Pump power and pipe length with Colebrook

Data. $Q = 0.25$ m³/s, pump power $P = 54\,642$ W, $D = 0.25$ m, $e = 0.26$ mm, $\mu = 0.0021$ N s/m², water $\rho = 1000$ kg/m³. Reservoir levels 12 m and 19 m, so the static lift is $19 - 12 = 7$ m. All losses are considered: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$ and friction.

(i) Length of the pipe

Pump head: $P = \gamma Q H_p$

$$H_p = \frac{54\,642}{9810(0.25)} = 22.280\ \text{m}$$

Energy equation between the reservoir surfaces: $H_p = 7 + h_L$, so the head loss is $h_L = 15.280$ m.

Velocity and Reynolds number:

$$V = \frac{0.25}{\pi(0.25)^2/4} = 5.093\ \text{m/s},\quad \frac{V^2}{2g} = 1.3220\ \text{m},\quad Re = \frac{\rho VD}{\mu} = \frac{1000(5.093)(0.25)}{0.0021} = 606305$$

Relative roughness $e/D = 0.00104$. Colebrook-White (solved by iteration):

$$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{e/D}{3.7} + \frac{2.51}{Re\sqrt f}\right) \;\Rightarrow\; f = 0.02031$$

Head loss: $h_L = \left(0.5 + f\dfrac{L}{D} + 1.0\right)\dfrac{V^2}{2g}$

$$15.280 = \left(1.5 + f\frac{L}{D}\right)(1.3220) \;\Rightarrow\; f\frac{L}{D} = 10.058$$ $$L = \frac{10.058\times 0.25}{0.02031} = 123.8\ \text{m}$$

(ii) Same power, loss of head doubled

"Loss of head doubled" means that, at the same flow, the system now loses twice as much head (for example a longer or rougher pipe). The head loss varies as $Q^2$, so for the original pipe

$$h_L = KQ^2,\qquad K = \frac{15.280}{0.25^2} = 244.5\ \text{s}^2/\text{m}^5$$

and for the new system $h_L' = 2KQ^2 = 489\,Q^2$ (taking $f$ as constant, since it changes very little). The pump head is then

$$H_p' = 7 + 489\,Q^2$$

For the same power $P = \gamma Q H_p'$:

$$9810\,Q\,(7 + 489\,Q^2) = 54\,642 \;\Rightarrow\; Q\,(7 + 489\,Q^2) = 5.570$$

Solving by trial: $Q = 0.200$ gives 5.31, $Q = 0.205$ gives 5.65, and $Q = 0.204$ gives 5.58. So

$$Q' \approx 0.204\ \text{m}^3/\text{s} = 204\ \text{l/s}$$

At this flow the head loss is $489(0.204)^2 = 20.3$ m and the pump head is $27.3$ m. (Allowing $f$ to vary with $Re$ in the Colebrook-White equation gives 203.6 l/s, practically the same.)

Comment. Doubling the loss cuts the discharge from 250 l/s to about 204 l/s, a fall of about 18%, for the same power. The 7 m static lift is a fixed requirement, but the pump must also supply the friction head. When friction rises, more of the same power is spent overcoming it and less water is delivered. The discharge falls by much less than the loss rises, because the head loss falls with the square of the flow. It would be wrong to keep the doubled loss of 30.56 m at the old flow and divide the power by $\gamma(7 + 30.56)$. That gives 148 l/s, which is too low, because the friction loss itself drops as the flow drops.

Answer: (i) $L \approx 124$ m ($f = 0.0203$); (ii) $Q' \approx 204$ l/s.