Question

Underground water is to be pumped by a 78 percent efficient 5-kW submerged pump to a pool whose free surface is 30 m above the underground water level. Determine (a) the water volume flow rate and (b) the pressure difference across the pump. Disregard friction losses and assume that the effect of kinetic energy correction factors to be negligible. Draw a sketch to illustrate the problem solution and show all your work (write formulae, substitutions with units, detailed calculations). (8 points) Pool 30 m Figure 2

please help...add sketch

Answer 1

A)

The sketch for the question is actually given in the question. I have attached the following version:

2 처 Pump LA

Applying Bernoulli's Equation between 1 and 2:

$\frac{P_1}{w}+\frac{V_1^2}{2g}+Z_1+H_{pump}=\frac{P_2}{w}+\frac{V_2^2}{2g}+Z_2$

where

• P1 = P2 = 0 (atmospheric pressures at the free surface)
• V1 = V2 = 0 (Free surface does not move)
• Z1 = 0 (considered as datum)
• Z2 = 30 m (given)
• H(pump) = pump head required = ?

$\frac{P_1}{w}+\frac{V_1^2}{2g}+Z_1+H_{pump}=\frac{P_2}{w}+\frac{V_2^2}{2g}+Z_2$

$0+0+0+H_{pump}=0+0+30=>H_{pump}=30\:m$

Now, for pump power, we have:

pump WQH pump Притр

where

• P(pump) = actual power of the pump = 5 kW = 5000 W
• w = Specific weight of the fluid (water) = 9810 N/m3
• H(pump) = 30 m
• $\eta _{pump}$ = Efficiency of pump = 78% = 0.78
• Q = Volumetric flow rate of water = ? (our answer)

pump WQH pump Притр

$5000=\frac{9810\times Q\times 30}{0.78}=>\mathbf{Q=0.01325\:m^3/s\:(Ans.)}$

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B)

Also, the Pressure Difference across the pump = $\Delta P=wH_{pump}=9810\times 30=294300\:Pa=>\mathbf{\Delta P=294.3\:kPa\:(Ans.)}$

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