Example 3.4
A converging-diverging nozzle (Fig. 3.7a) has a throat area
of 0.002 m2 and an exit area 0.008 m2 . Air stagnation
conditions are Po  1000kPa & To  500 K .
Compute (a) the design exit pressure and
mass flux and the exit pressure and mass flux if
(b) Pb  300kPa & (c) Pb  900kPa . Assume γ = 1.4
Ae
A
0.008
Solution
 e* 
4
At
0.004
A
A
Interpolating into supersonic Isentropic Table for A  4 we
Po
Po
1000
M

2
.
95
&

34
.
1
get e
 Pe 

 29.3kPa
Pe
34.1 34.1
Mass Flow Parameter (MFP)
MFP  0.68374 @ M  1
*
Prof. Dr. MOHSEN OSMAN
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 max To
m
P0 A
*

MFP
R
 0.0404
 max 500
m
 max  3.61kg / s
 0.040416  m
3
1000 x10 x0.002
If Pb  Pe  No supersonic flow.
(b) Pb  300kPa;
We are definitely far below the subsonic isentropic condition C in Fig. 3.7b but we may even be below condition F
with a normal shock in the exit plane.
It is condition g , where Pe  Pe,design  29.3kPa because
no shock has yet occurred. To find out, compute condition
F by assuming an exit normal shock with M 1  2.95 , that is
the design Mach number just upstream of the shock. From
Normal – Shock Table P  9.99 & P  9.99(29.3)  293kPa
2
P1
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2
Since this is less than the given Pb  300kPa, there is a normal
shock just upstream of the exit plane (condition E). The exit
flow is subsonic and equals the back pressure Pe  Pb  300kPa
 m
 max  3.61kg / s
Also
m
(c) Pb  900kPa, we compute M e andPe for condition (C) as a
comparison . Again A  4. For this condition, with a
A
subsonic M e estimated from subsonic Isentropic Table. We
get M e  0.147 & Pe  0.985
e
t
Po
Pe  985kPa Pb  900kPa
The given back pressure of 900 kPa is less than this value (of
985) corresponding roughly to condition (D) in Fig. 3.7b. Thus,
for this case, there is a normal shock just downstream of the
throat
Prof. Dr. MOHSEN OSMAN
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and the throat is chocked Pe  Pb  900kPa & m  m max  3.61kg / s
For this large exit-area ratio the exit pressure would have to be
larger than 985 kPa to cause a subsonic flow in the throat and a
mass flux less than maximum.
Governing Relations of the Normal-Shock Wave
.
x
y
.
V x  .............  V y
.
.
Normal-Shock Wave
V y2
Vx2
hx 
 hy 
 ho
2
2
1

m
  xV x   yV y
Continuity Equation
A
2
Energy Equation
“adiabatic, no work, steady”
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Momentum Equation
 (V y  V x )
( Px  Py ) A  m
Equation of state
Px   xV
h = h (s & ρ)
s  s ( P,  )
What is meant by Fanno Line ?
2
x
 Py   yV
2
y
3
4
5
Combination of
Fanno Flow and
Normal Shock
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If conditions of x are fixed, if we are required to calculate
conditions at y, Equations (1) and (2) and (4) define a locus of
states passing through x → this is called Fanno Line.
i.e.,
● Choose V y
● Calculate  y from 2
● Calculate h y from 1
● Calculate s y from 4
● repeat for other values of V y

m
& ho
Note: Fanno Line is a line of constant
A

Frictional effects are required to pass along Fanno Line.
Prof. Dr. MOHSEN OSMAN
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What is meant by Rayleigh Line?
● Similar to Fanno, but using equations (2), (3) & (5)

It is a line of constant m & F (friction coefficient)
A
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Fanno & Raleigh Lines
Since Normal Shock must satisfy equations (1), (2), (3),(4) and (5)
 Shock must lies on both Fanno and Rayleigh Lines !!
The points of maximum entropy s on both Fanno & Rayleigh
Lines are of sonic velocity.
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Normal Shock in a Perfect Gas
For a perfect gas use Mach number relations to find the Normal
Shock downstream conditions of all fluid properties M y , Py , T y ,
 y , Poy , A*y , ands y if you know that Tox  Toy  To .
What is meant by supersonic Pitot tube ?
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Where :
Poy
Poy
Poy
 1
Py

*
Px
Py
Px
Py
2
 1

M x2 
Px   1
 1
&
Py
 (1 
2

M y2 )  1
(  1) M x2  2
where......M 
2M x2  (  1)
2
y
Accordingl y...............................
Poy
Px
(

(
 1
2
2
x
M )

 1
2
 1
M x2 
)
 1
 1
1
 1
)
Which is known as Rayleigh Supersonic Pitot-tube Formula
Notes: You can use Normal-Shock Tables ….. to get M x .
Homework Assignment and submit next lecture in a report
Poy
Plot
...vs...M x ...... for :
Px
(a) Air (γ=1.4)
(b) Helium (γ=1.66)
Prof. Dr. MOHSEN OSMAN
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Example 4.1
A pitot tube is inserted into a supersonic air stream, and records
a pressure of 0.7 bar. The static pressure upstream of the tube is
0.15 bar, and the static temperature is 350 K.
Calculate the flow velocity upstream of the tube.
Solution
Poy
Px

P02
0.7

 4.667
P1
0.15
Interpolating into Normal-Shock Table
we get M x  1.8
V x  M x C x  M x (20.046 Tx )
V x  1.8(20.046 350 )  675m / s
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Supersonic Wind Tunnels
Major Types:
(a) Blowdown Wind Tunnels (open)
(b) Continuous-Flow Wind Tunnels (closed)
which consists of:
1) Convergent – Divergent (Supersonic) Nozzle
2) Test Section
3) Convergent-Divergent (Supersonic) Diffuser
4) Heat Exchanger
5) Compressor
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(a) Continuous Wind Tunnel (no diffuser)
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(b) Continuous Wind Tunnel with diffuser
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Wind Tunnel Start-Up Sequence
To provide a test section with supersonic flow requires a
converging–diverging nozzle. To operate economically, the
nozzle–test-section combination must be followed by a
diffusing section which also must be converging–diverging.
This configuration presents some interesting problems in flow
analysis. Starting up such a wind tunnel is another example of
nozzle operation at pressure ratios above the second critical
point. Figure 3.8 shows a typical tunnel in its most unfavorable
operating condition, which occurs at startup. A brief analysis of
the situation follows.
As the exhauster is started, this reduces the pressure and
produces flow through the tunnel. At first the flow is subsonic
throughout, but at increased power settings the exhauster
reduces pressures still further and causes increased flow rates
until the nozzle throat (section 2) becomes choked. At this
point the nozzle is operating at its first critical condition.
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As power is increased further, a normal shock is formed just
downstream of the throat, and if the tunnel pressure is
decreased continuously, the shock will move down the
diverging portion of the nozzle and pass rapidly through the
test section and into the diffuser. Figure 3.9 shows this
general running condition, which is called the most favorable
condition. We return to Figure 3.8, which shows the shock
located in the test section. The variation of Mach number
throughout the flow system is also shown for this case. This is
called the most unfavorable condition because the shock
occurs at the highest possible Mach number and thus the
losses are greatest. We might also point out that the diffuser
throat (section 5) must be sized for this condition.
Prof. Dr. MOHSEN OSMAN
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Figure 3.8 Supersonic tunnel at startup (with associated Mach number variation).
Due to the shock losses (and other friction losses), we know that pt 5 <
pt 2 , and therefore A5 must be greater than A2 . Knowing the testsection-design Mach number fixes the shock strength in this unfavorable
condition and A5 is easily determined from equation (pt 2 A2 = pt 5 A5 ). Keep
in mind that this represents a minimum area for the diffuser throat. If it is
made any smaller than this, the tunnel could never be started (i.e., we
could never get the shock into and through the test section). In fact, if A5
is made too small, the flow will choke first in this throat and never get a
chance to reach sonic conditions in section 2.
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Once the shock has passed into the diffuser throat, knowing that A5
> A2 we realize that the tunnel can never run with sonic velocity at
section 5. Thus, to operate as a diffuser, there must be a shock at this
point, as shown in Figure 3.9. We have also shown the pressure
variation through the tunnel for this running condition.
To keep the losses during running at a minimum, the shock in the
diffuser should occur at the lowest possible Mach number, which
means a small throat. However, we have seen that it is necessary to
have a large diffuser throat in order to start the tunnel. A solution to
this dilemma would be to construct a diffuser with a variable- area
throat. After startup, A5 could be decreased, with a corresponding
decrease in shock strength and operating power. However, the power
required for any installation must always be computed on the basis of
the unfavorable startup condition.
Although the supersonic wind tunnel is used primarily for
aeronautically oriented work, its operation serves to solidify many of
the important concepts of variable-area flow, normal shocks, and their
associated flow losses. Equally important is the fact that it begins to
focus our attention on some practical design applications.
Prof. Dr. MOHSEN OSMAN
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Figure 3.9 Supersonic tunnel in running condition (with
associated pressure variation).
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Wind Tunnel Start–Up Sequence
Draw the following sequence of Wind–Tunnel start-up !
(a) Initial Start-Up
(b) First Throat Sonic
(c) Shock in Diverging Section
(d) Shock in Test Section Entrance
(e) Shock Swallowed
(f) Shock-Free Deceleration with
Variable-Area Diffuser Throat
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Example 4.2
A continuous–duty supersonic wind tunnel is to be designed.
The test section specifications are a Mach number of 2, a static
pressure of 40 kPa, a static temperature of 250 K&area of 0.5
m2 . Determine the stagnation conditions required upstream of
the test section, the mass flow rate required, and the throat
area required. During the startup process, what is the
maximum stagnation pressure loss across the shock system?
Also, calculate the pressure loss during the steady-state
operating conditions. If a constant–area diffuser is used,
what is the minimum diffuser throat area? Sketch the startup
process on a T–S diagram.
Solution: Consider Figure 3.9 for fixed diffuser steady-state
operating conditions.
Prof. Dr. MOHSEN OSMAN
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A continuous–duty supersonic wind tunnel with:
M T  2.0,..........PT  40kPa,...........TT  250K ....... & ....... AT  0.5m 2
From Isentropic Table @ M=MT = 2.0, we get
PT
T
A
 0.1278,............ T  0.55556,............ T  1.6875
Po
To
A*
40
250
 313kPa,.............To 
 450 K
0.1278
0.55556
0.5
A* 
 0.2963m 2
1.6875
A * Po
MFP @ M  1
 m
 max 
m
x
To
R
Po 
 m
 max 
m
0.2963 x313
287 x 450
x 0.68473  0.1767 kg / s
Po max
occurs when shock takes place at maximum Mach
number; i.e. at MT
Prof. Dr. MOHSEN OSMAN
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From Normal-Shock Wave Tables @ M=MT = 2.0, we get
Poy
Pox
 0.72087
 Po max  Pox  Poy  Pox (1 
Poy
Pox
)
Po max  313(1  0.72087)  87.37 kPa
Ax* Poy
sin ce....... * 
 0.72087
Pox
Ay
0.2963
then............................................ A 
 0.411m 2
0.72087
*
y
Question 1: How do we calculate Po min (i.e., Po during the
steady–state operating conditions) ?
Question 2: Show start–up on T–s using Fanno–line !
Prof. Dr. MOHSEN OSMAN
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Solution to Question 1
For steady–state operating conditions, shock occurs at the
diffuser throat; i.e.
at A
0.411
*
x
A

0.2963
 1.3871077
A
from
Isentropic
Table
@
M x  1.7509
Ax*
& @ M  1.7509 from Normal–Shock Table
x
Poy
Pox
 1.3871
 0.8342
 Po min  Pox  Poy  Pox (1 
Poy
Pox
)
Po min  313(1  0.8342)  51.9kPa
If a variable throat diffuser is used, the throat must initially
posses at least an area of 0.411 m2 in order to swallow the
normal shock wave. The power to the compressor would be
Prof. Dr. MOHSEN OSMAN
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increased until the shock wave was swallowed. Once the shock
wave is swallowed, the variable–area throat would be reduced
to the area of the first throat, 0.2963 m2 . Simultaneously, the
back pressure would be increased to place the shock at the
diffuser throat. Ideally, the limit for a variable throat diffuser
would be shock–free flow as illustrated in Figure (C).
Solution to Question 2
T – S Diagram of wind tunnel start – up using Fanno – Line
Prof. Dr. MOHSEN OSMAN
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