Question

A centrifugal pump runs at 1500 revolution per minute (RPM) and delivers water against a head of 10 m. What would be the head developed if the pump is operated at 1650 RPM ?

• A. 11.0 m
• B. 12.1 m
• C. 13.31 m
• D. 10 m

Hint:

Remember that head is proportional to the square of the speed ($H \propto N^2$). A $10\%$ increase in speed ($1.1$ times) results in a $21\%$ increase in head ($1.1^2 = 1.21$ times).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The performance of a centrifugal pump at different speeds is governed by the Pump Affinity Laws.
Key Formula or Approach:
The affinity law relating pump head ($H$) to rotational speed ($N$) is: \[ \frac{H_1}{H_2} = \left( \frac{N_1}{N_2} \right)^2 \] This can be rewritten to find the new head ($H_2$): \[ H_2 = H_1 \times \left( \frac{N_2}{N_1} \right)^2 \]

Step 2: Detailed Explanation:
From the problem parameters:
- Initial speed ($N_1$) = $1500 \text{ RPM}$
- Initial head ($H_1$) = $10 \text{ m}$
- New speed ($N_2$) = $1650 \text{ RPM}$
Substitute these values into the affinity law formula: \[ H_2 = 10 \times \left( \frac{1650}{1500} \right)^2 \] Simplify the fraction inside the brackets: \[ \frac{1650}{1500} = \frac{165}{150} = \frac{11}{10} = 1.1 \] Now, calculate $H_2$: \[ H_2 = 10 \times (1.1)^2 \] \[ H_2 = 10 \times 1.21 = 12.1 \text{ m} \] Therefore, the head developed at $1650 \text{ RPM}$ is $12.1 \text{ m}$.

Step 3: Final Answer:
The developed head is $12.1 \text{ m}$, which corresponds to Option (B).