Temperature drop in gas pressure regulator for CNG

[Jezovuk]

Best regards! I' am mechanical engineer from Krajina and I have a task to construct gas pressure regulator for compressed natural gas. Here are some characteristics of the regulator:
Inlet pressure: 250 bar
Outlet pressure: 4 bar
Volume flow: 100 m^3/h
Inlet diameter: 1/2"
Outlet diameter: 1"
Gas is heated up to 80 C infront od the regulator by gas exchanger. I have following formula that circulates within the company and is considered to be "true" but I'm a bit sceptical about it:

to- ti= 0.4(pi- pr)+ (tr- ti)

to- gas temp.on heat excanger's outlet
ti- gas temp.on heat excanger's inlet (5 C)
pi- gas pressure on heat excanger's inlet
pr- gas pressure after reduction
tr- gas temp.after eduction
0.4(C/bar)- Joule Thomson's efect.

When I calculate temp.using that formula I get tr= -18.4 C.
I have also calculated temp.drop as adiabatic process by

T2= T1(p2/p1)^k-1/k and got T2= -108.255 C
k- adiabatic exponent (1.31 for nat.gas)

Can any of upper formula give me approximately real picture of process taking place inside the regulator?

 

[cloa]

By the way Krajina is part of Croatia.

 

[Jezovuk]

No.Krajina is in our heart, my friend.
But, this is neither the time nor the place for that discussion.

 

[ione]

The formula “that circulates within the company” is the definition of the Joule Thomson coefficient ?JT

to- ti= 0.4(pi- pr)+ (tr- ti)

(to – tr)/(pi – pr) = 0.4 = ?JT

http://en.wikipedia.org/wiki/Joule–Thomson_effect

I would also check the Joule Thomson coefficient for methane here

http://webbook.nist.gov/chemistry/fluid/

Please note the transformation that takes place through a reducing valve is not properly isenthalpic, which by definition implies that enthalpy remains constant during the whole process. It is more correct to define this process as a quasi-stationary process whose final enthalpy is equal to the initial enthalpy.

 

[ione]

Jezovuk,

How did you get -108.255 °C with the data you have reported in your OP?

T2 = T1* (P2/P1)^[(k-1)/k]

With k = 1.31 (be aware this value stands for methane at 1 barg and 25 °C ---> not your conditions)

T2 = 132.733 K = -140.417 °C

Moreover the problem is that what above stands for a reversible adiabatic transformation and for an ideal gas (which I doubt your case is).

 

[Jezovuk]

ione,

thank You for your time.
It was my mistake, Your result is correct.
I am aware of that, but I could not find adiabatic exponent of natural gas (k)for my condisions so I assumed that it is a ideal gas (k=cp/cv) and that it's change of state is adiabatic in order to get some starting, approximate data.
I realise now that is not the case and that result is completly wrong.

In Your previous message You said that (to – tr)/(pi – pr) = 0.4 = ?JT is a definition of Joule Thomson coefficient, but where is a reference to specific gas in that equation?
In my case: (to – tr)/(pi – pr)= 0.305 °C/bar.

 

[ione]

My hint was to check the value of the Joule-Thomson coefficient for methane at your reference conditions (pressure and temperature before expansion) using the NIST link. I am not sure the value you’ve reported (0.4 °C/bar) is correct.

Then the following formula

(deltaT/deltaP) at constant enthalpy = ?JT

(to-tr)/(pi-pr)= ?JT

allows you to compute the temperature tr of the expanded gas.