Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

850 | Thermodynamics

Differentiating, we have

Solving for dVgives

Combining this with the energy equation, we have

which is the equation for the Fanno line in differential form. At point a

(the point of maximum entropy) ds
0. Then from the second T dsrelation

(T ds
dhvdP) we have dh
vdP
dP/r. Substituting yields

Solving for V, we have

which is the relation for the speed of sound, Eq. 17–9. Thus the proof is

complete.

V
a

0 P

0 r

b

1 > 2

s

dP

r

V^2

dr

r

0 ¬¬at s
constant

dhV^2

dr

r

 0

dV
V

dr

r

r dVV dr
 0

EXAMPLE 17–9 Shock Wave in a Converging–Diverging Nozzle

If the air flowing through the converging–diverging nozzle of Example 17–7

experiences a normal shock wave at the nozzle exit plane (Fig. 17–35), deter-

mine the following after the shock: (a) the stagnation pressure, static pres-

sure, static temperature, and static density; (b) the entropy change across the

shock; (c) the exit velocity; and (d) the mass flow rate through the nozzle.

Assume steady, one-dimensional, and isentropic flow with k
1.4 from the

nozzle inlet to the shock location.

Solution Air flowing through a converging–diverging nozzle experiences a

normal shock at the exit. The effect of the shock wave on various properties

is to be determined.

Assumptions 1 Air is an ideal gas with constant specific heats at room

temperature. 2 Flow through the nozzle is steady, one-dimensional, and

isentropic before the shock occurs. 3 The shock wave occurs at the exit

plane.

Properties The constant-pressure specific heat and the specific heat ratio of

air are cp
1.005 kJ/kg · K and k
1.4. The gas constant of air is 0.287

kJ/kg 
K (Table A–2a).

Analysis (a) The fluid properties at the exit of the nozzle just before the

shock (denoted by subscript 1) are those evaluated in Example 17–7 at the

nozzle exit to be

P 01
1.0 MPa¬P 1
0.1278 MPa T 1
444.5 K¬r 1
1.002 kg>m^3

T = 444.5 K

Ma 1 = 2

P 01

1

1

1

= 1.0 MPa

P = 0.1278 MPa

r = 1.002 kg/m^3

Shock wave

12

m· = 2.86 kg/s

FIGURE 17–35

Schematic for Example 17–9.

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