exam1sol - 1 10.213 Chemical Engineering Thermodynamics...

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1 10.213 Chemical Engineering Thermodynamics Spring 2002 Test 1 Solution Problem 1 (50 points; 25 points for each part) Ethylene glycol [HOCH 2 CH 2 OH], also known as 1,2-ethanediol, is prepared by the hydrolysis of ethlene oxide [cyclic-CH 2 CH 2 O] in the reaction shown below: / O \ H 2 C—CH 2 ( g ) + H 2 O ( l ) Æ HOCH 2 CH 2 OH Consider a single-pass process where ethylene oxide is combined with water to generate ethylene glycol. In the process, the molar water to ethylene ratio entering the reactor is 5 and the reaction proceeds with 90% conversion of the entering ethylene oxide. The reactants (gaseous ethylene oxide and liquid water) enter the reactor at 80 ºC. The products from the reactor are separated by distillation in a second processing unit yielding a top stream that is water-rich (and also contains some ethylene oxide but no ethylene glycol) and a bottom stream (the product) that is ethylene glycol-rich. An analysis of the bottom stream shows that it has an ethylene glycol to water molar ratio of 10 and contains no ethylene oxide. For convenience, use the labels E for ethylene oxide, W for water, and G for ethylene glycol as your subscripts on flows. a) If the reactants are stored at 25 ºC and must be heated to 80 ºC before entering the reactor, i) determine the amount of heat that must be provided to the process per mole of ethylene oxide in the feed stream and ii) determine the amount of heat that must be provided to the process per mole of ethylene glycol produced in the final product stream . Solution i) Select as basis 1 mole of ethylene oxide (E). Thus the total flow into heater includes the 1 mol of E (vapor) and 5 mol of water (W) (liquid) as the molar ratio entering the reactor is 1:5. Need heat required to raise this stream from 25 ºC to 80 ºC. From Table C.1, C p ig /R for E = -0.385 + (23.463 x 10 -3 )*T – (9.296 x 10 -6 )*T 2 From Table C.2, C p /R for liquid W = 8.712 + (1.25 x 10 -3 )*T – (0.18 x 10 -6 )*T 2 For the mixture, the aggregate C p = 1*(C p for E) + 5*(C p for W) given their feed ratio Thus, for the mixture, C p /R= (-0.385 + 5*8.712) + (23.463+5*1.25)x10 -3 *T – (9.296+5*0.18)x10
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This note was uploaded on 11/03/2010 for the course CHEMICAL E 10.213 taught by Professor Blabla during the Spring '10 term at MIT.

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exam1sol - 1 10.213 Chemical Engineering Thermodynamics...

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