# Flow of a river power math question?

courtneywetts

## Homework Statement

The average flow rate in the Niagara River is 6.0 x 10^6 kg/s and the water drops 50 m over Niagara falls. If all this energy could be harnessed to generate hydroelectric efficiency at 90% efficiency, what would be the electric power output?

P=mgh

## The Attempt at a Solution

I did

(6.0 x 10^6 kg/s) (9.8 m/s^2) (50m) (.9)
= 2.65 x 10^9 kg*m^2/s^3

I think I did the math right? But am unsure of the units.
Is this the answer? Or should my answer be in watts?
If so how would I get this number to watts?

## Answers and Replies

Gold Member
• 1 person
Homework Helper
Gold Member
Good exercise in dimensional analysis.

What are the dimensions of each of your factors? (Example: dimension of kg/s = MT-1.)

M = mass
T = time
L = displacement

What is force F dimensionally ? (hint: F = ma).

Then, what is FL?
And finally FLT-1?

Of course, you can dispense with dimensional analysis & just use common sense, knowing that if you stick to the SI system you will wind up with Joules for energy, Watts for power, etc. But dimensional analysis is one of the most powerful and most neglected tools in your physics toolbox to check equations term-by-term, answer reality checks, etc. so I advise getting comfy with dimensional analysis.

Dimensional analysis can do much more than that, but that's another story.

Homework Helper
Gold Member
P=mgh

What is P? If P=Power then this is wrong.

Energy = mgh

Power (in Watts) is the rate of change of energy eg

Power = Δenergy (in Joules) / ΔTime (in seconds)

In other words

Power = mgh/t

However the problem gives the flow rate Fr (= 6.0 x 10^6 kg/s) which is equivalent to m/t. So your starting equation should

Power = Frgh

or to be complete

Power = Frgh * efficiency

As it happens that's actually what you calculated when you wrote..

(6.0 x 10^6 kg/s) (9.8 m/s^2) (50m) (.9)

You have the right answer but just need to understand how you got there :-)