normalshock_25
Consider the flow of air through the converging-diverging nozzle shown in the figure below. The flow begins at
stagnation conditions with p0 = 100 kPa (abs) and T0 = 300 K. The nozzle exit-to-throat area ratio is AE/AT = 1.688
with a throat area of AT = 1.0*10-4 m2.
p 0, T 0
pB
AT
AE
a.
b.
c.
Determine the back pressure at which the flow first becomes choked.
Determine the range of back pressures at which the flow at the exit is supersonic.
Determine the mass flow rate through the nozzle when the exit Mach number is 0.2.
Page 1 of 3
normalshock_25
SOLUTION:
The flow first becomes choked when the Mach number at the throat is equal to one (AT = A*) then goes back to
subsonic conditions. The area ratio for this case is
k−1
2 2( k+1)
AE AE
1 ⎛ 1+ k−1
2 Ma E ⎞
(1)
=
=
1.688
=
⇒ Ma E = 0.3721 .
⎟⎠
Ma E <1
A* AT
Ma E ⎜⎝ 1+ k−1
2
where k = 1.4. Since the flow is entirely isentropic, the back pressure ratio corresponding to this Mach number
may be found using,
k
pE
2 1−k
= 1+ k−1
Þ pE/p0 = 0.9088
(2)
2 Ma E
p0
(
)
Since the exit Mach number is subsonic, the exit pressure and back pressure are the same. Hence,
pB = pE.
Using the given inlet stagnation pressure and Eqs. (2) and (3),
pB = 90.8 kPa (abs)
(3)
(4)
The flow at the exit will be supersonic for back pressures less than the case when a normal shock wave stands at
the nozzle exit. The back pressure at which a normal shock stands at the exit may be found by noting that the flow
upstream of the exit will be entirely isentropic (and choked), with the Mach number just upstream of the shock at
the exit being supersonic. Hence,
k−1
2( k+1)
2
AE AE
1 ⎛ 1+ k−1
2 Ma E1 ⎞
=
= 1.688 =
⇒ Ma E1 = 2.000 .
*
⎜
⎟
Ma E1 >1
A
AT
Ma E1 ⎝ 1+ k−1
⎠
2
The pressure ratio just upstream of the shock at the exit may be found from the isentropic relations,
k
pE1
2 1−k
= 1+ k−1
Þ pE1/p01 = 0.1278 Þ pE1 = 12.78 kPa (abs)
2 Ma E1
p01
(
)
(5)
(6)
where p01 is the upstream stagnation pressure (the stagnation pressure decreases across the shock).
The static pressure ratio across the shock may be found using the normal shock relations,
pE 2
2k
k −1
Þ pE2/pE1 = 4.500
(7)
=
Ma 2E1 −
pE1 k + 1
k +1
so that the pressure just downstream of the shock is,
⎛ p ⎞⎛ p ⎞
(8)
pE 2 = ⎜ E 2 ⎟ ⎜ E1 ⎟ p01 Þ pE2 = 57.51 kPa (abs).
⎝ pE1 ⎠ ⎝ p01 ⎠
Note that the Mach number just downstream of the exit will be subsonic, so the downstream pressure will be
equal to the back pressure. Hence,
p B = pE 2 .
(9)
Thus, the range of back pressures for which the exit Mach number will be supersonic is,
pB < 57.51 kPa (abs).
(10)
Page 2 of 3
normalshock_25
Note that the Mach number downstream of the shock wave is,
Ma 2E 2 =
( k − 1) Ma 2E1 + 2
2kMa 2E1 − ( k − 1)
Þ MaE2 = 0.5774
(11)
and the stagnation pressure ratio across the shock wave is,
(
)
k
1
( k−1) ⎡
2
⎤ ( k−1)
p02 ⎡ k + 1 Ma E1 ⎤
k +1
⎥
⎢
⎥
Þ p02/p01 = 0.7209 Þ p02 = 72.1 kPa (abs)
=⎢
2
p01 ⎢ 2 + k − 1 Ma 2E1 ⎥
⎢
⎥
2kMa
−
k
−
1
E1
⎣
⎦
⎣
⎦
The isentropic stagnation pressure ratio at the downstream side of the shock is,
k
pE 2
1−k
2
= 1+ k−1
Þ pE2/p02 = 0.7978,
2 Ma E 2
p02
(
)
(
(
)
)
The back pressure for this case can be found by combining relations in the following manner,
⎛ p ⎞⎛ p ⎞
pb = pE 2 = ⎜ E 2 ⎟ ⎜ 02 ⎟ p01 = ( 0.7978 ) ( 0.7209 ) (100 kPa ) = 57.51 kPa ,
⎝ p02 ⎠ ⎝ p01 ⎠
which is exactly the same result found in Eqs. (8) and (9).
(12)
(13)
(14)
Given that the flow chokes at an exit Mach number of 0.3721 (found from Eq. (1)), the flow in the device must be
entirely subsonic when the exit Mach number is 0.2. Thus, the mass flow rate may be found from the isentropic
relations evaluated at the exit,
m = ρ EVE AE ,
(15)
where,
1
p
2 1−k
and ρ0 = 0 ,
(16)
ρ E = ρ0 (1+ k−1
2 Ma E )
RT0
VE = cE Ma E = kRTE Ma E ,
(
2
TE = T0 1+ k−1
2 Ma E
)
−1
(17)
,
(18)
⎛A ⎞
AE = ⎜ E ⎟ AT .
⎝ AT ⎠
Using the given data,
Þ r0 = 1.161 kg/m3
rE = 1.139 kg/m3
TE = 297.6 K
cE = 345.8 m/s
VE = 69.16 m/s
AE = 1.69*10-4 m2
−2
Þ m = 1.33*10 kg s
(19)
(20)
Note that this mass flow rate is less than the choked flow mass flow rate (since the flow isn’t choked).
Page 3 of 3