LECTURE 15
LECTURE 15
NORMAL SHOCK WAVE IN THE
NORMAL SHOCK WAVE IN THE
CLAPEYRON GAS
CLAPEYRON GAS
NORMAL SHOCK WAVE. HUGONIOT ADIABAT.
NORMAL SHOCK WAVE. HUGONIOT ADIABAT.
Gas dynamics admitts existence of strongly discontinuous
flow. The normal shock wave is the simplest example of
S0
such flow
S2
S1
ś�W� =� S0 �� S1 �� S2
Conservation laws for the NSW
(1) Mass r� vn ds =� 0 �� r�1u1 =� r�2u2
{�
��
1D case
ś�W�
2 2
(2) Linear momentum (r� vnv +� pn)ds =� 0 �� r�1u1 +� p1 =� r�2u2 +� p2
{�
��
1D case
ś�W�
k� p1 k� p2
2 2
1 1
(3) Energy +� u1 =� +� u2
(k� -�1)r�1 2 (k� -�1)r�2 2
Devide (2) by (1) &
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p1 p2 p2 p1
u1 +� =� u2 +� �� u1 -� u2 =� -� (4)
r�1u1 r�2u2 r�2u2 r�1u1
ć�
2k� p2 p1 ��
New form of (3) is (u1 -� u2)(u1 +� u2 ) =� -� (5)
k� -�1�� r�2 r�1 ��
Ł� ł�
Using (4) the Eq. (5) can be rewritten as
ć� p2 p1 �� 2k� p2 p1 ��
ć�
-� +� u2) =� -� (6)
��
r�2u2 r�1u1 ��(u1 k� -�1�� r�2 r�1 ��
Ł� ł� Ł� ł�
The LHS of (6) can be transformed using the mass conservation equation (1) &
p2 p2 p1 p1
LHS(6) =� u1 +� u2 -� u1 -� u2 =�
r�2u2 r�2 u2 r�1 u1 r�1u1
{� {�
=�r�1 u1 =�r�2 u2
p2 p2 p1 p1
=� +� -� -�
r�1 r�2 r�1 r�2
ć�
p2 p2 p1 p1 2k� p2 p1 ��
Thus, we get from (6) +� -� -� =� -� (7)
r�1 r�2 r�1 r�2 k� -�1�� r�2 r�1 ��
Ł� ł�
We multiply (7) by r�1 p1 and get the formula which involves only the ratios r�1 r�2 and
p2 p1 & .
r�1 k� +�1 r�2
-�1
p2 k� +�1 -�
r�2 r�1
k� -�1 k� -�1
=� =� (8)
r�2
k� +�1
p1 k� +�1 r�1
-�1 -�
r�2 r�1
k� -�1 k� -�1
We have obtained the formula describing the thermodynamic process which affects the
gas passing through the NSW. Note that it is different that the isentropic process! We
call the above formula the Hugoniot Adiabat (HA).
r�2 k� +�1
Let us analyse some properties of the HA. Note that for =� >� 1 we have the vertical
r�1
k� -�1
r�2
asymptot. If k� =� 1.4 then corrsponding ratio =� 6. Thus, at the shock wave the density of
r�1
gas always increases but never more than k� +�1 times.
k� -�1
k� +�1
For brevity we introduce y =� p2 p1 , x =� r�2 r�1 and a� =� . The formula (8) can be now
k� -�1
written as
a� x -�1
y(x) =�
a� -� x
Then, we have
(a� +�1)(a� -�1) a� +�1
ó� ó�
y (x) =� �� y (1) =� =� k�
(a� -� x)2 a� -�1
Note that in case of the isentropic process, we have y(x) =� xk� , so yó�(x) =� k� xk� -�1 and
yó�(1) =� k� . Moreover
2(a� +�1)(a� -�1)
ó�ó� ó�ó�
yHugoniot (x) =� �� yHugoniot (1) =� k� (k� -�1)
(a� -� x)3
ó�ó� ó�ó�
yisentropic(x) =� k� (k� -�1) xk� -�2 �� yisentropic (1) =� k� (k� -�1)
We see that the isentropic and normal shock wave adiabats fit very well to each other in
ó� ó�ó�
the vicinity x =� 1 (they have the same values of y(1), y (1) and y (1)). We say that these
lines are strictly tangent at x =� 1.
Rankin-Hugoniot and Poisson adiabats (k� = 1.4)
30
Physically it means that weak shock waves
Normal shock wave (Rankin-Hugoniot)
are nearly isentropic and
25
3
20
p2 p2 ć� r�2 ��
-� =� C -�1�� +� h.o.t
��
15
p1 Hugoniot p1 isentropic r�1 ł�
Ł�
.
10
5
Isentropic flow (Poisson)
0
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
r�2�/�r�1�
2
1
p /p
ENTROPY AND GASODYNAMIC RELATION ON THE NORMAL SHOCK.
ENTROPY AND GASODYNAMIC RELATION ON THE NORMAL SHOCK.
From the 1st Principle of Thermodynamics we have
dq 1 dp dT dp dT dp
ds =� =� (di -� ) =� cp -� =� cp -� R
T T r� T T r� T p
From the Clapeyron equation & .
p =� r� RT �� ln p =� ln r� +� lnT +� const
dp dr� dT dT dp dr�
Thus =� +� �� =� -�
p r� T T p r�
and the differential of (mass specific) entropy can be written as
ć� dp d r� �� dp dp d r�
ds =� cp �� -� -� R =� cv -� cp
��
p r� p p r�
Ł� ł�
After integration we get s =� cv ln p -� cp ln r� +� const =� cv ln( p r�k� ) +� const
{�
=�k�cv
Thus, the change of entropy between two thermodynamic states can be expressed as
follows
k�
D� s s2 -� s1 ć� �� ć� �� ć� �� ć� ��
p2 p1 p2 r�2
�� =� ln�� k� �� -� ln�� k� �� =� ln�� �� -� ln�� k� ��
cv cv r�2 r�1 p1 r�1
Ł� ł� Ł� ł� Ł� ł� Ł� ł�
Note that for the Hugoniot adiabat we have
k�
r�2 p2 ć� ��
r�2
<� 1 �� <� �� D� s <� 0
r�1 p1 Hugoniot �� r�1 ��
Ł� ł�
i.e., the shock wave cannot expand the gas (it would lead to entropy decrease which
contradics the 2nd Principle of Thermodynamics). Thus the ( nonrivial) shock wave
must be a compression wave! Indeed, in such case
k�
ć� ��
p2 r�2
>� �� D� s >� 0
p1 �� r�1 ��
Ł� ł�
-�1
r�2 ć� ��
r�2
We further observe that >� 1 and r�1u1 =� r�2u2 �� u2 =� u1 <� u1
�� ��
r�1 r�1
shock
Ł� ł�
wave
Using the energy equation (integral) we also conclude that
2 2
T2 -� T1 =� (u1 -� u2 ) / 2cp >� 0 �� T2 >� T1
i.e., after crossing the shock wave the gas warms up.
We would like to know what kind of flow exists at different sides of the NSW.
Writing the energy equation in the following form
k� p ć� k� p �� k� +�1
2
1 1
u2 +� =� u2 +� �� a*�
2 2
�� ��
(k� -�1)r� (k� -�1)r�
Ł� ł�M =�1 2(k� -�1)
we obtain the following
2
k� p1 k� +�1 a*�
1
u1 +� =�
2
(k� -�1)u1r�1 2(k� -�1) u1
2
k� p2 k� +�1 a*�
1
u2 +� =�
2
(k� -�1)u2r�2 2(k� -�1) u2
Subtracting the above equations we get
2 2
ć� ć�
2k� p1 p2 �� k� +�1 a*� a*� ��
u1 -� u2 +� -� =� -�
k� -�1�� r�1u r�2u2 �� k� -�1�� u1 u2 ��
Ł� Ł� ł�
1�4�1 3�
4�2�4�4�ł�
=� u2-�u1(Eq.4)
After some algebra we derive the Prandtl s Relation
u2 -� u1 2
2
u2 -� u1 =� a*� and u2 ą� u1 �� u1u2 =� a*�
u1u2
The immediate conclusion from PR is that u1 >� a*� and u2 <� a*�
What about the Mach number?
a2 k� +�1
2
1
The energy equation u2 +� =� a*�
2
k� -�1 2(k� -�1)
can be divided by the square of velocity to obtain
2
2 1 k� +�1 a*�
1+� =�
2
k� -�1 M k� -�1 u2
2
ć� �� k� +�1
u
After simple manipulations we have �� �� =�
2
a*�
Ł� ł�
k� -�1+�
2
M
From the above the following we infer that
u1 >� a*� �� M1 >� 1 and u2 <� a*� �� M2 <� 1
Thus, the flow in front of the NSW is always supersonic, while the flow behind it 
always subsonic.
u
The quantity l� =� is called the velocity coefficient. In contrast to the Mach number, the
a*�
velocity coefficient assumes values in the bounded interval, namely
k� +�1 k� -�1
lim l�1 =� , lim l�2 =�
M1��Ą� M1��Ą�
k� -�1 k� +�1
The Mach number of the flow behind the wave M2 can be expressed as the function of the
Mach number in front of the wave M1. To obtain this formula we use the Prandtl s
Relation &
u1 u2 k� +�1 k� +�1
=� 1 �� =� 1
a*� a*�
2 2
k� -�1+� k� -�1+�
2 2
M1 M2
After some algebra we get
2
2 +� (k� -�1)M1
M2 =� <� 1
2
2k� M1 -�k� +�1
Similarly, we can expressed the ratios of density, pressure and temperature values.
The density ratio can be evaluated as follows
-�1
r�2 u1 M1 a1 M1 ć� �� ć� ��
a a
(M1) =� (M1) =� (M1) =� (M1)�� �� [M2(M1)] >� 1
��
r�1 u2 M2(M1) a2 M2(M1) a0 �� a0
Ł� ł�is Ł� ł�is
To evaluate the pressure ratio (as the function of M1) we rewrite the momentum equation
in the following way
ć� u2 �� ć� k�u2 ��
2
p +� r�u2 =� p��1+� =� p��1+� =� p(1+�k� M ) =� const
��
p r� a2 ��
Ł� ł� Ł� ł�
Since the above expression has the same value at both sides of the shock wave, we get
2
p2 1+�k� M1
(M1) =� >� 1
2
p1 1+�k� M2 (M1)
The flow through the shock is adiabatic (total energy is conserved) and the total (or
stagnation) temperature T0 remains the same.
Thus, we can write
T T0 (M2 )
T2 (� )�
(M1) =� >� 1
T1 T T0 [M2(M1)]
(� )�
where the ratio T/T0 can be calculated from the formula derived in the Lecture 14
-�1
T
ć�1+� k� -�1 M ��
2
(M ) =�
�� ��
T0 2
Ł� ł�
NORMAL SHOCK WAVE (k�=�1�.�4�)� NORMAL SHOCK WAVE (k�=�1�.�4�)�
NORMAL SHOCK WAVE (k�=�1�.�4�)�
1
15 4.5
14
0.9
13 4
0.8
12
11 3.5
0.7
r�2�/r�1�
10
M2
0.6
9 3
8
0.5
7 2.5
0.4
6
T2/T1
5 2
0.3
4
1 2 3 4 5 6 7 8
3 1.5
M1
2
1 1
1 1.5 2 2.5 3 3.5 1 1.5 2 2.5 3 3.5
M1
M1
2
1
p / p
In the end, we will analyse what happens to the stagnation pressure. Conceptually, we
consider the process described as follows
p01 �� p1 �� p2 �� p02 <� p01
{� {� {�
acceleration shock deceleration
(isentropic) wave (isentropic)
We claim that the total (stagnation) pressure diminishes on the shock wave.
The justification of this fact goes as follows.
NORMAL SHOCK WAVE (k�=�1�.�4�)�
1
We know that the entropy of the gas increases on the
0.9
shock wave. The formula derived earlier can be written
for stagnation parameters, namely
0.8
s2 -� s1
0 <� =� ln p02 p01 -�k� ln r�02 r�01
(� )� (� )�
0.7
cv
Since the total temperature at both sides is the same
0.6
then
0.5
r�02 p02
0.4
T01 =� T02 �� T0 �� =�
Clapeyron
r�01 p01
Equation
0.3
D� s
Thus 0 <� =� (1-�k� )ln p02 p01 �� p02 p01 <� 1.
(� )�
0.2
cv
1 1.5 2 2.5 3 3.5
M1
02
01
p /p