17-41, 17-58, 17-61, 17-80, 17-81
17-41 Air flows through a duct. The state of the air and its Mach number are specified. The velocity and the stagnation pressure, temperature, and density of the air are to be determined.
Assumptions Air is an ideal gas with constant specific heats at room temperature.
Properties The properties of air at room temperature are R = 0.287 kPa.m
3 /kg.K and k = 1.4
(Table A-2a).
Analysis The speed of sound in air at the specified conditions is c
 kRT

( 1 .
4 )( 0 .
287 kJ/kg

K)(373.2
K)


1000
1 m
2
kJ/kg
/ s
2



387.2
m/s
Thus,
V

Ma
 c

( 0 .
8 )( 387 .
2 m/s) = 310 m/s
Also,
 
P
RT

200 kPa
( 0 .
287 kPa
 m
3
/kg

K)(373.2
K)

1 .
867 kg/m
3
Then the stagnation properties are determined from
AIR
T
0

T 1

( k

1 ) Ma
2
2

( 373 .
2 K) 1

( 1 .4
1 )( 0 .8)
2
2

421 K
P
0

P

0
T
0
T k /( k

1 )

( 200 kPa)
421.0
K
373.2
K
 


T
0
T


1 /( k

1 )

( 1 .
867 kg/m
3
)
1 .
4 /( 1 .
4

1 )
421.0
K
373.2
K

305 kPa
1 /( 1 .
4

1 )

2.52
kg/m
3
Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.
17-58 Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit.
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 properties of air are k = 1.4 and c p
= 1.005 kJ/kg·K (Table A-2a).
Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. i
150 m/s
AIR *
Ma = 1
1

T
0

T i
V i
2

2 c p

350 K

(150 m/s)
2
2

1 .
005 kJ/kg

K
1 kJ/kg
1000 m
2
/ s
2

361 .
2 K
And P
0

P i


T
0
T i

 k /( k

1 )

( 0 .
2 MPa)
361.2
K

1 .
4 /( 1 .
4

1 )
350 K

0 .
223 MPa
From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma = 1, we read T / T
0
= 0.8333, P / P
0
= 0.5283.
Thus,
T = 0.8333
T
0
= 0.8333(361.2 K) = 301 K and P = 0.5283
P
0
= 0.5283(0.223 MPa) = 0.118 MPa
Also, c i
 kRT i

( 1 .
4 )( 0 .
287 kJ/kg

K)(350 K)


1000
1 m
2
kJ/kg
/ s
2



375 m/s and Ma i

V i c i

150 m/s
375 m/s

0 .
40
From Table A-32 at this Mach number we read A i
/ A * = 1.5901. Thus the ratio of the throat area to the nozzle inlet area is
A *

A i
1

0.629
Discussion We can also solve this problem using the relations for compressible isentropic flow.
The results would be identical.
17-61 Air enters a converging-diverging nozzle at a specified pressure. The back pressure that will result in a specified exit Mach number is to be determined.
Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic.
Properties The specific heat ratio of air is k = 1.4 (Table A-2a).
Analysis The stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. It remains constant throughout the nozzle since the flow is isentropic,
P
0
= P i
= 0.5 MPa
From Table A-32 at Ma e
=1.8, we read P e
/ P
0
= 0.1740.
Thus, P = 0.1740
P
0
= 0.1740(0.5 MPa) = 0.087 MPa
Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical. i
V i

0
AIR e
Ma e
= 1.8
17-80 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry.
Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane.
2

Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,
P
01
= P i
= 2 MPa shock wave It is specified that A / A * =3.5. From Table A-32,
Mach number and the pressure ratio which corresponds to this area ratio are the Ma
1
=2.80 and
P
1
/ P
01
= 0.0368. The pressure ratio across the shock for this Ma
1
value is, from Table A-33, P
2
/ P
1
= i
V i

0
8.98. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be
P
2
=8.98
P
1
= 8.98

0.0368
P
01
= 8.98

0.0368

(2 MPa) = 0.661 MPa
AIR 1 2
P b
Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.
17-81 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry.
Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs.
Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,
P
01
= P i
= 2 MPa
It is specified that A / A * = 2. From Table A-32, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma
1
=2.20 and P
1
/ P
01
= 0.0935. The pressure ratio across the shock for this M
1
value is, from
Table A-33, P
2
/ P
1
= 5.48. Thus the back pressure, i
V i

0
AIR which is equal to the static pressure at the nozzle exit, must be
P
2
=5.48
P
1
= 5.48

0.0935
P
01
= 5.48

0.0935

(2 MPa) = 1.02 MPa
1 2 shock wave
P b
Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.
Alternate Solution for 17-81:
The wording of this question is ambiguous. In the previous solution, the assumption is that the shock remains at the exit plane, and the nozzle is changed so that A e
/ A * = 2. It would make more sense, given the wording, to consider that the nozzle remains the same, i.e. A e
/ A * = 3.5, and the shock moves upstream.
Consider location i to be the inlet, e to be the exit, x to be the location just upstream of the shock, and y to be the location just downstream. Then, A e
/ A * = 3.5 and A x
/ A * = A y
/ A * = 2. Since inlet
3

velocity is negligible:
P
0 , x

P
0 , i

P i

2 MPa
From Table A-15, using A x
/ A *:
M x
= 2.20
From Table A-16, using M x
:
P
0 , y
M y

P
0 , x
0

0 .
62812
.
54706
From Table A-15, using M y
:
A y
/ A y
* = 1.268 where A y
*is the area of the “hypothetical throat”, i.e., downstream of the shock, the flow behaves as if it were in a nozzle with throat area A y
*. Now, since P
0, e
= P
0, y
, all that is needed to find
P b
= P e
is the ratio P e
/ P
0, e
. However:
A e
A y
*

A
A y y
*

A
A y e and:
A e
A y

A e
A *

A *

A y
A e
A *


A y
A *



1

1 .
75 so:
A e
/ A y
* = 2.22
Finally, from Table A-15, using A e
/ A y
* and the fact that M e
< 1:
P e
/ P
0, e
= 0.9460 so:
P b
= P e
= P
0, e
( P e
/ P
0, e
) = P
0, y
( P e
/ P
0, e
) = P
0, x
( P
0, y
/ P
0, x
)( P e
/ P
0, e
) = 1.188 MPa
4