A water pump driven by an engine with an efficiency of 80\% raises water from an undergrou… — JAMB UTME

A water pump driven by an engine with an efficiency of $80\%$ raises water from an underground tank of depth $20\text{ m}$ and discharges it through a nozzle of cross-sectional area $10\text{ cm}^2$ at a steady speed of $10\text{ m/s}$ . Taking the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the minimum input power rating (in $\text{W}$ ) required for the engine?

Cevap: 3125 W

Cevap

The minimum input power rating required for the engine is

3125\text{ W}

The engine must supply power to lift

10\text{ kg}

of water per second through a vertical height of

20\text{ m}

while accelerating it to

10\text{ m/s}

. The useful power output is

2000\text{ W}

(potential)

+ 500\text{ W}

(kinetic)

= 2500\text{ W}

. Accounting for an engine efficiency of

80\%

, the total input power is

\frac{2500}{0.80} = 3125\text{ W}

Adım Adım Çözüm

Determine the mass of water discharged per unit time (mass flow rate).

\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}

Water is moving through a cross-sectional area at a constant velocity.

Calculate the useful output power required to lift the water and impart kinetic energy.

P_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}

The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.

Calculate the total input power using engine efficiency.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}

Efficiency is the ratio of useful power output to total power input.

Anahtar Kavram

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations

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