Calculate input power from the data

[asnali4u]

An electric motor with an efficiency of 78% drives water pump. What input power is required of the motor (in kW) to pump 22250 gals of the water from the lake to a storage tank 65 feet above the lake surface in 4 hours?
Attempt :
i know that efficiency = output/input
but i don't understand, how to calculate the output through this data

 

[LawrenceC]

We cannot assist without first seeing your attempt at a solution.

 

[LawrenceC]

Hint: Power is work done per unit time. Does that help?

 

[asnali4u]

Hint: Power is work done per unit time. Does that help?

Output = power = work/time
Here work will be force x distance?
To be honest i am not familiar with the unit "gal". My lecturer did not tell us about this.

 

[LawrenceC]

Gal is short for gallon. A gallon is exactly 231 in^3 in volume.

 

[asnali4u]

Gal is short for gallon. A gallon is exactly 231 in^3 in volume.

so how can i use volume or gal in the output equation?

 

[LawrenceC]

Gallons can be converted to cubic feet. Cubic feet of water can be converted to weight knowing the density of water in pounds/ft^3. If you know the weight and the height the water is lifted, you have the work done (ft-lbs). Factor in efficiency and time and come up with the answer you seek. You'll need to do some unit conversion to get to KW but that's easy.

 

[asnali4u]

Gallons can be converted to cubic feet. Cubic feet of water can be converted to weight knowing the density of water in pounds/ft^3. If you know the weight and the height the water is lifted, you have the work done (ft-lbs). Factor in efficiency and time and come up with the answer you seek. You'll need to do some unit conversion to get to KW but that's easy.

Thanks for the help. I will try this.