Pump head and power between two reservoirs

Given data

$Q = 0.06\ \text{m}^3/\text{s}$, $D = 0.2\ \text{m}$, $f = 0.02$, $L_s = 20\ \text{m}$, $L_d = 600\ \text{m}$, static lift $z = 30\ \text{m}$, $\sum K = 3$ (minor), exit $K = 1$, $\eta = 0.70$.

Velocity and velocity head

$$\begin{aligned} A &= \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2 \\ V &= \frac{Q}{A} = \frac{0.06}{0.03142} = 1.910\ \text{m/s} \\ \frac{V^2}{2g} &= \frac{1.910^2}{19.62} = 0.1859\ \text{m} \end{aligned}$$

Energy equation between the two reservoir surfaces

Both surfaces are at atmospheric pressure and velocity is nearly zero:

$$0 + 0 + H_p = z + h_f + h_{minor} + h_{exit}$$

Friction loss (total length 620 m):

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02 \times \frac{620}{0.2} \times 0.1859 = 11.53\ \text{m}$$

Minor losses: $3 \times 0.1859 = 0.558\ \text{m}$

Exit loss: $1 \times 0.1859 = 0.186\ \text{m}$

Head developed by the pump

$$H_p = 30 + 11.53 + 0.558 + 0.186 = 42.27\ \text{m}$$

Power

Hydraulic (water) power:

$$P_w = \rho g Q H_p = 1000 \times 9.81 \times 0.06 \times 42.27 = 24\,880\ \text{W} = 24.9\ \text{kW}$$

Shaft power:

$$P_s = \frac{P_w}{\eta} = \frac{24.88}{0.70} = 35.5\ \text{kW}$$

Answer: $H_p = 42.3\ \text{m}$; hydraulic power $= 24.9\ \text{kW}$; shaft power $= 35.5\ \text{kW}$