Sewage at a certain pumping station is raised vertically by 5.57 m at the rate of 1 860 000 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 3.00 kW runs the pump. Find its efficiency.
(a) To find the output mechanical power of the lift station, we can use the formula:
Power = (Density of sewage) * (Acceleration due to gravity) * (Volume flow rate) * (Height lifted)
The density of sewage is given as 1,050 kg/m^3, the acceleration due to gravity is approximately 9.8 m/s^2, the volume flow rate is 1,860,000 liters per day, and the height lifted is 5.57 m.
First, let's convert the volume flow rate from liters to cubic meters:
1 liter = 0.001 cubic meters
Volume flow rate = 1,860,000 liters/day * 0.001 cubic meters/liter
Now, we can calculate the power:
Power = (1,050 kg/m^3) * (9.8 m/s^2) * (1,860,000 liters/day * 0.001 cubic meters/liter) * (5.57 m)
Simplifying the units, we have:
Power = (1,050 kg/m^3) * (9.8 m/s^2) * (1.86 m^3/day) * (5.57 m)
Now, we can calculate the power:
Power = (1,050 kg/m^3) * (9.8 m/s^2) * (1.86 m^3/day) * (5.57 m)
= (1,050 kg/m^3) * (9.8 m/s^2) * (1.86 m^3/day) * (5.57 m)
≈ 97,724 W
Therefore, the output mechanical power of the lift station is approximately 97,724 Watts.
(b) To find the efficiency of the electric motor running the pump, we can use the formula:
Efficiency = (Output power / Input power) * 100%
The output power is the mechanical power calculated in part (a), which is approximately 97,724 Watts.
The input power is given as an average of 3.00 kW, which is equivalent to 3,000 Watts.
Now, we can calculate the efficiency:
Efficiency = (97,724 W / 3,000 W) * 100%
≈ 3.25%
Therefore, the efficiency of the electric motor running the pump is approximately 3.25%.