GENERALIZED CORRELATIONS FOR GASES - Volumetric Properties of Pure Fluids

Volumetric Properties of Pure Fluids

3.7 GENERALIZED CORRELATIONS FOR GASES

points in the table that represent the appropriate phase. The related Lee/Kesler correlation for predicting the vapor pressure curve itself is presented and discussed in Sec. 6.5.

The Lee/Kesler correlation provides reliable results for gases that are nonpolar or only slightly polar; for these, errors of no more than 2 or 3 percent are typical. When applied to highly polar gases or to gases that associate, larger errors can be expected.

The quantum gases (e.g., hydrogen, helium, and neon) do not conform to the same corresponding-states behavior as do normal fluids. Their treatment by the usual correlations is sometimes accommodated by use of temperature-dependent effective critical parameters.¹⁹ For hydrogen, the quantum gas most commonly found in chemical processing, the recom- mended equations are:

T c / K = _{__________}43.6 1 + _{______}21.8 2.016T

( for H 2 ) (3.54)

Pc / bar = _{__________}20.5 1 + _{______}44.2 2.016T

( for H 2 ) (3.55) Figure 3.11: The Lee/Kesler correlation for Z⁰ = F⁰(Tr, Pr).

0 0.2 0.4 0.6 0.8 1.0 1.2

P_r T_rr = 0.7

Z⁰

Compressed liquids (T_r<1.0) Two-phase

region

Gases 0.9

0.9 C 1.0

1.2 1.5

4.0

0.7

10.0 5.0

0.5 1.0

0.5 0.2

0.05 0.1

19J. M. Prausnitz, R. N. Lichtenthaler, and E. G. de Azevedo, Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd ed., pp. 172–173, Prentice Hall PTR, Upper Saddle River, NJ, 1999.

3.7. Generalized Correlations for Gases 105

V c / cm ³ ·mol ⁻¹ = _{__________}51.5 1 − _{______}9.91 2.016T

( for H 2 ) (3.56)

where T is absolute temperature in kelvins. Use of these effective critical parameters for hydrogen requires the further specification that ω = 0.

Pitzer Correlations for the Second Virial Coefficient

The tabular nature of the generalized compressibility-factor correlation is a disadvantage, but the complexity of the functions Z⁰ and Z¹ precludes their accurate representation by simple equations. Nonetheless, we can give approximate analytical expression to these functions for a limited range of pressures. The basis for this is Eq. (3.36), the simplest form of the virial equation:

Z = 1 + BP_{___}

RT = 1 + ₍ _{_}BP c RTc ⁾ P_{__} r

T r = 1 + B ˆ _{__}Pr

T r (3.57) Figure 3.12: Three-dimensional plot of the Lee/Kesler correlation for Z⁰ = F⁰(Tr, Pr) as given in Tables D.1 and D.3.

10–2 10–1

100 1001

0.2 0.4 0.6 0.8 1 1.2

Pr Tr Z0

The reduced (and dimensionless) second virial coefficient and the Pitzer correlation for it are:

B ˆ ≡ _{____}BP c

RT c (3.58) B ˆ = B ⁰ + ωB ¹ (3.59) Equations (3.57) and (3.59) together become:

Z = 1 + B ⁰ _{__}Pr

T r + ω B ¹ _{__}Pr T r

Comparison of this equation with Eq. (3.53) provides the following identifications:

Z ⁰ = 1 + B ⁰ _{__}P r

Tr (3.60)

and

Z ¹ = B ¹ P_{__} r T r

Second virial coefficients are functions of temperature only, and similarly B⁰ and B¹ are functions of reduced temperature only. They are adequately represented by the Abbott equations:²⁰

20These correlations first appeared in 1975 in the third edition of this book, attributed as a personal communication to M. M. Abbott, who developed them.

B ⁰ = 0.083 − _{_____}0.422

T _r^1.6 (3.61) B ¹ = 0.139 − _{_____}0.172

T _r^4.2 (3.62)

The simplest form of the virial equation has validity only at low to moderate pressures where Z is linear in pressure. The generalized virial-coefficient correlation is therefore useful only where Z⁰ and Z¹ are at least approximately linear functions of reduced pressure.

Figure 3.13 compares the linear relation of Z ⁰ to P_r as given by Eqs. (3.60) and (3.61) with values of Z⁰ from the Lee/Kesler compressibility-factor correlation, Tables D.1 and D.3 of App. D. The two correlations differ by less than 2% in the region above the dashed line of the figure. For reduced temperatures greater than T_r ≈ 3, there appears to be no limitation on the pressure. For lower values of T_r the allowable pressure range decreases with decreasing temperature. A point is reached, however, at T_r ≈ 0.7 where the pressure range is limited by the saturation pressure.This is indicated approximately by the left-most segment of the dashed line. The minor contributions of Z¹ to the correlations are here neglected. In view of the uncertainty associated with any generalized correlation, deviations of no more than 2% in Z ⁰ are not significant.

The relative simplicity of the generalized second-virial-coefficient correlation does much to recommend it. Moreover, temperatures and pressures of many chemical-processing

3.7. Generalized Correlations for Gases 107

C ˆ ≡ _{______}CPc ²

R ² T c ² (3.64) C ˆ = C ⁰ + ωC ¹ (3.65)

Figure 3.13: Comparison of correlations for Z⁰. The virial-coefficient correlation is represented by the straight lines; the full Lee/Kesler correlation, by the points. In the region above the dashed line the two correlations differ by less than 2%.

1.0

0.9

0.8

0.7

0.0 0.5 1.0 1.5 2.0 2.5

P_r T_r 0.8

0.9 1.0 1.1

1.3

1.5 1.8

2.4 4.0

Z⁰

operations lie within the region appropriate to the compressibility-factor correlation. Like the parent correlation, it is most accurate for nonpolar species and least accurate for highly polar and associating molecules.

Correlations for the Third Virial Coefficient

Accurate data for third virial coefficients are far less common than for second virial coefficients. Nevertheless, generalized correlations for third virial coefficients do appear in the literature.

Equation (3.38) may be written in reduced form as:

Z = 1 + B ˆ _{___}Pr

T r Z + C ˆ ₍ _{___}P r T r Z

)² (3.63)

where the reduced second virial coefficient B ˆ is defined by Eq. (3.58). The reduced (and dimensionless) third virial coefficient and the Pitzer correlation for it are:

An expression for C ⁰ as a function of reduced temperature is given by Orbey and Vera:²¹

C ⁰ = 0.01407 + _{________}0.02432

T r − _{_______}0.00313

T r ^10.5 (3.66) The expression for C¹ given by Orbey and Vera is replaced here by one that is algebraically simpler, but essentially equivalent numerically:

C ¹ = −0.02676 + _{________}0.05539

T r ^2.7 − 0.00242_{________}

Tr ^10.5 (3.67) Equation (3.63) is cubic in Z, and it cannot be expressed in the form of Eq. (3.53). With T_r and P_r specified, Z can be found by iteration. An initial value of Z = 1 on the right side of Eq. (3.63) usually leads to rapid convergence.

The Ideal-Gas State as a Reasonable Approximation

The question often arises as to when the ideal-gas state may be a reasonable approximation to reality. Figure 3.14 can serve as a guide.

21H. Orbey and J. H. Vera, AIChE J., vol. 29, pp. 107–113, 1983.

Figure 3.14: In the region lying below the curves, where Z⁰ lies between 0.98 and 1.02, the ideal-gas state is a reasonable approximation.

0.1

0.01

0.001 P_r

T_r Z⁰= 0.98

Z⁰= 1.02

0 1 2 3 4

3.7. Generalized Correlations for Gases 109

Example 3.10

Determine the molar volume of n-butane at 510 K and 25 bar based on each of the following:

(a) The ideal-gas state.

(b) The generalized compressibility-factor correlation.

(c) Equation (3.57), with the generalized correlation for B ˆ . (d) Equation (3.63), with the generalized correlations for B ˆ and C ˆ .

Solution 3.10

(a) For the ideal-gas state, V = RT_{___}

P = _{__________} ( 83.14 ) ( 510 )

25 = 1696.1 cm ³ ·mol ⁻¹ (b) With values of T_c and P_c given in Table B.1 of App. B,

Tr = _{_____}510

425.1 = 1.200 Pr = _{_____}25 37.96 = 0.659 Interpolation in Tables D.1 and D.2 then provides:

Z ⁰ = 0.865 Z ¹ = 0.038 By Eq. (3.53) with ω = 0.200,

Z = Z ⁰ + ωZ ¹ = 0.865 + ( 0.200 ) ( 0.038 ) = 0.873 V = ZRT_{____}

P = (________________ 0.873 ) ( 83.14 ) ( 510 )

25 = 1480.7 cm ³ ·mol ⁻¹

If Z¹, the secondary term, is neglected, Z = Z ⁰ = 0.865. This two-parameter corresponding-states correlation yields V = 1467.1 cm ³ ·mol ⁻¹ , which is less than 1% lower than the value given by the three-parameter correlation.

B ⁰ = −0.232 B ¹ = 0.059 Equations (3.59) and (3.57) then yield:

B ˆ = B ⁰ + ωB ¹ = −0.232 + ( 0.200 ) ( 0.059 ) = −0.220 Z = 1 + ( −0.220 ) 0.659_{_____}

1.200 = 0.879

from which V = 1489.1 cm ³ ·mol ⁻¹ , a value less than 1% higher than that given by the full Lee/Kesler compressibility-factor correlation.

(d) Values of C⁰ and C¹ are given by Eqs. (3.66) and (3.67):

C ⁰ = 0.0339 C ¹ = 0.0067 Equation (3.65) then yields:

C ˆ = C ⁰ + ωC ¹ = 0.0339 + ( 0.200 ) ( 0.0067 ) = 0.0352 With this value of C ˆ and the value of B ˆ from part (c), Eq. (3.63) becomes, Z = 1 + ( −0.220 ) ₍ _{_}0.659

1.200Z ₎ + ( 0.0352 ) ( _{______}0.659 1.200Z

) ²

Solution for Z yields Z = 0.876 and V = 1485.8 cm³·mol^–1. The value of V dif- fers from that of part (c) by about 0.2%. An experimental value for V is 1480.7 cm³·mol^–1. Significantly, the results of parts (b), (c), and (d) are in excellent agreement. Mutual agreement at these conditions is suggested by Fig. 3.13.

Example 3.11

What pressure is generated when 500 mol of methane is stored in a volume of 0.06 m³ at 50°C? Base calculations on each of the following:

(a) The ideal-gas state.

(b) The Redlich/Kwong equation.

Solution 3.11

The molar volume of the methane is V = 0.06 / 500 = 0.0012 m ³ ·mol ⁻¹ . (a) For the ideal-gas state, with R = 8.314 × 10 ⁻⁵ bar· m ³ ·mol ⁻¹ ·K ⁻¹ :

P = RT_{___}

V = ( 8.314 × 10 ⁻⁵ ) ( 323.15 )

__________________ 0.00012 = 223.9 bar (b) The pressure as given by the Redlich/Kwong equation is:

P = _{____}RT

V − b − _{______}a ( T₎

V ( V + b) (3.40) Values of b and a(T) come from Eqs. (3.44) and (3.45), with Ω, Ψ, and α ( T r ) = Tr ^−1/2 from Table 3.1. With values of T_c and P_c from Table B.1, we have:

T r = _{______}323.15

190.6 = 1.695

b = 0.08664 _________________(8.314 × 10 ⁻⁵ ) ( 190.6 ) 45.99 = 2.985 × 10 ⁻⁵ m ³ ·mol ⁻¹

a = 0.42748 ___________________________ ( 1.695 ) ^−0.5 (8.314 × 10 ⁻⁵ ) ² ( 190.6 ) ² 45.99 = 1.793 × 10 ⁻⁶ bar ·m ⁶ ·mol ⁻²

3.7. Generalized Correlations for Gases 111 Substitution of numerical values into the Redlich/Kwong equation now yields:

P = __________________0.00012 − 2.985 × 10 (8.314 × 10 ⁻⁵ ) ( 323.15 ) ⁻⁵ − __________________________0.00012(0.00012 + 2.985 × 10 1.793 × 10 ⁻⁶ ⁻⁵ ) = 198.3 bar (c) Because the pressure here is high, the full Lee/Kesler generalized compressibility- factor correlation is the proper choice. In the absence of a known value for P_r, an iterative procedure is based on the following equation:

P = _{____}ZRT

V = ___________________Z(8.314 × 10 ⁻⁵ ) ( 323.15 ) 0.00012 = 223.9 Z Because P = Pc Pr = 45.99 P r , this equation becomes:

Z = _{________}45.99 P r

223.9 = 0.2054 P r or P r = _{_______}Z 0.2054

One now assumes a starting value for Z, say Z = 1. This gives P_r = 4.68, and allows a new value of Z to be calculated by Eq. (3.53) from values interpolated in Tables D.3 and D.4 at the reduced temperature of T_r = 1.695. With this new value of Z, a new value of P_r is calculated, and the procedure continues until no significant change occurs from one step to the next. The final value of Z so found is 0.894 at P_r = 4.35. This is confirmed by substitution into Eq. (3.53) of values for Z⁰ and Z¹ from Tables D.3 and D.4 interpolated at P_r = 4.35 and T_r = 1.695. With ω = 0.012,

Z = Z ⁰ + ωZ ¹ = 0.891 + ( 0.012 ) ( 0.268 ) = 0.894 P = ZRT_{____}

V = __________________ ( 0.894 ) (8.314 × 10 ⁻⁵ ) ( 323.15 ) 0.00012 = 200.2 bar

Because the acentric factor is small, the two- and three-parameter compressibility- factor correlations are little different. The Redlich/Kwong equation and the generalized compressibility-factor correlation give answers within 2% of the experimental value of 196.5 bar.

Example 3.12

A mass of 500 g of gaseous ammonia is contained in a vessel of 30,000 cm³ volume and immersed in a constant-temperature bath at 65°C. Calculate the pressure of the gas by:

(a) The ideal-gas state;

(b) A generalized correlation.

SOLUTION 3.12

The molar volume of ammonia in the vessel is:

V = _{__}V ^t n = _{____}V ^t

m / ℳ = _{_________}30,000

500 / 17.02 = 1021.2 cm ³ ·mol ⁻¹

(a) For the ideal-gas state,

P = _{___}RT

V = ________________ ( 83.14 ) ( 65 + 273.15 ) 1021.2 = 27.53 bar

(b) Because the reduced pressure is low (P_r ≈ 27.53/112.8 = 0.244), the generalized virial-coefficient correlation should suffice. Values of B⁰ and B¹ are given by Eqs. (3.61) and (3.62). With T_r = 338.15/405.7 = 0.834,

B ⁰ = −0.482 B ¹ = −0.232 Substitution into Eq. (3.59) with ω = 0.253 yields:

B ˆ = − 0.482 + ( 0.253 ) ( −0.232 ) = −0.541 B = _{_____}B ˆ RTc

P c = __________________− ( 0.541 ) ( 83.14 ) ( 405.7 ) 112.8 = −161.8 cm ³ ·mol ⁻¹ By the second equality of Eq. (3.36):

P = _{____}RT

V − B = _____________ ( 83.14 ) ( 338.15 ) 1021.2 + 161.8 = 23.76 bar

An iterative solution is not necessary because B is independent of pressure. The c al- culated P corresponds to a reduced pressure of P_r = 23.76/112.8 = 0.211. Reference to Fig. 3.13 confirms the suitability of the generalized virial-coefficient correlation.

Experimental data indicate that the pressure is 23.82 bar at the given conditions. Thus the ideal-gas state yields an answer high by about 15%, whereas the virial-coefficient correlation gives an answer in substantial agreement with exper- iment, even though ammonia is a polar molecule.

Example 3.13

For n-butane at 470 K prepare a plot of compressibility factor Z as a function of pressure in bar, for pressures up to 200 bar, comparing the results from:

(a) The Redlich/Kwong equation of state.

(b) The Peng/Robinson equation of state.

(c) The 2-term virial equation, with the generalized correlation for B ˆ . (d) The 3-term virial equation, with the generalized correlations for B ˆ and C ˆ . (e) Data from the NIST Chemistry WebBook, which can be considered equivalent

to accurate experimental results.

Solution 3.13

From App. B, T_c = 425.1 K, P_c = 37.96 bar, and ω = 0.200. Thus, T_r = 470/425.1 = 1.1056

3.7. Generalized Correlations for Gases 113 (a) For the Redlich/Kwong equation of state, we compute:

q = Ψ_{__}

Ω T r^−3/2 = _{_______}0.42748

0.08664 1.105 6 ^−3/2 = 4.2442 β = Ω P _r

__ T _r = 0.08664

P∕37.96

_______ 1.1056 = 0.002064P with P in bar Then, for each pressure, we must solve:

Z = 1 + β − qβ Z − β

________

Z(Z + β) = 1 + 0.002064P − 0.008762P Z − 0.002064P

_______________ Z(Z + 0.002064P ) One way to do this for a series of pressures all at once is to formulate it as a mini- mization problem, minimizing the sum over all of the selected pressure values of

(1 + 0.002064 P_i− 0.008762 P_i Z_i− 0.002064 P_i

_________________ Z_i ( Z_i + 0.002064 P _i ) − Z_i ) ² This can be done in Microsoft Excel, for example, by computing the above expression for a list of pressures P_i , adding all of those cells, and then minimizing that sum by varying all of the Z_i . If, and only if, the equation that we were aiming to solve is satisfied at every pressure will this sum equal zero. Otherwise, it will be positive. The solver function in Excel is well suited to minimize the value in a single cell by varying values in many cells. Of course, many other approaches in many different software packages are possible. Doing the above produces the curve labeled R/K in the figure.

(b) For the Peng-Robinson equation of state, we compute:

α _PR ( T _r ; ω) = [1+ (0.37464 + 1.54226ω − 0.26992 ω ² )

(

1 − Tr ^−1/2

)

^]²

α _PR ( T _r ; ω) = [1 + (0.37464 + 1.54226 × 0.200 − 0.26992 × 0.200²) (_{1 − 1.1056}^−1/2)]²

= 1.0669

q = ___________ Ψα_PRΩ(TT_r_r; ω)= _______________ 0.45724 × 1.06690.07780 × 1.1056 = 5.6714 β = Ω P_r

__ T_r = 0.07780 P_{_______}∕37.96

1.1056 = 0.001854P with P in bar.

In this case, for each pressure we must solve:

Z = 1 + β − qβ Z − β

____________________________

(Z + (1 + ^√^__2 )β)(Z + (1 − ^√^__2 )β) Z = 1 + 0.001854P − 0.010513P Z − 0.001854P

___________________________Z + 0.004475P)(Z − 0.007678 P) ( The same solution strategies as in part (a) are applicable, and produce the curve labelled P/R in the figure.

B⁰ = 0.083 − _{_____}0.422

T _r^1.6 = 0.083 − _{________}0.422

1.1056^1.6 = −0.2764 B¹ = 0.139 − _{_____}0.172

T _r^4.2 = 0.139 − _{________}0.172 1.1056^4.2 = 0.0262 And

Z = 1 + (B⁰ + ωB¹) P_r

__ T_r = 1 + (−0.2764 + 0.200 × 0.0262) _{_______}P∕37.96 1.1056 With P in bar. This gives the straight line in the figure labeled 2-term.

(d) For the 3-term virial equation, with the generalized correlations for B ˆ and C ˆ , we must compute:

C⁰ = 0.01407 + _{_______}0.02432

T_r − _{_______}0.00313 T _r^10.5 C⁰ = 0.01407 + _{_______}0.02432

1.1056 − _{_________}0.00313 1.1056^10.5 = 0.03498 C¹ = −0.02767 + 0.05539_{_______}

T _r^2.7 − _{_______}0.00242 T _r^10.5 C¹ = −0.02767 + _{________}0.05539

1.1056^2.7 − _{_________}0.00242

1.1056^10.5 = 0.013724

B ˆ = B⁰ + ωB¹ = −0.2764 + 0.200 × 0.0262 = −0.2711

C ˆ = C⁰ + ωC¹ = 0.03498 + 0.200 × 0.01372 = 0.03772 And the 3-term virial equation in reduced form (Eq. 3.63) is:

Z = 1 + B ˆ P_r

___ T_rZ + C ˆ ( P_r

___ T_rZ) ² = 1 − 0.2711 P_{_______}∕37.96

1.1056Z + 0.03772 ( P_{_______}∕37.96 1.1056Z) ² Z = 1 − 0.00646 P

__ Z + 2.141 × 10⁻⁵ ( P

__ Z) ²

A solution strategy similar to that used in (a) and (b) can be applied, yielding the curve labeled 3-term in the figure.

(e) Finally, for this part, we simply download the data from the NIST WebBook at the selected pressures. The values of Z computed from the molar density down- loaded is plotted as the curve labelled NIST in the figure.

In this particular case, the Peng/Robinson equation gives excellent agreement with the NIST data. The Redlich/Kwong equation and the 3-term virial expansion also give reasonable results. Of course, the 2-term virial expansion can only work at relatively low pressures.

Dalam dokumen Introduction to Chemical Engineering Thermodynamics (Halaman 122-134)