Question

A pump is used to lift 500 kg of power from a depth of 80 m in 10 seconds.what is the power rating of the pump if its efficiency is 40%?

Answer 1

Hey!!

Here is your answer :-

As we know that, work = mass * displacement

therefore,    work =) 500 Х 80
                   work =)  400000 J
 

Now, Power = energy/time
           P = 500 Х 10 Х 80/10
           P= 40000 W
           P = 40 kW

Hence, power rating =  power / efficiency                      
                                 = 40/40/100                      
                                 =100 kW

Hope it helps ^_^

Answer 2

Hi,

Please see the attached file!

Thanks