An electric motor of 1hp is used to run water pump.the water pump takes 10 min to fill an overhead tank.the tank has a capicity of 800 litres and height 0f 15 meter .find the actual work done by electric motor to fill the tank.also find the efficiency of the system.
Answers
The efficiency of the system is 52.55%.
We know power = Work done/Time = mgh/t
Volume of water = 800L
Density of water = 1kg/L
So, mass (m) of water = volume × density = (800 × 1)kg = 800 kg
Acc. due to gravity 'g' = 9.8 m/s²
Height of tank from ground 'h' = 15m
Time taken to fill (t) = (10 × 60)s = 600s [1 min = 60s]
∴ Output Power = (800 × 9.8 × 15)/300 = 392 W
Work done = mgh = (800 × 9.8 × 15)J = 117600J
We know, 1 HP = 746W = Input power
So, efficiency = [(Output power/Input power) × 100]%
= [392/746 × 100]% = 52.55%
Power of pump, P = 1 hp = 746 W
time, t = 10 min = (10*60) s = 600 s
height, h = 15 m
mass, m = 800 kg (800 L; 1 L = 1 kg)
∴ Input Energy = P x t = 746 x 600 = 447600 J
Now, output energy = work done to lift water
∴ output energy or actual work done = mgh = 800 x 10 x 15 = 120000 J
So, now,
Efficiency = {Output Energy/Input Energy}*100
= {120000/447600}*100
∴ Efficiency = 26.8 %