Find the useful power used in pumping 3425 m³ of water per hour from a well 8 m deep to the surface, supposing 40% of the horsepower during pumping is wasted. What is the horsepower of the engine?
Answers
Answered by
23
Hey buddy,
● Answer- Horsepower = 170 hp
● Explaination-
# Given-
V/t = 3425 m^3/hr = 0.9514 m^3/s
h = 8 m
Power lost = 40%
# Solution-
Output power required here is-
Power = Work done/time
Po = mgh/t
Po = V×ρ×g×h/t
Po = 0.9514×1000×10×8
Po = 76112 W = 76.112 kW
Input power is given by
Pi = Po/η
Pi = 76112/60%
Pi = 126853 W
Horsepower is
P = Pi/746
P = 126853/746
P = 170 hp
Horsepower of the engine is 170 hp.
Hope that helps you...
● Answer- Horsepower = 170 hp
● Explaination-
# Given-
V/t = 3425 m^3/hr = 0.9514 m^3/s
h = 8 m
Power lost = 40%
# Solution-
Output power required here is-
Power = Work done/time
Po = mgh/t
Po = V×ρ×g×h/t
Po = 0.9514×1000×10×8
Po = 76112 W = 76.112 kW
Input power is given by
Pi = Po/η
Pi = 76112/60%
Pi = 126853 W
Horsepower is
P = Pi/746
P = 126853/746
P = 170 hp
Horsepower of the engine is 170 hp.
Hope that helps you...
Answered by
10
HEYA MATE HERE U GO.
Volume of water lifted , V = 3425 m^3
Density of water d = 1000kg/m^3
therefire mass of the water lifted , m = v × d = 3425× 10^3.
Height through which water was raised (h) = 8 m
Acceleration due to gravity (g) = 9.8 ms^-2
work done by the puml in one hour = mgh
= 3425 × 10^3 × 9.8 × 8 = 268520 × 10^3J
output power of the pump = workdone /time = 268520×10^3 J /3600s =74588.9 watt
output power of the pump in H.P = 74588.9/3600× 746 = 99.99H.P
Efficiency (n) Output/Input ==> 60/100 = 99.99/input power == 99.99/0.6 = 166.6 H.p
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