BLOCK I – COMPRESSIBLE FLOW
Solution of
Exercise 27
of the collection
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
1
Heading
The convergent-divergent nozzle as the one shown in the figure is considered. The
convergent part has a circular profile with a radius R=13 mm. The throat has a radius
of 6 mm. The divergent has a conical shape, with a half-angle of 2.5º, and the radius
at the exit is 15 mm.
R
a/2
Rg
Rs
Ld
The design conditions are 50 bar and 600 K of stagnation pressure and temperature
at the inlet of the nozzle.
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
2
Heading
It is asked:
1) Calculate the highest pressure at the outlet that makes that the critical conditions
are achieved in the throat.
2) Calculate the design outlet pressure.
3) Calculate the pressure at the exit if a shock wave appears in the middle of the
divergent.
4) Calculate the position of the shock wave if poutlet= 25 bar.
5) Calculate the lowest inlet pressure that makes the throat being at the critical
conditions when the stagnation temperature is kept constant, and the pressure at
the exit is kept at the design pressure.
6) Draw how the mass flow rate evolves as a function of the inlet stagnation pressure
assuming that the stagnation temperature and the outlet pressure are kept
constant.
Note: consider that the fluid is air, with R = 287 J/(kg·K) and  = 1.4.
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
3
Preliminary analysis of the exercise
Let’s break down the information given, as well as what we are asked to do:
 The geometry of the convergent-divergent nozzle is known. With the information
given, we can determine the area in each section of the nozzle.
 In the first question we are asked to find the highest outlet pressure that makes
having M=1 at the throat.
 In the second question we are asked to find the design outlet pressure.
10
9.9
9.8
9.75
9.721
Block I – Compressible flow
Illustrative example
(only valid
qualitatively)
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
4
Preliminary analysis of the exercise
Let’s break down the information given, as well as what we are asked to do:
 The next two questions correspond to intermediate situations (the outlet pressure
is placed between those of questions 1 and 2), and a shock wave will exist in the
divergent part of the nozzle.
 In question 3, the shock wave position is known, and we are told to find which the
outlet pressure is.
 In question 4, the information is reversed: now the outlet pressure is known, and
we are told to determine the position of the shock wave.
Illustrative example
(only valid
qualitatively)
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
5
Question 1
Even if this question can be solved analytically
(i.e., by equations), we are going to solve it using
the tables, because it is much easier and
powerful. Besides, it is the usual method that we
ask you to use in this kind of exercises.
 Since M=1 in the throat, A*=Athroat.
 By the geometry, we determine the area of the
outlet section, As.
 With As/A* (=6.25) we enter the table.
 We obtain ps/p* (=1.8814).
 p* can be found from p0, through the
expression
𝛾
𝑝0 =
 Thus, ps = 49.7 bar.
Block I – Compressible flow
𝑝∗
𝛾+1
2
𝛾−1
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
6
Question 2
As in the previous case, this question will be
solved using the tables.
 With As/A* (=6.25) we enter the table,
searching for the supersonic solution.
 We obtain ps/p* (=0.03046).
 Since p* is already known, we find that now
ps = 0.805 bar.
Block I – Compressible flow
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
7
Question 3
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
In this question, we are asked to find the outlet p
when the shock wave is at a given position
(middle of the divergent). To do so, several things
need to be considered:
 First, the mass flow rate is known, and is
equal to the chocked one (in fact, as long as
the outlet pressure is equal to or lower than
that of question 1, the mass flow rate is always
the same, since the conditions at the throat
are always the critical ones).
 To obtain p at the outlet section we need to
1
make use of the information available in this
section, which is A. For this we need A*. WithSupersonic flow
this information we can enter the table of the
isentropic flow and find p/p*. To obtain ps we
need to know p* of the flow downstream the
shock wave, and consequently we need to
know the stagnation p there.
Block I – Compressible flow
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
2
Subsonic flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
8
Question 3
To determine the stagnation pressure loss across
the shock wave, the key point is to find the
conditions we have just upstream the shock wave
(conditions 1), knowing that between the inlet of
the duct up to this section the flow is isentropic
(as already underlined, the flow is only NON
isentropic in the shock wave).
 With the geometrical information, we
determine A1 (=3.46 cm2).
 We calculate A1/A* (=3.06), and we enter the
table, searching for the supersonic solution
(we have always M>1 before the shock wave).
 We obtain M1, which is the parameter required
to enter the shock wave table: M1 = 2.63.
Block I – Compressible flow
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
9
Question 3
ShockDEwave
( = (
1.4)
TABLAS
ONDAtables
DE CHOQUE
=1.4)
Now we are going to characterize the flow
parameters at the other side of the shock wave
(which is also isentropic; the flow is only NON
isentropic in the shock wave).
 First, the stagnation T is the same (it is
invariant, independently if the flow is
isentropic or not).
 Second, we need to find the new stagnation
pressure (this parameter is not invariant
anymore). From M at the inlet of the shock
wave, M1, we obtain p20/p10 (=0.4477).
 Finally, we can obtain M2, which will be useful
to determine A*’ (M2 = 0.501).
Block I – Compressible flow
M1
1
1.02
1.04
1.06
1.08
1.1
1.12
1.14
1.16
1.18
1.2
1.25
1.3
1.35
1.4
1.45
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
M2
1
0.9805
0.9620
0.9444
0.9277
0.9118
0.8966
0.8820
0.8682
0.8549
0.8422
0.8126
0.7860
0.7618
0.7397
0.7196
0.7011
0.6684
0.6405
0.6165
0.5956
0.5774
0.5130
0.4752
0.4512
0.4350
0.4236
0.4152
p2/p1
1
1.0471
1.0952
1.1442
1.1941
1.2450
1.2968
1.3495
1.4032
1.4578
1.5133
1.6563
1.8050
1.9596
2.1200
2.2863
2.4583
2.8200
3.2050
3.6133
4.0450
4.5000
7.1250
10.3333
14.1250
18.5000
23.4583
29.0000
p20/p10
1
1.0000
0.9999
0.9998
0.9994
0.9989
0.9982
0.9973
0.9961
0.9946
0.9928
0.9871
0.9794
0.9697
0.9582
0.9448
0.9298
0.8952
0.8557
0.8127
0.7674
0.7209
0.4990
0.3283
0.2129
0.1388
0.0917
0.0617
T2/T1 r2/r1=c1/c2
1
1
1.0132
1.0334
1.0263
1.0671
1.0393
1.1009
1.0522
1.1349
1.0649
1.1691
1.0776
1.2034
1.0903
1.2378
1.1029
1.2723
1.1154
1.3069
1.1280
1.3416
1.1594
1.4286
1.1909
1.5157
1.2226
1.6028
1.2547
1.6897
1.2872
1.7761
1.3202
1.8621
1.3880
2.0317
1.4583
2.1977
1.5316
2.3592
1.6079
2.5157
1.6875
2.6667
2.1375
3.3333
2.6790
3.8571
3.3151
4.2609
4.0469
4.5714
4.8751
4.8119
5.8000
5.0000
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
10
Question 3
We need to determine A*’. There are two options:
 The first one is by using the isentropic flow
tables. In section 2 (just after the shock wave)
we already know the Mach number. From this
parameter we can obtain A2/A*’ (= 1.340).
Since A2=A1 (the width of the shock wave is
negligible), we can obtain the new value of A*’
(= 2.53 cm2).
 The other option would be from the equation
of the chocked flow, once the critical density is
known (which has changed because the
stagnation p has changed; c*, on the contrary,
has not changed, since the stagnation T is
constant).
 We determine As/A*’ (=2.798), and enter the
table searching for p/p* for the subsonic case
(= 1.832), from which ps = 21.67 bar.
Block I – Compressible flow
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
11
Question 4
In this question, things are reversed: we are told to determine the shock wave position
for a given value of the outlet pressure (25 bar). In this case, we are going to analyze
the information available in the outlet section:
 To begin with, as already repeated before, the mass flow rate is known, and is the
chocked one.
 The area is also known, as well as the pressure.
𝑚 = 𝑚∗ = 𝜌𝑠 𝑐𝑠 𝐴𝑠
 The equation for the mass flow rate is:
 Where c and r are:
𝑐𝑠 =
2𝑐𝑝 𝑇0 − 𝑇𝑠
𝑝𝑠
𝜌𝑠 =
𝑅 · 𝑇𝑠
 Consequently, an equation is found (once everything is substituted in the first

equation) where the only unknown is Ts (= 595.94 K).
From T0 and Ts, we can determine p0’ (the one of the flow downstream the shock
𝛾
wave) (= 25.6 bar):
𝑇 𝛾−1
𝑝0′ = 𝑝
Block I – Compressible flow
0
𝑇
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
12
Question 4
ShockDEwave
( = (
1.4)
TABLAS
ONDAtables
DE CHOQUE
=1.4)
Now the shock wave itself needs to be studied.
 The known data is p20/p10 (= 0.5120), from
which we obtain M1 (= 2.471).
Block I – Compressible flow
M1
1
1.02
1.04
1.06
1.08
1.1
1.12
1.14
1.16
1.18
1.2
1.25
1.3
1.35
1.4
1.45
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
M2
1
0.9805
0.9620
0.9444
0.9277
0.9118
0.8966
0.8820
0.8682
0.8549
0.8422
0.8126
0.7860
0.7618
0.7397
0.7196
0.7011
0.6684
0.6405
0.6165
0.5956
0.5774
0.5130
0.4752
0.4512
0.4350
0.4236
0.4152
p2/p1
1
1.0471
1.0952
1.1442
1.1941
1.2450
1.2968
1.3495
1.4032
1.4578
1.5133
1.6563
1.8050
1.9596
2.1200
2.2863
2.4583
2.8200
3.2050
3.6133
4.0450
4.5000
7.1250
10.3333
14.1250
18.5000
23.4583
29.0000
p20/p10
1
1.0000
0.9999
0.9998
0.9994
0.9989
0.9982
0.9973
0.9961
0.9946
0.9928
0.9871
0.9794
0.9697
0.9582
0.9448
0.9298
0.8952
0.8557
0.8127
0.7674
0.7209
0.4990
0.3283
0.2129
0.1388
0.0917
0.0617
T2/T1 r2/r1=c1/c2
1
1
1.0132
1.0334
1.0263
1.0671
1.0393
1.1009
1.0522
1.1349
1.0649
1.1691
1.0776
1.2034
1.0903
1.2378
1.1029
1.2723
1.1154
1.3069
1.1280
1.3416
1.1594
1.4286
1.1909
1.5157
1.2226
1.6028
1.2547
1.6897
1.2872
1.7761
1.3202
1.8621
1.3880
2.0317
1.4583
2.1977
1.5316
2.3592
1.6079
2.5157
1.6875
2.6667
2.1375
3.3333
2.6790
3.8571
3.3151
4.2609
4.0469
4.5714
4.8751
4.8119
5.8000
5.0000
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
13
Question 4
Finally, we focus now in the flow upstream the
shock wave:
 From M1 we find A1/A* (= 2.566).
 A* is that of the throat, since there is where
the critical conditions happen in the flow
upstream the shock wave. From this value, we
find A1 (= 2.90 cm2).
 From the value of the section area, we can
determine the radius (9.61 cm), as well as the
location (x = 82.76 mm from the throat, being
206.33 mm the length of the divergent; then,
the shock wave is located at around 40% of
the total length of the divergent).
Block I – Compressible flow
M
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.09
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.5
3
3.5
4
4.5
5
T/T*
1.200
1.200
1.200
1.200
1.199
1.199
1.198
1.198
1.198
1.190
1.179
1.163
1.143
1.119
1.093
1.064
1.033
1.000
0.966
0.932
0.897
0.862
0.828
0.794
0.760
0.728
0.697
0.667
0.533
0.429
0.348
0.286
0.238
0.200
p/p*
1.893
1.892
1.892
1.891
1.890
1.888
1.884
1.882
1.880
1.841
1.778
1.695
1.596
1.484
1.365
1.242
1.119
1.000
0.887
0.781
0.683
0.595
0.516
0.445
0.383
0.329
0.282
0.242
0.111
0.052
0.025
0.012
0.007
0.004
c/a*
0.011
0.022
0.033
0.044
0.055
0.066
0.088
0.099
0.109
0.218
0.326
0.431
0.535
0.635
0.732
0.825
0.915
1.000
1.081
1.158
1.231
1.300
1.365
1.425
1.482
1.536
1.586
1.633
1.826
1.964
2.064
2.138
2.194
2.236
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
A/A*
57.874
28.942
19.301
14.481
11.591
9.666
7.262
6.461
5.822
2.964
2.035
1.590
1.340
1.188
1.094
1.038
1.009
1.000
1.008
1.030
1.066
1.115
1.176
1.250
1.338
1.439
1.555
1.688
2.637
4.235
6.790
10.719
16.562
25.000
14
Question 5
What we are asked is to “calculate the lowest inlet pressure that makes the throat
being at the critical conditions when the stagnation temperature is kept constant, and
the pressure at the exit is kept at the design pressure”.
 Now the initial pressure will be reduced steadily. We are going to illustrate this in
the same example that we have already used for the previous questions.
 Consequently, what we are told is to find the value of p0 making that we are in the
same situation as in question 1, but now with an outlet p equal to the design p.
Illustrative example
(only valid
qualitatively)
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
15
Question 5
 The result can be easily found if the result of question 1 is normalized by p0:
Question 1
𝑝𝑠
= 0.994
𝑝0
Question 5
With 𝑝𝑠 =0.8046 bar
𝑝0 =0.8095 bar
Illustrative example
(only valid
qualitatively)
Block I – Compressible flow
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
16
Question 6
What we are asked is to “draw how the mass flow rate evolves as a function of the
inlet stagnation pressure assuming that the stagnation temperature and the outlet
pressure are kept constant”.
 As long as there are critical conditions at the throat (which is in most of the
operating range), the mass flow rate is chocked:
𝑚=
𝑚∗
=
𝜌∗ 𝑐 ∗ 𝐴∗
∗
with 𝑐 =
𝛾𝑅𝑇 ∗
∗
= 𝑐𝑡𝑒 and 𝜌 =
𝑝∗
𝑅𝑇 ∗
∝ 𝑝0
 Thus, the chocked mass flow rate will be proportional to p0. Then, the evolution will
be:
𝑚
ps = 0.8046 bar 0.8095 bar
(design p Q.2)
Block I – Compressible flow
(p0 Q.5)
𝑝0
AMPLIACIÓN DE ENERGÍA Y MÁQUINAS TÉRMICAS
17