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:
thermal expansion of liquid due to the change of its bulk temperature
thermal expansion of the vessel or pipe, assumed in the following as having the same temperature as the fluid
compressibility of liquid under the increase in pressure due to the constrained volume
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:
αfΔT
αvΔT
βΔP
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):
I don’t think so. Did you use search (magnifying glass icon) on the top right? If you have a specific problem you are working on, and have questions, we can help. We just need the scope of the problem and the necessary details. Remember the phase rule, when you add a phase (gas pocket) to the problem, a degree of freedom is added too, which means, more information is required than just temperature. You’ll need to figure out the volume compression of the gas pocket by the expanding liquid (compression ratio) and the temperature change. Then with an equation of state (EOS) or ideal gas law, the new pressure can be found.
It doesn’t sound like a “trapped liquid” problem to me. It sounds like homework, right?
Upload your solution and I’ll help.
You have 0.3 L liquid n-butane and 0.7 L of an unknown gas + n-butane vapor, because the vapor pressure of n-butane at 2 C < 6 barg. You need a n-butane phase diagram, tabulated vapor pressure data, a vapor pressure equation or something.
At 2 C, we know the partial pressure of the unknown gas + the vapor pressure (p*) of n-Butane = 7 bar absolute. Find p* of n-Butane at 2 C, and you can get the partial pressure of the unknown gas at 2 C = 7 bar_a - p* of n-Butane.
So, use Gay-Lussac’s law on the partial pressure of the unknown gas and look up p* (21 C) of n-Butane. Then, add them together. There’s your answer!
Considering your reasoning, which I follow and agree, the final pressure is independente of the gas volume given by the dip tube. We obtain the same result if the gas volume is 30%, 3% or 0.3%… I coulnd follow this line of thought because we should considering a problem between liquid/gas equilibrium and full trap liquid. My problem is finding this line and how to address it properly.
When considering gas bubbles, especially air bubbles or other gas not of the same species (vapor) of the liquid phase, you can also consider the solubility of the gas in the liquid phase (Henry’s law) which tends to reduce the peak pressure for the trapped volume.
One of the worst cases of “Pressure Increase Due to Thermal Expansion of a Trapped Liquid” was at the Williams Geismar La Olefins plant in 2013. Williams Olefin Plant
Watch the video.