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#68668 - 03/23/17 07:24 AM Fluid stopping force versus hammer loads
Michael_Fletcher Offline
Member

Registered: 01/29/10
Posts: 1025
Loc: Louisiana, US
Morning all,

Perhaps I'm over-analyzing this, but when you have a long line, let's give it the following parameters:

L = Length
D = Diameter
V = Line Speed
t = time to close the valve
Vs = speed of sound in fluid
a = fluid deceleration time = V/t
f = instantaneous closure multiplier (for a gate valve moving at constant speed, this would be 1.27)
ρ = density of fluid

Assume fluid is nearly incompressible.

Say that Vs*2*L
Surely if the pipe ends up moving as a result of this force, you'll see resistance at all the supports where it's touching, but how much of this force is being applied at the valve versus the supports in the system?

Or perhaps the water hammer calculations already calculate these forces, just in a circumlocutious manner? However, this seems like two different problems.

Hopefully someone sees this before today's webinar and can ask the presenter to touch on this.

Thanks.

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#68673 - 03/23/17 11:30 AM Re: Fluid stopping force versus hammer loads [Re: Michael_Fletcher]
Jim Wilcox Offline
Member

Registered: 12/20/99
Posts: 46
Loc: Calgary, AB, Canada
Hi Michael;

Transient analysis software will calculate the forces imparted by the fluid on the pipe, and when you apply these loads in CAESAR II, it will calculate the resulting loads on restraints & equipment nozzles as well as stresses and deflections...

"...how much of this force is being applied at the valve versus the supports in the system?"

All of it. Considering just the valve, this load shows up at the valve, and gets transmitted through the pipe to the rest of the system.
This is not similar to the scenario where an anchor takes some of a load and restraints take up the rest; these calculated loads are applied by the fluid on the pipe, and CAESAR II calculates the system response to the loads.

Hope this clarifies.
J.
_________________________
J.

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#68674 - 03/23/17 02:00 PM Re: Fluid stopping force versus hammer loads [Re: Michael_Fletcher]
Michael_Fletcher Offline
Member

Registered: 01/29/10
Posts: 1025
Loc: Louisiana, US
So if I understand the situation correctly, if you're attempting to stop a pipe full of incompressible fluid "slowly" you effectively will see Force = mass times deceleration at the valve, minus energy losses in the system, such as through compression of the fluid, expansion of the piping, and losses associated with hydraulic pressure losses, minus the pressure differential across the valve times its area.

In other words (and pulling numbers out of thin air), if I am attempting to stop 10,000,000 kg of water traveling at 1 m/s in a span of 50 seconds, that's 200kN. Which is some 50km at 20" pipe. This is correct?

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#68675 - 03/23/17 09:36 PM Re: Fluid stopping force versus hammer loads [Re: Michael_Fletcher]
Ibrahim Demir Offline
Member

Registered: 01/02/03
Posts: 255
Loc: Australia
Michael,

The attached spreadsheet that I prepared from the reference given might help.

Regards


Attachments
Surge_Rev2.zip (435 downloads)


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#68676 - 03/23/17 09:38 PM Re: Fluid stopping force versus hammer loads [Re: Michael_Fletcher]
Ibrahim Demir Offline
Member

Registered: 01/02/03
Posts: 255
Loc: Australia
Please note that the spreadsheet does not replace the transient analysis software, it is an estimate only.

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#68677 - 03/23/17 11:19 PM Re: Fluid stopping force versus hammer loads [Re: Michael_Fletcher]
Jim Wilcox Offline
Member

Registered: 12/20/99
Posts: 46
Loc: Calgary, AB, Canada
Hi Michael;

Slowly or quickly, your comment is generally correct, although the radial expansion of the pipe isn't a part of the software calculations (other than its consideration in how it affects wavespeed), and the fluid is considered incompressible.
In my own struggles to visualize this in the past, I developed an analogy of a train crashing into a wall. The lead car sees the load of all the cars behind it, and therefore sees the highest load; the last train car sees the smallest load (neglecting the reflection and all that). To translate this into a fluid scenario, the fluid at the face of the valve sees the highest pressure, as all the fluid behind it accelerates from V to zero. Now imagine a pipe full of liquid traveling at 9.81 m/s, which is then slowed to zero in one second. This is 1g, and the pressure distribution is exactly the same as a static vertical column of the same liquid of the same length (height); it also experiences a pressure from the acceleration of the fluid "behind" it. rho*g*h. As Einstein said: acceleration has the same effects whether the acceleration is caused by motion or by gravity.
This also supports what you say; the longer the column of liquid, the greater the effects of the acceleration.
I've seen a pressure chart from upstream of a 24" mainline oil pipeline that closed over about 8 minutes. Steady-state pressure was about 1800 kPa, and for about 4 minutes AFTER the valve was fully closed, the pressure increased to about 3700 kPa before a relief valve opened to prevent rupture. A very long column indeed, corresponding to my analogy of a very long vertical column, which kept getting longer as the stoppage in flow made its way down the line.
Interesting to see it in the software, but very interesting to see the real pressure-time data from a real system.
One complication of our moving fluid systems of course is that these pressures become dynamic, and they are like slinkys loading and unloading at each end of the line as the pressures travel up and down the line. The reflection part doesn't really come into play in this pipeline scenario, but it does in shorter piping as in my example.
_________________________
J.

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