1. Feb 1, 2010

### Marbig

Hi all,

I got the system (attached below) and I am trying to work out the following (at point x-outflow):

-net head of the system (I guess this isnt really at point x)
-flow rate
-pressure

I know the flowrate and pressure at point Y, I also know how high the water in the tank is at equilibrium (relative to point x) and I also know the frictional losses in the pipe. Now the ball on the tank is an equilibrium valve.

I have been told that the tank is a constant head tank, I know that means that it will maintain a constant head during the operation.

So Im a bit stuck on working out flow rate, pressure and net head at point x.

Here is what I think:
The net head of the system is basically the height between the water in the tank and point x

To work out the flow at point-X. I can apply bernoulli's equation between points y and x. I know Py, Vy,Hy...however all I know at x is Hx.

Thank you in advance for any assistance.

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2. Feb 1, 2010

### stewartcs

Assuming the flow is from the tank only...Just choose point 1 to be the surface of the fluid in the tank and point 2 to be location "x". Since the tank has a constant head the the flow rate will be constant at point "x".

The only other assumptions are that the pressure at point 1 (the surface of the fluid in the tank) is 0 psig and the velocity is 0 ft/s (due to the larger surface area as compared to the outlet area). If it is discharging to atmosphere then the pressure at point 2 (location "x") is also 0 psig.

EDIT: Note that if flow is from point "y" as well as the bottom of the tank you'll need to use some type of nodal analysis like Hardy-Cross method.

CS

3. Feb 1, 2010

### Marbig

Is it a safe to assume that the flow will occur only out the bottom of the tank? Does it work on the basis that if the water level in the tank drops the ball valve will move and allow a bit more water into the system from point Y and hence maintain the same head and flow rate at point X?

So if I calculate the flowrate and head at x based on the tank being full, this should be the same as when the water level in the tank has dropped slightly (during operation for example), because the flow from Y would compensate for this?

Last question, the net head that this system provides at X would simply be the height between the water level and point X minus any losses?

4. Feb 2, 2010

### stewartcs

You'll have to tell me how it works since it is your system! Unless of course this is a homework problem in which case the problem statement should be posted (and in the correct sub-forum).

However, it appears to have a parallel flow from what looks to be a three-way valve (i.e. the block at the top of the sketch). It's really hard to tell since those are non-standard drawing symbols. Anyway, without any further knowledge I'd treat it like parallel flow from point y to point x. In other words, one inlet at point y gets split and goes to the inlet of the tank and the outlet of the tank where it recombines.

Again it depends on how the system is supposed to work. I'm interpreting this statement as you saying that the flow is from the bottom of the tank only. So if that is how it is supposed to work (whether by design or problem statement) then yes. However, it doesn't appear that way from the sketch. It looks like a parallel flow to me.

Depends on what you mean by net head. If you mean the total energy in feet of fluid then use the general energy equation (modified Bernoulli that accounts for all energy additions and losses) to find the "total head" at point x.

CS