Thermodynamic Properties of Refrigerants
01-07-2017, 08:25 AM (This post was last modified: 01-11-2017 06:18 AM by Ángel Martin.)
Post: #1 Ángel Martin Senior Member Posts: 1,071 Joined: Dec 2013
Thermodynamic Properties of Refrigerants
Thermodynamic Properties of Refrigerants.
From the Author's Engineering Collection, included in the ETSII3 Module

Building on the "Principle of Corresponding States" program, this addition calculates the Thermodynamic properties Enthalpy and Entropy in the valid ranges for both saturated liquid & vapor dome and superheated vapor. The validity range is intended for pressures lower than the critical pressure PC.

These programs provide a replacement for the different refrigerant chart sheets – using a semi-empirical approach that combines the Martin Equation of State, the Antoine’s Equation, and polynomial expressions (in inverse T) for the isobaric specific heat capacity of the liquid and the gas - valid within the application range, defined as P < Pc.

The Martin Equation of State uses reduced magnitudes, and it’s cubic in the Volume:

Pr = Tr / [Zc.Vr – B ] – A / { Tr^n [ Zc.Vr – B + 1/8 ]^2 }

The data entry process involves all the parameters required, as obtained by the semi-empirical method used. The table below lists the parameters for the main refrigerant gases. Note that the parameters A, B, and N are dimensionless (“n” is not the number of moles). However all coefficients for the specific heats have dimensions - as required by the inversed polynomial expression: Cp = SUM ak/T^k ; with Cp [Energy/Mass*Temperature]

Equations and memory register map:-

Let Pr = P/Pc, Tr = T/Tc, and Vr = V/Vc the reduced pressure, temperature and volume over the corresponding critical values. Let To = 273 K, and Po = Pv(273).

The pressure is directly calculated by the EOS shown above, as:
Pr = Tr / [Zc.Vr – B ] – A / { Tr^n [ Zc.Vr – B + 1/8 ]^2 }

The specific volume is obtained as the maximum real root of the third degree equation given by:

a3.V^3 + a2.V^2 + a1.V + a0 = 0; where the coefficients are defined as:

a0 = A.B + (1/8 - B)^2 [ (Tr)^n+1 + B.Pr.(Tr)^n ]
a1 = (Zc/Vc). { A + Pr.(Tr)^n [3B^2 - B/2 + 1/64] - (1/8 - B)*2(Tr)^n+1 }
a2 = (Zc/Vc)^2 . [ (Tr)^n Pr.(1/4 - 3B) - (Tr)^n+1 ]
a3 = (Zc/Vc)^3 Pr (Tr)^n

The density of the saturated liquid is calculated by the formula:

Vs,l = Vsc. Vg (1-w.G) ; with the following auxiliary definitions:

Vsc = (R.Tc/PM.Pc).(0.292 – 0.0967 w)
G = 0.29607 – 0.09045 Tr – 0.048442 Tr^2

And Vg depends on the actual value of Tr, as follows:

• if: Tr < 0.2 or Tr > 1 => not in liquid phase
• if: 0.2 <= Tr < 0.8; then: Vg = 0.33593(1-Tr) + 1.51941 Tr^2 – 2.02512 Tr4^3 + 1.11422 Tr^4
• if: 0.8 <= Tr <= 1 ; then: Vg = 1 + 1.3 sqrt(1-Tr).log(1-Tr) – 0.50879(1-Tr) – 0.91534(1-Tr)^2

Enthalpy of Vapor: Defined as: H(T,P) = H*(T,P) + DisH(T,P)
- where H* is the value for an ideal gas and "Dis" indicates discrepancy from the perfect gas.

From the base expression for Enthalpy:
H(T,P) = U(T,P) – RT (Z-1) = U(T,P) – (P.V – R.T)

This is calculated using the Helmholtz Free Energy as auxiliary magnitude instead of the internal energy,
defined as: F = U – T.S ; thus: => U = F + TS

Hence the base equation results:
H(T,P)= [F(T,P) + T.S(T,P)] - Q(T,P); with: Q = P.V – R.T/PM

Moving along an isobaric line, we can relate these factors to the values in the saturated liquid - by adding the enthalpy of vaporization and the integral of the specific heat capacity (at constant pressure) between the boiling temperature and the final temperature. Doing that we get:

H(T,P) = [H*(Tb,P)-DisH(To,Po)] + Hvap + ITG {[Cpv]dT} + DisH(T,P)

and replacing the enthalpy by the free energy equivalents:

Hv(T,P) = {H*(Tb,P) + Hvap(Tb) + ITG[Cpv]dT } + [ Dis(F-Fo) - Dis(Q-Qo) + T.Dis(S-So) ]

The integral of the specific heat is easily obtained by direct integration:
ITG{Cpv} = ao(T-To) + a1.Ln(T/To) - a2(1/T - 1/To) - a3(1/T2 - 1/To2) /2 - a4(1/T3 - 1/To3) /3

The vaporization enthalpy is obtained using the Pitzer correlation shown below (per kg), where w is the acentric factor of the substance:
Hvap(To) = (R.Tc/PM) {10.95w.[(1-Tb/Tc)^0.456] + 7.08*(1-Tb/Tc)^0.354 }

and finally the discrepancies of free energy and entropy are calculated as follows, where Zo = P.V/R.T ; and: Zco = Pc.Vc/R.Tc ; in [mol]

DisF(T,P) = R.T.Ln(Z) + ITG{(P – R.T/V).dV} = (R.T/PM).Ln(PM.Zo) + Pc.Vc {A / [Zc.Trn.(Zc.Vr+1/8-B)] – Tr.[Ln(Vr)/PM.Zco - Ln(ZcVr –B)/Zc] }

DisS(T,P) = -∂[DisF]/∂T = (R/PM){ (n+1) + n.A.(PM.Zco/Zc) / [Zc.Vr + 1/8 - B].Tr^(n+1) – Ln(PM.Zo) + Ln[Vr / (Vr - B/Zc)^(PM.Zco/Zc)] }

Entropy of Vapor: Defined as: S(T,P) = S*(T,P) + DisS(T,P) ;

We'll follow the same approach as before, moving along the isobaric line at T=Tb (boiling temperature)l starting at the saturated liquid state.

Sv(T,P) = [S*(Tb,P)- DisS(Tb,P)] – R.Ln(P/Po) + ITG[(Cpv/T).dT] + DisS(T,P)

The integral of the specific heat over the temperature is easily obtained by direct integration of the inverted-polynomial expression:
ITG{Cpv/T} = a0.Ln(T/To) - a1(1/T - 1/To) - a2(1/T2 - 1/To2) / 2 - a3(1/T3 - 1/To3) /3 – a4(1/T4 - 1/To4) / 4

Using the expressions listed above, the final formula for the entropy is as follows:

Sv(T,P) = {S*(Tb,P) + Hvo(To)/To+ ITG[CPv/T]dT}–(R/PM).Ln(P/Po) + [DisS(T,P) – DisS(To,Po)]

Examples.

With the data from the table below you can test the program for a few commonly used refrigerants. Several refrigerant names ("R11", "R12", "R13", "R22", "NH3") can be entered at the initial prompt for an automated entry of all parameters - or alternatively each one of them is requested individually.

Code:
            R-11        R-12        R-13         R-22       R-113        R-717 Parameter   CFCl3       CF2Cl2      CF3Cl        CHClF2     CCl2FCClF2   NH3 ================================================================================​===== Zc          0.2766      0.2790      0.27739      0.267      0.2560       0.242004 A           0.421875    0.421875    0.421875     0.421875   0.421875     0.421875 B           0.05598     0.0578      0.0566       0.0488     0,0323       0.03 n          -1.1         0.900       0.60         1.00       0.75        -1.1 Tc (C)      197.99      112.00      28.7708      96.00      214.09       132.2808 Pc (kp/cm2) 44.6003     41.9613     39.460       50.300     34.80        115.0036 Vc (cm3/g)  1.8018      1.79168     1.7212       1.9041     1.7301       4.247 PM (g/mol)  137.38      120.90      104.47       86.50      187.39       17.032 w acentric  0.188       0.1760      0.180        0.215      0.252        0.253 ------------------------------------------------------------------------------------- Ln(Pv) = x1 + x2/T  ; with T in K and Pv in kp/cm2 x1         10.84436    10.1760      9.96485      10.6316    11.13964     11.71516 x2         3209.743    2469.5656    1901.3441    2460.1029  3568.1032    2803.591 ------------------------------------------------------------------------------------- Cp,v = SUM {ak / T^k} ; k= 0,1,..4  ; with T in K and Cp in kcal/kg.K a0,v        0.1886      0.25659     0.30905      0.3485      0.25732     0.95239 a1,v       -8.0924     -49.1350    -70.3763     -102.14     -15.946     -2.92521E2 a2,v       -3.8118E3    5405.60     1.0052E4     1.6546E4   -1.6728E4    6.2791E4 a3,v        4.7327E5   -239860.0   -5.6219E5    -1.003E6     4.9196E6   -4.725E6 a4,v        0.0000      0.0000      0.0000       0.0000     -4.1739E8    0.0000 -------------------------------------------------------------------------------------- Cp,l = SUM {ak / T^k} ; k= 0,1,..4 a0,l        0.56642     9.9361      11.190       14.4110     4.5917      0.24694 a1,l       -2.3715E2   -9.445E3    -8.4336E3    -1.3625E4   -5.058E3    -2.3079E4 a2,l        5.2605E4    3.4338E6    2.4332E6     4.9211E6    2.2046E6    8.4508E6 a3,l       -4.0107E6   -5.5272E8   -3.1189E8    -7.8839E8   -4.2602E8   -1.3705E9 a4,l        0.0000      3.3182E10   1.4964E10    4.7189E10   3.0626E10   8.2854E10

The programs work internally with the units reflected in the table but you can input and output the values in any other units. It comes without saying that these programs make extensive use of the Unit Management System (UMS), therefore the unit Conversion Module needs to be plugged into the calculator.

Example1.- Superheated Vapor.

For Freon-22, T= 25 deg C and P =0.2 atm
Vesp= 1.398 m3/kg
Hv = 324.646 kJ/kg
Sv = 1.440 kJ/kg.K
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