MIDTERM_2_Fall2007_SOLUTIONS

MIDTERM_2_Fall2007_SOLUTIONS - MECT 3345 MIDTERM...

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MECT 3345 MIDTERM EXAMINATION #2 11/14/07 PRINT NAME: ____________________SOLUTIONS____________________________ Instructions: Only pen, calculator and these sheets allowed on desk. Relevant handouts are attached at the end. Time limit 1 hour 30 minutes. Attempt to solve all problems. 1
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CONSTANTS: Avogadro’s Number: N A = 6.022 x 10 23 Boltzmann Constant: k = 1.38 x 10 -23 J/K = 8.617 x 10 -5 eV/K Universal Gas Constant: R = 8.314 J/mol K Temperature Conversion: T [K] T[ o C] + 273 Problem 1 (15 points) A chemical company is using a solid membrane of thickness L that is permeable to oxygen (by diffusion) to extract oxygen from a gas stream, as shown in the figure below. The pressures of the gases in the two chambers can be adjusted to control the concentrations c 1 and c 2 of oxygen at the right and left sides of the membrane, respectively. The operating conditions are T = 600 K, c 2 = 2c 1 , and the activation energy for oxygen diffusion in the membrane is Q d = 140 kJ/mol. The membrane operates under steady state conditions. As an engineer, you are asked to find the best solution for increasing the oxygen flow to the highest possible value. You have three options: 1.) Increase the pressure in chamber 2 such that c 2 =4c 1 2.) Increase the temperature to T = 700 K 3.) Replace the membrane by another one made of identical material but only half as thick as the current one. Which option would be your first choice? Which option would be your second choice? Justify your choices with calculations. Oxygen flow C 2 C 1 L Membrane Chamber 2 (gas 2, pressure 2) Chamber 1 (gas 1, pressure 1) 2
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Flux: dx dc RT Q exp D dx dc D j d ⎛ − = = 0 At steady state, linear concentration profile: L c c RT Q exp D j d 1 2 0 ⎛ − = Originally, c 2 =2c 1 : L c RT Q exp D j d 1 0 0 ⎛ − = 1.) After increasing c 2 = 4c 1 : L c RT Q exp D j d 1 0 1 3 ⎛ − = So the ratio is 3 0 1 = j j - the flux is increased three times. 2.) Flux at T o = 600 K: L c c RT Q exp D j d 1 2 0 0 0 ⎛ − = Flux at T 2 = 700 K: L c c RT Q exp D j d 1 2 2 0 1 ⎛ − = So the ratio is = = ⎛ − ⎛ − = K K molK J mol J T T R Q RT Q RT Q j j d d d 600 1 700 1 / 314 . 8 / 140000 exp 1 1 exp exp exp 0 2 0 2 0 2 After plugging in the numbers, 1 55 0 2 . j
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This test prep was uploaded on 04/15/2008 for the course MECE 3345 taught by Professor Goran during the Spring '08 term at University of Houston.

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MIDTERM_2_Fall2007_SOLUTIONS - MECT 3345 MIDTERM...

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