# Total mechanical energy

1. Aug 21, 2011

### TyErd

1. The problem statement, all variables and given/known data

In the diagram I have attached the initial question is to find the total mechanical energy. however in the solutions which I have placed under the diagram, they don't use the mass (m). So total mechanical energy i thought was = mgh + 0.5m$v^{2}$ but they use gh + 0.5$v^{2}$ . no mass. why

i think its something to do with the question asking to be in per unit mass but i dont see why.

2. Relevant equations

3. The attempt at a solution

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2. Aug 21, 2011

### cepheid

Staff Emeritus
They ask you for the energy per unit mass that is present, and so they divide both sides of the equation by the mass. It's nothing more complicated than that. It's a more useful quantity to consider in this situation than the amount of energy present, which depends upon how big a "parcel" of water you consider.

3. Aug 21, 2011

### cepheid

Staff Emeritus
Think of it this way: every kilogram of water that passes by carries with it 0.887 kJ of mechanical energy. You have the volume flow rate (which is the volume of water that flows by in 1 second). You can use that to determine the mass flow rate (the rate at which water flows by in kg/s). Combine that with your energy per unit mass, and suddenly you have the energy per second being generated (EDIT: or at least the energy per second that could be generated if all of that mechanical energy in the water flow could be transferred to some other system with 100% efficiency -- in a real hydro plant the efficiency is sure to be lower).

4. Aug 21, 2011

### quietrain

in essence,

per unit mass = mass = 1

so mgh = 1gh = gh

same with KE

5. Aug 22, 2011

### TyErd

thankyou guys. helps alot