Constant Temperature and Volume

2. Constant Temperature and Volume

Consider a fluid of mass m within a rigid tank of volume V that is immersed in a bath at a temperature T for which the real gas state equa- tion yields the P-v diagram presented in Figure

11 . The states F and G represent the saturated states at a specified temperature. At state K

( Figure 14 ) the fluid is in the form of slightly

(c )

sub–cooled vapor. We will divide the system into two parts 1 and 2, and disturb the volume such that portion 1 expands slightly to state K ″, while

portion 2 shrinks to K ′, but the total volume is fixed. Then, P K' >P K" and portion 2 will expand

back so that the system eventually reverts to its original state K.

(b )

Consider the fluid at State D (cf. Figure

14 ). If the fluid is disturbed so that the mother phase volume (say, portion 1) slightly expands to v D" while the embryo volume (portion 2) con- tracts to v D' with the total volume held fixed,

since v D" ≈v D and v D' «v D ,P D" ≈P D , while P D >

P D' . The mother phase is at a higher pressure than the embryo phase, which undergoes increasing

compression. The pressure in the embryo first (a ) decreases along path HJM (cf. Figure 11 ) but

starts increasing again along path MRL (the liq- uid path) until state L is reached at which the

pressure in the embryo equals P D . A disturbance

can occur at several locations inside the system creating a large number of drops that can combine

Figure 13: Variation in the value of g at to form a continuous liquid phase. However,

a specified value of T, but for different since liquid drops generally occupy a smaller

pressures. (a) P = P sat ; (b) P>P sat ; (c) P < volume than the gas phase in a fixed volume sys-

P sat .

tem, the entire vapor body may not condense, and a wet mixture may form. At equilibrium, there cannot be any potential gradients between the liquid and vapor phases so that g liquid =g vapor . At a specified temperature, this condition occurs at a particular value of the saturation pressure P sat , which allows us to determine the volume or quality x, i.e.,

v = xv g + (1–x) v f , or x = (v–v f )/(v g –v f ).

(47) The volume at a stable state v > v G . For the condition v N <v<v G , a metastable vapor

state exists and a mixture of vapor and liquid is formed, but at a higher quality. If v F <v<v M , then a metastable liquid state exists and a vapor-liquid mixture of lower quality is formed. The condition v M <v<v N is unstable and a mixture of vapor and liquid of medium quality is formed.

When phase transformation from a metastable state occurs at a specified temperature and volume, the Helmholtz energy is minimized. Figure 15 presents a plot of a¯ and P vs. v¯ . State D on the P- ¯v diagram (P = 140 bar, v = 0.1 m 3 kmole D -1 , T = 593) corresponds to the point Q on the ¯a- ¯v curve. A disturbance increases ¯a within the mother phase if the volume decreases and vice versa. However, the value of the Helmholtz energy increase is smaller than its decrease with the result that ¯a decreases (at fixed T and V). Conse- quently, the fluid eventually reaches

a vapor state and ¯a is minimized. At a specified temperature and vol- ume the pressure (i.e., P = 133 bar, v

3 = 0.1 m -1 kmole , T = 593 K) and quality adjust such that the Helm-

holtz energy is at a minimum value

D’

K’

(e.g., with vapor at State G where a¯

L D = 20.6 MJ and liquid at State F with K

a higher value of ¯a = 23.2 MJ).

Example 11 ” ”

k.

2.662 kmole of water are contained in a rigid tank of

volume 27 m 3 at 320ºC and

140 bar. What are the val- ues of ¯s, ¯u , and ¯a at

Figure 14. A Pressure-volume diagram for illustrating this state? If the tempera-

fluid stability along an isotherm. ture and volume are main- tained constant, what is the most stable equilibrium state, and what are the values of ¯s, ¯u , and ¯a at this state?

Assume that c Ρ = 5.96 kJ kg -1 K -1 . Solution

¯v = 2.662/27 = 0.0986 m 3 kmole -1 . Applying the RK equation at 593 K and 140 bar (State D in Figure 11 ),

P sat = 133 bar. Since P > P sat at 593 K, and P < P N = 155 bars the state is metastable.

We will use the method described in Chapter 7 to determine the fluid properties. For instance,

¯h

(320ºC, 0.0986 m -1 kmole ) = 44633 kJ kmole or 2477 kJ kg ,

3 -1

¯u (320C, 0.0986 m 3 kmole -1 )=h–P ¯v

= 43252 kJ kmole -1 or 2400 kJ kg -1 , and ¯s (320C, 0.0986 m 3 kmole -1 ) = 91.34 kJ kmole -1 or 5.07 kJ kg -1 . Thereafter,

¯a = ¯u – T ¯s = 43252–593 ×91.34 = –10,912 kJ kmole -1 or –606 kJ kg -1 , and

¯g = ¯h –T ¯s = –9,532 kJ kmole -1 . At a specified value of T and V, the value of ¯a decreases until it reaches a minimum. The fluid at state D transforms into a wet mixture, but g liquid =g vapor . Hence, P must equal 133 bar so that ¯a is minimized, i.e.,

f (320ºC, P new ) +x v g (320ºC, P new ) = 0.0986 m kmole The pressure P new = P sat , which is the saturation pressure at phase equilibrium (that equals 133 bar). Applying the RK equation at 593 K and 133 bar,

3 -1 ¯v = (1–x) v

v (320ºC, 133 bar) = 0.04275, m 3 kmole f -1 or 0.00237 m 3 kg -1 , and v g (320ºC, 133 bar) = 0.236 m 3 kmole -1 or 0.0131 m 3 kg -1 , i.e., x = (0.0986 – 0.04275)/(0.236 – 0.04275) = 0.289, and ¯a = ¯a (1–x) + ¯a

f g ×= –11,114 kJ kmole .

Remarks At equilibrium ¯a = –11,114 kJ kmole -1 , which is lower than the –10,912 kJ kmole -1 value at the metastable point at the same temperature and volume. The disturbance

creates a wet mixture. State K (593 K, 0.211 m 3 kmole -1 ) is a metastable state which, when disturbed, produces a wet mixture with a higher quality at the same temperature and volume, since the initial volume is larger. At specified values of T and P, g reaches two possible minimums. Such multiple states do not exist for the Helmholtz energy at specified values of T and v.