Thrust at pipe outlet

[PhilOos]

Dear All,

I am in the process of designing puprging pipe work for a power station. The piping blows down to atmosphere and I need to calculate the thrust at the outlet of the pipe.

I have an equation of the form
Force = density * VolumeFlow^2 * (1/Area_outlet - 1/Area_inlet)

The thing is, the exit velocity right before the exit is 436 [m/s] and at the outlet plane 625.3 [m/s] through as 0.387 [m] pipe. The figure I had in mind was about 20 tons, but the calcs show 9861 [N] force.

If I use the equation
Force = density * Volume_flow * Velocity
I get an answer of 22713.84 [N], which is a figure I can live with.

I don't know what I've missed...is the first equation applicable to the fluid or the pipe body? Anyway...I think I just answered myself.



Philip Oosthuizen
Company info:
SteinMuller Engineering Services

http://www.steinmuller.co.za/

 

[katmar]

Philip, I must start off with a disclaimer. I am a chemical engineer and I have been informed by some of the mechanical guys in the past that I should stick to the hydraulics side of piping. I am sure that once some of the mechanical engineers on the other side of the world wake up they will have some more advice for you, but in the meantime here is a fellow South African's take on the subject.

There must be two aspects to this problem. As the valve opens and the flow is rapidly increasing you have a dynamic situation that I have no idea how to calculate.

The second aspect will be the steady state situation, which it seems you were modeling with various equations. In my opinion, once you are at steady state you can easily calculate the force at the exit plane if you know the pressure at that plane because force = pressure x area. The pressure can only be the velocity head of the leaving steam - I cannot see anything else that would cause pressure here.

In pressure terms the velocity head is given by
pressure = density * velocity[sup]2[/sup] / 2

where pressure is in Pascal
density is in kg/m[sup]3[/sup]
velocity is in m/s

Guessing a density of 1.3 kg/m[sup]3[/sup] and using your velocity of 625.3 m/s and pipe diameter of 0.387 m gives a pressure of 254150 Pascal. If this pressure gives a density very different from the guessed value, you would have to guess a new value and try again. The area of the exit is 0.1176 m[sup]2[/sup], giving a force of 29895 N or about 3 tonne force.

I cannot reconcile this calculation back to either of the formulas you had. I look forward to seeing what the real experts come up with.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

It (OP) looks like a formula for calculating axial thrust at a reducer-expander due to the change in cross-sectional area.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[4Pipes]

If the flow is choked at the end of the pipe, you will need to include add a pressure x area term. The pressure being the pressure immediately upstream on the pipe side of the shock wave that you will get for choked flow.
Have a look at ASME B31.1 Appendix II for relief valves. The same principles should apply.

I am not sure what you mean by the 2 velocities. I don't know what the gas but the second looks more than M=1. This could only occur if you put a pipe reducer on the end of the pipe to increase its diameter with the small diameter choked.
The problem of putting a reducer any where near the end of the pipe to increase the pipe size is that you won't know how effective it is at increasing the velocity above M=1. I have never seen figues for an off the shelf pipe reducer.

I would guess you'd have to assume it was perfect to be safe. My suspicion is that you would be better having a constant diameter pipe straight through with several diameters of straight pipe at the end. The last few feet (few yards?) are where you get the high flow velocities if you look at compressible flow calcs. This will avoid high proportions of mach number at elbows which can only lead to more vibration.

You are not going to increase the mass flow by expanding the pipe near its exit but could increase the velocity, hence thrust and noise.

If there is any sudden loading, say using a quick opening valve to maximise the velocity for dirt blowing, you will also need to add a dynamic load factor. B31.1 uses a maximum of 2 for relief valves. Personally, I wouldn,t use less. Even though you won't have a flow change as sudden as a relief valve, I'd be looking for more to be comforable.

I was on a project many years ago which included blowing 800 m of 18" oxygen pipe by pressurising the 450 psig 40' storage bullets and letting rip. Very exiting, noisy, definitely choked as seen from the multiple shock waves. It definitly rattled around even in nominally steady state. Just like an F16 on take off but without the flame. You definitey don't want to skimp on the design. Its got to be agricultural.

 

[gcomyn]

Why does the velocity change? Is there a change in area of the outlet pipe?

 

[BigInch]

If that's something you might be doing often, consider installing silencers, as they reduce the velocity and also the noise (which will be horrendous at that velocity!!!).

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

As noted by 4Pipes (Mechanical), the exit pressure*exit area should be included. Pressure should be gage since the atmosphere is also pushing on the open pipe.

For unsteady flow one other term should be included. This is the Wave Force
= d/dt of the integral from 0 to L of (rho*A* V) dz

Where t is time, rho the density , A area, V velocity

rho, A and V are functions of the position z along a pipe of length L.

Regards

 

[chicopee]

figure out if velocity is choked:
If Pabs.up stream/ P abs.ambient >= ((K+1)/2)^(k/(k+1)) where k is specific heat ratio then flow is choked,
then Q=CA(gkdP)^1/2 (2/(K+1))^((K+1)/(k+2))
where Q= LB/sec; C=.72(about); A=sq.ft.; g=32.17 ft/sec^2;
d= lb/ft^3; P= absolute upstream pressure LB/sq.ft.
If non-choked then
Q=CA(2gdP)^.5 *(K/(K+1))^.5 *((PA/P)^(2/K) -(PA/P)^ ((K+1)/K))^(1/2))^.5
where PA = abs ambient or downstream pressure LB/sq.ft.

Use thrust equation on pipe F(lbs)=(Q/g)*V
where V=outlet velocity in ft/sec
Conversion to metric is up to you

 

[PhilOos]

Thanks guys! You’re replies were very helpful.

Just to try and clarify the issue around the outlet velocities. The large velocity (625 m/s) at the outlet is the velocity that will arise for the given mass flow rate, based on the pipe outlet temperature of 475 °C and the ambient pressure of 89 kPa.
The actual velocity at the outlet of the pipe is much lower 487 m/s because it is based on the outlet temperature and outlet pressure (392 kPa -> upstream of the shock).
The difference in velocity relates to the densities at the two conditions i.e. 0.2578 kg/m^3 and 1.0214 kg/m^3.

At the end I got a force of 34.6 kN for anyone that was wondering.


Philip Oosthuizen
Company info:
SteinMuller Engineering Services

http://www.steinmuller.co.za/

 

[4Pipes]

PhilOos,

I still don't understand the second higher velocity. How do you calculate it? If the last few metres of pipe is of constant diameter, it is only possible to get M=1 at the end of pipe. If you had an increase in diameter further up the pipe, this might choke and act as a nozzle thus increasing the velocity. However, the velocity would soon reduce back by forming shock waves down stream of the "nozzle" such that M=1 at the end of the following constant diameter section. M>1 can only exist at the end of a parallel pipe - at least that is my understanding.

 

[chicopee]

For exit velocity, I would base my thrust value on an exit velocity of Mach 1.

 

[sailoday28]

PhilOos (Mechanical)I hope you included the added effect of exit pressure*exit area in your computation for thrust.

regards

 

[CostasV]

Excuse me, if I am not corect, but I didn't noticed in the original post anything about the direction of flow. I mean, if it is a straight vertical open-end pipe, then, there is no force acting on the open-end. In this case, a foce is acting on the last bend, before the open-end.

There is, also, no force acting on the open-end, if the open-end is symmetrical as a T or a perforated type of (symmetrical) silencer.

Regards
Costas

 

[rjstephens]

Hi Philip,

Have a chat to Eric van Zyl & Alan Stewart on the subject. API 520 & B31.1 give guide lines as well as the Crocker & King, pipe break & associated thrust loads. CII has a good relief valve calculator in the Dynamic Analysis part of the program.

Cheers

Rob Stephens

(ex-Steinmuller)

 

[hacksaw]

B31.1 Appendix II (thanks 4Pipes) is specific to your question and covers all the contributing factors. Check it out.

 

[sailoday28]

Varsamidis (Mechanical) 11 Dec 06 2:57
Excuse me, if I am not corect, but I didn't noticed in the original post anything about the direction of flow. I mean, if it is a straight vertical open-end pipe, then, there is no force acting on the open-end. ????

Only if the exit pressure is ambient or atmospheric. If flow is choked, then we have another story.