Bernoulli's Equation: Deriving the Formula

[jderulo]

Hi

Can anyone advise how the following equation was derived.

http://uploadpie.com/PYLrD

 

[Mister T]

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

 

[jderulo]

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

 

[boneh3ad]

Have you attempted this yourself yet?

 

[jderulo]

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

 

[Chestermiller]

Hey Boneh3ad,

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

 

[jderulo]

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

 

[Chestermiller]

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

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

chet

 

[boneh3ad]

Hey Boneh3ad,

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

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.

 

[Chestermiller]

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.

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

Chet

 

[boneh3ad]

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

Chet

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 $Q$.

 

[256bits]

except the expression describes Q/w instead of just Q .

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.

 

[boneh3ad]

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.

It may be but it specifically says volumetric flow rate and leaves it out. You could certainly assume unit width but you'd have to say that in order for the units to make sense. It's probably just one of those things that the author overlooked as obvious but would have confused me as an undergraduate.