1. CHAPTER THREE NORMAL SHOCK WAVES
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
CHAPTER THREE
NORMAL SHOCK WAVES
We have seen that sound waves are caused by infinitesimally small
pressure disturbances, and they travel through a medium at the speed of
sound. We have also seen that for some back pressure values, abrupt
changes in fluid properties occur in a very thin section of a converging–
diverging nozzle under supersonic flow conditions, creating a shock wave.
It is of interest to study the conditions under which shock waves develop
and how they affect the flow.
3-1 Standing (stationary) normal shock waves
First we consider shock waves that occur in a plane normal to the
direction of flow, called normal shock waves. The flow process through the
shock wave is highly irreversible and cannot be approximated as being
isentropic.
Figure 3.1 Schlieren image of a normal shock in a Laval nozzle. The Mach number in the nozzle just upstream (to the left) of the shock wave is about 1.3. Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the shock.

2. Figure 3.3 The h-s diagram for flow across a normal shock
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Figure 3.2 Variation of flow properties across a normal shock.
Shockwave is an abrupt change over a very thin section of flow in
which the flow transitions from supersonic to subsonic flow. This abrupt
change in the flow causes a sudden drop in velocity to subsonic levels and a
sudden increase in pressure. Flow through the shock is highly irreversible;
and, thus, it cannot be approximated as isentropic.
Figure 3.3 The h-s diagram for flow across a normal shock

3.  Neglect potential , i.e., dz=O  Constant area , i.e., A1=A2
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Figure 3.4 shows a standing normal shock in a section of varying
area. We first establish a control volume that includes the shock region and
an infinitesimal amount of fluid on each side of the shock. In this manner
we deal only with the changes that occur across the shock. It is important
to recognize that since the shock wave is so thin (about 1O−6 m), a control
volume chosen in the manner described above is extremely thin in the x-
direction. This permits the following simplifications to be made without
introducing error in the analysis:
1.The shock wave is perpendicular to the flow streamlines.
2. The shock is so thin that it can be considered to take place at a
constant area.
3. The area on both sides of the shock may be considered to be the
same.
4. There is negligible surface in contact with the wall, and thus
frictional effects may be omitted.
We begin by applying the basic concepts of continuity, energy, and
momentum under the following assumptions:
 Steady one - dimensional flow
 Adiabatic, i.e., q=O
 No shaft work, i.e., ws
 Neglect potential , i.e., dz=O
 Constant area , i.e., A1=A2
Continuity
m = pvA (3.1)
p1 v1 A1 = p2 v2 A (3.2)
But area is constant, A1=A2
p1 v1 = p2 v (3.3)

4. Figure 3.4 Control volume for shock analysis.
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Figure 3.4 Control volume for shock analysis.
And in terms of Mach number:
p1 M1 = p2 M (3.4)
Energy
h01 + q = h02 + ws (3.5)
For adiabatic and no work, this becomes
h01= h (3.6) or
h = h (3.7)
Also
T01= T (3.8) k 1 2 V1 V2
From stagnation relations TO = T (1 + ( ) M ) , so: 2

5. / Fx = m(Vout,x - Vin,x ) = m(V2 - V1) (3.11)
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
k k k
T (1 + (k - 1) M2 )
Momentum
The x-component of the momentum equation for steady one-dimensional
flow is;
/ Fx = m(Vout,x - Vin,x ) = m(V2 - V1) (3.11)
With pressure force the only external forces acting on
the control volume, then
/ Fx = p1A1 - p2A2 = (p1 - p2 )A (3.12)
Thus the momentum equation in the direction of flow becomes
(p1 - p2)A = m(V2 - V1) (3.13)
Canceling the area and pv can be written as eitherp1 v1 or p2 v2 , then
p p 2 T
, M1 , = T
, M2,
(3.9) 2 2 T2
(1 + ( 2 ) M1 ) =
(3.10) 2 2
p1 + p1v1 =
p2 + p2v2
(3.14) p1 + RT
M1 kRT1 =
p2 + RT M2 kRT2
(3.15)
p1 (1 + kM1 ) =
p2 (1 + kM2 )
(3.16)
p (1 + kM1 ) =
(3.17) p1
(1 + kM2 )

6. We proceed to combine these equations above and derive an
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
In summary
p1M1 = <T (3.18) 2 k
T (1 + (k - 1) M2)
The above equations are the principle equations for a standing
normal shock, in addition to the foregoing assumptions. They called the
jump conditions and must be satisfied to preserve conservation of mass,
momentum and energy across the shock.
We proceed to combine these equations above and derive an
expression for M2 in terms of the information given.
Substitue eqs (3.18) and (3.19) into eq (3.2O) gives; k
= ; (3.21)
At this point notice that M2 is a function of only M1 and k . A trivial
solution of this is seen to be M1 = M2, which represents the degenerate case
p2M T2
p2 = (1 + kM1 ) (3.19)
T2 = (1 + ( 2 ) M1 ) (3.20)
p (1 + kM2 )
(1 + kM2 )M1
(1 + ( 2 ) M2 )
(1 + kM1 )M2
(1 + (k - 1
) M1 ) 2 2