Calculation of JT Effect for Gas Temperature Change 2

[DingWeiyang]

Hello All

I was suffered by this calculation for a long time.
is there anybody can help me on this?

Basis:
u=dT/dP

Case Assume:
A pressure regulator to reduce the gas online pressure from high pressure to desired pressure.

Known Conditions:
1.Gas Name: SiH4(example)
2.Inlet Pressure of Pressure Regulator: 10MPa(g)
3.Outlet Pressure of Pressure Regulator: 0.6MPa(a)
4.Inlet Temperature of Pressure Regulator: 20 deg. C

what's the temperature of regulator outlet? how to calculate?

Thanks
Ding

 

[delittle]

the value dT/dP is not constant over a range of pressures,
a better solution is to calculate H1(tin,Pin) and then t(H2,P2) where H2 = H1
for that you need a table with the properties (H) of Silane,
suppliers as AirLiquide can provide these tables.

As alternative you may run a software, for example Prode Properties returns about -70C (solving a flash H1 = H2 with P1 = 100 Bar.a, T1 = 15 C, P2 = 6 Bar.a for Silane) with a partial liquid fraction...
however these results may have limited practical meaning for many reasons which have been discussed in this forum,
see for example

http://www.eng-tips.com/viewthread.cfm?qid=374882

 

[Caloooomi]

I use NIST refprop for generating thermo data. To find the temperature after the reduction of pressure, I find the enthalpy of the gas at the first temperature and pressure condition. Assuming constant enthalpy, find the temperature at a lower pressure.

u_jt = (dt/dp)|h

 

[DingWeiyang]

thanks both.

Ideas you provide are appreciated, however, i expect someone can show me the calculation process.

regards
Ding

 

[apetri]

Ding,
to see the calculation process you need a mollier chart for Silane as suggested by Delittle, ask your supplier for that, with the mollier chart (or a table of values) things will be easy to understand...
by the way Refprop doesn't include Silane.

 

[Latexman]

Joule–Thomson effect

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[JungleTown]

Use a pressure enthalpy diagram.

 

[DingWeiyang]

Thanks for all replys, thanks for your kindly suggestions.

regards
Ding

 

[DingWeiyang]

Hi All

i found a calculation example from an engineering book named<PROCESS ENGINEERING and DESIGN USING VISUAL BASIC,2nd EDTION> where provide the manual calculation method for temperature calculation for JT Effect. but still can not figure out few things,

------example from book-------
Example 2.6
Estimate the Joule–Thomson cooling temperature if methane at 10,000 kPag and 20°C is expanded to the atmospheric pressure.
SOLUTION
This can be calculated by reducing the pressure in steps and estimating the temperature at the end of each step. The isobaric specific heat of a real gas is calculated using the procedure described in Example 2.5.
The PR EOS is used for this calculation.
Let the number of steps be 50; therefore, the reduction in pressure at the end of each step will be 200 kPa.
Constant parameters
• Critical pressure = 4640.68 kPa
• Critical temperature = 190.7 K
• Acentric factor = 0.0115
Step 1 calculation
• Calculated the value of mi (Equation 2.25i) = 0.3923
• Calculated the value of aci (Equation 2.25g) = 247.67
• Calculated the value of bi (Equation 2.25d) = 0.0266
• Compressibility (Z) = 0.81792
• Value of (∂Z/∂T)P = 0.00239
• Isobaric heat capacity of a real gas = 49.35 kJ/(kmol•K)
• Joule–Thomson coefficient (Equation 2.58) = 0.00342
• Drop in temperature (Equation 2.56) = 0.68°C
• Temperature after first step = 19.32°C
The above step is repeated 50 times to calculate the final temperature.
------end of example------

according to the process and numbers provided in book, i calculated the compressibility is only 0.48, which is completely different. also, i tried to use Newton-Raphson method, and the result is 0.812448.

is there anybody who can provide the calculation process about Z and Value of (∂Z/∂T)P? i have no idea to figure "Value of (∂Z/∂T)P" out.

i've uploaded the portion of the book which have the original numbers and methods.

regards
Ding

 

[apetri]

as said previously, in my opinion a more accurate solution would be to solve for
Hout = Hin
for that you need a procedure for calculating enthalpy
I do not know that book but you could use a tool such as Prode Properties to calculate the different values (Z, JT, H, etc.) and see if your values are correct...
Prode Properties exports Fg, H, S, V plus derivatives vs. W, P, T
with these it is very easy to calculate JT or similar properties

 

[DingWeiyang]

Hi Apetri

what i'm interesting now is to calculate it by manually or excel which will help myself to get the better understanding of thermodynamics

only reason i can't figure this out is that there shall have something miss i don't realize yet.

Thanks all the same

Regards
Ding

 

[apetri]

no problem,
but consider that the derivatives you are looking for JT

|dV|
|--|
|dT| P

are exported by Prode Properties as mentioned in my previous post,
anyway you may find more useful a good text discussing relationships among thermodynamic properties (Maxwell, Bridgeman etc...)

 

[SNORGY]

I usually say:

a = 27/64 * [Ru²Tc²/Pc]
b = Ru*Tc/(8*Pc)
µ = [1000/(M*Cp)]*[2*a/(Ru*T) - b]

With Pc in kPa, Tc in K, Cp in J/(kg*K), Ru in J/(kmol*K)

yielding µ in ^OK/kPa

 

[SNORGY]

Then

T₂ = T₁ - µ*(P₁ - P₂) in ^OK

 

[DingWeiyang]

Hello SNORGY

Thanks for the method you provided.
i tried you method, but the result seems not right to me.
Fluid NF3
Pin=10500kPa(a);
Pout=65kPa(a);
Tin=300K
R=8.314kJ/(kmol.K)
Tc=233.85K
Pc=4530kPa(a)
Cp=755J/(kg.K)(when 25 deg C)

Calculation result:
a=352.0636
b=0.05365
µ=0.0042
deltaT=44.7K

but the software provided the result is 64K(temperature drop)roughly.

is there something wrong in my calculation?
PS:i usually used Cp=A+BT+CT^2+DT^3+ET^4[J/(mol.K)] refer to Matheson Gas Tata Book, 7th Edition.

Thanks
Ding

 

[SNORGY]

Hi Ding.

You are getting the number I am getting - thanks for pointing out an error I made in my post; "Pc" should be in Pa absolute, not kPa absolute. "a" then works out to 352064.

There is nothing wrong in your calculation as nearly as I can tell.

The method I mentioned is quite sensitive to the Cp at upstream flowing conditions, so you want to use the Cp for NF3 at whatever the upstream temperature is, as you have done. I also checked the NF3 properties you used against what I could find, and they agree with what I found - except I came across a source that gives two values for Cp:

http://encyclopedia.airliquide.com/Encyclopedia.asp?GasID=67#GeneralData

<-- I believe the second value they report is really Cv

http://www.lindeus.com/internet.lg.lg.usa/en/images/high_purity_nitrogen_trifluoride138_28080.pdf

<-- as here

I am not sure how to rationalize the difference, apart from saying that the method I use is not as sophisticated as what proper thermodynamic software will use. For high pressure dense gas undergoing this magnitude of a pressure drop, certainly more scrutiny of the result is warranted.