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: Work with velocity V instead of discharge Q first.
Since $Q = AV$, we can rewrite the pump curve directly in terms of the pipe velocity V, using the pipe area:
\[ A = \frac{\pi D^2}{4} = \frac{3.14 \times 0.0625}{4} = 0.049 \text{ m}^2, \qquad Q = 0.049\,V \]
So the pump curve becomes:
\[ H = 140 - 9000(0.049V)^2 = 140 - 9000(0.002401)V^2 = 140 - 21.6\,V^2 \]

Step 2: Write the Darcy-Weisbach loss directly in terms of V.
\[ h_f = \frac{fL}{D}\cdot\frac{V^2}{2g} = \frac{0.04 \times 2000}{0.25}\cdot\frac{V^2}{2\times 9.81} = 320 \times \frac{V^2}{19.62} = 16.31\,V^2 \]
So the system curve, in terms of V, is:
\[ H_{system} = 35 + 16.31\,V^2 \]

Step 3: Equate the two curves in V.
\[ 140 - 21.6V^2 = 35 + 16.31V^2 \]
\[ 105 = 37.91\,V^2 \]
\[ V^2 = 2.77, \quad V = 1.665 \text{ m/s} \]

Step 4: Recover the head.
\[ H = 140 - 21.6(2.77) = 140 - 59.8 = 80.2 \text{ m} \]
(the small difference from the Q-based route is just rounding in the intermediate steps; both give the same operating head once carried to full precision)

Final Answer:
\[ \boxed{H \approx 80 \text{ m}} \]