Solutions Mannual for the fourth edition of
Gas Dynamics
Ethirajan Rathakrishnan
Preface
This manual gives the detailed solution for all the problems given at the end
of different chapters of the 3rd edition of Gas Dynamics. My sincere thanks to
my doctoral and masters students who helped me in checking and keying in the
solutions of this manual.
My sincere thanks to the Continuing Education Centre of Indian Institute of
Technology Kanpur for the financial support to prepare this manual.
E Rathakrishnan
i
ii
Contents
1 Some Preliminary Thoughts
1
2 Basic Equations of Compressible Flow
3
3 Wave Propagation
23
4 One-Dimensional Flow
25
5 Normal Shock Waves
79
6 Oblique Shock and Expansion Waves
119
7 Potential Equation for Compressible Flow
157
8 Similarity Rules
161
9 Two Dimensional Compressible Flows
165
10 Prandtl-Meyer Flow
169
11 Flow with Friction and Heat Transfer
173
12 MOC
205
13 Measurements in Compressible Flow
207
iii
Chapter 1
Some Preliminary Thoughts
1
2
Some Preliminary Thoughts
Chapter 2
Basic Equations of
Compressible Flow
2.1 In the reservoir, the air is at stagnation state. So, the entropy relation
would be
!
"
!
"
T02
p02
s2 − s1 = cp ln
− R ln
T01
p01
But, T01 = T02 for adiabatic process. Therefore,
!
"
p01
∆s = R ln
p02
!
"
p01
= R ln 1
= R ln 2
2 p01
=
198.933 J/(kg K)
Note: It should be noted that, for entropy only subscripts 2 and 1 are used;
since entropy is not defined like static or stagnation entropy.
2.2 Let the initial state be denoted by subscript 1 and expanded state by subscript 2.
(a) Since the cylinder is insulated, preventing any heat transfer what-so-ever,
the process is adiabatic. The governing equation for this process is given by
p1 Vγ1 = p2 Vγ2 = constant
(1)
Also, from ideal gas state equation
p1 V1
p2 V2
=
=R
T1
T2
3
(2)
4
Basic Equations of Compressible Flow
From Eqs. (1) and (2), we get
p1
=
p2
!
!
V2
V1
"γ
T1
#
10(γ−1) = 557.35 K
=
T1
T2
"γ/(γ−1)
Therefore,
T2
=
∆T
=
(b)
Work =
Also,
$
− 842.65 K
pdv =
$
pv γ = constant
dh −
$
du −
$
vdp
(3)
from equation (1)
Differentiating equation (1), we have,
pγv γ−1 dv + v γ dp = 0
Dividing throughout by v γ−1 and integrating, we get
$
$
pγdv + vdp = 0
$
vdp = −γw
(4)
Substituting equation (4) in equation (3) and simplifying, we get
(1 − γ) w
=
R∆T
w
=
R∆T
287 × (−842.65)
=
1−γ
(−0.4)
=
6.04 × 105 J/kg
Note: Since the process undergone is expansion from a high pressure, the work
removed is positive, i.e, work is done by the gas.
(c) Also, from equation (1)
p1
=
p2
Therefore,
!
V2
V1
"γ
= 101.4 = 25.1189
The pressure ratio = 25.1189
5
2.3 p1 v1γ = p2 v2γ , where v is specific volume, i.e. volume per unit mass = V/m.
Therefore,
!
"γ
!
"γ
V1
V2
p1
= p2
m1
m2
Also, V1 = V2 = V = volume of the tank.
!
"γ
m2
p 2 = p1
m1
! "1.4
1
= 5 × 105 ×
2
1.8946 × 105 Pa
=
From equation of state for a calorically perfect gas,
p1
p2
=
T2
=
ρ1 T1
ρ2 T2
! "!
"
p2
m1
T1
p1
m2
!
"
1.8946
× 2 × 500
5
=
378.92 K
=
2.4
p1
=
p2
(a) Therefore,
T2
!
T1
T2
"γ/(γ−1)
=
!
=
61/3.5 × 290 = 483.868 K
p2
p1
"(γ−1)/γ
T1
The change in the temperature is
∆T
=
=
T2 − T1 = 483.868 − 290
193.868 K
(b) By first law of thermodynamics, we have
du + d(pe) + d(ke) = dq + dw
6
Basic Equations of Compressible Flow
Here, velocity changes are neglected. Therefore,
d(ke) = 0
Also, assuming
d(pe) = 0
The first law of thermodynamics reduces to
du = dq + dw
But the process is isentropic, thus dq = 0. Therefore,
du
=
=
dw = cv ∆T = 717.5 × 193.868
1.39 × 105 J/kg
(c) The work done is negative, i.e. work is done on the gas. It has been
computed in (b) above.
2.5 Work done by the weight on the piston goes towards increasing the internal
energy of the gas. From the first law of thermodynamics
E2 − E 1 = Q + W
where, E, Q, and W are respectively the internal energy, heat transfered, and
work done. Since no heat is transfered, Q = 0. Therefore,
$
E2 − E1 = W = F . ds
where, F is force and ds is distance. At the new equilibrium position, the force
acting on the piston face is F = p2 Ap , Ap is the area of the piston face. The
distance traveled by the piston is ds = (V1 − V2 )/Ap , V1 and V2 are the initial
and final volumes. Thus we have,
E2 − E1
=
p2 . Ap (V1 − V2 )/Ap
=
−p2 (V2 − V1 )
For unit mass,
e2 − e1 = −p2 (V2 − V1 )
For calorically perfect gas, e = cv T . Therefore
!
"
RT2
RT1
cv (T2 − T1 ) = −p2
−
p2
p1
"
!
c v T2
T2
p2
−1
= − +
R T1
T1
p1
cv &
T2 %
cv
1+
=
+ λ (where λ = p2 /p1 )
T1
R
R
7
But
1
cv
=
. Thus,
R
γ−1
γ T2
γ − 1 T1
=
T2
T1
=
λ+
1
γ−1
1 + (γ − 1)λ
γ
Entropy change for a perfect gas can be written as
! "
! "
T2
p2
∆s = cp ln
− R ln
T1
p1
' (
) γ *
1 1 + (γ − 1)λ γ−1
∆s
= ln
R
λ
γ
' (
) γ *
1 1 + (γ − 1)λ γ−1
s2 − s1 = R ln
λ
γ
Let λ = 1 + $, where $ # 1. Therefore,
+
,
1 + (γ − 1)(1 + $)
∆s
γ
=
ln
− ln(1 + $)
R
γ−1
γ
+
,
1 + γ − 1 + $(γ − 1)
γ
ln
=
− ln(1 + $)
γ−1
γ
+
,
γ−1
γ
ln 1 +
$ − ln(1 + $)
=
γ−1
γ
Expanding the RHS, and retaining only upto second order terms, we get,
, !
+
"
∆s
γ
γ−1
(γ − 1)2 2
$2
=
$−
$
−
$
−
R
γ−1
γ
2γ 2
2
=
$−
=
$2
2γ
$2
$2
$2
γ−1 2
$2
$ −$+
=− +
+
2γ
2
2
2γ
2
Note: Work has been done by the weight which is equal to p2 Ap on the gas.
The weight has moved by a distance of ds. Therefore, ∆E = W . ds = p2 Ap . ds.
2.6 Since it is an open system,
Work done
= −cp (T2 − T1 )
8
Basic Equations of Compressible Flow
−cp
=
!
"
T2
− 1 T1
T1
-(
p2
p1
) γ−1
γ
.
=
−cp
− 1 T1
=
%
&
−1004.5 × 21/3.5 − 1 × 303
− 66.66 kJ/kg
=
2.7 Work done is given by
W
=
=
p (V2 − V1 ) = 101325 × 6 (2 − 0.3)
1.0335 MJ
since 1 atm = 101325 Pa.
2.8 The compression process is given as isentropic. Let subscripts 1 and 2 refer
to initial and final states, respectively. By isentropic process relation, we have
p1
ργ1
ρ2
=
p2
ργ2
=
!
=
p2
p1
"1/γ
ρ1 =
!
690
150
"1/1.3
× 1.5
4.85 kg/m3
2.9 As we know, the relation between temperature and pressure for isentropic
change of state may be written as
T2
=
T1
!
p2
p1
"(γ−1)/γ
where subscripts 1 and 2 refer to the initial and final states, respectively.
T2
=
=
T1
!
p2
p1
"0.4/1.4
519.9 K
! "0.286
7
= 298
1
9
2.10 By isentropic relation,
T2
=
T1
!
v1
v2
"(γ−1)
where subscripts 1 and 2 refer to the initial and final states and v is specific
volume. For air γ = 1.4.
Therefore,
T2
!
"0.4
v1
v2
=
T1
=
(30 + 273.15) (30)
=
0.4
= 1181.7 K
908.55◦ C
γ−1
γ
R, therefore, the gas constant R =
cp
γ−1
γ
0.4
× 1000 = 285.7 J/(kg K)
R=
1.4
2.11 (a) We have cp =
Also,
Ru
8314
=
M
M
where M is the molecular weight and Ru is universal gas constant. Thus,
R=
M=
8314
= 29.1
285.7
(b) By ideal gas state equation, we have
p1 V1 = mRT1
p2 V2 = mRT2
where subscripts 1 and 2 refer to initial and final states, respectively. But
p1 = p2 and therefore,
V2
V1
=
323.15
T2
50 + 273.15
=
=
T1
200 + 273.15
473.15
=
0.683
2.12 For an ideal gas, the speed of sound a may be expressed as
/
a = γRT
10
Basic Equations of Compressible Flow
where γ is the ratio of specific heats and R is the gas constant. For the given
gas,
R
=
Ru /M = 8314/29 = 286.7 J/(kg K)
Therefore,
400
=
γ
=
/
γ × 286.7 × 373.15
4002
= 1.5
286.7 × 373.15
The specific heat cp and cv can be written as
cp
=
γ
R
γ−1
cv
=
R
γ−1
=
(1.5/0.5) × 286.7
=
860.1 J/(kg K)
Therefore,
cp
cv
=
286.7/0.5
=
573.4 J/(kg K)
Note: The ratio of specific heats γ = cp /cv . For the present case γ = 1.5 =
860.1/573.4 is correct. This way the answer obtained for cp and cv may be
checked.
2.13 At the nozzle exit, V = 390 m/s and T = 28 + 273.15 = 301.15 K. The
corresponding speed of sound is
/
√
a =
γRT = 1.4 × 287 × 301.15
=
347.85 m/s
Thus,
M
=
V
390
=
a
347.85
=
1.12
11
By isentropic relation, we have
T0
T
T0
γ−1 2
M
2
=
1+
=
1 + 0.2 × 1.122 = 1.25
=
1.25 × 301.15 = 376.44 K
103.29◦ C
=
For the flow, the stagnation temperature is T0 = 376.44 K.
The static temperature
T = 92.5◦ C = 92.5 + 273.15 = 365.65 K
376.44
T0
=
= 1.03 = 1 + 0.2 M 2
T
365.65
Thus,
M2
=
0.03
= 0.15
0.2
M
=
0.387
This is the Mach number at the station where temperature is 92.5◦ C.
2.14 For hydrogen, the gas constant R = 8314/2.016 = 4124 J/(kg K).
By isentropic relation, we have
T2
T1
=
=
!
"(γ−1)/γ
p2
p1
! "0.286
1
= 0.573
7
Therefore,
T2
=
0.573 × T1
=
0.573 × 300 = 171.9 K
By energy equation, we have
h2 +
V22
= h1 , since V1 = 0
2
V22 = 2 (h1 − h2 )
12
But
Basic Equations of Compressible Flow
h = cp T ,
therefore,
V2 =
0
cp
2cp (T1 − T2 )
=
1.4
γ
R=
× 4124
γ−1
0.4
= 14434 J/(kg K)
Thus,
V2
=
/
=
2 × 14434 (300 − 171.9)
1923 m/s
The speed of sound is given by
/
√
a2 =
γRT2 = 1.4 × 4124 × 171.9
=
996.23 m/s
Thus,
M2 =
V2
1923
= 1.93
=
a2
996.23
The mass flow rate is given by
ṁ = ρ2 A2 V2
ρ2
=
p2
, by state equation
RT2
ρ2
=
101325
, since 1 atm =101325 Pa
4124 × 171.9
=
0.143 kg/m3
Thus,
ṁ
=
=
0.143 × 10 × 10−4 × 1923
0.275 kg/s
2.15 The given process is a polytropic process with index n = 1.32. Since air
is given as an ideal gas with constant specific heats, we have from isentropic
relations the change of entropy as
! "
! "
T2
p2
s2 − s1 = cp ln
− R ln
T1
p1
13
For a polytropic process,
T2
=
T1
!
p2
p1
" n−1
n
Combining these two equations, we obtain
+ !
"
, ! "
n−1
p2
s2 − s1 = cp
− R ln
n
p1
since cp =
γ
R for an ideal gas, the above equation may be written as
γ−1
! "
p2
(n − γ) R
s2 − s1 =
ln
n (γ − 1)
p1
With the given data,
R=
γ−1
0.4
cp =
× 1004 = 287 J/(kg K)
γ
1.4
Therefore,
s2 − s1
=
(1.32 − 1.4) 287 1100
ln
1.32 × 0.4
101
=
−103.8 J/(kg K)
Note: Since the entropy of the gas decreases for this internally reversible process,
heat must be removed from the gas. This is why the cylinder used for such a
compression process is usually water jacketed. Also, we know that the entropy of
the system and surrounding cannot decrease. But in this problem, the entropy
decreases. It should be noted that, what decreases is entropy of the system
alone and not the combined entropy of the system and surrounding.
2.16 For oxygen, molecular weight M = 32. The gas constant R = 8314/32 =
259.8 J/(kg K). Therefore,
cp
=
cv
=
γ
R = 909.3 J/(kg K)
γ−1
cp
= 649.5 J/(kg K)
γ
The increase in internal energy is
%u
=
=
cv (T2 − T1 ) = 649.5 (125 − 25)
64950 J/(kg K)
14
Basic Equations of Compressible Flow
The increase in enthalpy is
%h
= cp (T2 − T1 ) = 909.3 (125 − 25)
=
90930 J/(kg K)
2.17 Let the subscripts 1 and 2 refer to the inlet and exit states. At state 1, p1 =
100
kPa,
ρ1 = 1.175 kg/m3 . Therefore,
T1 =
At state 2,
p1
100 × 103
= 296.5 K
=
Rρ1
287 × 1.175
p2 = 500 kPa, ρ2 = 5.875 kg/m3 . Therefore,
T2 =
p2
500 × 103
= 296.5 K
=
Rρ2
287 × 5.875
Assuming air to be a perfect gas, the enthalpy difference is given by
h2 − h1
=
=
cp (T2 − T1 ) = cp (296.5 − 296.5)
0
2.18 The entropy change is given by
ds =
δq
T
where δq is the reversible heat addition per unit mass.
For an ideal gas
δq = dh − vdp
where dh is the enthalpy change and dh = cp dT
δq = cp dT − vdp
Using the above relation, we get
ds = cp
By state equation
v
dT
− dp
T
T
pv = RT ,
R
v
=
T
p
15
Therefore,
ds = cp
dT
R
− dp
T
p
Taking log of state equation and differentiating, we get
dT
dp dv
=
+
T
p
v
Substituting for
dT
is ds expression, we obtain
T
!
"
dp dv
dp
ds = cp
+
−R
p
v
p
ds = cp
But
cp − cv = R,
dv
dp
+ (cp − R)
v
p
cp − R = cv . Thus,
ds = cp
dv
dp
+ cv
v
p
For an isentropic change of state, we have
0 =
dv
v
cp dv
cv v
cp
But
cp
dv
dp
+ cv
v
p
=
−cv
=
−
dp
p
dp
p
cp
= γ. Therefore,
cv
γ
dp
dv
=−
v
p
Integrating both sides, we get
γ ln v
=
− ln p + constant
ln p + ln v γ
=
constant
ln (pv γ )
=
constant
or
pv γ = constant
16
Basic Equations of Compressible Flow
2.19 For an isothermal process, the change in entropy %s is given by
! "
p1
%s = s2 − s1 = R ln
p2
where the subscripts 1 and 2 stand for the initial and final states, respectively.
By state equation
p1 V1
=
mRT1
p2 V2
=
mRT2
But T1 = T2 , therefore, p1 V1 = p2 V2 = mRT2 . Hence,
p2
=
p1 V1
0.7 × 106 × 0.014
=
V2
0.084
=
0.117 MPa
Thus, the change of entropy is
%s
=
R ln
!
p1
p2
"
= 287 ln
!
0.7
0.117
"
513.4 J/(kg K)
=
2.20 Let subscripts 1 and 2 refer to initial and final states, respectively.
For process 1
p2 = 700 kPa.
By state equation, we have
p1 V
=
mRT1
p2 V
=
mRT2
Therefore,
T2
=
p2
700
× 308.15 = 616.3 K
× T1 =
p1
350
The change in entropy is given by
%s
=
cp ln
!
T2
T1
"
− R ln
!
p2
p1
"
For air cp = 1004.5 J/(kg K) and R = 287 J/(kg K). Thus,
!
"
!
"
616.3
700
%s = 1004.5 ln
− 287 ln
308.15
350
=
497.33 J/(kg K)
17
For 0.3 kg of air,
%s = 0.3 × 497.33 = 149.2 J/K
For process 2
pV1
=
mRT1
pV2 = mRT2
Therefore,
T2
=
V2
× T1
V1
Initial volume is given by
V1
=
0.3 × 287 × 308.15
mRT1
=
= 0.0758 m3
p
350 × 103
T2
=
0.2289
× 308.15 = 930.55 K
0.0758
The entropy change is given by
%s
!
T2
T1
"
=
cp ln
=
1110.163 J/(kg K)
= 1004.5 ln
!
930.55
308.15
"
For 0.3 kg of air,
%S = 0.3 × 1110.163 = 333 J/K
2.21 Given flow is adiabatic and frictionless and therefore, isentropic.
By energy equation, we have
h1 +
V12
V2
= h2 + 2
2
2
For perfect gas h = cp T , thus,
c p T1 +
V12
V2
= cp T2 + 2
2
2
For air cp = 1004.5 J/(kg K). Therefore,
1004.5 × 273.15 +
9002
3002
= 1004.5 T2 +
2
2
18
Basic Equations of Compressible Flow
Solving this, we get T2 = 631.54 K. Therefore, the temperature increase becomes
= T2 − T1 = 631.54 − 273.15
%T
=
358.39 K
By isentropic relation, we have
p2
=
p1
!
T2
T1
γ
" γ−1
Thus,
p2
=
p1
=
%p
T2
T1
"3.5
= 140
!
631.54
273.15
"3.5
= 2631 kPa
2.631 MPa
p2 − p1
=
=
2.22 For nitrogen,
!
2.491 MPa
R = Ru /M = 8314/28 = 297 J/(kg K).
For reversible process, the change in entropy may be expressed as
! "
! "
T2
p2
%s = cp ln
− R ln
T1
p1
But
T2 = T1 , therefore,
%s
=
=
− R ln
!
p2
p1
"
= − 297 ln
!
300
100
"
− 0.3263 kJ/(kg K)
Note: We know that the entropy of the system and surrounding cannot decrease.
But in this problem, the entropy decreases. It should be noted that, what
decreases is entropy of the system alone and not the combined entropy of the
system and surrounding.
2.23 By isentropic relation,
p2
=
p1
!
T2
T1
γ
" γ−1
19
where subscripts 1 and 2 refer to initial and final states. That is,
T2
T1
T2
=
=
=
!
p2
p1
" γ−1
γ
!
" 0.4
p2 1.4
T1
p1
!
"0.286
550
300
= 475.4 K
110
The change in enthalpy is (h2 − h1 ), and h = cp T . Therefore,
h2 − h1 = cp (T2 − T1 )
Also, cp =
γ
R
γ−1
and
R = 287 J/(kg K), for air.
cp =
1.4
× 287 = 1004.5 J/(kg K)
0.4
Thus,
h2 − h1
=
=
1004.5 (475.5 − 300)
176.19 kJ/kg
2.24 Let the initial and final states of air are designated by subscripts 1 and 2,
respectively.
By perfect gas state equation, we have
Dividing one by other, we get
p1 V =
mRT1
p2 V =
mRT2
p1
T1
=
p2
T2
The change in entropy for a perfect gas is given by
! "
! "
T2
p2
%s = cp ln
− R ln
T1
p1
! "
T2
= (cp − R) ln
T1
20
Basic Equations of Compressible Flow
Given,
T1
=
50 + 273.15 = 323.15 K
T2
=
125 + 273.15 = 398.15 K
cp
=
γ
R = 1004.5 J/(kg K)
γ−1
%s
=
(1004.5 − 287) ln
=
149.75 J/(kg K)
Thus,
!
398.15
323.15
"
2.25 By energy equation, we have
h0 = h +
V2
2
where subscript 0 refers to stagnation condition. Assuming air to be a perfect
gas, we can express h = cp T . Therefore,
c p T0
=
cp T +
T0 − T
=
V2
2 cp
V2
2
From standard atmosphere table, at 10000 m, T = 223.15 K. Therefore, the
speed of sound becomes
/
√
a =
γ R T = 1.4 × 287 × 223.15
=
299.436 m/s
The flight speed is
V = M a = 2 × 299.436 = 598.872 m/s
Thus, the temperature becomes
∆T
=
T0 − T =
=
178.52
598.8722
2 × 1004.5
21
2.26 Let subscripts 1 and 2 refer to the initial and final states. Given,
T1 = 15◦ C = 288.15 K,
v1 = 0.06 m3 ,
v2 = 0.12 m3 ,
By perfect gas state equation,
pv = RT
Therefore,
T1
T2
=
v1
v2
since p1 = p2 .
Thus,
T2
=
T1
288.15
× 0.12 = 576.3 K
v2 =
v1
0.06
=
303.15◦ C
p 1 = p2
22
Basic Equations of Compressible Flow
Chapter 3
Wave Propagation
23
24
Wave Propagation
Chapter 4
One-Dimensional Flow
4.1 Total temperature at “1” is T01 = 300 K. This much temperature is required
as
static
temperature at the test-section. Therefore,
TT
=
300 K
T0T
TT
=
1+
γ−1 2
M
2
T0T
300
=
1+
0.4
× 2.52
2
T0T
=
675 K
Hence, the temperature rise required,
∆T
=
=
T0T − TT = 675 − 300
375 K
4.2 Let the test-section conditions be denoted by subscript 2, and the sonic conditions
by
superscript *.
γ−1 2
M
2
T02
T2
=
1+
373
T2
=
1 + 0.2 × 22
T2
=
207.2 K
25
26
One-Dimensional Flow
Further, since we know that
T∗
T0
=
0.8333
T∗
=
373 × 0.8333 = 310.82 K
a∗
=
√
p0
=
3.12 × 101325 = 3.16 × 105
ρ0
=
p0
RT0
=
3.16 × 105
287 × 373
=
2.95 kg/m3
ρ∗
ρ0
=
0.634
ρ∗
=
1.8703 kg/m3
ṁ
= ρ∗ A∗ V ∗
=
=
1.4 × 287 × 310.82 = 353.4 m/s = V ∗
1.8703 × 80 × 10−4 × 353.4
5.288 kg/s
4.3 Let subscripts 0, 1, and e refer to stagnation state and states at the nozzle entrance
and
exit, respectively. We know that,
Te
= 0.8333
T0
This gives,
T0 =
293
= 351.6 K
0.8333
Also,
T0
T1
=
1+
γ−1 2
M
2
27
T1
=
351.6
1 + 0.2 × 9
=
125.57 K
pe
p0
= 0.5283
p0
=
0.8
= 1.5143 atm
0.5283
=
!
p0
p1
=
p1
1
γ−1 2
M
1+
2
1 + 0.2 × 32
=
1.5143
36.73
=
0.04123 atm
γ
" γ−1
23.5
√
1.4 × 287 × 293 = 343.11 m/s
Ve
= ae =
Ae
= 40 × 10−4 m2
ρe
=
pe
RTe
=
0.8 × 101325
= 0.964 kg/m3
287 × 293
ṁ
= ρe Ae Ve
= 0.964 × 40 × 10−4 × 343.11
=
1.323 kg/s
Note: It should be noted that the calculation made with equations for pressure,
density, and temperature in the problem can also be done using gas tables. In
fact that procedure will result in considerable time saving.
4.4 Given,
A1
=
0.6 × 0.4 = 0.24 m2
28
One-Dimensional Flow
A∗
=
0.3 × 0.4 = 0.12 m2
A2
=
h2 × 0.4 = 0.4 h2 m2
M2
=
2.5
T2
=
− 10◦ C
p2
=
0.15 atm
For M2 = 2.5, isentropic table gives
A2
= 2.6367
A∗
and
T2
= 0.4444
T02
This gives,
Thus,
A2 = 2.6367 × 0.12 = 0.3164 m2
h2
T02
=
0.3164
0.4
=
0.791 m
=
263
0.4444
=
592 K
A1
0.24
=2
=
A∗
0.12
A1
From subsonic part of isentropic table, for A
∗ = 2, we get
M2 = 0.3
and
T1
= 0.9823
T01
For isentropic flow, T01 = T02 . Therefore,
T1
=
=
V1
=
=
0.9823 × 592
581.5 K
/
M1 a1 = 0.3 γ R T
145 m/s
29
4.5 For M1 = 0.5, from isentropic table, we have
p1
p0
=
0.84
Therefore,
p2
p0
=
p2
p1
1.0
× 0.84
×
=
p1
p0
6.5
=
0.129
For p/p0 = 0.129, from isentropic table, we have
M2
=
2.0
A∗
A2
=
0.6
Mass flow rate is given by
ṁ
=
=
=
ρ1 V1 A1 = p1 M1 A1
3
γ
RT1
6.5 × 101325 × 0.5 × 0.016 ×
3
1.4
287.4 × 440
17.54 kg/s
4.6 If p0 is the stagnation pressure and A∗ is the critical throat area,
p∗
p0
=
A
A∗
=
1
1 + 0.2M 2
1
M
!
2−3.5
5 + M2
6
"3
With these relation we obtain,
p1
p2
=
A1
A2
=
!
"−3.5
1 + 0.2M12
1 + 0.2M22
!
"3
M2 5 + M12
M1 5 + M22
where subscripts 1 and 2 refer to entrance and throat of venturi. Substituting
proper values into the above two relations, we get
30
One-Dimensional Flow
1.5
1.2
=
4
3
=
!
"−3.5
1 + 0.2M12
1 + 0.2M22
!
"3
M2 5 + M12
M1 5 + M22
!
"3.0
M2 1 + 0.2M12
M1 1 + 0.2M22
=
These two simultaneous equations can be solved to get M1 and M2
!
4
3
Therefore,
!
1.5
1.2
1.5
1.2
"3.0/3.5
"3.0/3.5
M2
M1
=
5
4
=
From above two equations, we get,
=
!
=
M1
M2
4
3
!
1 + 0.2M12
1 + 0.2M22
"−3.0
! "6/7
5
= 1.61
4
"−3.5
1 + 0.2M12
1 + 0.2M22
M1
=
0.46
M2
=
0.74
4.7 p0 is the total pressure and p∞ is the static pressure.
1 2
ρV
q∞ =
2
p0 − p∞
=
490 mm of Hg = 0.653 × 105 Pa
p∞
=
(0.35 + 1.0132) × 105 = 1.3632 × 105 Pa
p0
=
(0.653 + 1.3632) × 105 = 2.0162 × 105 Pa
T0
=
25◦ C = 298 K
31
Therefore,
T∞
!
"0.286
p∞
T0
p0
!
"0.286
1.3632
298 ×
= 266 K
2.0162
=
=
But,
1
2
2
T∞ 1 + 0.2M∞
T0
=
2
M∞
=
T0 − T∞
298 − 266
= 0.602
=
0.2T∞
0.2 × 266
M∞
=
0.776
a∞
=
/
√
γRT∞ = 1.4 × 287 × 266 = 326 m/s
V∞
=
253 m/s
4.8
pi
=
1.0 atm
pf
=
6.0 atm
T1 = 290 K
For isentropic compression, we have
γ
! "γ ! " γ−1
ρ2
T2
p2
=
=
p1
ρ1
T1
(a)
Tf
=
Ti
!
pf
pi
γ
" γ−1
0.4
= (6) 1.4 = 1.67
Tf = 290 × 1.67 = 484 K
∆T = 484 − 290 = 194 K
(b) Change in the internal energy is given by,
de
= 1.39 × 105 (N m)/kg
=
cv dT =
287
× 194
0.4
32
One-Dimensional Flow
(c) Since the process is isentropic, from first law of thermodynamics, we have
de = δq + δw, and δq = 0. Hence, work imparted to the air becomes,
δw = δe = 1.39 × 105 N m/kg
4.9
a
=
M
=
/
γRT = 347.21 m/s
180
= 0.518
347.19
The maximum pressure that can be achieved is the isentropic stagnation pressure. Therefore,
p0
p
=
!
p0
=
1.013 × 105 × 1.201 Pa
γ−1 2
1+
M
2
γ
" γ−1
= 1.201
1.217 × 105 pa
=
4.10 (a) The critical pressure ratio for the sonic condition at the nozzle exit is
p∗
= 0.528
p0
If the nozzle is choked, then
p2
=
=
p∗ = 0.528 × 6.895 × 105
3.64 × 105 Pa
since, p2 > patm , the flow is choked. Hence, the pressure in the exit plane is
p2 = p∗
(b) The minimum stagnation pressure for choked flow occurs when
p2 = p∗ = patm
Thus,
p0min =
patm
p∗
=
= 1.92 × 105 Pa
0.528
0.528
33
(c) For p0 = 1.724 × 105 Pa, p∗ = 0.528 × p0 = 0.91 × 105 Pa. Since p∗ < patm
flow will not be choked. Further, subsonic flow is always correctly expanded,
hence,
p2 = patm = 1.014 × 105 Pa
4.11 Let the reaction force acting on the diffuser be R. Then,
R + p1 A1 − p2 A2
=
ṁV2 − ṁV1
R
=
ṁ(V2 − V1 ) − (p1 A1 − p2 A2 )
where subscripts 1 and 2 refer to diffuser inlet and exit, respectively. Given,
p1 = 0.35 × 105 Pa, V1 = 200 m/s, T1 = 230 K, ṁ = 25 kg/s
From the given data, we get
ρ1
=
p1
0.35 × 105
= 0.53 kg/m3
=
RT1
287 × 230
A1
=
ṁ
25
= 0.236 m2
=
ρ1 V1
0.53 × 200
200
= 0.66
1.4 × 287 × 230
From isentropic table for M1 = 0.66, we have
M1
=
√
p1
T1
= 0.7465,
= 0.9199
p01
T01
Thus,
p01
=
0.35 × 105
= 0.468 × 105 Pa
0.7465
230
= 250 K
0.9199
For M2 = 0.2, from isentropic table, we have
T2
p2
= 0.9725,
= 0.9921
p02
T02
T01
p2
=
=
0.9725 p01 (since for isentropic flow p01 = p02 )
=
0.45 × 105 Pa
T2
=
0.9921 T01 = 248 K
V2
=
√
M2 a2 = 0.2 1.4 × 287 × 248 = 63 m/s
34
One-Dimensional Flow
Therefore,
R
=
=
25 × (63 − 200) − (0.35 × 105 × 0.236 − 0.45 × 105 × 0.5)
−10.815 kN
4.12 Let subscripts 1 and 2 refer to entrance and exit of the tank. By energy
equation we have,
1
cp T1 + u21
2
=
1
cp T2 + u22
2
T1 − T2
=
u22 − u21
4 × 104 − 1 × 104
=
2cp
2 × 1004
≈
15◦
T2
=
T1 − 15◦
4.13
Work
u1
u2
Figure S4.13 Schematic of the work delivering machine.
Work delivered per unit mass is given by
u21 − u22
+ cp (T1 − T2 ) = work delivered
2
Given,
T1
=
373 K
u1 = 200 m/s
T2
=
288 K
cp = 1004 (N m)/(kg K)
Using this we get,
20000 −
u22
+ 1004 × 85 =
2
100000
35
20000 −
u22
+ 85340
2
=
u2
=
100000
103.3 m/s
When the machine is idling,
u22
2
=
u2idling
=
20000 + 85340 = 105350
459 m/s
4.14 Let jets “1” and “2” be denoted by the subscripts 1 and 2, and let T0
denote the temperature in the reservoir. For q = 0, adiabatic energy equation
gives,
! 2
"
! 2
"
u2
u1
+ c p T2 + m
+ cp T1 = 2mcp T0
m
2
2
This gives,
T0
=
u2 + u22
T1 + T2
+ 1
2
4cp
=
300 +
=
(1 + 9) × 104
4 × 1004
324.9 K
4.15 Let T0 denote the temperature in the tyre. Since the process is adiabatic,
we have
u2
2
=
c p T0
cp (T0 − T )
=
u2
2
u2
=
2cp (T0 − T )
cp T +
u =
=
√
2 × 1004 × 37
272.57 m/s
36
One-Dimensional Flow
4.16 At 15000 m, we have
Speed of sound
T
=
−56.5◦ C = 216.5 K
p
=
1.206 × 104 Pa
a
=
/
γRT
=
√
=
294.94 m/s
1.4 × 287 × 216.5
Speed of the airplane is,
u
=
800 km/hr = 800 ×
=
222.22 m/s
1000
3600
This gives the Mach number as
M
=
222.22
u
=
a
294.86
=
0.753
(a) Maximum possible temperature of the airplane skin will be the stagnation
temperature, at the nose of the airplane. Thus, it is the total temperature of
the air,
T0
T
=
1.1134
T0
=
1.1134 × 216.5
=
241.05 K
(b) Maximum possible pressure that can be felt by the airplane cannot exceed the stagnation pressure. This will be felt at the place where air comes to
complete rest, i.e., at the nose of the airplane and other similar places. Thus,
p0
= 1/0.6866
p
p0
=
1.206 × 104
0.6866
=
1.756 × 104 Pa
37
(c) Critical velocity of air relative to the airplane is
3
2
a0
a∗ =
γ+1
√
√
γRT0
γRT0
∗
√
= √
a =
1.2
1.2
=
284.1 m/s
(d)
Vmax
=
=
Vmax
=
=
3
3
2
ao
γ−1
γ+1 ∗
a
γ−1
√
6a∗
695.9 m/s
4.17
M = 0.6
Area = A
Figure S4.17 Schematic of convergent channel.
The mass flow rate is given by
ṁ
=
=
=
ρAV
/
p
A M γRT
RT
3
3
γ p
T0 1
√ MA
p0
R p0
T T0
From the problem and theory we know that,
38
One-Dimensional Flow
T0 = 550 K
p
p0
=
0.7840
p0 = 2 × 105 Pa
T
T0
=
0.9328
=
1.0354
=
4
3
T0
T
3
γ
R
ṁ = 29.188 kg/s
M = 0.6
=
=
/
√
γ
cp γ−1
γ
γ
cp (γ − 1)
1.4
= 0.0694
1017 × 0.4
This gives
A =
=
/γ
p
R p 0 p0
ṁ
0
T0 √1
M
T
T0
0.10125 m2
4.18
1 m2
2m2
Ae = 4 m 2
Figure S4.18 Schematic of convergent channel.
At the mouth of the duct
p
=
patm = 1.013 × 105 Pa
T0
=
288 K
(at sea level)
39
(a)
Maximum mass flow rate is given by
ṁmax
=
p
1
√ 0 A∗
24.741 T0
=
1 1.013 × 105
√
×1
24.741
288
=
241.26 kg/s
(b)
Ae
A∗
=
4,
A∗
= 0.25
Ae
For this area ratio, from isentropic table, we get
pe
= 0.0298
p0
Me
=
2.94,
pe
=
0.0298 × 1.013 × 105 Pa
=
3018.7 Pa
4.19
A∗
1 m2
p, T, ρ
M =2
Figure S4.19 Schematic of convergent channel.
Here we have to find ṁ, p∗ , T ∗ , ρ∗ , A, p, T , and ρ.
ρ0
ṁ
=
p0
7 × 105
kg/m3
=
RT0
287 × 313
=
7.8 kg/m3
=
0.6847 × p0 A∗
√
RT0
=
1599.133 kg/s
40
One-Dimensional Flow
Fluid properties at the throat
T∗
= 0.8333
T0
=⇒
T ∗ = 261 K
p∗
= 0.5283
p0
=⇒
p∗ = 3.7 × 105 Pa
ρ∗
= 0.6339
ρ0
=⇒
ρ∗ = 4.944 kg/m3
For the test-section Mach number 2, from isentropic table, we have
T
= 0.5556
T0
=⇒
T = 173.9 K
p
= 0.1278
p0
=⇒
p = 8.946 × 104 Pa
ρ
= 0.2300
ρ0
=⇒
ρ = 1.794 kg/m3
A
1
=
A∗
0.5926
=⇒
A = 1.6875 m2
4.20
60 m/s
245 m/s
1
2
Figure S4.20 Schematic of divergent channel.
V1
p1
ṁ
=
=
=
245 m/s
1 × 105 Pa
13.6 kg/s
(a)
ρ1
=
=
a1
=
p1
1 × 105
=
RT1
287 × 300
1.1614 kg/m3
/
γRT1 = 347.2 m/s
V2
T1
=
=
60 m/s
300 K
41
M1
=
245
= 0.7056 ≈ 0.706
347.2
p1
p0
= 0.7171
=⇒
T1
T0
= 0.909
=⇒
A1
=
ṁ
13.6
m2 = 0.0478 m2
=
ρ1 V1
1.614 × 245
A1
=
1
πD12
4
=
0.2467 m
=⇒
p0 = 1.39 × 105 Pa
T0 = 330 K
D1
(b) At the outlet, the energy equation is
c p T2 +
V22
= c p T0
2
Therefore,
T2
=
T0 −
a2
=
/
M2
=
V22
602
= 328.2 K
= 330 −
2cp
2 × 1004
γRT2 = 363.1 m/s
V2
= 0.165
a2
For M1 = 0.706, A1 /A∗ = 1.09 and for M2 = 0.165, A2 /A∗ = 3.57. Thus gives
A2 = 0.1566 m2 and D2 = 0.4465 m
(c) The rise in static temperature is T2 − T1 = 28.2 K
4.21
p02
=
2.0 × 105 Pa
p1
=
0.15 × 105 Pa
By Rayleigh supersonic pitot formula,
&1/(γ−1)
%
2γ
γ−1
2
M
−
1
γ+1
γ+1
p1
=
1 γ+1 2 2γ/(γ+1)
p02
M
2
1
42
One-Dimensional Flow
p02
p1
=
=
!
γ+1 2
Mi
2
γ+1 2
M1
2
"γ/(γ−1)
'
%
1
2γ
2
γ+1 M1
γ+1
2
2 M1
2γ
γ−1
2
γ+1 M1 − γ+1
−
γ−1
γ+1
1
* γ−1
Select M1 and calculate p02 for given p1 . Check that
&1/(γ−1)
p1
p01
is less than
p1
p02 .
Trial and error method gives M1 = 3.16 . Also, check p02 /p01 calculated with
normal shock relation and isentropic relation agree exactly. For this problem
p02 /p01 = 0.286.
4.22
ṁmax
=
241 kg/s
A
= 3
A∗
For this area ratio from isentropic table, M = 2.64. This is the Mach number
just upstream of the shock. Let the conditions just upstream and downstream
of the shock be represented by subscripts 1 and 2. From Normal shock table,
for M1 = 2.64, we have
p02
p2
M2 = 0.5
= 0.4452
= 7.9645
p01
p1
Therefore,
p02
=
0.4452 × p01 = 0.4452 × 1.0133 × 105 = 45110 Pa
For M2 = 0.5, from isentropic table, we have
A2
= 1.3398
A∗2
Therefore,
A∗2
=
3
A2
=
= 2.239 m2
1.3398
1.3398
Ae
A∗2
=
4
= 1.786
2.239
From isentropic table, for
Ae
= 1.786, we have
A∗2
Me
=
0.35
pe
p0e
=
0.9187
43
But p0e = p02 . Therefore,
pe
0.9187 × 45110
=
=
41442.5 Pa
4.23
p1 = 0.7 × 105 Pa
T1 = 300 K
A1 = 0.15 m
V1 = 240 m/s
2
V2 = 120 m/s
(a)
ṁ
p1
A1 V1
RT1
=
ρ1 A1 V1 =
=
0.7 × 105
× 0.15 × 240
287 × 300
=
29.26 kg/s
(b)
a1
=
M1
=
/
γRT1 = 347.5 m/s
V1
= 0.69
A1
Thus,
p1
p01
=
0.72735
T1
T01
=
0.9131
p01
=
0.9624 × 105 Pa
T01
=
328.6 K
Thus, stagnation pressure at the exit p02 = p01 = 0.9624 × 105 Pa.
(c) From (b) above, the stagnation temperature at the exit is T02 = T01 =
328.6 K
44
One-Dimensional Flow
(d)
1
cp T2 + V22
2
=
c p T0
T2
=
T0 −
A2
=
/
M2
=
V2
= 0.334
A2
p2
p02
=
0.9257
V22
1202 )
= 321.4 K
= 328.6 −
2cp
2 × 1004
γRT2 = 359 m/s
Therefore, the static pressure at the exit is p2 = 8.91 × 105 Pa
(e) Entropy change across the diffuser is zero since the flow is isentropic.
(f ) For M1 = 0.69,
the exit area,
A1
A∗
= 1.1018, and for M2 = 0.334,
A2 =
A2
A∗
= 1.850. This gives
A2 A∗
A1 = 0.252 m2
A∗ A1
4.24 The pressure, density, and temperature in the settling chamber can be
taken as stagnation quantities. Therefore,
p0 = 1.014 × 105 Pa,
ρ0 = 1.44 kg/m3 ,
T0 = 35 + 273 = 308 K
Since the effects of viscosity is neglected, the flow in the working section can be
treated
as isentropic. For M = 0.8, from isentropic table, we have
p
= 0.656,
p0
ρ
= 0.740,
ρ0
T
= 0.886
T0
Hence,
p
=
=
ρ =
=
0.656 p0 = 0.656 × 1.014 × 105
0.665 × 105 Pa
0.740 ρ0 = 0.740 × 1.144
0.847 kg/m2
45
T
=
=
0.886 T0 = 0.886 × 308
273 K
4.25
Shock
1 2
M1 = 3
Ath
Ae
Figure S4.25 Schematic of convergent-divergent channel.
Given,
Ae
= 11.91, M1 = 3.0
At
M1 = 3.0 gives, M2 = 0.4752 downstream of the normal shock, and pp02
=
01
0.3281. This gives, p02 = 2.2981 × 105 Pa. Since the flow is adiabatic, T02 =
T01 = 500 K. And since the flow is isentropic between the downstream of the
normal shock and the nozzle exit, T0e = T02 , and p0e = p02 .
A1
A1
A2
= 4.2346, and for M2 = 0.4752, A
For M1 = 3.0, A
∗ = A
∗ = 1.390(from
t
1
2
isentropic table). Also, since the shock is very thin, the area before and after
the shock can be taken as the same, i.e. A1 = A2 . Therefore,
! "
A
Ae
=
∗
A e
A∗2
=
Ae At A2
At A1 A∗2
=
3.9094
For this area ratio, from isentropic table, we get
Me
=
0.15
Te
T0e
=
0.9955
pe
p0e
=
0.9844
46
One-Dimensional Flow
Thus,
pe
=
2.2623 × 105 Pa
Te
=
497.8 K
4.26
M
=
2.5
p0
=
Ats
=
1 m2
T0
=
7 × 105 Pa
27 + 273 = 300 K
(a) For M = 2.5 we have,
A
A∗
=
A∗
=
2.637
0.38 m2
(b)
T∗
T0
=
0.833
T∗
=
0.833 × 300 = 249.9 K
=
− 23.1 ◦ C
(c)
M
=
a
=
V
=
V
a
/
γRT = 20.04 ×
√
T = 231.5 m/s
M a = 578.75 m/s
(d)
ṁ
A
=
p0
1
√ × A
24.743 T0
A∗
=
1
7 × 105
√
×
2.637
24.743 × 300
47
= 620
ṁ
=
620 kg/s
4.27 By definition, we have the pressure coefficient as
CP
=
CP∗
=
=
=
=
=
p − p∞
q∞
"
! ∗
p
−1
p∞
p∞
q∞
"
! ∗
p
2p∞
−1
p∞
2
γp∞ M∞
"
! ∗
p p0
2
−1
2
γM∞
p0 p∞
'!
*
1
2 γ
γ
"− γ−1
2 γ−1
2 + (γ − 1)M∞
γ+1
2
−1
γ
2
γM∞
2
2 γ−1
1

2 γ
2 γ−1
2  2 + (γ − 1)M∞
− 1
γ
1 γ+1 2 γ−1
γ
2
γM∞
× 2 γ−1
2
=
91
2
: γ
2
2 + (γ − 1)M∞
/(γ + 1) γ−1 − 1
2 /2
γM∞
4.28
p0 = 500 kPa, T0 = 30◦ C = 30 + 273.15 = 303.15 K
101325
The nozzle has to choke since pe /p0 =
= 0.203, which is well below
500 × 103
the critical pressure ratio of 0.528. Therefore,
ṁ = ṁmax =
0.6847 p0 ∗
√
A
RT0
where R = 287 m2 /(s2 K) for air. Thus,
ṁ
=
0.6847 × 500 × 103
√
× 0.5 × 10−4
287 × 303.15
=
0.058 kg/s
48
One-Dimensional Flow
4.29 The pressure ratio
0.1429, we get
pe
pe
1
= = 0.1429. From isentropic table, for
=
p0
7
p0
Me = 1.93,
Te
= 0.57307
T0
T0 = 180 + 273.15 = 453.15 K
Therefore, Te = 259.7 K. √
The speed of sound ae = γ RTe = 323 m/s.
The exit velocity is
Ve
=
Me ae = 1.93 × 323
=
623.4 m/s
4.30 The pressure ratio p/p0 across the nozzle is 1/5. This is well below the
critical pressure ratio and therefore the flow is chocked at the exit. The flow is
adiabatic and frictionless, therefore, the maximum mass flow rate is given by
ṁ
=
0.6847 × 5 × 105
√
× 6.5 × 10−4
287 × 288.15
=
0.774 kg/s
Thus, from isentropic table, for M = 1. we get
T
T0
=
T
=
0.83333
240.12 K
4.31 The mass flow rate may be expressed as
ṁ =
0.6847 p0
√
Ath
RT0
where p0 and T0 are the stagnation pressure and temperature, respectively.
The pressure ratio
we get
pe
pe
91.4
= 0.90495. From isentropic table, for
=
= 0.905,
p0
101
p0
Me = 0.38 and
Ae
= 1.6587
Ath
49
Therefore, Ath =
0.033
= 0.0199 m2 . Thus,
1.6587
ṁ
=
0.6847 × 101 × 103 × 0.0199
√
kg/s
287 × 293.15
=
4.74 kg/s
At the section with A = 0.022 m2 , the area ratio is
A
0.022
= 1.1055
=
Ath
0.0199
The corresponding Mach number is M = 0.69 and
p
= 0.72735
p0
Thus, the pressure at A = 0.022 m2 is
p = 73.5 kPa
4.32 Nozzle is adapted, therefore the exit pressure pe = p16000 at 16000 m
altitude. From standard atmospheric table, we get
p16000 = 10.299 kPa = pe
Therefore,
pe
p0
=
10299
, since 1 atm = 101325 Pa
15 × 101325
=
0.0067762
By isentropic relation,
p0
=
pe
1 + 0.2
!
γ−1
Me2
1+
2
Me2
=
!
Me
=
3.98
γ
" γ−1
1
0.0067762
"0.286
For Me = 3.98, from isentropic table, we get
Ae
Te
= 10.53 ,
= 0.23992
∗
A
T0
Te
=
=
0.23992 × (2600 + 273.15)
689.33 K
50
One-Dimensional Flow
The exit velocity becomes
Ve
= M e a e = Me
/
γRTe
√
= 3.98 1.4 × 287 × 689.33
=
Thrust
2094.6 m/s
= ṁe Ve = (ρe Ae Ve ) Ve
= ρe Ae Ve2
9000
By state equation, we have
pe
10299
= 0.052 kg/m3
=
RTe
287 × 689.33
ρe =
Therefore,
Ae =
Thus,
9000
0.052 × (2094.6)
A∗ = Ath =
2
= 0.0394 m2
0.0394
= 0.00374 m2
10.53
4.33 Given, V = 200 m/s and T = 15 + 273.15 = 288.15 K. The speed of sound
is given by
/
√
a =
γRT = 1.4 × 287 × 288.15
=
340.3 m/s
The Mach number becomes
M=
From isentropic table, for
p
p0
=
V
= 0.59
a
M = 0.59, we get
0.79013,
T
ρ
= 0.93491,
= 0.84514
T0
ρ0
Thus,
p0
=
101
0.79013
51
T0
ρ0
=
127.8 kPa
=
288.15
0.93491
=
308.2 K
=
p
1
ρ
=
×
0.84514
RT
0.84514
=
101 × 103
287 × 288.15 × 0.84514
=
1.445 kg/m3
ρ0 is also given by
ρ0
=
p0
127.8 × 103
=
R T0
287 × 308.2
=
1.445 kg/m3
Note: This problem may also be solved using the isentropic relations directly,
instead of table.
4.34 The sound speed at station 1 is
/
√
γRT1 = 1.4 × 287 × 303.15 = 349 m/s
a1 =
M1
=
V1
90.5
= 0.26
=
a1
349
For this flow with M1 = 0.26 to choke, the area ratio
tropic table) is
A1
required (from isenA∗
A1
= 2.317
A∗
The present area ratio
2.317 = 1.60.
6.9
A2
= 0.69. Therefore,
=
A1
10
A2
A2 A1
=
= 0.69 ×
∗
A
A1 A∗
For this area ratio, from isentropic table, we get
M2 = 0.4 ,
p2
T2
= 0.89561,
= 0.96899
p02
T02
For M1 = 0.26, from isentropic table, we get
p1
T1
= 0.95408,
= 0.98666
p01
T01
52
One-Dimensional Flow
p01
=
T01
=
For isentropic flow, T01 = T02
100
= 104.8 kPa
0.95408
303.15
= 307.25 K
0.98666
and p01 = p02 . Therefore,
p2
=
0.89561 × 104.8 = 93.86 kPa
T2
=
0.96888 × 307.25 = 297.72 K
4.35 For CO2 the molecular weight is 44 and γ = 1.3. The gas constant for
CO2 is
8314
= 189 J/(kg K)
R=
44
The stagnation pressure p0 = 6 atm and the back pressure pa = 1 atm. Therefore, the pressure ratio
pa
1
= = 0.167
p0
6
which is well below the critical pressure ratio of 0.54573, required for the flow
to choke. Hence, the flow is choked at the orifice. That is, M = 1 at the orifice.
From the isentropic table, for M = 1, we get
T
= 0.86957
T0
Thus,
T
=
=
0.86957 × (30 + 273.15) = 263.6 K
−9.55◦ C
This is the temperature with which CO2 comes out of the orifice. The mass
flow rate ṁ = ρ AV . By state equation,
ρ
=
0.54573 × 6 × 101325
p
=
RT
189 × 263.6
=
6.66 kg/m3
A
=
V
=
=
1
22
π 1 × 10−3
π
= × 10−6 m2
4
4
/
√
M a = M γ R T = 1 × 1.3 × 189 × 263.6
254.5 m/s
53
Thus,
ṁ
=
=
6.66 ×
π
× 10−6 × 254.5
4
0.00133 kg/s
Aliter:
The mass flow rate (for the gas with γ = 1.3) can also be expressed as
ṁ
=
0.6672 × p0 × Ath
√
R T0
=
0.6672 × 6 × 101325 ×
√
189 × 303
=
0.00133 kg/s
π × 10−6
4
4.36 Let subscripts 1 and 2 refer to inlet and exit of the nozzle.
V1
=
ṁ
0.7
=
ρ1 A1
8 × 12 × 10−4
= 72.92 m/s
M1
=
V1
V1
=√
a1
γ RT1
=
√
72.92
= 0.18
1.4 × 287 × 400
p1 = ρ1 RT1 = 8 × 287 × 400 = 918.4 kPa
From isentropic table, for M1 = 0.18, we get
p1
p01
=
0.97765
T1
T01
=
0.99356
p01
=
918.4
= 939.4 kPa
0.97765
T01
=
400
= 402.59 K
0.99356
54
One-Dimensional Flow
p2
ρ2 RT2 = 4 × 287 × 300
=
=
p2
p2
=
,
p02
p01
344.4 kPa
since the flow is isentropic. Therefore,
p2
344.4
= 0.3666
=
p01
939.4
p2
= 0.3666, we get
p01
From isentropic table, for
M2 = 1.29
Therefore,
V2
/
=
M2 a 2 = M2
=
√
1.29 1.4 × 287 × 300
=
γ R T2
447.87 m/s
Mass flow rate is
ṁ
=
ρ2 A2 V2
A2
=
ṁ
0.7
=
ρ2 V2
4 × 447.87
=
3.9 cm2
4.37 From energy equation, we have
T0
=
T1 +
V12
2 cp
where the subscripts 0 and 1 refer to stagnation and inlet conditions.
T0
=
400 +
=
405 K
1002
2 × 1004.5
55
By isentropic relation, we have
γ
! " γ−1
T0
p0
=
p1
T1
3
!
405
400
p0
=
200 × 10
pe
p0
=
150
= 0.718
209
"3.5
= 209 kPa
By isentropic relation, we have
p0
pe
=
1 + 0.2Me2
=
1
1 + 0.2 Me2
1
23.5
(1.393) 3.5
Solving we get,
The speed of sound
Me
=
0.7
T0
Te
=
1+
Te
=
405
= 368.85 K
1.098
ae =
γ−1 2
Me = 1.098
2
√
γRTe = 385 m/s.
Thus, the exit velocity is
Ve
=
=
Me ae = 0.7 × 385
269.5 m/s
By continuity, ρ1 A1 V1 = ρe Ae Ve .
or
p1
pe
A1 V1 =
Ae Ve
RT1
RTe
Ae
=
p1 T e V 1
A1
pe T 1 V e
=
1
22
π 75 × 10−3
100
200 368.85
×
×
×
150
400
269.5
4
=
0.00202 m2
56
One-Dimensional Flow
Thus,
de
3
=
=
4 × 0.00202
= 0.0507 m
π
50.7 mm
The mass flow rate is
ṁ
=
ρ1 A1 V1
=
22
1
p1 π 75 × 10−3
× 100
RT1
4
=
π × 752 × 10−6
200 × 103
×
× 100
287 × 400
4
=
0.77 kg/s
With the exit conditions,
ṁ
=
ρe Ae Ve
=
150 × 103
× 0.00202 × 269.5
287 × 368.85
=
0.77 kg/s
4.38 The flow process is described by the relation pVγ = constant. Per unit mass
of
air,
pv γ = constant. Therefore,
!
By state equation, we have
v1
p1 V1γ
=
p2 V2γ
"γ
=
p1
p2
v2
v1
pv = RT . Therefore,
=
R T1
287 × (400 + 273.15)
=
p1
3 × 106
=
0.0644 m3 /kg
Thus,
v2
=
v1
!
p1
p2
" γ1
= 0.0644
!
" 1
3 1.4
0.5
57
T2
=
0.2316 m3 /kg
=
0.5 × 106 × 0.2316
p2 v 2
=
R
287
=
403.4 K
(a) By energy equation
h2 +
V22
2
=
h1 +
V12
2
where V1 and V2 are the velocity at the entrance and exit of the nozzle. Also,
it is reasonable to assume that V1 is very small and hence can be taken as zero.
Thus,
V2
h2 + 2 = h1
2
Further, for air being a perfect gas, h = cp T . Therefore,
V22
=
2 (h1 − h2 ) = 2cp (T1 − T2 )
=
2 × 1004.5 (673.15 − 403.4)
V2
=
736 m/s
a2
=
/
=
402.6 m/s
γRT2 =
√
1.4 × 287 × 403.4
Therefore,
M2
=
V2
736
=
a2
402.6
=
1.83
(b) Since the flow at the nozzle exit is supersonic, at the throat the flow is
choked. That is, at the throat Mth = 1. From isentropic table, for M = 1, we
get
p∗
p0
=
0.52828
58
One-Dimensional Flow
T∗
T0
=
0.83333
p∗
=
0.52828 × 3 × 106 = 1.58484 MPa
T∗
=
0.83333 × 673.15 = 560.96 K
a∗
=
/
ṁ
=
ρ∗ Ath a∗
ρ∗
=
p∗
1.584 × 106
=
∗
RT
287 × 560.96
=
9.8 kg/m3
γRT ∗ = 474.76 m/s
Thus,
Ath
(c) The mass flow rate is
=
15.89
9.8 × 474.76
=
3415 mm2
ṁ = ρ2 A2 V2 .
ρ2 =
1
= 4.318 kg/m3
v2
Therefore,
ṁ =
=
4.318 × 5000 × 10−6 × 736
15.89 kg/s
4.39 First of all, let us check whether the flow is isentropic. The change in
entropy (assuming air to be calorically perfect gas) is
! "
! "
T2
p2
s2 − s1 = cp ln
− R ln
T1
p1
!
"
!
"
290
101
= 1004.5 ln
− 287 ln
300
200
=
162 J/(kg K)
59
Hence, the flow is nonisentropic.
Consider the inner surface of the nozzle as the control volume. Let F be the
force acting on the nozzle, due to the flow. By momentum balance, we have
p1 A1 − p2 A2 − F
=
ṁ (V2 − V1 )
V1
=
ṁ
ρ1 A1
ρ1
=
p1
200 × 103
=
RT1
287 × 300
=
2.32 kg/m3
=
5
= 10.78 m/s
2.32 × 0.2
V1
By energy equation, we have
h1 − h2
=
V2
V22
− 1
2
2
2 cp (T1 − T2 )
=
V22 − V12 = V22 − 10.782
2 × 1004.5 (300 − 290) + 10.782
=
V22
V2
=
142.15 m/s
The exit area is given by
A2 =
ṁ
ρ2 V2
where
ρ2
A2
=
p2
101 × 103
=
RT2
287 × 290
=
1.214 kg/m3
=
5
= 0.029 m2
1.214 × 142.15
Using these values in the momentum equation, we get
200 × 103 × 0.2 − 101 × 103 × 0.029 − F = 5 (142.15 − 10.78)
Thus,
F
=
40000 − 2929 − 656.85
60
One-Dimensional Flow
=
36414.15 N
4.40 Let us solve the problem by using relations and also by using gas tables.
The speed of sound at the flight altitude is given by
/
/
a =
γRT = 1.4 × 287 × (−15 + 273.15)
=
322.1 m/s
The flight speed V = M a = 0.8 × 322.1 = 257.68 m/s.
The stagnation enthalpy h0 is given by
h0 = h +
For an ideal gas h0 = cp T +
V2
2
V2
and therefore,
2
c p T0 = c p T +
or
T0 = T +
V2
2
V2
2cp
Thus,
2
T0
=
258.15 +
=
291.2 K
(257.68)
2 × 1004.5
By isentropic relation, we have
p0
p
=
p0
=
=
Using gas tables
" γ
T0 γ − 1
T
!
"3.5
291.2
44
258.15
!
67.08 kPa
61
For M = 0.8, from isentropic table, we get
p
T
= 0.65602
= 0.88652
p0
T0
Thus,
p0
=
44
= 67.07 kPa
0.65602
T0
=
258.15
= 291.2 K
0.88652
Note: From the above solution it is seen that, use of the gas tables has tremendous advantage over using equations directly.
4.41 Static temperature T = 216 K. Total temperature T0 = 40 + 273.15 =
313.15 K. From isentropic relation, we have
T0
T
1.45
γ−1 2
M
2
= 1 + 0.2 M 2
= 1+
M2
=
0.45
= 2.25
0.2
M
=
1.5
The pressure ratio p0 /p given by isentropic relation is
p0
p
=
=
p0
=
!
9
γ−1 2
1+
M
2
1 + 0.2 × 1.52
γ
" γ−1
:3.5
= 3.671
(3.671)(0.55) = 2.019 atm
Note: This problem can be solved by using gas tables instead of solving the
isentropic relations, as follows.
For
T
= 0.689655, from isentropic tables, we get
T0
M = 1.5 and p/p0 = 0.2724
Therefore,
p0
=
0.55
= 2.019 atm
0.2724
62
One-Dimensional Flow
4.42 By energy equation, we have
T01
V12
2cp
=
T02 = T1 +
=
(70 + 273.15) +
=
348.13 K
1002
2 × 1004.5
Similarly,
T2
T02 −
=
4002
= 268.49 K
2 × 1004.5
/
√
γRT2 = 1.4 × 287 × 268.49
348.13 −
=
a2
V22
2cp
=
=
328.45 m/s
=
V2
400
=
a2
328.45
=
1.218
Therefore,
M2
Vmax
/
=
=
2 cp T01 =
√
2 × 1004.5 × 348.13
836.3 m/s
By isentropic relation,
Thus,
p01
=
M1
=
p01
=
p02
=
1
23.5
p1 1 + 0.2 M12
V1
= 0.269
γRT1
23.5
1
2.5 1 + 0.2 × 0.2692
= 2.63 atm
√
%
&3.5
2
p2 1 + 0.2 × (1.218)
= 1.24 atm
p02
1.24
= 0.471
=
p01
2.63
63
4.43 Given,
p0 = 350 kPa, T0 = 420 K
Ae = 0.22 m2 , Ve = 525 m/s
By energy equation we have
T0 = Te +
Te
= T0 −
=
Ve2
2cp
Ve2
5252
= 420 −
2 cp
2 × 1004.5
282.8 K
√
Therefore, the speed of sound at nozzle exit ae = γRTe . That is,
√
ae = 1.4 × 287 × 282.8 = 337.01 m/s
Thus,
Me
pe
ρe
ṁ
=
Ve
525
=
ae
337.01
=
1.56
=
p0
(1 +
3.5
0.2Me2 )
=
350
4.007
=
87.35 kPa
=
pe
87.35 × 103
=
RTe
287 × 282.8
=
1.076 kg/m3
=
=
ρe Ae Ve = 1.076 × 0.22 × 525
124.3 kg/s
From isentropic table, for Me = 1.56, we get
Ae
= 1.219
A∗
Thus,
A∗ =
0.22
= 0.18 m2
1.219
64
One-Dimensional Flow
4.44 The speed of sound at 1 is
/
√
a1 =
γRT1 = 1.4 × 287 × 315
= 355.76 m/s
The Mach number at 1 is M1 =
V1
150
= 0.42.
=
a1
355.76
Now, for M1 = 0.42, the corresponding area ratio A1 /A∗ may be calculated using
the area–Mach number relation for isentropic flow, or the value of A1 /A∗ may
be read directly from isentropic table. From isentropic table for M1 = 0.42, we
get
A1 /A∗ = 1.52891
Thus,
A1 = 1.52891 × 25 = 38.22 cm2
The mass flow rate ṁ = ρ1 A1 V1
ρ1
=
p1
152 × 103
=
RT1
287 × 315
=
1.681 kg/m3
Therefore,
ṁ = 1.681 × 38.22 × 10−4 × 150 = 0.9637 kg/s
4.45
T0 = 40◦ C = 40 + 273.15 = 313.15 K,
√
The exit velocity Ve = 200 m/s = Me γRTe .
γ = 1.4
By isentropic relation we have
23.5
1
p0
= 1 + 0.2 Me2
pe
By energy relation, we have
Ve2
2cp
T0
=
Te +
Te
=
313.15 −
=
/
=
343.25 m/s
The speed of sound is
ae
2002
= 293.24 K
2 × 1004.5
γRTe =
√
1.4 × 287 × 293.24
65
The Mach number at the exit is
Me
p0
pe
=
Ve
200
=
ae
343.25
=
0.584
1
=
1 + 0.2 × 0.5832
23.5
= 1.259
For the convergent nozzle, pexit has to be equal to the atmospheric pressure, since
the subsonic flow exiting a convergent nozzle will always be correctly expanded.
pe
Thus,
=
ρgh = 13.6 × 103 × 9.81 × 22.7 × 10−3
=
3028.54 Pa (gauge)
=
101325 + 3028.54 = 104353.54 Pa (absolute)
pa = 104353.54 Pa
The storage pressure is given by
p0
=
1.259 × 104353.54
=
131381.11 Pa
=
224.7 mm of Hg (gauge)
Note: The standard atmospheric pressure is patm = 101325 Pa = 760 mm of
Hg.
4.46 Let the first and second sections are represented by subscripts 1 and 2,
respectively. At section 1,
M1
V1
365
=√
γRT1
1.4 × 287 × 305.15
=
√
=
1.04
From isentropic table, for
M1 = 1.041, we get
p1
p0
=
0.50389
T1
T0
=
0.82215
66
One-Dimensional Flow
Therefore, p0 = 158.76 kPa,
T0 = 371.16 K.
At station 2,
p2
p0
From isentropic table, for
120
= 0.7559
158.76
=
p2 /p0 = 0.7559, we get
T2
= 0.92428
T0
M2
=
T2
=
343.06 K = 69.91◦ C
a2
=
/
Therefore,
0.64
γRT2 = 371.27 m/s
V2 = M2 a2 = 237.6 m/s
4.47 For M = 3.0, from isentropic table, we get
p
= 0.027224,
p0
T
= 0.35714,
T0
A
= 4.23456
A∗
Thus,
A∗
=
0.05
= 0.0118 m2
4.23456
p0
=
0.2 × 101325
= 744.4 kPa
0.027224
T0
=
300
= 840 K
0.35714
The mass flow rate is given by
ṁ
=
0.6847 p0 A∗
√
RT0
=
0.6847 × 7.444 × 105 × 0.0118
√
287 × 840
=
12.25 kg/s
4.48 The velocity at nozzle entrance is very low. Hence, the pressure and temperature at the entrance can be taken as the stagnation pressure and stagnation
temperature. That is,
p0 = 1 MPa,
T0 = 300 K
67
From isentropic table, for M = 2, we have
p
= 0.1278,
p0
T
= 0.55556,
T0
ρ
= 0.23005
ρ0
Therefore,
p
=
0.1278 × 1 × 106 = 127.8 kPa
T
=
0.55556 × 300 = 166.67 K
ρ =
0.23005 × ρ0 = (0.23005) (p0 /RT0 )
"
106
= 2.672 kg/m3
287 × 300
√
√
a = γRT = 1.4 × 287 × 166.67 = 258.78 m/s. There0.23005
V
=
M a = 517.56 m/s
ṁ
=
ρAV = 2.672 × 0.15 × 517.56
The speed of sound
fore,
=
4.49 Given,
!
=
207.43 kg/s
p0 = 3.5 MPa, T0 = 500◦ C = 773.15 K
pe = 0.7 MPa,
ṁ = 1.3 kg/s
At the nozzle exit, the pressure ratio is
pe
0.7
= 0.2
=
p0
3.5
From isentropic table, for this pressure ratio, we get
Me = 1.71 ,
Te
Ae
= 0.63099,
= 1.3471
T0
Ath
Therefore,
Te = 0.63099 × 773.15 = 487.85 K
The corresponding speed of sound ae is
/
√
ae = γRTe = 1.4 × 287 × 487.85 = 442.74 m/s
68
One-Dimensional Flow
Ve = Me ae = 757 m/s
The exit velocity
ρe =
pe
, by state equation
RTe
Therefore,
0.7 × 107
= 4.9 kg/m3
287 × 487.85
ρe =
The mass flow rate
ṁ = ρe Ae Ve . Thus,
Ae
Ath
4.50 Given,
=
ṁ
1.3
=
ρe Ve
4.9 × 757
=
3.5 cm2
=
3.5
Ae
=
1.3471
1.3471
=
2.6 cm2
T0 = 25◦ C = 298.15 K, Ae = 15 cm2 .
Applying momentum analysis to the control volume considered, we get
F = 100 N = ρe Ae Ve2
Assuming air to be an ideal gas,
Ae ρe Ve2 =
Ae pe 2
Ae γpe 2
Ve =
V = Ae γpe Me2
RTe
γRTe e
Also,
γ Ae pe Me2 = 100 N
The nozzle exit flow is subsonic and hence has to be correctly expanded with
pe = patm = 101325 Pa. Thus,
Me2
=
100
= 0.47
1.4 × 15 × 10−4 × 101325
Me
=
0.69
From isentropic table, for Me = 0.69, we get
Te
= 0.91306,
T0
pe
= 0.72735
p0
69
Thus,
Te
p0
Ve
=
298.15 × 0.91306 = 272.23 K
=
−0.92◦ C
=
101325
= 139.3 kPa
0.72735
=
1.37 atm
=
Me a e = Me
=
0.69 ×
=
/
γRTe
√
1.4 × 287 × 272.23
228.2 m/s
4.51 Given, ṁ = 1 kg/s,
pe = 101.325 kPa.
The mass flow rate may be expressed as
ṁ =
0.6847 p0 ∗
√
A
RT0
Thus,
∗
A
=
=
=
At the exit,
√
RT0
0.6847 p0
√
287 × 325
0.6847 × 700 × 103
6.37 cm2
pe
101325
=
= 0.14475
p0
700 × 103
From isentropic table, the corresponding Mach number Me and temperature
ratio are the following.
Me
=
1.92
Te
T0
=
0.57561
Te
=
0.57561 × 325 = 187 K
70
One-Dimensional Flow
The speed of sound ae =
Thus,
√
Ve
γ RTe = 274.1 m/s.
Me ae = 1.92 × 274.1
=
=
526.27 m/s
4.52 Assuming air to be a perfect gas, we have γ = 1.4 and R = 287 J/(kg K).
The speed of sound at nozzle inlet is
/
√
γRT1 = 1.4 × 287 × 300
a1 =
= 347.2 m/s
The inlet Mach number is
M1 =
V1
100
= 0.29
=
a1
347.2
From isentropic table, for M1 = 0.29, we get
p1
p01
=
0.94329
T1
T01
=
0.98346
p01
=
101325
= 107416.6 Pa
0.94329
T01
=
300
= 305 K
0.98346
The flow is supersonic at the exit, therefore at the throat M ∗ = 1. For M = 1,
from isentropic table we have
p∗
p0
=
0.52828
T∗
T0
=
0.8333
Thus,
p∗
=
=
0.52828 × 107416.6,
56746 Pa
since p∗0 = p01
71
=
T∗
0.56 atm
0.8333 × 305
=
=
254 K
Since the flow is isentropic,
T02
=
T01 = 305, K
p02
=
p01 = 107416.6 Pa
=
1.06 atm
The mass flow rate is
ṁ
=
ρ1 A1 V1 = ρt At Vt
=
p1
101325
× 5 × 10−4 × 100
A1 V1 =
RT1
287 × 300
=
0.0588 kg/s
At
=
0.0588
ρt Vt
ρt
=
pt
56746
=
RTt
287 × 254
=
0.78 kg/m3
=
at =
=
319.5 m/s
Thus,
Vt
Therefore,
At =
/
γRTt =
√
1.4 × 287 × 254
0.0588
= 2.36 cm2
0.78 × 319.5
4.53 Let subscripts 01 and 02 refer to properties at reservoirs 1 and 2, respectively. Pressure p02 is the back pressure for the nozzle. Therefore, the nozzle
pressure ratio (NPR) becomes
p02
3
= = 0.5
p01
6
72
One-Dimensional Flow
Also,
!
p∗
=
p01
2
γ+1
"(γ)γ−1
= 0.528
where p∗ is the pressure corresponding to sonic speed. The NPR is less than
the critical pressure ratio of 0.528. Therefore, the nozzle must experience sonic
condition at the exit. Thus, the mass flow through the nozzle ṁ is
ṁ = ρ∗ V ∗ A∗
where ρ∗ and V ∗ are the density and velocity at M = 1. From isentropic table,
for M = 1, we get
T∗
T01
=
0.8333
Thus,
T∗
=
0.8333(30 + 273.15) = 252.6 K
V∗
=
a∗ =
=
318.58 m/s
/
γRT ∗ =
√
1.4 × 287 × 252.6
From state equation, we have
ρ∗
=
p∗
6 × 101325 × 0.528
=
∗
RT
287 × 252.6
=
4.428 kg/m3
=
ṁ
0.5
=
ρ∗ V ∗
4.428 × 318.58
=
0.0003544 m2 = 3.544 cm2
Thus,
A∗
Aliter:
This problem can also be solved without going into the details at the nozzle
exit, as follows.
We know that in terms of reservoir pressure and temperature, the mass flow
rate ṁ is given for a gas with γ = 1.4 as
0.6847
p01 A∗
ṁ = √
RT01
73
Thus,
A∗
=
/
ṁ
RT01
0.6847 p01
=
0.0003543 m2 = 3.543 cm2
4.54 The overall pressure ratio
pe /p0 =
100 × 103
= 0.1.
106
This is well below the critical pressure ratio of 0.528. Therefore, the flow is a
choked flow.
The area ratio
Ae /Ath = 13.5/8 = 1.6875.
From isentropic table, for Ae /Ath = 1.6875, the corresponding exit Mach number
Me = 2.0
4.55 Given, p0 = pa = 101325 Pa and T0 = 288.15 K. For the flow to choke,
pb /p0 ≤ 0.528. Therefore, the pressure in the tank pb has to be
pb ≤ 0.528 × 101325 ≤ 53.5 kPa
This implies that pbmax = 53.5 kPa.
The mass flow rate is given by
ṁ
=
0.6847 p0 Ath
√
RT0
=
0.6847 × 101325 π
√
× (0.04)2
4
287 × 288.15
=
0.303 kg/s
4.56 Given, Ae /Ath = 4 and Me > 1. Therefore, the supersonic solution of
the area Mach number relation is the answer for the problem. That is, the
supersonic value of M given by the following equation is the exit Mach number.
Ae
1
=
Ath
M
!
5 + M2
6
"3
This equation may be solved directly to get Me or the value of Me corresponding
to Ae /Ath = 4 can be read directly from isentropic table. From isentropic table,
we get
74
One-Dimensional Flow
Me = 2.94 and
pe
= 0.029795
p0
This is the pressure ratio required.
4.57 (a) For Me = 1.63, from isentropic table, we have
Ae
pe
= 1.275
= 0.22501
A∗
p0
For correct expansion, pe = pb . Therefore,
pb = 0.22510 × p0 = 0.22501 × 10 × 101325 Pa
since 1 atm = 101325 Pa.
Thus,
pb = 228 kPa
(b) The flow will remain supersonic at the exit for all back pressures below
228 kPa
(c) For choking, M = 1 at the throat. After choking the flow can expand as a
subsonic flow at the divergent portion of the nozzle. From isentropic table, for
Ae /A∗ = 1.275, we have
Me = 0.54 p/p0 = 0.82 (approx)
Therefore, the nozzle will remain choked for all back pressures below
pb = 0.82 × 10 × 101325 = 830.87 kPa
4.58 From normal shock theory, we have
1
2
2 M12 − 1
ρ2
= 1+ 2
ρ1
M1 (γ − 1) + 2
M2
=
'1
2 1
2 *1/2
γM12 + 1 − M12 − 1
2 (γM12 + 1) − (γ + 1)
where subscripts 1 and 2 refer to states upstream and downstream of the shock,
respectively. Given, M1 is very large. Also, γ = 1.4 for air. The density ratio
can be written as
&
%
1
2
1
−
ρ2
M12
=1+
ρ1
(γ − 1) + M22
1
when M1 → ∞, we get
75
ρ2
ρ1
Similarly,
=
1+
1
2
=1+
γ−1
1.4 − 1
=
1+
2
=1+5
0.4
=
6
%
& %
& 1/2
γ + M12 − 1 − M12
1

&
% 1
M2 = 
γ+1
1
2 γ + M2 − M2
1
In the limit M1 → ∞, it reduces to
M2
=
=
3
γ−1
=
2γ
1
3
0.4
2.8
0.378
4.59 From isentropic table, for M = 2.0, we get
p
= 0.1278
p0
Therefore,
p
0.1278 × 3 × 101325 = 38848 Pa
=
since, 1 atm = 101325 Pa.
Given, 101325 Pa = 760 mm Hg, therefore,
p
=
38848
× 760
101325
=
291.384 mm of Hg
This is absolute pressure. The gauge pressure, shown by the manometer, will
be
pg
=
=
pabs − patm = 291.384 − 760
− 468.61 mm
The negative sign indicates that the measured pressure is below the atmospheric
pressure or subatmospheric.
76
One-Dimensional Flow
4.60 The difference between the measured pressures is 500 mbar. That is,
∆p = 0.5 bar
We know that,
1 bar = 105 Pa
Therefore,
∆p = 0.5 × 105 Pa
Also,
∆p = ρair g h
where h is the vertical height climbed. Therefore,
h =
0.5 × 105
9.81 × 1.1
=
4633.49 m
4.61 Let subscript ‘0& refer to stagnation state. Given that the total pressure
of air is p0 . For maximum velocity, the limiting pressure is zero.
Assuming the flow as incompressible, by Bernoulli equation, we have
1
p + ρV 2 = p0
2
For Vmax , p = 0, thus,
1 2
ρV = p0
2
Also, for incompressible flow, ρ = ρ0 , thus,
3
p0
Vmax, incomp. = 2
ρ0
For compressible flow, the Bernoulli equation, we have
Therefore,
γ p0
γ p V2
+
=
γ−1ρ
2
γ − 1 ρ0
Vmax, comp. =
That is,
Vmax, comp.
=
=
=
3
3
3
2
γ p0
γ − 1 ρ0
γ
Vmax, incomp.
γ−1
1.4
Vmax, incomp.
1.4 − 1
1.87 Vmax, incomp.
77
That is, the error in treating this compressible flow as incompressible is 87
percent.
4.62 Let the inlet and the exit of the nozzle be denoted by subscripts 1 and 2,
respectively.
(a) By energy equation, we have
h1 +
V12
2
= h2 +
V22
2
3025 × 103 +
602
2
= 2790 × 103 +
V22
= 473600
V2
=
=
ρAV
=
A1 V1
v1
=
0.1 × 60
0.19
=
31.58 kg/s
688.2 m/s
(b) The mass flow rate is
ṁ
(c) Mass flow rate is also given by
ṁ = ρ2 A2 V2 =
A2 V2
v2
Thus,
A2
=
ṁ v2
V2
=
31.58 × 0.5
688.2
=
0.0229 m2
V22
2
78
One-Dimensional Flow
Chapter 5
Normal Shock Waves
5.1
Vg
500 − Vg
500 m/s
(a) Stationary observer
(b) Observer moving with the shock
Figure S5.1
Mach number of the air stream, M1 is given by
500
M1 = √
= 1.51
1.4 × 287 × 273
From Normal shock table, for M1 = 1.51, we have
p2
p1
=
500 m/s
2.493,
T2
= 1.327
T1
Therefore,
p2
=
1.745 atm
T2
=
362.27 K
V2
V1
=
1
500 − Vg
=
500
1.879
500 − Vg
=
266.1 m/s
Vg
=
Also,
233.9 m/s
79
80
Normal Shock Waves
Since the velocity of the observer does not affect the static properties,
pb
=
p2 = 1.745 atm
Tb
=
T2 = 362.3 K
The Mach number of the flow behind the shock wave is
Mb
Vg
233.9
=
381.5
γRTb
=
√
=
0.613
From isentropic table, for M = 0.613, we have
p
pt
=
0.7760,
T
= 0.9301
Tt
Therefore, after passage of shock the stagnation pressure is
p tb
T tb
=
1.745
0.7760
=
2.25 atm
=
362.3
0.9301
=
389.5 K
Note:
• For a stationary observer the stagnation temperature after passage of the
wave is greater than that before passage of the wave.
• For an observer sitting on the wave, however, there is no change of stagnation temperature across the wave.
81
5.2
up
u2 = C s − up
Cs
2
u1 = C s
1
Figure S5.2 Shock wave motion in a tube.
/
a1
=
γRT1 = 347 m/s
u1
u2
=
Cs
(γ + 1)M12
=
Cs − u p
(γ − 1)M12 + 2
M1
=
Cs
a1
Therefore,
Cs
Cs − u p
=
Cs2
+2
a21
% &2
s
(γ + 1) C
a1
% &2
s
(γ − 1) C
+2
a1
=
(γ + 1)
(γ − 1)Cs2 + 2a21
=
(γ + 1)Cs2 − (γ + 1)up Cs
2Cs2 − (γ + 1)up Cs − 2a21
=
0
up Cs − a21
=
0
γ + 1 up
M1 − 1
2 a1
=
0
(γ − 1)
Cs2 −
!
γ+1
2
M12 −
"
M1
=
=

Cs
(Cs − up )
a21
1  γ + 1 up
±
2
2 a1
1.19
4!
γ + 1 up
2 a1
"2

+ 4
Positive sign is taken here, since M1 cannot be less than 1. Hence, Cs = M1 a1 =
413 m/s . From Normal shock table, for M1 = 1.19, pp21 = 1.485. Thus, the
pressure on the face of the piston is p2 = 1.485 × 1.0133 × 105 = 1.505 × 105 Pa
82
Normal Shock Waves
5.3
up
2
1
Figure S5.3a The flow field.
(a)
p2
p1
=
a1
=
!
" 2γ
γ − 1 |up | γ−1
1−
2 a1
/
√
γRT1 = 20.04 300
=
347 m/s
Therefore,
p2
p1
=
!
p2
=
0.606p1
=
120
1 − 0.2
347
"7
= 0.606
0.606 atm
(b)
up
u2 = C s − up
Cs
2
1
Figure S5.3b Schematic of the flow field.
u1
u2
=
Cs
Cs − u p
=
(γ + 1)M12
(γ − 1)M12 + 2
u1 = C s
83
M12 −
(γ − 1)M12 + 2
=
(γ + 1)M12 − (γ + 1)
γ + 1 up
M1 − 1
2 a1
=
0
γ + 1 up
2 a1
=
1.2 ×
up
M1
a1
120
= 0.415
347
Therefore, M12 − 0.415 M1 − 1 = 0. Solving for positive value of M1 , we get
M1
=
1.228
p2
p1
=
1+
=
1.595
2
2γ 1 2
M1 − 1
γ+1
Therefore,
p2 = 1.595 atm
5.4
Cs
up
Cs
u1 = C s + up
Gas at
rest
u2 = 0
u2 = C s
Figure S5.4 Schematic of the flow field.
The velocity of the wave relative to the pipe = Cs . Velocity of air entering the
normal shock wave relative to the shock wave is
u1
= Cs + up
M1
=
Cs + u p
a1
u1
u2
=
(γ + 1)M12
2 + (γ + 1)M12
84
Normal Shock Waves
Cs
=
Cs + u
Cs
=
M1 a 1 − u p
From this we get
(γ + 1)M12
2 + (γ − 1)M12
M12
γ+1
−
2
!
up
a1
"
=
M1 a 1
M1 a 1 − u p
=
M1
u
M1 − a1p
M1 − 1
=
0
a1
=
/
γRT1 = 347 m/s
Solving for M1 , we get M1 = 1.2924, taking only the positive sign, since M1 is
supersonic. Hence,
Cs
=
M1 a1 − up = 1.2924 × 347 − 150
=
298.5 m/s
From shock tables, for M1 = 1.294, we have
p2
p1
T2
= 1.185
T1
=
1.775,
=
1.775 × 1.5 × 105
Therefore,
p2
=
T2
=
=
2.66 × 105 Pa
1.185 × 300
355.5 K
Also, since the gas is at rest,
5.5
p02
=
p2
T02
=
T2
85
Vp
1
u2 = C s
u1 = Cs + Vp
Cs
2
Figure S5.5 Schematic of the flow field.
(γ + 1)M12
2 + (γ − 1)M12
u1
u2
=
u1
= Cs
M1
Cs
a1
=
= Cs − Vp
u2
Therefore,
(γ + 1)
%
2 + (γ − 1)
Cs
a1
%
&2
Cs
a1
=
&2
=
+
,
Cs Cs
Vp
(γ + 1)
−
a1 a1
a1
Solving for
Cs
a1 ,
Cs
a1
Cs
a1
−
Vp
a1
2 + (γ − 1)
!
Cs
a1
"2
we get,
Cs
=
a1
In the limit as
=
Cs
Cs − u p
Vp
a1
!
γ+1
4
→ ∞,
Cs
a1
"
'
!
"2 ! "2 *1/2
Vp
Vp
1 γ+1
+ 1+
a1
4
2
a1
→ ∞. In the limit as
Vp
a1
5.6
a4
Cs
3
up = 300 m/s
4
→ 0,
Cs
a1
→ 1.
86
Normal Shock Waves
Figure S5.6a Schematic of the flow field.
(a)
up
=
p3
p4
=
300 m/s
!
" 2γ
γ − 1 |up | γ−1
1−
2 a4
!
" 2.8
300 0.4
1 − 0.2 ×
360
! "7
5
6
! "7
5
× 1 = 0.2791 atm
6
+
,
γ − 1 |up |
5
1−
=
2 a4
6
=
=
Therefore, a3 =
5
6
p3
=
a3
a4
=
× 360 = 300 m/s. Slope of the terminating characteristic is
dx
dt
=
C3 = a4 −
=
360 −
γ+1
|up |
2
2.4
× 300 = 0
2
For the pressure on the face of the piston,
!
. 12
"2γ
p2
γ+1
−1
p2
γ−1
p3
p3 + γ+1
up
=
a3
γ
up
=
a3 = 300 m/s
Therefore,
!
p2
p3
"2
!
"2
p2
−1
p3
=
γ
− 3.68
p2
+ 0.72
p3
=
0
2
Therefore,
p2
p3
=
3.473
p3
=
0.2791 atm
' p2
p3
+
γ−1
γ+1
2γ
γ+1
*
87
Therefore, pressure at the piston face p2 = 3.473 × 0.2791 = 0.969 atm
(b) The velocity of the shock with respect to a stationary observer is
C s = a3
+
γ − 1 γ + 1 p2
+
2γ
2γ p3
, 12
= 529.88 m/s
Therefore, time for the shock to hit the terminating characteristic after the
piston has stopped is
30
= 0.0566 s
t1 =
529.88
(c)
t
0
Piston
path
300 m/s
dx/dt = 0
t
a4
ock
300 m/s
Pis
ton
dx
/dt
pa
th
=|
up
|
Terminating
characteristic
dx
/d
t=
Sh
Cs
x
30 m
Figure S5.6c
5.7 Given, MS = 5.0, T1 = 27c ircC = 300 K, p1 = 0.01 atm, T4 = 300 K.
From normal shock table, for MS = 5, we have
p2
p1
p4
p1
=
29.0
=
4
", −2γ
+
!
γ4 −1
2γ1 MS2 − (γ1 − 1)
1
γ4 − 1
1−
MS −
γ1 + 1
γ4 + 1
MS
=
2.26 × 106
Cs
0
p2
p3
88
Normal Shock Waves
p4
=
2.26 × p1 = 2.26 × 104 atm
Note: Instead of above equation for p4 /p1 , equation (5.57) also can be used for
solving this problem.
5.8
(a) p2 /p1 = 29. Therefore, static pressure behind the shock is, p2 = 29p1 =
0.29 atm. From normal shock table for M1 = 5, we get
T2
= 5.8
T1
Therefore, static temperature behind the shock T2 is,
T2 = 5.8 × 300 = 1740 K
(b) Particle speed u2 is given by,
u2
a2
M2
=
!
"
2
1
a 1 MS −
γ1 + 1
MS
=
1388.71 m/s
=
20.04 ×
=
836 m/s
=
1388
836
=
1.66
/
T2
From isentropic tables, for M2 = 1.66, we have
T2
p02
=
0.6447
p2
p02
=
0.215
Thus,
T02 = 2699 K and p02 = 1.349 atm
(c) Testing time available is
∆t
=
t2 − t1 =
8
8
−
u2
Cs
89
u2
Cs
=
1388.71 m/s
Ms a1 = 5 × 347 = 1735.9 m/s
=
t
dx
/d
t
=
u
2
t2
t1
dt =
dx/
Cs
x
8m
Figure S5.8c
Therefore,
∆t
=
8
8
−
1388 1735
=
1.15 × 10−3 s
(d)
µ
sin−1
=
=
1
1
= sin−1
M2
1.66
37 deg
5.9
up
2
up
2
2
Cs
1
uR
up + uR
5
uR
5
90
Normal Shock Waves
Figure S5.9 Schematic of the flow field.
(a)
uR
Cs
=
uR up
up Cs
up + uR
uR
=
ρ5
,
ρ2
=
ρ5
ρ1
×
ρ1
ρ2
=
ζ
η
from continuity
Therefore,
up
uR
ζ −η
η
=
Also, (refer to Example 5.5)
up
Cs
=
1
(1 − )
η
From this, we get
uR
Cs
=
!
"
1
η
× 1−
ζ −η
η
=
η−1
ζ −η
(b)
p2
p1
=
=
"
ρ1
1+
1−
ρ2
!
"
1
1 + γ1 MS2 1 −
η
γ1 MS2
!
Similarly,
p5
p2
=
=
!
"
2
(up + uR )
ρ2
×
1
−
a22
ρ5
!
"2 !
"
a2 u p + u R
η
1 + γ2 12
1−
a2
a1
ζ
1 + γ2
91
"2 !
"
up
uR Cs
η
+
1−
a1
Cs a1
ζ
!
"2 !
"
η−1
ζ −η
γ1 p1 ρ2
η−1
MS +
MS
= 1 + γ2
ρ1 γ2 p2 ζ − η
η
ζ
a21
a22
= 1 + γ1
p1 (η − 1)2 ζ
M2
η
p2
η2
ζ −η S
Therefore,
p5
p1
=
=
=
=
p5 p2
p2 p1
(η − 1)2 ζ
p2
+ γ1 MS2
p1
η
ζ −η
!
"
(η − 1)2 ζ
1
1 + γ1 MS2 1 −
+ γ1 MS2
η
η
ζ −η
!
"
η−1
(η − 1)ζ
1 + γ1 MS2
1+
η
(ζ − η)
=
1 + γ1 MS2
η − 1 η(ζ − 1)
η
ζ −η
=
1 + γ1 MS2
(η − 1)(ζ − 1)
(ζ − η)
(c)
Therefore,
!
= 1 + γ2
h2
h1
=
h5
h2
=
!
"
γ1 − 1 2
1
MS 1 − 2
1+
2
η
"2 !
"
!
γ2 − 1 up + uR
ρ2
1+
1 − 22
2
a2
ρ5
=
γ2 − 1 a21 2 (η − 1)2
ζ2
1+
M
S
2 a22
η2
(ζ − η)2
h2
=
a22
γ2 − 1
h1
=
a21
γ1 − 1
a21
a22
=
γ 1 − 1 h2
γ 2 − 1 h1
!
η2
1− 2
ζ
"
92
Normal Shock Waves
Therefore,
γ1 − 1 h1 2 (η − 1)2 ζ + η
M
2 h2 S η 2
ζ −η
h5
h2
=
1+
h5
h1
=
h5 h2
h2 h1
=
=
=
h2
γ1 − 1 2 (η − 1)2 ζ + η
MS
+
h1
2
η2
ζ −η
!
"
γ1 − 1 2 η − 1
ζ +η
1+
MS 2
(η + 1) + (η − 1)
2
η
ζ −η
1 + (γ1 − 1)MS2
(η − 1)(ζ − 1)
η(ζ − η)
5.10 Given that, the stagnation pressure and temperature are
p0 = 1 atm = 101325 Pa,
T0 = 15◦ C = 288.15 K
From isentropic table, for M = 3.0, we have
p
= 0.02722
p0
Therefore, the static pressure at the test-section is
p
=
0.02722 p0
=
0.02722 × 101325
=
2758 Pa
The pitot pressure measured by a pitot tube placed in the test-section is the
pressure behind a normal shock.
From normal shock table, for M = 3.0, we have
p02
= 0.3283
p01
Thus, the pressure that a pitot tube at the test-section will measure is
p02
= 0.3283 p01
= 0.3283 × 101325
=
33.265 kPa
93
5.11 From normal shock table, for M1 = 2.5, we have
p2
= 7.125,
p1
T2
= 2.1375,
T1
ρ2
= 3.3333,
ρ1
p02
= 8.5261
p1
Therefore,
M2
=
p2
=
(7.125)(1) = 7.125 atm
ρ2
=
(3.3333)(1.225) = 4.083 kg/m3
p02
=
(8.5261)(1) = 8.5261 atm
T2
=
(2.1375)(T1 )
0.51299
From state equation, we have
T1
=
101325
p1
=
ρ1 R
1.225 × 287
=
288.2 K
Therefore,
T2
=
(2.1375)(288.2) = 616.03 K
a2
=
/
V2
=
M2 a2 = (0.51299)(497.51)
=
γ R T2 = 497.51 m/s
255.22
From isentropic table, for M2 = 0.51299, we have
T2
= 0.950
T02
Therefore,
T02
=
616.03
T2
=
0.950
0.950
=
648.45 K
94
Normal Shock Waves
5.12 For nitrogen, molecular weight is 28.02 and γ = 1.4. Thus, the gas constant
is
γ
R = 1.4
R = 8314/28.02 = 297 J/(kg K) and cp = γ−1
0.4 × 297 = 1039.5 J/(kg
K).
The speed of sound upstream of the shock is
/
√
a =
γ R T = 1.4 × 297 × 303
=
354.95 m/s
Therefore, the upstream Mach number M1 is
M1
V1
923
=
a1
354.95
=
= 2.6
Also, since the flow process across a normal shock is adiabatic, T01 = T02 . Now,
from normal shock table, for M1 = 2.6, we have
p2
= 7.72
p1
M2 = 0.504
T2
= 2.2383
T1
Therefore,
p2 = 7.72 × 300 = 2.316 MPa
a2
V2
=
/
=
531.03 m/s
=
M2 a2 = 0.504 × 531.03
=
γRT =
√
1.4 × 297 × 2.2383 × 303
267.64 m/s
When the flow slows down isentropically from V1 to V2 , by energy relation, we
have
T2
= 678.32 K
V12 − V22
2cp
=
T1 +
=
303 +
9232 − 267.642
2 × 1039.5
95
By isentropic relation, we have
p2
p1
=
!
T2
T1
γ
" γ−1
=
!
=
16.786
678.32
303
"3.5
Thus,
p2 = 5.036 MPa
5.13 For blunt nosed model at Mach 3, there will a detached bow-shock standing
in front of the nose. This shock can be approximated to a normal shock at the
nose of the model around the stagnation point. Therefore, pressure at the
stagnation point is the total pressure behind the normal shock.
From normal shock table, for M1 = 3, we have
p02
= 0.32834
p01
Thus,
p02
=
0.32834 p01 = 0.32834 × 10
=
3.2834 atm
=
332.69 kPa
The flow process across the shock is adiabatic. Hence, T02 = T01 . Therefore,
T02 = 315 K
After the normal shock, the flow decelerates isentropically to stagnation condition at the nose. Hence, the stagnation density ρ02 can be expressed as
ρ02
=
p02
332690
=
R T02
287 × 315
=
3.68 kg/m3
5.14 Let us make the shock stationary and look at the field. The flow field with
stationary shock will look like that shown in Fig. s5.14.
96
Normal Shock Waves
T1 = 300 K
p1 = 101 kPa
p2 = 5000 kPa
V2
V1
Shock
Figure S5.14
p2
p1
=
49.5
=
1+
=
1 + 1.167 M12 − 1.167
M12
=
49.5 + 0.167
= 42.55
1.167
M1
=
6.52
2
2γ 1 2
M1 − 1
γ+1
Thus,
The velocity becomes
V1
=
M1 a1 = 6.52
=
2374.16 m/s
√
1.4 × 287 × 330
By normal shock relation, we have
M2
=
'
=
0.4
γ−1
2
2 M1
M12 − γ−1
2
1+
γ
*1/2
1
2
2
2(γ − 1) γ M12 + 1 1 2
48.41
× 41.5 = 9.205
M1 − 1 = 1 +
2
2
(γ + 1) M1
244.86
T2
T1
=
1+
T2
=
9.205 × 330 = 3037.65 K
The speed of sound downstream of the shock is
/
√
γ R T2 = 1.4 × 287 × 3038.34
a2 =
=
1104.9 m/s
97
Thus, the velocity V2 is
V2
Vg
=
M2 a2 = 0.4 × 1104.8
=
441.92 m/s
=
Cs − Vg
=
2374.16 − 441.92
=
1932.24 m/s
Aliter
The above problem can also be solved using gas tables, as follows.
From normal shock tables, for pp21 = 49.5, we have
M1
≈
6.5
M2
≈
0.4
T2
T1
≈
9.2
T2
≈
3036 K
a2
=
1104.47 m/s
V2
=
M2 a2 = 0.4 × 1104.47
=
441.8 m/s
=
√
1.4 × 287 × 330
=
364.13 m/s
=
M1 a1 = 6.5 × 364.13
=
2366.84 m/s
a1
V1
98
Normal Shock Waves
= 2366.84 − 441.8
Vg
= 1925.04 m/s
5.15 Upstream of the shock, the speed of sound a1 is
/
√
γ RT1 = 1.4 × 287 × 300
a1 =
=
The Mach number M1 =
347.2 m/s
V1
412
= 1.19.
=
a1
347.2
From isentropic table, for M1 = 1.19, we get
p1
T1
= 0.41778,
= 0.77929
p01
T01
Therefore,
p01
T01
=
92
p1
=
0.41778
0.41778
=
220.21 kPa
=
300
T1
=
0.77929
0.77929
=
384.96 K
From normal shock table, for M1 = 1.19, we get
p2
T2
p02
= 1.4854,
= 1.1217,
= 0.99372, M2 = 0.84846.
p1
T1
p01
Therefore,
p2
T2
=
1.4854 × 92
=
136.66 kPa
=
1.1217 × 300
=
336.51 K
99
T02
=
T01 = 384.96 K , since flow process across a shock is adiabatic
p02
=
0.99372 × 220.21
=
218.83 kPa
The speed of sound is a2 =
fore,
V2
√
=
=
γRT =
√
1.4 × 287 × 336.51 = 367.7 m/s. There-
M2 a2 = 0.84846 × 367.7
311.98 m/s
The entropy rise across a normal shock is given by
!
"
p01
∆s = R ln
p02
!
"
220.21
= 287 ln
218.83
=
1.8042 J/(kg K)
5.16 Let the subscripts 1 and 2 refer to conditions upstream and downstream
of the normal shock, respectively. From normal shock table, for M1 = 2.5, we
get
p2
p02
= 7.1250,
= 8.5261.
p1
p1
The static pressure in the flow just downstream of the shock is,
p2 = 7.125 × 1.0 = 7.125 atm
If a normal shock has to be positioned at the nozzle exit, the back pressure to
which the nozzle discharges has to be equal to the total pressure downstream
of the shock. The total pressure downstream of the shock is
p02
=
=
8.5261 × 1.0 = 8.5261 atm
863.91 kPa
i.e. the back pressure has to be 863.91 kPa to position a normal shock at the
nozzle exit.
100
Normal Shock Waves
5.17 From isentropic table, for Me = 2.5, we have
pe
= 0.058528,
p0e
Ae
= 2.637
A∗
where subscripts 0, e, and ∗ refer to stagnation, exit and throat, respectively.
The throat area is
A∗
=
4
Ae
=
2.637
2.637
1.517 cm2
1
= 17.09 atm
p0e =
0.058528
For the normal shock, the upstream Mach number is 1.5. From isentropic table,
for M1 = 1.5,
=
A1
= 1.176,
A∗
p1
= 0.2724,
p01
T1
= 0.68966
T01
Thus,
A1
=
1.176 × 1.517 = 1.784 cm2
p1
=
0.2724 × 17.09 = 4.655 atm
T1
=
0.68966 × 500 = 344.83 K
From normal shock table, for M1 = 1.5,
p2
= 2.40583,
p1
T2
= 1.3202
T1
p02
= 3.4133,
p1
M2 = 0.70109
The back pressure required is p02 , thus,
p02 = 3.4133 × 4.655 = 15.89 atm
Downstream of the shock, the flow is isentropic upto the nozzle exit. However,
for this flow the effective throat area is not the same as A∗ , since p02 < p01 .
Let the effective throat area downstream of shock to be A∗2 . From isentropic
table, for M2 = 0.70,
A2
= 1.09437
A∗2
where A2 is the area at the shock location, which is same as A1 . Thus,
A∗2
=
1.784
= 1.63 cm2
1.09437
Ae
A∗2
=
4
= 2.454
1.63
101
From isentropic table for
Ae
A2 ∗
= 2.454, we have
T2
= 0.98,
Te
Me = 0.245
Thus,
Te = 0.98 × 500 = 490 K
The speed of sound is
ae =
√
1.4 × 287 × 490 = 443.71 m/s
Ve
= Me × ae = 0.245 × 443.71
The flow velocity is
=
108.71 m/s
5.18 Since the Mach number upstream of the shock is 2.32, the area ratio
corresponding to this Mach number will give the area at the shock location.
(a) For M1 = 2.32, from isentropic table,
A1
= 2.23
A∗
Thus, area at shock location is A1 = 2.23 × 5 = 11.15 cm2
(b) The Mach number downstream of the shock M2 given by normal shock
table for M1 = 2.32 is
M2 = 0.53
For M2 = 0.53, from isentropic table, we have
A2
= 1.29
A∗
Since A1 = A2 = area at the shock location, we have
A∗2 =
11.15
A2
=
= 8.64 cm2
1.29
1.29
Therefore,
12.5
Ae
= 1.447
=
A∗2
8.64
From isentropic table, for
Ae
= 1.447, the exit Mach number Me = 0.45
A∗2
102
Normal Shock Waves
(c) For the given nozzle, the area ratio
Ae
is
Ath
Ae
Ae
12.5
= 2.5
= ∗ =
Ath
A
5
Ae
= 2.5, we have
A∗
p2
= 0.064261
Me = 2.44 and
p02
From isentropic table, for
For complete isentropic flow, p02 = p0 = 700 kPa. Thus,
44.98 kPa.
p2 = 0.064261×700 =
The back pressure range for the flow to be completely isentropic is pb ≤ 44.98 kPa .
5.19 Given, T1 = 22 K and T01 = 400 K, where subscripts 1 and 2 refer to
conditions ahead of and behind the shock, respectively.
For
T1
22
= 0.055, from isentropic table,
=
T01
400
M1 = 9.2
From normal shock table, for M1 = 9.2, we have
M2
=
T2
T1
=
T2
=
0.3893
17.4
382.8 K
5.20 Upstream of the normal shock, the Mach number M1 is
M1
=
V1
500
=√
a1
1.4 × 287 × 300
=
1.44
From normal shock table for M1 = 1.44, we get
M2 = 0.72345,
p2
= 2.2525,
p1
T2
= 1.2807,
T1
Thus,
p2
=
2.2525 × 100 = 225.25 kPa
p02
= 0.9476
p01
103
T2
=
1.2807 × 300 = 384.21 K
V2
=
M2 a2 = 0.72345 ×
=
√
1.4 × 287 × 384.21
284.25 m/s
The entropy increase is
∆s
=
R ln
!
p01
p02
"
= 287 ln (1.0552)
15.432 J/(kg K)
=
5.21 The flow velocity at nozzle entrance is low. Therefore, the pressure and
temperature of the flow at the entry can be treated as the stagnation quantities.
Thus,
T01 = 300 K
p01 = 1 MPa,
From isentropic table, for M1 = 2, we have
p1
= 0.1278
p01
and
T1
= 0.55556
T01
Therefore,
p1
=
(0.1278)(1) = 0.1278 MPa
T1
=
(0.55556)(300) = 166.67 K
From normal shock table, for M1 = 2, we get
p2
T2
p02
= 4.5,
= 1.6875,
= 0.72087
p1
T1
p01
and
M2 = 0.57735
Thus,
p2
=
4.5 × 0.1278 = 0.5751 MPa
T2
=
1.6875 × 166.67 = 281.3 K
p02
=
0.72087 × 1 = 0.72087 MPa
a2
=
√
1.4 × 287 × 281.3 = 336.19 m/s
104
Normal Shock Waves
V2
M2 a2 = 0.57735 × 336.19
=
=
194.1 m/s
5.22 Given, p0 = 101 kPa and T0 = 30 + 273.15 K. A normal shock at the
nozzle exit implies that the entire nozzle flow is isentropic and also the flow is
choked at the throat.
The area ratio is
Ae
0.0724
= 2.896
=
∗
A
0.025
From isentropic table, for
Ae
A∗
= 2.896, we have
p1
= 0.050115,
p0
M1 = 2.6,
T1
= 0.42517
T0
This is the Mach number upstream of the shock. Thus,
M1
=
p1
=
0.050115 × 101 = 5.06 kPa
T1
=
0.42517 × 303.15 = 128.9 K
p01
=
2.6
101 kPa
Let the conditions downstream of the shock be referred to by subscript 2. From
normal shock table, for M1 = 2.6, we get
M2
p2
p1
=
0.504
T2
= 2.2383,
T1
= 7.72,
p2
=
39.06 kPa
T2
=
288.5 K
p02
=
46.47 kPa
p02
= 0.46012
p01
105
5.23 Given, p0 = 200 kPa and T0 = 350 K. Let subscripts 1 2 and 3 refer
to locations upstream and downstream of the shock wave and the nozzle exit,
respectively.
Ath = A∗1 = 0.2 m2
At the shock location, A1 = 0.6 m2 , thus,
A1
0.6
= 3.0
=
∗
A1
0.2
From isentropic table, for
A1
= 3.0, we get
A∗1
M1 = 2.64,
p1
= 0.04711
p01
Up to the shock the flow in the nozzle is isentropic and therefore, p01 = p0 .
Thus,
p01
=
p1
=
200 kPa
200 × 103 × 0.04711
=
9.422 kPa
where, p1 and p01 are the static and total pressures, respectively, ahead of the
shock. Let subscript 2 refer to condition behind the shock.
Now, from normal shock table, for M1 = 2.64, we have
p2
= 7.9645,
p1
M2 = 0.50,
p02
= 0.44522
p01
Thus,
p2
=
=
p02
=
=
7.9645 × 9.422
75.04 kPa
0.44522 × 200
89.04 kPa
Also, from normal shock theory we have (Section 4.6)
p01
A2 ∗
=
= 2.25
p02
A1 ∗
106
Normal Shock Waves
Thus,
A∗2 = 2.25 × 0.20 = 0.45 m2
A∗2 may also be obtained from M2 . From isentropic table, for M2 = 0.5,
A2
= 1.34
A∗2
Thus,
0.6
= 0.448 m2
1.34
This A∗2 is the equivalent throat area for the flow downstream of the shock.
Therefore, at the nozzle exit,
A2 ∗ =
A3
0.8
= 1.786
=
∗
A3
0.448
Now, from isentropic table, for
M3 = 0.35,
A3
= 1.786,
A∗3
p3
= 0.91877,
p03
T3
= 0.97609
T03
Here, p03 = p02 and T03 = T02 = T0 . Thus,
p3
=
p03
=
T3
=
0.91877 × 89.04 = 81.81 kPa
89.04 kPa
0.97609 × 350 = 341.63 K
For area ratio A3 /A∗3 the subsonic solution from the isentropic table was used
since after a normal shock the flow becomes subsonic and this flow is further
decelerated in the divergent portion of the duct.
5.24 Given, p01 = 5 atm and p02 = 3.6 atm. Therefore,
p02
3.6
= 0.72
=
p01
5
Assuming γ = 1.4, from normal shock table, for
M1 = 2.0,
p02
= 0.72, we have
p01
p2
= 4.5
p1
Now, from isentropic table, for M1 = 2.0, we get
p1
= 0.1278
p01
107
Hence, the pressure just behind the normal shock at the nozzle exit is
p2
=
=
4.5 p1 = 4.5 × 0.1278 × 5
2.8755 atm
5.25 The pitot tube will read the actual total pressure in a subsonic stream.
But in a supersonic flow, the pressure measured by a pitot probe is the total
pressure downstream of a detached shock which stands at the nose of the pitot
tube. Therefore, it is essential to find out whether the flow is subsonic or supersonic.
It can be easily seen from the isentropic relations that for M = 1, the pressure
0.95
p
=
= 1.8 atm.
ratio pp0 = 0.528. Hence, p0 =
0.528
0.528
Thus, when p0 < 1.8 atm, the flow is subsonic, and when p0 > 1.8 atm, the flow
is supersonic.
(i) p0 = 1.1 atm. The flow is subsonic and hence the pitot tube is measuring
the actual total pressure of the flow.
p
0.95
= 0.8636
=
p0
1.1
From isentropic table, for
p
= 0.8636, we get
p0
M = 0.465
(ii) p0 = 2.5 atm. The flow is supersonic and the pitot tube measures p02
behind a normal (detached) shock.
p02
2.5
= 2.63
=
p1
0.95
p02
From normal shock table, for
= 2.63, we get
p1
M = 1.275
(iii) p0 = 10 atm. The flow is supersonic.
p02
10
= 10.526
=
p1
0.95
p02
From normal shock table, for
= 10.526, we have
p1
M = 2.79
108
Normal Shock Waves
5.26 The shock will be only at the divergent portion of the nozzle, since only
after the throat the flow becomes supersonic.
The Mach number M1 just upstream of the shock will be given by the area ratio
Ashock
.
Ath
Ashock
2000
=2
=
Ath
1000
From isentropic table, for area ratio 2, we have
M1 = 2.2 ,
p1
= 0.093522
p01
Up to the shock, the stagnation pressure does not change and therefore,
p01
=
200 kPa
p1
=
200 × 0.093522 = 18.7 kPa
Now, from normal shock table, for M1 = 2.2, we get
p2
= 5.48,
p1
M2 = 0.55,
p02
= 0.62814
p01
where subscript 2 refer to condition just downstream of the shock. Therefore,
p2
=
5.48 × 18.7 = 102.48 kPa
p02
=
0.62814 × 200 = 125.63 kPa
For M2 = 0.55, from isentropic table, we get
A2
= 1.255
Ath2
where Ath2 is the throat area required for the flow downstream of the shock to
choke.
A2
Ath2 =
1.255
2
But A2 = A1 = Ashock = 2000mm , therefore,
Ath2 =
Thus,
2000
= 1593.63 mm2
1.255
Ae
3000
= 1.8825
=
Ath2
1593.63
For this area ratio, from isentropic table (subsonic part), we get
Me = 0.325
109
Therefore,
1
p02
pe
=
pe
=
125.63
p02
=
1.076
1.076
=
116.76
1 + 0.2 × 0.3252
23.5
= 1.076
The pressure loss occurs only across the shock and the loss of pressure ∆p0 is
∆p0
=
=
p01 − p02 = (200 − 125.63)
74.37 kPa
5.27 Let the subscripts i, 1, 2 and e refer to the inlet, just upstream and just
downstream, and the nozzle exit, respectively. It can be shown for this flow
that,
(1)
p01 A∗1 = p02 A∗2
For Mi = 2.0, from isentropic table, we get
Ai
= 1.6875
A∗i
Therefore,
A1
A∗1
=
A1 Ai
Ai A∗1
=
A1 Ai
,
Ai A∗i
=
2 × 1.6875 = 3.375
since A∗1 = A∗i
For this area ratio, from isentropic table, we get
M1 = 2.76
For M1 = 2.76, from normal shock table, we have
p02
= 0.40283
p01
Using this in Eq. (1), we get
A∗1
= 0.40283
A∗2
110
Normal Shock Waves
Ae
A∗2
=
Ae Ai A∗1
Ai A∗1 A∗2
=
4 × 1.6875 × 0.40283 = 2.7191
For this area ratio, from isentropic table (subsonic solution), we get
Me = 0.22
The exit pressure pe may be expressed as
pe
=
pe p02 p01
pi
p02 p01 pi
=
pe p02 p0i
pi ,
p02 p01 pi
=
0.96684 × 0.40283 ×
=
since p01 = p0i
1
× 80
0.1278
243.8 kPa
Thus, the back pressure required is 243.8 kPa.
5.28 When a pitot tube is placed in a supersonic stream, there will be a detached shock standing at its nose. At the nose where the pressure tap is located,
the shock may be treated as a normal shock and hence what the pitot tube measures is the pitot pressure p02 downstream of the shock.
The wall pressure measured by a pressure tap may be treated as the actual
static pressure of the stream. Thus we may take the static pressure upstream
of the shock as p1 = 112 kPa. Thus,
p02
2895
= 25.848
=
p1
112
p02
= 25.848, we get
p1
T2
M1 = 4.44 ,
= 4.7706
T1
Now from isentropic table, for M1 = 4.44, we get
From normal shock table, for
T1
T01
=
0.2023
T1
=
0.2023 × 500 = 101.15 K
a1
=
/
γ R T1 = 201.6 m/s
111
Thus,
V1
=
M1 a1 = 895.1 m/s
5.29 At 10,000 m altitude, from standard atmospheric table, we have
p = 26.452 kPa, T = 223.15 K
During steady–state operation, mass flow through the test–section is given by
ṁ
/
p
A M γRT
RT
=
ρ AV =
=
√
π(0.25)2
26.452 × 103
×
× 2.4 1.4 × 287 × 223.15
287 × 223.15
4
=
14.57 kg/s
The stagnation temperature during steady–state operation is
T0 =
T0
T
T
For,
M = 2.4,
T0
1
=
= 2.152
T
0.46468
Thus,
T0 = 2.152 × 223.15 = 480.22 K
For the present geometry of fixed angle diffuser, the optimum condition for
steady state operation is a normal shock at the diffuser throat. The diffuser
throat area is
A∗
A
A∗ =
A
From normal shock table, for M1 = 2.4, M2 = 0.52.
From isentropic table, for M2 = 0.52, we have
A2
= 1.3
A∗2
Here A∗2 is the area of the second throat and A is the area of test-section.
A2
Therefore, looking in to the corresponding supersonic Mach number for ∗ =
A2
1.3, we get
M = 1.66
112
Normal Shock Waves
This is the Mach number just upstream of the second throat with a shock. From
normal shock table, for M1 = 1.66, we have
p02
= 0.872
p01
This pressure loss must be compensated by the compressor.
(a) The power required for the compressor is given by
Power = h0 − hi = Cp (T0 − Ti )
where the subscripts 0 and i refer to outlet and inlet conditions. For an isentropic
compression,
! " γ−1
p0 γ
T0
=
Ti
pi
*
'! " γ−1
p0 γ
T0 − Ti = Ti
−1
pi
'!
*
"0.286
1
= 480.22
− 1 = 19.18 K
0.872
Thus, the power input required for mbox mass of air becomes
Power = 1004.5 × 19.18 = 19.266 kJ/kg
The total power required for the compressor is
ṁ 19266
Power =
746
=
376.3 hp
This is the running power required for the compressor.
(b) During start-up, M1 = 2.4 in the test-section. The corresponding total
pressure ratio across the normal shock is
p02
= 0.5401
p01
The isentropic work required for the compressor during start-up to drive the
shock out of the test-section per mbox mass of air is
'!
*
"0.286
1
Power = Cp T0i
−1
0.5401
'!
*
"0.286
1
= 1004.5 × 480.22
−1
0.5401
=
92929.87 J/kg
113
The power required to start the tunnel is
Power
=
ṁ × 92929.87
14.57 × 92929.87
=
746
746
=
1815 hp
5.30 Let subscripts 1 and 2 refer to conditions just upstream and downstream
of the normal shock. The Mach number just upstream of the shock is 1.85.
From isentropic table, for M1 = 1.85, we get
p1
A1
= 0.1612, ∗ = 1.495
p01
A1
p1 = 0.1612 × 400 = 64.48 kPa
A1 = 1.495 × 10 = 14.95 cm2
Now from normal shock table, for M1 = 1.85, we have
p02
= 0.79023
p01
M2 = 0.60,
Again from isentropic table, for M1 = 0.6, we get
A2
= 1.1882
A∗2
But A2 = A1 = 14.95 cm2 . Therefore,
A∗2 =
14.95
= 12.58 cm2
1.1882
We want the Mach number and pressure at a state 3 where A3 = 2 A1 =
2 × 14.95 = 29.9 cm2 .
Downstream of the shock, A∗2 = A∗3 . Thus,
A3
A∗3
=
29.9
= 2.377 ≈ 2.4
12.58
From isentropic table, for this area ratio (from subsonic solution),
M3 = 0.25
and
p3
= 0.95745
p03
But p03 = p02 = 316.1 kPa. Therefore,
p3 = 0.95745 × 316.1 = 302.65
Note: This problem has been solved using isentropic table and hence the results
obtained are only approximate. For exact results, we have to use the actual
relations.
114
Normal Shock Waves
5.31 (a) Given, ṁ = 0.9 kg/s, p0 = 4×101325 = 405300 Pa, T0 = 30+273 = 303
∆p
= 2.
K and
p1
The pressure ratio across the shock is
p2 − p 1
=
p1
p2
p1
2
=
2+1
=
3
p2
For
= 3, from normal shock table,
p1
M1 = 1.65, M2 = 0.654,
T2
= 1.4228
T1
From isentropic table, for M1 = 1.65,
T1
A
= 0.6475, ∗ = 1.292
T0
A
Therefore,
T1
T2
=
0.6475T0
=
0.6475 × 303
=
196.19 K
=
1.4228T1
=
1.4228 × 196.19
=
279.14 K
The mass flow rate is 0.9 kg/s, therefore,
ṁ
=
Ath
=
=
=
0.9 =
0.6847 p0 Ath
√
RT0
√
0.9 RT0
0.6847 p0
√
0.9 287 × 303
0.6847 × 405300
9.56 cm2
115
Therefore, the exit area becomes
1.292 Ath = 1.292 × 9.56
=
Ae
12.35 cm2
=
(b) The exit velocity is
V2
=
M2 a 2 = M2
=
0.654 ×
=
√
/
γRT2
1.4 × 287 × 279.14
219 m/s
5.32 a) Given, (p2 −p1 )/p1 = 3.5, where p1 and p2 are the pressures ahead of and
behind the shock. Therefore, the pressure ratio across the shock is p2 /p1 = 4.5.
For p2 /p1 = 4.5, from normal shock table,
M2 = 2,
ρ2
= 2.67
ρ1
Therefore, the shock speed becomes
Cs
=
M1 a 1 = 2 ×
=
2×
=
694.4 m/s
√
/
γRT
1.4 × 287 × 300
The piston speed is given by
Vp
=
=
=
=
=
V1 − V2 = V1
!
!
V2
1−
V1
"
ρ1
Cs 1 −
ρ2
!
"
1
694.4 × 1 −
2.67
694.4 × (1 − 0.375)
434 m/s
"
116
Normal Shock Waves
(b) Given (p2 − p1 )/p1 = 7. Therefore, p2 /p1 = 8.
For p2 /p1 = 8, from normal shock table,
ρ2
M2 = 2.65,
= 3.5047
ρ1
Therefore, the shock speed becomes
√
Cs = 2.65 × 1.4 × 287 × 300
=
920.05 m/s
=
!
920.05 × 1 −
=
920.05 × 0.7147
Thus,
Vp
=
1
3.5047
"
657.56 m/s
5.33 (a) Let subscripts 1 and 2 refer to states ahead of and behind the shock,
respectively and subscript e refer to the nozzle exit.
Given,
p01 − p02
× 100 = 12.4
p01
1−
p02
p01
=
0.124
p02
p01
=
0.876
From normal shock table, for p02 /p01 = 0.876,
M1 = 1.65 , M2 = 0.654,
T2
= 1.4228
T1
(b) For M1 = 1.65 from isentropic table, we have
T1
T01
=
0.6475
T1
=
0.6475 × T01
=
0.6475 × 330
=
213.7 K
117
Therefore,
T2 = 1.4228 × 213.7 = 304 K
By energy equation, we have
he +
Ve2
2
=
h0e
c p Te +
Ve2
2
=
cp T0e
But the flow process across the shock is adiabatic, therefore, T01 = T02 = T0e .
Hence the flow speed at the nozzle exit becomes
Ve
=
=
=
0
/
2 × cp (T0e − Te )
2 × 1004.5 × (330 − 300)
245.5 m/s
The flow speed behind the shock is
V2
=
M2 a 2 = M2 ×
=
0.654 ×
=
√
/
γ R T2
1.4 × 287 × 304
228.57 m/s
(c) The mass flow rate is
ṁ
=
0.6847 p01 Ath
√
RT01
=
0.6847 × (5 × 101325) × (5 × 10−4 )
√
287 × 330
=
0.5636 kg/s
118
Normal Shock Waves
Chapter 6
Oblique Shock and
Expansion Waves
6.1 Given, M1 = 2 and β = 40◦ , therefore, M1n = 2.0 sin 40 = 1.29. From
normal shock tables, for M1n = 1.29, we have
p2
p1
=
1.775
T2
p1
=
1.185 and
M2n
=
0.7911
Therefore,
p2
T2
=
0.5 × 1.775 × 105
=
0.8875 × 105 Pa
=
=
1.185 × 273
323.5 K
For the adiabatic process, Tt1 = Tt2 . From isentropic tables, for Mach 2, we
have T1 /Tt = 0.556. Thus,
T1t = T2t =
273
= 491 K
0.556
From isentropic table, for T2 /T2t = 0.659, we have
M2 = 1.61
119
120
Oblique Shock and Expansion Waves
Also,
u2
w2
=
0.791
0.791
= 0.4913
=
M2
1.61
β−θ
=
29.43
=⇒ θ
=
10.57 deg
sin (β − θ) =
Thus,
the wedge angle = 2θ = 21.14◦
6.2 For M1 = 2.0, from isentropic table, we have ν1 = 26.38◦ . Prandtl-Meyer
function after expansion is
ν2 = ν1 + θ = 26.38 + 5 = 31.38 deg
Therefore, for ν2 = 31.38◦ the corresponding Mach number from isentropic table
is M2 ≈ 2.18 . From isentropic tables for M2 = 2.18, we have
p2
p02
T2
T02
ρ2
ρ02
=
0.0965;
=
0.5127;
=
0.1882;
p2
= p1 ×
=
T2
ρ2
6.3 (a)
p1
p01
T1
T01
ρ1
ρ01
0.0965
0.0965
= 98 ×
0.1278
0.1278
74 kPa
= 300 ×
0.5127
0.5556
=
276.8 K
=
74 × 103
287 × 276.8
=
0.9315 kg/m3
=
0.1278
=
0.5556
=
0.2300
121
pe = 1 atm
Me = 3
p0 = 70 × 105 kPa
Figure S6.3a
M1 = 3.0
=⇒
p1
= 0.02722
p01
With p1 = pe = patm , p01 =
1.01325 × 105
= 37.2 × 105 Pa.
0.02722
Therefore, the supply pressure for which oblique shock wave will first appear in
the exhaust jet is
≤ 37.2 × 105 Pa
(b)
0.45 D
p0 = 70 × 105 kPa
0.1 D
0.45 D
D
97 D
1.0
0.45 D
D
Figure S6.3b
sin β
=
0.45
= 0.41
1.097
M1n
=
M1 sin β = 3 × 0.41 = 1.23
p2
p1
=
1.5984; p2 = pe
122
Oblique Shock and Expansion Waves
Therefore,
p1
M1
=
atm
pe
=
1.5984
1.5984
=
0.634 × 105 Pa
=
3.0 =⇒
p1
= 0.02722
p01
Therefore,
p01
=
0.634 × 105
0.02722
=
23.3 × 105 Pa
Therefore, minimum supply pressure for obtaining the desired test region is
23.3 × 105 Pa.
(c)
pe = 1 atm
M1 = 3
p0 = 70 × 105 kPa
1
2
Figure S6.3c
M1
p2
p1
p1
p01
=
3.0
p1
=
=
10.33
p2
=
=
0.02722
p01
=
patm
= 0.098105 Pa
10.333
pe = patm
0.098 × 105
= 3.6 × 105 Pa
0.02722
Therefore, the minimum supply pressure for which a normal shock will appear
at the nozzle exit is
3.6 × 105 Pa
Note: The static pressure after the shock has to be equal to the back pressure,
namely the atmospheric pressure. This is because, subsonic jets are always
correctly expanded. Thus, the total pressure of this subsonic flow is higher than
123
its static pressure. Hence, the flow will move some distance downstream of the
nozzle exit before coming to rest.
6.4 The relation between θ and β is
1
2
2 cot β M12 sin2 β − 1
tan θ =
M12 (γ + cos 2β) + 2
For θ = 15◦ , M1 = 2.0, solution by iteration yields β = 79.8◦ . This is the
strong shock solution.
(a)
β = 79.8 deg
(b)
p2
p1
=
2γ
γ−1
M 2 sin2 β −
γ+1 1
γ+1
=
4.354
(c)
T2
T1
=
p2 ρ2
/
p1 ρ1
=
1.662
(d)
β−θ
=
64.8◦
ρ2
ρ1
=
tan β
tan (β − θ)
=
2.615
(e)
M12 sin2 (β − θ)
=
γ−1
2
2
2 M1 sin β
γ−1
M12 sin 2 β − 2
1+
γ
=
1 + 0.2 × 4 × (sin 79.8◦ × sin 79.8◦ )
1.4 × 4 × (sin 79.8◦ )2 − 0.2
=
1.775
= 0.34
5.224
124
Oblique Shock and Expansion Waves
M22
=
0.34
0.34
= 0.415
=
(sin 64.8)2
0.819
M2
=
0.644
Weak shock
(a) Solving θ − β − M relation we can obtain βweak = 45.3◦
(b)
p2
p1
=
2γ
γ−1
M 2 sin2 β −
γ+1
γ+1
=
2.191
(c)
T2
T1
=
1.267
(d)
β−θ
=
45.3 − 15 = 30.3 deg
ρ2
ρ1
=
tan 45.3◦
tan 30.3◦
=
1.729
(e)
M2 = 1.446
Aliter:
Using Normal shock tables:
M1n = M1 sin β , 1.97
Strong shock solution
lution
Weak shock so-
125
γ = 1.4
M1n = 1.97
M1n = 2sin 45.3 deg = 1.42
γ = 1.4
p2
p1
=
4.361
M2n
=
0.7314
T2
T1
=
1.663
p2
p1
=
2.186
M2n
=
0.583
ρ2
ρ1
=
1.724
ρ2
ρ1
=
2.622
T2
T1
=
1.268
M2n
=
M2 sin (β − θ)
M2
=
M2
=
M2 n
sin (β − θ)
M2n
sin (β − θ)
=
0.7314
= 1.45
sin 30.3
=
0.583
= 0.644
sin(79.8 − 15)
Note: The solution obtained with oblique shock relations may also be obtained
using oblique shock tables, which will result in considerable time saving.
6.5
2
1
Figure S6.5
From isentropic tables, for M1 = 2.0, we have
p1
= 0.1278. Since,
pt1
0.75
= 0.0599.
0.1278 ×
1.60
pt1 = pt2
for this isentropic expansion,
For this pressure ratio, from isentropic table, we get
M2 = 2.48 .
p2
=
pt2
126
Oblique Shock and Expansion Waves
From isentropic table, the Prandtl-Meyer function for Mach 2.0 and 2.48, respectively are
ν1
=
26.38◦
ν2
=
38.655◦
Therefore, the flow turning angle becomes ν12 = ν2 − ν1 = 12.275◦
6.6 (a) Angles for which the oblique shock remains attached to the wedge (from
oblique shock table) are
At
At
M1
θmax
=
=
2.0
22◦
M1
θmax
= 3.0
=
34◦
θd
15◦
25◦
M1min
1.65
2.11
40◦
4.45
(b)
6.7
M1 = 3.5
θ
M2
45◦
θ
Figure S6.7
M1
=
3.5
β
=
45 deg
127
28.158◦
=⇒ θ
=
M1n
=
M1 sin β = 3.5 sin 45 = 2.47
M2n
=
0.51592
M2
=
M2n
sin (β − θ)
=
1.78
6.8
M2
=
4.0
ν2
=
65.785◦
ν2
=
ν1 + |∆θ|
ν1
= ν2 − |∆θ|
Therefore,
(a)
|∆θ|
=
60 deg −15 deg = 45 deg
Therefore,
ν1
=
=⇒ M1
=
65.785 deg −45 deg = 20.785 deg
1.8022
(b)
|∆θ|
=
60 deg −30 deg = 30 deg
Therefore,
ν1
=
=⇒ M1
=
65.785 deg −30 deg = 35.785 deg
2.360
128
Oblique Shock and Expansion Waves
(c)
|∆θ|
=
0 deg
Therefore,
nu1
=
ν2 = 65.785 deg
For this value of Prandtl Meyer function, from isentropic table, we get M1 =
4.0 .
(d)
|∆θ|
=
60 deg +15 deg = 75 deg
ν1
=
65.785 deg −75 deg = − 9.215 deg
A negative ν is not possible. The flow downstream can exist only upto |∆θ| =
65.785 deg for which ν1 = 0 =⇒ M1 = 1.0.
6.9
p1 , M 1
p2 ,
M2
θ
Figure S6.9
For M1 = 2.0, from isentropic table, we have p1 /p0 = 0.1278. Therefore,
p2
p1
1
p2
=
×
= × 0.1278 = 0.0639
p0
p1
p0
2
For p2 /p0 = 0.0639, from isentropic table, we have M2 = 2.444 .
M1
=
2.0 =⇒ ν1 = 26.38 deg
M2
=
2.444 =⇒ ν2 = 37.81 deg
ν2
=
ν1 + |∆θ| = ν1 + θ2
θ2
=
ν2 − ν1 = 37.81 − 26.38
=
11.43 deg
129
6.10
M 1 , p1
p0
θ
Me
Figure S6.10
(a)
=⇒
and
M1
=
3.0
p1
p0
=
0.02722
pe
p0
=
1.01325 × 105
70 × 105
=
0.0145 <
p1
p0
Hence, there will be expansion at the nozzle exit to reduce the pressure from p1
to pe . From isentropic table, for pp0e = 0.145 =⇒ Me = 3.431.
M1 = 3.0 =⇒ ν1 = 49.75◦
Me = 3.431 =⇒ νe = 57.42◦
For expansion,
νe = ν1 + |∆θ|
Therefore,
|∆θ| = 57.42 − 49.75 = 7.67◦
(b) For θe = 0,
pe = p1 = 1.0133 × 105 Pa
pe
p1
=
= 0.02722
p0
p0
Therefore, stagnation pressure required for θe = 0 is
νe = ν1 ,
p0
=
=
1.0133 × 105
0.02722
37.226 × 105 Pa
130
Oblique Shock and Expansion Waves
6.11 From isentropic table, for M1 = 1.5, we have
p1
p01
=
0.2724
T1
T01
=
0.6897
Therefore,
p2
p01
=
p2
p1
×
p1
p01
=
0.13
× 0.2724 = 0.118
0.30
=⇒ M2
=
2.05
T2
T02
=
0.5433
(a) From Prandtl-Meyer function table, we have
M1
=
1.5
=⇒ νM1
=
12◦
M2
=
2.05
=⇒ νM2
=
27.75◦
The deflection angle required is
θ = νM1 − νM2 = −15.75◦
(b)
M2 = 2.05
(c)
T2
6.12
=
T2
T01
0.5433
× 350
×
× T1 =
T02
T1
0.6897
=
275.7K
131
#1
1
#2
2
3
A1
#3
A2
4
A3
5◦
10◦
15◦
A4
Figure S6.12
Flow deflection in regions (2), (3) and (4) as measured from flow deflection in
region (1) are − 5 deg, − 15 deg and − 30 deg.
For expansion,
ν2 = ν1 + |∆θ|
Region
(1)
(2)
(3)
(4)
|∆θ|
(deg)
0
5
10
15
ν
(deg)
26.38
31.38
41.38
56.38
M
2.000
2.187
2.60
3.370
µ
(deg)
29.928
27.200
22.600
17.260
Fan angle (deg)
ξ1 = µ1 − (µ2 − ∆θ)
7.73
14.60
20.34
A
A∗
1.688
1.980
2.896
6.012
Therefore,
A1 : A2 : A3 : A4
=
=
1.688 : 1.980 : 2.896 : 6.012
1 : 1.173 : 1.716 : 3.562
6.13 From oblique shock table, for M1 = 2.4 and θ = 7 deg, we get β =
30.25 deg.
M1n
=
=
M1 sin β
2.4 sin 30.25 = 1.21
From normal shock table for M1n = 1.21, we have
p2
p1
T2
T1
a2
a1
p02
p01
=
1.539
=
1.134
=
1.065
=
0.9918
132
Oblique Shock and Expansion Waves
M2n
=
M2
=
From normal shock table, for
p3
p2
T3
T2
a3
a2
p03
p02
M3
p2
T2
p3
T3
0.830
M2n
= 2.12
sin (β − θ)
M2 = 2.12, we have
=
5.077
=
1.7875
=
1.337
=
0.6649
=
=
0.5583
1.57 × p1 = 0.77 × 105 Pa
1.134 × 280 = 317.52 K
4.881 p2 = 3.91 × 105 Pa
1.7875 × 317.52 = 567.57 K
=
=
=
(a)
ṁ =
ρ3 A3 V3
Therefore,
Ai
=
ρ3
=
V3
=
=
=
A3 =
ṁ
ρ3 V3
p3
RT3
2.4 kg/m3
M3 a3 = 0.5583 × 477.6
266.64 m/s
Hence,
Ai
=
=
20
2.385 × 266.64
0.03125 m2
(b) At the exit, Me = Ve /ae . By energy equation, we have
Ve2
a2e
a2
+
= 0e
2
γ−1
γ−1
133
ae
=
a0e
=
0
a20e − 0.2Ve2
/
a03 = a01 = 20.04 T01
From isentropic table, for M1 = 2.4,
0.061
T1
T01
=
T01
=
Thus, a0e
=
Hence ae
=
Me
=
0.4647
280
= 602.5K
0.4647
√
20.04 602.5 = 492 m/s
/
4922 − 0.2 × 302 = 492 m/s
30
= 0.061
492
For Me = 0.061,
pe
p03
Te
T03
For M3
p3
p03
T3
p03
=
0.9975
=
0.9993
=
0.5583
=
0.809
=
0.941
Therefore,
pe
=
=
=
Te
=
ρe
=
=
Hence, Ae
=
pe p03
p3
p03 p3
0.9975
× 3.91 × 105
0.809
4.82 × 105 Pa
0.9993
× 567.57 = 602.7 K
0.940
pe
4.75 × 105
=
RTe
287 × 595
2.79 kg/m3
20
= 0.240 m2
2.78 × 30
for Me =
134
Oblique Shock and Expansion Waves
6.14 (a)
For normal shock diffuser, at M1 = 3.0,
p02
p01
= 0.3283
(b) For the wedge shaped diffuser,
M1 = 3.0, θ = 8◦
Therefore,
β
=
25.5◦
M2
θ
=
=
2.6
8◦
=
=
29.0◦ ,
2.26
For M2 = 2.6 and θ = 8◦ , we have
β
M3
Therefore,
M1n
M2n
=
=
M1 sin β
3.0 sin 25.5
=
=
1.3
2.6 sin 29 = 1.26
From Normal shock table, for M1n = 1.3,
p02
p01
M2n
p03
p01
M3
p04
p03
=
0.9794,
=
1.26
=
0.986,
=
2.26
=
0.60105
Therefore, the overall stagnation pressure ratio is
p04
= 0.9794 × 0.986 × 0.60105 = 0.5804
p01
Note: The first diffuser with a single normal shock, the pressure loss is 67.2%
but for the wedge shaped diffuser it is only 41.96%.
135
6.15 The wave pattern over the plate will be as shown in the Fig. S6.15a.
(a)
M1
1
2
θ=α
β=
42 ◦
3
Figure S6.15a
M1
=
2.0
β
=
42 deg
=⇒ θ
=
α = 12.3◦
(b)
1
2
4
M1
β
δ
Slipstream
3
5
Figure S6.15b
M1n = M1 sin β = 1.338
For M1 = 1.338, from normal shock table,
for M1 = 2.0,
p3
p1
M3n
Therefore, p3
=
1.928
=
=
0.7664
1.928 atm
From isentropic table,
M3
=
ν1
|∆θ|
=
=
For expansion,
ν2
=
=
The corresponding Mach number from isentropic table is
M3n
= 1.55
sin (β − θ)
26.38◦
12.3◦
ν1 + |∆θ|
26.38 + 12.3 = 38.68◦
136
Oblique Shock and Expansion Waves
M2
For M1
p1
p0
=
=
2.48
2.0
=
0.1278
For M2
p2
p0
Therefore,
p2
=
2.48
=
0.0604
=
p2 p0
p1
p0 p1
=
0.473 atm
(c)
Trial
M2n
1
2
Trial
δ
β
= M2 sin β
12 deg
34
1.387
12.5 deg
34.5
1.405
M4
p4
p1
p4
0.744
0.740
2.09
2.12
0.989
1.003
p5(1)
0.041
ν5 = ν3 + |∆θ|
M5
p5
p0
p3
p5
p5
p5(2)
∆p
p4(2)
p4(1)
12◦
12.5◦
δ
Fig. S6.15c
∆p − 0.041
δ − 12
For ∆p
δ
For which p4 = p5 = 1 atm
6.16
=
=
=
−0.002 − 0.041
0.5
0
12.5◦
1
2
25.38
1.96
25.88
1.98
0.136
0.132
0.534
0.521
1.03
1.005
137
1
2
M1
15◦
Figure S6.16
For θ = 15◦ , M1 = 3.0 =⇒ β = 32.2 deg.
M1n
=
M1 sin β = 1.60
M2n
=
0.6684
p2
p1
=
2.82
ρ2
ρ1
=
2.03
T2
T1
=
1.388
p02
p01
=
0.8952
From isentropic table, for M1 = 3.0, we have
p1
p01
=
0.0272
T1
T01
=
0.3571
Therefore,
p1
=
1.904 × 105 Pa
T1
=
107.13 K
ρ1
=
a1
=
p1
= 6.19 kg/m3
RT1
/
γRT1 = 207.5 m/s
Using ratios of flow properties, we obtain,
138
Oblique Shock and Expansion Waves
(a)
p2
=
5.37 × 105 Pa
ρ2
=
12.57 kg/m3
T2
=
148.7 K
p02
=
62.66 × 105 Pa
(b)
(c)
M2
a2
V2
6.17
λu
=
=
=
λl
=
=
CL
=
=
=
M2n
= 2.26
sin (β − θ)
/
=
γRT2 = 244.4 m/s
=
= M2 a2 = 552.3 m/s
!
dz
dx
"
u
1
− α 0 ≤ x ≤ 0.3c
3
1
− − α 0.3c ≤ x ≤ c
!7 "
dz
dx l
−α
$ c
2
−
(λu + λl ) dx
βc 0
"
+$ 0.3c !
1
2
−
− α dx
βc 0
3
"
$ c !
1
+
− − α dx
7
0.3c
,
$ c
+
−αdx
0
"
+!
2
1
−
− α 0.3c
βc
3
139
−
2
[−2αc]
βc
4α
β
=
π
4
= 0.049365
α = 2 deg, CL = − √ ×
8 90
α = 0 deg, CL = 0 and for
CD
=
=
=
=
=
CD
=
λu
λl
=
=
λc
=
=
=
λc
=
=
Cmac
=
2
βc
$
"
1
+ α 0.7c − αc ]
7
−
=
For
!
c
0
1
2
λ2u + λ2l dx
"2
$ 0.3c !
1
2
[
− α dx
βc 0
3
"2
$ c !
$ c
1
+
−α2 dx ]
− − α dx +
7
0.3c
0
!
"2
1
2
π
[
−
0.3c
βc
3 90
!
"2
% π &2
1
π
+
+
0.7c +
c]
7 90
90
2
[0.02672 + 0.02212 + 0.0012185]
β
2
× 0.05006
β
0.0354
λt + λc − α
−λt + λc − α
(λu + λl )
+α
11 2
2
3 −α−α
+α
2
1
6
1 1
2
−7 − α − α
+α
2
1
−
14
4
βc2
$
c
λc xdx
0
0 ≤ x ≤ 0.3c
0.7c ≤ x ≤ c
140
Oblique Shock and Expansion Waves
=
=
=
=
xcp
c
=
=
xcp
=
$ c
$ 0.3c
1
1
4
xdx
−
xdx ]
[
βc2 0
6
14
0.3c
+
<,
1
4
1 ; 2
2
2
(0.3c)
−
−
(0.3c)
c
βc2 12
28
,
+
1
4 0.09
√
−
(1 − 0.09)
28
8 12
−0.0354
Cmac
1
+
CL
2
(−0.0354) 1
+
−
0.049365
2
−
1.217 c
6.18
1
2
M1 = 3
3
1"
2
"
3"
Figure S6.18
Upper Surface
For M1 = 3.0, from isentropic table, we have
ν1
=
49.757 deg
Therefore,
ν2
=
49.757 + 12 = 61.757 deg
For ν2 = 61.757◦ , from isentropic table, we get
M2
=
3.71
Flow Mach number normal to the oblique shock is
M2n
=
M2 sin β
where β is shock angle. From oblique shock chart for M2 = 3.71 at θ = 12 deg,
we have
β2
=
26 deg
141
Therefore,
=
M2n
3.71 sin 26 = 1.626
From normal shock table, for M2n = 1.63, we get
M3n
=
M3
=
0.6596 Therefore,
0.6596
M3n
=
sin (β − θ)
sin 14
=
2.726
Lower Surface
For M1 = 3.0, and θ = 12◦ , from oblique shock chart, we get
M 1# n
=
M1 sin β
β 1#
=
29.25 deg
Therefore,
=
M 1# n
3 sin 29.25 = 1.47
From normal shock table, for M1# n = 1.47, we have
M 2# n
=
0.712
Therefore,
M 2#
0.712
= 2.4
sin (29.25 − 12)
=
From isentropic table, for M2# = 2.4, we have
ν 2#
=
36.75 deg
Therefore,
ν 3#
=
36.75 + 12 = 48.75 deg
The corresponding Mach number is
M 3#
6.19
=
2.95
142
Oblique Shock and Expansion Waves
1
2
12◦
M1 = 3
3
10◦
4
2"
3"
4"
Figure S6.19
θ12
ν2
=
=
M2
p2
p01
θ23
=
3.105
=
0.02326
=
20 deg
ν3
M3
p3
p01
θ34
β34
=
=
71.75 deg
4.493
=
0.003484
=
=
22 deg
33.5 deg
M3n4
p4
p3
p4
p01
M4n
=
2.48
=
7.009
=
0.0245
=
M4
=
θ12#
$12#
=
=
0.5149
0.5149
= 2.58
sin11.5
22 deg
40 deg
Comparing
6.20
p4
p01
and
2 deg
51.75 deg
p4 #
p01 ,
M1n
p 2#
p1
p 2#
p01
p02#
p01
M2# n1
=
1.93
=
4.179
=
0.115
=
0.7535
=
M 2#
=
ν 2#
=
=
0.5899
0.5899
= 1.91
sin18
23.85
20
ν 3#
M 3#
p3
p02#
p 3#
p01
θ 3# 4#
ν 4#
=
=
43.87 deg
2.71
=
0.0423
=
0.0319
=
=
2 deg
45.87 deg
M4
p 4#
p02#
p 4#
p01
=
2.8058
θ
2# 3#
=
0.0366
=
0.0276
it can be seen that, the slipstream is very weak.
143
p2
M1
p3
10◦
p1
Figure S6.20
M1
p1
=
=
3.0
1.0133 × 105 Pa
θ
β
=
=
M2
p2
p1
ν2
=
2.505
=
2.054
=
39.24◦
θ3
ν3
10◦
27.38◦
M3
M2
p2
p02
M3
p3
p03
= 20◦
= ν 2 + θ3
=
=
39.24◦ + 20◦
59.24◦
=
=
3.545
2.505
=
0.05808
=
3.545
=
0.0123
Therefore,
p3
p1
=
=
L
=
CL
=
=
=
=
=
=
p2 p3 p02
p1 p03 p2
0.01230
= 0.435
2.055 ×
0.05808
;
;
c<
c<
−p2 1 ×
− p3 1 ×
+ p1 [1 × c]
2
2
L
q1 c
L
γ
2
2 M1 p1 c
2 L
γM12 cp1
,
+ !
"
p2
2
p3 1
+
1
+
−
γM12
p1
p1 2
,
+
2
1
1 − (2.055 + 0.435)
1.4 × 9
2
− 0.0389
144
Oblique Shock and Expansion Waves
Drag, D
=
=
CD
=
=
=
=
c
c
p2 tan 10◦ − p3 tan 10◦
2+
2
,
p1 c p2
p3
−
tan 10◦
2 p1
p1
D
1
2
2 γM1 p1 c
+
,
p2
1
p3
−
tan 10◦
γM12 p1
p1
1
(1.62 × 0.1763)
1.4 × 9
0.02267
6.21 Let subscripts 1 and 2 refer to states upstream and downstream of the
expansion fan. We know that the flow across the Prandtl-Meyer expansion fan
is isentropic. Therefore, p01 = p02 . From isentropic table, for M1 = 2.0, we
have
p1
= 0.1278, µ1 = 30◦ ,
and
ν1 = 26.37◦
p01
Therefore,
p2
= 0.5 =
p1
p2
p02
p1
p01
Thus,
p2
= 0.5 × 0.1278 = 0.0639
p02
p2
Again from isentropic table, for
= 0.0639, we have
p02
M2 = 2.44,
µ2 = 24.19◦ ,
and
ν2 = 37.71◦
Thus, the flow turning angle is
θ = ν2 − ν1 = (37.71 − 26.37) = 11.34◦
For the first Mach line the angle relative to the freestream is µ1 = 30◦ . For
the last Mach line the angle relative to the freestream is
µ2 − θ = 24.19 − 11.33 = 12.86◦
6.22 From oblique shock table, for M1 = 2.2 and θ = 5◦ C, we have
β = 31.09719
and
p2
= 1.3397
p1
145
M1n = M1 sin β = 1.136. From normal shock table, for M1n = 1.136, we have
p02
= 0.99726
p01
6.23 Assuming air to be a perfect gas, γ = 1.4. From oblique shock table, for
M1 = 3.0 and θ = 8◦ , from gas tables, we have the shock angle as
β = 25.61◦
p2
= 1.7953, where p2 is the pressure behind the shock, which is also the
p1
pressure at the cone surface. Thus,
Also,
p2
=
0.05 × 1.7953 = 0.0898 atm
=
0.0898 × 101.325
=
9.1 kPa
6.24 Basically the given stream passes through an expansion fan and a oblique
shock at the convex and concave corners, respectively, as shown in Fig. S6.24.
1
2
3
15◦
15◦
Figure S6.24
From isentropic table, for M1 = 3.0, we have
ν1 = 49.757◦ ,
Given, θ1 = 15◦ , thus,
p1
= 0.02722
p01
ν2 = ν1 + 15◦ = 64.757◦
For ν2 = 64.757◦ , from isentropic tables, we have
M2 = 3.92,
p2
= 0.0073316
p02
From oblique shock chart, for M2 = 3.92 and θ2 = 15◦ , we have β = 28◦ .
Therefore,
M2n = M2 sin β = 3.92 sin 28 = 1.84
146
Oblique Shock and Expansion Waves
From normal shock table, for M2n = 1.84, we have
p3
M3n = 0.6078,
= 3.7832
p2
Thus,
M3
=
0.6078
M3n
=
sin (β − θ)
sin (28 − 15)
=
2.7
p3 =
p3 p2
p1
p2 p1
Now,
p2
p1
=
p2 p02 p01
p02 p01 p1
=
p2 p01
p02 p1
since p01 = p02 across an expansion fan. Therefore,
p2
p1
=
0.0073316
0.02722
=
0.269346
Thus,
p3
=
=
(3.7832)(0.269346)(1)
1.019 atm
6.25 By Ackerets theory,
CL = /
Also,
CL =
1
2
4α
2 −1
M∞
L
γ M2 A p
where A = 1 × 1 = 1 m2 is the wing area. Thus,
/
2 −1
C L M∞
α =
4
√
0.228 1.62 − 1
=
4
=
0.0712 radians = 4◦
147
Again by Ackerets theory,
CD = /
Thus,
4 α2
= α CL
2 −1
M∞
CD = (0.228)(0.0712) = 0.0162
The aerodynamic efficiency of the plate is
CL
0.228
= 14
=
CD
0.0162
6.26 For M1 = 2.4 and β = 33◦ , from oblique shock table, we get
θ = 10◦
M1n = M1 sin β = 1.31
From normal shock table, for M1n = 1.31, we have
p2
= 1.8354
p1
T2
T1
=
1.1972
M2n
=
0.78093
M2
=
0.78093
M2n
=
sin (β − θ)
sin 23◦
=
2.00
6.27 The pressure ratio across an oblique shock is given by
2
p2
2γ 1 2 2
M1 sin β − 1
=1+
p1
γ+1
That is,
p2
−1
p1
=
p2 − p1
=
=
p2 − p1
ρ1
=
2
2γ 1 2 2
M1 sin β − 1
γ+1
!
"
2γ
1
p1 M12 sin2 β − 2
γ+1
M1
!
"
2γ
u21
1
u2
2
p1 2 sin β − 2 , since M12 = 21
γ+1
a1
M1
a1
!
"
2 γ p1 u21
1
sin2 β − 2
γ + 1 ρ1 a21
M1
148
But
Oblique Shock and Expansion Waves
γp
ρ
= a2 , from isentropic relations. Therefore,
p2 − p1
4
=
1
2
γ
+
1
ρ
u
2 1 1
"
!
1
sin2 β − 2
M1
6.28 Given, p2 /p1 = 5. From normal shock table, for p2 /p1 = 5, we get the Mach
number normal to the oblique shock at the compression corner as M1n = 2.1.
But,
M1n = M1 sin β
where β is the shock angle. Thus,
β
=
=
"
M1n
M1
!
"
2.1
−1
sin
= 44.427◦
3.0
sin−1
!
From oblique shock chart, for M1 = 3.0 and β = 44.427◦ , we get
θ = 25.5◦
6.29 From oblique shock table, for M1 = 2 and θ = 10◦ , we get
β = 39.3◦
Therefore, the Mach number normal to the shock becomes
M1n = M1 sin β1 = 1.27
From normal shock table, for M1n = 1.27, we get
M2n = 0.80167
Therefore,
M2 =
M2n
= 1.64
sin (β1 − θ)
For M2 = 1.64 and θ = 10◦ , from oblique shock chart, we get
β3 = 49.5◦
Thus, the angle of the reflected shock relative to the flat wall is
φ = β3 − θ = 49.5 − 10 = 39.5◦
M2n = M2 sin 49.5 = 1.25
149
From normal shock table, for M2n = 1.25, we have
M3n = 0.81264
Thus,
M3
=
0.81264
M3n
=
sin (β3 − θ)
sin 39.5◦
=
1.28
6.30 For M1 = 2 and θ = 7◦ , from oblique shock table, we get
M2 = 1.75
For M2 = 1.75, from oblique shock table, we have
θmax = 18◦
This is the maximum of θ up to which the second shock will remain attached.
6.31 The given flow is through an oblique shock. Therefore, there are two
solutions possible for the flow. One is called the weak solution, for which the
flow downstream of the shock will continue to be supersonic with a reduces Mach
number. The second is strong solution for which the Mach number downstream
of the shock will be subsonic.
I. Weak solution
Let subscripts 1 and 2 refer to conditions upstream downstream of the shock.
For M1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have
β = 45.34 deg and M2 = 1.45
The Mach number normal to the shock M1n = M1 sin β.
M1n = 2 sin 45.34 = 1.42
Now, the shock may be treated as a normal shock with upstream Mach number
1.42.
From normal shock relations, the change in entropy across the shock is given by
!
"
p01
∆s = R ln
p02
From normal shock table, for M1n = 1.42, we have
p01
= 1.049
p02
150
Oblique Shock and Expansion Waves
Thus,
%s
=
287 ln (1.049)
=
13.73 J/(kg K)
II. Strong solution
For M1 = 2 and flow turning angle θ = 15◦ , from oblique shock table, we have
β
=
79.83◦
M2
=
0.64
M1n = 2 sin (79.83◦ ) = 1.97
From normal shock table, for M1n = 1.97, we get
p01
= 1.36
p02
Thus,
%s
=
287 ln (1.36)
=
88.25 J/(kg K)
The shock will be attached up to θmax = 22.97◦
6.32 Since the flow is turned at a compression corner, the problem is effectively getting the flow field downstream of a oblique shock. From oblique shock
table(or chart), for M1 = 3.0 and θ = 10◦ , we have
β = 27.38◦
The Mach normal to the shock is
M1n = M1 sin β = 1.38
From normal shock table, for M1n = 1.38, we get
M2n = 0.74829,
p2
= 2.0551,
p1
T2
= 1.2418,
T1
Thus,
M2
=
M2n
sin (β − θ)
p02
= 0.96304
p01
151
p2
=
0.74829
sin (17.38◦ )
=
2.5
= (2.0551)(1)
=
T2
2.0551 atm
= (1.2418)(200)
=
248.36 K
For M1 = 3, from isentropic table,
p1
= 0.027224,
p01
p02
=
p02 p01
p1
p01 p1
=
(0.96304) (1/0.027224) (1)
=
T02
T1
= 0.35714
T01
35.37 atm
T01
T1
T1
=
T01 =
=
200
0.35714
=
560 K
6.33 Note that for a given Mach number and turning angle, the oblique shock
can have a weak and a strong solutions. Further, the Mach number downstream
of the shock is supersonic for weak solution and subsonic for the strong solution.
Let us solve this problem with oblique shock table as well as with oblique shock
charts.
From oblique shock table, for γ = 1.4 and M1 = 3.0 and θ = 10◦ , we have
For weak solution
152
Oblique Shock and Expansion Waves
β = 27.38256◦
M2 = 2.505 ,
p2
= 2.0545
p1
For strong solution
β = 86.40836◦
M2 = 0.48924 ,
p2
= 10.292
p1
From oblique shock chart I, for M1 = 3.0 and θ = 10◦ , we have
β = 27.4◦
as weak solution
β = 86.3◦
as strong solution
Using oblique shock chart II, the pressure ratio across and Mach number downstream of the shock are obtained as
For weak solution
M2 = 2.5,
p2
= 2.05
p1
M2 = 0.49,
p2
= 10.3
p1
For strong solution
From the above solutions the elegance of oblique shock tables is obvious. Further, the solutions obtained with oblique shock charts are only approximate
whereas, the oblique shock table gives accurate results.
6.34 (a) From oblique shock table, for M1 = 2.2 and θ = 6◦ , we have
β = 31.98◦
Therefore, the Mach number normal to the shock is
M1n = M1 sin β = 1.16
From normal shock table, for M1n = 1.16, we have
p02
= 0.99605,
p01
Therefore,
M2 =
M2n = 0.86816
M2n
= 1.98
sin (β − θ)
From normal shock table, for M2n = 1.98, we have
p03
= 0.73021
p02
153
Thus,
p03
p02
=
p03 p02
p02 p01
=
(0.73021)(0.99605) = 0.727
Thus,
Pressure loss
= (1 − 0.727) × 100%
=
27.3 per cent
(b) From oblique shock table, for M2 = 1.98 and θ = 6◦ ,
β = 35.8◦
Therefore, the Mach number normal to the shock is
M2n = M2 sin β = 1.1582 ≈ 1.16
From normal shock table, for M2n = 1.16, we have
p03
= 0.99605,
p02
Therefore,
M3 =
M3n = 0.86816
M3n
= 1.7468
sin (β − θ)
From normal shock table, for M3 = 1.75, we have
p04
= 0.83457
p03
Thus,
p04
p01
=
p04 p03 p02
p03 p02 p01
=
(0.83457)(0.99605)(0.99605) = 0.8279
Thus,
Pressure loss
=
=
(1 − 0.8279) × 100%
17.2 per cent
6.35 For M1 = 2.4 and β12 = 30◦ , from oblique shock chart, we have
θ12 = 6.2◦
154
Oblique Shock and Expansion Waves
The Mach number normal to the incident shock is
M1n = M1 sin β12 = 2.4 × sin 30◦ = 1.2
From normal shock table, for M1n = 1.2, we have
M2n = 0.84217
Therefore,
M2
=
M2n
sin (β − θ)
=
0.84217
sin (30 − 6.2)
=
2.09
For M2 = 2.09 and θ23 = 6.2◦ , from oblique shock chart, we have
β23 = 34◦
The Mach number normal to the reflected shock is
M2rn = M2 sin β23 = 2.09 sin 34◦ = 1.17
From normal shock table, for M2rn = 1.17, we get
M3n
=
0.86145
M3
=
M3n
sin (β − θ)
=
0.86145
sin (34 − 6.2)
=
1.84
Hence the Mach numbers upstream and downstream of the reflected shock are
2.09 and 1.84, respectively.
6.36 Let the state ahead of and behind the first shock be represented by subscripts 1 and 2, respectively, and those behind the second shock by 2 and 3,
respectively.
Given M1 = 2.3 and θ1 = 8◦ . For this Mach number and turning angle, from
oblique shock table, we get
β1 = 32.42◦ , M2 = 1.99
155
Therefore,
M1n
=
M1 × sin β
=
2.3 × sin (32.42◦ )
=
1.23
From normal shock table for Mn1 = 1.23,
p02
= 0.9896
p01
The pressure loss caused by the second shock is
"
!
1
1
p02
(1 − 0.9896)
=
1−
2
p01
2
=
Thus,
0.0052
p03
= 1 − 0.0052 = 0.9948
p02
For p03 /p02 = 0.9948, from normal shock table,
M2n = 1.18
Thus,
β2
!
=
sin−1
=
36.37◦
M2n
M2
"
= sin−1
!
1.18
1.99
"
For M2 = 1.99 and β2 = 36.37◦ , from oblique shock chart I,
θ2 ≈ 7◦
156
Oblique Shock and Expansion Waves
Chapter 7
Potential Equation for
Compressible Flow
7.1 (a) We know from Eq. (7.6) that,
T
ds
dn
uζ +
=
uζ + cp
T ds =
i.e., T0 ds
T0
ds
dn
ds
dn
=
dT0
dn
(1)
1
dh − dp
ρ
1
dp0
ρ0
=
dh0 −
=
cp dT0 −
=
cp
Multiplying Eq. (1) by
T0
dh0
dn
=
1
dp0
ρ0
dT0
1 dp0
−
dn
ρ0 dn
T0
, we get
T
T0
T0 dT0
uζ + cp
T
T
dn
(3)
From Eqs. (2) and (3), we get
cp
1 dp0
dT0
−
dn
ρ0 dn
(2)
=
T0
T0 dT0
uζ + cp
T
T
dn
157
158
Potential Equation for Compressible Flow
i.e., −
1 dp0
ρ0 dn
T0
uζ +
T
=
!
=
1+
!
"
T0
dT0
− 1 cp
T
dn
"
γ−1 2
γ − 1 2 dT0
M uζ +
M cp
2
2
dn
7.2 For supersonic flows,by Eq. (7.37), we have
1
or
2 ∂ 2 φ 1 ∂φ ∂ 2 φ
2
− 2 =0
M∞
−1
−
∂x2
r ∂r
∂r
∂ 2 φ 1 ∂φ
− β2 = 0
+
∂r2
r ∂r
/
2 − 1. The solution for the above equation, valid over a slender
where, β = M∞
body of revolution with a closed nose and arbitrary (smooth) meridional section
(Ref. Liepmann and Roshko) is,
! "
$ x
2
U
U
S && (ξ) ln (x − ξ)dξ
φ(x, r) = − S & (x) ln
−
2π
βr
2π 0
where, S(x) = πR2 is the cross sectional area of the body at x and ξ = x − βr.
Therefore,
! "
,
+
$ x
2
∂φ
S & (x)
1 d
&
= U∞ −
ln
S (ξ) ln (x − ξ)dξ
u = U∞
−
∂x
2π
βr
2π dx 0
∂φ
R dR
and, v = U∞
= U∞
i On the body, r = R, the pressure coefficient is
∂r
r dx
given by
S && (x)
Cp =
ln
π
!
2
λR
"
1 d
+
π dx
For the given body, R = $x3/2
$
x
0
&&
S (x) ln (x − ξ)dξ −
0≤x≤1
Therefore, S = πR2 = π$2 x3
S & (x)
=
dS(x)
= 3π$2 x2
dx
S && (x)
=
6π$2 x
dR
dx
=
3 1/2
$x
2
!
dR
dx
"2
159
Substituting the above relation into the CP expression, we obtain
Cp
=
Cp
$2
=
Now consider the term
6
d
dx
$
!
"
$ x
2
9
2 d
ξ ln (x − ξ) dξ − $2 x
+
6$
dx 0
4
β$x3/2
+ ! "
,
$ x
2
9
d
6x ln
ξ ln (x − ξ) dξ − x
− ln x3/2 + 6
β$
dx 0
4
6$2 x ln
=x
d
dx
x
ξ ln (x − ξ)dξ
0
=
Now use the rule,
ξln (x − ξ) dξ. Express this as
0
i6
$
,
+ $ x
$ x
d
(x − ξ) ln (x − ξ) dξ +
x ln (x − ξ) dξ
−
dx
0
0
a
0
f (a − t)dt =
$
a
f (t)dt
0
Therefore, we have
6
d
dx
$
x
ξ ln (x − ξ) dξ
0
−6
=
d
−6
dx
=
−6
=
lim ξ → 0 both
since
d
6
dx
$
ξ2
2
+$
x
0
+!
ξ ln ξ dξ − x
$
"
2 x
ξ2
ξ
ln ξ −
2
4
0
x
ln ξ dξ
0
,
− x (ξ ln ξ −
ξ ln (x − ξ)dξ
,
+
d x2
x2
ln x −
− x2 ln x + x2
dx 2
4
=
&
%
x x
−6 x ln x + − − x − 2x ln x + 2x
2
2
=
6x ln x − 6x
Hence,
Cp
$2
=
=
Cp
$2
=
x
ξ)0
ln ξ and ξ ln ξ tend to zero. Therefore,
x
0
d
dx
+ ! "
,
2
9
6x ln
− ln x3/2 + 6x ln x − 6x − x
β$
4
! "
2
33
3
6x ln
x
− 6x ln x + 6x ln x −
β$
2
4
6x ln
%
2
β%
&
− 3x ln x −
33
4
x
,
160
Potential Equation for Compressible Flow
Drag Coefficient
The drag coefficient can be expressed as,
CD S(L)
=
S&
=
!
3
$
L
Cp (x)S & (x)dx
0
3π$2 x2
Therefore,
CD S(L)
3π$4
$
=
0
+
"
2
33 3
3
− 3x ln x − x dx
6x ln
β$
4
! 4
"
,L
6 4
x
x4
2
33 x4
x ln
−3
ln x −
−
4
β$
4
16
4 4 0
+
,
L
L 11
3 3
2
L L ln
− ln L + −
2
β$
2
8
8
=
=
S(x)
L
π$2 x3
=
Therefore,
S(L)
=
π$2 L3
Thus,
But, β =
√
CD π$2 L3
3π $4
=
+
,
L
L 11
3 3
2
L L ln
− ln L + −
2
β$
2
8
8
With L
=
1, we get,
CD
3$2
=
CD
=
+
,
10
3
2
−
ln
2
β$
8
+
,
2
15
2 3
ln
−
3$
2
β$
8
M 2 − 1 = 1 and $ = 0.1. Therefore,
CD
=
=
0.03 [1.5 ln 20 − 1.875]
0.0786
Chapter 8
Similarity Rules
8.1 By Prandtl-Glauert rule, we have
!
dCL
dα
dCL
dα
"
(dCL /dα)inc
/
2
1 − M∞
!
"
dCL
= 0.108 deg
dα M∞ =0.2
=
=
inc
Therefore,
dCL
dα
M∞
0.2
0.3
0.4
0.5
0.6
0.7
0.75
1 dC 2
L
dα
/
=
1 dC 2
0.108
per degree
2
1 − M∞
L
measured
0.108
0.113
0.115
0.124
0.130
0.127
0.100
dα
P −G rule
0.1080
0.1132
0.1178
0.1247
0.1350
0.1512
0.1633
8.2 For hypersonic flow, M∞ tan (θ + α) ≥ 0.5.
When (θ + α) is small, M∞ (θ + α) ≥ 0.5.
where, θ is the half angle at the vertex of the model and α is the angle of attack.
For hypersonic similarity,
M∞1 (θ1 + α1 )
M∞1
=
=
M∞2 (θ2 + α2 )
10
161
162
Similarity Rules
α1
θ1
=
=
3 deg
3 deg
M∞1 (θ1 + α1 )
=
10 × (3 + 3) ×
π
>1
180
Hence the flow is hypersonic. When the test Mach number is small, the angle
involved will be large.
(a)
M∞2
θ2
=
=
3.0
12 deg
M∞1 tan (θ1 + α1 )
M∞2 tan (θ2 + α2 )
=
=
10 tan 6 deg = 1.05
3 tan (12 + α2 ) = 1.05
=
=
0.35
19.3
tan (12 + α2 )
12 + α2
Therefore,
α2 = 7.3 deg
(b)
M∞3
=
3.0
θ3
M∞3 tan (θ3 + α3 )
tan (3 + α3 )
=
=
=
3 deg
3.0 tan (3 + α3 ) = 1.05
0.35
3 + α3
=
19.3
Therefore,
α3 = 16.3 deg
8.3 Hypersonic similarity factor K = M θ. In order to apply the results of
model to the missile, K should be same for the model and the missile. That is,
K 1 = K2
that is,
M 1 θ 1 = M2 θ 2
Given, M1 = 12, θ1 = 4 deg and M2 = 2.5.
Therefore,
θ2
=
M1 θ 1
M2
163
=
12 × 4
2.5
=
19.2 deg
That is, the semi-vertex angle of model = 19.2deg
8.4 For supersonic flow past thin wedge with semi-wedge angle δ, the linearized
theory yields,
! "
2 dz
CPu =
β dx u
=
2
2
tan δ ≈ δ
β
β
Therefore,
(γ + 1)1/3 CP
δ 2/3
=
=
or
2δ 1/3 (γ + 1)1/3
/
2 −1
M∞
4
2/3
[(γ + 1) δ]
2
2 −1
M∞
9
:
2 1/3
CP (γ + 1) M∞
δ 2/3
where, χ =
=
2
χ
2
M∞
−1
2 ]3/2
[δ(γ+1)M∞
∗
8.5 (a) Given M∞
= 0.3. Therefore, by Equation (8.57), we have the minimum
Cp over the profile as
'!
*
"3.5
2 + 0.2.32
2
Cp =
−1
1.4 × 0.32
2.4
'!
*
"3.5
2.018
= 15.873 ×
−1
2.4
=
−7.22
∗
(b) Given M∞
= 0.4. Therefore,
Cp
=
2
1.4 × 0.42
'!
2 + 0.2.42
2.4
"3.5
*
−1
164
Similarity Rules
=
8.929 ×
=
−3.94
'!
2.032
2.4
"3.5
*
−1
Chapter 9
Two Dimensional
Compressible Flows
9.1 The linearized perturbation velocity potential equation for two-dimensional
supersonic flow Eq.(9.1) is
β 2 φxx − φyy = 0
(1)
/
2 − 1 and φ is the perturbation velocity potential. Equation
where β = M∞
(1) may also be written as,
∂2φ
1
∂2φ
−
=0
2 − 1 ∂y 2
∂x2
M∞
(2)
2
Keeping in mind that (M∞
− 1) > 0 for supersonic flow, it can be visualized
that equation (2) is also of the form of the classical wave equation. Hence, a
solution to equation (2) can be expressed as,
/
/
2 − 1y) + g(x +
2 − 1y)
M∞
(3)
φ(x, y) = f (x − M∞
For the problem under consideration, only left running waves are present and
therefore,
/
2 − 1y) = 0
g(x + M∞
Thus,
φ(x, y) = f (x −
and
/
2 − 1y)
M∞
<
∂φ ; #
= f (x − βy) (−β)
∂y
By boundary condition, at the wall
∂φ
∂y
w
= U∞ dy
dx
x
x
x
dyw
= k(1 − ) − k = k − 2k
dx
l
l
l
165
(4)
(5)
166
Two Dimensional Compressible Flows
Therefore,
%
∂φ
x&
= U∞ k − 2k
∂y
l
Substituting this into Eq.(5), we get,
%
#
x&
= −βf (x − βy)
U∞ k − 2k
l
#
U∞ %
x&
f (x) = −
k − 2k
β
l
(6)
(7)
Integrating Eq.(7) with respect to its argument, [Note that the argument is
(x − βy), but with y = 0], we have
!
"
U∞
x2
f (x) = −
kx − k
+ constant
(8)
β
l
At x = 0, f = 0, which gives constant = 0. Therefore,
!
"
U∞
x2
f (x) = −
kx − k
β
l
(9)
Since f (x) is defined throughout the flow, not just at the wall, and because it
has the form of Eq.(8a), where x represents the argument of f , Eq.(4) can be
written as
&
%
x−βy
k
φ(x, y) = f (x − βy) = − U∞
β (x − βy) 1 −
l
9.2
λu
=
=
λL
=
CL
=
=
=
!
dz
dx
"
=
u
1
−α
7
0 ≤ x ≤ 0.7c
1
− −α
0.7c ≤ x ≤ c
3
! "
dz
= −α
dx l
$ c
2
−
(λu + λL ) dx
βc 0
"
"
,
+$ 0.7c !
$ c !
$ c
1
2
1
−
− α dx +
−αdx
− − α dx +
βc 0
7
3
0.7c
0
"
!
"
,
+!
1
2
1
− α 0.7c −
+ α 0.3c − αc
−
βc
7
3
167
=
β
=
α
=
4α
β
/
M 2 − 1 = 2.29
2 deg =
2π
= 0.0349
180
Thus,
CL
CD
=
4 × 0.0349
2.29
=
0.06096
=
2
βc
$
c
0
'!
1
2
λ2u + λ2L dx
1
−α
7
"2
!
1
+α
3
"2
=
2
βc
=
2
[0.0081583 + 0.0406787 + 0.001218]
2.29
=
2 × 0.050055
2.29
=
0.7c +
2
0.3c + α c
*
0.04372
9.3
0 ≤ x ≤ 0.3c
λu
=
=
λl
0.3c ≤ x ≤ c
"
0.1
−α
0.3
0.333 − α
"
!
0.03
−α
0.3
0.1 − α
!
=
=
α
=
2 deg = 2
β
=
√
λu
=
=
λl
=
=
π
≈ 0.035 radian
180
8 = 2.83
"
0.1
−α
0.7
−0.143 − α
"
!
0.03
−α
−
0.7
−0.043 − α
!
−
168
Two Dimensional Compressible Flows
φ(x, y)
=
f (x − βy)
∂φ
|
∂y y=0
=
−βf & (x) = U∞ λ
Therefore,
U∞
λ
β
f & (x − βy)
f & (x)
=
φx (x, y)
=
φx (x, 0)
=
f & (x) = −
Cp
=
−2
−
U∞
λ
β
φx
λ
=2
U∞
β
For 0 ≤ x ≤ 0.3c
Cpu
=
2
=
C pl
=
0.298 × 2
λu
=
β
2.83
0.211
2
=
0.065 × 2
λl
=
β
2.83
0.046
For 0.3 ≤ x ≤ c
C pu
=
2
− 0.1258
=
C pl
=
=
0.178 × 2
λu
=−
β
2.83
2
0.078 × 2
λl
=−
β
2.83
−0.0551
Chapter 10
Prandtl-Meyer Flow
10.1 Let subscripts 1 and 2 refer to conditions upstream and downstream of
the Prandtl-Meyer fan.
p0
=
33.5 × 105 Pa
M1
=
2.0
From isentropic table, for M1 = 2.0,
ν1
=
26.38 deg
p1
p0
=
0.1278
p2
p0
=
1.0133
= 0.03025
33.5
where subscript 2 refer to conditions downstream of the expansion fan. From
p2
isentropic table, for
= 0.03025,
p0
M2
=
2.93
ν2
=
48.388 deg
But ν2
=
ν1 + ∆θ
Therefore,
∆θ
=
=
48.388 − 26.38
22 deg
169
170
Prandtl-Meyer Flow
That is, after initial expansion, the flow direction is 22 deg with respect to nozzle
axis.
10.2 For M1 = 2.3 and θ = 12 deg, from oblique shock chart, β = 36.5 deg.
M1n
=
M1 sin β = 2.3 × sin 36.5 deg
=
1.368
For M1n = 1.368, normal shock table gives: M2n = 0.7527, and
Thus,
M2
T2
=
M2n
sin (β − θ)
=
1.815
=
1.235 × 500
=
617.5K
T2
T1
= 1.235.
For M2 = 1.815, from isentropic table, ν2 = 21 deg. Also, |θ| = 24 deg. Therefore,
ν3
=
ν2 + |θ| = 21 + 24
=
45 deg
For ν3 = 45 deg, isentropic table gives, M3 = 2.764 , and
fore,
T3
= 0.396. ThereT0
T3
T3 T02
=
T2
T02 T2
But T02 = T03 , thus, we have
T3
T2
=
0.396
= 0.657
0.603
T3
=
0.657 × 617.5
=
405.7 K
For M3 = 2.764 and θ = 12 deg, oblique shock chart gives, β = 32 deg, and
M3n = 2.764 × sin 32 = 1.465. Now, for M3n = 1.465, the normal shock table
171
gives M4n = 0.7157, and
T4
= 1.2938. Thus,
T3
M4
=
T4
=
0.7157
= 2.09
sin 20 deg
524.9 K
Note: In this problem, oblique shock table may be used in places where oblique
shock chart is used.
10.3 Let the subscripts 1 and 2 refer to conditions upstream and downstream
of expansion.
M1
=
2.0
p1
p01
=
0.1278
For isentropic compression p02 = p01 . Therefore,
!
"
760 − 240
p2
= 0.1278 ×
= 0.0642
p02
760 + 275
=⇒ M2
=
2.44
=⇒ ν1
=
26.38 deg
=⇒ ν2
=
37.71 deg
or, the flow has been turned through ν2 − ν1 = 11.33 deg .
Note: Compare this problem with 6.8.
172
Prandtl-Meyer Flow
Chapter 11
Flow with Friction and
Heat Transfer
11.1
M∗ = 1
M 1 , p1
p∗
p01
p02
Figure S11.1
This problem has to be solved by trial and error. The pressure ratio
written as
Given,
p1
can be
p∗
p1 p01 p02
p1
=
p∗
p01 p02 p∗
p01
p02
= 10. We want ∗ = 1.893 (critical pressure ratio). Thus,
p02
p
p1
p∗
=
18.93
p1
p01
or
p1 /p∗
p1 /p01
=
18.93
p1
p1
and
for different M1 and check for the ratio to be 18.93. For
∗
p
p01
that, use the relations
Calculate
p1
p∗
=
1
M1
!
1.2
1 + 0.2M12
173
" 12
174
Flow with Friction and Heat Transfer
p1
p01
=
Trial
!
1+
M1
1
2
3
4
5
0.1
0.06
0.055
0.059
0.058
γ
"− γ−1
γ−1 2
M1
2
%
&
p1 p1
p∗ / p01
10.944/0.99
18.25/0.977
19.90/0.998
18.56/0.998
18.88/0.998
=
=
=
=
=
11.05
18.29
19.95
18.60
18.92
It is seen that M1 = 0.058 is the required Mach number. Now use the relation
'
*
γ+1
2
M
4f¯L∗
1 − M12
γ+1
1
2
=
ln
+
2
D
γM12
2γ
1 + γ−1
2 M1
to calculate L∗ ,
4f¯L∗
D
L∗
!
0.0040368
1.0006728
=
211.6 + 0.857 ln
=
211.6 − 4.72 = 206.88
=
206.88 × 0.1
4 × 0.005
=
1034.4 m
p1
=
T1
p2
ṁ
=
=
=
"
11.2
L =
f¯ =
3.5 × 105 mboxP a
300 K
1.4 × 105 mboxP a
0.09 kg/s
600 m
0.004
ṁ
=
ρ1 V1 A
π
p1 /
γRT1 M1 D2
RT1
4
=
0.090
√
3.5 × 105
π
× 20.04 300 × × M1 D2
287 × 300
4
=
0.090
M1 D 2
=
8.12 × 10−5
175
For an isothermal flow with friction, it can be shown that
'
! "2 *
! "2
p2
p2
4f L
1
=
1−
+ ln
2
D
γM1
p1
p1
'
*
!
!
"2
"2
1.4
1.4
4 × 0.004 × 600
1
=
1
−
+
ln
D
γM12
3.5
3.5
9.6
D
=
0.6
1
− 1.832
M12
From Eqs. (1) and (2), we obtain
9.6
D
=
9.6
D
= 9.10 × 107 D4 − 1.832
0.6
D4 − 1.832
(8.12)2 × 10−10
Solving equation (3), We obtain D = 0.0402 m .
11.3 Energy equation gives,
h+
V2
= h0 = constant
2
Also, for a perfect gas,
V2
a2
+
γ−1
2
!
"
γ−1 2
2
M
a 1+
2
!
"
γ−1 2
M
T1 1 +
2
=
a20
γ−1
=
a20
=
T0
From continuity equation for a perfect gas,
G =
=
=
ṁ
= ρ1 V1 = ρ1 a1 M1
A
p1 /
γRT1 M1
RT1
3
γ 1
√ p1 M1
R T1
From Eq. (2), we have
T1
=
γp21 M12
RG2
176
Flow with Friction and Heat Transfer
Using this in Eq. (1), we get
!
"
γ p21 M12
γ−1 2
M1
1+
R G2
2
!
"
γ−1 2
M1
M12 1 +
2
M1
!
γ−1 2
M1
1+
2
=
=
" 12
=
T0
R G2
T0 2
γ
p1
4
R G
T0
γ p1
Now,
G
=
ṁ
ṁ∗
=
A
A
=
Γ
A∗
√
p0
γRT0 A
where
Γ
=
γ
!
2
γ+1
γ+1
" 2(γ−1)
Using Eq. (4) in Eq. (3), we get,
M1
!
γ−1 2
M1
1+
2
" 12
=
4
=
Γ p0 A∗
γ p1 A1
R
Γ
A∗ 1
T0
p0
γ γRT0 A1 p1
i.e.
!
"1
γ
γ−1 2 2
M1 1 +
M1
Γ
2
p1
p0
=
A
A∗
=
=
p0 A∗
p1 A1
2.38 × 105
= 0.0354
67.3 × 105
!
"2 !
"2
D
0.0127
=
= 2.082 = 4.33
D∗
0.0061
Therefore,
p1 A
p0 A∗
=
0.0354 × 4.33 = 0.1533
177
=⇒ M1
!
=⇒
4f L∗
D
"
=
2.51
=
0.4341
=
4.85 × 105
= 0.072
67.3 × 105
1
p2
p0
Therefore,
=⇒
p2
A
× ∗
p0
A
=
0.072 × 4.33 = 0.312
=⇒ M2
=
1.53
=
0.147
!
4f L∗
D
"
2
Therefore,
!
"
=
0.4341 − 0.147
∆L
D
=
29.60 − 1.75 = 27.85
4f L∗
D
Hence, the average friction coefficient is
f¯ =
=
0.2871
4(27.85)
0.002577
11.4 (a)
Ae /A∗ = 2
1
p0
T0
Figure S11.4a
pb
Shock
M =1
A∗
L
178
Flow with Friction and Heat Transfer
Normal shocks can exist only for M1 > 1. Therefore, the interpretation that a
normal shock stands in the throat implies M = 1 at the throat and downstream
is subsonic. Therefore, At = A∗ .
A∗
Ae
0.50
M1
0.306
ρ1
ρ0
T1
T0
a1
a0
p1
p0
0.9547
0.9816
0.9907
0.9371
p0
7 × 105
= 8.13 kg/m3
=
RT0
287 × 300
/
√
γRT0 = 1.4 × 287 × 300 = 347 m/s
ρ0
=
a0
=
ρ1
=
(0.9547)ρ0 = (0.9547) × (8.13) = 7.7617 kg/m3
a1
=
(0.9907)a0 = (0.9907) × (347) = 343.8 m/s
p1
=
(0.9371)p0 = (0.9371) × (7 × 105 ) = 6.5597 × 105 Pa
% ¯ ∗&
4f L
p1
M1
D
p∗
M1
0.306
5.0776
3.547
p1
= 1.849 × 105 Pa
3.547
Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L.
p∗
p2
p∗
=
pB
2.8 × 105
=
= 1.514
p∗
1.849 × 105
% ¯ ∗&
=
p2
p∗
1.514
Therefore,
4f¯L
D
4f L
D
M2
0.220
=
! ¯ "
! ¯ "
4f L
4f L
−
D M1
D M2
=
5.0776 − 0.220
=
G =
=
=
4.8576
ρ1 V1 = ρ1 a1 M1
(7.7617)(343.8)(0.306)
816.5 kg/m2 s
179
(b)
1
Shock
M =1
A∗
L
Figure S11.4b
Upstream of normal shock wave
A∗
Ae
p
p0
ρ
ρ0
T
T0
a
a0
M
0.5
0.09396
0.1847
0.5088
0.7133
2.197
p
=
ρ =
=
T
a
=
=
(0.09396)(7 × 105 ) = 0.6577 × 105 Pa
(0.1847)(8.13)
1.5 kg/m3
(0.5088)(300) = 152.6 K
(0.7132)(347) = 247.5 m/s
Downstream of normal shock wave
This zone is referred by subscript 1.
M
2.197
p1
p
5.465
ρ1
ρ
2.947
T1
T
1.854
a1
a
1.362
M1
0.5475
Therefore,
p1
ρ1
=
=
(5.465)(0.6577 × 105 ) = 3.5943 × 105 Pa
(2.947)(1.5) = 4.4205 kg/m3
T1
a1
=
=
(1.854)(152.6) = 282.9 K
(1.362)(247.5) = 337.1 m/s
% ¯ ∗&
4f L
p1
M1
D
p∗
M1
0.5475
0.7451
1.9434
180
Flow with Friction and Heat Transfer
Thus,
3.5943 × 105
p1
=
= 1.85 × 105 Pa
1.9434
1.9434
Since p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L.
p∗ =
p2
pB
2.8 × 105
=
=
= 1.514
p∗
p∗
1.85 × 105
% ¯ ∗&
p2
p∗
4f L
D
1.514
Therefore,
4f¯L
D
=
=
=
G
M2
0.220
! ¯ "
! ¯ "
4f L
4f L
−
D M1
D M2
0.7451 − 0.220
0.5251
=
ρ1 V1 = ρ1 a1 M1
=
(4.4205)(337.1)(0.5475)
=
815.9 kg/m2 s
(c)
1
Shock
M =1
A∗
Figure S11.4c
Properties at station 1 are the same as properties in Part-b.
Upstream of normal shock wave
Therefore,
p1
ρ1
=
=
0.6577 × 105 Pa
1.5 kg/m3
T1
a1
M1
=
=
=
152.6 K
247.5 m/s
2.197
181
M1
2.197
Thus, p∗ =
%
4f¯L∗
D
&
p1
p∗
M1
0.36011
0.35566
p1
= 1.849 × 105 Pa.
0.35566
Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pa at L downstream
of normal shock wave.
From the diagram in Fig. s11.4d, it is clear that the Mach number Mx at
point x jumps.
h
pb
=
0
28
a
kP
*
1
s
Figure S11.4d
The normal shock pressure jump and the Fanno line equation are given by
=
2γ
γ−1
Mx2 −
γ+1
γ+1
px
p∗
=
1
Mx
2.8 × 105
1.849 × 105
=
Let
=
p3
px
f (Mx )
-
1+
γ+1
2
γ−1
2
2 Mx
.1/2
2
1 1
1.1667Mx2 − 0.1667
Mx
2
1 1
1.1667Mx2 − 0.1667
Mx
!
!
1.2
1 + 0.2Mx2
1.2
1 + 0.2Mx2
Use Secant method to find Mx as follows.
!
"
xj − xj−1
xj+1 = xj − f (xj )
f (xj ) − f (xj−1 )
Trial 1
x1
=
1.6
"1/2
"1/2
− 1.5143
182
Flow with Friction and Heat Transfer
=
x0
1.5
x2
=
1.6 − 0.0559
x2
=
1.53
f (x2 )
=
0.000752
!
1.6 − 1.5
0.0559 − (−0.0236)
x3
=
1.53 − 0.000752
x3
=
1.53,
!
1.53 − 1.6
0.000752 − 0.0553
Therefore, Mx = 1.53.
M2
! ¯ ∗"
4f L
D
M2
1.53
0.14699
Therefore,
4f¯L
D
=
! ¯ "
! ¯ "
4f L
4f L
−
D M1
D M2
=
0.36011 − 0.14699
=
G
0.21312
=
ρ1 V1 = ρ1 a1 M1
=
(2.197)(247.5)(1.5)
=
815.6 kg/(m2 s)
11.5 The speed of sound a1 is given by,
a1
=
=
M1
=
=
/
"
√
γRT1 = 20.04 333.3
366 m/s
V1
73.15
=
a1
366
0.2
"
183
From tables Rayleigh flow tables, for M1 = 0.2, we have
T01
T0∗
T1
T∗
p01
p∗0
p1
p∗
V1
V∗
=
0.17355
=
0.20611
=
1.2346
=
2.2727
=
0.09091
From isentropic tables, for M1 = 0.2, we have
p1
= 0.97250
p01
Therefore,
p01
=
=
T1
T01
=
0.5516 × 105
0.9725
5.672 × 104 Pa
0.99206
Therefore,
T01
=
333.3
= 336 K
0.99206
(a)
T02
∆h
=
=
stagnation temperature after combustion
cP (T02 − T01 )
Therefore,
=
T02
=
=
∆h
cP
1395.5 × 103
336 +
1004.5
1725.2 K
T01 +
(b)
T02
T01
=
1725.2
= 5.1345
336.0
184
Flow with Friction and Heat Transfer
T02
T0∗
=⇒ M2
T2
T∗
p02
p∗0
p2
p∗
V2
V∗
=
T01 T02
T0∗ T01
=
(0.17355)(5.1345) = 0.89109
=
0.68
=
0.98144
=
1.0489
=
1.4569
=
0.67366
(c)
T2
=
T2 T ∗
T1
T ∗ T1
=
0.98144
× 333.3
0.20661
=
1583.2 K
(d)
p02
∆p0
=
p02 p∗
p01
p∗ p01
=
1.0489
× 5.672 × 104 = 4.819 × 104 Pa
1.2346
=
p02 − p01 = (4.819 − 5.672) × 104
−8530 Pa
=
(e)
s2 − s1
=
=
=
cp ln
!
T02
T01
"
− R ln
!
p02
p01
"
1004.5 ln (5.1345) − 287 ln
1690.1 J/kg.K
!
4.819
5.672
"
185
(f )
V2
=
V2 V ∗
V1
V ∗ V1
=
0.67366
× 73.15
0.09091
=
542.1 m/s
(g) The initial conditions can be maintained till the flow is choked at the duct
T01
exit after combustion. That is, M2 = 1 and TT02∗ = 1 and since ∗ = 0.17355,
0
T0
and T01 = 336K, we have
T02
=
T02 T0∗
T01
T0∗ T01
=
336.0
0.17355
= 1936 K
Maximum heat of reaction q is given by
q
=
=
cP ∆T0 = 1004.5(1936 − 336)
1607.2 kJ/kg
11.6 The flow process is adiabatic and therefore, it can be treated as a Fanno
flow. The velocity at station 1 is
/
V 1 = M1 a 1 = M1 γ R T 1
=
√
0.2 1.4 × 287 × 300 = 69.44 m/s
From Fanno flow table, for M1 = 0.2, we have
p1
= 5.4554,
p∗
T1
= 1.1905,
T∗
V1
= 0.21822,
V∗
p01
= 2.9635
p∗0
Again, from Fanno flow table, for M2 = 0.5, we have
p2
= 2.1381,
p∗
T2
= 1.1429,
T∗
V2
= 0.53452,
V∗
p02
= 1.3398
p∗0
186
Flow with Friction and Heat Transfer
Thus,
p2
T2
V2
p02
=
p2 p∗
2.1381
× 5 × 101325
p1 =
p∗ p1
5.4554
=
198.558 kPa
=
T2 T ∗
1.1429
× 300
T1 =
∗
T T1
1.1905
=
288 K
=
V2 V ∗
0.53452
× 69.44
V1 =
V ∗ V1
0.21832
=
170.09 m/s
=
p02 p∗0
1.3398
× 520.951
p01 =
p∗0 p01
2.9635
=
235.52 kPa
11.7 For, M1 = 0.2, from isentropic table, we have,
T1
= 0.99206
T01
Therefore,
72 + 273.15
T1
=
= 347.9 K
0.99206
0.99206
Since the tube is perfectly insulated, T01 = T02 , thus,
T01 =
T02 = 347.9 K
The initial density is
ρ1
=
p1
2 × 101325
=
R T1
287 × 345.15
=
2.046 kg/m3
since, 1 atm = 101325 Pa. Thus, the mass flow rate through the tube is
ṁ
=
ρ1 A1 V1 = ρ1 A1 M1 a1
=
2.046 × (0.1 × 0.1) × 0.2 ×
/
γ R T1
187
=
√
0.004092 1.4 × 287 × 345.15
=
1.524 kg/s
Now, assuming a control volume between the sections 1 and 2, we can write the
force balance equation as
(p1 − p2 ) A + F = ρ1 A1 V1 (V2 − V1 )
where F is the frictional drag and A is the cross-sectional area of the tube.
For M2 = 0.76, from isentropic table, we have
T2
= 0.89644
T02
Therefore,
T2
=
0.89644 T02 = 0.89644 × 347.9
= 311.87 K
a2
=
√
1.4 × 287 × 311.87
= 354 m/s
V2
= M2 a2 = 0.76 × 354
= 269.04 m/s
For obtaining p2 , let us use the Fanno flow table. For M1 = 0.2 and M2 = 0.76,
respectively, from Fanno flow table, we have
p1
= 5.4554,
p∗
p2
= 1.3647
p∗
Therefore,
p2
=
p2 p∗
p1
p∗ p1
=
1.3647
× 2 = 0.5 atm
5.4554
Thus,
F
=
ρ1 A1 V1 (V2 − V1 ) − (p1 − p2 ) A
188
Flow with Friction and Heat Transfer
=
1.524(269.04 − 74.52) − (2 − 0.5) × 101325 × (0.1 × 0.1) = 296.45 − 1519.88
=
− 1223.43 N
Hence,
Drag = 1223.43 N
11.8 Given, L/D = 50, V1 = 195 √
m/s, T1 = 310 K and Me = 1. The speed
of sound at the entrance is a1 = γRT1 . For carbon dioxide, γ = 1.3 and
molecular weight is 44, thus,
R
8314
Ru
=
M
44
188.95 J/(kg K)
=
=
Therefore,
a1
=
=
M1
=
=
√
1.3 × 188.95 × 310
275.95 m/s
V1
195
=
a1
275.95
0.71
From Fanno flow table for γ = 1.3, for M1 = 0.71, we have
4 f Lmax
D
=
0.20993
Thus,
f
=
=
0.20993
50 × 4
0.00105
11.9 From Fanno flow table, for M2 = 0.8, we have
4 f L∗2
D
p2
p∗
T2
T∗
4 f L∗1
D
=
0.07229
=
1.2893
=
1.0638
=
=
4 f L∗2
4 fL
+
D
D
4 × 0.005 × 51
0.07229 +
= 40.87229
0.025
189
From Fanno flow table, for
4 f L∗
1
D
= 40.87229, we have
M1
=
0.13
p1
p∗
=
8.457
T1
T∗
=
1.1959
p1
=
p1 p∗
8.457
×1
p2 =
∗
p p2
1.2893
=
6.56 atm
=
T1 T ∗
1.1959
× 270
T2 =
T ∗ T2
1.0638
=
303.52 K
T1
11.10 By energy equation, we have
V12
2
Treating air as a perfect gas, we can express h = cp T and hence
h0 = h1 +
c p T0 = c P T1 +
V12
2
For air cp = 1004.5 J/(kg K). Therefore,
T1
a1
=
T0 −
=
!
=
=
V12
2cp
"
1352
K = 349.9 K
2 × 1004.5
/
√
γRT1 = 1.4 × 287 × 349.9
359 −
374.95 m/s
Therefore,
M1 =
V1
135
= 0.36
=
a1
374.95
(a) For M1 = 0.36, from Fanno flow table, we have
4f Lmax
= 3.1801
D
190
Flow with Friction and Heat Transfer
Thus,
Lmax
=
3.1801 × 5 × 10−2
4 × 0.02
=
1.99 m
This is the minimum length of the tube for the flow to choke.
(b) For L2 = 0.6 m,
!
4f L
D
"
=
2
!
4f L
D
"
1
−
!
4f L
D
"
12
where 1 and 2 stands for the inlet and exit of the tube
!
"
4f L
4 × 0.02 × 0.6
= 3.1801 −
D 2
5 × 10−2
3.1801 − 0.96 = 2.22
=
The corresponding Mach number, from Fanno flow table, is
M2 ≈ 0.41
and
For M1 = 0.36, from Fanno flow table,
p02
p02
= 1.5587
p∗0
p02
= 1.7358. Therefore,
p∗0
=
p02 p∗0
p01
p∗0 p01
=
1.5587
× 135
1.7358
=
121.23 kPa
For M1 = 0.36, from isentropic table,
p1
p01
=
0.91433
p1
=
0.91433 × 135 = 123.43 kPa
ρ1
=
p1
123.43 × 103
=
RT1
287 × 349.9
=
1.229 kg/m3
191
Thus, the mass flow rate through the tube ṁ is
ṁ
ρ1 A1 V1 = 1.229 × 135 ×
=
=
π × 52 × 10−4
4
0.326 kg/s
11.11 For hydrogen, M = 2.016 and γ = 1.4
R=
8314
= 4124 J/(kg K)
2.016
At the tube inlet, the Mach number is
M1
=
V1
V1
200
=√
=√
a1
γRT1
1.4 × 4124 × 303
=
0.15
From Fanno flow table, for M1 = 0.15, we get
4f¯Lmax
D
p1
p∗
=
27.932
=
7.2866
Thus, the tube length required for the flow to choke is
Lmax
pexit
=
27.932D
4f¯
=
27.932 × 25 × 10−3
4 × 0.03
=
5.82 m
p1
7.286
=
p∗ =
=
250
7.2866
=
34.31 kPa
11.12 At the pipe entrance, Mach number M1 =
V1 = 200 m/s
V1
a1 ,
where
192
Flow with Friction and Heat Transfer
and
a1
=
/
=
349 m/s
γRT1 =
√
1.4 × 287 × 303.15
Thus,
200
= 0.57
349
From Fanno flow table, for M1 = 0.57, we have
M1 =
4f Lmax
= 0.62288
D
Therefore,
Lmax =
0.62288 × 20 × 10−3
= 0.156 m
4 × 0.02
Thus, the length of the pipe at which the flow would be sonic is 15.6 cm
11.13 For methane, γ = 1.3,
R
M1
=
8314
Ru
=
M
16.04
=
518 J/(kg K)
=
V1
V1
=√
a1
γRT1
=
√
=
0.0538
25
25
=
464.2
1.3 × 518 × 320
From Fanno flow equations, we have
4f¯L∗1
D
=
=
=
1 − M12
(γ + 1) M12
γ+1
2
ln 1
+
2
2
γM1
2γ
2 1 + γ−1
2 M1
!
"
0.00666
0.997
+ 0.8846 ln
= 265.16 − 5.046
0.00376
2
260.114
The pipe length at which the flow chokes is
L∗
=
260.114 × 25 × 10−3
4 × 0.004
=
406.4 m
193
Again by Fanno flow relations, we have
'
* 12
p1
1
γ+1
2 2
1
=
p∗
M1 2 1 + γ−1
M1
2
=
1
0.0538
=
19.93
!
2.3
2
" 12
Thus,
p∗
T1
T∗
=
1 × 106
p1
=
19.93
19.93
=
50.18 kPa
=
γ+1
2.3
2=
1
2
2
2 1 + γ−1
M
1
2
= 1.15
Thus,
T∗
V∗
=
320
T1
=
1.15
1.15
=
278.16 K
=
a∗ =
=
/
γRT ∗ =
√
1.3 × 518 × 278.26
432.87 m/s
11.14 For argon, γ = 1.67. The given flow is a Fanno flow. From Fanno flow
table, for M1 = 0.6, we have
"
!
4f Lmax
= 0.4908
D
1
p1
p∗
=
1.76336
T1
T∗
=
1.1194
For the given duct,
4f L
4 × 0.02 × 1.1194
=
= 0.2984
D
0.3
194
Flow with Friction and Heat Transfer
Also,
4f L
D
=
=
!
4f Lmax
D
"
1
−
!
4f Lmax
D
"
2
0.2984
where subscript 2 refers to duct exit. From the above equation, we have
!
"
4f Lmax
= 0.4908 − 0.2984
D
2
For
!
4f Lmax
D
"
=
0.1924
= 0.1924, from Fanno flow table, we get
2
M2
=
0.73
p2
p∗
=
1.426
T2
T∗
=
1.084
Thus,
p2
T2
=
p2 p∗
1.426
× 90
p1 =
∗
p p1
1.76336
=
72.81 kPa
=
T2 T ∗
1.084
× 300
T1 =
T ∗ T1
1.1194
=
290.62 K
11.15
%L
=
D
4f
-!
4f l
D
"
M1
−
!
=
D
(14.533 − 0)
4f
=
50 × 10−3
× 14.533
4 × 0.006
=
30.277 m
4f l
D
"
M2
.
195
11.16
A =
π × 0.12
= 78.539 × 10−4 m2
4
ρ =
p
= 1.94 kg/m3
RT
Therefore, the inlet velocity becomes
V
=
ṁ
= 177.2 m/s
ρA
M
=
0.49
From Fanno flow table, for M = 0.49, we have
4f l
D
=
1.1539
p
p∗
=
2.1838
T
T∗
=
1.145
Thus,
L
p∗
T∗
=
1.1539 × 40.1
4 × 0.006
=
4.8 m
=
1.8 × 105
2.1838
=
82.4 kPa
=
323.15
= 282.23 K
1.145
=
9.08◦ C
At half way before the chocking location, L = 2.4 m. Thus,
4f L
4 × 0.006 × 2.4
=
= 0.576
D
0.1
From Fanno flow table, for
4f L
D
p
p∗
= 0.576, we get
=
1.8282
196
Flow with Friction and Heat Transfer
T
T∗
=
p
=
150.6 kPa
T
=
44.19◦ C
1.1244
11.17 (a) From Fanno flow table, for M = 0.2, we have
4f Lmax
= 14.533
D
where Lmax the distance from the Mach 0.2 location at which the Mach number
becomes unity. Therefore,
Lmax
=
14.533 × 0.05
4 × 0.00375
=
48.44 m
(b) This problem has to be solved by finding the chocking location for initial
Mach numbers 0.2 and 0.6.
LM =0.6 = (Lmax )M =0.2 − (Lmax )M =0.6
From Fanno flow table, for M = 0.6, we have
4f Lmax
D
=
0.49082
Lmax
=
0.49082 × 0.05
= 1.64 m
4 × 0.00375
Thus,
LM =0.6 = 48.44 − 1.64 = 46.8 m
11.18 (a) Given, T0 = 380 K. By energy equation, we have
h0 = h1 +
V12
2
where h0 is the stagnation enthalpy and h1 and V1 are the static enthalpy and
velocity, respectively, at the duct entrance. Treating air to be a perfect gas, we
197
have, h0 = cp T0 and h1 = cp T1 . Therefore, the energy equation becomes,
T0
=
T1 +
V12
2 cp
T1
=
T0 −
V12
302
= 380 −
2 cp
2 × 1004.5
=
379.6 K
since cp = 1004.5 J/(kg K) for air. The speed of sound is
/
a1 = γ R T1 = 390.54 im/s
M1 =
V1
30
≈ 0.08
=
a1
390.54
From Fanno flow table, for M1 ≈ 0.08, we have
!
"
4f L
= 106.72
D
1
Thus,
D
=
4 × 0.02 × 55
m
106.72
=
4.12 cm
(b) The inlet velocity V1 = 90 m/s. Therefore,
T1
=
380 −
a1
=
/
M1
=
902
= 376 K
2 × 1004.5
γ R T1 = 388.7 m/s
90
= 0.23
388.7
From Fanno flow table, for M1 = 0.23, we have
4f L
D
=
10.416
D
=
4 × 0.02 × 55
1046
=
0.422 m
198
Flow with Friction and Heat Transfer
(c) The inlet velocity V1 = 425 m/s. Therefore,
T1
=
380 −
a1
=
/
4252
= 290.1 K
2 × 1004.5
γ R T1 = 341.4 m/s
425
= 1.24
341.4
From Fanno flow table, for M1 = 1.24, we have
M1
=
4f L
D
=
0.04547
D
=
4 × 0.02 × 55
0.04547
=
96.77 m
Note: For the supersonic flow at the duct entrance, the diameter comes out to
be more than the length, for the present data.
11.19 The hydraulic diameter of the duct is
Dh
=
4 × 0.03 × 0.03
4 × cross-sectional area
=
perimeter
4 × 0.03
=
0.03 m
At the duct entrance the flow velocity is V1 = 1000 m/s. The local speed of
sound is
/
√
a1 =
γRT1 = 1.4 × 287 × 350
= 375 m/s
Thus, the Mach number at duct entrance is
M1
=
V1
1000
= 2.67
==
a1
375
For M1 = 2.67, from Fanno flow table, we have
!
"
4f Lmax
= 0.46619
D
1
Therefore, the duct length required to decelerate the flow to Mach 1.0 is
Lmax
=
0.46619 × 0.03
4 × 0.0025
=
1.40 m
199
11.20 For mass flow rate ṁ to be maximum, the exit Mach number M2 should
be unity, i.e. the flow is choked. For the given duct,
4f L
D
=
4 × 0.023 × 0.25
0.025
=
0.92
4f L
D ,
For this value of
if the flow at the exit has to choke, from Fanno flow table,
we
have
M1 = 0.52. From isentropic table, for M1 = 0.52, we have
p1
= 0.83165
p01
T1
T01
=
0.94869
p1
=
0.83165 × 50 × 101325 = 4.21 MPa
T1
=
0.94869 × 320 = 303.58 K
Therefore,
ṁmax
=
ρ1 A1 V1 = ρ1 A1 M1 a1
=
48.32 ×
=
/
22
π1
25 × 10−3 × 0.52 × γRT1
4
√
0.012334 1.4 × 285 × 303.58
=
4.3 kg/s
Further,
p∗
p1
=
1
= 0.487
2.0519
Thus,
p∗
=
0.487 × 4.21 = 2.05 MPa
Therefore, the mass flow rate will remain maximum for the back pressure range
0 < pb < 2.05 MPa
11.21 (a) Let subscripts 1 and 2 refer to duct entry and exit conditions. The
duct length ∆L required to accelerate the from Mach 0.2 to 0.5 can be determined from
! ∗"
! ∗"
fL
fL
f ∆L
=
−
D
D M =0.2
D M =0.5
200
Flow with Friction and Heat Transfer
where f is friction factor, D is duct diameter and M is Mach number. From
Fanno flow table, we have
! ∗"
fL
= 14.533
D M =0.2
! ∗"
fL
= 1.0691
D M =0.5
Thus,
0.025∆L
30 × 10−3
∆L
=
=
14.533 − 1.0691
16.157 m
(b) From Fanno flow table, we have
! ∗"
fL
D M =1
≈
0
Thus,
f ∆L
D
=
14.533
∆L
=
14.533 × 30 × 10−3
0.025
=
17.44 m
11.22 For the given pipe, we have
4f Le
De
=
4 × 0.02 × 18
5 × 10−2
=
28.8
For 4fDLe e = 28.8, from Fanno flow table, Me = 0.15. The speed of sound at pipe
exit is
√
ae = 1.4 × 287 × 468 = 433.74 m/s
The exit velocity and density are
Ve
=
0.15 × 433.74 = 65.062 m/s
ρe
=
pe
101325
= 0.754 kg/m3
=
RTe
287 × 468
201
Therefore,
ṁ
=
=
ρe Ae Ve = 0.754 ×
π × 0.052
× 65.062
4
0.096 kg/s
11.23 Let subscripts 1 and 2 refer to conditions at inlet and exit of the tube,
respectively. Given, pe = 0 Pa, T01 = 300 K and p01 = 6 × 101325 Pa, since 1
atm = 101325 Pa.
2
is much lower than the choking limit of 0.48667(for γ =
(a) L = 0 m and pp01
1.67), therefore, the flow is choked. Thus, the mass flow rate is
22
1
0.7266 × 6 × 101325 π 30 × 10−2
√
ṁ =
4
RT01
For argon, molecular weight is 39.944, gas constant R is 208 J/(kg K) and
γ = 1.67. Therefore, the mass flow rate becomes
ṁ = 125 kg/s
(b) L = 2.22 m. Therefore,
4f Lmax
D
=
4 × 0.005 × 2.22
30 × 10−2
=
0.148
From Fanno flow table, for γ = 1.67 and
4f Lmax
D
= 0.148, we have M1 = 0.71.
Now, from isentropic table, for M1 = 0.71(γ = 1.67), we have
p1
= 0.67778
p01
T1
T01
=
0.85552
p1
=
6 × 101325 × 0.67778 = 412056.35 Pa
T1
=
300 × 0.85552 = 256.66 K
The density at the inlet is
ρ1
=
p1
412056.35
=
RT1
208 × 256.66
=
7.72 kg/m3
202
Flow with Friction and Heat Transfer
Thus, the mass flow rate is
ṁ
=
ρ1 A1 V1 = ρ1 A1 M1 a1
=
7.72 ×
=
√
π0.32
× 0.71 1.67 × 208 × 256.66
4
115.68 kg/s
11.24 Given, p1 /p2 = 15 and p1 = 150 atm. Therefore,
p2 =
150
= 10 atm
15
where p1 is the storage tank static pressure and p2 is the settling chamber static
pressure. Let p2 be the pressure at which the flow chokes, i.e. p2 = p∗ . Thus,
p1
= 15
p∗
From Fanno flow table, for p1 /p∗ = 15, we have
4f L∗
D
=
140.66
Thus,
L∗
=
140.66 × 0.1
4 × 0.005
=
703.3 m
That is, the pipe will choke at a length of 703.3 m.
11.25 For the given piping system,
p01
600
= 13.33
=
p4
45
This pressure ratio is much more than the pressure ratio required for the flow
to choke. Therefore, at the exit, the Mach number M4 can be taken as unity.
Also,
!
"
4f L
4 × 0.013 × 1.1
=
D B
3.10 × 10−2
=
1.845
203
For this value, from Fanno Table, we get M3 ≈ 0.43. From isentropic table, for
M3 = 0.43, we have
A3
A∗
=
1.5
Also,
A2
A3
=
!
A2
A2 ∗
=
A2 A3
,
A3 A3 ∗
"2
6
3.1
= 3.75
since A2 ∗ = A3 ∗
Thus,
A2
A2 ∗
3.75 × 1.5 = 5.625
=
From isentropic table, for AA22∗ = 5.625, we have M2 ≈ 0.1 and from from Fanno
flow table, for M2 = 0.1, we have
4f L
= 66.922
D
For tube A,
!
4f L
D
"
=
4 × 0.015 × 0.9
6 × 10−2
=
0.9
A
Also,
!
4f L
D
"
=
1−2
!
4f L
D
"
1
−
!
4f L
D
"
2
Therefore,
!
4f L
D
"
=
1
!
4f L
D
"
+
1−2
!
4f L
D
=
"
= 0.9 + 66.922
2
67.822
%
&
From Fanno flow table, for this 4fDL = 67.822, we have M1 ≈ 0.1. From
isentropic table, for M1 = 0.1, we get
ρ1
ρ01
=
0.99502
T1
T01
=
0.998
204
Flow with Friction and Heat Transfer
The stagnation density is
ρ0
=
p0
600 × 103
=
RT0
287 × 390
=
5.36 kg/m3
Thus,
ρ1
=
0.99502 × 5.36 = 5.33 kg/m3
T1
=
0.998 × 390 = 389.22 K
a1
=
/
V1
=
M1 a1 = 39.55 m/s
A1
=
π × 0.062
= 28.27 × 10−4 m2
4
γRT = 395.5 m/s
Thus, the mass flow rate is
ṁ
=
=
ρ1 A1 V1 = 5.33 × 28.27 × 10−4 × 39.55
0.596 kg/s
Chapter 12
MOC
205
206
MOC
Chapter 13
Measurements in
Compressible Flow
13.1 (a)
p
= (760 − 500) = 260 mm of Hg
p0
= (760 − 350) = 410 mm of Hg
p
p0
=
260
= 0.634
410
From isentropic table the corresponding Mach number is M = 0.835
(b)
p0
=
760 + 275 = 1035 mm of Hg
p
p0
=
260
= 0.251
1035
From isentropic table the corresponding Mach number is M = 1.56
13.2
pgauge
=
3.6 × 104 Pa
patm
=
0.756 × 9.81 × 13.6 × 103
=
1.0086 × 105 Pa
207
208
Measurements in Compressible Flow
=
50 cm of Hg = 0.5 × 9.81 × 13.6 × 103
=
6.67 × 104 Pa
T0
=
300 K
p0,gauge
=
1.027 × 105 Pa
p0abs
=
(1.027 + 1.0086) × 105 Pa
p0 − p
Therefore,
ρ0 =
p0
RT0
=
2.0356 × 105
287 × 300
=
2.36 kg/m3
(a) By compressible Bernoulli equation, we have
γ p
u2
+
2
γ−1ρ
=
γ p0
γ − 1 ρ0
=
!
p
p0
=
!
1.3686 × 105
2.0356 × 105
ρ
=
0.753 × 2.36 = 1.78 kg/m3
u2
2
=
3.5
u
=
ρ
ρ0
" γ1
!
1
" 1.4
= 0.753
2.0356 1.3686
−
2.36
1.78
"
× 105
256.06 m/s
(b) For incompressible flow, ρ = ρ0 . Therefore,
u =
=
4
2 (p0 − p)
=
ρ
237.55 m/s
3
2 × 6.67 × 104
2.364
209
Note: The error committed in assuming the flow to be incompressible in this
problem is 7.23%.
13.3 From isentropic table, for M1 = 0.9, we have
p1
= 0.5913
p0
For M2 = 0.2, again from isentropic table, we have
p2
= 0.9725
p0
Therefore,
p2
p1
=
0.9725
= 1.6448
0.5913
p2
=
1.6447 × 4.15 × 105 = 6.826 × 105 Pa
Therefore,
p2 − p1 = (6.826 − 4.15) × 105 = 2.676 × 105 Pa
13.4 (a) T = 500◦ C = 773 K. Therefore,
√
a =
1.4 × 287 × 773 = 557.3 m/s
M
=
400
V
=
= 0.718
a
557.3
From isentropic table, for M = 0.718, we have
p0
=
=
p0
p
= 1.4098. Therefore,
1.4098 × 1.01325 × 105
1.428 × 105 Pa
(b)
T
=
−50◦ C = 273 − 50 = 223 K
a
=
√
M
=
p0
=
p0
=
1.4 × 287 × 223 = 299.33 m/s
400
= 1.336
299.33
<3.5
;
2
p 1 + 0.2 (1.336)
2.949 × 105 Pa
210
Measurements in Compressible Flow
Note: Note the time saving in using tables instead of actual relations.
13.5 (a) From Standard atmospheric table, at 10, 000 m altitude, T∞ = 223.15 K.
The speed of sound is given by,
√
a∞ = 1.4 × 287 × 223.15 = 299.44 m/s
The aeroplane velocity is
V∞ =
900
= 250 m/s
3.6
The flight Mach number is
M∞ =
V∞
= 0.835
a∞
From isentropic table, for M∞ = 0.835, we get
T0
= 1.139
T∞
Thus, the temperature at the stagnation region is
T0 = 1.139 × 223.15 = 254.17 K
(b) The temperature caused by impact is given by,
∆T = T0 − T = 254.17 − 223.15 = 31.02 K
13.6 The test section Mach number is M = 4. At 1650 m, from standard
atmospheric table, ρ0 = 1.0425 kg/m3 . From isentropic table, for M = 4, we
have
ρ
= 0.0277
ρ0
Therefore, the test-section density is
ρts = 0.0277 × 1.0425 = 0.0289 kg/m3
13.7 The probe measures the stagnation temperature T0 as 100◦ C. That is,
T0 = 100 + 273.15 = 373.15 K
By energy equation, we have
h0
=
h+
V2
2
211
c p T0
=
cp T +
T0
=
T+
T
=
T0 −
V2
2
V2
2 cp
V2
2 cp
where T is the actual temperature(static temperature) of the air.
For air cp = 1004.5 J/(kg K). Therefore,
T
2502
= 373.15 − 31.11
2 × 1004.5
=
373.15 −
=
342.04 K = 68.9◦ C
13.8 Let the subscripts ‘TS’, ‘i’ and ‘0’ refer to the test-section, nozzle inlet and
stagnation state, respectively.
Given MTS = 2.5 and pTS = 100 kPa. From isentropic table, for MTS = 2.5,
pTS
ATS
= 0.0585,
= 2.637
p0
Ath
where Ath is the nozzle throat area.
Therefore,
Ai
Ath
p0
ATS
Ath
=
2
=
2 × 2.637
=
5.274
=
100
0.0585
=
1.71 MPa
For Ai /Ath = 5.274, from isentropic table (subsonic solution),
Mi ≈ 0.11 ,
pi
= 0.9916
p0
212
Measurements in Compressible Flow
Thus, the pressure at the nozzle inlet is
pi
=
0.9916 × 1.71
=
1.696 MPa