Basic Equation for Region

This section contains all details relevant for the use of the basic equation of region 3 of IAPWS-IF97. Information about the consistency of the basic equation of this region with the basic equations of regions 1, 2, and 4 along the corresponding region boundaries is summarized in Section 10. The auxiliary equation for defining the boundary between regions 2 and 3 is given in Section 4. Section 11 contains the results of computing-time comparisons between IAPWS-IF97 and IFC-67. The estimates of uncertainty of the most relevant properties can be found in Section 12.

The basic equation for this region is a fundamental equation for the specific Helmholtz free energy f This equation is expressed in dimensionless form, 0 = fl(RT), and reads

where S = p/p*, r = T*/T with p* = pc , T* = Tc and R, Tc, and pc given by Eqs. (1), (2), and (4). The coefficients n and exponents and J of Eq. (28) are listed in Table 30.

In addition to representing the thermodynamic properties in the single-phase region, Eq. (28) meets the phase-equilibrium condition (equality of specific Gibbs free energy for coexisting vapor + liquid states; see Table 31) along the saturation line for T> 623.15 K to Tc . Moreover, Eq. (28) reproduces exactly the critical parameters according to Eqs. (2) to (4) and yields zero for the first two pressure derivatives with respect to density at the critical point.

Table 30. Numerical values of the coefficients and exponents of the dimensionless Helmholtz free energy for region 3, Eq. (28)

i

Ii

Ji

ni

i

Ii

Ji

ni

1

0

0

0.106 580 700 285 13 x 101

21

3

4

- 0.201 899 150 235 70 x 101

2

0

0

- 0.157 328 452 902 39 x 102

22

3

16

- 0.821 476 371 739 63 x 10"2

3

0

1

0.209 443 969 743 07 x 102

23

3

26

- 0.475 960 357 349 23

4

0

2

- 0.768 677 078 787 16 x 101

24

4

0

0.439 840 744 735 00 x 10"1

5

0

7

0.261 859 477 879 54 x 101

25

4

2

- 0.444 764 354 287 39

6

0

10

- 0.280 807 811 486 20 x 101

26

4

4

0.905 720 707 197 33

7

0

12

0.120 533 696 965 17 x 101

27

4

26

0.705 224 500 879 67

8

0

23

- 0.845 668 128 125 02 x 10"2

28

5

1

0.107 705 126 263 32

9

1

2

- 0.126 543 154 777 14 x 101

29

5

3

- 0.329 136 232 589 54

10

1

6

- 0.115 244 078 066 81 x 101

30

5

26

- 0.508 710 620 411 58

11

1

15

0.885 210 439 843 18

31

6

0

- 0.221 754 008 730 96 x 10"1

12

1

17

- 0.642 077 651 816 07

32

6

2

0.942 607 516 650 92 x 10"1

13

2

0

0.384 934 601 866 71

33

6

26

0.164 362 784 479 61

14

2

2

- 0.852 147 088 242 06

34

7

2

- 0.135 033 722 413 48 x 10"1

15

2

6

0.489 722 815 418 77 x 101

35

8

26

- 0.148 343 453 524 72 x 10"1

16

2

7

- 0.305 026 172 569 65 x 101

36

9

2

0.579 229 536 280 84 x 10"3

17

2

22

0.394 205 368 791 54 x 10"1

37

9

26

0.323 089 047 037 11 x 10"2

18

2

26

0.125 584 084 243 08

38

10

0

0.809 648 029 962 15 x 10"4

19

3

0

- 0.279 993 296 987 10

39

10

1

- 0.165 576 797 950 37 x 10^3

20

3

2

0.138 997 995 694 60 x 101

40

11

26

- 0.449 238 990 618 15 x 10"4

All thermodynamic properties can be derived from Eq. (28) by using the appropriate combinations of the dimensionless Helmholtz free energy and its derivatives. Relations between the relevant thermodynamic properties and $ and its derivatives are summarized in Table 31. All required derivatives of the dimensionless Helmholtz free energy are explicitly given in Table 32.

Table 31. Relations of thermodynamic properties to the dimensionless Helmholtz free energy 0 and its derivatives a when using Eq. (28)

Property

Relation

Specific internal energy u = f- T{d fid T)p

Specific enthalpy h = f - T(d fid T) + p (d ft dp )T

Specific isochoric heat capacity Cv = (d Uid T)p

Specific isobaric heat capacity Cp = (dh id T) p

Phase-equilibrium condition (Maxwell criterion)

R Ytt

" d0 '

" d 20 "

/A

~d0 '

/A

' d20 "

/A

1 d20 "

dS _

S

S

dSdT

Table 32. The dimensionless Helmholtz free energy equation and its derivatives a according to Eq. (28)

Table 32. The dimensionless Helmholtz free energy equation and its derivatives a according to Eq. (28)

40 40

= n / S+X ni Ii S1 -1 TJi $SS = -n / S2+Xni /i(/i-1) S1 -2 TJi i=2 i = 2 40 40

<pT = 0+X ni S1 Jii tJi -1 $TT = 0+X n S1 Ji (Ji -1) tJi -2

0 0

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