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cee111-fa10-mt2-Bill Nazaroff-soln

Course: CE100 13972, Spring 2011
School: Berkeley
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111: CE Environmental Engineering MID TERM EXAM #2 3 November 2010 SOLUTION 1. Concepts in environmental transport (a) Ja = U C; Ja = advective flux (g m-2 h-1); U = fluid speed (m h-1); C = concentration (g m-3) (b) The parameter is the turbulent diffusivity and it has units of L2 T-1 (e.g., m2 s-1). (c) It is the particle Reynolds number: Rep = dp U p / . (d) The time required for acceleration is very short...

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111: CE Environmental Engineering MID TERM EXAM #2 3 November 2010 SOLUTION 1. Concepts in environmental transport (a) Ja = U C; Ja = advective flux (g m-2 h-1); U = fluid speed (m h-1); C = concentration (g m-3) (b) The parameter is the turbulent diffusivity and it has units of L2 T-1 (e.g., m2 s-1). (c) It is the particle Reynolds number: Rep = dp U p / . (d) The time required for acceleration is very short (s to ms, commonly) compared with typical time scales to settle through distances of environmental significance. 2. Transformations in flow reactors (a) Write material balance expressions for each of the three species in the reactor and solve for the steady-state concentrations. (Or, use the MB approach for [A] and solve the rest by stoichiometry.) d(V[A])/dt = Q [A]in Q [A] k[A]V = 0 [A] = [A]in (1/1+k) = 2.5 mM d(V[B])/dt = Q[B]in Q[B] k[A]V = 0 [B] = [B]in [A] k = 7.5 mM d(V[C])/dt = Q[C]in Q[C] + k[A]V = 0 [C] = [C]in + [A] k = 17.5 mM (b) Solve the problem using the batch reactor on a conveyor belt approach for [A] and then applying stoichiometry for [B] and [C]. d[A]/dx = -k[A] [A]() = [A]in exp(-k) = 1.84 mM change in [B] across the reactor is the same as the change in [A], so [B] 6.84 = mM loss of [A] corresponds to gain of [C], so [C] = 18.3 mM 3. Transient response in reactors (a) Approach: Write MB relationship on [A] considering separately the periods t t* and t t*. Solve the relationship for the first period in steady state. Use that result as the initial condition for the second period. Apply the dC/dt = S LC problem and solution for the second period. Results: First period (t < t*): d([A]V)/dt = QA1 Q[A] k[A]V = 0 [A] = A1/(1+k) Second period: d([A])/dt = (Q/V)A2 (Q/V + k) [A] with [A](t*) = A1/(1+k). Note that for second period, L = /(1+k) and the final steady state concentration is [A]ss = A2/(1+ k). [A] A2 (1+k!) A1 (1+k!) time t* t* + ! (1+k!) CE 111: Environmental Engineering MID TERM EXAM #2 3 November 2010 (b) Approach: By conceptual reasoning, recognize that the shape of the inlet concentration profile will be preserved at the outlet and that the time of the sharp rise will occur at t = t*+ . Use batch reactor on a conveyor belt concept to determine that the outlet concentration for times before the sharp rise will be [A] = A1 exp(-k) and that after the sharp rise it will be [A] = A2 exp(-k). [A] A 2 " exp(-k!) A1 " exp(-k!) time t* t* + !
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Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
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Berkeley - CE100 - 13972
Berkeley - CE100 - 13972
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