Calculating Water Pump Efficiency and Work Rate

[p.mather]

Hello,

Was wondering if anybody could help with the following question as I am struggling, and this area is not my strong point. Note this is not homework purely self-study, out of curiosity to help me understand the subject area in more detail.

In each minute a pump draws 2.4m^3 of water from a well 5m below ground, and issues it at ground level through a pipe of cross sectional area 50cm^2. Find a) the speed at which the water leaves the pipe, b) the rate at which the pump is working. If in fact the pump is only 75% efficient (i.e. 25% of the power is lost in the running of the pump), find the rate at which it must work.

A solution with some commentary would be absolutely brilliant.

Many thanks to whom can help.

 

[androidx219]

(a) volume flow rate = (2.4/60) SI units
should be equal to (AREA*speed) = (50 cm^2)*speed = (0.25*speed) SI units
therefore speed = {2.4/(60*0.25} m/s
by LAW OF CONSERVATION OF VOLUME (MASS, assuming incompressible liquid)

(b) Now for calculating power, take density of water as 1000 Kg/m^3
density * volume gives mass.
so amount of mass pumped per second can be obtained
this multiplied with acceleration due to gravity and height(5m) would give you the amount of power required to pump water.
Divide your result by 0.75 to get actual answer.

 

[p.mather]

Thanks for you help.

so when working a) i seem to get ... 0.16 however i believe the answer is 8m/s

should i have done 2.4/(60*.05)=0.8 however have i messed my units up somewhere?

Thanks again.

 

[androidx219]

ok, that's my usual silly mistakes. i don't know why i squared 50 and made it 0.25. Obviously you are correct.

Thanks

 

[androidx219]

OK, 50 cm^2 = 0.005 m^2 and not 0.05 m^2 . Got your point. The answer is 8 m/s.