- #26

Chestermiller

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Yes I know but would still like to know what is wrong with my reasoning. It's simply finding p1 by Bernoulli from just to the left of the S1/S2 junction to the outlet, Bernoulli again from the top of the tank to the outlet, then saying F = p1(S1 - S2).

Hi Rude Man,

I think I can explain it to your satisfaction. Let me walk it through. I agree that the force F is solely that acting on the upstream side of the junction, and that the area of the upstream side is (S

_{1}-S

_{2}). You made the approximation that the pressure on the upstream side of the junction is P

_{1}, and, on that basis, you calculated that the force as 8N, rather than the 6N determined exactly (for an ideal fluid) from the overall momentum balance. Therefore, your estimate of the average pressure on the upstream junction surface must have been too high. The average pressure on the junction must be lower than P

_{1}. But, why?

Well, in order for the flow to get into the smaller S

_{2}section, the velocity profile no longer remains flat, and, in addition, the flow develops a radial velocity component in order to allow convergence into the smaller channel. As a result, the average vector velocity is higher in the region immediately approaching the junction than further upstream. From Bernoulli, this results in a lower pressure. Another way of looking at this is to consider the very outer streamline. The pressure variation along this streamline determines the pressure distribution on the junction surface. The particles that travel along this streamline have to travel a larger distance than particles that travel down the center of the channel. But they catch up with the particles moving down the center of the channel, so they must be going faster. Again, this results in reduced pressure. The net effect is that, in the problem considered, the average pressure on the junction surface is only about 3/4 P

_{1}.

It think that, to flesh this out more quantitatively, we would have to solve the detailed pde's.

Chet