calculate the electric energy consumed by a 2 HP pump if it is used for 2 hours . If the pump is used to fill an overhead tank , which is at a mean height of 10m , find the quantity of water lifted by the pump. (Take g = 10ms–²)
Answers
Given,
Power of pump = 2 HP or 1492 watt { 1 HP = 746 watt }
Time, t = 2 hours
Electrical Energy = Power ( in kW ) × Time ( in Hr )
==> E = 1492 / 1000 × 2
==> Energy consumed by the pump in 2 hours = 2.984 kWh
Also, the motor pumps water at a height of 10 m then the quantity of water lifted is and the electrical energy consumed by the motor is used in lifting the water at the height of 10 m so Potential energy would be same as of the energy consumed for 2 hours.
So,
==> Potential energy = 1492 × 2 × 60
We have to convert time into seconds because Power is energy consumed per second.
==> Potential energy = 179,040 Joules
Also, We know,
Potential energy = mgh
So,
==> m ( quantity of water ) = Potential energy / gh
==> m = 179040 / (10*10) { take, g = 10 m/s² }
==> m = 1790.4 kg
Since, 1 kg = 1 litre
So, 1790.4 litres of water will be lifted.