Pump horsepower, TEL and HGL

Given: $Q = 0.12$ m³/s, $d = 0.15$ m, $h_f = 27V^2/2g$, efficiency 75 %, lower reservoir surface at 120 m, upper reservoir surface at 150 m. The pump delivers to the upper reservoir, and both surfaces are open to the atmosphere with negligible velocity.

Velocity and losses

$$\begin{aligned} A &= \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2\\ V &= \frac{Q}{A} = \frac{0.12}{0.017671} = 6.791\ \text{m/s}\\ \frac{V^2}{2g} &= \frac{6.791^2}{19.62} = 2.350\ \text{m}\\ h_f &= 27\times 2.350 = 63.46\ \text{m} \end{aligned}$$

Pump head (energy equation between the two surfaces)

$$\begin{aligned} z_1 + H_p &= z_2 + h_f\\ H_p &= (150 - 120) + 63.46 = 93.46\ \text{m} \end{aligned}$$

Power

$$\begin{aligned} P_{water} &= \gamma Q H_p = 9.81\times 0.12\times 93.46 = 110.02\ \text{kW}\\ P_{input} &= \frac{110.02}{0.75} = 146.69\ \text{kW}\\ &= \frac{146.69\times 1000}{746} = 196.6\ \text{hp} \end{aligned}$$

Answer: pump head 93.46 m; power needed to drive the pump $\approx$ 146.7 kW, i.e. about 197 hp (water power 110 kW, or 147.5 hp).

TEL and HGL

The pump's position along the pipe is not given, so take the suction pipe as short, with all of the loss $h_f = 63.46$ m (pipe friction plus the exit loss) in the delivery pipe.

 El. (m)
 213.46 +   T                         T = TEL
 211.11 +   H  .                      H = HGL (2.35 m below TEL)
        |   |     ' .
        |   |         ' .   both fall uniformly by friction
 152.35 +   |              ' .  T     TEL at the pipe outlet
 150.00 +   |                   H=========  upper reservoir
        |   |
 120.00 +=T=P  pump P: TEL and HGL rise by Hp = 93.46 m
 117.65 +  H
        +----------------------------------> distance along pipe

• The TEL starts at 120 m at the lower reservoir surface (no velocity). Where the water enters the pipe, the HGL drops below the TEL by $V^2/2g = 2.35$ m (to 117.65 m).
• At the pump, both the TEL and the HGL rise suddenly by $H_p = 93.46$ m: the TEL to 213.46 m and the HGL to 211.11 m.
• In the delivery pipe, the TEL slopes down uniformly with friction, and the HGL runs parallel to it, 2.35 m below (uniform diameter).
• At the outlet the HGL equals the upper reservoir level, 150 m. The TEL is 152.35 m there and drops by the exit loss $V^2/2g = 2.35$ m to 150 m. The total fall of the TEL is 63.46 m.