Temperature rise due across a Pressure control valve

[SimonCurson]

hello,

I am calculating the heat loss down some lagged supply lines, so I can create atemperature profiles for the lines. Some one has asked me how the change in pressure along the lines will affect the temperature. What method should be used to calculate the temperature after a pressure drop. Can the equation pV=nRT be used as in P1V1/T1=V2V2/T2, or an equation with constant entropy? Any help would be appreciated.

Regards,

Simon

 

[quark]

This topic is well discussed in thread391-109449

Regards,

 

[MortenA]

As the refernce tread will show you - you need to know if your gas is ideal (air is close) and to include the JT effect.

Best regards

Morten

 

[sailoday28]

For any fluid, under adiabatic conditions, the stagnation enthalpy remains constant Ho=h +V^2/2

Knowing the difference in velocity (Kinetic energy) allows calc of change in enthalpy.

For a perfect gas the change in enthalpy is the intergal of Cp dt.

 

[jsummerfield]

O'Day 28 sailor,
Does this always mean that the temperature drops instead of rises across a pressure control valve as the posted question implies???

Being a simple control systems engineer, I understand that compressing gas increases the temperature and dropping the pressure reduces the temperature. Further, if you do work as typically in a turbo-expander compressor, the temperature drop is greater. I never much cared for that thermodynamic stuff. Seriously, can the temperature increase across a control valve?

John

 

[25362]

To jsummerfield, yes. You probably heard of throttling processes aka J-T isenthalpic (free) expansions. The entropy rises since the process is highly irreversible.

Whenever the coefficient [&mu;] = ([&part;]T/[&part;]P)_H is positive (the most common case) gases cool on expansion. When the [&mu;]<0, gases heat on expansion.

The temperature at which [&mu;] = 0, i.e., the gas is neither heated or cooled on free expansion, is called its inversion temperature.

On a P[&uarr;], T[&rarr;] ortogonal diagram the locus of the inversion points looks like an inverted U. Meaning that at a given pressure a gas may have two inversion temperatures, converging towards one single value at a pressure corresponding to the apex of the inverted U.

CO₂, for example, at 50 bar, has T_L=243 K and T_U =1290 K. The inversion temperatures at 200 bar, are 266 K and 1205 K, respectively. They coincide at 608 K and 884 bar.

For air at 25 bar, these temperatures are 114 K and 641 K.
Within this range air will cool on free expansion.

For normal hydrogen all low and high inversion temperatures are below 0^oC, thus it would heat on expansion at ambient temperatures (!). I/o to liquify it, it must be first refrigerated down to, below 90 K to enable cooling by a throttling process. Helium would require a pre-cooling down to about 20 K!

I took these values from Perry's VI Ed.

 

[sailoday28]

25362 (Chemical)
Please see my post. The process you are describing is for constant enthalpy. If adiabatic there is a change in enthalpy depending on the change in KE across the valve.

If change in velocity is negligible then use of the JT coef is appropriate.