A motor is to be used to lift water from a depth of 10m & discharging it on ground level at a speed 10m/s. If 2 kg of water is lifted per second & the efficiency of motor is 40%. The minimum power of the motor should be nearly
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To find the minimum power of the motor, we calculate the total work done per second (power output) to lift and eject the water, then account for the motor's efficiency.
The power output, P_out, has two parts:
1. Power to overcome gravity: mgh = (2 kg/s)(10 m/s²)(10 m) = 200 W.
2. Power to provide kinetic energy: (1/2)mv² = (1/2)(2 kg/s)(10 m/s)² = 100 W.
Total P_out = 200 W + 100 W = 300 W.
Given efficiency η = 40% = 0.4, the input power from the motor is P_in = P_out / η.
Final Answer: 750 W