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Operating Equations and Analysis

In Figure (5.3), 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 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 (5.1) for 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.

This table was generated by Potto-GDC (in HTML)

Normal Shock		Input: Py/Px		k = 1.32
Mx	My	Ty/Tx	ρy/ρx	Py/Px	P0y/P0x
2.7245	0.476422	2.111	3.93596	8.30879	0.381089

**Figure:** `The ratios of the static properties of the two sides of the shock.`

To illustrate the use of the above equations, an example is provided.
$\begin{examl} Air flows with a Mach number of \begin{rawhtml} Mx=... ...f the shock. Assume that \begin{rawhtml} k=1.4\end{rawhtml}. \end{examl}$
Solution

Analysis:
First, the known information M_x=3, P_x=1.5[bar] and T=273 K. Using these data, the total pressure can be obtained (through an isentropic relationship Table (4.2), 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 downstream of the normal shock.

$\displaystyle {P_x \over P_{0x} } = 0.0272237 \Longrightarrow P_{0x} = 1.5/0.0272237 = 55.1 [bar]$

$\displaystyle c_x = \sqrt{k R T_x} = \sqrt {1.4 \times 287 \times 273} = 331.2 m/sec$

Normal Shock		Input: Mx		k = 1.4
Mx	My	Ty/Tx	ρy/ρx	Py/Px	P0y/P0x
3	0.475191	2.67901	3.85714	10.3333	0.328344

$\displaystyle U_x = M_x \times c_x = 3\times 331.2 = 993.6 [m/sec]$

Now the velocity downstream is determined by the inverse ratio of ρ_y = 993.6 / 3.85714 = 257.6 [m/sec].

U_y=993.6/3.85714=257.6[m/sec]

P_0y = ( P_0y / P_0x ) * P_0x = 0.32834 * 55.1 [bar] = 18.09 [bar]

Subsections

Next: The Limitations of the Up: Normal Shock Previous: Prandtl's Condition Index

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