Biomedical Engineering Reference
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C 2
C 2 H 5 OH: m EtOH (5%)
C 6 H 10 O 5 : 2 kg
C 6 H 10 O 5 :
H 2 O: 2 kg
H 2 O:
Yeast:
Yeast: 10 g
FIGURE E3-5 Schematic diagram showing the control volume or system of interest for mass balance.
To produce 1 mol of alcohol, 1 mol of CO 2 is also produced and released, while half a mole of
glucose-equivalent of starch is consumed.
46:07 :
m EtOH ¼ 44:01 :
m CO 2
m EtOH
46:07
m CO 2 ¼ 44:01
¼ 44:01
46:07
m EtOH
Overall mass balance, Eqn (3.96) :
0
0
In - Out + Generation = Accumulation
Which leads to: m 0 ¼
2
þ
2
þ
0.01 kg
¼
In
¼
Out
¼
m CO 2 þ
m mix
Therefore:
m CO 2 ¼ 4:01 44:01
m mix ¼
m 0
m EtOH
46:07
We also know that
m EtOH
m mix
5
%
¼ u EtOH ¼
It then leads to:
m EtOH
u EtOH ¼
0 44:01
46:07
m
m EtOH
Solve for m EtOH , we obtain
0 u EtOH
1 þ 44:01
46:07
m
4:01 0:05
1 þ 44:01
m EtOH ¼
u EtOH ¼
kg
¼ 0:1914
kg
46:07 0:05
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