GATE 2023 BT Q.53: solution and answer

GATE 2023 · BTQ.53 · 2 marks · NAT

Q.53 The following schematic diagram shows a chemostat with cell recycle

[Figure]

where $F_{0}$ and $F_{r}$ are the volumetric flow rates (in $L h^{- 1}$ ) of feed and recycle streams, respectively. $X_{1}$ , $X_{0}$ and $X$ are the cell concentrations (in $g L^{- 1}$ ) in the reactor, recycle-stream and product-stream, respectively. If $\frac{X _{0}}{X _{1}} = 1.5$ , $\frac{F _{r}}{F _{0}} = 0.7$ and $X_{1}$ is $7.3 g L^{- 1}$ , the value of $X$ (in $g L^{- 1}$ , rounded off to one decimal place) is ________.

GATE 2023 BT question 53: A stirred-tank reactor (vessel with a liquid surface near the top, a central shaft and an impeller at the bottom). — Figure from the GATE 2023 BT paper. It is part of the question.

Answer and solution

ANSWER

$X = X_{1} (1 + α - α C) = 7.3 \times 0.65 = 4.745 \approx 4.7 g L^{- 1}$ .

Official answer: 4.6 to 4.9Source: Official answer key, GATE 2023 BT (IIT Kanpur)

Symbols: $F_{0}$ = feed flow rate, which is also the product flow rate ( $L h^{- 1}$ ). $F_{r}$ = recycle flow rate ( $L h^{- 1}$ ). $X_{1}$ = cell concentration in the reactor ( $g L^{- 1}$ ). $X_{0}$ = cell concentration in the recycle stream. $X$ = cell concentration in the product stream. Define the recycle ratio $α = F_{r} / F_{0} = 0.7$ and the concentration factor $C = X_{0} / X_{1} = 1.5$ .
The reactor is well mixed, so the stream leaving it, $F_{0} + F_{r}$ , carries cells at the reactor concentration $X_{1}$ . That stream enters the cell separator. Two streams leave the separator: the recycle, $F_{r}$ at $X_{0}$ , and the product, $F_{0}$ at $X$ . The flows balance: $F_{0} + F_{r}$ in, $F_{r} + F_{0}$ out.
Take a steady-state cell balance on the separator alone. Cells do not grow or die in the separator, so cells in = cells out: $(F_{0} + F_{r}) X_{1} = F_{r} X_{0} + F_{0} X$
Divide by $F_{0}$ and substitute $X_{0} = C X_{1}$ : $(1 + α) X_{1} = α C X_{1} + X \Rightarrow X = X_{1} (1 + α - α C)$
Substitute the numbers: $1 + α - α C = 1 + 0.7 - 0.7 \times 1.5 = 1.7 - 1.05 = 0.65$ .
$X = 0.65 \times 7.3 = 4.745 g L^{- 1} \approx 4.7 g L^{- 1}$ (to one decimal place). As a check, $X$ is less than $X_{1}$ , which is what we expect: the separator sends concentrated cells ( $X_{0} > X_{1}$ ) back to the reactor, so the product stream leaving it is thinner.

KEY IDEA

All you need is a cell balance around the cell separator. Nothing grows there, so at steady state the cells coming in with the reactor outflow, $(F_{0} + F_{r}) X_{1}$ , must equal the cells leaving in the recycle, $F_{r} X_{0}$ , plus those leaving in the product, $F_{0} X$ . No growth kinetics ( $μ$ , $D$ , $K_{s}$ ) are needed. Dividing by $F_{0}$ gives $X = X_{1} (1 + α - α C)$ , with $α = F_{r} / F_{0}$ and $C = X_{0} / X_{1}$ .

WATCH OUT FOR

Three mistakes come up often. (1) Giving the stream into the separator a flow of $F_{0}$ instead of $F_{0} + F_{r}$ . That gives $X = X_{1} (1 - α C) = - 0.37 X_{1}$ , a negative concentration, which is impossible. (2) Writing the balance over the reactor instead. That brings in growth and needs $μ$ , which the question does not give. (3) Getting the sign wrong on the recycle term, e.g. $X = X_{1} (1 + α + α C) = 2.75 X_{1}$ . With recycle concentrated at the separator, the product concentration must come out below $X_{1}$ , not above it.

From the GATE 2023 BT paper. Every question of this paper · Download the full paper · All 24 Bioreactor operation questions, with notes

GATE 2023 BT Q.53: solution and answer

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