GATE 2019 BT Q.52: solution and answer

GATE 2019 · BTQ.52 · 2 marks · NAT

Phenolic wastewater discharged from an industry was treated with Pseudomonas sp. in an aerobic bioreactor. The influent and effluent concentrations of phenol were 10,000 and 10 ppm, respectively. The inlet feed rate of wastewater was $80 L h^{- 1}$ . The kinetic properties of the organism are as follows:

Maximum specific growth rate $(μ_{m}) = 1 h^{- 1}$

Saturation constant $(K_{S}) = 100 mg L^{- 1}$

Cell death rate $(k_{d}) = 0.01 h^{- 1}$

Assuming that the bioreactor operates under 'chemostat' mode, the working volume required for this process is ________ L (rounded off to the nearest integer).

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Answer and solution

ANSWER

The required working volume is $V = \frac{88000}{89} \approx 989 L$ .

Official answer: 970 to 1010Source: Official answer key, GATE 2019 BT (IIT Madras)

Symbols: $μ$ = specific growth rate, $μ_{m}$ = maximum specific growth rate, $K_{S}$ = Monod saturation constant, $k_{d}$ = specific cell death rate, $F$ = feed rate, $V$ = working volume, $D = F / V$ = dilution rate, $S$ = phenol concentration in the reactor, $X$ = biomass concentration. A chemostat is well mixed, so the effluent concentration equals the concentration inside the reactor: $S = 10 ppm = 10 mg L^{- 1}$ (for a dilute aqueous solution, $1 ppm \approx 1 mg L^{- 1}$ ).
Monod growth rate at the reactor concentration: $μ = \frac{μ _{m} S}{K _{S} + S} = \frac{1 \times 10}{100 + 10} = \frac{1}{11} = 0.0909 h^{- 1}$
Biomass balance at steady state with a sterile feed and cell death: $\frac{d X}{d t} = μ X - k_{d} X - D X = 0 \Rightarrow D = μ - k_{d}$
Dilution rate: $D = \frac{1}{11} - 0.01 = \frac{100 - 11}{1100} = \frac{89}{1100} = 0.08091 h^{- 1}$
Working volume: $V = \frac{F}{D} = \frac{80}{89/1100} = \frac{88000}{89} = 988.76 L$
The influent phenol concentration ( $10, 000 ppm$ ) fixes how much biomass is formed ( $X = Y_{X / S} (S_{0} - S)$ ), not the residence time, so it does not enter the volume. Rounded to the nearest integer, $V \approx 989 L$ .

KEY IDEA

At steady state in a chemostat with cell death, the biomass balance gives $D = μ - k_{d}$ , where $μ$ is the Monod rate at the effluent (reactor) concentration. Once you have $D$ , the volume is $V = F / D$ . The inlet concentration $S_{0}$ only sets how much biomass is formed. It does not change the dilution rate.

WATCH OUT FOR

Ignoring death and taking $D = μ$ gives $V = 880 L$ , which is outside the key. Adding death with the wrong sign ( $D = μ + k_{d}$ ) gives about $793 L$ . Putting the influent $10, 000 ppm$ into the Monod equation instead of the effluent $10 ppm$ is also wrong, because in a well-mixed chemostat the cells only ever see the outlet concentration.

From the GATE 2019 BT paper. Every question of this paper · Download the full paper · All 24 Bioreactor operation questions, with notes

GATE 2019 BT Q.52: solution and answer

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