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Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

(OP)
I am witnessing a customer system failure when they capture 20°F chilled 30% glycol between two ball valves, and allow the captured solution to heat up as much as 70°F. I realize that an expansion tank is in order, but I need to calculate the pressure change to justify it to them. There are many tools to calculate the volume change, but I have not found anything to calculate the pressure generated. Any and all help is most greatly appreciated.

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

(OP)
BTW - This is a 2" copper system, using Press-connection fittings that are separating, if this is of any value.

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

The fittings are separating, and the customer needs calculations to believe something is wrong?

Idiots.
 

Mike Halloran
Pembroke Pines, FL, USA

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

You need to calculate two things.  One is the effective "spring constant" of the trapped pipe section, i.e. how its internal volume changes w.r.t. pressure.  The second is to find the bulk modulus for the 30% glycol solution (this might be tricky, but you could fudge by just assuming it's the same as pure water).  Set (Total volume change of fluid due to temperature) = (vol. change of pipe) + (vol. change of fluid due to pressure), and iterate to a solution.

Mike's answer is much simpler.

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

The pipe volume increases with temperature, roughly equal to the original volume x 3 x the thermal expansion rate of the pipe material.  There is also a volumetric thermal expansion coefficient of the liquid, probably greater than the pipe, so pressure increases as temperature increases, since there is an ever decreasing net volume per degree difference remaining between the pipe and the fluid and the fluid experiences a net compression.  You can solve directly, the Bulk Modulus of the fluid at the final temperature divided by the net change in the unrestrained expanded volumes of the fluid's volume minus the steel's will give you the pressure.  I think that water's 320,000 psi bulk modulus would have the highest bulk modulus, most organics being around about 2/3's of that (but not sure about 30%glycol mix... say 3/4ths ?).

We are more connected to everyone in the world than we've ever been before, except the person sitting next to us.  Lisa Gansky

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

(OP)
btruebllod and BigInch, thank you for your insight. Your suggestions are directed at the increase in volume - do I assume that the glycol mix is 100% non-compressing, and then look at the calculated pressure necessary to change the pipe volume and assume that is the pressure created?

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

IMHO, a copper piping system with press fittings is not a good choice here.

A thin-walled steel system with bends may be more suitable.

   

RE: Pressure generated by a 30% Ethylene Glycol on a closed thermal sys?

1.) Let both the pipe and the liquid expand unrestrained according to their temperature change,
pipe volume is the original volume x it's expansion coefficient x 3.  The water volume is the original volume x the bulk thermal expansion coefficient, expansion for liquids is defined in that manner (the linear expansion coefficient is already multiplied by 3).
2.) calculate the net change in volume, the volume of glycol-water should be greater, so that indicates that it will be compressed back into the now expanded volume of the pipe.
3.) Take the difference between the two,
4.) Now squeeze the glycol-water mix (compress it) back to the now expanded volume of pipe.
5.) Calculate the pressure increase using the mix's bulk modulus at 70F.

The above is actually considering the expansion effects only due to temperature, neglecting the expansion of the pipe volume with pressure, but you could include that for even more accruacy.  

That expansion due to pressure at yield/per unit length of pipe  = Area1 * (1+2*SMYS/E).
SMYS is the specified minimum yield stress of the pipe.
E is Young's modulus
Area1 is the original pipe area.
 

We are more connected to everyone in the world than we've ever been before, except the person sitting next to us.  Lisa Gansky

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