Basics of Fluid Mechanics, 2014a

11.5. NORMAL SHOCK 413
or in a dimensionless form
11.5.3 Operating Equations and Analysis
M ∗ 1 M ∗ 2 = c ∗2 (11.102)
In Figure 11.12, the Mach number after the shock, M y , and the ratio of the total
pressure, P 0y /P 0x , are plotted as a function of the entrance Mach number. The working
equations were presented earlier. Note that the M y has a minimum value which depends
on the specific heat ratio. It can be noticed that the density ratio (velocity ratio) also
has a finite value regardless of the upstream Mach number.
The typical situations in which these equations can be used also include the
moving shocks. The equations should be used with the Mach number (upstream or
downstream) for a given pressure ratio or density ratio (velocity ratio). This kind of
equations requires examining Table (11.3) for k =1.4 or utilizing Potto-GDC for for
value of the specific heat ratio. Finding the Mach number for a pressure ratio of 8.30879
and k =1.32 and is only a few mouse clicks away from the following table.
To illustrate the use of the
above equations, an example is provided.
Example 11.12:
Air flows with a Mach number of
M x =3, at a pressure of 0.5 [bar] and
a temperature of 0 ◦ C goes through a
normal shock. Calculate the temperature,
pressure, total pressure, and velocity
downstream of the shock. Assume
that k =1.4.
Solution
Analysis:
First, the known information are
M x = 3, P x = 1.5[bar] and T x =
273K. Using these data, the total
120.0
110.0
100.0
90.0
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
Shock Wave relationship
P y
/P y
, ρ y
/ρ x
and T y
/T x
as a function of M x
P y
/P x
T y
/T x
ρ y
/ρ x
0.0
1 2 3 4 5 6 7 8 9 10
Fri Jun 18 15:48:25 2004
M x
Fig. -11.13. The ratios of the static properties of
the two sides of the shock.
pressure can be obtained (through an isentropic relationship in Table (11.2), i.e., P 0x
is known). Also with the temperature, T x , the velocity can readily be calculated. The
relationship that was calculated will be utilized to obtain the ratios for the downstream
P x
P 0x
of the normal shock. =0.0272237 =⇒ P 0x =1.5/0.0272237 = 55.1[bar]
c x = √ kRT x = √ 1.4 × 287 × 273 = 331.2m/sec
M x M y
T y
T x
ρ y
ρ x
P y
P x
P 0y
P 0x
3.0000 0.47519 2.6790 3.8571 10.3333 0.32834