Basics of Fluid Mechanics, 2014a

11.5. NORMAL SHOCK 409
Energy equation (11.72) can be converted to a dimensionless form which can be expressed
as
(
T y 1+ k − 1 ) (
2
M y = T x 1+ k − 1 )
2
M x (11.81)
2
2
It can also be observed that equation (11.81) means that the stagnation temperature
is the same, T 0y = T 0x . Under the perfect gas model, ρU 2 is identical to kPM 2
because
ρU 2 =
M 2
⎛{ }} ⎞{
ρ
{}}{
P
⎜
U 2
⎟
RT ⎝kRT⎠ kRT = kPM2 (11.82)
} {{ }
c 2
Using the identity (11.82) transforms the momentum equation (11.71) into
P x + kP x M x 2 = P y + kP y M y
2
(11.83)
Rearranging equation (11.83) yields
P y
= 1+kM x 2
P
2
(11.84)
x 1+kM y
The pressure ratio in equation (11.84) can be interpreted as the loss of the static
pressure. The loss of the total pressure ratio can be expressed by utilizing the relationship
between the pressure and total pressure (see equation (11.27)) as
(
P
P y 1+ k − 1 ) k
2 k − 1
M y
0y
2
=
P 0x (
P x 1+ k − 1 ) k
2 k − 1
M x
2
(11.85)
The relationship between M x and M y is needed to be solved from the above set of
equations. This relationship can be obtained from the combination of mass, momentum,
and energy equations. From equation (11.81) (energy) and equation (11.80) (mass)
the temperature ratio can be eliminated.
(
Py M y
P x M x
) 2
=
1+ k − 1 2
M x
2
1+ k − 1 2
M y
2
(11.86)