The pump motor, developing some power, raises 200 m ^ 3 of water to a height

The pump motor, developing some power, raises 200 m ^ 3 of water to a height of 10 m in 5 minutes. The engine efficiency is 40%. Find the engine power.

V = 200 m3.

ρ = 1000 kg / m3.

h = 10 m.

t = 5 min = 300 s.

g = 10 m / s2.

Efficiency = 40%.

Nz -?

The efficiency of the pump, which raises the water, shows what percentage of the spent mechanical work Az of the pump, when the water rises, goes into the useful work Ap.

Efficiency = Ap * 100% / Az.

The useful work of the pump Ap is expressed by the formula: Ap = m * g * h, where m is the mass of the raised water, h is the height of the rise of water, g is the acceleration of gravity.

The expended work Az of the pump will be expressed by the formula: Az = Nz * t, where Nz is the power that the pump develops, t is the water rise time ..

Efficiency = m * g * h * 100% / Nc * t.

Nz = m * g * h * 100% / efficiency * t.

We express the mass of water m, which was raised by the formula: m = ρ * V, where ρ is the density of the water, V is the volume of the raised water.

Nz = ρ * V * g * h * 100% / efficiency * t.

Nz = 1000 kg / m3 * 200 m3 * 10 m / s2 * 10 m * 100% / 40% * 300 s = 166666.7 W.

Answer: when the water rises, the pump develops a power of Ns = 166666.7 W.




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