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Grade: 12th pass

                        

A water pumps lifts water from a level 10m below the ground .water is pumped at rate of 30 kg/min with negligible velocity.calculate the minimum horsepower the engine should have to do thisthis.

4 years ago

Answers : (1)

Vikas TU
12280 Points
							
Since mass is leaving ccontinuosly its a variable mass system problem and hence,
F = dP/dt eqn. can be used 
where P is momentum.
Hence,
F = d(mv)/dt = > vdm/dt + mdv/dt => vdm/dt +  ma
since dm/dt = 30 kg/min but velocity is negligible that is v = 0.
Hence,
F = m*g
and work would be = F x 10 => 100m.
To cover 10 m distance time taken would be = > root(2) seconds.
Now with rate 30 kg in a minute
it would lift in 1.414 seconds the mass would be = > 30*1.414/60 = >0.707106 kg.
 
Hence work done = > 100 * 0.707106 = > 70.7106 Joule.
Horse Power = > 0.09478 HP.
 
4 years ago
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