Fomula estimating pressure increase vs pump rate in fixed volume

[mhldrillpipe]

Gentlemen and laydies

Looking for a formula to predict simple pressure increase within a cylinder as a function of fluid pumped into the cylinder (pump rate).

suspect pressure to be dependent on size of fixed volume, fluid compressibility (oil in this case), and amount of fluid pumped.....(?)

hope you can help

many thanks

 

[BigInch]

Its dependent on the bulk modulus (dP/dV) of the liquid.

Bulk Modulus [PSI] of Water at 15 psia
and the indicated temperatures
32ºF 68ºF 120ºF 200ºF
292000 320000 332000 308000

Initial Cylinder Vol @ P1=15 psia and 60ºF
Vcyl1 (@P1) = [π] * r^2 * 1 in long

Elongation of the Cylinder Hoop e = Sh * [π] * D/E where D = diameter of the cylinder
E = Young's modulus of the cylinder
Sh = hoop stress of the cylinder,
Sh = P2 * diam / 2 / wallthickness
P2 = new pressure

New Diam of Cylinder @ P2, D2 = ([π] * D + e)/[π]
New Cylinder Volume, Vcyl2= [π] * D2^2/4

Balance the change in volume of water with the change in volume of the cylinder,

dVwater = dVcylinder

change in cylinder volume
dVcyl = Vcyl2 - Vcyl1 = [π] * D2^2/4 -[π] * D1^2/4

change in water volume
dVwater = (P2-P1)/ BulkModulus
== change in Cylinder Vol

(P2 - P1) / BulkModulus = (Vcyl2-Vcyl1)= [π] * D2^2/4 -[π] * D1^2/4

P2 = P1 + BulkModulus * ([π] * D2^2/4 -[π] * D1^2/4 )
P2 = P1 + BulkModulus * ([π] * [([π] * D + e)/[π]]^2/4 -[π] * D1^2/4 )

I think that's it. At least for a thin-walled short length of cylinder. (and No contraction of length due to Poisson ratio effect)

Don't let the temperature change while you're doing this.

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[sailoday28]

If this is a transient, then the inlet flow rate will vary with time.
First consider the conservative approach where the entering fluid is basically a moving piston suddenly accelerated from at rest to a velocity of (volume flow rate/face area of piston)
The change in pressure at the moving piston face is then determined by the "hammer equations" or method of characteristics.
change in pressure N/sq meter= u*rho*c
where u = initial velocity of piston m/s
rho = mass density kg/cu m
c= sound velocity of fluid m/s
In reality, the fluid will not suddenly enter and entrance conditions will have to be determined from other upstream boundary conditions.

 

[BigInch]

The pressure inside the cylinder at any time t is independent of the time it took to develop.

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[sailoday28]

BigInch
The pressure inside the cylinder is a function of the inflow rate.
I don't know what you are "talking about". Is there something wrong in my presentation? If so, please spell it out.

 

[BigInch]

Nothing is wrong with your statement. Maybe its just me, but I don't see why assuming any time dependency is necessary, given the OP's question.


If one is examining a system over time, including transient factors such as flow transients may be necessary, but still it would only be relavent to do so if the flow in a subject system varies with time and the variable of interest exhibits some dependence on time. The original post made no mention of pressure or flow(rate) in that system being a function of time, therefore I think it is an illogical extension to assume it is. Although a change in anything implies time is passing, a given function of interest may not necessarily exhibit time dependency. It brings me to wonder why you did not mention temperature, as a change in temperature (transient heat flowrate to/from the control volume) could also just as easily be assumed to be just as relavent in its effects on pressure vs volume as would be a change in mass flow into or out of a control volume. In this case, the introduction of a massflow-time dependence needlessly overcomplicates the issue at hand, just as would an introduction of another function involving heat-time dependence. There are a nearly infinite number of "what ifs" I could mention and we could go on for a month or two discussing them all. Does the Youngs modulus creep over time? There is variation with Bulk Modulus with temperature. There is creep in the value of Young's modulus. There is strain hardening. Did the cylinder material yield? Is there a chemical reaction between fluid and cylinder material over time?
*/ end rant

Is there some other reason that you thought consideration of transient flows needed to be included? Can you give its relevancy to the original post's question? Sorry if it seems like I'm giving you a hard time. Hope you understand its not intended to be personal... just a general rant about the endless "what ifs" that are always popping up around here, of which I'm sure I'm guilty myself... /one hand slapping other/ Have a good day.

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[BigInch]

Am I guilty? Although not mentioned in the OP, the Title of the Post, "Fomula estimating pressure increase vs pump rate in [COLOR=white red]fixed volume[/color]", implies the OP believes the formula should not even have a volume dependency. So did I unnecessarily assume a volume dependence?

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[sailoday28]

BingInch
I find it hard to see how one can have a steady state flow problem when one is pumping into a fixed closed volume.
For steady state, and assuming the fluid is only slightly compressible, you could go to the pump curves and find the change in pressure.

The simple model that I have given is with the approximation of rho*c being constant and a completely rigid system with a moving piston or wave front and no friction. If you disagree with the assumptions of rigity, rho*c =constant, etc. then use the equations below.

For a rigid system integration of the "hammer equations are necessary du +- dp/(rho*c) =0 (1)
AND Tds/dt= 0 (2)
where T = temperature, s = entropy and t=time
For additional one dimensional effects, (1) should include friction
and (2) heat transfer and friction.
It is much more complicated to include movement due to elongation.
Proper use of (1) with appropriate boudary conditions should give good results. Clearly for one end of the cylinder, velocity =0. For the inlet end, I have only suggested that the velocity= Q/A. Knowing that boundary condition will cetainly help further improve the results.

 

[mhldrillpipe]

Thanks guys!!!

I think for my purpose (pressure testing of casing, a 2000 m 9 5/8 ID steel tube cemented against formation) the simple formula: V (assuming fixed), mud compressibility at an assumed constant temp and given pump rate of pumping unit is sufficient....(?).

so that is: deltaV = Cp*deltaP*V

your discussion is much appreciated

 

[BigInch]

I thought so. Please excuse our hijack....

Now Sailoday, you know I wouldn't, couldn't or even think of disagreeing with your formulation. I'm simply saying, "Not necessary". Like hundreds of engineering problems, formulas are not the way to the quickest and "bestest" answers. Its not just math and physics, its using what you need from math and physics, et alii to arrive at a solution to a practical problem. That may only incude the physics while you neglect the chemistry entirely or the other way around. And Sailoday, that's just it; IMO, Its not a pumping problem, and its not a pumping problem that needs to include transient effects either; you made it one. Pressure vs volume are not necessarily dependent on time, unless you extend the definition of BM=dP/dV and the scope of the problem to include any dP/dt, dV/dt terms, which in this case [/]wasn't necessary. As I asked before, why did you not choose to include dTemperature/dt too? If you examine a problem over an infitesimal period of time, even transient pressures disappear (as would transient temperatures). Simply; examine the problem over an infitesimal period of time (whenever you can do so) and you will have the quickest, clearest and "bestest" engineering solution.

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[BigInch]

P.S. I'm willing to share equally in the guilt. As I finally realized, "fixed volume" meant fixed volume of the cylinder, which would have really simplified the solution. For me, not even a cemented in well casing is a "fixed" volume. I happen to believe concrete and steel have a Youngs modulus and un-reinforced concrete cracks in tension, but then I'm too sensitive in those areas. I should have guessed it wasn't necessary from the title... much earlier than I finally realized.

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[sailoday28]

BigInch
There is nothing wrong with "what ifs", if they are reasonable and can be modeled.

BigInch (Petroleum) 31 May 07 1:51
Nothing is wrong with your statement. Maybe its just me, but I don't see why assuming any time dependency is necessary, given the OP's question.

Well, maybe as you state you don't see why. But my approach would give an answer,-- if the user modeled the problem correctly.

Do you think all the proper what if's were done on the 1985 shuttle launch? And on the SSN "Thersher"?

Again, find something wrong with the "what if", then you can justify not using it.

 

[BigInch]

There are usually many ways to solve a problem, some more costly than others. In that respect, the simplest is always the best, although perhaps sometimes not the most complete. I don't think PD/2t tells the real story either, but with the right safety factor it works pretty well.

I don't understand how the examples you mention apply. Challanger o-ring risk was known, but improperly evaluated on launch day. I am not familiar with the cause of the Thresher sinking... screen doors?

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[sailoday28]

BigInch
What I can't get across is that if you don't understand a model, or don't care to understand it, then don't discount its validity.
Simpest may not be the best if it is not valid.

 

[BigInch]

P=nRT/V where's the time?

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[sailoday28]

BigInch (Petroleum) 3 Jun 07 1:59
P=nRT/V where's the time?


The above is an equation of state and is not time dependent.
It may be applied to conservation of mass, momentum or energy, which for transients are time dependent.
I presume you won't be satisfied with my explanation and others might help you out if you pose your question on another thread.

 

[BigInch]

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[dcasto]

Once the OP clarified that he was pressure testing a vessel (pipe) it's all clear. The pipe gets bigger with pressure, the fluid smaller and the rate they change is a function of mainly Pressure and therefore volume and time comes into play by volume/time. In reallity, you'll just pump it up and watch the dead weight tester.

 

[BigInch]

deltaV = Cp*deltaP*V.

No time in that one either. If you want to complicate the problem, go ahead and add it.

Really, IMO its the same as asking how much volume some hole contains and then saying V depends on how fast you can dig up the dirt.

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[sailoday28]

BigInch
deltaV=
Did this change occur while time stood still?
And if you think time stood still----
Lets end this, if you don't get it, you don't get it.