Convergent-Divergent Nozzles and Shock Waves - Desklib

Electrical power 1
AERODYNAMICS
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Electrical power 2
SECTION 1 – CONVERGENT-DIVERGENT NOZZLESQuestion 1.1
Explain the difference between stagnation temperature and static
temperature.
[Stagnation temperature is literature reservoir temperature for the stream fluid
which is being measured. In most cases it measured by the help of thermal probe.
While the static temperature is that which is measured through a help of a sensor
moving at the same speed of the fluid. In some cases it is referred to as
temperature without velocity effects].
Question 1.2
Under what conditions will the flow through a convergent-divergent nozzle become
choked? What happens to the mass flow rate when exit pressure is decreased on a
choked nozzle?
[The fluid flow in the nozzle is chocked due to continues reduction in a back
pressure in order to extent that more reduction is not capable to move the point far
from the throat. The mass flow rate will decrease as the pressure at the exit
decrease.]
Question 1.3
If a supersonic flow is run through a convergent duct will its velocity:
a) Decrease
b) Increase
c) Remain constant
d) It depends on the upstream Mach number
e) It depends on the upstream stagnation pressure and temperature.
[Increase]
Question 1.4
A convergent-divergent nozzle is designed to operate with isentropic flow with an
exit Mach number, ME. The flow in the nozzle is supplied from a reservoir of air with
a static pressure of PR and a static temperature of TR and the nozzle has a throat
area, AT, as specified in the table below.

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Part a
[Given data
Exit Mach number ME =2.1
Ar= 0.6
Reservoir pressure is 610
Tr = 400
PEC= 560
Now let the exit area of the nozzle be as in the following equation
Aexit
A∗¿ ¿ = 1
Mexit ¿
Aexit
A∗¿ ¿ = 1
2.1 ¿
Aexit
A∗¿ ¿ = 1
2.1 ¿
Aexit
A∗¿ ¿ =0.476 ¿
Aexit
A∗¿ ¿ =0.476 ¿
Aexit
A∗¿ ¿ =0.476 [3.857 ]

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Aexit
A∗¿ ¿ = 1.726
Part b
Now we calculate the exit area of the nozzle .
Area of throat is AT= 0.6
From Aexit
A∗¿ ¿ = 1.726
Aexit = A* × 1.726
A* = 0.6
Aexit = 0.6 × 1.726
Aexit = 1.03556 m2
Part c
Now we obtain the exit pressure at which the nozzle will become chocked
This is obtained by the use of the following equation
P∗¿
Po ¿=( 2
r +2 ¿
r
r −1
Stagnation pressure / Reseivor static pressure i.e
P∗¿
Po ¿=( 2
3.4 ¿
1.4
1.4−1
P∗¿
Po ¿= 0.15610
P* = 0.15610×610
P* = 95.221 kPa
Part d
The mass flow rate for the exit pressure of 0kpa is
Pexit = 0
So there will be no mass flow rate
M = 0kg/sec
Part e

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Now we obtain the mass flow rate for exit pressure of 560k Pa
We have mi= P0Mexit A √ r
RT 0 (1+ r−1
2 ) M 2 ¿
r+1
2(r−1)
After substitution we obtain the following;
M= 610× 1.03556×2 √ 1.4
8.314 ×400 [(1+ 0.4
2 ) 2.12 ¿ ¿¿−3
M= 1263.38 × 0.02051 × [5.292¿−3
M= 1263.38 × 0.02051 ×0.0067474
M= 0.17483 kg / Sec
Question 1.5
A small supersonic wind tunnel, consists of a convergent-divergent nozzle designed
to be operated using a bottle of compressed helium gas (γ = 5/3, R = 2.077
kJ/kg/°K). The test section has a diameter, d, and the tunnel is designed to operate
at Mach number, M, with a static pressure, P, and a static temperature, T, specified
in the table below.

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T 0
T = 1+ ( r−1
2 ¿ m2
P 0
P = [1+( r−1
2 ¿ m2 ¿
r
r−1
ρ 0
ρ = [1++( r−1
2 ¿ m2 ¿
r
r−1
At the given dimension
⍴= ρ
RT = 30× 1000
2077 ×185 = 30000
384245 = 0.0780kg/m3
Part a
m=⍴ AV = ⍴ ( π d2
4 )m√rRT
m= 0.0780 ( π 152 10−6
4 ) 1.5 √ 5
3 ×2.077 ×100017.5
m= 1.61
Part b
Area = A/At = 1/m ( 2+(r −1)(m2 )
r +1 ) r +1
2(r−1)
π ( 152 )
4
π ( d t2 )
4
= 1
1.5 ( 2+ ( 1.667−1 ) (1. 52 )
1.6667+1 ¿
1.6667+ 1
1.6667−1¿ ¿
152
d t2 = 1.1483 or simply
dt= 13.997≅ 14 mm