Cumene Energy Balance

Distillation Column –1 :1 :To find the temperature at which the product stream is fed to distillation column –At P 1 = 25 atm, T 1 = 200 °CAt P 2 =1 atm T 2 = ?Cp avg at 100 °C = 0.6126 x 163.42 +0.0095 x 243.76 + 0.1673 x 107.01+0.2105 x 79.47T 2 = T 1 (P 2 /P 1 )R/Cp avg8.314 / 137.05=100(1/25)=82.26 0 c= 137.05 J/gm moleThis is further cooled to 25 °C and fed to the distillation column.F=1749.72 kmoles/hrD=368 kmoles/hrW=1381.72 kmoles/hrEnthalpy of vapor that goes as overhead :Hv= Latent heat of vaporisation + sensible heatAs propane is the major constituent that goes with the overhead, taking λ and Cp valuesof Propane,Hv =V [λ + Cp (Tb –To )]Assuming a reflux ratio of 0.5, we have R=L/D =0.5L=0.5 D =0.5 x 368 x 44 =8096 kg/hrV=L+D=8096+16192 =24288 kg/hrTaking reference temperature as the temperature at which feed enters,19

T 0 =25 °C ; T b = 42.1 °C , Cp =2.41 KJ/kg °Cλ = 0.4251 KJ/gm =425.1 KJ/kgThere fore Hv =24288 [425.1 + 2.41 ( 42.1 –25 )]=11.3257 x 10 6 KJ/hrH D =DCp(T b –T 0 )=16192 x 2.41 ( 42.1 –25 )=6.673 x 10 5 KJ/hrH L =L Cp (T b –T 0 )=8096 x 2.41 (42.1 –25)=3.336 x 10 5 KJ/hrTaking enthalpy balance around the condenser,Hv= Q c +H D +H L11.3257 x 10 6 = Q c +6.675 x10 5 +3.336x 10 5Q c = 10.325 x 10 6 KJ/hrCooling water requirement :Let us assume inlet and exit water temperature as 25°C and 45 °CCp=4.18 KJ/kg °CThere fore Q c = m steam CpdT10.325 x 10 6 = m steam x 4.18x 20m=123.5 x 10 3 kg/hrTotal enthalpy balance :H F + Q B = H D + Q C + H W20

To find H W :H W =WCp avg (T b –T 0 )By using p i = X i P i and checking P t = 760 mm Hg we found T b = 137 0 CCp avg = 0.776 x 176.32 + 0.01199 x 257.11 + 0.2120 x 110.73= 174 J/mole K= 174 kJ/kmole KM avg = 111.72 kg/kmoleTherefore Cp avg = 174 / 111.72=1.5575 KJ/kg KH w = 1381.72 x 1.5575(137-25) x 111.72= 26.927 x 10 6 KJ/hrH F = 0 [ because T F = T 0 ]Q B =H D + Q C + H W - H F= 6.673 x 10 5 + 10.325 x 10 6 +26.927 x 10 6 -0=37.92 x 10 6 KJ/hrSaturated steam required :Q B = m steam 37.92 x 10 6 = m steam x 2256.9m steam = 16801.5 kg/hr21

H L = L Cp (T b –T 0 )= 7984.7 x 1.2246 (80.1 –137 )= -0.55637 x 10 6 KJ/hrH v = Q C + H L +H D54.45 x 10 6 = Q C –0.55637 x 10 6 – 1.1127 x 10 6Q C = 56.12 x 10 6 KJ/hrCooling water requirement :Let us assume inlet and exit water temperature as 25°C and 45 °CCp=4.18 KJ/kg °CThere fore Q c = m steam CpdT54.45 x 10 6 = m steam x 4.18 x 20m steam = 67.128 x 10 4 kg/hrTotal enthalpy balance :H F + Q B = H V + Q C +H wTo find H W :w = 138190 Kg/hrT b = T F for distillation column –3= 153.4 °CCp avg § &S RI &XPHQH= 1.91 KJ/kg °CH w = 138190 x 1.91(153.4 –137)= 3.0774 x 10 6 KJ/hrH F = 0 [ because T F = T 0 ]23

Q B = 54.94 x 10 6 + 65.11 x 10 6 +4.06 x 10 6 -12.245 x 10 6= 11.46 x 10 7 KJ/hrSaturated steam required :Q B = m steam 11.46 x 10 7 = m steam x 2256.9m steam = 50.81 x 10 3 kg/hrDistillation column – 3 :F = 138190 kg/hrD = 129051 kg/hrw = 9139 kg/hrEnthalpy of vapor that goes at the top:$V &XPHQH LV WKH PDMRU FRQVWLWXHQW WKDW JRHV ZLWK WKH RYHUKHDG WDNLQJ DQG &Svalues of Cumene,H v 9> &S7 b –T 0 ) ]Taking reference temperature T 0 =T F = 153.4 °CB.P. of Cumene at 1 atm = 152.4 °C RI &XPHQH FDOJP= 312.1264 KJ/kgCp of Cumene vapor at 152.4 °C = 0.4047 cal/gm °KV = D + L = 129051 + 68655.1= 1.6931 KJ/kg °K24

=197706.1 kg/hrH v = 197706.1[ 312.1264 + 1.6931 ( 152.4 –153.4)]= 61.3745 x 10 6 KJ/hrH D = D Cp (T b –T 0 )= 129051 x 1.6931(152.4 –153.4)= -0.218496 x 10 6 KJ/hrH L = L Cp(T b –T 0 )= 68655.1 x 1.6931(152.4 –153.4)= -0.116239 x 10 6 KJ/hrH v = Q C + H D +H L61.3745 x 10 6 = Q C –0.218496 x 10 6 -0.116239 x 10 6Q C = 61.71 x 10 6 KJ/hrCooling water requirement :Let us assume inlet and exit water temperature as 25°C and 45 °CCp=4.18 KJ/kg °CThere fore Q c = m steam CpdT61.71 x 10 6 = m steam x 4.18 x 20m steam = 73.8148 x 10 3 kg/hrTotal enthalpy balance :H F + Q B = H V + Q C +H wTo find H W :W = 9139 kg/hrH w = W Cp avg (T b –T 0 )25

T b at x w = 0.2934 =184.5 °CCp avg at 184.5 °C = 0.013x 214.1952 + (1 –0.013) x 288.93= 287.9584 J/mole °K= 2.88795 KJ/kg °KH w = 9139 x 2.8795(184.5 –153.4)= 81.84 x 10 4 KJ/hrH F = 0 [ because T F = T 0 ]Q B = H V + Q C + H W - H F= 61.3745 x 10 6 + 73.8143 x 10 3 + 81.84 x 10 4=62.2667 x 10 6 KJ/hrSaturated steam required :Q B = m steam 62.2667 x 10 6 = m steam x 2256.9m steam = 27589.5 kg/hr26