Reactive Force for a Piping Discharge 2

[sshep]

Friends,

What is the formula for calculating the reactive force from a liquid discharge? I am expecting a function of density, flow, velocity squared, etc and am asking for some quick help.

This is the story: I come back from holiday to find a piping wreck at one of our cooling towers. The story I get is that a pump was started using a "minimum flow line" back to the tower. This is a 12" line without any restriction running from a pump discharge (rated at 3050 cubic meters/minute) and about 4 or 5 meters up the side of the tower. It has two horizontal bends at the top that seem to match a reactive twisting action of the piping wreckage. The story that I got from the RCA team was that some sort of a hammer occured, but I think the answer is much simpler. The team did not consider a reactive force acting on a long moment arm. I made an inquiry to the manager as to whether the piping anchor at the top of the line might have failed, and was surprised to be told that the piping was not anchored. There was supposed to be a U-bolt anchor, but apparently it was never installed. This makes me even more suspicious of a reactive force causing the piping wreck. Anyway, this is the story, so any help with a formula or comment is appreciated. If I calculate the force, I can get a mechanical engineer to comment on whether my theory is plausible.

best wishes always,
sshep

 

[Compositepro]

Review a basic physics book (you can find this on the web, too). F=ma a=dv/dt F= d(mv)/dt F= v(dm/dt)

F=force m=mass a=acceleration d=derivative(change in)

ma= momentum

At elbows force is caused by acceleration.

At nozzles (jets) force is due to the change in momentum.

 

[sshep]

I had really hoped for something a little more useful than force = mass * acceleration, but thanks anyway.

Per API520 relief pipe reaction force calculation:
F=W^2/(12.96*A)*(x/rhog+(1-x)/rhol)+A/1000*(Pe-Pa)
where
W=flow, kg/hr
A= pipe exit area, mm^2
x= vapor mass frac
rhog, rhol= density gas and liquid, kg/m^3
Pe= pressure at exit, kPaa
Pa= ambient pressure, kPaa

So it would be a simple plug and chug, except for the exit pressure (Pe). How is this calculated?

You would think I should know that sort of detail after 25years, except that in most problems I am solving for flow using an exit pressure from the pipe equal to atmospheric pressure- i.e. Pe=Pa and the only residual energy is kinetic. The only cases where it has ever come up for me is when the flow is at a sonic limit (choked flow), and even then I am usually left wondering what exit pressure is reasonable. Is that what the (Pe-Pa) term means, and if so how do I calculate Pe? Any help is appreciated.

best wishes,
sshep

 

[Latexman]

Pe is usually 1 velocity head more than atmospheric pressure. K_{exit loss} = 1.0

Good luck,
Latexman

 

[katmar]

sshep, I think you would be more likely to get the answer in the Pipelines, Piping and Fluid Mechanics forum but in the meantime I would make a first approximation using Latexman's approach. If you have a typical cooling water discharge pressure of around 300 kPag you could get a flow of 74 m³/minute and a velocity of 17 m/s. The velocity head would be about 140 kPa. If this is the pressure in the plane through the open discharge, then applying it to the pipe cross sectional area gives a force of 10 kN or about 1000 kgf. My instinct tells me that hanging 1000 kg off a 6 m long piece of 12" pipe would bend it. Let's see what the mechanical guys in the Pipelines forum come up with.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sshep]

Thanks! I get a whopping reactive force from both terms in the equation using "only" the rated flow of 3050cum/h, which is 11.6m/s, but the flow could be much higher given rated discharge pressure of 550kPad. The pump probably tripped on high current even as the piping was wrecking itself.

Katmar, I am ending my day and don't normally work in metric units. How do you convert from velocity to kPa- i.e. from velocity head to pressure?

best wishes,
Sean

 

[ione]

Velocity head: v^2/2g

with v in m/s and g = 9.81 m/s^2. Then convert from metres of H20 to kPa. You can use this valid free converter

http://www.katmarsoftware.com/uconeer.htm

 

[rconner]

Wow, 17 m/sec sounds like a pretty high flow velocity in general for a pumped water line (just curious, do you know why this was apparently designed or operated like this?)

 

[hacksaw]

Starting any large pump under flowing conditions is sure to cause problems even if it is a bypass line and properly supported.

Take a look at your operations manual