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Temperature drop in gas pressure regulator for CNG

Temperature drop in gas pressure regulator for CNG

Temperature drop in gas pressure regulator for CNG

(OP)
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?

RE: Temperature drop in gas pressure regulator for CNG

By the way Krajina is part of Croatia.

RE: Temperature drop in gas pressure regulator for CNG

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

RE: Temperature drop in gas pressure regulator for CNG

The formula "that circulates within the company" is the definition of the Joule Thomson coefficient μJT

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

(to – tr)/(pi – pr) = 0.4 = μJT

http://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_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.
 

RE: Temperature drop in gas pressure regulator for CNG

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).
 

RE: Temperature drop in gas pressure regulator for CNG

(OP)
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 = μJT 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.

RE: Temperature drop in gas pressure regulator for CNG

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 = μJT

(to-tr)/(pi-pr)= μJT

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

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