To calculate the volumetric flow rate ($Q$) of the pump, we'll use the given formula:

$Q = \frac{P \cdot \eta}{\Delta P}$

Where:

• $P$ = power output (in watts, W),
• $\eta$ = efficiency (as a decimal),
• $\Delta P$ = differential pressure (in pascals, Pa).

Given values:

• Power output ($P$) = 0.75 hp,
• Efficiency ($\eta$) = 65% = 0.65,
• Differential pressure ($\Delta P$) = 120 Pa.

First, we need to convert the power from horsepower (hp) to watts (W). The conversion factor is:

$1 \, \text{hp} = 745.7 \, \text{W}$

So:

$P = 0.75 \, \text{hp} \times 745.7 \, \text{W/hp} = 559.275 \, \text{W}$

Now, we can calculate the volumetric flow rate $Q$:

$Q = \frac{P \cdot \eta}{\Delta P} = \frac{559.275 \, \text{W} \times 0.65}{120 \, \text{Pa}}$

Let's compute that.The volumetric flow rate ($Q$) of the pump is approximately $3.03 \, \text{m}^3/\text{s}$.

Would you like further details or explanations on this calculation?

Here are five related questions:

1. How does pump efficiency affect the volumetric flow rate?
2. How can power losses impact the overall performance of the pump?
3. What is the significance of differential pressure in a pump system?
4. How would the flow rate change if the efficiency of the pump were higher?
5. What happens to the volumetric flow rate if the differential pressure increases?

Tip: Always check the units during calculations, as incorrect conversions can lead to significant errors.