# 11.5.1: Solution of the Governing Equations

Equations (1), (2), and (3) can be converted into a dimensionless form. The reason that dimensionless forms are heavily used in this book is because by doing so it simplifies and clarifies the solution. It can also be noted that in many cases the dimensionless equations set is more easily solved. From the continuity equation (1) substituting for density, $$\rho$$, the equation of state yields

$\dfrac{P_x }{ R\, T_x } \,U_x = \dfrac{P_y }{ R\, T_y } \, U_y \label{shock:eq:continutyNonD} \tag{7}$
Squaring equation (7) results in
$\dfrac{ {P_x }^{2} }{ R^{2} \,{T_x}^2}\, {U_x}^{2} = \dfrac{ {P_y }^{2} }{ R^{2} \,{T_y}^2}\, {U_y}^{2} \label{shock:eq:massNonD0} \tag{8}$
Multiplying the two sides by the ratio of the specific heat, k, provides a way to obtain the speed of sound definition/equation for perfect gas, $$c^2 = k\,R\,T$$ to be used for the Mach number definition, as follows:

$\dfrac{ {P_x }^{2} }{ T_x \underbrace{k\, R\, {T_x}}_ ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/11:_Compressible_Flow_One_Dimensional/11.5_Normal_Shock/11.5.1:_Solution_of_the_Governing_Equations), /content/body/p[3]/span, line 1, column 4 _{{c_y}^2} } {U_y}^{2} \label{shock:eq:massNonD1} \tag{9}$

Note that the speed of sound is different on the sides of the shock. Utilizing the definition of Mach number results in $\dfrac{ {P_x}^{2} }{ T_x } {M_x}^{2} = \dfrac{ {P_y}^{2} }{ T_y } {M_y}^{2} \label{shock:eq:massNonD2} \tag{10}$   Rearranging equation (10) results in $\dfrac{T_y }{ T_x} = \left( \dfrac{ P_{y} }{ P_{x}} \right)^{2} \left( \dfrac{M_y }{ M_x} \right)^{2} \label{shock:eq:nonDimMass} \tag{11}$
Energy equation (3) can be converted to a dimensionless form which can be expressed as
$T_y \left( 1 + \dfrac{k-1 }{ 2}\, {M_y}^{2} \right) = T_x \left( 1 + \dfrac{k-1 }{ 2}\, {M_x}^{2} \right) \label{shock:eq:energyDless} \tag{12}$
It can also be observed that equation (12) means that the stagnation temperature is the same, $${T_0}_y = {T_0}_x$$. Under the perfect gas model, $$\rho\, U^{2}$$  is identical to $$k\, P\, M^{2}$$ because
$\rho U^{2} = \overbrace{P \over R\,T}^{\rho} \overbrace{\left( {U^2 \over \underbrace{k\,R\,T}_{c^2}}\right)} ^{M^2} k\,R\,T = k\, P\, M {2} \label{shock:eq:Rindenty} \tag{13}$
Using the identity (13) transforms the momentum equation (2) into
$P_x + k\, P_x\, {M_x}^{2} = P_y + k\, P_y\, {M_y}^{2} \label{shock:eq:Punarranged} \tag{14}$
Rearranging equation (14) yields
$\dfrac{P_y }{ P_x} = \dfrac{1 + k\,{M_{x}}^2 }{ 1 + k\,{M_{y}}^2} \label{gd:shock:eq:pressureRatio} \tag{15}$
The pressure ratio in equation (15) 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 (??)) as
$\dfrac ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/11:_Compressible_Flow_One_Dimensional/11.5_Normal_Shock/11.5.1:_Solution_of_the_Governing_Equations), /content/body/p[4]/span, line 1, column 4 \label{shock:eq:combineDimMassEnergy} \tag{17}$
Combining the results of (17) with equation (15) results in
$\left( \dfrac{1 + k\,{M_{x}}^2 }{ 1 + k\,{M_{y}}^2} \right)^{2} = \left( \dfrac{ M_x }{ M_y }\right)^{2} { 1 + \dfrac{ k-1 }{ 2} {M_x}^{2} \over 1 + \dfrac{ k-1 }{ 2} {M_y}^{2}} \label{shock:eq:toBeSolved} \tag{18}$
Equation (18) 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
$M_y = M_x \label{shock:eq:Msolution1} \tag{19}$
It can be observed that equation (18) is biquadratic. According to the Gauss Biquadratic Reciprocity Theorem this kind of equation has a real solution in a certain range which will be discussed later. The solution can be obtained by rewriting equation (18) as a polynomial (fourth order). It is also possible to cross–multiply equation (??) and divide it by $$\left({M_x}^2- {M_y}^2\right)$$ results in
$1 + \dfrac{k -1 }{ 2} \left({M_{y}}^2+ {M_{y}}^2 \right) - k \,{M_{y}}^2\, {M_{y}}^2 = 0 \label{shock:eq:generalSolution} \tag{20}$
Equation (20) becomes

Shock Solution

$\label{shock:eq:solution2} {M_y}^2 = \dfrac{ {M_x}^2 + \dfrac{2 }{ k -1} } {\dfrac{2\,k }{ k -1}\, {M_x}^2 - 1 } \tag{21}$

The first solution (19) 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 (15) and equation (21) provides the pressure ratio as only a function of the upstream Mach number as
\begin{align*}
{P_y \over P_x} = \dfrac{2\,k }{ k+1 } {M_x}^2 - \dfrac{k -1 }{ k+1} \qquad \text{or}
\end{align*}

Shock Pressure Ratio

$\label{shock:eq:pressureMx} \dfrac{P_y }{ P_x} = 1 + \dfrac{ 2\,k }{ k+1} \left({M_x}^2 -1 \right ) \tag{22}$

The density and upstream Mach number relationship can be obtained in the same fashion to became

Shock Density Ratio

$\label{shock:eq:densityMx} \dfrac{\rho_y }{ \rho_x} = \dfrac{U_x }{ U_y} = \dfrac{( k +1) {M_x}^{2} }{ 2 + (k -1) {M_x}^{2} } \tag{23}$

The fact that the pressure ratio is a function of the upstream Mach number, $$M_x$$, provides additional way of obtaining an additional useful relationship. And the temperature ratio, as a function of pressure ratio, is transformed into

Shock Temperature Ratio

$\label{shock:eq:temperaturePbar} \dfrac{T_y }{ T_x} = \left( \dfrac{P_y }{ P_x} \right) \left( \dfrac{\dfrac{k + 1 }{ k -1 } + \dfrac{P_y }{ P_x}} { 1+ \dfrac{k + 1 }{ k -1 } \dfrac{P_y }{ P_x}} \right) \tag{24}$

In the same way, the relationship between the density ratio and pressure ratio is

Shock $$P-\rho$$

$\label{shock:eq:densityPbar} \dfrac{\rho_x }{ \rho_y} = \dfrac{ 1 + \left( \dfrac{k +1 }{ k -1} \right) \left( \dfrac{P_y }{ P_x} \right) } { \left( \dfrac{k+1}{ k-1}\right) +\left( \dfrac{P_y }{ P_x} \right)} \tag{25}$

which is associated with the shock wave.

Fig. 11.12 The exit Mach number and the stagnation pressure ratio as a function of upstream Mach number.

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.