EXPLORATION SUB
module uses previously discussed equations and concepts to explore
solutions of advanced level problems. The topics include:
Determination of back pressure ranges for the seven regimes of flow in
diverging nozzles
Normal shock in nozzles and diffusers
Measurement of Mach number in supersonic flow
Calculation of nozzle thrust
Determination of Flow Regimes for a Given
Converging Diverging Nozzle Geometry (
Let us illustrate this with an area ratio (
) equal to 2.20
From isentropic gas tables one gets two Mach numbers at exit namely
(subsonic regime, curve c) and
= 2.20 (supersonic regime, curve d). The
corresponding pressures at exit
= 0.939 and
= 0.094. The pressure
level e corresponds to a case in which a normal shock appears at the nozzle exit.
Upto nozzle exit, and before the normal shock the pressure is
. After the shock
the pressure in
. To determine
the normal shock table. For (
2.20 (before the shock at exit), the Mach number after the shock is (
= 0.547.
651
.
0
ox
oy
P
P
. From isentropic gas table, at (ME)
= 0.547,
214
.
0
oy
y
p
P
Since
814
.
0
e
o
e
P
P
, and
094
.
0
d
o
d
P
P
530
.
0
65
.
0
814
.
0
od
oe
oe
e
od
e
P
P
P
P
P
P
or
530
.
0
o
e
P
P
Following are the possible flow regimes:
Subsonic isentropic flow throughout (0
< 0.939)
Subsonic (converging), sonic (throat) and subsonic (diverging) isentropic flow
= 0.939)
Normal shock in diverging section (0.53 <
Normal shock at the exit section (
= 0.53)
Oblique shocks outside the nozzle (over expanded nozzle flow case) (0.09 <
Subsonic (converging), sonic (throat) and supersonic (diverging) isentropic
= 0.09)
Subsonic (converging), sonic (throat) and supersonic (diverging) isentropic
under expanded nozzle flow. Further expansion outside the nozzle (
Normal Shocks in the Nozzles and Diffusers
Statement: Consider a converging
diverging nozzle handling air (shown in figure).
= 1000K,
= 1000 kPa
A shock occurs in the diverging section where area
: Flow properti
at the exit section.
Since shock occurs in the diverging section, flow must be as follows:
2
0004
.
0
0008
.
0
*
A
A
x
From Isentropic Flow Tabl
= 2.20 (Mach number ahead of the shock)
From shock table
= 0.547 (Mach number behind the shock)
x
y
o
o
P
P
/
7165
.
6
x
o
P
P
y
48
.
5
x
y
P
P
8569
.
1
x
y
T
T
Since
1000
o
o
P
P
x
kPa, (up to shock wave)
628
1000
62814
.
0
x
x
y
y
o
o
o
o
P
P
P
P
= 2.20, (From isentropic flow table)
9035
.
0
x
o
x
P
P
508
.
0
x
o
x
T
T
x
o
x
P
P
0935
.
0
= 0.0935
1000 = 93.5 kPa
508
508
.
0
x
o
x
T
T
(Note: T
1000
oE
oy
ox
T
T
38
.
512
5
.
93
48
.
5
x
x
y
y
P
P
P
P
= 628 kPa (Isentropic flow after shock)
Calculate Exit Conditions
628
y
E
o
o
P
P
1000
o
o
o
oE
T
T
T
T
x
y
= 0.547, determine
*
/
A
A
y
from isentropic flow tabl
new
y
A
A
*
/
(Note A* has changed across the shock).
2
2
*
000632
.
0
265
.
1
0008m
.
0
265
.
1
m
A
A
y
new
From the shock location (downstream side) to the exit section, (
remains
constant.
Now form the ratio
new
E
A
A
*
5822
.
1
000632
.
0
001m
.
0
2
*
A
A
new
E
om isentropic flow table
40
.
0
E
M
E
o
E
E
o
E
P
P
P
P
895
.
0
895
.
0
562
628
895
.
0
E
P
969
.
0
E
o
E
T
T
1000
969
.
0
969
.
0
E
o
E
T
T
969
E
T
: Why does A* change across a normal shock?
1
2
1
2
1
1
*
k
k
k
R
k
T
P
A
m
o
o
This equation is derived from mass conservation equation given below by setting
M = 1 and
1
2
1
2
*
2
1
1
k
k
o
o
M
k
M
R
k
T
P
A
m
Note that
decreases across the shock, but
remains fixed. From Eq. (A),
increases across shock since
m
remains constant.
Measurement of Mach Number
Subsonic Flow
Mach number at any point in the flow field can be measured by using a pitot
probe or a pitot
static tube probe. This is illustrated in Figure 1.
Since pitot tube is small and is designed to generate small disturbances in the flow
an be neglected and flow can be assumed to be
isentropic. The right hand limb of the manometer measures stagnation pressure
) while the left hand limb measures static pressure
. Since the two are related
through the isentropic relationship
,
2
1
1
1
2
k
k
o
M
k
P
P
one can solve for
1
1
2
1
k
k
o
P
P
k
M
From manometer equation, one gets
gL
P
P
m
o
P
gL
P
P
m
o
1
Equation (2) can be substituted in Eq. (1) to obtain Mach number M.
Example
Determine Mach number and air velocity (V) at a point in the flow field where P = 100 kPa, T = 300K
and pitot
tube manometer deflection 100 mm of mercury.
m
m
m3
kg
gL
P
P
m
o
1
.
0
2
sec
8
.
9
13500
kPa
P
P
P
o
23
.
13
13230
1323
.
0
100
23
.
13
1
P
P
o
1323
.
1
P
P
o
425
.
0
1
1323
.
1
4
.
0
2
1
1
2
286
.
0
1
k
k
o
P
P
k
M
Speed of sound
300
287
4
.
1
KRT
C
= 347 m/s. Flow velocity
347 =
147 m/sec.
Supersonic Flow
Introduction of a pitot
tube probe in supersonic flow leads to a large disturbance and a shockwave
standing in front of the probe (Fig.
The flow behind the shockwave is subsonic (Fig. 2), and the stagnation pressure
2
o
P
) is less than
1
o
P
(stagnation pressure ahead of the shock). Note that
2
o
P
1
P
are the measured pressures.
1
2
2
1
2
2
P
P
P
P
P
P
o
o
1
2
2
2
1
1
2
k
k
o
M
k
P
P
1
1
2kM
2
1
1
2
k
k
P
P
Shockwa
2
o
P
1
1
2k
1
2
2
1
2
1
2
2
M
k
k
M
M
Across the shock.
Substituting Eqs. (4)
(6) in Eq. (3) one gets,
1
1
2
1
1
2
1
1
1
1
1
2kM
2
1
2
k
k
k
o
k
k
k
M
K
P
P
Equation (7) is known as the Rayleigh formula; and is tabulated in the shock table.
A pitot
tube probe is used to measure Mach number in a supersonic air flow. The
static pressure and temperature are respectively 50 kPa and
C. The difference
between stagnation pressure
2
o
P
1
P
is measured as 300 kPa. Determine the
Mach number.
1
2
P
P
o
2
o
P
300 + 50 = 350 kPa
7
50
350
2
2
P
P
o
From the shock table, M
(behind the shockwave) is 0.514.
Calculation of Nozzle Thrust
The nozzle thrust can be calculated by applying the Newton’s second law to the
control volume surrounding the nozzle.
Thrust
can be expressed
a
E
E
i
E
P
P
A
V
V
m
T
is the back pressure.
Note that for subsonic flows
. For underexpanded flows,
overexpanded flow
consequently the flow conditions may be such that pressure thrust component may
be zero, negative or positive.