Joule-Thomson Calculation Using a Mollier Diagram and Steam Tables

Steam at 400 bar and 500ºC undergoes a Joule-Thomson expansion to 1 bar. Determine the temperature of the steam after the expansion using

a. Fig. 3.3-1a

b. Fig. 3.3-1b

c. The steam tables in Appendix A.III

Question

Joule-Thomson Calculation Using a Mollier Diagram and Steam Tables

Steam at 400 bar and 500ºC undergoes a Joule-Thomson expansion to 1 bar. Determine the temperature of the steam after the expansion using

a. Fig. 3.3-1a

b. Fig. 3.3-1b

c. The steam tables in Appendix A.III

Accepted Answer

(Since only one thermodynamic state variable—here the final temperature—is unknown, from the discussion that precedes this illustration we can expect to be able to obtain a solution to this problem.)

We start from Illustration 3.2-3, where it was shown that

[latex]\hat{H}_{1}=\hat{H}\left(T_{1}, P_{1} ight)=\hat{H}\left(T_{2}, P_{2} ight)=\hat{H}_{2}[/latex]

for a Joule-Thomson expansion. Since [latex]T_{1} ext { and } P_{1} ext { are known, } \hat{H}_{1}[/latex] can be found from either Fig. 3.3-1 or the steam tables. Then, since [latex]\hat{H}_{2}\left(=\hat{H}_{1} ight.[/latex], from the foregoing) and [latex]P_{2} ext { are known, } T_{2}[/latex] can be found.

a. Using Fig. 3.3-1a, the Mollier diagram, we first locate the point P = 400 bar = 40 000 kPa and T = 500ºC, which corresponds to [latex]\hat{H}_{1}[/latex] = 2900 kJ/kg. Following a line of constant enthalpy (a horizontal line on this diagram) to P = 1 bar = 100 kPa, we find that the final temperature is about 214ºC.

b. Using Fig. 3.3-1b, we locate the point P = 400 bar and T = 500 º C (which is somewhat easier to do than it was using Fig. 3.3-1a) and follow the curved line of constant enthalpy to a pressure of 1 bar to see that [latex]T_{2}[/latex] = 214ºC.

c. Using the steam tables of Appendix A.III, we have that at P = 400 bar = 40 MPa and T = 500ºC, [latex]\hat{H}[/latex] = 2903.3 kJ/kg. At P = 1bar = 0.1MPa, [latex]\hat{H}[/latex] = 2875.3 kJ/kg at T = 200ºC and [latex]\hat{H}[/latex] = 2974.3 kJ/kg at T = 250 º C. Assuming that the enthalpy varies linearly with temperature between 200 and 250ºC at P = 1 bar, we have by interpolation

[latex]T=200+(250-200) imes \frac{2903.3-2875.3}{2974.3-2875.3}=214.1^{\circ} C[/latex]

[ The Aspen Plus [latex]^R[/latex] simulation for this illustration available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 3>3.4-1. The results using the IAPWS-95 model (equations on which the steam tables are based) give the following results:

[latex] ext { exit temnerature }=48885 K =2157^{\circ} C[/latex]

Using the more approximate Peng-Robinson equation of state (discussed in Chapter 5), the result is

[latex] ext { exit temperature }=484.87 K =211.7^{\circ} C[/latex]

This result is in reasonable agreement with that obtained with the more accurate IAPWS-95 model. ]

Comment

For many problems a graphical representation of thermodynamic data, such as Figure 3.3-1, is easiest to use, although the answers obtained are approximate and certain parts of the graphs may be difficult to read accurately. The use of tables of thermodynamic data, such as the steam tables, generally leads to the most accurate answers; however, one or more interpolations may be required. For example, if the initial conditions of the steam had been 475 bar and 530ºC instead of 400 bar and 500ºC, the method of solution using Fig. 3.3-1 would be unchanged; however, using the steam tables, we would have to interpolate with respect to both temperature and pressure to get the initial enthalpy of the steam.

One way to do this is first, by interpolation between temperatures, to obtain the enthalpy of steam at 530ºC at both 400 bar = 40 MPa and 500 bar. Then, by interpolation with respect to pressure between these two values, we obtain the enthalpy at 475 bar. That is, from

[latex]\begin{aligned}\hat{H}\left(40 MPa , 500^{\circ} C ight) &=2903.3 kJ / kg & \hat{H}\left(50 MPa , 500^{\circ} C ight) &=2720.1 kJ / kg \\hat{H}\left(40 MPa , 550^{\circ} C ight) &=3149.1 kJ / kg & \hat{H}\left(50 MPa , 550^{\circ} C ight) &=3019.5 kJ / kg\end{aligned}[/latex]

and the interpolation formula

[latex]\Theta(x+\Delta)=\Theta(x)+\Delta \frac{\Theta(y)-\Theta(x)}{y-x}[/latex]

where Θ is any tabulated function, and x and y are two adjacent values at which Θ is available, we have

[latex]\begin{aligned}\hat{H}\left(40 MPa , 530^{\circ} C ight) &=\hat{H}\left(40 MPa , 500^{\circ} C ight)+30 imes \frac{\hat{H}\left(40 MPa , 550^{\circ} C ight)-\hat{H}\left(40 MPa , 500^{\circ} C ight)}{550-500} \&=2903.3+30 imes \frac{3149.1-2903.3}{50}=2903.3+147.5 \&=3050.8 kJ / kg\end{aligned}[/latex]

and

[latex]\hat{H}\left(50 MPa , 530^{\circ} C ight)=2720.1+30 imes \frac{3019.5-2720.1}{50}=2899.7[/latex] kJ / kg

Then

[latex]\begin{aligned}\hat{H}\left(47.5 MPa , 530^{\circ} C ight) &=\hat{H}\left(40 MPa , 530^{\circ} C ight)+7.5 imes \frac{\hat{H}\left(50 MPa , 530^{\circ} C ight)-\hat{H}\left(40 MPa , 530^{\circ} C ight)}{50-40} \&=3050.8+\frac{7.5}{10} imes(2881.7-3050.8)=2924.0 kJ / kg\end{aligned}[/latex]

Question 3.4.1: Joule-Thomson Calculation Using a Mollier Diagram and Steam ...

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