GATE 2025 BT Q.44: solution and answer

GATE 2025 · BTQ.44 · 2 marks · MCQ

For a specific bioreactor configuration, the power requirement for a Rushton-turbine impeller agitating an unaerated Newtonian fluid in the turbulent regime will be

(A)
proportional to the stirring speed of the impeller
(B)
proportional to the square of the stirring speed of the impeller
(C)
proportional to the cube of the stirring speed of the impeller
(D)
inversely proportional of the stirring speed of the impeller

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Answer and solution

ANSWER

(C) proportional to the cube of the stirring speed of the impeller. In turbulent flow the power number of a Rushton turbine is constant, so $P = N_{P} ρ N_{i}^{3} D_{i}^{5} \propto N_{i}^{3}$ .

Official answer: CSource: Official answer key, GATE 2025 BT (IIT Roorkee)

Define the power number $N_{P} = \frac{P}{ρ N _{i}^{3} D _{i}^{5}}$ . Here $P$ is the ungassed power input ( $W$ ), $ρ$ the fluid density ( $kg m^{- 3}$ ), $N_{i}$ the stirring speed ( $s^{- 1}$ ) and $D_{i}$ the impeller diameter ( $m$ ).
The impeller Reynolds number is $R e_{i} = \frac{ρ N _{i} D _{i}^{2}}{μ}$ , where $μ$ is the viscosity of the Newtonian fluid ( $Pa s$ ). The flow is fully turbulent above roughly $R e_{i} \approx 1 0^{4}$ .
In the turbulent regime the power curve goes flat, so $N_{P}$ no longer depends on $R e_{i}$ . For one fixed geometry (a baffled vessel with a Rushton turbine) it settles at a constant value, $N_{P} \approx 5$ to $6$ .
Since $N_{P}$ is constant, rearrange the definition: $P = N_{P} ρ N_{i}^{3} D_{i}^{5}$
For a specific configuration $D_{i}$ is fixed, and for the same fluid $ρ$ is fixed. So $N_{P}$ , $ρ$ and $D_{i}^{5}$ are all constants, and $P \propto N_{i}^{3}$ .
For contrast, take the laminar regime ( $R e_{i} < 10$ ). There $N_{P} = \frac{k _{1}}{R e _{i}}$ , with $k_{1}$ a geometry constant, so $P = k_{1} μ N_{i}^{2} D_{i}^{3} \propto N_{i}^{2}$ . That is the relation in option (B), and it does not hold in the turbulent regime.
So the power requirement is proportional to the cube of the stirring speed of the impeller. The answer is (C).

KEY IDEA

The power number $N_{P} = P / (ρ N_{i}^{3} D_{i}^{5})$ is constant in the turbulent regime (about 5 to 6 for a baffled Rushton turbine). So for a fixed impeller diameter and fluid density, ungassed power goes as $N_{i}^{3}$ (and as $D_{i}^{5}$ ). In the laminar regime $N_{P} \propto 1/ R e_{i}$ , so power goes as $N_{i}^{2}$ instead.

WATCH OUT FOR

Choosing (B) by carrying over the laminar result $P = k_{1} μ N_{i}^{2} D_{i}^{3} \propto N_{i}^{2}$ into the turbulent regime. Another mistake is treating $N_{P}$ as if it still changes with speed in turbulent flow. Once $R e_{i}$ is above about $1 0^{4}$ , $N_{P}$ is flat, and all of the speed dependence sits in the $N_{i}^{3}$ of its definition.

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GATE 2025 BT Q.44: solution and answer

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