1) 25 (πD2 / 4) 12 πD (t –c)

= 5.925 x - 11.85 Adding all, fdw = 17.351 x - 0.344 Calculation of stress due to wind loads : Assume wind pressure

= 25 psi

fwx

= 15.89 deff. x2 / d02 (ts – c)

def

= insulated tower + vapor line = (51.2 + 6) + (12+6) = 75.2 – in

fwx

= 2.43x2

Calculation of combined stresses under operating conditions Upwind side : ft (max) = fwx + fap - fdx ft (max) = 2.43 X2 + 1003.13 - 17.351 X i.e.

= (12650) (0.85)

Solving, x = 67 ft Downwind side : fc (max)=fwx - fap + fdx fc (max)

= 2.43 X2 + 17.351 X - 1003.474

From elastic stability, fc

= 1.5 × 106 (t/r) = 10986.3 psi ≤ 1/3 (yield pressure = 40,000)

Solving x = 66.76 ft. If credit is taken for the stiffening effect of tray support rings, a higher allowable compressive stress will result. Therefore tt = ts + (Ay / dy) ty → equivalent thickness of shell, inches Ay → c.s.a of one circumferential stiffener, in tx = ts (since no longitudinal stiffeners are used) The tray support rings are 2 ½ × 2 ½ × 3/8 -in angles Ay

= 1.73 sq – in

= 18” (tray spacing)

= 0.1875 + 1.73/18 = 0.2836 –in

But fc = 1.5 × 106 (ty tx)1/2 / r

≤ 1/3 y.p.

12,011.8 ≤ 1/3 (40000) 2.43 x2 + 17.351 x – 12.0131

= 0

Solving, x = 66.83 ft. Stress conditions are satisfied for a tower height of 42.65 –in. We choose 6 courses of 8 –ft wide and 5/16 –in plate. So the actual tower height = 6 × 8

= 48 ft.

Design of Flanges Design pressure = 16.17 psi Flange material

- ASTM – A – 201, grade B.

Bolting material

- ASTM A – 193, grade B-7

Gasket material

- Stainless steel

Shell outside dia

= 51.18 –in

Shell thickness

= 5/16 –in

Allowable stress of flange material

= 15,000 psi

Allowable stress of bolting material

= 20,000 psi

Calculation of gasket width do/di = (Y – pm / (y – p (m+1))) ½ Y, min. design seating stress = 3700 m, gasket factor = 2.75 do/di

= 1.002

Assume di of gasket to be 53 –in Then do = 53.06 –in Min. gasket width = (53.06 – 53)/2 = 0.028 –in Therefore use a ½ -in gasket width Then mean gasket diameter G = 53.5 in Then do = 54 –in

Calculation of bolt loads Load to seat gasket, Wm2 = Hy = bπ GY bo = 0.5/2 = 0.25 = b Wm2 = Hy = 1,55,470 lb Load to keep joint under operation,Hp= 2bπGpm = 7474lb Load from internal pressure, H

= (πG2/4)p = 36,350 lb

Total operating load, Wm1 = H + Hp = 46,541.7 lb Wm2 > Wm1 Controlling load is 1,55,470 lb Calculation of minimum bolting area Am2 = Wm2 / fb fb = 20,000 psi = 7.7735 sq –in By using shape constants, K = A/B = O.D. of flange / shell dia = 1.24 A = 63.46 –in Taking 5/4 –in bolt size whose root area = 0.202 sq –in, no. of bolts is n

2y πG × gasket width Root area ×fb

= 154 bolts Ab, actual total c.s.a of bolt

= (154) (0.202) = 31.108 sq –in

Moment Computations For bolting up condition (no internal pressure) Design load is given by, W

= (Ab + Am) f /2 = 3, 8, 8815

The corresponding level arm is given by : hG

= (C – G) / 2 = 11.5

C → O.D. of gasket + 2 × dia of bolt = 56.5 in

Flange Moment : Ma

= WhG = 5,83,225.5 lb

For operating condition, W = Wm2 For HD ; HD = 0.785 B2p = 33,249.13 lb The lever arm hD = (C – B) / 2 = 2.66 in Moment MD = HD × hD = 88442.7 in – lb HG

= Hy – H = 1,19,120 lb

= 1.5

Moment

MG = HG × hG = 1,78,680 in-lb

HT is given by HT = H – HD = 3100 lb The corresponding lever arm, hT = (hD + hG) / 2 = 2.08 –in Moment

MT = HT × hT = 6448 in –lb

∴ The summation of moments for operating condition Mo

= MD + MG + MT = 2,73,570.7 in-lb

Therefore the operating moment is controlling and Mmax = 2,73,570.7 in-lb

Calculation of flange thickness : t = (y. Mmax / f.B) ½ For K = 1.24, from figure, y = 10 t = 1.63 -in = 41.5 mm

Stress due to seismic loads Total dead wt. stress fdw = 17.351 × - 0.344

X, Actual tower length = 48” ∴ fdw = 832,504 psi ∴ The total wt. of dead loads is = fdw × c.s.a. of tower ∑ Wx = 48 ft = fdw × π × d × ts = 41829.9 lb Wavg = 41829.9 / 48 = 871/46 lb/ft From table 9.3, C = 0.04 Msx

= 4CWX2 [3H – X] / H2 = 4 × 0.04 × 4182.9 × 482 [ 3 × 60 – 48] / 602 = 5,65406.4 in-lb

Corresponding stress is – fsx = Msx / π r2 (ts – c) = 1,466.524 psi At X = 48.0 ft wind load stress is Fwx = 2.43 X2 = 5,598 psi < fsx ∴ Wind loads are controlling rather than seismic.

10. MECHANICAL DESIGN-MINOR EQUIPMENT Horizontal condensor using M.V. Joshi Shell side : 1 –4 passes Material – carbon steel Corrosion allowance = 3mm Working pressure

= 0.1 N/mm2

Design pressure

= 0.11 N/mm2

Permissible stress for carbon steel = 95 N/mm2 Tube side : Working pressure = 0.5 N/mm2 No. of tables – 688 do = 19.05 mm di = 15.748 mm Length, L = 4.88 m Design pressure = 0.55 N/mm2 Shell thickness → ts

= PD / (2fJ + P) = 0.47 mm

Minimum thickness of shell must be 6.3 mm, with corrosion allowance, let ts=8mm Head thickness → Shallow dished and Torispherical head th = PRcW / 2 fJ Rc, Crown radius W, stress intensification factor Rx, Knuckle radius = 0.06Rc W = ¼ [3 + (Rc/Rk)½] = 1.77 Rc = 1, Rk = 0.06 tn = (0.11 x 686 x 1.77)/(2x 95x1) = 0.858mm

from IS 4503 – 1967 min head thickness = 10mm Baffles : No. of baffles = 5 Thickness = 6mm

IS : 4503 – 1967

Tie Rods and spaces From IS – 4503 – 1967, for shell dia = 686mm Dia of rod = 12mm No. of tie rods = 6 Flanges Flange material, IS 2004 – 1962, Class R Bolting steel

Æ 5% G M steel o

Gasket material – asbestos composition Allowable stress of bolting material = 138 MN/m2 Determination of gasket width do/di = 1.002 with y = 25.5 MN/m2 m = 2.75 Assumed gasket width = 1.6mm Let di of gasket = 7.4mm Then do =0.7054m Taking a gasket width of 0.01m = N do = 0.724 mean gasket dia, G= di + N = 0.705m Estimation of bolt loads Load due to design pressure,H =πG2 P/4 = 0.043 MN Load to keep joint tight under operation, Hp = πG (2b)mp = 0.0082MN Total operating load, Wm = H + Hp = 0.0512MN Load to seat gasket under botten up condition Wm2 is controlling

Wm2 = πGB = 0.3456MN

Calculation of minimum bolting area, Am2 = Wm2 / S = 0.3456/138 = 2.5 x 10-3m2 Total flange moment Mo = W1a1 + W2a2 + W2a3 a1 = (C-B)/2 = 0.037M a3 = (C-G)/2 = 0.0275M a2 = (A1 + A3)/2 = 0.03225m Mo = 1.8054 x 10-3 MN For bolting up condition Mg = Wa3 W = (Am2 + Ab)Sg /2 Ab = 44 (1.54 x 10-4) = 6.76x10-3 m2 W = 0.639MN Mg = 0.0176MN Mg > Mo; Mg is controlling Mg = M Calculation of flange thickness, t = (My/BS Fo )½ For K = 1.163; y = 15

Æ S = 100 Fo

Bolt circle dia, C = O.D gasket + 2 x bolt dia C = 0.724 + 2 x 0.018 = 0.76m No. of bolts = 44, bolt dia = 18mm Calculation of flange O.D A = C + bolt dia + 0.02 = 0.798m

Check of gasket width

AbSg/πGN = 42.22 <2y Hence, satisfied.

Flange moment computations For operating conditions, Wo = W1 + W2 + W3 W1 = πB2P/4 = 0.04066

B – shell dia

W2 = H-W1 = 0.00234 MN Hp, gasket load = 0.0082MN Then t = 0.062m Standard flange thickness = 50mm Tube sheet thickness tTS = FG ( 0.25P/f)½

f = 95, F = 1

= 0.0268m = 27mm. With corrosion allowances, tTS = 30mm Nozzle : shell side Select inlet x outlet dia

= 100mm

Vent

= 50mm

Drain

= 50mm

Opening for relief valve = 75mm Nozzle thickness tn = PD/ (2fJ – P) J = 1, for seamless pipe tn = 5.9836 x 10-2mm using a corrosion allowance of 3mm tn = 4mm

Tube side

Inlet and outlet dia = 100mm Nozzle thickness tn = 3.6mm With corrosion allowance, tn = 8mm

Support for the vessel – saddle Material

Æ low carbon steel

POLLUTION CONTROL AND SAFETY The first accounts of the poisonous action of ‘methylated spirits’ were published in 1855. However the number of cases of poisoning increased only after production of low – order methanol. Methanol vapour is taken up in an amount of 70 – 80% by lungs. The compound is distributed throughout body fluids and is largely oxidized to formaldehyde and then to formic acid.

This leads to

hyperacidity of blood which is ultimately responsible for methanol poisoning. Methanol has a slight irritant action on the eyes, skin and mucous membranes in humans. Chronic methanol poisoning is characterized by damage to visual and central nervous systems. The treatment of acute oral methanol poisoning should be initiated as quickly as possible with following measures. 1. Administration of ethanol – because ethanol has a greater affinity for alcohol dehydrogenase than methanol, oxidation of methanol is inhibited; production of formaldehyde and formic acid is suppressed. 2. Gastric lavage 3. Hemodialysis 4. Treatment with alkali 5. Administration of CNS stimulants 6. Drinking larger volumes of fluid 7. Eye bandage; eyes should be protected against light 8. Patient should be kept warm

Occupational Health: No special precautions need be taken when handling methanol since it is not corrosive, caustic or environmentally harmful. However, absorption through skin constitutes danger, and methanol should be prevented from coming in direct contact with skin.

Appropriate workplace hygiene measures should be adopted if methanol is handled constantly.

Rooms in which methanol is stored or handled must be

ventilated adequately. Gas testing tubes can be used to measure the concentration in air. Respirators must be worn if substantially high concentrations are present. Filter masks can be used only for escape or life saving purposes because they are exhausted very quickly. Respirators with a self contained air supply and heavy duty chemical protective clothing should be used for longer exposures to high methanol concentrations.

PLANT LAYOUT The management of equipment and facilities specified from process flow sheet considerations is a necessary requirement for accurate pre construction cost estimation or for future design involving piping, structural and electrical facilities. Careful attention to the development of plots and elevation plans will point out unusual plant requirements and therefore, give reliable information about building and site costs required for precise pre – construction cost accounting. Rational design must include arrangements such as processing areas, storage areas and handling areas in efficient coordination and with regards to such factors are given below. 1. New site development or addition to a previously developed site. 2. Future expansion 3. Economic distribution of services – water, process steams power and gas. 4. Weather condition 5. Safety consideration – possible hazards of fire, explosions and fumes 6. building code requirements 7. Waste disposal problems 8. Sensible use of floor and elevation space.

Some points to be considered in plant layout are, • Effluent plant is located at the end of the premises • Administration buildings, canteens are located near the entrance of the industry where they will not interfere with production and its is convenient to contact the people outside the industry. • Laboratory and workshops are placed in the position form where it is easy to communicate with all other departments.

• Location of services like power plant, cooling water, pump house, and switch house are done such that their usage is not hindered and they are easily accessible in case of fire. • Pipelines laid are minimal and human safety is taken into account. • Storage layout: storage facilities for raw materials and products may be located in isolated areas or in adjoining areas. Hazardous materials become a decided menace to life and should be isolated when stored. Storage tanks must be separated to facilitate suitable quantity. Process water may be drawn from river, wells or purchased from local authority. Electrical power will be needed at all sites. So plant should be located close to a cheap source of power. A competitively priced fuel must be available for steam and power generation. • Effluent disposal: Effluent disposal should be according to the Indian standards. The appropriate authorities must be consulted during the initial site survey to determine the standards that must be met. • Local community considerations: The local community must be able to provide adequate facilities for the plant personnel: schools, banks, housing and recreational cultural facilities etc... Also the plant should be located so that the local community is not harmed. The proposed plant must fit in with and be acceptable to the local community. • Availability of suitable land: Sufficient suitable land be available for the proposed plant and for future expansion. The land should be ideally flat, well drained and have suitable load bearing capacity. A full site evaluation should be made to determine the need for pining or other special foundations. It should also be available at low cost. • Political and strategic consideration: Capital grants, tax concessions and other incentives provided by governments to direct new investment to preferred locations, such as areas of high un-employment should be the overriding considerations in the site selection.

• Climate: Adverse climatic conditions at a site will increase costs. Abnormally low temperatures will require the provision of additional insulation & special heating for equipment & pipe runs. Stronger structures will be needed at locations subject to high winds or earthquakes.

COST ESTIMATION & PLANT ECONOMICS Methaonol Plant Capacity=100 TPD Cost in 1971=Rs 3.6 x 108 ; Cost index ,year 1971=132[4 ] Cost index,year 2002=402 Fixed capital investment=3.6 x 108 x402/132 i.e,FCI=Rs.10.96x 107 This is nothing but the present cost of the plant. Using Peter & Timmerhaus

Estimation of total investment cost: a.

Direct cost: A. i.Purchased equipment cost(PEC) Taking a 25% FCI,PEC=Rs. 2.74 x 108 ii. Installation cost Taking a 35% PEC,this is=0.822 x 108 iii. Instrumentation & Control cost Taking a 15% PEC,this cost=Rs.0.411x 108 iv Piping,installed cost Taking a 50% PEC,this is=1.37x 108 Rs v. Electrical & installed cost Taking a 25%PEC, this is=Rs.0.685 x 108 B. Building,process,auxillary Taking a 45% PEC,this is=1.233 x 108 Rs. C. Services- facilities & yard improvement Taking a 75%,this is =2.055 x 108 Rs. D. Land Taking a 6%,this is =0.1644 x 108 Rs. Then Total Direct cost=9.4804 x 107 Rs. Which is around 86% of FCI b. Indirect costs A. Engineering & supervision cost: Taking a 15% of D.C. this is=1.42206 x 108 Rs. B. Construction expenses & contractor’s fee; Taking a 10%,this=0.94804 x 108 C. Fixed Capital Investment=Direct cost+ Indirect cost i.e, new FCI=12.7273 x 108 Rs. D. Working capital Taking 15% TCI, WC=0.15 TCI Total Capital Investment,TCI=FCI+WC Solving,TCI=14.975 x 108 WC=2.246 x 108

Estimation of Total Product Cost: 1. Manufacturing Cost=Direct Product cost+ Fixed charges + Plant overhead cost A. Fixed Charges: Depreciation=1.30972 x 108 Rs. Local taxes-taking a 2.55 of FCI, is =0.318 x 108 Rs. Insurance – taking a 0.7% of FCI, is =0.089 x 108 Rs. Rent =0.13974 x 107 Rs. Then total Fixed Charges= 1.85646 x 108 Rs. B. Diect Production cost: We know FC is about 10-20% of TPC.Taking a 15%,TPC=12.3764 x 108 Rs. Raw materials-taking a 25% of TPC, is=3.0941 x 108 Rs.

Operating lobour- taking a 15% of TPC, is = 1.856 x 108 Direct supervisory & clerical labour-taking a 15% ,is = 0.278 x 108 Rs. Utilities –taking a 15% of TPC,is=Rs.1.856 x 108 Maintenance & repairs-taking a 5%,is = 0.6364 X 108 Operating supplies-taking a 15%,is = 0.0955 x 108 Laboratory charges-taking a 15% of OL,is=0.2784 x 108 Rs. Total Direct Production cost=8.0944 x 108 Rs. Which is about 60.54% of TPC C. Plant Overhead Costs: Taking a 8% of TPC, this is=0.99 x 108 Rs.. Therefore total manufacturing cost=10.9409 x 108 Rs. 2.

General Expenses = Administrative cost+ Distribution & selling cost + research development cost Administrative cost-taking a 4% of TPC,is= 0.495 x 108 Distribution & Selling cost- taking a 12% of TPC,is = 1.485 x 108 Rs. Research & Development cost- a 5% of TPC= 0.61882 x 108 Rs. Then General E xpenses=2.599 x 108 Rs.

3. Total Product Cost = Manufacturing cost + General Expenses = 13.5399 x 108 Rs. This is greater than the assumed value(12.3764 x 108)

PROFIT ANALYSIS: A. Earnings: Total gross earnings= Total income – TPC = 1.2 x 109 Rs. Calculation of total income: Annual working days=340 Annual methanol prodution =100 tonnes Wholesale price of methanol = Rs.85/kg Annual total income= 2.55 x 109 Rs. B. Annual Rate of Return=gross profit/TCI =48.4 % Gross profit = gross earnings – tax Tax = 30% of gross earnings C. Pay-back period = 100/R.O.R = 2 years