Where Kv=0.05
Mass flow rate(m)=ρus ⇒s=(П/4)×D2
=m/ρu Top: (u1)=K1*(( ρL- ρv1)/ ρv1)0.5=1.905m/sec=6858 m/hr S1=V1/ (u1* ρ v1)= 797.08/(6858*0.0323)=3.6 m2. Bottom: (u2)=K2*(( ρL- ρv2)/ ρv2)0.5=1.027 m/sec=3697.2 m/hr S2=V2/ (u2* ρ v2)= 170.426/(3697.2*0.026)=1.772 m2
So, higher cross sectional area S1 is used for further calculations. 10.4.12.Diameter of the column
Enriching section [4]: ρ v1=0.703 kg/m3
=0.0323 kmol/m3
Vapor rate Q=V1/molar density=797.08/(0.0323*3600)=6.855 m3/s. ρL2=1021.45 kg/m3
Liquid rate q=327.938 kmol/hr*92.75/1021.45=8.3*10-3 m3/s Net tower cross sectional area An=3.6 m2
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Tentatively choose a weir length W=0.7T. Table 6.1[4]:the tray area used by one downspout=8.8%
At=3.6/(1-0.088)=3.947 m2
Tower diameter T=(4*At/π)0.5=(4*3.947/π)0.5=2.242 m, say 2.24 m Corrected At = πT2/4 = π2.242/4 = 3.94 m2.
W = 0.7*T = 0.7*2.24 = 1.568 m
Downspout cross sectional area Ad=0.088*3.94=0.3467 m2
For the design area taken by tray support + disengaging and distributing zones [4]=0.222m2. Active perforated area Aa=At-2*Ad-0.222=3.94-2*0.3467-0.222=3.0246 m2
q/W=6.26*10-3 m3/m.s is ok. Weir crest h1 and weir height hw:
Let h1=25mm=0.025 m.h1/T=0.025/2.24=0.011 T/W=1/0.7=1.429 Eq:6.34[4]:Weff/W=(1.4292-((1.4292-1)0.5+2*0.011*1.429)2)0.5=0.967 Eq.6.33 [4]:h1=0.666(6.26*10-3*0.967)2/3=0.0207m So take h1=0.0207m, h1/T=0.0207/2.24=0.0092 T/W=1/0.7=1.429 Eq:6.34[4]:Weff/W=(1.4292-((1.4292-1)0.5+2*0.0092*1.429)2)0.5=0.972 Eq.6.33 [4]:h1=0.666(6.26*10-3*0.972)2/3=0.0206m.is ok So, final h1=0.0207m.
Set weir height haw=50mm=0.05m. Dry pressure drop hD:
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A0=0.0555*Aa=0.0555*3.0246=0.1679 m2. V0=Q/A0=6.855/0.1679=40.83 m/s.
Viscosity of gas is taken as µG=0.012cP=1.25*10-5 kg/m.s
Hole Reynolds number [4] =d0*V0* ρ v1/ µG=0.0045*40.83*0.703/1.25*10-5=10333.5 So, fanning friction factor for NRe=10333.5 is f=0.0075,g=9.807 m2/s,l=.002l.Eq.6.36, hD=0.1096
Hydraulic head hL:
Va=Q/Aa=6.855/3.0246=2.266 m/s. Z=(T+W)/2=(2.24+1.568)/2=1.904 m Eq 6.38:hL=0.0209 m
Residual pressure drop hR Eq.6.42 [4]:hR=(6*0.04*1)/(1021.42*0.0045*9.807)=0.00532 m Total gas pressure drop hG:
Eq.6.35 [4]:hG=hD+hL+hR=0.1096+0.0209+0.00532=0.13582 m Pressure loss at liquid entrance h2:
The area for liquid flow under apron =0.025*W=0.025*1.568=0.0392 m2. Since this is smaller than Ad,Ada=0.0392 m2.
Eq. 6.43[4]:h2= (q/Ada) 2*1.5/g= (0.00827/0.0392)2*1.5/9.807=0.00681 m. Backup in downspout: Eq.6.44 [4]; h3=hG+h2=0.13582+0.00681=0.14263 m. Check on flooding: hW+h1+h3=0.05+0.025+0.14263=0.21763 m
For tower diameter T between 1 and 3 m, tray spacing t=0.60m. hW+h1+h3=0.21763, which is well below t/2=0.30m.
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Weeping velocity: for W/T=0.7, the weir is set 0.3296T=0.738 m from the centre of the tower. Therefore Z=2*0.738=1.476 m.
All other quantities in Eq 6.46 have been evaluated and then the equation yields VOW=7.47 m/s.
The tray will not weep excessively until the gas velocity through the holes V0 is reduced ti close to this value.
Entrainment:
ρv1=0.703 kg/m3=0.0323 kmol/m3
vapour rate Q=V1/molar density=797.08/(0.0323*3600)=6.855 m3/s. ρL2=1021.45 kg/m3
Liquid rate q=327.938 kmol/hr*92.75/1021.45=8.3*10-3 m3/s L’=q* ρL=0.0083*1021.45=8.478 kg/s
G’=Q* ρ v1=6.855*0.703=4.819 kg/s V/VF=0.8, (ρ v1/ ρL)0.5*L’/G’=0.046
Fig 6.17: E=0.07.The recycling of liquid resulting from such entrainment is too small to influence the tray hydraulics appreciably.
Stripping Section: ρ v2=PM/RT=2.4175 kg/m3=0.026 kmol/m3 M2=∑x*M=(93.1262*.995+18.0152*.005)=92.75 V2=797.08-626.654=170.426 kmol/hr. Q=170.126/(3600*0.026)=1.82 m3/s ρL2=1021.45 kg/m3
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Net tower cross sectional area An=1.772m2
Tentatively choose a weir length W=0.7T. Table 6.1[4]:the tray area used by one downspout=8.8%
At=1.772/(1-0.088)=1.943 m2
Tower diameter T=(4*At/π) 0.5=(4*1.943/π)0.5=1.573 m, say 1.57 m Corrected At=πT2/4= π1.572/4=1.936 m2.
W=0.7*T=0.7*1.57=1.099 m
Downspout cross sectional area Ad=0.088*1.936=0.1704 m2
For the design area taken by tray support + disengaging and distributing zones=0.222m2. Active perforated area [4] Aa=At-2*Ad-0.222=1.936-2*0.1704-0.222=1.3732 m2
q/W=7.55*10-3 m3/m.s is ok. Weir crest h1 and weir height hw:
Let h1=25mm=0.025 m.h1/T=0.025/1.57=0.016 T/W=1/0.7=1.429 Eq:6.34[4]:Weff/W=(1.4292-((1.4292-1)0.5+2*0.011*1.429)2)0.5=0.951 Eq.6.33 [4]:h1=0.666(7.55*10-3*0.951)2/3=0.0265m So take h1=0.0265 m, h1/T=0.0265/1.57=0.0169 T/W=1/0.7=1.429 Eq:6.34[4]:Weff/W=(1.4292-((1.4292-1)0.5+2*0.0169*1.429)2)0.5=0.948 Eq.6.33 [4]:h1=0.666(7.55*10-3*0.972)2/3=0.0266m.is ok So, final h1=0.0265m.
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Dry pressure drop hD:
Eq.6.37 [4]c0=1.09*(d0/l)0.25=1.09*(0.0045/0.002)0.25=1.335 A0=0.0555*Aa=0.0555*1.3732=0.0762 m2.
V0=Q/A0=1.82/0.0762=23.88 m/s.
Viscosity of gas is taken as µG=0.012cP=1.25*10-5 kg/m.s
Hole Reynolds number=d0*V0* ρ v1/ µG=0.0045*23.88 *2.4175 /1.25*10-5=18473.57 So, fanning friction factor for NRe=10333.5 is f=0.0065,g=9.807 m2/s,l=.002l.Eq.6.36[4], hD= 0.133 m
Hydraulic head hL:
Va=Q/Aa=1.82/1.3732=1.333 m/s. Z= (T+W)/2= (2.24+1.568)/2=1.3345 m Eq.6.38:hL=0.0253 m
Residual pressure drop hR Eq.6.42 [4]:hR= (6*0.04*1)/(1021.42*0.0045*9.807)=0.00532 m Total gas pressure drop hG:
Eq.6.35 [4]:hG=hD+hL+hR=0.133+0.0253+0.00532=0.16362 m Pressure loss at liquid entrance h2:
The area for liquid flow under apron =0.025*W=0.025*1.568=0.0275 m2. Since this is smaller than Ad,Ada=0.0275 m2.
Eq. 6.43[4]:h2= (q/Ada) 2*1.5/g= (0.00827/0.0275)2*1.5/9.807=.0139 m. Backup in downspout: Eq.6.44; h3=hG+h2=0.16362 +.0139 =0.1775 m. Check on flooding: hW+h1+h3=0.05+0.0265+0.1775 =0.254 m
For tower diameter T between 1 and 3 m, tray spacing t=0.60m. hW+h1+h3=0.254, which is well below t/2=0.30m.
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Therefore the chosen t is satisfactory.
Weeping velocity: for W/T=0.7, the weir is set 0.3296T=0.517 m from the centre of the tower. Therefore Z=2*0.517=1.034 m.
All other quantities in Eq.6.46 have been evaluated and then the equation yields VOW=2.901 m/s.
The tray will not weep excessively until the gas velocity through the holes V0 is reduced ti close to this value.
Entrainment: ρ v2=PM/RT=2.4175 kg/m3=0.026 kmol/m3 M2=∑xM=(93.1262*.995+18.0152*.005) =92.75 V2=797.08-626.654=170.426 kmol/hr. Q=170.126/(3600*0.026) =1.82 m3/s ρL2=1021.45 kg/m3
Liquid rate q=327.938 kmol/hr*92.75/1021.45=8.3*10-3 m3/s L’=q* ρL=0.0083*1021.45=8.478 kg/s
G’=Q* ρ v1=1.82*2.4175=4.4 kg/s V/VF=0.8, (ρv2/ ρL)0.5*L’/G’=0.094
Fig 6.17: E=0.042.The recycling of liquid resulting from such entrainment is too small to influence the tray hydraulics appreciably.
The minimum column diameter for tray columns is typically 0.75 m; otherwise, packed columns are used.
The maximum diameter of the column can be quite large – up to 5 m – although it may be decided to operate 2 or more separate columns in place of an otherwise large diameter single column.
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As the column diameter decreases, the vapour velocity increases for a given vapour flow rate.
The minimum column diameter is based upon the maximum vapour velocity that causes excessive entrainment and flooding.
The maximum column diameter is based upon maintaining a high enough velocity to prevent excess weeping.
The operating vapour velocity, and hence actual column diameter, is specified as a fraction of the flooding vapour velocity – typically 0.65 to 0.90.
The final consideration is column cost – a larger diameter column is more expensive than a smaller diameter column, although economies of scale enter
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11. COST ESTIMATION AND ECONOMICS
An acceptable design must present a process that is capable of operating under conditions which yields profit. Capital must be allocated for the direct plant expenses such as those for raw materials, labour and equipment. Besides direct expenses many other indirect
expenses are also incurred e.g. administration, salaries, product distribution cost, cost for interplant communications.
A capital investment is required for any industrial process and determination of necessary investment for any process consists of fixed capital investment for physical equipment and facilities in the plant plus working capital which must be available to pay the salaries keep raw materials and products on the hand and handle other specified items required a direct cash outlay.
Thus in analysis of cost in industrial processes, capital investment cost, manufacturing costs and general expenses including income taxes must be taken into consideration. Here method chosen is the annual cost method because of the following advantages:
1. Effects of seasonal variations are soothed out.
2. Plant on stream time or equipment operating factor is considered.
3. It permits more rapid calculation of operating cost at less than full capacity.
4. It provides a convenient way of considering infrequently occurring but large expenses such as annual turn around cost.
Cost of producing aniline per annum in 2002 = $ 5.5x106 Chemical plant index for the year 2002 = 76.2
Chemical plant index for the year 2012 = 402 Therefore cost of plant in 2012 =
Cost in 2002x {(cost index in 2012)/ (cost index in 2002)} = 5.5x106(402/76.2) = $ 29.016x106
Or Rs. 143.63 crores (1$ = 49.5 Rs)
Therefore fixed capital cost = FCC = Rs 143.63 crores Total capital investment = TCI = FCC + Working capital
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DISTRIBUTION OF CAPITAL COST Table 17:
INDIRECT COST % Of FCC COST (crores of Rs.) Table-18:
ESTIMATION OF TOTAL PRODUCT COST