Figure 8.7 shows a process on the P-V diagram with an arbitrary process path, wherein a system undergoes a change from its initial state to a final state along the path shown by the solid line.
Figure 8.7 An arbitrary thermodynamic process.
As stated earlier in the chapter, the types of computational problems, at the elementary level, involve estimations of the changes in the thermodynamic properties. For those properties that are state functions, the approach to solving these problems and determining the changes in the thermodynamic quantities involves postulating an alternative path from the initial state of the system to its final state, as shown by the dashed lines in Figure 8.7. The system first undergoes a constant volume process at Vinitial, labeled 1, followed by a constant pressure process at Pfinal, labeled 2. Computations of thermodynamic property changes along these constant volume and constant pressure paths present a much more manageable challenge as compared to those along the actual path followed by the process. The generalized principle of this approach is shown Figure 8.8 [6]. Appropriately designed, it is possible to determine the changes in the required quantities in each of the segments constituting the postulated alternate path. The overall change in the transformation of the system from the initial to final state is simply the sum of the changes in these individual segments.
Figure 8.8 Solution strategy for thermodynamic problems—path formulation, abstraction, and interpretation of results.
Source: Prausnitz, J. M., R. M. Lichtenthaler, and E. G. de Azevedo, Molecular Thermodynamics of Fluid-Phase Equilibria, Third Edition, Prentice Hall, Upper Saddle River, New Jersey, 1999.
Next, as mentioned earlier, the problem of determining equilibrium involves estimating the fugacities of the components. Appropriate mathematical expressions are available for various situations to determine the fugacities from the volumetric properties of the substances. The following examples provide an idea of the computational problems in chemical engineering thermodynamics.
EXAMPLE 8.2.1 VOLUMETRIC FLOW RATE OF METHANE
Calculate the volumetric flow rate of methane to the natural gas burner described in problem 7.2.1 using both the ideal gas law and van der Waals EOS. Assume that the pressure is atmospheric. What is the error in using the ideal gas law if the van der Waals EOS correctly describes the volumetric behavior of methane?
Solution
The molar volume, according to the ideal gas law, follows:
Since the molar flow rate is 1 mol/s, the volumetric flow rate of 24.436 L/s is obtained using the ideal gas law. It should be noted that the temperature is expressed in units of kelvin (K) and pressure in atmosphere (atm). If the specific volume of the gas is desired in the units of liters per mole (L/mol), then the value of the universal gas constant R used for the calculation is 0.082 L atm/mol K.
The values of constants a and b in the van der Waals equation are obtained from the following [5]:
The critical temperature and pressure for methane are 190.6 K and 45.6 atm, respectively.
The values of a and b calculated from these data and the resulting van der Waals EOS at the given pressure and temperature are as follows:
Equation E8.3 is a cubic equation in vvdw, which can be solved using techniques described in the earlier chapters. Using the Goal Seek function in Excel yields a molar volume value of 24.387 L/mol (the value obtained from the ideal gas law was used as the initial guess for obtaining the solution).8 The volumetric flow rate is then equal to 24.387 L/s. The percentage of error if the ideal gas volume is used is 0.2%, which is quite low under these conditions.
8. A cubic equation should have three roots, meaning that there are three values of v that satisfy equation E8.3. However, only one root is positive and real; the other two are complex and are not valid solutions for the molar volume.
It is instructive to see how the error changes when the conditions are changed. If the pressure is changed to 200 atm, then the ideal gas molar volume is 0.122 L/mol, yielding a volumetric flow rate of 122 mL/s. The molar volume, according to the van der Waals EOS, is computed to be 0.099 L/mol, yielding a volumetric flow rate of 99 mL/s. Assuming that the van der Waals EOS yields a more accurate estimate of the actual volume, an engineer will incur an error of approximately 23% if this calculation is performed assuming ideality of gas behavior. An error of this magnitude is unacceptable in design and operation of the process, reinforcing the need for an EOS that can accurately predict the volumetric behavior of the substances.
EXAMPLE 8.2.2 EQUILIBRIUM MOISTURE CONTENT OF EXHAUST GASES
Determine the quantity of water that will condense when the hot exhaust gas of problem 7.2.2 is cooled to 25°C.
Solution
Figure 8.9 is a schematic representation of the physical process occurring as the hot exhaust gases exchange heat with the cooling water and are cooled in turn to 25°C (298 K). Some of the water condenses, resulting in two outlet streams—one gas and the other liquid—exiting the process. Neither CO2 nor N2 undergo any condensation at this temperature, and the liquid phase consists only of water. However, some water is present in the gas stream as well, as not all of the water can condense. The component material balance readily yields the moles of water condensing m = (2 − n) mol/s, leaving n as the only unknown in the problem.
Figure 8.9 Condensation of water vapor.
The concept of equilibrium is used to solve for n. The two outlet streams are in equilibrium with each other, and hence, according to equation 8.9, the fugacity of the water vapor in the gas phase must equal the fugacity of the condensed water. The fugacity of the water vapor is equal to its partial pressure, and the fugacity of the liquid water is equal to the vapor pressure (also called saturation pressure) at its temperature.9 The vapor pressure of water at 25°C (298 K) is 23.8 mm Hg (from thermodynamic data sources), and assuming that outlet streams are exhausted to the atmosphere (pressure 760 mm Hg):
9. This is the idealization of the system behavior; however, it is an excellent approximation for the system under consideration here. In many other situations, nonideality of mixtures must be accounted for.
The term in front of 760 represents the mole fraction of water vapor in the gas. Solving equation E8.4, we get the following:
n = 0.275 mol/s and m = 1.725 mol/s
The condensate flow rate, therefore, is 1.725 × 18 (molecular weight of water) = 31 g/s, or the volumetric flow rate is ~1.86 L/min. The mole fraction of water vapor in the gas phase is found to be 0.031.
EXAMPLE 8.2.3 DIFFUSION EQUILIBRIUM IN A MEMBRANE SYSTEM
A closed chamber at 40°C is separated into two equal parts by a membrane permeable only to helium. The left part is filled with 0.99 mol of ethane and 0.01 mol helium. The right part has 0.01 mol helium and 0.99 mol nitrogen. What are the mole fractions of helium in each part of the chamber at equilibrium? The final pressures were 37.5 bar in the left part and 84.5 bar in the right part of the chamber.
Solution
The initial schematic of the system is shown in Figure 8.10.
Figure 8.10 Helium transport across a membrane.
The solution to this problem requires the application of equation 8.9 to helium:
The superscripts denote the left and right parts of the chamber.
It should be noted that helium is the only component present on both sides of the membrane, and there is no escaping tendency on the part of ethane or nitrogen from the part of the chamber where they exist. The fugacity is expressed in terms of the total pressure of the chamber P, the mole fraction of helium in the chamber (yHe), and the fugacity coefficient ϕHe:
Equation E8.6 expresses fugacity of helium on the left part of the chamber. A similar equation can be written for its fugacity on the right part of the chamber. The fugacity coefficient is obtained from the EOS for the mixture. When virial EOSs (see problem 8.8) are used to describe the volumetric behavior of the mixtures on either side of the chamber, the resulting equation is as follows:
Here B is the second virial coefficient, the two numbers in the subscript of B indicating two components with the numbers 1, 2, and 3 referring to components helium, ethane, and nitrogen, respectively, and the superscript L and R referring to the left and right parts of the chamber. Comparison of equations E8.6 and E8.7 indicates that the exponential terms in the second equation are the fugacity coefficients. The values of various virial coefficients (in cm3/mol) are B11 = 17.17, B12 = 24.51, B22 = –169.23, B23 = –44.39, B33 = –1.55, and B13 = 21.97.
Equation E8.7 contains two unknowns: the mole fractions of helium on the left and right sides of the chamber. The material balance on helium allows us to express both these mole fractions in terms of moles of helium on one of the sides of the chamber. If there are 
moles of helium on the left side of the chamber at equilibrium, then we get the following:
Equations E8.7, E8.8, and E8.9 form a system of three equations in three unknowns that can be solved using any appropriate software.
Solution (using Excel)
The solution using Excel is shown in Figure 8.11.
Figure 8.11 Excel solution to example 8.2.3.
The values of the pressures of the two parts, the gas constant, temperature, and the virial coefficients are entered in cells A2 through I2. A guess for the value of the moles of helium on the left part of the chamber is entered in cell A8. Cells B8 through F8 contain the values of moles of helium on the right part of the chamber, the mole fractions on the left and right parts of the chamber, and the fugacities on the left and right sides of chambers. An objective function equal to the difference in the two fugacities is entered in cell G8 (=E8-F8). Then the Excel Goal Seek function is used to find the solution by telling it to set the value of cell G8 equal to zero by manipulating cell A8. As can be seen from the figure, nearly 0.005 mole of helium diffuses from the right side to the left side of the membrane. The fugacity of helium at equilibrium is 0.478 bar.
The exercise for obtaining the solution to the problem using Mathcad and confirming that the same result is obtained is left to the reader.
These three examples provide an idea of the nature of computational problems in chemical engineering thermodynamics. The student will encounter such problems in the junior year of the study and beyond in his/her career.
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