1.INTRODUCTION •
Methanol, CH3 OH , M= 32.042 32.042 , also termed termed methyl methyl alcohol or
carbinol , is one
of the the most most important important chemical chemical raw materials materials .
discovery in late 1600s , methanol has grown to become 21 chemical with over 12 × 10
6
st
From its
largest commodity
metric tonnes annually produced in the world .
Methanol has been called wood alcohol ( or wood spirit) because it was abtained commercially commercially from destructive 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 synthesi s gas intermediate. ethanol can & is , however being produced from such alternative alternative feedstocks as as coal & residual residual fuel oil [2]
Table 1. Specifications for Pure Methanol
Property
Grade A
Grade A A
methanol content,wt %,min
99.85
99.85
acetone & aldehydes,ppm,max
30
30
acetone,ppm,max
20
ethanol,ppm,max
10
acid ( as acetic acid),ppm,max acid),ppm,max
30
30
water content,ppm,max content,ppm,max
1500
1000
specific gravity,20/20 gravity,20/20 °C
0.7928
0.7928
permanganate time,min
30
30
odor
characteristic
distillation range at 101 Kpa
1°C,must
charactristic
include color,platinum-cobalt color,platinum-cobalt scale,max
64.6°C
appearance
5
residual on evaporation,g/100 ml
clear-
carbonizable impurities
colorless 0.001 30
1°C,must include 64.4°C 5 clearcolorless 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 solvents[2].General physical properties of methanol are presented in table 2.
Table 2.Physical 2.Physi cal 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 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 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
36 o
Autoignition temperature C o
470
Flash point C
12
Surface tension , dyn/cm
22.6
Specific heat o
1.370
o
2.533
Of vapor at 25 C,J/gK Of liquid at 25 C.J/gK
-166.81
o
Vapor pressure at 25 C, Kpa
16.96
Solubility in water
miscible
Density at 25 C ,g’ ,g’cm
o
3
0.78663
Refractive Refractive index,n
1.3284
Viscosity of liquid,cP
0.541 o
Dielectric constant at 25 C
32.7 o
Thermal conductivity at 25 C ,W/mK
0.202
Chemical Properties: Methanol is the simplest aliphatic alcohol. As a typical
representative representative of this
class of substances,
its reactivity reactivity is determined determined
by
functional hydroxyl hydroxy l groups.Reactions groups.R eactions of methanol take place via cleavage of the C-O or O-H bond & are characterized characterized by the substitution of the –H or –OH group. In In contrast to higher aliphatic
alcohols however,
β -elimination with
formation of a multiple bond cannot occur. Important Important industrial reactions of methanol methanol include following a. Dehydrogenation Dehydrogenation & Oxidative dehydrogenation dehydrogenation b. Carbonylation c. Esterification with organic or inorganic acids & acid derivatives. derivatives. d. Etherification e. Addition to unsaturated bonds. f. Replacement Replacement of hydroxyl groups us es.[2] 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’ 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 : → CH3OH
CO + 2H2
∆H=-26.4
Kcal
Methanation reactions: CO + 3H2
→
2CO + 2H2
∆H=
CH4 + H2O →CH4
∆H=
+ CO2
-50 Kcal
-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
zinc,chromium,manganese, maintained at 300-375
°C
&
contains
a
mixed
oxides.The
catalyst
of
temperature
is
or
aluminium
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 & x W = 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.
Enriching Section
Reflux L
D, XD
Stripping Section
L
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 o
Assumptions:The reactant stream is heated to a temperature of 100 C by heat exchange with product stream.Reaction takes place at temperature of 340C.The temp. of product o
Gases reduces 200 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 = [mC P∆T]CO + [mCP∆T]H2 CO,CP = 30.142 kJ/kmolesK H2 , CP=29.98 kJ/kmolesK o
o
∆T = 100 –25 =75 C ; 25 C = reference temperature
then Qi = 1702472.244 kJ/hr
3
Heat generated = ∑(∆H)n = 20101 x 10 kJ/kmolesK
Heat o/p, Qo = ∑[mCP∆T] CP, kJ/kmolesK CH3OH
69.86
CO
30.83
H2
29.346
CH4
53.67
H2O
36.507
CO2
48.135
o
∆T = 340 – 200 = 140 C
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
25°C
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 o
log mean temperature difference = 31.4775 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; X D / (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 th
Feed enters at 11 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 Top
Bottom
Stripping section Top
Bottom
Liq. Kmoles/hr
102.055
102.055
240.415
240.415
Vap. Kmoles/hr
232.244
232.244
232.244
232.244
(M)liq kg/kmol
31.962
31.146
31.146
18.1404
(M)vap kg/kmol
31.962
31.538
31.538
18.1404
X
0.9973
0.939
0.939
0.01
Y
0.9973
0.967
0.967
0.01
Tliq
64.512
64.981
64.981
92.501
Tvap
64.571
67.201
67.201
98.418
Liq kg/hr
3261.88
3178.6
7487.9
4361.13
Vap kg/hr
7422.98
7324.5
7324.5
4212.9
3
453.56
484.58
484.58
978.25
3
1.1536
1.1295
1.1295
0.5928
0.02216
0.02095
0.04936
0.02548
PL kg/m Pv kg/m
(L/G (PG /PL)
0.5
Surface tension for mixture is evaluated as follows (Perry)
σmix ¼
= ψ w σw + ψ o σo ¼
¼
σw, surface tension of pure water = 71.4 dynes/cm σo, surface tension of pure methanol = 22.6 dyn/cm ψ w is defined by relation, log (ψ w) = q
log
1 - ψ w
q
(XwVw) (XwVw +XoVo)
1-q
XoVo
T 3
Vw, Molar volume of water = 18.09 cm /mol 3
Vo, molar volume of methanol = 40.98 cm /mol For methanol, q = 1 Xo = 0.9675; Xw = 0.03285 Solving,
Ψw = 7.5 x 10-3; Ψo = 0.9925
+ 0.441 q
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 d h
= 5.0 mm
Hole pitch lp
= 15 mm
Tray thickness t T
= 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 t s = 18”, Csb, flood = 0.295 ft/s Unf
= (Csb, flood) ( σ )
0.2
(ρL – ρv)
0.5
ρr
20 0.2
= (0.295) (22.827)
(484.58 – 1.1295)
20
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
=
3
7324.5
= 1.801 m /s
3600 × 1.1295 2
Net area, An = 1.801 / 1.52 = 1.1786 m Assume Lw/Dc = 0.75
θc =
2 sin
-1
Lw
Weir length
Dc
Plate diameter
(Lw /Dc) = 97.2°
Ac = 0.785 Dc
2
Downcomer area, Ad
2
= 0.785 Dc
× θc 360 2
= 0.08795 Dc
-
0.75 4
2
Dc cos 97.2 2
An
= Ac - Ad 2
1.1786= 0.69705 Dc Dc
= 1.3 m
Lw
= 0.975 m ≈ 1m
Then Lw /Dc = 0.77 2
Ac
= 1.3266 m
Ad
= 0.1486 m
2 2
Active area, Aa = Ac – 2 Ad = 1.029 m
2
Perforated area, Ap = Aa – Acz – A2z = 0.8039 m 2
Calming zone, Acz = 0.134 Ac = 0.1459 m
2
Waste peripheral zone, A wz = 0.06 Ac = 0.07959 m 2
Then Ah = 0.08039 m
No. of holes, nh = 0.08039 /
π (0.005)2 =
4094.23
4 Weir height, h 2 = 50 mm (assumption)
Weeping check :
Head loss through holes, hd = k 1 + k 2
PG Uh
2
PL 2
k 1 = 0; k 2 = 50.8 /Cv
For Ah / Aa = 0.07812 and tt / dL = 0.6, Cv = 0.74; K2 = 92.77 (Sieve trays) 3
Vol. flow of vap at top = 1.7874 m /s Uh (at top)
= 1.7874 / 0.08039 = 22.34 m/s
Uh (at bottom)
= 1.801 / 0.08039
Then hd (at top)
= 92.768 × (1.1536 / 453.56) × 22.34
= 22.4 m/s
= 146.79 mm clear liquid hd (at bot.)
= 108.49 mm (min)
2
hσ = 409 (σ /dLρL) = 4.117 mm clear liq how = Fw (664) (q/Lw) q = 18.22 1
2.5
q / Lw
×
-4
3
2/3 1
10 m /s; q = 28.882 GPM 2.5
= 28.882 / (3.2802) = 1.4814
Lw / Dc = 0.77 For these values, Fw = 1.005 how = 1.005 × 664 (18.22 × 10 / 1.0) = 9.955 mm -4
2/3
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
Segmental weir hit
hσ
Head loss due to bubble formation
hds
Dynamic seal
hhg
Hydraulic gradient
ht
Total pressure drop across plate
D.C. Flooding Check hds = hw + how + (hhg / 2) (hhg neglected)
q (max. at top) = 19.97 × 10 1
2.5
For q / Lw
-4
3
m /s = 31.667 GPM
= 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 = 1
he =
0.59 ;
φt = 0.2
β .hds = 1
hf = he /
35.78 mm
φt =
178.9 mm
had = 165.2 (q/Ada) C
2
Ada - Min. flow area under down comer.
13 to 38 mm
Let C = 1 – in = 25 mm Lap = hds - C = 35.64 mm Ada = Lw × hap = 0.03564 m
2
had = 0.5187 mm 1
ht = hd (top) + he = 146.79 + 35.78 = 165.75 mm D.C. Back-up hdc = ht + hw + how + hhg + had = 24.36 mm
Taking φdc = 0.5 ; h
1
dc
S = 243.6 / 0.5 = 487 mm
hdc < ts. NO FLOODING.
STRIPPING SECTION
ts
= 500 mm
dh
= 5.0 mm
lp
= 15 mm ∆ pitch
tT
= 3 mm
Ah / Ap = 0.1 (L/G) (f G / PL)
0.5
= 0.04936 (max. at top)
Csb , flood = 0.28 ft/s Surface tension is calculated to be 38.289 dyne / cm = Unf
= 6.596 ft/s
Un
= 0.8 Unf = 5.277 ft/s
σ mix
= 1.6084 m/s Vol. flow rate of vapour at top (max) An
2
= 1.1199 m
2
= 1.8013 m /s
Assuming a Lw / Dc = 0.75;
θc
= 97.2°C.
Dc is found to be 1.267 m. We assume D c = 1.3 m for simplification of design and to bring down tower cost.
≈
Lw
= 0.955
1m
Lw / Dc = 0.77
Ac
= 1.3266 m
2
Acz
= 0.1459 m
Ad
= 0.1486 m
2
Ah
= 0.08039 m
nh
= 4094.23
Ap
= 0.8039 m
Aa
= 1.029 m
Awz
= 0.07959 m
2
2 2
2 2
Weeping Check Ah / Aa = 0.07812 ; tT / dh = 0.6
Then Cv = 0.74 ; K2 = 92.77 3
Vol. flow of vapour at top = 1.8013 m /s 3
Vol. flow of vapour at bot. = 1.9741 m /s 2
Ah = 0.08039 m
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. -4
3
1
q (bottom) = 12.3836 x 10 m /s ; q = 19.63 GPM 1
2.5
For q / Lw
= 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 -4 3 q (max. at top) = 42.92 × 10 m /s
1
q = 680.4 GPM 1
2.5
For q / (Lw) = 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
φ = 0.21 ; β = 0.61
Fga = 1.104 1
he
= 43.048 mm
hf
= 215.34 mm
c
= 25 mm, hap = 45.57 mm
Ada
= 0.04557 m
hda
= 1.4655 mm
ht
= 151. 547 mm
2
Actual D.C. Back-up hdc = 223.58 mm 1
hdc
= hdc / φdc
φdc =
= 447.165 < ts
NO FLOODING
0.5
Column Efficiency a. Enriching Section
Calculation of EOG : h w = 50 mm Avg. vapour rate = 7373.74 kg/hr Avg. density = 1.14155 kg/m
3
2
Aa = 1.029 m ; Ua (top) = 1.737 m/s Df = (Dc + Lw)/2 = 1.15 m Avg. liquid rate = 3220.24 kg/hr 3
Avg. liquid density = 469.07 kg/m q = 19.97 × 10 m /sec -4
3
w = q/Df = 17.36 × 10 m /sec. m -4
3
0.5
Ng = (0.776 + 0.00457 hw – 0.238 UaPg + 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 p. ;
(µg)H2O = 1200 x 10 p.
-3
y1 = 0.98215
ρg =
-8
y2 = 0.01785
1.153 kg/m
3 -3
Dg = DAB =
10
T
1.75
[ 1/ µA + 1/ µB ]
P [ (Σ γ i A)
+ (Σγ iB)
0.33
Σ γ iA
= 29.9
Σ γ iB
= 9.44
DAB
= 29.38 × 10
Ng
= 0.977
NL
= KL. a. θL
KL a
= (3.875 × 10
-6
8
0.5
0.33 2+
]
2
m /s
× Dl) 0.5
(0.4 UaPg
0.5
+ 0.17)
Liquid diffusivity Calculation -8 0.5 –6 Dl = ( 7.4 × 10 (θ MB) T ) (nB γ A)
φ CH3OH
= 1.9
MB
= 32.04
T
= 337.5 K
nB
= 0.34 cp
γ A
= 14.8 cm / gmol, molar volume of water
3
De = DAB = 113.7 × 10 m /s -10
KL . a = 1.259 (sec)
θL
-1
= hL . Aa / (1000 q) 1
hL = hL = 35.78 mm Aa
2
= 1.029 m
2
q
= 19.95 × 10 m / sec
θL
= 18.43 sec
-4
∴ NL =
3
23.2115
mtop
= 0.241
mbot
= 0.35
Gm / Lm 2.27
λ t
= 0.547
λ b
= 0.7945
Nog
= [ (1/Ng) + (λ / NL) ]
EOG = 1 – e
-NOG
λ =
0.7877
-1
= 0.9436
= 0.6115
Murphee Plate Efficiency, E MV ZL = DcCos (θc /2) = 0.8597 m
θL
= 18.43 sec
DE
= 6.675 × 10
-3
×
Ua
1.44
+ 0.92 × 10 x hL – 0.00562 -4
2
= 0.01246 m /s 2
NPe, Peclet No. = Ze / DE For λ EOG = 0.4816
θL =
and
3.218 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) ] 0.5
For 80% flooding and (L/G) (P G / PL) = 0.0215,
ψ
= 0.14
Ea
= 0.6063
Eoc
= 0.5773
Nact
= Nth / Eoc
= 19.05 = 19 trays
∴ Tower height
= ts
× Nact
= 9.5 m
Stripping Section Cal. of EOG : hw = 50 mm
Avg. vapour rate = 5768.7 kg/hr 3
Avg. density = 0.7964 kg / m Ua
= 1.7507 m/s
Df
= 1.15 m
Avg. liquid rate
= 5924.15 kg/hr 3
Avg. density = 731.415 kg/m q
= 27.6518 × 10 m /s
w
= 24.045 × 10 m /sec m
Nsg
= 0.429
-4
3
-5
3
Properties of gas mixtures after evaluation are as follows : -5
Mg
= 1.29 x 10 Pas
Pg
= 0.9358 kg/m
Dg
= 32.14 × 10 m /s
∴ Ng
= 0.9563
3
-6
2
Liquid diffusivity, D e = 39.6 × 10 m /s -10
θL
= 16.03 sec
∴ NL =
16.832
mtop = 0.35 mbot
= 4.44
Gm / Lm = 0.978
2
λ t
= 0.342
λ b
= 4.32
Nog
= 0.844
λ =
2.3425
EOG = 0.5701 EMV :
ZL
= 0.8597
DE
= 0.0133 m /s
NPe
= 3.47
2
For λ EOG = 1.335 and NPe = 3.47 Emv / EOG = 1.5
∴ Emv =
0.855
EOC : 0.5
For 80% flooding and (L/G) (P G / PL) = 0.04936, Ea
= 0.8033
Eoc
= 0.859
NA
= 6.98 = 7 trays
ψ =
0.07
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 2
Assume overall h.t. coeff. Ud = 500 W/m k 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
2
= 0.05987 m /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 1 –4 passes Shell dia = 27” - 686 mm Corrected H.T. area = 138.768 m Corrected Uod
2 2
= 460.43 W/m . K
Shell side mean temp.
= 64.571 °C
Tube side mean temp.
= 32.5 °C 2
Assuming a condensing film transfer co-efficient of 1500 W/m K, (64.571 – Tw) A (1500) = (64.571 – 32.5) A (460.4) Tw
= 54.726°C, Wall temp.
Tf
= (64.571 + 54.726) / 2) = 59.6485°C, film temp.
Tube side velocity :
∆ pitch
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/ m ; µ = 0.59 cp; K = 0.163 W/mk. 3
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. Γ h
Γ h → the tube loading, condensate flow per unit length of the tube. Γ h
= 1.157 / (4.88 × 48) = 4.93 × 10 kg/ms
ρv
= 1.87 kg/m
ho
= 1296 W/m K
-6
3 2
Tube side : T = 32.5°C 3
P
= 994 kg / m
µ
= 0.724 cp
K
= 0.62 w/mk
(NRe) = DiVP / µ (NPo) = NNu
µ Cp / K
= 2.9228 = 4.87
= 0.023 (NRe)
0.8
0.4
(NPr) = 161.949
hi Do / K = 161.949 1/Uod = 1/1295.5 + 19.05 / (15.746 15.748) × 1 / (2 × 45) + 5 × 10
×
6375) + 19.03
×
10
-3
× ln
-4
Dirt factor is assumed. Uod
2
= 597.65
W/m K > Uad assumed
Pressure drop : Bell’s method, using Perry Tube side – cooling water.
(NRe) = 29228 (NPr) = 4.87 2
hi
= 6375 W/m K
∆PL
= 2fLVt
f
= 0.079 (NRe)
2
∴ ∆ PL = ∆ PE
ρt / D1 –0.25
= 0.006
6739.85 N/m 2
= 2.5 [Pt Vt / 2]
∆ Ptotal
2 2
= 2264.45 N/m
= Np [ ∆PL + ∆PE ] = 36.012 KN/m
2
Shell Side : Methanol vapours
ρvap (64.571 °C) → µvap → Let
3
1.93 kg/m
0.0135 cp
Nb, No. of baffle = 5 Ls, baffle spacing = L / (Nb + 1) = 0.81
Sm
1
1
= [ (P = Do) Ls] (Ds/P ) 2
= 0.1136 m Vs
= 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,
(19.05 /
= Ds (1 – 2 (lc /Ds) / PP = 12.47
∆P in cross flow section ∆Pc
= [ b. f K . W
b
= 2 × 10
∆Pc
= 0.469 KPa
-3
2
× Nc / ρt. Sm2 ] [µw / µb] 0.14
f K
= 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 : B = 5 × 10
∆Pw
2
= bw (2 + 0.6 Ncw) / Sm. Sw. P.
-4
Sw, area of flow through window = 150 – inch Swt, area occupied by tubes = (N /8) (1-Fc) t Ls / Ds = 0.3, Fc = 0.55 Swt
2
= 0.009798 m
∴ ∆Pw =
2
0.831 KN/m
(∆P)t = 2∆PE + (Nb – 1) ∆Pc + Nb. 2
= 8.923 KN/m
∆ Pw
2
π Do2
9. MECHANICAL DESIGN Major : Distillation Tower – 2,
Shell diameter
= 1300 mm = 51.18” = 4.265
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
1
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 Accessories
= 3 – in on column
→
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) d t (ρ /1728) 2
= 346.89 lb = 157 kg
Calculation of axial stress Assume di = d0 = 5.18 – in
Axial stress, f ap = Pd/4 (ts – c) = 1003.13 psi
Calculation of dead weight ƒ dead wt. shell = 3.4x
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 (1.5 – 1.0 ) 40 2
2
= 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 (πD / 4) 2
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
2
2
f wx
= 15.89 deff. x / d0 (ts – c)
def
= insulated tower + vapor line = (51.2 + 6) + (12+6) = 75.2 – in 2
f wx
= 2.43x
Calculation of combined stresses under operating conditions Upwind side : f t (max) = f wx + f ap - f dx 2
f t (max) = 2.43 X + 1003.13 - 17.351 X i.e.
= (12650) (0.85)
Solving, x = 67 ft Downwind side : f c (max)=f wx - f ap + f dx f c (max)
2
= 2.43 X + 17.351 X - 1003.474
From elastic stability, = 1.5 × 10 (t/r) 6
f c
= 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 t t = ts + (Ay / dy) ty Ay
→
equivalent thickness of shell, inches
→ 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
dy
= 18” (tray spacing)
ty
= 0.1875 + 1.73/18 = 0.2836 –in
But f c = 1.5 × 10 (ty tx) / r 6
12,011.8
≤ 1/3 y.p.
1/2
≤ 1/3 (40000)
2
2.43 x + 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, Wm 2 = Hy = bπ GY bo = 0.5/2 = 0.25 = b Wm2 = Hy = 1,55,470 lb Load to keep joint under operation,H p= 2bπGpm = 7474lb Load from internal pressure, H
= ( πG /4)p 2
= 36,350 lb
Total operating load, Wm 1 = H + Hp = 46,541.7 lb Wm2 > Wm1 Controlling load is 1,55,470 lb Calculation of minimum bolting area Am2 = Wm2 / f b f b = 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 ×f b
= 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
= (C – G) / 2 = 11.5
→
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 B
2
p
= 33,249.13 lb The lever arm h D = (C – B) / 2 = 2.66 in Moment MD = HD
× hD
= 88442.7 in – lb HG
= Hy – H = 1,19,120 lb
hG
= 1.5
Moment
MG = HG × hG = 1,78,680 in-lb
HT is given by H T = H – HD = 3100 lb The corresponding lever arm, h T = (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 M max = 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 f dw = 17.351
×
- 0.344
X, Actual tower length = 48”
∴ f dw = ∴
832,504 psi
The total wt. of dead loads is = f dw
× c.s.a. of tower
∑ Wx = 48 ft =
f dw × π × d × ts
= 41829.9 lb Wavg = 41829.9 / 48 = 871/46 lb/ft From table 9.3, C = 0.04 Msx
2
= 4CWX [3H – X] / H
2
= 4 × 0.04 × 4182.9 × 48 [ 3 × 60 – 48] / 60 2
2
= 5,65406.4 in-lb Corresponding stress is – f sx = Msx / π r (ts – c) 2
= 1,466.524 psi At X = 48.0 ft wind load stress is 2
Fwx = 2.43 X = 5,598 psi < f sx
∴ 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 2
Working pressure
= 0.1 N/mm
Design pressure
= 0.11 N/mm
2 2
Permissible stress for carbon steel = 95 N/mm Tube side : 2
Working pressure = 0.5 N/mm No. of tables – 688 do = 19.05 mm di = 15.748 mm Length, L = 4.88 m
2
Design pressure = 0.55 N/mm 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 ½
W = ¼ [3 + (Rc/Rk) ] = 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
5% G Mo steel
Gasket material – asbestos composition Allowable stress of bolting material = 138 MN/m
2
Determination of gasket width 2
do/di = 1.002 with y = 25.5 MN/m 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 = πG P/4 = 0.043 MN 2
Load to keep joint tight under operation, Hp = πG (2b)mp = 0.0082MN Total operating load, W m = H + Hp = 0.0512MN Load to seat gasket under botten up condition Wm2 is controlling
Wm 2 = πGB = 0.3456MN
Calculation of minimum bolting area, -3
2
Am2 = Wm2 / S = 0.3456/138 = 2.5 x 10 m 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 -4
Ab = 44 (1.54 x 10 ) -3
2
= 6.76x10 m 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
SFo = 100
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, W o = W1 + W2 + W3 W1 = πB P/ 4 2
= 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, t TS = 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 -2
tn = 5.9836 x 10 mm 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
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 10 8 ; Cost index ,year 1971=132[4 ] Cost index,year 2002=402 Fixed capital investment=3.6 x 10 8 x402/132 i.e,FCI=Rs.10.96x 10 7 This is nothing but the present cost of the plant. Using Peter & Timmerhaus
Estimation of total investment cost: a.
b.
Direct cost: A. i.Purchased equipment cost(PEC) Taking a 25% FCI,PEC=Rs. 2.74 x 10 8 ii. Installation cost Taking a 35% PEC,this is=0.822 x 10 8 iii. Instrumentation & Control cost Taking a 15% PEC,this cost=Rs.0.411x 10 8 iv Piping,installed cost Taking a 50% PEC,this is=1.37x 10 8 Rs v. Electrical & installed cost Taking a 25%PEC, this is=Rs.0.685 x 10 8 B. Building,process,auxillary Taking a 45% PEC,this is=1.233 x 10 8 Rs. C. Services- facilities & yard improvement Taking a 75%,this is =2.055 x 10 8 Rs. D. Land Taking a 6%,this is =0.1644 x 10 8 Rs. Then Total Direct cost=9.4804 x 107 Rs. Which is around 86% of FCI Indirect costs A. Engineering & supervision cost: Taking a 15% of D.C. this is=1.42206 x 10 8 Rs. B. Construction expenses & contractor’s fee; Taking a 10%,this=0.94804 x 10 8 C. Fixed Capital Investment=Direct cost+ Indirect cost i.e, new FCI=12.7273 x 10 8 Rs. D. Working capital Taking 15% TCI, WC=0.15 TCI
Total Capital Investment,TCI=FCI+WC Solving,TCI=14.975 x 10 8 WC=2.246 x 10 8
Estimation of Total Product Cost: 1. Manufacturing Cost=Direct Product cost+ Fixed charges + Plant overhead cost A. Fixed Charges: Depreciation=1.30972 x 10 8 Rs. Local taxes-taking a 2.55 of FCI, is =0.318 x 10 8 Rs. Insurance – taking a 0.7% of FCI, is =0.089 x 10 8 Rs. Rent =0.13974 x 107 Rs. Then total Fixed Charges= 1.85646 x 10 8 Rs. B. Diect Production cost: We know FC is about 10-20% of TPC.Taking a 15%,TPC=12.3764 x 10 8 Rs. Raw materials-taking a 25% of TPC, is=3.0941 x 10 8 Rs.
