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Home > Forums > Mechanical Engineers > Activities > Pipelines, Piping and Fluid Mechanics engineering > FAQs

Pipelines, Piping and Fluid Mechanics engineering FAQ

Expansion and flexibility

Pressure increase due to thermal expansion of a trapped liquid by prex
faq378-1339
Posted: 25 May 07

The calculation of pressure increase due to thermal expansion of a liquid fully filling, without any gas bubbles or pockets, a metallic enclosure, may be treated as follows.
The phenomena to be considered are:
1)thermal expansion of liquid due to the change of its bulk temperature
2)thermal expansion of the vessel or pipe, assumed in the following as having the same temperature as the fluid
3)compressibility of liquid under the increase in pressure due to the constrained volume
4)increase in volume of the vessel under the increased pressure of the fluid.
The corresponding contributions to the relative change in volume δV/V of fluid or of the containing space may be evaluated as follows:
1)αfΔT
2)αvΔT
3)βΔP
4)ΔPD/tE
where:
-αf is the volumetric coefficient of thermal expansion of the fluid
-αv is the volumetric coefficient of thermal expansion of the vessel (= three times the linear one)
-ΔT is the change in temperature
-β is the compressibility factor of the fluid, or the relative change in volume per unit change in pressure
-ΔP is the change in pressure of the fluid (it is what we seek)
-D,t and E are respectively the diameter, the thickness and the elastic modulus of the vessel (or pipe)
By equating the change in volume of the fluid to the change in volume of the vessel one gets:
αfΔT-βΔP=αvΔT+ΔPD/tE
or
ΔP=(αf-αv)ΔT/(β+D/tE)

Let's take as an example water as the entrapped fluid and carbon steel for the container, at temperatures not far from room temperature.
We have:
αf=210x10-6 ¦C-1
αv=36x10-6 ¦C-1
β=4x10-4 MPa-1
E=2x105 MPa
D/t will of course widely vary according to the dimensions and design pressure of the container; we may assume a variation between 100 for a light vessel or pipe and 10 for a quite heavy vessel, the value of 0 corresponding to an infinitely rigid container.
With these figures we get (per degree C of change in temperature):
ΔP=0.19 MPa =~ 2 bar for the light container
ΔP=0.39 MPa =~ 4 bar for the heavy container
ΔP=0.44 MPa =~ 4.4 bar for the rigid container
 

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