# Fluid Mechanics

1. Mar 17, 2009

### Jason03

I have completed the first part of the problem below. Im checking to make sure that I have the right thought process...

Also the next part of the question asks if the inlet pressure to the pump is -2.36 psi compute the power delivered by the pump to the water.....How would I approach that?

http://i674.photobucket.com/albums/vv106/jason03_2009/question.jpg [Broken]

Last edited by a moderator: May 4, 2017
2. Mar 19, 2009

### minger

The power delivered by the pump is a time rate input of energy. You can approach this by looking at the energy equation in rate form.
$$\frac{d}{dt}\left[m\left(u + \frac{V^2}{2}+gz\right)\right]_V = \sum \dot{m}_{in} \left( h + gz + \frac{V^2}{2}\right)_{in} - \sum \dot{m}_{out} \left( h + gz + \frac{V^2}{2}\right)_{out} + \dot{W}_V - \dot{Q}_V$$
Where the V subscript denotes over the control volume. For steady state operation, there can be no time rate of change of mass or energy, so the term on the left disappears. Also, for steady-state operation, you can assume steady adiabatic flow, so:
$$\dot{Q} = \frac{dQ}{dt} = 0$$
We can then substitute
$$h = u + P\nu$$
into the equations. Assume that the change in internal energy is negligible and this should get you nearly there.

edit: They are hard to see, but make sure you notice that those are mass flows on the right-hand side.

3. Mar 19, 2009

### Jason03

Ok....I did some calculations below using the formulas from my textbook which makes it slighlty more simplified.....but is my thought process ok?

http://i674.photobucket.com/albums/vv106/jason03_2009/part2.jpg [Broken]

Last edited by a moderator: May 4, 2017
4. Mar 19, 2009

### Jason03

Last edited: Mar 19, 2009