Pressure distribution along convergent- divergent Nozzle

SAIF AL-DIN ALI MADI
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
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[Fluid Laboratory II]
University of Baghdad
Name: - Saif Al-din Ali -B-

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TABLE OF CONTENTS
ABSTRACT.........................................................................I
OBJECTIVE........................................................................II
INTRODUCTION..............................................................III
THEORY..........................................................................V
APPARATUS...................................................................VI
Calculations and results................................................VI
DISCUSSION ...............................................................VII

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Experiment Name:-
Pressure distribution along convergent- divergent Nozzle
1. Abstract
This aim of this practical was to investigate compressible flow in a
convergent-divergent nozzle. Different flow patterns that influence
the results of the investigation are also explored. The different
pressure distributions that occur at varying lengths in the nozzle
were also recorded and analyzed
2. OBJECTIVE
1. The study effect pressure distribution along nozzle by using
mercury manometer to measure the head pressure (P) along
down surface nozzle, the distance between point and other
point (25 mm).
2. Compare your result with theoretical result.
3. Introduction:-
1.1 NOZZLE -:
A nozzle is often a pipe or tube of varying cross sectional area, and it can
be used to direct or modify the flow of a fluid (liquid or gas). Nozzles are
frequently used to control the rate of flow, speed direction, mass, shape,
and/or the pressure of the stream that emerges from them. Three types
of nozzle are convergent, divergent, convergent-divergent nozzle

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1.2 CONVERGENT- DIVERGENT NOZZLE:
The nozzle was developed by
Swedish inventor Gustaf de Laval
in 1897 for use on an impulse
steam turbine. A de Laval nozzle
(or convergent-divergent nozzle,
CD nozzle or con-di nozzle) is a
tube that is pinched in the middle,
making an hourglass-shape. The
cross sectional area first
decreases from its entrance to the
throat and then again increases from throat to the exit. This case is used
in the case where the back pressure is less than the critical pressure. Also,
in present day application, it is widely used in many types of steam
turbines and also in modern rocket engine and supersonic jet engines
1.3 SPEED OF SOUND:
An important consequence of compressibility of the fluid is that the
disturbances introduced at some point in the fluid propagate at finite
velocity. The velocity at which these disturbances propagate is known as
“acoustic velocity/speed of sound”. Mathematically, it is represented as
below;

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1.4 MACH NUMBER :-
The Mach number is the ratio of flow velocity after a certain limit of the
sounds speed. The formula that represents it is:
M = u/c
The Mach number is M,
Based on the limits the local flow velocity is u, the speed of
sound in that medium is c. To explain it simply, the speed of
sound can be equated to Mach 1 speed.
2.FLOW REGIME CLASSIFICATION
2.1 SUBSONIC FLOW:
The subsonic flow region is on the right of the incompressible flow
region. In subsonic flow, fluid velocity (c) is less than the sound velocity
(a) and the mach number in this region is always less than unity.
i.e. m = c/a = 1.
Eg: passenger air craft
2.2 SONIC FLOW:
If the fluid velocity (c) is equal to the sound velocity (a), that type of flow
is known as sonic flow .In sonic flow Mach number value is unity.
i.e. M = c/a = 1 ⇒ c =a.
Eg: nozzle throat
2.3 SUPERSONIC FLOW:
The supersonic region is in the right of the transonic flow region. In
supersonic flow, fluid velocity (c) is more than the sound velocity (a) and
the Mach number in this region is always greater than unity.
i.e. M = c/a >1.
Eg: military air crafts
2.4 HYPERSONIC FLOW:
In hypersonic flow region, fluid velocity (c) is much greater than sound
velocity (a).In this flow, Mach number value is always greater than 5.
i.e. M = c/a >5. Eg: rockets

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2.5 TRANSONIC FLOW:
If the fluid velocity close to the speed of sound, that type of flow is
known as transonic flow .In transonic flow, Mach number value is in
between 0.8 and 1.2. i.e.0.8 < M < 1.2.
For understanding the working principle of convergent-divergent type
of nozzles, first we need to look the working principle of only
convergent type of nozzles. In these type of nozzles the area of the
nozzle reduces gradually in the direction of flow. The pressure at intake
is called stagnation pressure and the pressure at exit is called back
pressure. The value of back pressure can never be more than 1 in case
of a nozzle. As we start reducing the back pressure we observe that flow
velocity and mass flow rate also starts increasing, but this will happen
upto a certain limit, after which no increase in velocity or mass flow rate
takes place. This situation is known as choked i.e no further increase in
mass flow rate takes place whatever be the back pressure now. This
situation takes place at a particular mach number i.e at mach number
'1'. But the case is not the same when we use a divergent nozzle just
after the convergent. Actually the principle reverses i.e when we attach
a divergent nozzle just after the convergent nozzle our flow speed starts
increasing with the decrease in back pressure and also the mass flow
rate. And therefore in this type of nozzles we can reach to the speeds
above sonic i.e supersonic.

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4. Theory
Following assumptions are taken under consideration during
modeling the present case study:
1. Nozzle is chocked.
2. The pressure during nozzle is beggar, this due to velocity
of supersonic.
3. One dimension.
4. Isentropic process
𝑨
𝑨 ∗
=
𝟏
𝑴
𝟐
𝑲 + 𝟏
[ 𝟏 +
𝑲 − 𝟏
𝟐
𝑴 𝟐]
𝑲+𝟏
𝟐(𝑲−𝟏)
𝐏
𝐏𝐨
= [𝟏 +
𝐊 − 𝟏
𝟐
𝐌 𝟐]
𝐊
𝐊−𝟏
£=
𝑲−𝟏
𝟐
𝐌𝐞 𝟐
(𝟏+
𝐊−𝟏
𝟐
𝐌 𝟐[𝟏−(
𝐏𝐨
𝐏
)
𝐊−𝟏
𝐊
]
Where:
A = Cross-section area (m2)
K = Specific heat ratio (Cp/Cv)
M = Mach Number
Po = pressure (special in supersonic Nozzle)
P = pressure (special in supersonic Nozzle)
* = the critical stat, (Mach Number 1).
£ = efficiency

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5. APPARATUS

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6. Calculations and results
Theory :-
S = 152 – (YU + YL)
Find the s along nozzle and represented value minimum (S) in throat
this name (S min.)
Where:- S = distance (mm) S*= 59.2 mm
X(mm) Yu YL
0 10.7 10.7
25 24.4 20.7
50 37.5 25
75 47.1 24.5
100 58.2 25.5
125 66.2 25.5
150 67.7 25.5
175 63.4 25.5
200 59 25.5
225 55.2 25.5
250 50.2 25.5
275 45 25.5
300 40.1 25.5
325 37 25.5
350 35 25.5
375 33.5 25.5
400 31.3 25.5
425 30.3 25.5
450 29.3 25.5
475 28.4 25.5
500 28.5 25.5
525 28 25.5
550 27 25.5
575 25.4 25.5
600 25.4 25.5

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1. S = 152 – (YU+YL) = 152 – (10.7+10.7)  S = 130.6
𝑺
𝑺∗
=
𝟏𝟑𝟎.𝟔
𝟓𝟗.𝟐
= 2.206
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
= 2.206
Mth = 0.27
__________________________________________
2. S = 152 – (YU+YL) = 152 – (24.4+20.7)  S = 106.9
𝑺
𝑺∗
=
𝟏𝟎𝟔.𝟗
𝟓𝟗.𝟐
= 1.805
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
= 1.805
Mth = 0.34
__________________________________________
3. S = 152 – (YU+YL) = 152 – (37.5+25)  S = 89.5
𝑺
𝑺∗
=
𝟖𝟗.𝟓
𝟓𝟗.𝟐
= 1.511
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.511
Mth = 0.43

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__________________________________________
4. S = 152 – (YU+YL) = 152 – (47.1+24.5)  S = 80.4
𝑺
𝑺∗
=
𝟖𝟎.𝟒
𝟓𝟗.𝟐
= 1.358
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.358
Mth = 0.49
__________________________________________
5. S = 152 – (YU+YL) = 152 – (58.2+25.5)  S = 68.3
𝑺
𝑺∗
=
𝟔𝟖.𝟑
𝟓𝟗.𝟐
= 1.153
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.153
Mth = 0.63
__________________________________________
6. S = 152 – (YU+YL) = 152 – (66.2+25.5)  S = 60.3
𝑺
𝑺∗
=
𝟔𝟎.𝟑
𝟓𝟗.𝟐
= 1.0185
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.0185
Mth = 0.86

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__________________________________________
7. S = 152 – (YU+YL) = 152 – (67.7+25.5)  S = 58.8
𝑺
𝑺∗
=
𝟓𝟖.𝟖
𝟓𝟗.𝟐
= 0.9932
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=0.9932
Mth = 1
__________________________________________
8. S = 152 – (YU+YL) = 152 – (63.4+25.5)  S = 63.1
𝑺
𝑺∗
=
𝟔𝟑.𝟏
𝟓𝟗.𝟐
= 1.065
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.065
Mth = 1.29
__________________________________________
9. S = 152 – (YU+YL) = 152 – (59+25.5)  S = 67.5
𝑺
𝑺∗
=
𝟔𝟕.𝟓
𝟓𝟗.𝟐
= 1.1402
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.1402
Mth = 1.45

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__________________________________________
10. S = 152 – (YU+YL) = 152 – (55.2+25.5)  S = 71.3
𝑺
𝑺∗
=
𝟕𝟏.𝟑
𝟓𝟗.𝟐
= 1.204
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.204
Mth = 1.54
__________________________________________
11. S = 152 – (YU+YL) = 152 – (50.2+25.5)  S = 76.3
𝑺
𝑺∗
=
𝟕𝟔.𝟑
𝟓𝟗.𝟐
= 1.2888
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.2888
Mth = 1.65
__________________________________________
12. S = 152 – (YU+YL) = 152 – (45+25.5)  S = 81.5
𝑺
𝑺∗
=
𝟖𝟏.𝟓
𝟓𝟗.𝟐
= 1.3766
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.3766
Mth = 1.74

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__________________________________________
13. S = 152 – (YU+YL) = 152 – (40.1+25.5)  S = 86.4
𝑺
𝑺∗
=
𝟖𝟔.𝟒
𝟓𝟗.𝟐
= 1.4594
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.4594
Mth = 1.81
__________________________________________
14. S = 152 – (YU+YL) = 152 – (37+25.5)  S = 89.5
𝑺
𝑺∗
=
𝟖𝟗.𝟓
𝟓𝟗.𝟐
= 1.511
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.511
Mth = 1.87
__________________________________________
15. S = 152 – (YU+YL) = 152 – (35+25.5)  S = 91.5
𝑺
𝑺∗
=
𝟗𝟏.𝟓
𝟓𝟗.𝟐
= 1.5456
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.5456
Mth = 1.89

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__________________________________________
16. S = 152 – (YU+YL) = 152 – (33.5+25.5)  S = 93
𝑺
𝑺∗
=
𝟗𝟑
𝟓𝟗.𝟐
= 1.5709
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.5709
Mth = 1.92
__________________________________________
17. S = 152 – (YU+YL) = 152 – (31.3+25.5)  S = 95.2
𝑺
𝑺∗
=
𝟗𝟓.𝟐
𝟓𝟗.𝟐
= 1.608
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.608
Mth = 1.94
__________________________________________
18. S = 152 – (YU+YL) = 152 – (30.3+25.5)  S = 96.2
𝑺
𝑺∗
=
𝟗𝟔.𝟐
𝟓𝟗.𝟐
= 1.625
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.625
Mth = 1.95

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__________________________________________
19. S = 152 – (YU+YL) = 152 – (29.3+25.5)  S = 97.2
𝑺
𝑺∗
=
𝟗𝟕.𝟐
𝟓𝟗.𝟐
= 1.641
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.641
Mth = 1.97
__________________________________________
20. S = 152 – (YU+YL) = 152 – (28.4+25.5)  S = 98.1
𝑺
𝑺∗
=
𝟗𝟖.𝟏
𝟓𝟗.𝟐
= 1.657
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.657
Mth = 1.98
__________________________________________
21. S = 152 – (YU+YL) = 152 – (28.5+25.5)  S = 98
𝑺
𝑺∗
=
𝟗𝟖
𝟓𝟗.𝟐
= 1.655
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.655
Mth = 1.983

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__________________________________________
22. S = 152 – (YU+YL) = 152 – (28+25.5)  S = 98.5
𝑺
𝑺∗
=
𝟗𝟖.𝟓
𝟓𝟗.𝟐
= 1.6638
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.6638
Mth = 1.985
__________________________________________
23. S = 152 – (YU+YL) = 152 – (27+25.5)  S = 99.5
𝑺
𝑺∗
=
𝟗𝟗.𝟓
𝟓𝟗.𝟐
= 1.6807
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.6807
Mth = 1.995
__________________________________________
24. S = 152 – (YU+YL) = 152 – (25.4+25.5)  S = 101.1
𝑺
𝑺∗
=
𝟏𝟎𝟏.𝟏
𝟓𝟗.𝟐
=1.7077
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.7077
Mth = 2.01

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__________________________________________
25. S = 152 – (YU+YL) = 152 – (25.4+25.5)  S = 101.1
𝑺
𝑺∗
=
𝟏𝟎𝟏.𝟏
𝟓𝟗.𝟐
=1.7077
Since
𝑺
𝑺∗
=
𝑨
𝑨∗
From table (B.2) at
𝑨
𝑨∗
=1.7077
Mth = 2.01
__________________________________________

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X(mm) Yu YL S(mm) S/S* M P/PO
0 10.7 10.7 130.9 2.206 0.27 0.9506
25 24.4 20.7 106.9 1.805 0.34 0.9231
50 37.5 25 89.5 1.511 0.43 0.8806
75 47.1 24.5 80.4 1.358 0.49 0.8486
100 58.2 25.5 68.3 1.153 0.63 0.7654
125 66.2 25.5 60.3 1.0185 0.86 0.6170
150 67.7 25.5 58.8 0.9932 1 0.5282
175 63.4 25.5 63.1 1.065 1.29 0.3658
200 59 25.5 67.5 1.1402 1.45 0.2927
225 55.2 25.5 71.3 1.204 1.54 0.2570
250 50.2 25.5 76.3 1.2888 1.65 0.2183
275 45 25.5 81.5 1.3766 1.74 0.1907
300 40.1 25.5 86.4 1.4594 1.81 0.1714
325 37 25.5 89.5 1.511 1.87 0.1563
350 35 25.5 91.5 1.5456 1.89 0.1515
375 33.5 25.5 93 1.5709 1.92 0.1447
400 31.3 25.5 95.2 1.608 1.94 0.1402
425 30.3 25.5 96.2 1.625 1.95 0.1381
450 29.3 25.5 97.2 1.641 1.97 0.1339
475 28.4 25.5 98.1 1.657 1.98 0.13184
500 28.5 25.5 98 1.655 1.983 0.13182
525 28 25.5 98.5 1.6638 1.985 0.129817
550 27 25.5 99.5 1.6807 1.995 0.129805
575 25.4 25.5 101.1 1.7077 2.01 0.12583
600 25.4 25.5 101.1 1.7077 2.01 0.12583

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Experiment:-
Find different pressure (∆h) from mercury manometer and the
following:-
Po – P = ƴ∆H
Where
∆h = m.hg
Po=0.76 ƴ hg
Atmospheric pressure = 760 mhg
𝑷
𝑷𝒐
= 𝟏 −
∆𝒉
𝟎. 𝟕𝟔
After calculating the value of
𝑃
𝑃𝑜
from the isentropic table, find the Mexp
x ho (cm)
0 10.2
50 16.5
100 23.3
150 35.4
200 43.2
250 58.2
300 61.5
350 62.2
400 63.8
450 63.4
500 56..4
550 61.6
600 59.6

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the atmospheric have changed from 0.76 mHg to 76 cmHg
The value was taken from tables of compressible flow function
to the closest value tabulated.
__________________________________________
∆h = h-ho = 10.2- 2.9 = 7.3 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
7.3
76
= 0.903
𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 ( 𝐵. 2) 𝑎𝑡
𝑃
𝑃𝑜
= 0.903
𝑀𝑒𝑥 = 0.38
__________________________________________
∆h = h-ho = 16.5 - 2.9 = 13.6 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
13.6
76
= 0.821
𝑃
𝑃𝑜
= 0.821
𝑀𝑒𝑥 =0.54
__________________________________________
∆h = h-ho = 23.3- 2.9 =20.4 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
20.4
76
= 0.7315
𝑃
𝑃𝑜
= 0.7315
𝑀𝑒𝑥 = 0.68
__________________________________________

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∆h = h-ho = 35.4- 2.9 = 32.5 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
32.5
76
= 0.572
𝑃
𝑃𝑜
= 0.572
𝑀𝑒𝑥 = 0.93
__________________________________________
∆h = h-ho = 43.2- 2.9 = 40.3 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
40.3
76
= 0.469
𝑃
𝑃𝑜
= 0.469
𝑀𝑒𝑥 = 1.10
__________________________________________
∆h = h-ho = 58.2- 2.9 = 55.3 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
55.3
76
= 0.272
𝑃
𝑃𝑜
= 0.272
𝑀𝑒𝑥 = 1.50
__________________________________________
∆h = h-ho = 61.5 - 2.9 = 58.6 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
58.6
76
= 0.228
𝑃
𝑃𝑜
= 0.228
𝑀𝑒𝑥 = 1.62

7/1/2019 23 | P a g e
∆h = h-ho = 62.2- 2.9 = 59.3 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
59.3
76
= 0.2197
𝑃
𝑃𝑜
= 0.2197
𝑀𝑒𝑥 = 1.65
__________________________________________
∆h = h-ho = 63.8- 2.9 = 60.9 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
60.9
76
= 0.1986
𝑃
𝑃𝑜
= 0.1986
𝑀𝑒𝑥 = 1.71
__________________________________________
∆h = h-ho = 63.4 - 2.9 = 60.5 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
60.5
76
= 0.2039
𝑃
𝑃𝑜
= 0.2039
𝑀𝑒𝑥 = 1.70
__________________________________________
∆h = h-ho =56.4 - 2.9 = 53.5 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
53.5
76
= 0.296
𝑃
𝑃𝑜
= 0.296
𝑀𝑒𝑥 = 1.44
_______________________________________

7/1/2019 24 | P a g e
∆h = h-ho = 61.6- 2.9 = 58.7 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
58.7
76
= 0.2276
𝑃
𝑃𝑜
= 0.2276
𝑀𝑒𝑥 = 1.62
__________________________________________
∆h = h-ho = 59.6 - 2.9 = 56.7 cm
𝑃
𝑃𝑜
= 1 −
∆ℎ
76
= 1 -
56.7
76
= 0.2539
𝑃
𝑃𝑜
= 0.2539
𝑀𝑒𝑥 = 1.55
__________________________________________
X ho(cm) ∆h(cm) P/Po Mexp.
0 10.2 7.3 0.903 .38
50 16.5 13.6 0.821 0.54
100 23.3 20.4 0.7315 0.68
150 35.4 32.5 0.572 0.93
200 43.2 40.3 0.469 1.10
250 58.2 55.3 0.272 1.50
300 61.5 58.6 0.228 1.62
350 62.2 59.3 0.2197 1.65
400 63.8 60.9 0.1986 1.71
450 63.4 60.5 0.2039 1.70
500 56..4 53.5 0.296 1.44
550 61.6 58.7 0.2276 1.62
600 59.6 56.7 0.2539 1.55

7/1/2019 25 | P a g e
7. DISCUSSION
1. What is the condition inside Nozzle?
Analysis of gas flow in de Laval nozzles
The analysis of gas flow through de Laval nozzles involves a number of
concepts and assumptions:
 For simplicity, the gas is assumed to be an ideal gas.
 The gas flow is isentropic (i.e., at constant entropy). As a result, the flow
is reversible (frictionless and no dissipative losses), and adiabatic(i.e.,
there is no heat gained or lost).
 The gas flow is constant (i.e., steady) during the period of
the propellant burn.
 The gas flow is along a straight line from gas inlet to exhaust gas exit (i.e.,
along the nozzle's axis of symmetry)
 The gas flow behaviour is compressible since the flow is at very
high velocities (Mach number > 0.3)

2. Illustrate the back pressure Pb on the flow inside nozzle?
Pressure Distribution and Choking in a Converging Nozzle
Consider a converging nozzle as shown in figure (a) below,

7/1/2019 26 | P a g e
Figure (b) shows the pressure ratio p/po along the length of the nozzle.
Where pis the static pressure. The inlet conditions of the gas are at the
stagnation state ( popo, ToTo) which are constants. The pressure at the exit
plane of the nozzle is denoted by PEPE and the back pressure
is PBPB which can be varied by the adjustment of the valve.
Case (i): As shown in fig (b), case (i); the pressure Po is
throughout, i.e. Po=PE=PB. There will be no flow through the
nozzle.
Case (ii): As we decrease PB gradually, the flow rate will increase.
The pressure will decrease in the direction of the flow as shown in
fig (b), case (ii). The pressure PE at the exit plane of the nozzle shall
remain equal to PB as long as the maximum discharge condition is
not reached.
The flow rate is directly proportional to mass flow rate,
so as the flow rate increases mass flow rate will also
increase.
Case (iii): Fig (b), case (iii); illustrates the pressure distribution in the
maximum discharge situation. The flow rate/mass flow rate has
attained its maximum valve, i.e. when Mach, Ma = 1 is achieved and
the nozzle is said to be choked. PE is equal to p∗ (pressure for Mach 1
flow). Since the nozzle does not have a diverging section, further reduction
in PBPB will not accelerate the flow to supersonic condition. As a result, PE
will continue to remain at p∗ even though PB is lowered further.
Case (iv): Since PB is less than p∗ the flow leaving the nozzle has to expand
to match the lower back pressure as shown in fig (b), case (iv); This expansion
is three-dimensional and the pressure distribution can’t be predicted by one-
dimensional theory.

7/1/2019 27 | P a g e
Pressure Distribution in a Converging-Diverging Nozzle
Consider the flow in a convergent-divergent nozzle, as shown below,
Similar to converging nozzle, when the pressure Po is throughout, i.e.
Po=PE= PB. There will be no flow through the nozzle.
Case (i): Slight decrease in back pressure will start the flow which will
be completely subsonic, the pressure distribution is shown by curve (i).
Case (ii): When PB is lowered in such a way the choking/sonic
condition is reached at the throat, the flow rate becomes maximum for a
given nozzle and the stagnation condition. The pressure distribution is
shown in curve (ii). The maximum mass flow rate will be
achieved here.
Case (iii) & (iv): On further reduction of the back pressure PB, the
flow upstream of the throat does not respond, the pressure distribution
upstream of the throat does not change beyond this. However, the
further reduction in back pressure will make the flow supersonic in the
diverging section, but then adjust to the back pressure by means of a
normal shock standing inside the nozzle; as shown in curve (iii) and (iv).
In such cases, the position of the shock moves downstream as PB is
decreased, and for curve (IV) the normal shock stands right at the exit
plane. If the back pressure is set at (VI), the flow will be
isentropic throughout the nozzle, and supersonic at nozzle
exit. Nozzles operating at PB (corresponding to curve (VI)) are
Said to be at design conditions.
Case (v): If back pressure is further reduced, then oblique shock will be
formed at the exit of the nozzle to match the back pressure, as shown in
curve (v). However the exit pressure will remain the same as design exit
pressure PEPE, which was achieved in curve (iv) at the exit of the nozzle

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3. The effect of back pressure Pb on the mass
flow rate?
Case (ii): When PB is lowered in such a way the choking/sonic
condition is reached at the throat, the flow rate becomes maximum
for a given nozzle and the stagnation condition. The pressure
distribution is shown in curve (ii). The maximum mass flow rate
will be achieved here.
When the back-pressure pb is equal to the supply chamber pressure, there
is, of course, no flow through the duct. As the back-pressure is decreased,
the mass flow rate through the duct and the Mach number at the duct exit
increase, the exit plane pressure pe being equal to the back-pressure pb.
This continues until Me reaches a value of 1. Further decreases in the back-
pressure have no effect on the flow in the duct, the adjustment from pe to
pb in this situation taking place through expansion waves outside the duct.
The mass flow rate is thus limited, i.e., the flow is “choked,” as a result of
friction. The variation of the mass flow rate with back-pressure is
illustrated Since once the flow is choked the Mach number always has a
value of 1 at the exit of the duct, if the length of the duct is changed, the
Mach number at the discharge of the nozzle
and the mass flow rate will change.

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4. Draw the relation between the (Mexp-Mth) and X.
0
0.5
1
1.5
2
2.5
0 100 200 300 400 500 600 700
X
M
the relation between the (Mexp-Mth) and X
Mexp. Mth
X Mexp. Mth
0 0.38 0.27
50 0.54 0.43
100 0.68 0.63
150 0.93 1
200 1.1 1.45
250 1.5 1.65
300 1.62 1.81
350 1.65 1.89
400 1.71 1.94
450 1.7 1.97
500 1.44 1.983
550 1.62 1.995
600 1.55 2.01

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5. Draw the relation between the (P/Po)th-(P/Po)exp) and X.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 100 200 300 400 500 600 700
X
P/Po
the relation between the (P/Po)th-(P/Po)exp) and X
P/Po (exp) P/PO (Th)
X P/Po
(exp)
P/PO (Th)
0 0.903 0.9506
50 0.821 0.8806
100 0.7315 0.7654
150 0.572 0.5282
200 0.469 0.2927
250 0.272 0.2183
300 0.228 0.1714
350 0.2197 0.1515
400 0.1986 0.1402
450 0.2039 0.1339
500 0.296 0.13182
550 0.2276 0.129805
600 0.2539 0.12583

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6. Compare the result that you have found from experimental
and theoretic value and suggest about develop system.
Through the charts we notice a little difference between
the practical and theoretical result of human errors and
inaccuracy of taking readings and the process of
calibration of the continuous device and vibrations
Suggest about developing the device:-
Introducing the machine to the student graduation
project process of maintenance and calibration

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