GATE 2022 BT Q.60: solution and answer

GATE 2022 · BTQ.60 · 2 marks · NAT

Liquid-phase mass transfer coefficient ( $k_{L}$ ) is measured in a stirred tank vessel using steady-state method by sparging air. Oxygen uptake by the microorganism is measured. The bulk concentration of $O_{2}$ is $1 0^{- 4} mol L^{- 1}$ . Solubility of $O_{2}$ in water at $25^{\circ} C$ is $1 0^{- 3} mol L^{- 1}$ .

If the oxygen consumption rate is $9 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$ , and interfacial area is $100 m^{2} / m^{3}$ , the value of $k_{L}$ is ________ $cm s^{- 1}$ .

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Answer and solution

ANSWER

$k_{L} = 1 cm s^{- 1}$ .

Official answer: 1Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

Steady-state balance. In the steady-state method the dissolved-oxygen level in the broth stops changing. Oxygen then enters the liquid exactly as fast as the cells use it, so the oxygen transfer rate (OTR) equals the oxygen uptake rate (OUR): $k_{L} a (C^{*} - C_{L}) = OUR$ Here $k_{L}$ is the liquid-film mass transfer coefficient, $a$ is the gas-liquid interfacial area per unit liquid volume, $C^{*}$ is the saturation (solubility) concentration of $O_{2}$ , and $C_{L}$ is the bulk dissolved $O_{2}$ concentration.
Driving force. $C^{*} - C_{L} = 1 0^{- 3} - 1 0^{- 4} = 9 \times 1 0^{- 4} mol L^{- 1}$ .
Volumetric coefficient. $k_{L} a = \frac{OUR}{C ^{*} - C _{L}} = \frac{9 \times 1 0 ^{- 4} mol L ^{- 1} s ^{- 1}}{9 \times 1 0 ^{- 4} mol L ^{- 1}} = 1 s^{- 1}$
Interfacial area in cm units. $a = 100 m^{2} / m^{3} = 100 m^{- 1}$ . Since $1 m = 100 cm$ , $a = \frac{100}{100 cm} = 1 cm^{- 1}$ .
Separate $k_{L}$ . $k_{L} = \frac{k _{L} a}{a} = \frac{1 s ^{- 1}}{1 cm ^{- 1}} = 1 cm s^{- 1}$ This is the same as $0.01 m s^{- 1}$ . The answer is $1 cm s^{- 1}$ .

KEY IDEA

At steady state, oxygen transfer equals oxygen uptake: $k_{L} a (C^{*} - C_{L}) = OUR$ . Divide the OUR by the driving force to get $k_{L} a$ . Then divide by the specific interfacial area $a$ to get $k_{L}$ .

WATCH OUT FOR

There are three common slips. (1) Using $C^{*}$ alone as the driving force instead of $C^{*} - C_{L}$ , which gives $k_{L} a = 0.9 s^{- 1}$ . (2) Leaving $a$ as $100 m^{- 1}$ , which gives $k_{L} = 0.01 m s^{- 1}$ , then entering 0.01 even though the blank asks for $cm s^{- 1}$ . (3) Getting the m-to-cm conversion of $a$ upside down, which gives $100 cm^{- 1}$ .

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GATE 2022 BT Q.60: solution and answer

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