Question

1. Two reactors, one of them a CMBR and the other a CSTR, are operating side by side in parallel. Each is treating one half of a total flow rate of 800 gal/min containing a substance that enters at a concentration of 80 mg/L and experiences first-order degradation with a k value of 0.1/min.

The reactors have identical volumes of 8000 gal, so they also have equal liquid retention times of 20 min.

(a) What is the concentration of the substance leaving the CMBR?

(b) What is the concentration of the substance leaving the CSTR?

(c) Why are the two concentrations as different as they are?

2. The total flow drops to 400 gal/min, the CMBR is shut down (so all of the flow is going through the CSTR), and the influent concentration of the substance jumps to 120 mg/L.

(a) What will be the new, long-term effluent C from the CSTR?

(b) What will be the effluent concentration 5 minutes after the change?

(c) 20 minutes after the change?

Answer 1

a)

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b) and c)

CF = CO / [(1+Kθ₁)* (1+Kθ₂)* (1+Kθ₃)]

15 = 150/ [(1+0.4θ) *(1+0.4*2θ) *(1+0.4*θ)]

Rearrange equation,

1+1.6θ + 0.8θ² +0.13θ³ = 10

Trial and error solution to get the

Mean residence time

θ₁ = θ₃ = 2.25 hrs

θ₂ =2 θ =2*2.25 = 4.5 hrs

Volume of the reactors

V1= V3 = V = Qθ = (100gal/min) * (60 min/hr) * 2.25 hr =13500 gal

V2 = 2V = Qθ₂ = (100gal/min) * (60 min/hr) * 4.5 hr = 27000 gal

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C1 = [CO/ (1+Kθ₁)] = 150/ (1+0.4*(2.25)) = 79 mg/L

C2 = [C1/ (1+Kθ₂)] = 79/ (1+0.4*(4.5)) = 28 mg/L

C3 = [C2/ (1+Kθ₃)] = 28/ (1+0.4*(2.25)) = 15 mg/L

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r1 = [(CO – C1) / θ₁] = [(150-79) / 2.25] = 32 mg/L-h

r2= [(C1 – C2) / θ₂] = [(79-28) / 4.5] = 32 mg/L-h

r3 = [(C2 – C3) / θ₃] = [(28-15) / 2.25] = 5.8 mg/L-h