# Bernoulli's Equation

1. Oct 16, 2015

### jderulo

Hi

Can anyone advise how the following equation was derived.

Last edited by a moderator: May 7, 2017
2. Oct 16, 2015

### Mister T

Before we try to answer your question, you'll have to convince us that this is not a homework assignment.

3. Oct 16, 2015

### jderulo

It is part of a document of examples not part of an assignment that's why I have all the answers

4. Oct 16, 2015

Have you attempted this yourself yet?

5. Oct 18, 2015

### jderulo

Yes but cannot fathom the 1/D wher eit came from

6. Oct 18, 2015

### Staff: Mentor

Have you noticed that the given answer is not dimensionally correct. They left out the width of the channel (if the really mean that Q is the volumetric flow rate).

Chet

7. Oct 18, 2015

### jderulo

I took the expression as meaning the velocity - I know it states for Q but it does not multiply by area anywhere.

8. Oct 18, 2015

### Staff: Mentor

$\frac{Q}{wD}=$ velocity at the left of the figure, where w is the width of the channel. So the area is wD.

chet

9. Oct 18, 2015

Yes. I was able to reproduce the formula from the problem with the added $w$ term included, but I am not 100% convinced that the assumptions used to get there make a whole lot of sense to me at the moment.

10. Oct 18, 2015

### Staff: Mentor

Me neither, if you are referring to the dip in the upper surface.

Chet

11. Oct 18, 2015

I was referring to the fact that you have to assume that the pressure differenc causing acceleration is wholly explained by the slight change in hydrostatic pressure due to $\Delta h$. The more I think about it, though, the more that makes sense. That assumption gives the same answer as the original post, except the expression describes $Q/w$ instead of just $Q$.

12. Oct 18, 2015

### 256bits

I would think a unit width is implied, which makes the expression easier to work with.

Multiply by the whole width to obtain the total flow in the channel.

13. Oct 18, 2015