Question

A fed-batch operation is initiated by feeding substrate solution at 1 L h\textsuperscript{-1 containing 50 g L\textsuperscript{-1} substrate. The reactor initially contains 50 g biomass. The biomass yield $Y_{X/S}^{M}$ is 0.4 $\frac{g\;biomass}{g\;substrate}$. Assuming quasi-steady state, the maximum biomass after 5 h of feeding is _________ g.}

Hint:

In fed-batch systems without growth limitation, total biomass = initial biomass + (yield × total substrate fed).

Answer 1

Correct Answer: 150

Feed rate: 1 L h\textsuperscript{-1}
Substrate concentration in feed: 50 g L\textsuperscript{-1}
Substrate fed per hour:
\[ 50 \times 1 = 50\;g\;substrate/h \]
For 5 hours, total substrate fed:
\[ 50 \times 5 = 250\;g\;substrate \]
Biomass yield coefficient:
\[ Y_{X/S}^M = 0.4\;\frac{g\;biomass}{g\;substrate} \]
Biomass formed from 250 g substrate:
\[ X_{formed} = 0.4 \times 250 = 100\;g \]
Initial biomass = 50 g
Total biomass:
\[ X_{total} = 50 + 100 = 150\;g \]
Thus, the maximum biomass after 5 hours is:
\[ \boxed{150} \]