How to calculate the value of specific enthalpy of R134a in vapor state?

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Why does the calculated  enthalpy value of R134a in vapor state differ from the value shown in the following diagram? 

Accepted Answer

MathWorks Support Team
MathWorks Support Team on 30 Dec 2021
Edited: MathWorks Support Team on 22 Nov 2021
Since you are interpolating the value of specific internal energy "u", you are observing this difference. 
You can compute specific enthalpy "h" from the specific internal energy "u", pressure "p", and specific volume "v":
>> u_sat_v = interp1( r134aPropertyTables.p, r134aPropertyTables.vapor.u_sat, p)
>> v_sat_v = interp2( r134aPropertyTables.p, r134aPropertyTables.vapor.unorm, r134aPropertyTables.vapor.v, p, 1)
>> h_sat_v = u_sat_v + p*1e3*v_sat_v
To get "v_sat_v", I interpolated the table at unorm = 1, where "unorm" is what we call the normalized internal energy, because that corresponds to the vapor saturation line. 
For liquid, specific volume is very small, so specific enthalpy and specific internal energy are very close in value.
In general, please be careful about comparing values of enthalpy, internal energy, and entropy from different sources because they are state properties, which means they can have an arbitrary reference value.
  1 Comment
Abdel karim Nakhwe
Abdel karim Nakhwe on 26 Apr 2022
Edited: Abdel karim Nakhwe on 26 Apr 2022
can u please tell how to interpolate u and v at superheated state knowing pressure and temperature?

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