An overhead tank of capacity 1000 litres has to be filled in 1/2 an hour using a water pump. The tank is kept at a height 10m above ground and the water level is 10m below ground. The opening of the inlet pipe inside the tank is 1.11cm2. Assuming the efficiency of the motor to be 60%, the electric power used is (Neglect viscosity):

Explanation

Determine Total Height and Mass of Water: The total height (H) the water needs to be lifted is the sum of the height of the tank above ground and the depth of the water level below ground.H=10 m+10 m=20 mThe mass (m) of the water is calculated from its volume (1000 litres = 1 m3) and density (1000 kg/m3).m=1 m3×1000 kg/m3=1000 kgCalculate Water Velocity: The time to fill the tank is 30 minutes, which is 1800 seconds. The flow rate (Q) is the volume divided by time.Q=1m3/1800 s≈0.000556 m3/sThe area (A) of the inlet pipe is 1.11 cm², which converts to 0.000111 m². The velocity (v) is the flow rate divided by the area.v=Q/A=0.000556 m³s-1 /0.000111 m²≈5.005 m/sCalculate Total Work Done: The total work is the sum of the potential energy (for lifting the water) and kinetic energy (for the water's final speed).Potential Energy (PE) = mgh=1000 kg×9.8 m/s²×20 m=196,000 JKinetic Energy (KE) = 1/2mv²=0.5×1000 kg×(5.005 m/s)²≈12,525 JTotal Work (W) = PE+KE=196,000 J+12,525 J=208,525 JDetermine Electric Power: The ideal power (Pideal ) is the total work divided by the time taken.Pideal =W/t=208,525 J/1800 s≈115.85 WGiven the motor's efficiency (η) is 60% (or 0.60), the electric power (Pelectric ) is the ideal power divided by the efficiency.Pelectric= Pideal /η=115.85 W/0.60≈193.08 W