Exit Pipe Loss/Pressure in Pump System 1

Hi tmo42

I am trying to understand your system layout, I think your system is like this:-

the pump feeds the tank and maintains a certain level within the tank, the tank then feeds down to the outlet pipe
and the head in the tank maintains the flow , am I correct?.
If I am correct can you please provide more information about the system ie pipe sizes and lengths, relative height
of tank to pump outlet etc, what fluid are you pumping and what flow rate do you require and what pressures do you have for the system already.
I was thinking a Bernoulli's equation analysis may be more appropriate for your needs.
If you can provide the above information I may be able to help you further.

regards

desertfox

Hmm let's see if I can clarify. My main concerns are how to model the tank, and what to do about the exit loss.

The tank is a gas bladder accumulator with a certain precharge pressure, and a initial stored pressure. There's one inlet to the tank. The tank is connected off the branch line of a tee.

During startup, the flow out from the tank provides flow until the pump discharge provides enough pressure to cause flow (Assumed a check valve is that line.), at which point, once the pump's discharge flow is greater than the output flow, the tank begins to fill again (At steady state, the discharge flow and output flow would naturally then be equal).

It's basically a system found in a residential house to provide water, only at this step I'm modelling a general version, with only one path from the pump to the outlet (which could be a sink, shower head, etc).

Right now, I'm simulating 1" pipe from pump to the tee, and 3/4" pipe for the rest of the system. I'm using custom-made blocks in the Simulink/Matlab package made by The MathWorks.

Currently I'm using a non-iterative approximation to the Colebrook-White eq. to find friction factor, finding head loss as h=[f*L/D+sum(Minor Losse coeffs)] * V^2/(2g). I'm using P=I*dQ/dt for fluid inertia, and Q=C*dP/dt for the capacitance of the tank.

At time 0 of the simulation, it is assumed that a valve at the output is immediately opened to a certain loss coefficient, and the system discharges into the air.

The eventual goal is to design a controller to keep the system pressure constant.

I'm confused as to how to determine the pressure at the outlet. I've been treating it as 1 atm or 0 psig and finding the pressure drop from the tee to outlet to be tank initial pressure - zero, then calculating flow due to that pressure differencew and working backwards to solve for each part of the system. But this causes there to be 1 or 2 psi in the pipe immediately before the outlet. This method of thinking doesn't seem to be correct to me, so I was wondering how I should model the system at its at the outlet (again, which would typically be a sink, howerhead, garden hose, etc) with respect to output pressure.

Hopefully that wasn't too longwinded, and that I gave all teh information needed.

Thanks for any help.

Tmo42:

The pressure at the outlet is atmospheric plus velocity pressure. The static pressure is zero but since there is flow there is velocity pressure. V2/2g. There is no other pressure. There is no pipe from the outlet onward so there is now friction. There is now liquid level above the outlet to put pressure on the outlet, so there is only velocity pressure. See Bernoulli's Therom. There are four terms, Elevation (static head), velocity pressure, friction loss and pressure head (in this case atmospheric pressure which is zero).

The pressure at the outlet will be atmospheric plus velocity pressure. The excess pressure you are seeing is the velocity pressure.

At start up the pressure in the tank in the motive force for flow. When the valve opens the flow will be at what ever quantity will produce a pressure drop equal in the pipe to the pressure in the tank minus velocity pressure. Some of the pressure energy is changed to velocity and the rest of the pressure energy is dissipated in friction loss in the pipe.

When the pump starts the flow from the pump will be added to the flow from the tank as the pressure and flow from the pump builds the flow from the tank will diminish until the pump pressure overcomes the pressure in the tank. At this point the pressure from the pump will force flow back into the tank until the pressure in the tank and the pump pressure are equal.

Your methodology is correct. Since you know the pressure in the tank the flow will be what ever produces that pressure drop minus the velocity pressure. Don't forget that some of the pressure is used to accelerate the water from zero velocity to the velocity in the pipe. This takes some of the pressure energy so all of it is not available to overcome the friction in the pipe.

The system pressure cannot be kept constant with this setup! Since you are using the stored energy in the tank the pressure in the tank will go down as soon as you open the outlet valve. The flow will also go down.

If you need the pressure at the outlet to be constant you can add a self contained pressure control valve like a Cash valve. This valve will vary the total pressure drop in the system and maintain a certain pressure at the outlet. Since the piping system is open to atmosphere (a constant pressure) and the pipe system does not change by maintaining a constant pressure down stream of the pressure control valve the flow will also be constant.

You will need to have some extra pressure in the tank and from the pump to overcome the added pressure drop through the valve.

Hope this helps!!

Thanks a bunch. I think I understand exactly where I was having a problem. I was forgetting about velocity pressure.

For clarification, I didn't mean "system pressure" to be constant, rather I wanted to hold pressure constant at a certain point in the system.

If you or anyone else has something to add, though, I'll be appreciative.

Also, one thing I am concerned with still is how to model the capacitance of a tank. I've been using... C = Vo/(1.4*Po) (Vo = initial air volume, Po=Precharge Pressure.

DLANDISSR:

Bernouli's equation only works within the confines of a vessel conveying flow like a pipe. When the fluid is no longer restrained, there is no component of pressure left. The fluid leaving the pipe has velocity but not as V^2/2g. It only exists a V.

The pressure inside a pipe exists in two forms, pressure head (EGL)and velocity head and the velocity head is a component of the pressure head and can never exceed this value in a system of flows. Fluid flows by the HGL which is the EGL-Velocity head since velocity head does not contribute to the flow in the system.

Please don't mind my response, I take my fluids too seriously some times.

BobPE

BobPE:
Thanks BobPE. I agree. Perhaps I did not say is just right. When I said there is only v^2/2g when the flow leaves the pipe I am saying this occurs at the end of the pipe when the length is infinitely small. Out side the pipe it is just V as you said.

tmo42: I don't have any data on how to model the tank Capacitance.

I've got the block diagram down for the most part. I had just needed information about the discharge of a system. I'm squared away now, except for the capcitance of the tank which I am not sure about. I ahve another thread that is covering that, however.

Thanks