Question

Raw sewage is raised vertically by 5.49m in the amountof

1.89*106 liters each day . if the average density ofthe
sewage is 1.050 kg/m3 , and the waste enters andleaves
the pump through pipes of equal diameter and atmosphericpressure ,
What is the power output of the shift stationpump.

Answer 1

P = 1235.7646 W

Step-by-step explanation:

Given data:

height of raised sewage = 5.49 m

rate of sewage =1.89*10^6 lt/day

density of sewage = 1.050 kg/m^3

power is written as

`P = (work)/(time)`

work = m g h

`m - mass = ( \rho V)`

h -height

mass lifted per day
`= ( 1.89*10^6)(1.050)`

= 1984500 kg

time = 24 hours* 3600 seconds per hour

power
`P = (( 1984500)(9.8)( 5.49))/((24)(3600))`

P = 1235.7646 W