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Rearrange:
\( k_s + S = \mu S \)
\( S(\mu - 1/\theta) = k_s/\theta \)
\( S = \dfrac{k_s}{(\mu\theta - 1)} \)
Mass balance on S
\( Q(S_0 - S) + V r_s = 0 \)
\(-r_s = \dfrac{S_0 - S}{\theta} = \dfrac{\mu X}{Y} \)
\( X = \dfrac{Y(S_0 - S)}{\mu \theta} \)
\( \mu = 1 \)
\( X = Y(S_0 - S) \)
c) \( \tau = \tau_{min},\ S = S_0 \)
\( 1/\tau_{min} = 2d(250)/(50 + 250) \)
\( \tau_{min} = 0.6d \)
b) for \( S = 25 \text{ mg/L} \)
\( \tau(S) = \dfrac{k_s + S}{\mu S} = \dfrac{50 + 25}{2(25)} d = 1.5d \)
\( V = 10^4 \text{ m}^3/d \times 1.5d = 1.5 \times 10^4 \text{ m}^3 \)
c) \( X(1.5d) = X = 0.6(250 - 25) = 135 \text{ mg COD}_x/\ell \)
\( \dfrac{dX}{d\tau} = \hat{\mu}X \dfrac{S}{(K_s + S)} \qquad \text{where } X_0 > 0 \)
numerical solution for \( 5 \le X_0 \le 50 \)
\( X_0 \) (mgCODx/l)
5
10
25
50
\( \tau_{PFR} (S=25 \text{ mgCODx/l}) \) (d)
1.9
1.6
1.1
0.8
(See numerical solution tables and graphs following)
Problem 1d, comparison of ideal CSTR and PFR results from numerical solution of coupled mass balances for S and X, assuming values for \( X_0 \) 5, 10, 25, and 50 mgCODx/l for \( S \) effluent = 25 mg/L COD
\( \tau_{CSTR} \) (d)
1.5
1.5
1.5
1.5
Note that for \( X = 10 \text{ mgCODx/l} \) coupled mass balance solution is close to assuming \( X_{avg} = 67.5 \text{ mgCODx/l} \)
\( X_{avg} = 135/2 = 67.5 \text{ mg CODx/l} \)
\[
\int_{S_0}^{S} \frac{K_s + S}{S} \, dS
=
-\left(\frac{\hat{\mu}X}{Y}\right)\int_0^\tau d\tau
\]
\( K_s \ln S + (S - S_0) = -\hat{\mu} X_{avg}\tau \)
\[
\tau = \dfrac{Y}{\hat{\mu}X_{avg}}
\left(
K_s \ln \dfrac{S_0}{S} + S_0 - S
\right)
\]
If using this note that \( \tau \) is very sensitive to \( X_{avg} \)
for \( X_{avg} = 67.5 \) and \( S = 25 \text{ mg/L COD} \)
\[
\tau_{PFR} =
\dfrac{0.6}{2(67.5)}
\left(
50\ln\left(\dfrac{250}{25}\right) + (250 - 25)
\right)
= 1.5d
\]
\(\approx \tau_{CSTR}\) from 1b
for \( X_{avg} = 100 \)
\[
\tau_{PFR} =
\dfrac{0.6}{2(100)}
\left(
50\ln(10) + 225
\right)
= 1.0d \ (33\% \text{ less})
\]
Alternative solution to 1d, solve for \( \tau \), assuming \( X_{avg} \) constant
\( X = 135/(2.5, 67.5 \text{ mg/l}) \)
\[
\tau_{pf} =
\dfrac{Y}{\hat{\mu}X_{avg}}
\left(
K_s \ln(S_0/S) + (S_0 - S)
\right)
\]
\[
\tau_{cst} = 1/\mu = 1/((K_s+S)/S)
\]
\( V = 1.8d(300 \text{ m}^3/d) = 540 \text{ m}^3 \)
\[
X = \dfrac{\theta}{\tau}Y\frac{(S_0 - S)}{(1 + b\theta)}
\]
\( X = 1.8(10^4 \text{ m}^3/d)(0.6)\dfrac{225}{540 \text{ m}^3(1 + 0.1(1.8))} \)
\( X = 3800 \text{ mgCODx}/\ell \)
c. compare
CSTR recycle \( \tau_{90\%}(d) \) \( V_{90\%}(d) \) \( X_{90\%}(\text{mgCODx}/\ell) \)
recycle 0.054 \( 1.5\times10^4 \) 135
no recycle 1.5 540 3800
\( V_{norecycle}/V_{recycle} = 28 \), \( X_{recycle}/X_{norecycle} = 28 \)
Cell recycle increases biomass by factor of 28
and reduces required reaction time
and volume by factor of 28.
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