Basis:- 1 Hr. Operation

ENERGY BALANCES ACROSS THE FEED PREHEATING EQUIPMENT:The heating of cold 2-butanol feed is to be completed in three stages in order to avoid large heat losses. The cold feed is first preheated to its boiling point using steam as heating medium and then vaporized in a thermosyphon reboiler utilizing the heat contained in the reaction products. The vapor, which has entrained liquid is removed from the knock out drum, and then heated to reaction temperature using flue gas.

COLD FEED PREHEATER:Total feed = 23412.14+2605.77 kg = 26017.91 kg 26017.91 kg per hour of 2-butanol is to be preheated from 250C it boiling point 107.50C. Heat load on preheater, Q = 26017.91 x 1.497(380.5 – 288)

= 3.603 x 106 KJ The heating medium is used is dry saturated steam at 420K Steam requirement = 3.603 x 106 2123.4 = 1696.8 kg/hr 2-BUTANOL VAPORIZER:2-butanol feed is vaporized at 107.50C using reaction products. Heat load on vaporizer is , Q = M x λ = 26017.91 x 557.43 kj = 14.5032 x 106 kj/hr Average Cp of reaction products = 2.3 KJ / Kg 0k 14.5032 x 106 = 26017.91 x 2.304 (642 – T)

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T = 400 K

FIRST SUPER HEATER:2-butanol vapors is to be heat4ed from 380.5 k to 573k . using flue gas which enters at 673 k & cools to 423k. Cp = 2.038 kj/kg k. Heat load on super heater, Q = M x Cp x (T2 – T1) = 26017.91 x 2.038 (573 – 380.5) Q = 10.207 x 106 kj/hr Mass flow rate of flue gas = 10.2072 x 106 1.2 x ( 673-423) = 3.4024 x 104 kg/hr

SECOND SUPER HEATER:It raises the temperature of vapor from 573 to 773 k . using flue gas which enters at 873k abd cools to 623k Cp = 2.674 kj/kg k Heat load on super heater Q = M x Cp x (T2 – T1) = 26017.91 x 2.674 (773 – 573) = 13.9144 x 106 kj/hr Flue gas requirement, 13.9144 x 106 1.195 x (873- 623) = 4.6575 x 103 kg/hr =

ENERGY BALANCE ACROSS THE REACTOR:The feed enters the reactor at 773k and leaves at 663k. The conversion obtained in the reactor is 89.1%. The heat balance includes two terms i.e , sensible heat change due to cooling of reaction mixture and heat consume due to reaction. Q = M x Cp x(T2 – T1) + M x ∆H Mean Cp = 2.68 kj/kg k

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∆H = 73900 kj/kgmol Total 2-butanol feed = 23412.14 + 2579.71 kg = 25991.85 kg/hr = 351.24 kmol/hr

Q = 26017.91 x 2.68 x (663- 773) + 351.24 x .891 x 73900 = 15.457 x 106 It is assume that a flue gas is available at 800k & leaving 750k. Cp = 1.195 kj/kg k Flue gas requirement, 15.457 x 106 1.195 x (800-750) = 25.869 x 104 kg/hr =

ENERGY BALANCE ACROSS THE CONDENSER :The vapor enters condenser at 1270C (400k) at 1 bar pressure. Liquid and the vapor leaving the condenser is in equilibrium. At the more mole fraction x = 0.88 of the condensate we get both liquid and vapor temperature from the T- x- y diagram and temperature are 299.6 k. = liquid temperature., 335.7 k = vapor temperature We take cooling water enters at 240C and leaving at 420C CpH2 = 14650 J/ kg k Cpalcohol = 1760 J/ kg k CpMEK = 1664 J/ kg k Cpmix = 0.867 x 1664 + 0.1089 x 1760 + 0.0241 x 14650 = 1987.42 J/kg. Heat loss to reduction in vapor temperature from 400k to 335.7k Q1= 26017.91 x 1.98742 (400- 335.7) = 3.325 x 106 kj/ hr Heat loss due to condensation Q2 = 2591.34 x 689.19 + 18931.44 x 486.11 = 10.989 x 106 kj/hr Heat loss due to further cooling of a part of vapor is Q3 = 21522.78 x 1.987 x (335.7 – 299.6)

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= 1.544 x 106 kj/hr Total heat lost = Q = Q1 + Q2 + Q3 = (3.325 + 10.989 + 1.544) x 106 kj/hr = 15.858 x 106 kj/hr Mass flow of cooling water = 15.858 x 106 4.187 x (42- 24) = 2.104 x 105 kg / hr

ENERGY BALANCE ACROSS THE ABSORBER :Heat of condensation of MEK = 3707.072 x 443.14 = 1.643 x 106 kj/hr

Heat of condensation of alcohol is = 233.488 x 560 = 0.1308 x 106 kj/hr Heat of solution = (3707.072 + 233.488) x 0.35 = 0.001379 x 106 kj/hr

Heat loss in cooling gasses from 62.7 to 270C = (74.98 x 1.47 + 8.392 x 1.53 + 624.723 x 14.65) (62.7 – 27) = 0.3311 x 106 kj/hr Total heat released = 2.1063 x 106 kj/hr Heat...