Chapter 29 : Substrate-Limiting Microbial Fermentation

Example 29.3 pageno : 639

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import math 

# Variables
CAo = 6.            # glucose solution
CM = 0.4            #kg/m3
V = 1.              #m3
k = 4.

# Calculations and Results
N = math.sqrt(1+(CAo/CM));
kt_op = N/(N-1);
C_by_A = 0.1;
t_op = kt_op/k;
v_op = V/t_op;

#The feed rate of glumath.cose
FAo = v_op*CAo;
print " The feed rate of glumath.cose is %.1f"%(FAo) ,
print "kg/hr"

#Max consumption rate of glumath.cose is
XA = N/(N+1);
c_max = FAo*XA;
print " Max consumption rate of glumath.cose is %.1f "%(c_max),
print "kg/hr"

#Max production rate of E.coli is
Cc_op = (C_by_A)*CAo*N/(N+1);
Fcmax = v_op*Cc_op;
print " Max production rate of E.coli is %.2f"%(Fcmax),
print "kg/hr"
 The feed rate of glumath.cose is 18.0 kg/hr
 Max consumption rate of glumath.cose is 14.4  kg/hr
 Max production rate of E.coli is 1.44 kg/hr
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