The tasks that remain are to select a particular equation of state, take the appropriate derivatives, make the substitutions, develop compact expressions, and add up the change in properties. The good news is that many years of engineering research have yielded several preferred equations of state (see Appendix D) which can be applied generally to any application with a reasonable degree of accuracy. For the purposes of the text, we use the Peng-Robinson equation or virial equation to illustrate the principles of calculating properties. However, many applications require higher accuracy; new equations of state are being developed all the time. This means that it is necessary for each student to know how to adapt the departure function method to new situations as they come along.
The following example illustrates the procedure with an equation of state that is sufficiently simple that it can be applied with either the density-dependent formulas or the pressure-dependent formulas. Although the intermediate steps are a little different, the final answer is the same, of course.
Example 8.2. Real entropy in a combustion engine
A properly operating internal combustion engine requires a spark plug. The cycle involves adiabatically compressing the fuel-air mixture and then introducing the spark. Assume that the fuel-air mixture in an engine enters the cylinder at 0.08 MPa and 20°C and is adiabatically and reversibly compressed in the closed cylinder until its volume is 1/7 the initial volume. Assuming that no ignition has occurred at this point, determine the final T and P, as well as the work needed to compress each mole of air-fuel mixture. You may assume that 
for the mixture is 32 J/mole-K (independent of T), and that the gas obeys the equation of state,
PV = RT + aP
where a is a constant with value a = 187 cm3/mole. Do not assume that CV is independent of ρ. Solve using density integrals.
Solution
The system is taken as a closed system of the gas within the piston/cylinder. Because there is no flow, the system is irreversible, and reversible, the entropy balance becomes
showing that the process is isentropic. To find the final T and P, we use the initial state to find the initial entropy and molar volume. Then at the final state, the entropy and molar volume are used to determine the final T and P.
This example helps us to understand the difference between departure functions at fixed T and V and departure functions at fixed T and P. The equation of state in this case is simple enough that it can be applied either way. It is valuable to note how the ln(Z) term works out. Fixed T and V is convenient since the volume change is specified in this example, and we cover this as Method I, and then use fixed T and P as Method II.
This EOS is easy to evaluate with either the pressure integrals of Section 8.6 or the density integrals of Section 8.5. The problem statement asks us to use density integrals.a First, we need to rearrange our equation of state in terms of Z = f (T, ρ). This rearrangement may not be immediately obvious. Note that dividing all terms by RT gives PV/RT = 1 + aP/RT. Note that Vρ = 1. Multiplying the last term by Vρ, Z = 1 + aZρ which rearranges to
Also, we find the density at the two states using the equation of state,
Method I. In terms of fixed T and V, 
; 
Method II. In terms of T and P,
Since the departure is zero, it drops out of the calculations.
. However, since we are given the final volume, we need to calculate the final pressure. Note that we cannot insert the ideal gas law into the pressure ratio in the last term even though we are performing an ideal gas calculation; we must use the pressure ratio for the real gas.
Now, if we rearrange, we can show that the result is the same as Method I:
This is equivalent to the equation obtained by Method I and T2 = 490.8 K.
Finally, 
W = ΔU = (U – Uig)2 + CVΔT –(U – Uig)1 = 0 + CVΔT – 0 = 6325 J/mole
a. The solution to the problem using pressure integrals is left as homework problem 8.7.
Examples 8.1 and 8.2 illustrate the procedures for deriving and computing the impacts of departure functions, but the equations are too simple to merit broad application. The generalized virial equation helps to broaden the coverage of compound types while retaining a simple functional form.
Example 8.3. Compression of methane using the virial equation
Methane gas undergoes a continuous throttling process from upstream conditions of 40°C and 20 bars to a downstream pressure of 1 bar. What is the gas temperature on the downstream side of the throttling device? An expression for the molar ideal gas heat capacity of methane is CP = 19.25 + 0.0523 T + 1.197E-5 T2–1.132E-8 T3; T [≡] K; CP [≡] J/mol–K
The virial equation of state (Eqns. 7.6–7.10) may be used at these conditions for methane:
Z = 1 + BP/RT = 1 + (B0 + ωB1)Pr/Tr
where B0 = 0.083 – 0.422/Tr1.6 and B1 = 0.139 – 0.172/Tr4.2
Solution
Since a throttling process is isenthalpic, the enthalpy departure will be used to calculate the outlet temperature.
ΔH = 0 = H2–H1 = (H2 – H2ig) + (H2ig – H1ig) – (H1 – H1ig)
The enthalpy departure for the first and third terms in parentheses on the right-hand side can be calculated using Eqn. 8.29. Because Z(P,T), we use Eqn. 8.29. For the integrand, the temperature derivative of Z is required. Recognizing B is a function of temperature only and differentiating,
Inserting the derivative into Eqn 8.29,
We can easily show by differentiating Eqns. 7.8 and 7.9,
Substituting the relations for B0, B1, dB0/dTr and dB1dTr into Eqn. 8.32 for the departure functions for a pure fluid, we get
For the initial state, 1,
Assuming a small temperature drop, the heat capacity will be approximately constant over the interval, CP ≈ 36 J/mole-K.
For a throttle, ΔH = 0 ⇒ (H – Hig)2 + 36(T2 – 40) + 287 = 0.
Trial and error at state 2 where P = 1 bar, T2 = 35°C ⇒ –13 + 36(35 – 40) + 287 = 94.
T2 = 30°C ⇒ –13 + 36(30 – 40) + 287 = –87
Interpolating, T2 = 35 + (35 – 30)/(94 + 87)(–94) = 32.4°C, another trial would show this is close.
Finally, the Peng-Robinson model is sufficiently sophisticated to permit broad application, but the derivations are quite tedious. It may be helpful to see how simple the eventual computations are before getting overwhelmed with the mathematics. Hence, the first example below simply shows how to obtain results based on the computer programs furnished with the text. The subsequent examples confront the derivation of the departure function formulas that appear in the computer programs for the Peng-Robinson equation.
Preos.xlsx or PreosPropsMenu.m.
Example 8.4. Computing enthalpy and entropy departures from the Peng-Robinson equation
Propane gas undergoes a change of state from an initial condition of 5 bar and 105°C to 25 bar and 190°C. Compute the change in enthalpy and entropy.
Solution
For propane, Tc = 369.8 K; Pc = 4.249 MPa; ω = 0.152. The heat capacity coefficients are given by A = –4.224, B = 0.3063, C = –1.586E-4, D = 3.215E-8. We may use the spreadsheet Preos.xlsx or PreosPropsMenu.m. If we select the spreadsheet, we can use the PROPS page to calculate thermodynamic properties. Using the m-file, we specify the species ID number in the function call to PreosPropsMenu.m and find the departures in the command window. We extract the following results:
For State 2:
For State 1:
Ignoring the specification of the reference state for now (refer to Example 8.8 on page 320 to see how to apply the reference state approach), divide the solution into the three stages described in Section 8.1: I. departure Function; II. ideal gas; III. departure function.
The overall solution path for H2 – H1 is
Similarly, for S2 – S1 =
The three steps that make up the overall solution are covered individually.
Step I. Departures at state 2 from the spreadsheet:
Step II. State change for ideal gas: The ideal gas enthalpy change has been calculated in Example 2.5 on page 60.
The ideal gas entropy has been calculated in Example 4.6 on page 151:
Step III. Departures at state 1 from the spreadsheet:
The total changes may be obtained by summing the steps of the calculation.
ΔH = –1490 + 8405 + 401 = 7316 J/mole
ΔS = –2.292 + 6.613 + 0.708 = 5.029 J/mole-K
We have laid the foundation for deriving and applying departure functions given an equation of state. What remains is to organize our efforts to enable convenient and accurate calculations for the myriad of applications that we may encounter. Since these types of computations must be repeated many times, it is worthwhile to implement a broadly applicable (“one size fits all”) thermodynamic model and construct a computer program that facilitates input and output. The derivations and programming are more tedious, but they repay the invested effort through multiple applications. The examples below illustrate the derivations for the Peng-Robinson model, but several alternative models could have formed the basis for broad application. You should be able to demonstrate your mastery of the model equations underlying the programming by performing equivalent derivations with alternative models like those illustrated in the homework problems
The final result of this example is incorporated into Preos.xlsx and PreosPropsMenu.m
Example 8.5. Enthalpy departure for the Peng-Robinson equation
Obtain a general expression for the enthalpy departure function of the Peng-Robinson equation.
Solution
Since the Peng-Robinson equation is of the form Z(T,ρ), we can only solve with density integrals.
where da/dT is given in Eqn. 7.18. Inserting,
We introduce F(Tr) as a shorthand.
Also, B ≡ bP/RT ⇒ bρ = B/Z and A ≡ aP/R2T2 ⇒ a/bRT = A/B. Note that the integration is simplified by integration over bρ (see Eqn. B.34).
The final result of this example is incorporated into Preos.xlsx and PreosPropsMenu.m
Example 8.6. Gibbs departure for the Peng-Robinson equation
Obtain a general expression for the Gibbs energy departure function of the Peng-Robinson equation.
Solution
The answer is obtained by evaluating Eqn. 8.26. The argument for the integrand is
Evaluating the integral (similar to the integral in Example 8.5), noting again the change in integration variables,
Making the result dimensionless,
It is often valuable to recognize simplifications that may circumvent the tedium. If two models are similar, you can reuse the part of the derivation that is equivalent. If thermodynamic identities can be used to substitute major portions of a derivation, so much the better. Productive engineers should be aware of opportunities to leverage their time efficiently, as illustrated below.
Example 8.7. U and S departure for the Peng-Robinson equation
Derive the departure functions for internal energy and entropy of the Peng-Robinson equation. Hint: You could start with Eqns. 8.22 and 8.23, or you could use the results of Examples 8.5 and 8.6 without further integration as suggested by Eqn. 8.20 and Eqn. 8.21.
Solution
By Eqn. 8.20, the U departure can be obtained by dropping the “Z – 1” term from Eqn. 8.35. We may immediately write:
By Eqn. 8.21, the entropy departure can be obtained by the difference between the enthalpy departure and Gibbs energy departure, available in Eqns. 8.35 and 8.36. Then, we may immediately write
8.8. Reference States
If we wish to calculate state changes in a property, then the reference state is not important, and all reference state information drops out of the calculation. However, if we wish to generate a chart or table of thermodynamic properties, or compare our calculations to a thermodynamic table/chart, then designation of a reference state becomes essential. Also, if we need to solve unsteady-state problems, the reference state is important because the answer may depend on the reference state as shown in Example 2.15 on page 81. The quantity HR – UR = (PV)R is non-zero, and although we may substitute (PV)R = RTR for an ideal gas, for a real fluid we must use (PV)R = ZRRTR, where ZR has been determined at the reference state. We also may use a real fluid reference state or an ideal gas reference state. Whenever we compare our calculations with a thermodynamic chart/table, we must take into consideration any differences between our reference state and that of the chart/table. Therefore, to specify a reference state for a real fluid, we need to specify:
Pressure
Temperature
In addition we must specify the state of aggregation at the reference state from one of the following:
1. Ideal gas
2. Real gas
3. Liquid
4. Solid
Further, we set SR = 0, and either (but not both) of UR and HR to zero. The principle of using a reference state is shown in Fig. 8.4 and is similar to the calculation outlined in Fig. 8.2 on page 303.
Figure 8.4. Illustration of calculation of state changes for a generic property M using departure functions where M is U, H, S, G, or A. The calculations are an extension of the principles used in Fig. 8.2 where the initial state is designated as the reference state.
Ideal Gas Reference States
For an ideal gas reference state, to calculate a value for enthalpy, we have
where the quantity in parentheses is the departure function from Section 8.5 or 8.6 and 
may be set to zero. An analogous equation may be written for the internal energy. Because entropy of the ideal gas depends on pressure, we must include a pressure integral for the ideal gas,
where the reference state value, 
, may be set to zero. From these results we may calculate other properties using relations from Section 6.1: G = H – TS, A = U – TS, and U = H – PV.
Real Fluid Reference State
For a real fluid reference state, to calculate a value for enthalpy, we adapt the procedure of Eqn. 8.5:
For entropy:
Changes in State Properties
Changes in state properties are independent of the reference state, or reference state method. To calculate changes in enthalpy, we have the analogy of Eqn. 8.5:
To calculate entropy changes:
Example 8.8. Enthalpy and entropy from the Peng-Robinson equation
Propane gas undergoes a change of state from an initial condition of 5 bar and 105°C to 25 bar and 190°C. Compute the change in enthalpy and entropy. What fraction of the total change is due to the departure functions at 190°C? The departures have been used in Example 8.4, but now we can use the property values directly.
Solution
For propane, Tc = 369.8 K; Pc = 4.249 MPa; and ω = 0.152. The heat capacity coefficients are given by A = –4.224, B = 0.3063, C = –1.586E-4, and D = 3.215E-8. For Preos.xlsx, we can use the “Props” page to specify the critical constants and heat capacity constants. The reference state is specified on the companion spreadsheet “Ref State.” An arbitrary choice for the reference state is the liquid at 230 K and 0.1 MPa. Returning to the PROPS worksheet and specifying the desired temperature and pressure gives the thermodynamic properties for V,U,H, and S.
The changes in the thermodynamic properties are ΔH = 7315J/mole and ΔS = 5.024J/mole-K, identical to the more tediously determined values of Example 8.4 on page 314. The purpose of computing the fractional change due to departure functions is to show that we understand the roles of the departure functions and how they fit into the overall calculation. For the enthalpy, the appropriate fraction of the total change is 20%, for the entropy, 46%.
Example 8.9. Liquefaction revisited
Reevaluate the liquefaction of methane considered in Example 5.5 on page 213 utilizing the Peng-Robinson equation. Previously the methane chart was used. Natural gas, assumed here to be pure methane, is liquefied in a simple Linde process. The process is summarized in Fig. 8.5. Compression is to 60 bar, and precooling is to 300 K. The separator is maintained at a pressure of 1.013 bar and unliquefied gas at this pressure leaves the heat exchanger at 295 K. What fraction of the methane entering the heat exchanger is liquefied in the process?
Figure 8.5. Linde liquefaction schematic.
Solution
Before we calculate the enthalpies of the streams, a reference state must be chosen. The reference state is arbitrary. Occasionally, an energy balance is easier to solve by setting one of the enthalpies to zero by selecting a stream condition as the reference state. To illustrate the results let us select a reference state of H = 0 at the real fluid at the state of Stream 3 (6 MPa and 300 K). Because state 3 is the reference state, the H3 = 0. The results of the calculations from the Peng-Robinson equation are summarized in Fig. 8.6.
Figure 8.6. Summary of enthalpy calculations for methane as taken from the files Preos.xlsx (above) and PreosPropsMenu.m below.
The fraction liquefied is calculated by the energy balance: m3H3 = m8H8 + m6H6; then incorporating the mass balance: H3 = (1 – m6/m3)H8 + (m6/m3)H6.
The throttle valve is isenthalpic (see Section 2.13). The flash drum serves to disengage the liquid and vapor exiting the throttle valve. The fraction liquefied is (1 – q) = m6/m3 = (H3 – H8)/(H6 – H8) = (0 – 883)/(–12,954 – 883) = 0.064, or 6.4% liquefied. This is in good agreement with the value obtained in Example 5.5 on page 213.
Example 8.10. Adiabatically filling a tank with propane
Propane is available from a reservoir at 350 K and 1 MPa. An evacuated cylinder is attached to the reservoir manifold, and the cylinder is filled adiabatically until the pressure is 1 MPa. What is the final temperature in the cylinder?
Solution
The critical properties, acentric factor and heat capacity constants, are entered on the “Props” page of Preos.xlsx. On the “Ref State” page, the reference state is arbitrarily selected as the real vapor at 298 K and 0.1 MPa, and HR = 0. At the reservoir condition, propane is in the one-root region with Z = 0.888, H = 3290 J/mol, U = 705 J/mol, and S = –7.9766 J/mol-K. The same type of problem has been solved for an ideal gas in Example 2.16 on page 82; however, in this example the ideal gas law cannot be used. The energy balance reduces to Uf = Hin, where Hin = 3290 J/mol. In Excel, the answer is easily found by using Solver to adjust the temperature on the “Props” page until U = 3290 J/mol. The converged answer is 381 K. In MATLAB, the dialog boxes can be used to match U = 3290 J/mol by adjusting T. In the MATLAB window, note that the final T is shown in the “Results” box. The initial guess is preserved in the upper left.
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