Transcript Example 3.1 Air flows from a reservoir where P = 300 kPa

Example 3.1
Air flows from a reservoir where P = 300 kPa and T = 500 K
through a throat to section 1 in Fig. 3.4, where there is a normal –
shock wave.
Compute
(a) P1
(b) P2
(c) Po2
(d) A*2
(e) Po3
(f) A*3
(g) P3
(h) To3
(i) T3
Air
Reservoir
P o1 = 300 kPa
T o1 = 500 K
12
1 ft2
2 ft2
Prof. Dr. MOHSEN OSMAN
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3ft2
1
A shock wave cannot exist unless M1 is supersonic; therefore the
flow must have accelerated through a throat which is sonic
A t  A1  1 ft
*
2
We can now find the Mach no. M1 from the known isentropic area
A
2 ft
ratio

 2 .0
2
1
*
A1
1 ft
2
A1
From isentropic Table at A  2 we get M1 = 2.20
The pressure P1 follows from the isentropic table P
*
1
o1
 P1 
300 kPa
 28 . 06 kPa
 10 . 7
P1
10 . 7
The pressure P2 is now obtained from M1 and the normal-shock
P
Table P  5 . 48  P2  5 . 48 ( 28 . 06 )  154 kPa
In similar manner, for M1 = 2.2, from Normal Shock Table
P
A
 0 . 628 and
 1 . 592 . Then Po 2  0 . 628 ( 300 )  188 kPa
P
A
& A 2*  1 .592 (1 ft 2 )  1 .592 ft 2
2
1
02
*
2
o1
*
1
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The flow from section 2 to 3 is isentropic ( but at higher entropy
than the flow upstream of the shock)
Po 3  Po 2  188 kPa
A 3  A 2  1 . 592 ft
*
*
2
Knowing A 3* , we can now compute P3 by finding M 3 and without
bothering to find M 2 ( which is equal to 0.547 ).
2
A
3
ft
3
The area ratio at section 3 is

 1 . 884
*
A3
1 . 592 ft
2
Then, since M 3 is known to be subsonic because it is downstream
of a normal shock, from isentropic Table, we get M 3  0 .33
The pressure P3 then follows from P  1 . 078
o3
P3 
188
 174 kPa
P3
1 . 078
Meanwhile, the flow is adiabatic throughout the duct; thus
T o 3  T o 2  T o 1  500 K
Prof. Dr. MOHSEN OSMAN
sin ce
To3
T3
 1 . 022  T 3  489 K
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Example # 3.2
An explosion in air,   1 .4 , creates a spherical shock wave propagating radially into still air at standard conditions. At the instant shown
in figure the pressure just inside the shock is 200 psia. Estimate
(a) The shock speed C & (b) the air velocity V just inside the shock
V1  a
P1  200 psia
Speed of sound
In U.K. units
a 1  49
T1
T1  520 R
POW
atmospheric
conditions
P = 14.7 psia
T = 520 R
T1
in Rankine
V  V1  V 2
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Solution
(a) In spite of the spherical geometry the flow across the shock
waves normal to the spherical wave front; hence the
normal- shock relations (3.2) to (3.11) apply.
Fixing our control volume to the moving shock, we find that
the proper conditions to use in Fig. 3.1 are :
a  V 1 , P1  14 . 7 psia , T1  520 R
V  V 1  V 2 , P2  200 psia ,
The speed of sound outside the shock is C  49 T  49 520  1117 ft / s
We can find M 1 from the known pressure ratio across the shock
1
P2
P1

200 psia
1
 13 . 61
14 . 7 psia
From Normal-Shock Table, we get M
1
 3 . 436
a  V 1  M 1 C 1  3 . 436 (1117 ); a  V 1  3840 ft / s
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(b) To find V 2 , we need the temperature or sound speed inside
the shock. Since M 1 is known, from Normal-Shock Table for
M1 = 3.436, we get T  3 . 228
T
T2 = 3.228(520) = 1679 R
At such a high temperature we should account for non-perfectgas energy equation that: C T  1 V  C T
2
1
2
p
2
p
o
V 2  2 C p (T1  T 2 )  V 1
2
2
V 2  2 ( 6010 )( 520  1679 )  ( 3840 )
2
2
 V 2  903 ft / s
Notice that we did that without bothering to compute M2 ,
which equals 0.454, or C  49 T  49 1679  2000 ft / s
Finally, the air velocity behind the shock is
2
2
V  V 1  V 2  3840  903  2940 ft / s
Thus a powerful explosion creates a brief but intense blast wind as
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it passes.
Operation of Converging and Diverging Nozzles
By combining the isentropic flow and normal-shock relations
plus the concept of sonic throat chocking, we can outline the
characteristics of converging and diverging nozzles.
Converging Nozzle
Fig. 3.6 Operation of a Converging Nozzle
(a) Nozzle geometry showing characteristic pressures;
(b) Pressure distribution caused by various back pressures;
(C) Mass flux versus back pressure.
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Converging Nozzle
First consider the converging nozzle sketched in 3.6 a. There
is an upstream reservoir at stagnation pressure Po . The flow is
induced by lowering the downstream outside, or back, pressure
Pb , resulting in the sequence of states a to e shown in Fig. 3.6
b and c.
For a moderate drop in Pb to states a and b, the throat pressure
is higher than the critical value P* which would make the
throat sonic.
The flow in the nozzle is subsonic throughout, and the jet exit
pressure Pe equals the back pressure Pb . The mass flux is
predicted by subsonic isentropic theory and is less than the
critical value  as shown in Fig. 3.6 c.
m max
For condition ( c ), the back pressure exactly equals the critical
pressure P* of the throat. The throat becomes sonic, the jet exit
flow is sonic, Prof. Dr. MOHSEN OSMAN
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Pe = Pb , and the mass flux equals its maximum
value
1
from
m max  
OR
m max
2
(
2
 1
To
(  1)
)
2 (  1 )
A* o
RT o
 0 . 68374 ( for   1 . 4 )
Po A *
The flow upstream of the throat is subsonic
everywhere and predicted by isentropic theory
A( x)
based on the local area ratio
and Isentropic
A*
Table.
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Finally, if Pb is lowered further to conditions (d) or (e)
below P* the nozzle cannot respond further because it is
chocked at its maximum throat mass flux. The throat
remains sonic with Pe = P*, and the nozzle pressure
distribution is the same as in state c , as sketched in Fig.
3.6 b.
The exit jet expands supersonically so that the jet
pressure can be reduced from P* down to Pb. The jet
structure is complex and multi-dimensional and not
shown here. Being supersonic, the jet cannot send any
signal upstream to influence the chocked flow conditions
in the nozzle.
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Operation of Converging and Diverging Nozzle
Converging – Diverging Nozzle
Figure 3.7 Flow through converging-diverging nozzle under variable back pressure.
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Now consider the converging-diverging nozzle sketched
in Fig. 3.7a. If the back pressure Pb is low enough, there
will be supersonic flow in the diverging portion and a
variety of shock-wave conditions may occur, which are
sketched on Fig. 3.7b. Let the back pressure be gradually
decreased. For curves A and B in Fig. 3.7b, the back
pressure is not low enough to induce sonic flow in the
throat, and the flow in the nozzle is subsonic throughout.
The pressure distribution is computed from subsonic
isentropic area-change relations, e.g., Isentropic Table.
A
A
For curve C the area ratio A exactly equals A for a
subsonic Me in isentropic Table. The throat becomes
sonic, and mass flux reaches a maximum in Fig. 3.7c.
e
e
*
t
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The remainder of the nozzle flow is subsonic, including
the exit jet, and Pe = Pb .
Now
jump
for
a
moment
to
curve
H.
Here
P
is
such
that
b
P
A
for a
P exactly corresponds to the critical area ratio
A
supersonic Me in the table.
The diverging flow is entirely supersonic, including the
jet flow, and Pe = Pb . This is called the design pressure
ratio of the nozzle and is the back pressure suitable for
operating a supersonic wind tunnel or an efficient rocket
b
e
*
o
nozzle.
Now back up and suppose that Pb lies between curves C
and H, which is impossible according to purely isentropic
flow calculations. Then back pressure D to F occur in Fig.
3.7b. The throat remains chocked at the sonic value, and
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we can match Pe = Pb by placing a normal shock at just
the right place in the diverging section to cause a subsonic
diffuser flow back to the back-pressure condition.
The mass flux remains at maximum in Fig. 3.7c. At back
pressure F, the required normal shock stands in the duct
exit . At back pressure G no single normal shock can do
the job, and so the flow compresses outside the exit in a
complex series of oblique shocks until it matches Pb .
Finally, at back pressure I, Pb is lower than the design
pressure H but the nozzle is chocked and cannot respond.
The exit flow expands in a complex series of supersonic
wave motion until it matches the low back pressure.
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