midterm1_key - Solutions Problem 1 a For first-order...

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Unformatted text preview: Solutions. Problem 1. a) For first-order reaction kinetics sec / 3 . 10000 min 10 2.5 1 sec / 3 . 10000 min 10 2.5 k 1 k Da 1 Da X 1 3- 1 3- A L L L L × × + × × = + = + = − − τ τ =0.58 b) Consider the non-steady state design equation for CSTR, where we can have dt dN r F F A A A A0 = + − V Since in liquid phase with constant density, we have dt dC kC- C C A A A A0 τ τ = − or equivalently τ τ τ A0 A A C C k 1 dt dC = + + with the initial conditions C A,t=0 =(1-0.58)×0.12 mol/L=0.05 mol/L, new τ =10000L/(0.7×0.3*60 L/min)=793.7 min Therefore, integrate this equation we can have ) k 1 exp( C ]} k 1 exp[ 1 { k 1 C ) ( C t A, A0 A t t t τ τ τ τ τ + − + + − − + = = Therefore the conversion at time t is ] k 1 exp[ C C ]} k 1 exp[ 1 { k 1 1 1 ) ( X A0 t A, A t t t τ τ τ τ τ + − − + − − + − = = After 60 min of changing flow rate, 598 . ] k 1 exp[ C C ]} k 1 exp[ 1 { k 1 1 1 min) 60 ( X A0 t A, A = + − − + − − + − = = t t τ τ τ τ τ The new steady state fractional conversion is...
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midterm1_key - Solutions Problem 1 a For first-order...

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