A pump with Q = 0.05 m^3/s, head H_p = 60 m; ρ = 1000 kg/m^3; g = 9.81; compute hydraulic power P_h.

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Multiple Choice

A pump with Q = 0.05 m^3/s, head H_p = 60 m; ρ = 1000 kg/m^3; g = 9.81; compute hydraulic power P_h.

Explanation:
Hydraulic power from a pump is the energy transferred to the fluid per unit time, which depends on how much fluid is moved, the height it is lifted, and the fluid’s weight. The formula is P_h = ρ g Q H_p. Substituting ρ = 1000 kg/m^3, g = 9.81 m/s^2, Q = 0.05 m^3/s, and H_p = 60 m gives ρ g = 9,810 N/m^3, and P_h = 9,810 × 0.05 × 60 = 9,810 × 3 = 29,430 W = 29.43 kW. So the hydraulic power is 29.43 kW. (This is the hydraulic output; the actual pump input power would be higher if efficiency is considered.)

Hydraulic power from a pump is the energy transferred to the fluid per unit time, which depends on how much fluid is moved, the height it is lifted, and the fluid’s weight. The formula is P_h = ρ g Q H_p. Substituting ρ = 1000 kg/m^3, g = 9.81 m/s^2, Q = 0.05 m^3/s, and H_p = 60 m gives ρ g = 9,810 N/m^3, and P_h = 9,810 × 0.05 × 60 = 9,810 × 3 = 29,430 W = 29.43 kW. So the hydraulic power is 29.43 kW. (This is the hydraulic output; the actual pump input power would be higher if efficiency is considered.)

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