AER-4027 Design of Nozzles and Intakes
Mid-Term Exam, Take Home (110 Marks), Start (April 1, Tue), Due (April 8, Tue)
Compare the shown symmetric 2D expanding duct profiles. The inlet flow is subsonic, from Left to Right, and
the flow inside each duct is isentropic. Consider the following criteria for your comparison:
(1) Subsonic Inlet Flow Characteristics, (2) Velocity Distribution, (3) Pressure Distribution, (4) Critical Reynolds
Number, (5) B.L. Behavior and Flow Stability, (6) Separation Risk, (7) Losses in Percent from Inlet to Exit
(mention loss measure you are using), (8) Advantages, (9) Applications.
[A] Bell
Upper
Profile
y ( x) =
[B] Truncated Cone
[C] Horn
y ( x) = 1+ x
y ( x) = e ( )
3 1
− cos ( x )
2 2
Mention the limiting operating conditions for flow inside a convergent-divergent nozzle.
Question 3 (20 Marks): Internal Flow [Ventilation System]
Question 1 (20 Marks): Introduction and Review [Duct Flows]
#
Question 2 (10 Marks): Internal Flow [Quasi 1-D Flow]
A fan is to produce a constant air speed of throughout the pipe loop shown. The 3-m-diameter pipes are smooth,
and each of the four elbows has a loss coefficient of 0.30. Determine the power that the fan adds to the air.
ln 2 x
Plot
Question 4 (30 Marks): Nozzle
(a) Consider a nozzle with efficiency  n between the inlet and any station downstream. Find the functions,
P0
m T0
= f 2 , M t = f3
= f1 ,
A P0
P
Show that M = 1 does not occur at the nozzle minimum area (throat).
(b) Consider a supersonic nozzle constructed with an exit to throat area ratio of 2. The nozzle is supplied with
low-speed air at 7 atm and 60 oC. The overall nozzle efficiency from inlet to exit is 90%, but the flow is
isentropic to the throat. Calculate the pressure, velocity, and Mach number at the exit and compare with the
corresponding values for isentropic flow.
f1 (N , M ) , f 2 (N , M ) , and f3 (N ) in the expressions,
Repeat your comparison if the flow is from Right to Left. Organize your work in table format.
Flow from Left to Right
#
Comparison Criteria
1
Subsonic Inlet Flow Characteristics
2
Velocity Distribution
9
Applications
[A]
[B]
[C]
Question 5 (30 Marks): Diffuser
Consider the shown inviscid two-dimensional airflow.
(a) Find the minimum Mach number, M 1 , for this case to exist.
(b) Find M 2 , M 3 , M 4 , and the ratios P2 / P1 , P3 / P1 , and P4 / P1 .
(c) Find the length L in terms of the height H corresponding to this M 1 .
Flow from Right to Left
#
Comparison Criteria
1
Subsonic Inlet Flow Characteristics
2
Velocity Distribution
[A]
[B]
[C]
L
H
9
1
2
4
3
Applications
1
2
3
Question 3
• Power added by fan:
P = ṁ ∗ g ∗ hL = 406.5 Kw
4
Question 4
A) Nozzle Efficiency
Actual K.E
h1 − h 2 a
=
Isen. K.E
h1 − h2 s
ηN =
ηN =
• Given:
∵M = √
– stream velocity V=40 m/s
– pipe diameter D=3m
∴ ηN =
– #Num of Elbows =4
V
→ V 2 = M 2 γRT
γRT
Ma2 γRTa
M2
, Ta = Ts → ηN = a2
Ms2 γRTs
Ms
– loss coefficient per elbow K =0.3
∴ Ma =
– Smooth pipe (i.e Moody charts could be used)
πD2
= 7.07m2
4
ṁ = ρAV = 339.29Kg/s
• Head Loss
from Moody chart we can see that
f = 0.015
hf = f ∗
L V
∗
= 24.45m
D 2g
• Minor Loss
η N ∗ Ms ≡ f 3
√
ηN
γ
γ
γ − 1 2 γ−1
Po
γ−1 2
= (1 +
Ma )
Ms ηN ) γ−1 ≡ f1
= (1 +
P
2
2
r
√
γ+1
ṁ To
γ
γ − 1 2 − 2(γ−1)
Ma (1 +
Ma )
=
APo
R
2
r
√
γ+1
γ√
ṁ To
γ−1 2
Ms ηN )− 2(γ−1) ≡ f2
∴
=
ηN ∗ Ms (1 +
APo
R
2
Function
f1
2
√
∴ Math =
L = 2(20 + 10) = 60m
A=
Va2
Vs2
f2
f3
Formula
γ
pγ√
R
2
γ−1
(1 + γ−1
2 Ms η N )
as Mach number at throat is function in Nozzle efficiency so
hm =
X
V2
K∗
= 4 ∗ 0.3 ∗ 81.5 = 97.8m
2g
ηN ̸= 1 → Mth < 1
• Total head loss
hL = hf + hm = 122.25m
7
γ+1
ηN ∗ Ms (1 + γ−1
Ms2 ηN )− 2(γ−1)
√ 2
ηN
8
5
B)
Question 5
given that
Ae
= 2, low air speed @ P = 7atm, T = 60o C, ηN = 90%
Ath
∵ air at low speed → Po = 7atm, To = 60 + 273 = 333K
∵ N ozzle is subsonic → Pb = Pcr3
To
Ae
Po
= 2 isen.table
−−−−−−→M = 2.2, P = 10.6927, T = 1.968
Ath
p
∴ Ps = 0.65465atm, Ts = 169.2073K, Vs = Ms ∗ γRT = 573.63708m/s
∴
using f1 &f2 obtained at point a
γ−1 2
Po
= (1 +
Ms η N )
P
2
Ma =
Va = Ma
√
p
check → Va =
γ
γ−1
= 8.96232 → Pa = 0.78104atm
we divide the problem into through regions:
• From 1-2 : Oblique Shock with deflection angle of 10 deg
• From 2-3 : Reflected Oblique Shock with deflection angle of 10 deg
• From 3-4 : Reflected Oblique Shock with deflection angle of 10 deg
A) Minimum Inlet Mach number
The constrain on inlet Mach number that specifies minimum inlet Mach number is that the inlet
Mach number to reflective wave at 3-4 is the minumum for the deflection angle of θ = 10o
From Fig. of M1 − M2 − θ relation below
ηN ∗ Ms = 2.0871
γRT = 544.199988 m/s
√
ηN ∗ Vs = 544.199988 m/s
Figure 1: M1 − M2 − θ relation
we can see that:
@ θ = 10o M3min = 1.421
Oblique shock from 2-3
from Fig.1
@ θ = 10o , M3 = 1.421 → M2 = 1.771
9
10
AER-4027 Design of Nozzles and Intakes
Mid-Term Exam, Take Home (110 Marks), Start (April 1, Tue), Due (April 8, Tue)
Compare the shown symmetric 2D expanding duct profiles. The inlet flow is subsonic, from Left to Right, and
the flow inside each duct is isentropic. Consider the following criteria for your comparison:
(1) Subsonic Inlet Flow Characteristics, (2) Velocity Distribution, (3) Pressure Distribution, (4) Critical Reynolds
Number, (5) B.L. Behavior and Flow Stability, (6) Separation Risk, (7) Losses in Percent from Inlet to Exit
(mention loss measure you are using), (8) Advantages, (9) Applications.
[A] Bell
Upper
Profile
y ( x) =
[B] Truncated Cone
[C] Horn
y ( x) = 1+ x
y ( x) = e ( )
3 1
− cos ( x )
2 2
Mention the limiting operating conditions for flow inside a convergent-divergent nozzle.
Question 3 (20 Marks): Internal Flow [Ventilation System]
Question 1 (20 Marks): Introduction and Review [Duct Flows]
#
Question 2 (10 Marks): Internal Flow [Quasi 1-D Flow]
A fan is to produce a constant air speed of throughout the pipe loop shown. The 3-m-diameter pipes are smooth,
and each of the four elbows has a loss coefficient of 0.30. Determine the power that the fan adds to the air.
ln 2 x
Plot
Question 4 (30 Marks): Nozzle
(a) Consider a nozzle with efficiency  n between the inlet and any station downstream. Find the functions,
P0
m T0
= f 2 , M t = f3
= f1 ,
A P0
P
Show that M = 1 does not occur at the nozzle minimum area (throat).
(b) Consider a supersonic nozzle constructed with an exit to throat area ratio of 2. The nozzle is supplied with
low-speed air at 7 atm and 60 oC. The overall nozzle efficiency from inlet to exit is 90%, but the flow is
isentropic to the throat. Calculate the pressure, velocity, and Mach number at the exit and compare with the
corresponding values for isentropic flow.
f1 (N , M ) , f 2 (N , M ) , and f3 (N ) in the expressions,
Repeat your comparison if the flow is from Right to Left. Organize your work in table format.
Flow from Left to Right
#
Comparison Criteria
1
Subsonic Inlet Flow Characteristics
2
Velocity Distribution
9
Applications
[A]
[B]
[C]
Question 5 (30 Marks): Diffuser
Consider the shown inviscid two-dimensional airflow.
(a) Find the minimum Mach number, M 1 , for this case to exist.
(b) Find M 2 , M 3 , M 4 , and the ratios P2 / P1 , P3 / P1 , and P4 / P1 .
(c) Find the length L in terms of the height H corresponding to this M 1 .
Flow from Right to Left
#
Comparison Criteria
1
Subsonic Inlet Flow Characteristics
2
Velocity Distribution
[A]
[B]
[C]
L
H
9
1
2
4
3
Applications
1
2
Question 1 Introduction and Review [Duct Flows]
Cairo University
Faculty of Engineering
4th Year Aerospace Department
Design of Nozzles and Intakes
AER-402
Take-home Midterm (Solution)
Solution
Duct Profiles
Submitted to:
Dr. Hisham El-Banna
Eng. Ahmed Hafez
Complete Comparative Table (Flow: Left to right)
Name: Sofian Magdy Mohamed Hassen
Sec: 1
BN: 22
#
1
Criterion
Subsonic Inlet Flow
[A] Bell
Sharp contraction may disturb
flow; mild pressure rise
[B] Truncated Cone
Steep entrance; possible
separation at sharp
corner
[C] Horn
Sudden contraction; flow
acceleration at inlet
Velocity
Distribution
Sharp velocity drop (from ~0.53 to
~0.25)
Gradual velocity drop (to
~0.22)
Pressure
Distribution
Critical Reynolds
Number
B.L. Behavior &
Stability
Separation Risk
Sharp pressure rise (~11 kPa or
10.5%)
Reduced to ~ 4.8 × 10⁵
Moderate rise (~12 kPa or
11.5%)
Standard ~ 4.5 × 10⁵
Thick B.L. near inlet; moderate
instability
Moderate at inlet
Moderate B.L. growth;
decent attachment
Moderate to high
2.1%
2.8%
8
Losses (Stagnation
Pressure Loss %)
Advantages
9
Applications
Compact deceleration zone; decent
recovery
Reverse nozzles, subsonic diffuser
exits
Easy to fabricate; robust
across Re
HVAC exhaust diffusers
2
3
4
5
6
7
Strong deceleration (to
~0.20); high gradient at
inlet
Large initial jump (~14 kPa
or 13.5%)
Low ~ 4.2 × 10⁵
Rapid B.L. growth; poor
flow attachment
Very high at inlet due to
steep contraction
5.0%
Very compact; aggressive
deceleration
Shock tubes, pulsed
decelerators
Question 2 (10 Marks): Internal Flow [Quasi 1-D Flow]
Mention the limiting operating conditions for flow inside a convergent-divergent nozzle.
Solution
Determining Flow Regimes in a C-D Nozzle Based on Pressure Ratios
The operating conditions inside a convergent-divergent nozzle are governed by:


2
Velocity
Distribution
3
Pressure
Distribution
Critical Reynolds
Number
B.L. Behavior &
Stability
Separation Risk
4
5
6
7
8
Losses (Stagnation
Pressure Loss %)
Advantages
9
Applications
[A] Bell
Smooth entry curvature; excellent
flow attachment
Smooth acceleration; nearly linear
Mach rise to ~0.53
0.3% (minimal)
[B] Truncated Cone
Sharp corner; moderate
attachment
Moderate acceleration;
potential stagnation
zones
Slightly non-uniform drop
(~11%)
Standard transition (~5 ×
10⁵)
Risk of mild separation at
edges; minor instability
Moderate; linear slope
may cause detachment
1.6% (mild loss)
Low loss, stable; best for subsonic
isentropic flow
Jet engines, rocket nozzles,
turbines, wind tunnels
Easy to manufacture,
compact
HVAC ducts, subsonic
intakes
Gradual pressure drop (~9%)
Slight delay in transition due to
shape (~6 × 10⁵)
Excellent: thin B.L., no separation,
stable flow
Very low; ideal diffuser profile
𝐴
𝑡ℎ
Step-by-Step Guide to Finding Critical Pressure Ratios

Criterion
Subsonic Inlet Flow
𝑜
The geometry of the nozzle, particularly the area ratio 𝐴 𝑒 (exit area to throat area)
First Critical Pressure Ratio — End of Isentropic Subsonic Expansion
Complete Comparative Table (Flow: Left to right)
#
1
𝑃
The pressure ratio 𝑃𝑏 (back pressure to stagnation pressure)
[C] Horn
Rapid area growth; gentle
entry
Exponential acceleration;
steep velocity gradient
Fast pressure drop (~13%);
mild shocks possible
Early transition likely (~4.5
× 10⁵)
Thickening B.L. near outlet;
susceptible to unsteadiness
High risk due to aggressive
area change
3.2% (significant due to
separation)
High expansion in small
space
Compact supersonic
nozzles, architecture wind
deflectors

𝐴
Use isentropic flow tables or equations for subsonic flow at the given area ratio 𝐴 𝑒
𝑡ℎ
Obtain the corresponding pressure ratio:
(
𝑃𝑏
)
𝑃𝑜 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙1
Third Critical Pressure Ratio — Fully Isentropic Supersonic Flow


𝐴
Use isentropic flow tables for supersonic flow at the same area ratio 𝐴 𝑒
𝑡ℎ
Obtain:
Supersonic Mach number at exit, M3
Pressure ratio:
(
Second Critical Pressure Ratio — Normal Shock at Exit


𝑃𝑏
)
𝑃𝑜 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙3
Use normal shock tables or equations at M3 (found in Step 2).
Determine the post-shock static pressure P2 and compute the ratio:
(
𝑃𝑏
𝑃2
)
=
𝑃𝑜 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙2 𝑃𝑜
Then for a given back pressure and stagnation pressure the flow in the nozzle operates as following
table:
Nozzle Flow Regimes Based on Back Pressure Pb
No.
Back pressure condition
1.
𝑃𝑏 = 𝑃𝑡𝑜𝑡𝑎𝑙
-
No flow (no pressure gradient)
-
Subsonic flow throughout the nozzle
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2 < 𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1
-
Sonic flow at the throat
subsonic elsewhere
Subsonic everywhere except sonic at throat
Normal shock forms in divergent section
Normal shock occurs exactly at nozzle exit
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3 < 𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2
-
Supersonic flow exits nozzle
over expansion oblique shock outside
Ideal case fully expanded flow
𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3
-
Under expanded flow
expansion waves format exit
2.
3.
4.
5.
6.
7.
8.
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1 < 𝑃𝑏 < 𝑃𝑡𝑜𝑡𝑎𝑙
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3
Mach number Limits for Nozzle Flow Conditions
No.
Back pressure condition
1.
-
2.
𝑃𝑏 = 𝑃𝑡𝑜𝑡𝑎𝑙
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1 < 𝑃𝑏 < 𝑃𝑡𝑜𝑡𝑎𝑙
-
3.
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1
-
4.
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2 < 𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 1
-
5.
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2
6.
𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3 < 𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2
-
7.
𝑃𝑏 = 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3
8.
𝑃𝑏 < 𝑃𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 3
Flow behavior
Flow behavior
Flow behavior
𝑀= 0
-
0< 𝑀 < 1
-
M= 1 at throat
𝑀 < 1 elsewhere
-
M= 1 at throat
Normal shock forms in
divergent section
M= 1 at throat
𝑀 > 1 elsewhere
𝑀 < 1 outside
𝑀 > 1 at exit
-
-
𝑀 > 1 at exit
-
-
𝑀 > 1 at exit
-
No pressure difference;
fluid at rest
Entire nozzle (converging
diverging) has subsonic
flow
Flow is choked; further
decrease in Pb leads to
supersonic flow
Shock in divergent
section, part supersonic
and part subsonic
-
Normal shock exactly at
nozzle exit
-
Exit pressure lower than
ambient; oblique shocks
form outside
Exit pressure equals
ambient; ideal expansion
Exit pressure higher than
ambient; expansion fans
form outside
Question 3 (20 Marks): Internal Flow [Ventilation System]
For smooth turbulent and Re=8e6, we can get from graph that f=0.0083
A fan is to produce a constant air speed of throughout the pipe loop shown. The 3-m-diameter pipes
are smooth, and each of the four elbows has a loss coefficient of 0.30. Determine the power that the
fan adds to the air.
Solution
ℎf = ∑ f𝑖
Givens:
-
Calculating minor loss
Velocity v= 40 m/s
Pipe diameter D=3 m
Smooth pipe
Loss coef per elbow Ki=0.3
𝑖
ℎm = ∑ 𝐾𝑖
𝑖
Total head loss
-
𝐿𝑖 𝑉𝑖2
2 ∗ (10 + 20)
402
= 0.0083 ∗
∗
= 13.5372 m`
𝐷𝑖 2 𝑔
3
2 ∗ 9.81
Step-by-Step Solution
ℎm = ∑ f𝑖
𝑖
Calculating major loss
ℎf = ∑ f𝑖
𝑖
𝐿𝑖 𝑉𝑖2
𝐷𝑖 2 𝑔
Where f is the friction factor coefficient obtained from Moody diagram based on Re
Calculating Re:
Re =
𝜌VD 1.225 ∗ 40 ∗ 3
=
= 8.215 ∗ 106
𝜇
1.81e − 5
This is a turbulent flow since Re>500000
We will go to moody chart and get friction factor for smooth turbulent
𝑉𝑖2
402
= 4 ∗ 0.3 ∗
= 97.8593 m
2 𝑔
2 ∗ 9.81
𝐿𝑖 𝑉𝑖2
𝑉𝑖2
𝐿𝑖
𝑉𝑖2
+ 𝐾𝑖
= ∑(f𝑖 + 𝐾𝑖 )
𝐷𝑖 2 𝑔
2 𝑔
𝐷𝑖
2 𝑔
𝑖
ℎL = ℎ𝑓 + ℎ𝑚 = 13.5372 + 97.8593 = 111.3965 m
Power needed by the fan
Calculate the volume flow rate
𝑄 =𝐴∗𝑉 =
𝑊𝑓𝑜 = 𝜌 ∗ 𝑔 ∗ 𝑄 ∗ ℎ𝑠
𝜋
𝜋
∗ 𝐷2 ∗ 𝑉 = ∗ 32 ∗ 40 = 282.7433 𝑚3
4
4
𝑊𝑓𝑜 = 𝜌 ∗ 𝑔 ∗ 𝑄 ∗ ℎ𝑠 = 1.225 ∗ 9.81 ∗ 282.7433 ∗ 111.3965
𝐖𝐟𝐨 = 𝟑𝟖𝟕. 𝟓 𝐊𝐖
Question 4 (30 Marks): Nozzle
a) Consider a nozzle with efficiency n  between the inlet and any station downstream. Find the functions,
Show that 1 M= does not occur at the nozzle minimum area (throat).
b) Consider a supersonic nozzle constructed with an exit to throat area ratio of 2. The nozzle is supplied with
low-speed air at 7 atm and 60 C. The overall nozzle efficiency from inlet to exit is 90%, but the flow is isentropic
to the throat. Calculate the pressure, velocity, and Mach number at the exit and compare with the
corresponding values for isentropic flow.
Normalized Mass Flow Rate 𝑓2 (𝜂𝑛 , 𝑀)
Mass Flow Definition
𝑚𝑜
𝑃
𝑃𝑀√𝛾
=𝜌𝑢=
∗ 𝑀 ∗ √𝛾𝑅𝑇 =
𝐴
𝑅𝑇
√𝑅𝑇
Express in Terms of Stagnation Properties (From 𝑓1 )
𝑃=
Substituting:
Solution
Simplifying exponents:
For isentropic flow
𝛾
𝛾−1
𝑃𝑜
𝛾−1
= (1 +
∗ 𝑀2 )
𝑃
2
Non-isentropic correction nozzle efficiency can be written as follows
From isentropic relations
ℎ𝑜 − ℎ
𝑇𝑜 − 𝑇
𝜂𝑛 =
=
ℎ𝑜 − ℎ𝑠 𝑇𝑜 − 𝑇𝑠
𝛾−1
𝑃𝑜 𝛾
𝑇𝑜
=( )
𝑃
𝑇𝑠
Substituting into efficiency definition:
𝜂𝑛 =
Solving for T
Using modified isentropic relation:
Final expression
,
𝛾−1
𝑃𝑜 𝛾
𝑇𝑠 = 𝑇𝑜 ∗ ( )
𝑃
𝑇𝑜 − 𝑇
𝛾−1
𝑇𝑜
𝛾−1
= 1 + 𝜂𝑛 ∗
∗ 𝑀2
𝑇
2
𝑓1 (𝜂𝑛 , 𝑀) = (1 + 𝜂𝑛 ∗
𝛾−1
𝛾
∗ 𝑀2 )^
𝛾−1
𝛾
(1 + 𝜂𝑛 ∗
𝑇𝑜
𝛾−1
2
2 ∗𝑀 )
1
−
2(𝛾−1)
𝑚𝑜 𝑃𝑜 𝑀√𝛾
𝛾−1
=
∗ (1 + 𝜂𝑛 ∗
∗ 𝑀2 )
𝐴
2
√𝑅𝑇𝑜
Normalized Form:
𝛾+1
−
2(𝛾−1)
𝑚𝑜 𝑀√𝛾
𝛾−1
=
∗ (1 + 𝜂𝑛 ∗
∗ 𝑀2 )
𝐴
2
√𝑅
Throat Mach number 𝑓3 (𝜂𝑛 , 𝑀)
Differentiate f2 with respect to M:
𝛾−1
𝑃 𝛾
𝑇 = 𝑇𝑜 ∗ [1 − 𝜂𝑛 ∗ (1 − ( ) )]
𝑃𝑜
, 𝑇=
𝛾+1
Let 𝐶 = 1 + 𝜂𝑛 2 𝑀2
𝛾−1
𝑃 𝛾
𝑇𝑜 − 𝑇𝑜 ∗ ( 𝑃𝑜 )
𝛾
2
𝛾−1
(1 + 𝜂𝑛 ∗ 2 ∗ 𝑀2 )
𝑚𝑜 𝑃𝑜 𝑀√𝛾
=
∗
𝛾
𝐴
√𝑅𝑇𝑜
𝛾−1
𝛾−1
(1 + 𝜂𝑛 ∗ 2 ∗ 𝑀2 )
Point a (Proofs)
Stagnation pressure ratio 𝑓1 (𝜂𝑛 , 𝑀)
𝑃𝑜
𝛾−1
𝛾−1
(1 + 𝜂𝑛 ∗ 2 ∗ 𝑀2 )
Simplify:
Solve for Mt
Final expression
𝛾+1
𝑑
𝛾 − 1 2 −2(𝛾−1)
[𝑀 (1 + 𝜂𝑛
𝑀 )
]=0
𝑑𝑚
2
−
𝛾+1
𝐶 2(𝛾−1) − 𝑀 ∗
𝛾+1
𝛾+1
−
−1
∗ 𝐶 2(𝛾−1) ∗ 𝜂𝑛 (𝛾 − 1)𝑀 = 0
2(𝛾 − 1)
1−
𝜂𝑛 ∗ (𝛾 + 1) 2 −1
𝑀𝑡 𝐶
2
𝑀𝑡 = √
𝑓3 (𝜂𝑛 ) = √
1
𝜂𝑛
2
1 𝛾−1
∗( −
)
𝛾 + 1 𝜂𝑛 𝛾 + 1
Another Approach
Point b (Calculations)
REFERENCE CASE (ISENTROPIC FLOW SOLUTION)
𝛾+1
2∗(𝛾−1)
𝐴𝑒
1
2
𝛾−1
=
∗(
∗ (1 +
∗ 𝑀𝑒2 ))
𝐴𝑡ℎ 𝑀𝑒
𝛾+1
2
Given that Nozzle efficiency
Start with 𝑀𝑒 = 2.2
Isentropic relations give:
𝛾
−
𝛾−1
𝑃𝑒
𝛾−1
= (1 +
∗ 𝑀𝑒2 )
= 0.0935
𝑃𝑜
2
𝑃𝑒 = 0.0935 ∗ 7 = 0.6545 𝑎𝑡𝑚
1
𝛾−1
𝑇𝑒
= (1 +
∗ 𝑀𝑒2 ) = 0.508
𝑇𝑜
2
𝑇𝑒 = 0.508 ∗ 333.15 = 169.2 𝑘
𝑉𝑒 = 𝑀𝑒 ∗ √𝛾𝑅𝑇𝑒 = 2.2 ∗ √1.4 ∗ 287 ∗ 169.2 = 573.5 𝑚/𝑠
NON-ISENTROPIC FLOW SOLUTION
Substitute with 𝛾𝑛 = 0.9
Proof that M=1 doesn’t occur at Throat When n<1
From 𝑓3 (𝜂𝑛 ) :
• When 𝜂𝑛 = 1 , 𝑀𝑡ℎ = 1
• When 𝜂𝑛 < 1 , 𝑀𝑡ℎ < 1
2
1
0.4
𝑀𝑡 = √2.4 ∗ (0.9 − 2.4) = 0.887 (Subsonic)
Pressure
𝛾
Temperature
−
𝛾−1
𝑃𝑡ℎ
𝛾−1
2
= (1 + 𝜂𝑛
∗ 𝑀𝑡ℎ
)
= 0.6289
𝑃𝑜
2
𝑇𝑡ℎ
𝛾−1
2
= (1 + 𝜂𝑛
∗ 𝑀𝑡ℎ
) = 0.847
𝑇𝑜
2
The nozzle efficiency is defined as:
𝜂𝑛 =
,
,
𝑃𝑡 = 0.6289 ∗ 709275 𝑝𝑎
𝑇𝑡ℎ = 0.847 ∗ 333.15 − 291.822 𝑘
ℎ𝑜 − ℎ
𝑇𝑜 − 𝑇𝑒
=
ℎ𝑜 − ℎ𝑒,𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑇𝑜 − 𝑇𝑒,𝑖𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐
Solving for actual exit temperature:
0.9 =
333.15 − 𝑇𝑒
, 𝑇𝑒 = 333.15 − 0.9 ∗ 163.95 = 185.6 𝑘
333.15 − 169.2
Actual exit velocity
𝑉𝑒 = √2 𝐶𝑝 (𝑇𝑜 − 𝑇𝑒 ) = √2 ∗ 1004.5 ∗ (333.15 − 185.6) = 544.6
Actual exit Mach number
𝑀𝑒 =
𝑉𝑒
√𝛾𝑅𝑇𝑒
𝑚
𝑠
= 1.99
Question 5 (30 Marks): Diffuser
Consider the shown inviscid two-dimensional airflow.
(a) Find the minimum Mach number, 1 M, for this case to exist.
(b) Find Mach numbers and pressure ratios
(c) Find the length L in terms of the height H corresponding to this M1.
For the exit pressure, we account for stagnation pressure loss:
𝛾
𝛾−1
𝑃𝑜,𝑒𝑥𝑖𝑡
𝛾−1
= (1 +
∗ 𝑀𝑒2 )
= 7.85
𝑃𝑒
2
𝑃𝑜,𝑒 = 𝜂𝑛 ∗ 𝑃𝑜 = 6.3 𝑎𝑡𝑚
,
𝑃𝑒 =
6.3
= 0.803 𝑎𝑡𝑚
7.85
Solution
We divide the problem into through regions:
Final Results
Parameter
Mth
Me
Pe
Ve
Te
Isentropic
1
2.2
66300 pa
573.8 m/s
169.3 k
Non-isentropic
0.887
1.99
91496 pa
573.8 m/s
169.3 k
• From 1-2: Oblique Shock with deflection angle of 10 deg
• From 2-3: Reflected Oblique Shock with deflection angle of 10 deg
• From 3-4: Reflected Oblique Shock with deflection angle of 10 deg
A) Minimum Inlet Mach number
The constrain on inlet Mach number that specifies minimum inlet Mach number is that the inlet Mach number
to reflective wave at 3-4 is the minimum for the deflection angle of θ = 10
From the figure we can see that: at θ =10 𝑀3min = 1.42
From M1 −M2 −θ relation we can see that: at θ =10 𝑀3min = 1.42
Second Shock (2 3):
• Deflection: 𝜃2 = 10
• Shock angle: 𝛽2 = 45
• Normal Mach: 𝑀2𝑛 = 𝑀2 ∗ sin(45) = 1.77 ∗ 𝑠𝑖𝑛(45) = 1.25
𝑃3
𝑃
𝑃
𝑃
= 1.65 , 𝑃3 = 𝑃2 ∗ 𝑃3 = 2.89
𝑃2
1
1
2
0.81
𝑀3𝑛
= sin(45−10) = 1.42
Downstream Mach: 𝑀3 =
sin(𝛽2 −𝜃2 )
From Normal shock waves tables:
,
𝑀2𝑛 = 0.81
Second Shock (3 4):
• Deflection: 𝜃3 = 10
• Shock angle: 𝛽3 = 67
Oblique shock from 3-4
From Fig.1
Oblique shock from 2-3
From Fig.1
Oblique shock from 1-2
From Fig.1
• Normal Mach: 𝑀3𝑛 = 𝑀3 ∗ sin(67) = 1.42 ∗ 𝑠𝑖𝑛(67) = 1.31
@θ = 10 , M3 = 1.421 → M4 = 0.9
@θ = 10 , M3 = 1.421 → M2 = 1.77
@θ = 10 , M2 = 1.77 → M1 = 2.15
Then the minimum inlet Mach number to ensure that M3 is the Mmin at θ = 10:
B) Find Mach numbers and pressure ratios
∴ 𝑀1min = 2.15
First Shock (1 2):
• Deflection: 𝜃1 = 10
• Shock angle: 𝛽1 = 37
• Normal Mach: 𝑀1𝑛 = 𝑀1 ∗ sin(37) = 2.15 ∗ 𝑠𝑖𝑛(37) = 1.29
𝑃2
= 1.75 ,
𝑀2𝑛 = 0.79
𝑃1
𝑀2𝑛
0.79
Downstream Mach: 𝑀2 = sin(𝛽 −𝜃 ) =
= 1.77
sin(37−10)
1
1
From Normal shock waves tables:
𝑃4
𝑃4
𝑃
𝑃
𝑃
= 1.8 ,
= 4 ∗ 3 ∗ 2 = 5.2
𝑃3
𝑃1
𝑃3 𝑃2 𝑃1
0.9
𝑀
Downstream Mach: 𝑀4 = sin(𝛽4𝑛−𝜃 ) = sin(67−10) = 0.9
3
3
From Normal shock waves tables:
C) The length L in terms of the height H corresponding to this M1
Using shock angles 𝛽1 = 37 , 𝛽2 = 45 , 𝛽3 = 67 :
First segment (L1)
𝐿1 =
𝐻
= 1.327 𝐻
𝑡𝑎𝑛(37)
Second segment (L2): Using Law of Sines:
𝐿2 = 1.4142 ∗
Total length:
𝑠𝑖𝑛(67)
= 1.854 𝐻
𝑠𝑖𝑛(45)
𝐿𝑡𝑜𝑡𝑎𝑙 = 𝐿1 + 𝐿2 = 1.327 𝐻 + 1.854 𝐻
𝑳𝒕𝒐𝒕𝒂𝒂𝒍 = 𝟑. 𝟏𝟖𝟏 𝑯
,
𝑀3𝑛 = 0.9
Final results
Quantity
M1
M2
M3
M4
P2/P1
P3/P1
P4/P1
L
Value
2.15
1.77
1.42
0.9
1.75
2.89
5.2
3.18 H1