Cubical Expansion Coefficient for Themal Expansion Calculations 4

[BE1980]

I am currently sizing a PSV which is situated on the discharge of a heat exchanger. The relieving fluid is cooling medium i.e. 75%potable water/25% MEG. Based on API standards, I have sized a rupture disc to prevent against tube rupture and process gas breakthrough. Therefore,the only credible scenario for sizing the PSV is the thermal expansion case. However, I am unsure of what value I should use for my cubical expansion coefficient for the 75%potable water/25% MEG cooling medium. The calculated gravity is ~5 oAPI and API521 suggests using a coefficient of 0.0004oF-1. However, if the cooling medium is 100% water, API suggests a cubical expansion coefficient of 0.0001oF-1. As this value is directly proportional to the required relief rate, it has a large effect on the outcome. This may mean a larger valve than required will be installed.

As the mixture 75%potable water/25% MEG, can I take a weighted mean for the cubical expansion coefficient i.e. say 0.000175oF-1? Does anyone know a standard which documents cubical expansion coefficients for MEG/water mixtures?

 

[25362]

Read thread124-83163 and thread378-55160.

 

[hacksaw]

thermal relief being what it is, how can you claim that it drastically affects the valve sizing? What areas are you comming up with?

 

[BE1980]

I agree that the thermal relief rate will not drastically affect valve sizing.

However, the required relief rate is directly proportional to the cubical expansion coefficient.

Q = (B*H)/(500*G*C) API 521

where Q= Thermal relief rate, B = cubical expansion coefficient, H= heat transfer rate, G= specific gravity of the fluid and C= specific heat of the fluid.

If I assume 100% water, API 521 states I should use a cubical expansion coefficient of 0.0001oF-1.The calculated gravity for the cooling medium is ~5 oAPI and API521 suggests using a coefficient of 0.0004oF-1. This will give me 4x the required relief rate which in turn gives me 4x the required area.

Taking B=0.0004oF-1, I have calculated an orifice size of 0.08sqin. The basis is conservative and acceptable. Assuming 100% water will probably only marginally reduce the size of the relief value due to the standard sizes provided by vendors.

However, I have another similar scenario. I am back-calculating to determine if a selection of PSVs have been sized correctly for the required relief rate. The same issue has arisen. The PSV has sufficient capacity based on 100% water (B=0.0001oF-1) but does not have sufficient capacity based on the 75%water 25%TEG mixture (B=0.0004oF-1). I have very limited information from the initial design calculations. If I can determine the actual expansion coefficient for the cooing medium it may prevent a PSV change out.

 

[Montemayor]

BE1980:

If your tube rupture scenario is your worse case (capacitiy-wise), then you have protected the heat exchanger shell side. Don't waste any more time trying to calculate the expansion coefficient and the amount or rate of shell-side liquid that could determine the theoretically correct orifice size for the thermal expansion case. It isn't worth the effort.

The instantaneous rate of liquid expansion will be safely taken care of by a 1/2" or 3/4" thermal relief valve - and you don't require any calculations or documentation as per OSHA regulations. I guarantee it; I've done this hundreds of times on safety relief valve projects here in the Texas Gulf Coast for well-known and world-recognized chemical processing plants.



Art Montemayor
Spring, TX

 

[BenThayer]

i basically agree with Montemayor that this is an exercise that is typically not needed.

but to approximate the volumetric coefficient of thermal expansion, see the 5th edition of Perry's pg. 3-227, eqn. 3-3 which is:

beta = (rho1^2-rho2^2)/(2*(t2-t1)rho1*rho2)

if you have density information of the materials you are using over an expected temperature range (typically the case), you can approximate it with the above. "rho" is my way of saying density ("rho^2" would be "density squared") and "t" is the corresponding temperature.

i don't think the 6th edition has this equation.

 

[25362]

To BenThayer.

I don't have access to Perry VII. Nevertheless, I've checked that formula in view of [β] = (1/V) ([Δ]V/[Δ]T).

If we carry on by replacing 1/[ρ][sub]1[/sub]=V[sub]1[/sub] and 1/[ρ][sub]2[/sub]=V[sub]2[/sub], and instead of 1/V we write 1/V[sub]ave[/sub], we get the above-quoted formula assuming 1/V[sub]ave[/sub]=[ρ][sub]ave[/sub]=([ρ][sub]1[/sub]+[ρ][sub]2[/sub])/2.

I think a more correct approach would be using

1/V[sub]ave[/sub]=1/[(1/[ρ][sub]1[/sub]+1/[ρ][sub]2[/sub])/2]​

With a final simpler result:

[β] = ([Δ][ρ]/[Δ]T)/[ρ][sub]ave[/sub]​

BTW, I've checked both formulas for water over a span of 1[sup]o[/sup]C and over a [Δ]T = 10[sup]o[/sup]C, and found that both give equal results up to the third significant figure.

Would you care to check the formula I got ?