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temperature drop across regulator
3

temperature drop across regulator

temperature drop across regulator

(OP)

I am struggling trying to calculate the temperature of the outlet air from a regulator which is reducing from 350 bar to 17 bar.  I have read a long thread from this forum on a related subject but can't apply it to my problem.  Can anyone help.

thanks

Jamie

RE: temperature drop across regulator

You can approximate this to an adiabatic process to get the lowest possible temperature. As soon as the pipe exits the regulator it extracts heat from the atmosphere and its temperature rises. T1/T2 = (P2/P1)(1-k)/k where both temperature and pressure should be in absolute scale.

Steam expansion is a different story.

Regards,

RE: temperature drop across regulator

No stars yet, please. Actually the upstream pressure you indicated is close to air liquification process. You should check the Joule-Thompson effect. If the upstream air temperature is above the inversion temperature then air gets heated up upon expanison.

RE: temperature drop across regulator

2
I guess Jamie Massie should have also mentioned the starting air temperature to assess whether, and how much, cooling would be expected, which comes to show that quark is right (as usual).

Perry VI gives the envelope conditions for expanding air to heat or cool on a J-T expansion. It follows that at 350 bar as long as air is below 445 K (172oC) it will cool on a isenthalpic J-T expansion.

RE: temperature drop across regulator

(OP)
starting temperature will be 0deg C.  

Before asking the question I did the T1/T2 = (P2/P1)^(1-k)/k calc giving me T2=113K.  When colleagues expressed disbelief that it would be so cold, I had a re-think and realised that isentropic expansion was probably not the correct process for this.  I now think that it is an isenthalpic process which means the calc is much harder.  Is this correct ?

As you may have guessed, this is not really my field

thanks

Jamie

RE: temperature drop across regulator

Your 113K is the lower limit for the gas temp.  If it is a two stage regulator you have to consider that.  If the flow rate is low then you will pick up quite a bit of heat from the environment.
But yes, I have frozen regulators.  The gas can get cold.

= = = = = = = = = = = = = = = = = = = =
Corrosion never sleeps, but it can be managed.
http://www.trenttube.com/Trent/tech_form.htm

RE: temperature drop across regulator

(OP)
it is a single stage Tescom 1200 series Dome Loaded Regulator.  Flowrate about 70cubic metres/min (at 1 atm)

Jamie

RE: temperature drop across regulator

Using thermodynamic tables for compressed air, no calculation would be needed. After an isenthalpic expansion from 350 bar and 0oC to 17 bar, the gas would come near -50oC. An interesting discussion can be found in Thread391-78680.


RE: temperature drop across regulator

If there is negligible heat transfer, you should take into account the kinetic energy of the fluid.  the change in enthalpy (hout-hin)=(vin^2-vout^2)/2   were v is velocity.
If change in KE is negligible, the isenthalpic is sat.

RE: temperature drop across regulator

Quick aside

Is it Thomson or Thompson?

Thomson as in Joule-Thomson, as in Lord Kelvin. My research has shown it to be Thomson, but there are as many references to Thompson.

RE: temperature drop across regulator

In short:

William Thomson (or Thomsen) later Lord Kelvin, an Irish-born scientist, was the first to study the cooling of gases on expansion into a vacuum, by the mid 19th century, realizing that heat is related with molecular motion.

Benjamin Thompson, an American-born scientist working as director of the Bavarian arsenal by the end of the 18th century, noted the large amount of heat produced in the process of boring cannons. He, then, concluded that heat is not a conserved fluid. He suggested that heating was associated with mechanical work by the boring tool.

By about half a century later the British physicist James P. Joule confirmed Thompson's observations, and quantified the relation between heat and energy (mechanical, electrical and chemical), bringing thermal phenomena under the powerful conservation-of-energy law. The SI energy unit is named after Joule in recognition of this major synthesis in physics.
 

RE: temperature drop across regulator

I assume the regulator is controlling flow, not overall pressure drop.  Therefore, the largest change in kinetic energy will be when flow across regulator is highest.

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