Procedure for Determining Thermodynamic Properties

1. Procedure for Determining Thermodynamic Properties

Thermodynamic properties can be determined, once the state equation, critical con- stants, and corresponding ideal gas properties are known. Some useful formulas are listed be- low and some thermodynamic data is listed in Table 2 . The reference conditions should be specified. For example for water, the reference condition is generally specified as that of the saturated liquid at the triple point. The choice of reference conditions is arbitrary. Here,

v R ´ = v/v c ´, v c ´ = RT c /P c

v R = v/v c , Z=P R v R ´/T u

= – u Res c,R R /(RT c ) = (u o (T) –

u(T,P))/RT c ,

h c,R = (h o (T) – h(T,P))/RT c ,

s c,R = (s o (T,P) – s(T,P))/R,

c P,c R = (c P (T,P) – c P,o (T))/R,

c v,c,R = (c v (T,P) – c v,o (T))/R, µ JT,R = µ JT c p /v c ´,

g c,R = (g o (T,P) – g(T,P))/RT c ,

a c,R = (a o (T,P) – a(T,P))/RT c φ = f/P = exp((g(T,P)– g o (T,P))/RT)

f T g with g =g and T inv with µ JT =0

sat

The constants are used in the formula c po Figure 23: Schematic illustration of a method = of determining the thermodynamic properties

A o +B o T+C o T –2 +D o T 2 +E T 3 o . In the

of a material using a P–h diagram.

h(873,60

h fg )

Figure 24: Schematic illustration of the determination of enthalpy of a vapor or a real gas with respect to the values at the reference condi- tion.

range 0 K < T < 1000 K, the maximum error is less than 8%. oo. Example 41

Determine the thermodynamic properties of water at 250 bar and 600ºC using the RK equation of state. Assume that c = 28.85 + 0.01206 T + 1.002

×10 K –1 ,h fg,ref = 2501.3 kJ kg –1 ,P c = 220.9 bar, and T c = 647.3 K. The reference conditions are those for the saturated liquid at its triple point, i.e., 273.15 K and 0.006113 bar.

5 /T 2 kJ kmole –1

p,o

The molecular weight of water is 18.02 kg kmole –1 . Solution

A schematic diagram of the procedure followed is illustrated in on a P–h diagram in Figure 25 and Figure 24 . First, the reference condition is selected at which h f =s f = 0 (e.g., the saturated liquid state at the triple point of water at point A in the figure). At the reference condition h = 0 + h

(point B in Figure 25 and Figure 24 ). Therefore, s g =h fg /T = 2501.3

g fg = 2501 kJ kg

÷273 = 9.17 kJ kg –1 K ,

The entropy of the saturated vapor at 273.15 K and 0.0061 bar is 9.17 kJ kg –1 K –1 above the entropy of saturated liquid at same temperature and pressure.

For the vapor at 273 K and 0.0061 bar (i.e., at P R,ref = 0.000028 and T R,ref = 0.422) the reduced correction factor (h o – h)/RT c ≈ 0, since the pressure is low and the intermo- lecular attraction forces are weak. Therefore,

h = h = 2501 kJ kg o –1 at the triple point (point B in Figure 25 and Figure 24 ). Similarly, s = s = 9.17 kJ kg –1 K –1 o or 165.2 kJ kmole –1 K –1 . The values of h o (873 K) and s o (873 K, 250 bar) can be obtained using the specific heat relations for an ideal gas, i.e.,

Figure 25: Schematic illustration of an entropy calculation starting from a temperature of absolute zero; C: critical point.

h o –1 = 3706 kJ kg = 66782 kJ kmole –1 .

This corresponds to the point D. Similarly,

s –1 o (873 K, 250 bar) = 6.47 kJ kg K or 116.6 kJ kmole K .

The ideal gas internal energy u o at 873 K can be determined using the relation u –1

o =h o – RT = 66782 – 8.314 × 873 = 59,524 kJ kmole . The correction or residual factors at 873 K and 250 bar can be obtained. From charts

we see that at P R = 1.13, T R = 1.35, Z = 0.845. Therefore, (h o – h)/RT c = 0.735, (u o – u)/RT c = 0.526, and (s o – s)/R = 0.389. Consequently,

u = 3146 kJ kg –1 , and s = 6.29 kJ kg K . Thereafter, the enthalpy can be determined using the relation h = u + Pv = u + ZRT.

× (8.314 ÷ 18.02) × 873 = 3486 kJ kg –1 ,

which is represented by point C. Other properties can be similarly obtained.