Question:
A centrifugal pump running at 500 rpm delivers 60 liters/minute with a head of 50 m. At the same efficiency, if the rotational speed is increased to 1000 rpm, the discharge rate and head would respectively be:
A centrifugal pump running at 500 rpm delivers 60 liters/minute with a head of 50 m. At the same efficiency, if the rotational speed is increased to 1000 rpm, the discharge rate and head would respectively be:
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Remember the pump Affinity Laws: $Q \propto N$, $H \propto N^2$, and $P \propto N^3$. Doubling speed doubles flow, quadruples head, and increases power by eight times.
Updated On: Aug 29, 2025
- 120 liters/minute and 200 m
- 120 liters/minute and 100 m
- 60 liters/minute and 200 m
- 60 liters/minute and 100 m
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The Correct Option is A
Solution and Explanation
- Centrifugal pump performance follows Affinity Laws:
1. Discharge $Q \propto N$
2. Head $H \propto N^2$
where $N$ = rotational speed.
- Initial condition: $N_1 = 500$ rpm, $Q_1 = 60$ L/min, $H_1 = 50$ m.
- New condition: $N_2 = 1000$ rpm. \[ Q_2 = Q_1 \times \frac{N_2}{N_1} = 60 \times \frac{1000}{500} = 120 \; \text{L/min} \] \[ H_2 = H_1 \times \left(\frac{N_2}{N_1}\right)^2 = 50 \times \left(\frac{1000}{500}\right)^2 = 50 \times 4 = 200 \; \text{m} \] - Thus, at 1000 rpm, the pump delivers 120 L/min discharge with 200 m head.
1. Discharge $Q \propto N$
2. Head $H \propto N^2$
where $N$ = rotational speed.
- Initial condition: $N_1 = 500$ rpm, $Q_1 = 60$ L/min, $H_1 = 50$ m.
- New condition: $N_2 = 1000$ rpm. \[ Q_2 = Q_1 \times \frac{N_2}{N_1} = 60 \times \frac{1000}{500} = 120 \; \text{L/min} \] \[ H_2 = H_1 \times \left(\frac{N_2}{N_1}\right)^2 = 50 \times \left(\frac{1000}{500}\right)^2 = 50 \times 4 = 200 \; \text{m} \] - Thus, at 1000 rpm, the pump delivers 120 L/min discharge with 200 m head.
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