Compressible Flow Free Textbook

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Formal Model

Equations (5.1), (5.2), and (5.3) 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 (5.1) substituting for density, ρ, the equation of state yields

$\displaystyle {P_x \over R T_x } U_x = {P_y \over R T_y } U_y$

(5.8)

Squaring equation (5.8) results in

$\displaystyle { {P_x }^{2} \over R^{2} {T_x}^2} {U_x}^{2} = { {P_y }^{2} \over R^{2} {T_y}^2} {U_y}^{2}$

(5.9)

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, c²=kRT to be used for the Mach number definition, as follows:

$\displaystyle { {P_x }^{2} \over T_x \underbrace{k R {T_x}}_{{c_x}^2} } {U_x}^{2} = { {P_y }^{2} \over T_y \underbrace{k R {T_y}}_{{c_y}^2} } {U_y}^{2}$

(5.10)

Note that the speed of sound on the different sides of the shock is different. Utilizing the definition of Mach number results in

$\displaystyle { {P_x}^{2} \over T_x } {M_x}^{2} = { {P_y}^{2} \over T_y } {M_y}^{2}$

(5.11)

Rearranging equation (5.11) results in

$\displaystyle {T_y \over T_x} = \left( { P_{y} \over P_{x}} \right)^{2} \left( {M_y \over M_x} \right)^{2}$

(5.12)

Energy equation (5.3) can be converted to a dimensionless form which can be expressed as

$\displaystyle T_y \left( 1 + {k-1 \over 2} {M_y}^{2} \right) = T_x \left( 1 + {k-1 \over 2} {M_x}^{2} \right)$

(5.13)

It can also be observed that equation (5.13) means that the stagnation temperature is the same, T_0y= T_0x . Under the perfect gas model, ρU² is identical to k P M² because

$\displaystyle \rho U^{2} = \overbrace{P \over RT}^{\rho} \overbrace{\left( {U^2 \over \underbrace{kRT}_{c^2}}\right)} ^{M^2} kRT = k P M {2}$

(5.14)

Using the identity (5.14) transforms the momentum equation (5.2) into

$\displaystyle P_x + k P_x {M_x}^{2} = P_y + k P_y {M_y}^{2}$

(5.15)

Rearranging equation (5.15) yields

$\displaystyle {P_y \over P_x} = {1 + k{M_{x}}^2 \over 1 + k{M_{y}}^2}$

(5.16)

The pressure ratio in equation (5.16) 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 (4.11)) as

$\displaystyle {{P_0}_y \over {P_0}_x} = { P_y \left( 1 + {k-1 \over 2} {M_y}^{2... ...ver k-1} \over P_x \left( 1 + {k-1 \over 2} {M_x}^{2} \right) ^ {k \over k-1} }$

(5.17)

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 (5.13) (energy) and equation (5.12) (mass) the temperature ratio can be eliminated.

$\displaystyle \left( {P_y M_y \over P_x M_x }\right)^{2} = { 1 + { k-1 \over 2} {M_x}^{2} \over 1 + { k-1 \over 2} {M_y}^{2}}$

(5.18)

Combining the results of (5.18) with equation (5.16) results in

$\displaystyle \left( {1 + k{M_{x}}^2 \over 1 + k{M_{y}}^2} \right)^{2} = \left(... ...}\right)^{2} { 1 + { k-1 \over 2} {M_x}^{2} \over 1 + { k-1 \over 2} {M_y}^{2}}$

(5.19)

Equation (5.19) 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

$\displaystyle M_y = M_x$

(5.20)

It can be observed that equation (5.19) is biquadratic. According to the Gauss Biquadratic Reciprocity Theorem this kind of equation has a real solution in a certain range^5.3which will be discussed later. The solution can be obtained by rewriting equation (5.19) as a polynomial (fourth order). It is also possible to cross-multiply equation (5.19) and divide it by (M_x² - M_y²) results in

$\displaystyle 1 + {k -1 \over 2} \left({M_{y}}^2+ {M_{y}}^2 \right) - k {M_{y}}^2 {M_{y}}^2 = 0$

(5.21)

Equation (5.21) becomes

$\displaystyle {M_y}^2 = { {M_x}^2 + {2 \over k -1} \over {2k \over k -1} {M_x}^2 - 1 }$

(5.22)

The first solution (5.20) 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_x ; P_x = P_x . 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, . Utilizing equation (5.16) and equation (5.22) provides the pressure ratio as only a function of the upstream Mach number as

$\displaystyle {P_y \over P_x} & = {2k \over k+1 } {M_x}^2 - {k -1 \over k+1}$

$\displaystyle {P_y \over P_x}$

$\displaystyle = 1 + { 2k \over k+1} \left({M_x}^2 -1 \right )$

(5.23)

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

$\displaystyle {\rho_y \over \rho_x} = {U_x \over U_y} = {( k +1) {M_x}^{2} \over 2 + (k -1) {M_x}^{2} }$

(5.24)

The fact that the pressure ratio is a function of the upstream Mach number,

, provides additional way of obtaining an additional useful relationship. And the temperature ratio, as a function of pressure ratio, is transformed into

$\displaystyle {T_y \over T_x} = \left( {P_y \over P_x} \right) \left( {k + 1 \over k -1 } + {P_y \over P_x} \over 1+ {k + 1 \over k -1 } {P_y \over P_x} \right)$

(5.25)

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

$\displaystyle {\rho_x \over \rho_y} = { 1 + \left( {k +1 \over k -1} \right) \l... ...P_x} \right) \over \left( k+1 \over k-1\right) +\left( {P_y \over P_x} \right)}$

(5.26)

which is associated with the shock wave.

**Figure:** `The exit Mach number and the stagnation pressure ratio as a function of upstream Mach number.`

Subsections

Next: The Maximum Conditions Up: Solution of the Governing Previous: Informal Model Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21