The volumetric flow rate to be pumped is 4 L/s.The required useful pump head = 131 m The efficiency of the pump is to be 79 %.The density and viscosity of water at anticipated operation conditions are taken to be 1000 kg/m3 and 0.00131 kg/m-s, respectively.Determine the power requirement of the pump in kW.
Question
The volumetric flow rate to be pumped is 4 L/s.The required useful pump head = 131 m The efficiency of the pump is to be 79 %.The density and viscosity of water at anticipated operation conditions are taken to be 1000 kg/m3 and 0.00131 kg/m-s, respectively.Determine the power requirement of the pump in kW.
Solution
To calculate the power requirement of the pump, we can use the formula for hydraulic power:
P = ρ * g * Q * H / η
where: P is the power requirement, ρ is the density of the fluid (water in this case), g is the acceleration due to gravity, Q is the volumetric flow rate, H is the pump head, and η is the efficiency of the pump.
First, we need to convert the units of the given values to be consistent with the formula.
The volumetric flow rate Q = 4 L/s = 0.004 m^3/s (since 1 L = 0.001 m^3) The pump head H = 131 m The density of water ρ = 1000 kg/m^3 The efficiency of the pump η = 79% = 0.79 The acceleration due to gravity g = 9.81 m/s^2
Substituting these values into the formula gives:
P = 1000 kg/m^3 * 9.81 m/s^2 * 0.004 m^3/s * 131 m / 0.79 P = 6540.44 W
Since 1 kW = 1000 W, the power requirement of the pump is 6.54 kW.
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