Question

Example #5: GIVEN The pump shown in Fig. E5 25a adds 10 horsepower to the water as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the head loss is 15 ft. Section (2) FIND Determine (a) the flowrate and Control volume (b) the power loss associated with this flow. 30 ft Flow Section (1) Pump Flow

Answer 1

(a) Apply steady flow energy equation for incompressible flow, between points on sections 1 and 2: 1 1 Pa +ż pv? + pghị + APpump = P2 +ż pv+ Pgh2 + APlos Where, P1 = P2 = Pa = 1 atm V1 V2 = 0 → Speed of flow at sections 1 and 2 hu = 0 → Height of section 1 hzi = 30 ft = 9.144 m → Height of section 2 → Pressure generated due to pump pump ΔΡloss → Pressure loss :: APpump = pghz + AP loss ... (1) The APpump can be found from power input to pump Epump Epump = APpump XQ Where, Еритр = 10 hp = 7456 W Q → Flow rate of water :: 7456 = APpump ,XQ :: AP, pump = 7456 0 The AP Ploss can be found from the head loss Ahloss APloss = - pg Ahloss Where, Ahloss = 15 ft 4.572 m Water density p= 103 kg/m3 g = 9.81 m/s2 Gravitational acceleration APLoss 103 x 9.81 X 4.572 AP loss 44851.32 Pa .(3) Substitute results 2 & 3 in equation 1: 7456 .. 103 x 9.81 x 9.144 + 44851.32 ::( = 0.05541 m/s

:: Q = 1.9569 ft²/s (b) The power loss associated with the flow is: Eloss = AP, loss X Q Use equation (3) for API loss substitution and the final result of (a) for Q substitution Eloss 44851.32 x 0.05541 - Eloss ~ 2485.33 W :: Eloss ~ 3.33 hp