Question

What is the power requirement for pumping 450 litres of water per minute against a head of 50 m, assuming a pump efficiency of 65%? What size (SHP) of electric motor is required?

• A. 4.93
• B. 7.50
• C. 8.50
• D. 9.20

Hint:

In metric units, the constant is 75: \( \text{HP} = \frac{Q \cdot H}{75 \cdot \eta} \) where \( Q \) is in liters per second and \( H \) is in meters.
Using this directly: \( \text{HP} = \frac{7.5 \times 50}{75 \times 0.65} = 7.69 \text{ HP} \), which is rounded to the standard 7.5 HP motor.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the shaft horsepower (SHP) of the pump, we must first calculate the theoretical water horsepower (WHP) required to lift the given discharge against the specified head, and then divide by the pump's mechanical efficiency ($\eta$).
Key Formula or Approach:
1. Convert discharge to liters per second: $Q \text{ (lps)} = \frac{Q \text{ (lpm)}}{60}$
2. Water Horsepower (WHP in metric HP):
\[ \text{WHP} = \frac{Q \text{ (lps)} \times H \text{ (m)}}{75} \] 3. Shaft Horsepower (SHP):
\[ \text{SHP} = \frac{\text{WHP}}{\eta} \]

Step 2: Detailed Explanation:
Given values:
- Discharge ($Q$) = $450 \text{ lpm}$
- Head ($H$) = $50 \text{ m}$
- Efficiency ($\eta$) = $65\% = 0.65$
Convert discharge to liters per second ($Q$):
\[ Q = \frac{450}{60} = 7.5 \text{ lps} \] Calculate Water Horsepower ($\text{WHP}$):
\[ \text{WHP} = \frac{7.5 \text{ lps} \times 50 \text{ m}}{75} = \frac{375}{75} = 5.0 \text{ HP} \] Calculate Shaft Horsepower ($\text{SHP}$):
\[ \text{SHP} = \frac{5.0 \text{ HP}}{0.65} \approx 7.69 \text{ HP} \] Taking into account the standard commercially available sizes of electric motors, a motor with a rating of ______7.50 HP______ is selected.

Step 3: Final Answer:
The required size of the electric motor is 7.50 HP, corresponding to option (B).