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864 | Thermodynamics

EXAMPLE 17–13 Extrema of Rayleigh Line

Consider the T- sdiagram of Rayleigh flow, as shown in Fig. 17–54. Using

the differential forms of the conservation equations and property relations,

show that the Mach number is Maa
1 at the point of maximum entropy

(point a), and Mab
1/ at the point of maximum temperature (point b).

Solution It is to be shown that Maa
1 at the point of maximum entropy

and Mab
1/ at the point of maximum temperature on the Rayleigh line.

Assumptions The assumptions associated with Rayleigh flow (i.e., steady

one-dimensional flow of an ideal gas with constant properties through a con-

stant cross-sectional-area duct with negligible frictional effects) are valid.

Analysis The differential forms of the mass (rV
 constant), momentum

[rearranged as P+ (rV)V
constant], ideal gas (P
rRT), and enthalpy

change ( h
cp T) equations can be expressed as

(1)

(2)

(3)

The differential form of the entropy change relation (Eq. 17–40) of an

ideal gas with constant specific heats is

(4)

Substituting Eq. 3 into Eq. 4 gives

(5)

since

cpR
cvSkcvR
cvScv
R/(k1)

Dividing both sides of Eq. 5 by dTand combining with Eq. 1,

(6)

Dividing Eq. 3 by dVand combining it with Eqs. 1 and 2 give, after rear-

ranging,

(7)

Substituting Eq. 7 into Eq. 6 and rearranging,

(8)

Setting ds/dT
0 and solving the resulting equation R^2 (kRTV^2 ) 
0 for

Vgive the velocity at point ato be

Va
2 kRTa¬and¬Maa
(9)

Va

ca

2 kRTa

2 kRTa

 1

ds

dT

R

T 1 k 12



R

TV^2 >R

R^21 kRTV^22

T 1 k 121 RTV^22

dT

dV

T

V



V

R

ds

dT

R

T 1 k 12



R

V

dV

dT

ds
cp

dT

T

Ra

dT

T



dr

r

b
 1 cpR 2

dT

T

R

dr

r

R

k 1

dT

T

R

dr

r

ds
cp

dT

T

R

dP

P

P
rRT S dP
rR dTRT dr S

dP

P

dT

T



dr

r

P 1 rV 2 V
constant S dP 1 rV 2 dV
 0 S

dP

dV

rV

rV
constant S r dVV dr
 0 S

dr

r



dV

V

1 k

1 k

smax

Tmax

Ma  1

 0

Ma  1



T

a

a

b

b

ds

dT

s

dTds
^0

FIGURE 17–54

The T-sdiagram of Rayleigh flow

considered in Example 17–13.

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