The horsepower of a pump of efficiency 80%, which sucks up water from 10 m below ground and ejects it through a pipe opening at ground level of area 2 cm2 with a velocity of 10 m/s, is about:(1 hp = 746 W)

Engineering

Physics

BernoulIis Equation and Equation of Continuity

Question

The horsepower of a pump of efficiency 80%, which sucks up water from 10 m below ground and ejects it through a pipe opening at ground level of area 2 cm² with a velocity of 10 m/s, is about:
(1 hp = 746 W)

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The horsepower required is calculated from the power output, which is the rate of kinetic energy imparted to the water. The mass flow rate is ρAv, where density

ρ = 1000kg/m³ area A = 2 × 10^–4 m², and velocity v = 10m/s. The kinetic energy per second (power output) is $\frac{1}{2} \times ρAv \times v^{2}$ . The pump must also provide power to lift the water against gravity, ρAvgh, where h = 10m. The total power output is the sum of these. With 80% efficiency, the power input is (Power Output)/0.8. Convert this to horsepower (1 hp = 746 W).

Final Answer: 0.5 hp