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Sewage at a certain pumping station is raised vertically by 5.49 m at the rate of 1 890 000 liters each day. The sewage, of density 1 050 kg/m3, enters and leaves the pump at atmospheric pressure and through pipes of equal diameter. (a) Find the output mechanical power of the lift station. (b) Assume an electric motor continuously operating with average power 5.90 kW runs the pump. Find its efficiency.

Respuesta :

Answer:

Given that

h= 5.49 m

Q= 1890000 liter per day

1 day = 24 hr

1 day = 24 x 3600 s

1 day = 86400 s

1 liter = 10⁻³ m³

Q= 1890000 liter per day

Q= 1890000/86400  m³/s

Q=21.87 x 10⁻³ m³/s

The mechanical power P

P = ρ Q g h                          

By putting the values

P = 1050 x 21.87 x 10 x 5.49 x 10⁻³ W

P =  12.6006 KW

Given that

Output power ,P'= 5.9 KW

Efficiency η

η = P'/P

η = 5.9/12.6

η = 0.468

η = =46.8%

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