Water is pumped from a lake to a storage tank 20 m above at a rate of 95 L/s while consuming 22.3 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.

(a) η = 0.835 = 83.5%

(b) ΔP = 196000 Pa = 196 KPa

Explanation:

(a)

The useful output of the pump-motor system will the power required to pump the water to the storage tank. That power is given as:

P = ρghQ

where,

P = Power to pump = ?

ρ = density of water = 1000 kg/m³

g = 9.8 m/s²

h = height = 20 m

Q = Volume Flow Rate = (95 L/s)(0.001 m³/1 L) = 0.095 m³/s

Therefore,

P = (1000 kg/m³)(9.8 m/s²)(20 m)(0.095 m³/s)

P = 18620 W = 18.62 KW

So, now the efficiency is given as:

η = Desired Output/Required Input

where,

η = overall efficiency = ?

Desired Output = Power to Pump = 18.62 KW

Required Input = Electric Power = 22.3 KW

Therefore,

η = 18.62 KW/22.3 KW

η = 0.835 = 83.5%

(b)The pressure difference between inlet and the outlet is given by the formula:

ΔP = ρgh

where,

ΔP = Pressure Difference = ?

ρ = density of water = 1000 kg/m³

g = 9.8 m/s²

h = height = 20 m

Therefore,

ΔP = (1000 kg/m³)(9.8 m/s²)(20 m)

ΔP = 196000 Pa = 196 KPa

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