Help with this problem please You are using a two-stage hemostat to culture yeas
ID: 164399 • Letter: H
Question
Help with this problem please
You are using a two-stage hemostat to culture yeast and harvest product. the first stage is used for cell production (volume=300L), and the second stage is used for product formation(volume=400L). You can assume the yeast growth follows Monod equation. the feeding flow rate is 50L/hour and initial substrate concentration in the feed is 5g/l. What is cell mass concentration and substrate concentration at the outlet of the first reactor (assume no cell death and product formation)? If you assume there is no cell growth in the second reactor, what is the substrate and product concentration at the outlet of the second reactor? (mu max=0.2/hr, Ks=0.2g/L, Y_x/S=0.4, qp=0.05/h, and Y_p/S=0.7)Explanation / Answer
Stage 1
S1 = Ks D / (Mm-D) D=V/Q V= volume of culture Q = Flow rate D=300/50=6
= 0.2 * 6 / 0.2 -6 = 0.20 g
X1 = Yx/s * (So-S1) = 0.4 (5-0.20) = 1.92
Stage 2 - Product formation
M2 = (Mmax * S2) / Ks + S2 = D2 (X2-X1)/X2 --------------(1) D2=400/50=8
S2 = S1 (M2 X2 / D2 Yx/s) + (qp X2 / D2 Yp/s) -------------(2)
Solve 1 and 2
Substitute all the values and solve it
Eqn1 becomes 1.6X2-15.2S2=3.04 ----------(3) Multiply eqn 3 with -10 -16X2 + 152S2 = - 30.4
Eqn 2 becomes 16X2 + 6S2 = 31.6 --------(4) 16X2 + 6S2 = 31.6
--------------------------------------
158S2 = 1.2
S2=0.007g
X2= 31.5
Product.
FP2=V2qpX2 F - Volume in 2nd reactor
400 P = 50 * 0.005 * 31.5
P 2= 0.19
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.