Summary of Relations

5. Summary of Relations

These are four important relations, namely

( ∂ T / ∂ v= ) s −∂∂ ( P / s ) v , ( ∂ T / ∂ P ) s = ( ∂∂ v / s ) P , ( ∂∂ P / T= ) v ( ∂∂ s / v ) T , and ( ∂∂ v / T= ) P −∂∂ ( s / P ) T .

Even though the relations were derived by using thermodynamic relations for closed systems, the derivations are also valid for open systems as long as we follow a fixed mass.

Figure 4: Thermodynamic mnemonic dia- gram.

For a point function, say, P = P(T,v), it can be proven that

(29) Equation (29) is useful to obtain the derivative ( ∂v/∂T) P if state equations are available for the

( ∂P/∂T) v ( ∂T/∂v) P ( ∂v/∂P) T = –1, i.e., ( ∂v/∂T) P =–( ∂P/∂T) v /( ∂P/∂v) T

pressure (e.g., in the form of the VW equation of state). Likewise, if u = u(s,v), ( ∂u/∂s) v ( ∂s/∂v) u ( ∂v/∂u) s = –1.

Using the relation ∂u/∂s = T, ∂s/∂v = P/T, and ∂v/∂u = –P, we find that, as expected, T (P/T)(–1/P) = –1.

e. Example 5 Show that both the isothermal expansivity β P = (1/v)( ∂v/∂T) P and the isobaric com-

pressibility coefficient β T = – (1/v)( ∂v/∂P) T tend to zero as T → 0. Solution

Example 1 shows that ( ∂s/∂T) v → 0 and (∂s/∂P) v → 0 as T → 0. From the fourth of the relations,

( ∂v/∂T) P = –( ∂s/∂P) T , so that ( ∂v/∂T) P → 0. Similarly, using the third relation and the cyclic relations it may be shown that

( ∂P/∂T) v = –(( ∂v/∂T)/(∂v/∂P)) = (∂s/∂v) T → 0 as T → 0. Since ∂v/∂T → 0, it is appar- ent that ∂v/∂P → 0 as T → 0.

Remark The experimentally measured values of β P and β T both tend to 0 as T → 0. Using the

relations the reverse can be shown, i.e., ( ∂s/∂T) v → 0 and (∂s/∂P) v → 0.

f. Example 6 When a refrigerant is throttled from the saturated liquid phase using a short orifice, a two-phase mixture of quality x is formed. We are asked to determine the choking flow conditions for the two-phase mixture, which occurs when the mixture reaches

the sound speed (c 2 =–v 2 ( ∂P/∂v) s ). We must also derive an expression for the speed of sound in a two-phase mixture. Assume ideal gas behavior for the vapor phase and

that the liquid phase is incompressible. Solution

During the elemental expansion of a two phase mixture of a specified quality x from P to P + dP, and v to v + dv,

dh = Tds + v dP, and (A) du = Tds – Pdv.

(B) Since,

dh f = d(u f +Pv f )

For incompressible liquids,

dh f = du f +v f dP.

For a two phase mixture of vapor and liquid,

dh = x dh g + (1 – x) dh f = x dh g + (1 – x)(du f +v f dP).

Assuming ideal gas behavior for the vapor phase, and if du f = cdT, then

(C) Similarly,

dh = x c p,o dT + (1 – x)(c f dT + v f dP).

(D) Considering constant entropy in Eqs. (A) and (B), using Eqs. (C) and (D), dividing by

du = x c vo dT + (1 – x) c f dT.

dT, we obtain the relation

(dP/dT) s = (x c p,o + (1 – x) c f )/(v – (1 – x)v f ), i.e.,

(E)

(F) Dividing Eq. (E) by Eq. (F), we obtain the relation –(dP/dv) s = (x c p,o + (1 – x)c f )(P/(v – (1 – x)v f ))/(x c vo + (1 – x)c f ).

(dv/dT) s = –(x c vo + (1 – x)c f )/P.

(G) Using the definition of the sound speed,

c 2 = –v 2 (dP/dv) s , where v = xv g + (1 – x)v f ,

(H) Eq. (G) can be written as

(I) Since v g = RT/P,

c 2 =v 2 (xc p,o + (1 – x)c f ) P/((v – (1 – x)v f )(xc vo + (1 – x)c f .

v = (xRT/P+ (1 – x)v f ), and

c 2 = RT(x + (1 – x)(Pv f /(RT))) 2 (xc p,o + (1 – x)c f )/(x(xc vo + (1 – x)c f )). (J) If x =1, then, as expected,

x, quality

Figure 5: Sound speed of a two phase mixture for R –134a Figure 5: Sound speed of a two phase mixture for R –134a

c 2 = RT(Pv f /RT) 2 /x → ∞.

Using tabulated values for R–134a,

c f = 1.464 kJ kg –1 K –1 , and c p,o at 298 K = 0.851 kJ kg –1 K –1 .

Using the values for P = 690 kPa, R = 0.08149 kJ kg –1 K ,c vo = 0.7697 kJ kg K , v f = 0.000835 m 3 kg –1 , and γ = 1.1. For the conditions x = 1, T = 298 K,

c = 163.4 m s –1 .

A plot for c with respect to x is illustrated in Figure 5 . The expression for the speed of sound in a solid–vapor mixture is similar except that c f must be replaced by c S , spe- cific heat of solid.