Compressibility Charts (Principle of Corresponding States)

5. Compressibility Charts (Principle of Corresponding States)

We now illustrate how compressibility charts can be constructed by employing Eq. (3) and the RK equation of state, Eq. (37). We define

(42) Using Eq. (42) in Eq. (37) and writing in terms of reduced variables we obtain the expression:

v R ’ = v/v c ’, where v ′= c RT P c / c .

( v R ′− . 0 08664 ) Tvv 05 R . R ′ ( ′+ R . 0 08664 )

Therefore, for given values of P R and T R , v ′ R can be obtained. Thereafter, the compressibility factor is determined from the relation Z = P v / R T by inserting the reduced variables, since

(44) It is useful to eliminate v ′ R in Eqs. (43) and (44) and obtain a relation for Z RK (based on the RK equation of state) in terms of P R and T R , i.e.,

Z=P R P c v ′ R v ′ c /( R T R T c )=P R v ′ R /T R .

Z RK T R / P R − . 0 08664 Z RK T R / PZ R ( RK T R / P R + . 0 08664 )

This relation can be simplified in order to obtain Z (= Z RK ) in terms of P R and T R , namely, Z 3 – Z 2 + (a´ – b´ 2 – b´) Z – a´ b´ = 0, where

(46) and b´ = 0.08664 P R /T R .

a´ = 0.4275 P R /T 2.5 R ,

(47) The appendix presents explicit solutions for the three roots of Z. As an illustration, Figure 6a

contains a compressibility chart for Z vs P R with T R as a parameter for an equation of state for which Z C =0.2801. Figure 7 presents a compressibility chart for Z vs P R with T R as a parameter

for an equation of state for which Z C =0.2801.

For sake of illustration, we consider water at a pressure of 250 bar, a temperature of 873 K. Therefore, P R = 250 ÷220.9 = 1.132, T R = 873 ÷647.3 = 1.349, so that a´ = 0.329 and b´ =

0.0727. Consequently, the value of (a´ – b´ 2 – b´) is 0.151, and a´ b´ = 0.01665, and a single real root exists for Eq. (46), i.e., Z = 0.845 (point A, Fig. 6a ). Likewise, if P = 133 bars and T = 593 K, P R = 0.601, T R = 0.916, a´ = 0.320, b´ = 0.0569, (a´ – b´ 2 – b´) = 0.260, and a´ b´ = 0.01824. There are now three roots for the equation, i.e.,, Z 1 = 0.115 (for a liquid–like solution, point L), Z 2 = 0.249 (unstable solution, point M; see Chapter 10), and Z 3 = 0.632 (vapor–like solution, point V).

e. Example 5 What is the value of Z RK at P R = 1.5 and v ′ R = 0.45?

What is the value of Z RK at P R = 1.5 and T R = 1.15?

Figure 7: A compressibility chart for Z vs. P R with T R as a parameter for an equation of or state for which Z C =0.2801 (from R. Sonntag, C. Borgnakke, and G. J. Wiley, Funda- mentals of Classical Thermodynamics, 5th Ed. John Wiley &Sons, 1998, p 763. With

permission.). What is the value of Z for CH 4 at T = 219.3 K and P = 69.6 bars according to the RK

equation? What is the value of Z for N 2 at T = 145.1 K and P = 50.85 bars according to the RK

equation? Solution

= 3.2029 – 1.6409 = 1.56. By interpolating we find that Z RK = 0.59. The compressibility charts give the same re- sult. For the second problem assume Z RK = 0.6. The LHS = 1.5.

The RHS = 1.15 ÷ (0.6 × 1.15 ÷ 1.5 – 0.08664) – 0.42748 × 1.5 ÷ (1.15 1.5 × 0.6 × (0.6 × 1.15 ÷ 1.5 + 0.08664)) = 3.0801 – 1.5853 = 1.494. Hence, the LHS ≈ RHS. P R = 69.6 ÷ 46.4 = 1.5, T R = 219.3 ÷ 190.7 = 1.15, and, therefore, Z RK = 0.6. P R = 50.85 ÷ 33.9 = 1.5, T R = 145.1 ÷ 126.2 = 1.15, and, therefore, Z RK = 0.6.

Remarks It is seen that at specified P R and T R Z RK is equal for all real gases.

f. Example 6 Determine the value of v and compare it with that obtained from the steam tables, the value of v ′ R , and that of Z RK for water at a pressure of 250 bars and a temperature of

600ºC. Solution

For water P c = 220.9, and T c = 647.3 K. Using Eq. (41)

a = 0.42748 R 2 T 2.5/ c P c = 0.42748 ×(0.0814 bar m 3 kmole –1 K –1 ) 2 ×(647.3 K) 2.5 ÷220.9 bar

= 142.59 bar m 6 K 1/2 kmole –2

b = 0.08664 RT c /P = 0.08664 × 0.08314 bar m 3 kmole –1 K c –1 × 647.3 K ÷ 220.9 bar = 0.0211 m 3 kmole –1 . Therefore, for RK equation (37) 250 bar = (0.08314 bar m 3 kmole –1 K –1 × 873 K) ÷ ( v – 0.0211)

– 142.59 bar m 6 ÷(873 1/2 v ( v + 0.0211)).

Solving for v , we obtain three real solutions. Selecting the largest of the three values, which corresponds to a vapor–like solution, v = 0.246 m 3 kmole –1 . ∴ v = 0.246 m 3 kmole –1 ÷ 18.02 kg kmole –1 = 0.01361 m 3 kg –1 .

The steam tables give a value of 0.014137, a difference of –3.7%.

3 v –1 ′

c =T c /P c = 0.08314 bar m kmole K × 647.3 K ÷ 220.9 bars = 0.244 m kmole . Since, v R ′ = v / v ′ c = 0.246 ÷ 0.244 = 1.008, and P R = 1.132, T R = 1.349, using Eq.

(43), v R ′ = 1.008. Now, (P v )

RK =Z RK

R T. Since,

v RK = 0.246 m kmole ,

Z = 250 bar × 0.246 m 3 kmole –1 ÷ (0.08314 bar m RK 3 kmole –1 K –1 × 873 K) = 0.845. Remarks

The reduced parameters P R = P/P c = 250 ÷ 220.9 = 1.132, and T R = T/T c = 873 ÷ 647.3 = 1.349. A value of v ′ R = 1.007 can be obtained using Eq. (43). Thereafter,

since v ′ c = 0.244, v = 0.246 m 3 kmole –1 .

Equation (46) is a representation of the principle of corresponding states, which states that the compressibility factor for all gases is the same at specified values of P R and T . The factor evaluated with two parameter equation of state is normally denoted as Z (0) R . Using Eq. (48) we can now plot Z RK vs. P R using v ′ R as a parameter. At fixed tem-

peratures, as the pressure of a gas increases from very low values, its volume de- creases, the product Pv ≈ constant, and, hence, initially Z ≈ 1. As the volume de-

1.2 2.5 (Boyle Isotherm)

Sat. Vap

T R = 1.0

0.4 Sat. Liq

Figure 8: Simplified Z chart illustrating various regimes. Most of the plots were generated using RK equation.

creases, the pressure increases due to frequent molecular collisions of molecules. However, the intermolecular attractive forces reduce the pressure so that (Pv) < (Pv) ideal gas and, consequently, the value of Z decreases. Beyond a certain pressure, further pressure increments produce smaller and smaller reductions in the volume (i.e, liquid volumes) so that the value of Pv (or Z) again increases. Therefore, the com- pressibility factor passes through a minimum value with respect to P R (e.g., point B in Figure 8 at T R =1.2), since the finite body volume dominates pressure effects at high pressures and overwhelms the intermolecular attractive forces. At low pressures, the variation in the values of Z with pressure has both negative and positive slopes. The temperature T R at which ( ∂Z/∂P R ) = 0 is known as the Boyle temperature. At higher

pressures Z = 1 at a particular value of P R , once T R is fixed, i.e., the decrease in pres- sure due to attractive forces at this condition equals the increase in pressure due to more frequent collisions within a smaller free volume available for molecular motion. The Z–minimum condition, the Boyle temperature, and the Z = 1 condition are dis- cussed below. Table 1 contains a comparison of values of Z from charts and other state equations.

Table 1: Comparison of values of Z from charts and other state equations.

% error Z % error

2 1.3 0.7 0.65 -7.1

0.7 0 0.69 -1.4

2 3 0.96 0.92 -4.2

0.95 -1

0.95 -1

4 1.3 0.69 0.75 8.7 0.7 1.4 0.66 -4.3

4 2 0.96 0.91 -5.2

0.95 -1

% error Z % error