how do I find max. flow through 3" pipe. 2

[AndrewUK]

Hi I'm new here and was after some help.

Been a while since I've done any fluid flow.

I have a 3" pipe that falls 6" into a sump for a tank, the connection has various bends. The sump is then pumped up back into the tank.
My problem is I think the pumps are too powerful for the return flow. So I need to find out the maximum flow due to gravity through the
3" pipe. How do I do this.

 

[quark]

You may find these discussions helpful.

thread378-119269
thread378-119917
thread378-116505

Regards,

 

[AndrewUK]

cheers for that it showed my flowed to be the same as I had calc. i.e. 10.9 L/s

I used the link to take into account the pipe lengths and fittings from the tank exit to final exit into the sump.

It shows the pressure drops, but how should I calc the fluid flow drop?

I'm trying to balance the system out with the pumps but unitl my ultr sonic flow meter arrives I trying to match thr pumps into the tank with the exit flow.

 

[TD2K]

What do you mean the fluid flow drop?

I realize you included the pipe fittings in getting your answer but I also looked at what the maximum flow of water I could get through a 3" sch 40 pipe draining due to gravity. The maximum flow is where the static head is all being consumed by the frictional losses which I calculated is about 700 gpm or about 45 L/sec.

 

[AndrewUK]

The head is a weir feed from a 5000 litre tank.

The weir maintains a head of about 12" from the centre of the exit pipe. The exit pipe being 3" diameter.

If this flow due to gravity is say 45L/sec then how will this change when it flows through 5m of horizontal pipe work and 0.5 meter of vertical pipe with 7 (90degrees ) conrners and a ball type valve. Am I right in thinking this pipe work and valve will restrict the flow and therfore the exit flow will be less than calculated above.

 

[quark]

You might have missed my post dated 26th in the thread378-116505. What you should do is to calculate the flowrate w.r.to 12" head and then calculate frictional losses with the flowrate. Now your actual head that is causing flow will be the initial static head - frictional losses. Go on doing it till you get the require flowrate and corresponding resistance. You have to do some trial and error(or iteration).

Regards,

 

[AndrewUK]

cheers
from the 1st spread sheet i get. densisty being 1.1

Exit Velocity, V: 2.45 m/s
Volume Flowrate: 10.9 l/s
Mass Flowrate: 12.0 kg/s

this give me my inital static head conditions, I've entered this into the 2nd spreed sheet and this gives me a mass flow rate of 719kg/hr = 10.89 l/s doesn't seem a big drop!?

Thats with equ. of 22.4 m of pipe and a 0.02 height change.

I missing something here but can't work out what. I would have thought that the flow rate would reduce when the pipe work was introduced.

 

[AndrewUK]

went through some of my texts last night.

Answered my own question I belive, after reading up on the Continuity Equation.

That is the flow leaving the header tank which enters the pipe work and travels through the valves will be the same as the flow that exits.

The only changes would be pressure and velocity.

Think I need to refresh myself on fuild props.

Can anyone confirm this?

 

[quark]

No, not exactly. I fear the more we explain the more you will be confused. Still, I try my bit. Assume that you connected a pipe of 12" length vertically upwards to the outlet nozzle of the tank. Now flow in the pipe will be zero as the head available is consumed to overcome overall resistance in the pipe. Now convert the vertical 12" piping to a horizontal piping which has a frictional drop of 12". You will not have flow in this case also.

The maximum flowrate in this setup will be what you get at the tank outlet nozzle(assuming no piping is connected) and minimum flowrate will be zero. Your actual flowrate lies somewhere in between the maximum flowrate and zero. This is not an easy calculation and you have to do some trial and error methods. Your vertical pipe of 0.5m makes the procedure much difficult.

What you have to do is to check flowrate assuming 12" head available at the tank and get the flowrate. Based on the flowrate check the pressure drop in the pipeline. Now this resistance is overcome by the static head of fluid in the tank and the 0.5meters of downward piping(which equals 19.685inches). If the resistance is morethan (12+19.685)inches then reduce the static head by 0.5 inches and do the calculation again. Suppose at x"static head of liquid in the tank you get y" of pressure drop in the pipeline and if 19.685+(12-x) = y then stop the calculation and you got your flowrate. With 0.98 Cd(which is too high), you may get around 34056L/hr with a liquid column height of 9". Frictional loss is 21.7192", liquid column height is 9" and 0.5m pipe equals 19.685". So [19.685+(12-9)] = 22.685", which is close to 21.7192". You can get better value if you try further.

Regards,

 

[quark]

Still worse, my above calculation is true if and only if you provide 12" constant liquid column height in the tank. Otherwise, it varies as the liquid level goes down and the average flowrate will be half that of what I calculated.