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influence of temperature on pressure in pipeline

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alizafari1

Petroleum
Mar 17, 2011
9
IN PIPING HYDROSTATIC TEST, RECOGNIZING LEAKAGES IS BASED ON PRESSURE DROP AFTER HOLDING TIME, BUT IN CASES, PRESSURE GAUGE SHOWS PRESSURE DROP BECAUSE OF DECREASING WATER TEMPERATURE DUE TO AMBIENT TEMP IN HOLDING TIME; AND WE CAN’T UNDERSATAND ABOUT LEACKAGE CERTAINLY. NOW I NEED A FORMULA TO CALCULATE PRESSURE DROP BECAUSE OF TEMPERATUE DROP. PLEASE SHOW ME THE WAY.
 
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Very Roughly the change is around 100 psi for 1[°]F for steel pipelines.

From "BigInch's Extremely simple theory of everything."
 
Alizafari1,
Please don't shout, all caps is hard for some of us to read.

The relationship that BigInch gave you doesn't translate well. It is a good idea to convert Celsius to Fahrenheit, get your pressure change and convert that to your pressure units. When people try to convert in one step they often mess it up (it is not an empirical relationship, it is based on the bulk modulus of water and coincidentally works out to a number close to 100 psi per degree F so it can be translated, but people keep messing it up).

David
 
I didn't think anything required translation. Are you saying I should have said 100 psi for 1 F[°]

From "BigInch's Extremely simple theory of everything."
 
It will be much higher for buried pipelines, since the longitudinal thermal growth of the pipe is restrained by soil friction, the water will expand in 3 dimensions, but the pipe in only the radial direction (2 of the XYZ dimensions), hence the difference between the volume of the expanded water and the not-so expanded "vessle" would be much greater than what is shown in the FAQ.

From "BigInch's Extremely simple theory of everything."
 
BigInch,
No I was just saying that people who disdain the use of Imperial units often try to turn that relationship into kPa per degree C and I've seen several get it wrong. The relationship you quoted is really easy to remember, the metric version isn't so much with the round numbers.

David
 
I think that it is possible to quantify that "much higher". If you have constraints in your pipeline (i.e. buried pipe) which prevent pipe from expanding in three orthogonal directions, then use the same formula and replace the pipe volumetric coefficient of thermal expansion with the linear coefficient of thermal expansion (which is roughly one third of the volumetric one).
 
Still not so, because you have to let the pipe expand in the right orthagonal directions, ie. only the radial directions. If you use the 3[α], 2[α], 1[α] method, you don't get the right proportions of total pipe volume. You must let the pipe expand in the radial directions only, hence the expanded new volume would be [π] * ([α] * r * [δ]T)[sup]2[/sup]/4 * Nonexpanded_Length_of_Pipeline. Or possibly * the ratio of any unrestrained pipeline length to its total length.

From "BigInch's Extremely simple theory of everything."
 
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