Entropy During Phase Change

4. Entropy During Phase Change

Consider the case of a boiling liquid. Since the pressure and temperature are generally unchanged during a phase transformation, applying Eq. (43),

ds = dh/T – vdP/T = dh/T. (55) Integrating the expression between the saturated liquid and vapor states

(56) Generalizing for any change from phase α to β,

s g –s f = (h g –h f )/T = h fg /T.

s α –s β =h αβ /T (57) m. Example 13

The entropy of water at T tp = 0ºC ,P TP = 0.611 kPa, is arbitrarily set to equal zero, where the subscript tp refers to the triple point. Using this information, determine:

s(liquid, 100ºC) assuming c = 4.184 kJ kg –1 K –1 . Compare your results with values tabulated in the Steam tables ( Tables A-4 ). s(sat vapor, 0ºC, 0.611 kPa) assuming h

fg = 2501.3 kJ kg K . The entropy generated if the water at 0ºC and 0.611 kPa is mechanically stirred to form vapor at 0ºC in an adiabatic blender.

Figure 26: P–v diagram for water. s(393 K, 100 kPa) assuming c p,0 = 2.02 kJ kg –1 K –1 and that steam behaves as an ideal

gas. Solution

Applying Eq. (53),

K . Since s(0ºC) = 0, s(100ºC) =1.306 kJ kg K . –1 –1 From the Table A-4A , s(100ºC) = 1.3069 kJ kg K , which is very close. Applying Eq. (56) to the vaporization process at the triple point,

s(373) – s(273) = 4.184 ln (373/273) = 1.306 kJ kg –1

s –s = 2501.3÷273 = 9.16 kJ kg –1 K –1 g f . Since,

s f (273 K, 0.611 kPa) = 0, s g (273 K, 0.611 kPa) = 9.16 kJ kg –1 K –1 . (A )

ds – δq/T b = δσ. Since δq = 0, ds = δσ. Integrating this expression,

s g –s f = σ. (B)

Using Eq. (A) and (49b), with s f (0ºC, 0.611 kPa) = 0

g –s f = 9.16 kJ kg K = σ. (C) s(393 K, 100 kPa) – s(273 K, 0.611) =

s –1

2.02 ln (393 –1 ÷273) – (8.314÷18.02)ln (100÷0.611) = – 1.616 kJ kg K , or

s(393, 100 kPa) = 9.16 – 1.616 = 7.54 kJ kg –1 K .

Conventional Steam tables (e.g., –1 Table A-4A ) yield a value of 7.467 kJ kg K .

Pc

u=Const

P=Const

h=Const h=Const

Figure 27: T-s diagram of a pure fluid. Remarks

For estimating entropy in vapor phase at low pressures, one can use ideal gas tables

0 ( 0 Tables A-12 ) also, e.g., s(393,100) – s (273,0.611)= (s (393) – R ln (100/100)) – (s (273) – R ln (0.611/100)) where P ref = 100 kPa

The stirring process is irreversible. Therefore, viscous dissipation converts mechani- cal energy into thermal energy. The heat vaporizes the liquid, and increases the en- tropy.

n. Example 14 Determine the enthalpy of water at 25ºC and 1 bar (i.e., at point A of Figure 26 ) if the

enthalpy of saturated liquid (at point F) at that temperature is known. Solution

From the Steam tables ( –1 A-4A )P = 0.03169 bar and h

sat

f = 104.89 kJ kg for saturated

liquid water at 25ºC (at point F). Since c = constant for incompressible liquids,

ds = c dT/T (A) Along the 25ºC isotherm (curve FA), Eq. (A) illustrates that ds = 0, and the process is

isentropic. Since, dh = Tds + vdP, for this case

dh = v dP.

Upon integrating between points F and

,P 1 ,P 1 A,

,P

h A (25 C, 1 bar) – h F (25ºC, 0.03169 bar) = ∫v dP ≈ v f

(25ºC, 0.03169) (P A Figure 28: Illustration of the Gibbs-Dalton law. –P F ) =

1.0029 –1 ×10 × (100 – 3.169) = 0.09711 kJ kg , and

h –1

A (25 C, 1 bar) = 104.89 + 00.09711= 104.987 kJ kg .

Remarks Since the enthalpy values are virtually insensitive to pressure, one can assume that h A ≈h F , i.e., the enthalpy of a compressed liquid at given temperature and pressure is ap-

proximately that of the saturated liquid at that temperature. If pressure at point A is 25 bar, h –1

–1 A = 104.89 + 2.504 = 107.394 kJ kg . Use of Table A-4 yields a value of 107.2 kJ kg , which is very close, with the difference being due

to the assumption of constant specific volume.

a. T–s Diagram We are now in a position to discuss the representation of the states of a pure fluid on a T–s diagram. For instance, we may arbitrarily assign a zero entropy to liquid water at its triple point (i.e., point B of Figure 27 ). For incompressible liquids, we can assume that s(0.01ºC, 1

bar) ≈ s(0.01ºC,0.006 bar) = 0. If the water is again heated from 0.01ºC to 100 C at 1 bar, it is possible to evaluate the values of s, and those of s f (100ºC, 1 bar), (point F) using Eq. (53), and

s g (100ºC, 1 bar) (point G) using Eq. (56). If the vapor behaves as an ideal gas (which is gener- ally true at lower pressures) the entropy may be evaluated using either of Eqs. (47c) or (50a) (Point S) . In this manner, the behavior of a substance can be characterized at lower pressures on the T–s diagram, as illustrated by the curve BFGS in Figure 27 at 1 bar. Thereafter, by changing the pressure, entropy values can be obtained at higher pressures. Since the ideal gas assumption is flawed at elevated pressures, Eqs. (47c) or (49) must be modified. This will be discussed further in Chapters 6 and 7. As is apparent from the path A–C–D, an inflection oc-

curs in the slope of the isobar (at the critical pressure P c ) at the critical point C, i.e., ( ∂T/∂s) Pc =

0 at this point. Also illustrated on the diagram are isometric, isenthalpic and isoquality lines.