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

crosses the normal shock. Normal shock waves are discussed in

Section 17–5.

4.When PEPb0, the flow in the diverging section is supersonic,

and the fluid expands to PFat the nozzle exit with no normal shock

forming within the nozzle. Thus, the flow through the nozzle can be

approximated as isentropic. When Pb
PF, no shocks occur within

or outside the nozzle. When PbPF, irreversible mixing and expan-

sion waves occur downstream of the exit plane of the nozzle. When

PbPF, however, the pressure of the fluid increases from PFto Pb

irreversibly in the wake of the nozzle exit, creating what are called

oblique shocks.

844 | Thermodynamics

EXAMPLE 17–7 Airflow through a Converging–Diverging Nozzle

Air enters a converging–diverging nozzle, shown in Fig. 17–28, at 1.0 MPa

and 800 K with a negligible velocity. The flow is steady, one-dimensional, and

isentropic with k
1.4. For an exit Mach number of Ma 
2 and a throat

area of 20 cm^2 , determine (a) the throat conditions, (b) the exit plane condi-

tions, including the exit area, and (c) the mass flow rate through the nozzle.

Solution Air flows through a converging–diverging nozzle. The throat and

the exit conditions and the mass flow rate are 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.

Properties The specific heat ratio of air is given to be k
1.4. The gas

constant of air is 0.287 kJ/kg 
K.

Analysis The exit Mach number is given to be 2. Therefore, the flow must

be sonic at the throat and supersonic in the diverging section of the

nozzle. Since the inlet velocity is negligible, the stagnation pressure and

stagnation temperature are the same as the inlet temperature and pressure,

P 0 
1.0 MPa and T 0 
800 K. The stagnation density is

(a) At the throat of the nozzle Ma 
1, and from Table A–32 we read

Thus,

Also,

517.5 m/s

V*
c*
 2 kRT*


B

1 1.4 21 0.287 kJ>kg#K 21 666.6 K2a

1000 m^2 >s^2

1 kJ>kg

b

r*
0.6339r 0 
 1 0.6339 21 4.355 kg>m^32 
2.761 kg/m^3

T*
0.8333T 0 
 1 0.8333 21 800 K 2 
666.6 K

P*
0.5283P 0 
 1 0.5283 21 1.0 MPa 2 
0.5283 MPa

P*

P 0

0.5283 ¬

T*

T 0

0.8333 ¬

r*

r 0

0.6339

r 0 

P 0

RT 0

1000 kPa

1 0.287 kPa#m^3 >kg#K 21 800 K 2

4.355 kg>m^3

At = 20 cm^2

T 0 = 800 K

Mae = 2

P 0 = 1.0 MPa

Vi ≅ 0

FIGURE 17–28

Schematic for Example 17–7.

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