Question

A centrifugal pump is delivering water from an underground tank to an overhead reservoir against a static head of 35 m through a 2 km long, 250 mm diameter pipe. The head-discharge characteristic of the pump is given by
\[ H = 140 - 9000 Q^2 \]
where H is the head (in m) generated by the pump and Q is the discharge (in m³/s) of the pump.

Neglecting all minor losses, the head (in m) generated by the pump is (rounded off to the nearest integer).

Use: Darcy-Weisbach friction factor f = 0.04
Acceleration due to gravity = 9.81 m/s²
\(\pi\) = 3.14

Hint:

Write the friction head loss $h_f$ purely in terms of Q using the pipe's area, add the static head to get the system curve, then equate it with the pump's H-Q curve to find the operating point.

Answer 1

Correct Answer: 80

Step 1: Understanding the Question.
The pump must supply a head equal to the static lift plus the friction head loss in the pipe, and this required head must also match the head the pump is capable of giving at that same discharge. This is the operating point where the pump curve meets the system curve.

Step 2: Write the system head curve in terms of Q.
Pipe area:
\[ A = \frac{\pi D^2}{4} = \frac{3.14 \times (0.25)^2}{4} = 0.049 \text{ m}^2 \]
Darcy-Weisbach friction loss, written directly in terms of Q (since $V = Q/A$):
\[ h_f = \frac{fLQ^2}{2gDA^2} = \frac{0.04 \times 2000}{2 \times 9.81 \times 0.25} \times \frac{Q^2}{(0.049)^2} = 16.31 \times \frac{Q^2}{0.002407} = 6775.7\,Q^2 \]
So the system curve is:
\[ H_{system} = 35 + 6775.7\,Q^2 \]

Step 3: Equate the pump curve and the system curve.
\[ 140 - 9000Q^2 = 35 + 6775.7Q^2 \]
\[ 140 - 35 = (9000+6775.7)Q^2 \]
\[ 105 = 15775.7\,Q^2 \]
\[ Q^2 = 0.006656, \quad Q = 0.0816 \text{ m}^3/\text{s} \]

Step 4: Find the head generated.
Substitute back into the pump curve:
\[ H = 140 - 9000(0.006656) = 140 - 59.9 = 80.1 \text{ m} \]

Final Answer:
Rounded to the nearest integer, the head generated by the pump is 80 m.
\[ \boxed{H = 80 \text{ m}} \]