Entropy ds Relation

1. Entropy ds Relation

Any thermodynamic property for a simple compressible substance can be expressed as a function of two other independent thermodynamic properties. Therefore, for instance,

s = s (v,T), i.e.,  ∂ s 

ds =

 (30) ∂ v  T

dv +

 ∂ T  v

dT .

Eliminating ( ∂s/∂T) v with Eq. (14), we obtain ds =  ∂ s 

 dT 

 (31) ∂ v  T

dv + c v

 T 

where c v = f(T,v). Using the third relation (cf. Eq. 22) the above expression can be expressed in terms of measurable properties, namely,

dT

ds =

 (32) ∂ T  v

dv + c v

 T 

Likewise, considering s = s(T,P), ds = ( ∂s/∂T) P dT + ( ∂s/∂P) T dP. Substituting from Eq. (17b) for the first term on the RHS of the above equation, and using the

fourth relation Eq. (26) for its second term, we obtain the expression ds = c p dT/T – ( ∂v/∂T) P dP

(33) Both Eqs. (32) and (33) provide relations for ds in terms of the measurable properties P, v, and

T. Equation (32) is suitable with a state equation of the form P = P(T,v) (e.g., the Van der Waals equation), while Eq. (33) is more conveniently used when the state equation is ex- pressed in the form v = v(T,P).

a. Remarks Since ( ∂s/∂T) v = c v /T (and c v > 0 when T > 0), the gradient ( ∂s/∂T) v (taken along an isometric curve) has a finite and positive slope for all T > 0. Similarly, the gradient ( ∂s/∂T) P =c P /T has positive values (along isobars), since c P > 0 when T > 0.

Along an isotherm, ds T =–( ∂v/∂T) P dP T (cf. Eq. 33). Since in the vicinity of T = 0 K, ( ∂v/∂T) P = 0, this implies that s is independent of pressure at a temperature of abso- lute zero.

Incompressible solids and liquids undergo no volumetric change, i.e., dv = 0. Since, Eq. (32) states that ds = c v dT/T, and c v =c p = c = constant for this case,

s = c ln T + constant. Consider Eq. (33), ds = c p dT/T – ( ∂v/∂T) P dP = c p dT/T – β P v dP,

(34) where β P = (1/v)( ∂v/∂T) P (and, likewise, β T = –(1/v)( ∂v/∂P) T ). (In Chapter 6 we have

learned that for liquids and solids the following relation relates the volume to the two compressibilities, namely, v/v ref = exp( β P (T–T ref )– β T (P–P ref )).) For isentropic processes, ds = 0. and Eq. (34) yields the relation

dT/T = ( β p v dP)/c p , so that ( ∂T/∂P) s =T( ∂v/∂T) P /c P =Tv β P /c P =T β P /c p ´,

(35) where c 3

p ´=c p /v, kJ/K m . If c P , β P and v are all approximately constant, then by inte- grating Eq. (34), we can obtain isentropic relations for solids and liquids, i.e.,

ln T/T ref =( β P ) v(P–P ref )/c P . (36) If β P ≈ 0, the temperature remains constant as the pressure changes. In case β P > 0

(i.e., the liquid or solid expands upon heating), the temperature increases during com- pression and decreases during expansion. Likewise, if β P < 0 (e.g., for rubber, or wa- ter between 0 and 4ºC at 1bar), the temperature decreases with an increase in pres- sure.

Alternately, using Eq. (32), ds = c v dT/T + ( ∂P/∂T) v dv. Since ( ∂P/∂T) v = –( ∂v/∂T) P /( ∂v/∂P) T = β P / β T ,

ds = c v dT/T + ( β P / β T ) dv. For isentropic processes, ds = 0, and

( ∂T/∂v) s =–T( ∂P/∂T) v /c v or ( ∂T/∂v) s = –T β P /( β T c v ),

where both β T and β P are almost constant for most solids and liquids. Therefore, dv s = –dT s β T c v /(T β P ).

(37) The term β T is related to the stress–strain relation in solids. From the above relation

we find that when dv s < 0 (i.e., during isentropic compression) dT s > 0 if β P > 0 (i.e., the material expands upon heating) which implies that the temperature rises. On the we find that when dv s < 0 (i.e., during isentropic compression) dT s > 0 if β P > 0 (i.e., the material expands upon heating) which implies that the temperature rises. On the

the fact that h fg and v fg are finite, while the temperature remains constant during the boiling process). However, we can use other relations, such as T ds + v dP = dh to obtain the relation ds = dh/T during isobaric vaporization which on integration yields s fg =h fg /T

g. Example 7 Weights are gradually placed on an insulated copper bar in order to compress it to 1000 bars. If the compression is adiabatic and reversible (i.e., the material reverts to its original state once the load is removed) determine: The change in the solid temperature. The internal energy change. The temperature after the load is removed. Assume that T

–7 –1 –4 3 o

= 250 K, β P = 48 ×10 K , β T = 7.62 ×10 bar , v = 1.11 ×10 m

= 0.372 kJ kg K , and c v = 0.364 kJ kg K . Treat Cu as simple com- pressible substance Solution Since the process is adiabatic and reversible we will use the relation

( T ∂T/∂P) s = Tv β P /c P or ( ∆T/∆P) s =T o v β P /c P .=

3 {250K - ×1.1×10 m kg

1 ×48x10 -6 K -1 } /0.372kJkg -

K -6 = 3.548x10 K/Kpa and ∆P = 100 × 1000 Kpa. Hence Stretched

1 -1

dT s = 0.36 K and T will rise to 250.36 K.

Figure 6: The bending of a beam.

Applying the First law to an adiabatic reversible process du s = – Pdv s . Recall from Eq. (37) that

dv s = –dT s β T c v /(T β P ), i.e., du s = {P β T c v /(T β P )}dT s Integrating and assuming that P is not a function of temperature and remaining at an

average value of 500 bar. du = {500 bar × 7.62×10 –7 bar –1 × 0.364 kJ kg s –1 K –1

÷ (250 K × 48×10 –6 K –1 )} 0.36 K = 0.00416 kJ kg –1 .

The temperature after the load is removed is 250 K, since the process is reversible.

h. Example 8 Obtain a relation for ds for an ideal gas. Using the criterion for an exact differential show that for this gas c v is only a function of temperature.

Solution For an ideal gas

P = RT/v. (A) Using Eq. (A) and Eq. (32), we obtain ds = R (dv/v) + c v (dT/T).

(B) Comparing Eq. (B) with the relation dZ = Mdx + Ndy, and using the criterion for an

exact differential we obtain ∂{(c v /T)/ ∂v} T = ∂ {(R/v)/∂T} v = 0.

since at constant volume (R/v) is not a function of temperature (or pressure). There- fore, the term ∂{(c v /T)/ ∂v} T is not a function of v and, at most, is a function of tem- perature alone.

i. Example 9 Obtain a relation for ds for a gas that follows the RK equation of state.

Solution For an RK gas, a state relation of the form

P = RT/(v – b) – a/(T 1/2 v(v + b)) (A) can be used. Using Eqs. (A) and (32), we obtain the expression ds = c v (dT/T) + (R/(v – b) + a/(2(T 3/2 v(v + b))).

(B) Remark

We can show that c v is a function of both v and T for an RK gas by using the criterion for an exact differential.

j. Example 10

A VW gas is used as the working fluid in an ideal power cycle. A relation between T and v is required for an isentropic process (data for c vo (T) is available). If v 1 = 0.006 m 3 kg –1 ,T 1 = 200 K, the compression ratio v 1 /v 2 = 3, determine the values of T 2 and P 2 if the gas is air.

Solution Recall that

ds = c v dT/T + ( ∂P/∂T)vdv, i.e., ds = cv dT/T + R dv/(v–b) (A)

Using the criterion for an exact differential, ( ∂c v / ∂v) T = ∂ [{R/(v – b)}/∂T] v = 0. This implies that c v is not a function of volume and is a function of temperature alone,

i.e., c v =c vo (T). Since ds = 0 for the ideal cycle, upon integrating Eq. (A), ∫c vo dT/T = – R ln(v – b) + C.

(B) Since, s o = ∫c p,o dT/T, we define (s´) o (T) = ∫c vo dT/T = ∫(c p,o – R)dT/T = s o – R ln T.

(C) We use Eqs. (B) and (C) to obtain the relation

(s´) o = – R ln {(v–b)} + C´. Therefore,

(D) Simplifying,

(s o

2 ´) – (s 1 ´) = R ln((v 1 – b)/(v 2 – b)).

2 ´) /R – (s 1 ´) /R) = (exp (s 2 /R)/T 2 )/(exp(s 1 /R)/T 1 ) = (v 2 – b)/(v 1 – b). (E) Upon defining v r = exp (s o /R)/T, Eq. (E) can be written in the form

exp ((s o

v r2 /v r1 = (v 2 – b)/(v 1 – b).

(F) Values of v r are usually tabulated. Once the volume ratio v 2 /v 1 is specified, T 2 can be

determined from Eq. (F). Using the VW equation of state, we can then determine P 2 . Since, v 1 =0.006 m 3 kg –1 at T 1 = 200 K, the VW equation yields P 2

= 121.3 – 45.3 = 76 bar. At T= 200 K, v r1 = 1707. We will use the relation

v r2 /v r1 = (v 2 – b)/(v 1 – b),

÷ 28.97 = 0.00127 m 3 kg –1 . Therefore,

and the values v 1 = 0.006 m 3 kg –1 , b = 0.0367

v 2 = 0.006 ÷ 3 = 0.002 m 3 kg –1 , and

v r2 /v r1 = (0.002 – 0.00127) ÷ (0.006 – 0.00127) = 0.154, so that v r2 = 1707 × 0.154 = 262.9.

The tabulated values indicate that at v r2 = 263, T 2 = 423 K.

Finally, using the VW equation of state P

2 = 0.08314 × 423 ÷ (0.002 × 28.97–0.0367) – 1.368 ÷ (0.002 × 28.97) = 1656 – 408= 1248 bar. Remarks

If a = b =0, Eq. (E) represents the state relation for ideal gas. For an ideal gas, the re-

lation yields v r2 = 469, T 2 = 335 K, and P 2 = 480 bar.

Applying Eq. (B) and assuming c vo to be constant, c v ln(T 2 /T 1 ) = R ln((v 1 – b)/(v 2 – b)), which can be simplified and written in the form

(G) where k = c p,o /c vo .

T 2 /T 1 = [(v 1 – b)/(v 2 – b)] (k–1) ,

Using (G) in VW equation of state

(P – a/v 2 )(v – b) k = (P – a/v 2 )(v – b) 2 k 2 2 1 1 1 (P – a/v 2 )(v – b) k = (P – a/v 2 )(v – b) 2 k 2 2 1 1 1

measurable properties of a simple compressible substance. Show that c p /c v =k= β T / β s . Determine a relation for the sound speed for an ideal gas.

Determine a relation for the sound speed for a VW gas. Solution Recall that the speed of sound

c 2 = –v 2 ( ∂P/∂v) s = v/ β s

ds = 0 =c v dT/T + ( ∂P/∂T) v dv, and (A) ds = 0 = c p dT/T –( ∂v/∂T) P dP.

(B) We multiply Eq. (A) by (T/c v ) and Eq. (B) by (T/c p ) and then subtract one of the re-

sulting relations from the other to obtain ( ∂P/∂T) v (T/c v ) dv s +( ∂v/∂T) P (T/c p ) dP s = 0, or

(C) ( ∂P/∂v) s =–k( ∂P/∂T) v /( ∂v/∂T) P , where

(D) k(T,v) = c p (T,v)/c v (T,v).

(E) Applying the expression for the speed of sound c 2 = –v 2 ( ∂P/∂v) s = v/ β s in Eq. (D),

c 2 =v 2 k(T,v)( ∂P/∂T) v /( ∂v/∂T) P . (F) Using the cyclical rule ( ∂P/∂v) T ( ∂v/∂T) P ( ∂T/∂P) v = –1

(G) we obtain

( ∂v/∂T) = – (∂P/∂T)/(∂P/∂v) (H) Substituting from Eq. (H) in Eq. (F),

(I) With c 2 = v/ β s , in Eq. (I) v/ β s = k(T,v) v/ β T , or k(T,v) = β T / β s . In the case of ideal gases,

2 c 2 = – k(T,v) v ( ∂P/∂v)

T .= k(T,v) v/ β T

2 2 2 2 k = – (c 2 /v )/(–RT/v )=c /RT or c = kRT. (J)

Typically we denote c as c 0 for ideal gases.

For a VW gas,

2 ∂P/∂v = – RT/(v – b) 3 + 2a/v (K) Thereafter, combining Eqs. (I) and (K)

c 2 = k(T,v) v 2 (RT/(v – b) 2 + a/v 3 )

(L)

Remarks If, in the VW state relation, a = b= 0, the expression reduces to the sound speed for an ideal gas. In that case, k = k(T). High pressures often develop within the clearance space in turbine seals, and gas leaks are governed by the resulting choked flow conditions. The value of the sound speed through a real gas is required in order to evaluate this condition. In the case of liquids and solids, a very large pressure is required to cause a small change in the volume so that ( ∂P/∂v) T → ∞ and, consequently, c → ∞. Therefore,

sound travels at faster speeds in liquids and solids.

2 Applying Eq. (L) to the case of an ideal gas, c 2

= k(T) RT, and dividing Eq.(I) by c o we obtain the expression,

(c 2 /c 2 o 2 ) = – (k(T R ,v R ´)/(k o (T R )T R )) v R ( ∂P R / ∂v R ´) TR .