Test2Sol - CE 561, Exam 2, December 11, 2009 This exam...

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CE 561, Exam 2, December 11, 2009 This exam consists of three questions, each with multiple parts. You should be careful not to get stuck on one part. If you do not know how to do a problem, move on and return to it if you have time at the end. If you cannot find the numerical answer to a problem, explain how you would find the answer if you had more time. You may use three pages (2-sided) of hand-written notes. Carefully explain any assumptions you make, clearly indicate what part of what problem you are working on, and define the symbols that you use. The point value of each sub-part is indicated – budget your effort accordingly. There are 100 points total. Good luck. 1. The reversible, exothermic reaction A B, with rate = k 1 C A k 2 C B is to be carried out in solution in a well-mixed adiabatic batch reactor. At the start of each batch, the reactor contains 100 L of a solution of A at a concentration of 5 mol per liter ( C Ao = 5 mol/L, C Bo = 0) and a temperature of 300 K. Emptying, cleaning, and re-filling the reactor between batches requires 2 hours. The reactor contents have a density of 1 kg per liter and a specific heat of 4200 J/(kg K). The enthalpy of reaction is -42 kJ/mol. The forward and reverse rate constants are given by k 1 = 5×10 5 exp(-5000/ T ) hr -1 k 2 = 10 12 exp(-10000/ T ) hr -1 where T is the temperature in Kelvins. (a) Write the species mole balance and energy balance equations for the reactor and derive relationships that allow you to write the temperature and both species concentrations in terms of a single concentration or conversion variable. (5 pts.) The species mole balances are 12 A AB B dC r kC dt dC r kC dt = −= + = += And the energy balance is ( ) ˆˆ pp dT H H r dt CC ρρ −∆ = = Adding the species balances, we obtain ( ) 0 d dt From which 5 mol/L A B Ao CCC = If we define the conversion x as Ao A Ao x C = Then we have
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( ) 5 1 mol/L C 5 mol/L A B Cx x = = This is one way of writing all of the concentrations in terms of a single conversion variable. Likewise, we can combine the species A mole balance with the energy balance to obtain ( ) ˆˆ ˆ Ao A pp p HC dC dT H H dx r dt dt dt CC C ρρ ρ −∆  = = −=   From which ( ) ˆ Ao o p TT x C = + Evaluating the second term gives ( ) 42000*5 50 K ˆ 1*4200 Ao p C = = So, T = 300 + 50 x (b) Compute the maximum conversion of A to B that can be obtained in this adiabatic batch reactor for the conditions given. (10 pts.) The equilibrium conversion is attained when the net reaction rate is zero. Thus, we want to set k 1 C A k 2 C B = 0. Substituting into this the expressions for the rate constants in terms of T gives 5×10 5 exp(-5000/ T ) C A = 10 11 exp(-10000/ T ) C B Writing the temperature and all of the concentrations in terms of the fractional conversion x gives ( ) ( ) 5 12 7 5000 10000 5 10 exp *5 1 10 exp *5 300 50 300 50 5000 5 10 1 exp 300 50 xx x −− × ++ = ×− + Iterating on the above version of the equation didn’t lead to convergence, nor did any other rearrangement that I tried. When all else fails, we can use Newton’s method. Writing our equation as ( ) ( ) ( ) ( ) 6 6 22 5000 2 10 exp 0 1 300 50 1 250000 5000 ' 2 10 exp 300 50 1 300 50 x fx x = = −+ = ×+ +
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Test2Sol - CE 561, Exam 2, December 11, 2009 This exam...

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