Basics of Fluid Mechanics, 2014a

410 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
Combining the results of (11.86) with equation (11.84) results in
(
2
) 2
1+kMx
2
=
1+kM y
(
Mx
) 2 1+ k − 1 2
M x
2
M y
1+ k − 1 2
M y
2
(11.87)
Equation (11.87) is a symmetrical equation in the sense that if M y is substituted with
M x and M x substituted with M y the equation remains the same. Thus, one solution is
M y = M x (11.88)
It can be observed that equation (11.87) is biquadratic. According to the Gauss Biquadratic
Reciprocity Theorem this kind of equation has a real solution in a certain
range 7 which will be discussed later. The solution can be obtained by rewriting equation
(11.87) as a polynomial (fourth order). It is also possible to cross–multiply equation
(11.87) and divide it by ( M x 2 − M y
2 ) results in
1+ k − 1
2
Equation (11.89) becomes
(
My 2 + M y
2 ) − kM y 2 M y 2 =0 (11.89)
M y 2 =
Shock Solution
M x 2 + 2
k − 1
2 k
k − 1 M x 2 − 1
(11.90)
The first solution (11.88) is the trivial solution in which the two sides are identical and
no shock wave occurs. Clearly, in this case, the pressure and the temperature from
both sides of the nonexistent shock are the same, i.e. T x = T y ,P x = P y . The second
solution is where the shock wave occurs.
The pressure ratio between the two sides can now be as a function of only a
single Mach number, for example, M x . Utilizing equation (11.84) and equation (11.90)
provides the pressure ratio as only a function of the upstream Mach number as
P y
=
2 k
P x k +1 M x 2 − k − 1
k +1
or
Shock Pressure Ratio
P y
=1+
2 k ( 2 Mx − 1 )
P x k +1
(11.91)
7 Ireland, K. and Rosen, M. ”Cubic and Biquadratic Reciprocity.” Ch. 9 in A Classical Introduction
to Modern Number Theory, 2nd ed. New York: Springer-Verlag, pp. 108-137, 1990.