Question

no wrong ans or I'll report

Answer 1

$\huge\green{ASSALAMUALAIKUM!}$


Mass rate= 500g/min.
=8.33g/sec.

Height to wich the water need to be raised
=2.5m

acceleration due to gravity =9.8m/s^

POWER= mass rate*gravity*height

8.33*10 to power minus 3*9.8*2.5

power= .204 W

in horse power

power=.000273 hp

hence the power needed to lift the water

.000273 hp.

273/100000= 273*10 to power minus six

2.73*10 to the power minus four...

$\huge\orange{INSHAALLAH it will help you!}$

Answer 2

ANSWER.

FORMULA USED



$water \: house \: power = \frac{tdh \times q \times sg}{3960}$


TDH= Total dyinamic head (height)



q= Flow rate


sg = Specific. gravity



GIVEN



$q = 500 \frac{g}{m}$


Sg = Specific gravity of water is 1


1meter=3.28ft

TDH=2.5m= 2.5×3.28ft=8.20ft



so, now use in formula



$water \: house \: power = \frac{tdh \times q \times sg}{3960}$




$water \: house \: power = \frac{8.20ft\times 500g\times 1}{3960}$





$water \: house \: power = \frac{4100}{3960}$






$water \: \: house \: \: power = 1.035$




$minimum \: \: \: house \: power \: required \: to \: run \: the \: pump = \frac{whp}{pump \: efficiency}$



AS WE KNOW PUMP Efficiency (n)


0<n<1



so let efficiency of water pump = 0.44


then




$minimum \: \: \: house \: power \: required \: to \: run \: the \: pump =$



$= \frac{whp}{efficiency \: of \: pump \: }$





$= \frac{1.0355}{0.44} = 2.53 \: hp$



HENCE


Minimum house lower required to pump is 2.53 hp


IMPORTANT POINT



ANS WILL BE DIFFERENT IF PUMP EFFICIENCY CHANGED