GATE 2024 BT Q.56: solution and answer

GATE 2024 · BTQ.56 · 2 marks · NAT

Ethanol is produced in a 10,000 L stirred bioreactor using an impeller of diameter 1 m. The density and viscosity of fermentation broth are $1000 kg m^{- 3}$ and 1 cp, respectively. The data relating the Power number and Impeller Reynolds number is given below:

| Reynolds number | 1-5 | 5-500 | $> 1 0^{5}$ |
|---|---|---|---|
| Power number | 70 | 10 | 5 |

Using the above data, the power required for the stirrer to operate at 300 rpm is ________ kW (Answer in integer).

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Answer and solution

ANSWER

The stirrer needs 625 kW. At 300 rpm the impeller Reynolds number is $5 \times 1 0^{6}$ , which is above $1 0^{5}$ , so $N_{P} = 5$ and $P = N_{P} ρ N^{3} D_{i}^{5} = 6.25 \times 1 0^{5} W$ .

Official answer: 625Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

Symbols: $N$ is the impeller speed, $D_{i}$ the impeller diameter, $ρ$ the broth density, $μ$ the broth viscosity, $R e_{i}$ the impeller Reynolds number, $N_{P}$ the Power number and $P$ the power drawn by the stirrer.
Convert to SI units: $N = 300 rpm = \frac{300}{60} = 5 s^{- 1}$ ; $D_{i} = 1 m$ ; $ρ = 1000 kg m^{- 3}$ ; $μ = 1 cp = 1 0^{- 3} Pa s$ .
Impeller Reynolds number: $R e_{i} = \frac{ρN D _{i}^{2}}{μ} = \frac{1000 \times 5 \times 1 ^{2}}{1 0 ^{- 3}} = 5 \times 1 0^{6}$
Because $R e_{i} = 5 \times 1 0^{6} > 1 0^{5}$ , the flow is fully turbulent, so the Power number comes from the last column of the table: $N_{P} = 5$ .
Definition of the Power number: $N_{P} = \frac{P}{ρ N ^{3} D _{i}^{5}}$ , so $P = N_{P} ρ N^{3} D_{i}^{5}$ .
$P = 5 \times 1000 \times 5^{3} \times 1^{5} = 5 \times 1000 \times 125 = 6.25 \times 1 0^{5} W$
$P = 625 kW$ . The reactor volume (10,000 L) is not needed for this calculation.

KEY IDEA

First use the impeller Reynolds number $R e_{i} = ρN D_{i}^{2} / μ$ to decide which flow regime you are in, and so which Power number to read from the table. Then rearrange $N_{P} = P / (ρ N^{3} D_{i}^{5})$ to get the power. In fully turbulent flow $N_{P}$ is constant, so $P$ goes as $N^{3} D_{i}^{5}$ .

WATCH OUT FOR

Using the speed in rpm (300) rather than rev/s (5): this gives a power $6 0^{3} = 216, 000$ times too large. A second mistake is leaving the viscosity as 1 instead of converting 1 cp to $1 0^{- 3} Pa s$ . That gives $R e_{i} = 5000$ , which falls in the gap between 500 and $1 0^{5}$ where the table gives no value. The 10,000 L volume is a distractor: power is fixed by $N$ , $D_{i}$ , $ρ$ and $N_{P}$ alone.

From the GATE 2024 BT paper. Every question of this paper · Download the full paper · All 24 Bioreactor operation questions, with notes

GATE 2024 BT Q.56: solution and answer

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