# Homework Help: Derivation of pressure differences in a stream.

1. Oct 24, 2012

### chingel

1. The problem statement, all variables and given/known data
I tried to reach the Bernoulli principle this way:
Two pipes are connected, one has a cross-sectional area of $S_{1}$ and speed of $v_{1}$; $S_{2}$ and $v_{2}$ for the other. The pipes are horizontal, the connecting wall between them at the crossing from one pipe to the other is vertical, water flows from the first pipe to the second, the first one is larger.

3. The attempt at a solution
In any second, a mass of fluid $m=S_{2}v_{2}\rho=S_{1}v_{1}\rho$ has gone from the larger pipe to the smaller and its velocity has changed by $dv=v_{2}-v_{1}$. Therefore the change in momentum is $F=dmv=S_{2}v_{2}\rho(v_{2}-v_{1})$. This force comes from the differences in pressure between the two regions and since the size of the connection between them is $S_{2}$, because that is the area of the smaller pipe, then the difference in force at the connection is $dF=S_{2}(P_{1}-P_{2})$. Equating the two forces and doing some simplifying, I get that $\frac{P_{1}}{\rho}+v_{1}v_{2}=\frac{P_{2}}{\rho}+v_{2}^{2}$, which is clearly the wrong answer.

In Wikipedia, there is the derivation using conservation of energy:

http://en.wikipedia.org/wiki/Bernoulli's_principle#Derivations_of_Bernoulli_equation

However, I would like to know where did I go wrong and why did it give the wrong answer.

2. Oct 24, 2012

### haruspex

I think your problem is that you are assuming the pressure would be uniform in each section of pipe. The calculation, if valid at all, considers the pressures either side of the change in width, as though the pressure change were sudden. In practice, the pressure would change over some volume approaching the boundary.
Bernoulli's equation is derived for a different scenario. Flows are laminar, and the cross-section changes slowly enough that the pressure at some point along the pipe is meaningful.
You could try to rescue your approach by considering the area change to be just small delta. I notice that the difference between the equation you get and Bernoulli's is that between a factor of v2 and one of (v1+v2)/2. That might disappear as a second order small quantity.

3. Oct 25, 2012

### chingel

Yes good idea, if the force due to the pressure differences were not $F=S_{2}/(P_{1}-P_{2})$, but instead it were $F=S_{2}/(P_{1}-P_{2})*\frac{2v_{2}}{v_{1}+v_{2}}$ then I would get the right answer. But since the extra term is very near one when the speeds are similar, I consider only small changes in v and integrate. Not that $F=S_{2}v_{2}\rho(v_{2}-v_{1})=S_{2}(P_{1}-P_{2})$ as previously, but instead $v_{2}\rho*dv=-dP$, having divided the S2 out. This gives $\frac{dP}{dv}=-v\rho$. Then $P=-\frac{v^2}{2}\rho+C$, from where $P+\frac{v^2}{2}\rho=C$.

It was wrong to assume the force difference is just the connecting area times the pressure difference, because for example how would I know if the force exerted by the perpendicular wall at the connection point is exactly enough to counteract the larger area the pressure in the first pipe acts on. I assumed that the force difference is just the connecting area times pressure difference, thinking that the extra cross-sectional area the pressure in the larger pipe acts on is exactly counteracted by the perpendicular walls, which might not be the case because the flow there is probably different than further away from the connection.

But now I wonder why exactly is it right for small changes in diameter that the force is the connection size times the pressure difference, if it isn't right for larger changes?

4. Oct 28, 2012

### haruspex

Because you localised the area change right at the constriction, while the pressure change in reality would not be that localised.