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Normal Shock in Variable Duct Areas

In the previous two chapters, the flow in a variable area duct and a normal shock (discontinuity) were discussed. A discussion of the occurrences of shock in flow in a variable is presented. As it is was presented before, the shock can occur only in steady state when there is a supersonic flow. but also in steady state cases when there is no supersonic flow (in stationary coordinates). As it was shown in Chapter 5, the gas has to pass through a converging-diverging nozzle to obtain a supersonic flow.

**Figure:** `The flow in the nozzle with different back pressures.`
$\begin{figure}\centerline{\includegraphics{cont/shockVariable/nozzle}} \end{figure}$

In the previous chapter, the flow in a convergent-divergent nuzzle was presented when the pressure ratio was above or below the special range. This Chapter will present the flow in this special range of pressure ratios. It is interesting to note that a normal shock must occur in these situations (pressure ratios).

In Figure (6.1) the reduced pressure distribution in the converging-diverging nozzle is shown in its whole range of pressure ratios. When the pressure ratio, $\PB$ is between point ``a'' and point ``b'' the flow is different from what was discussed before. In this case, no continuous pressure possibly can exists. Only in one point where $\PB=P_b$ continuous pressure exist. If the back pressure, $\PB$ is smaller than a discontinuous point (a shock) will occur. In conclusion, once the flow becomes supersonic, only exact geometry can achieve continuous pressure flow.

In the literature, some refer to a nozzle with an area ratio such point b as above the back pressure and it is referred to as an under-expanded nozzle. In the under-expanded case, the nozzle doesn't provide the maximum thrust possible. On the other hand, when the nozzle exit area is too large a shock will occur and other phenomenon such as plume will separate from the wall inside the nozzle. This nozzle is called an over-expanded nozzle. In comparison of nozzle performance for rocket and aviation, the over-expanded nozzle is worse than the under-expanded nozzle because the nozzle's large exit area results in extra drag.

The location of the shock is determined by geometry to achieve the right back pressure. Obviously if the back pressure, $\PB$ , is lower than the critical value (the only value that can achieve continuous pressure) a shock occurs outside of the nozzle. If the back pressure is within the range of to than the exact location determines that after the shock the subsonic branch will match the back pressure.

**Figure:** `A nozzle with normal shock`
$\begin{figure}\centerline{\includegraphics{cont/shockVariable/nozzleWithShock}} \end{figure}$

The first example is for academic reasons. It has to be recognized that the shock wave isn't easily visible (see Mach's photography techniques). Therefore, this example provides a demonstration of the calculations required for the location even if it isn't realistic. Nevertheless, this example will provide the fundamentals to explain the usage of the tools (equations and tables) that were developed so far.
$\begin{examl} A large tank with compressed air is attached into a converging-di... ... the back pressure (point \lq\lq \textbf{a}'' and point \lq\lq \textbf{b}''). \end{examl}$
Solution

Since the key word ``large tank'' was used that means that the stagnation temperature and pressure are known and equal to the conditions in the tank.

First, the exit Mach number has to be determined. This Mach number can be calculated by utilizing the isentropic relationship from the large tank to shock (point ``x''). Then the relationship developed for the shock can be utilized to calculated the Mach number after the shock, (point ``y''). From the Mach number after the shock, , the Mach number at the exit can be calculated utilizing the isentropic relationship.

It has to be realized that for a large tank, the inside conditions are essentially the stagnation conditions (this statement is said without a proof, but can be shown that the correction is negligible for a typical dimension ratio that is over 100. For example, in the case of ratio of 100 the Mach number is 0.00587 and the error is less than %0.1). Thus, the stagnation temperature and pressure are known and . The star area (the throat area), $A^{*}$ , before the shock is known and given as well.

$\displaystyle {A_{x} \over A^{*}} = {6 \over 3} = 2$

With this ratio $(A/A^{*}= 2)$ utilizing the Table (5.1) or equation (4.48) or the GDC-Potto, the Mach number, is about 2.197 as shown table below:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{} \times P_0}$*
2.1972	0.50877	0.18463	2.0000	0.09393	0.18787

With this Mach number, the Mach number, can be obtained. From equation (5.22) or from Table (4.2) $M_y \cong 0.54746$ . With these values, the subsonic branch can be evaluated for the pressure and temperature ratios.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.1972	0.54743	1.8544	2.9474	5.4656	0.62941

From Table (4.2) or from equation (4.11) the following Table for the isentropic relationship is obtained

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{} \times P_0}$*
0.54743	0.94345	0.86457	1.2588	0.81568	1.0268

Again utilizing the isentropic relationship the exit conditions can be evaluated. With known Mach number the new star area ratio, ${A_y}/{A^{*}}$ is known and the exit area can be calculated as

$\displaystyle {A_e \over A^{*}} = {A_e \over A_y} \times {A_y \over A^{*}} = 1.2588 \times {9 \over 6 } = 1.8882$

with this area ratio, ${A_e \over A^{*}}= 1.8882$ , one can obtain using the isentropic relationship as

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{} \times P_0}$*
0.32651	0.97912	0.94862	1.8882	0.92882	1.7538

Since the stagnation pressure is constant as well the stagnation temperature, the exit conditions can be calculated. P_exit = ( P_exit P_0 ) ( P_0 P_y) ( P_y P_x) ( P_x P_0) P_0

$\displaystyle P_{exit} = 0.92882\times \left( {1 \over 0.81568 } \right) \times 5.466 \times 0.094 \times 4 \cong & 2.34[Bar]$

The exit temperature is

$\displaystyle T_{exit} = \left({ T_{exit} \over T_{0}} \right) \left( {T_{0} \o... ...y}\right) \left( {T_{y} \over T_x}\right) \left( {T_{x} \over T_0}\right) T_0$

$\displaystyle T_{exit} = 0.98133 \times \left( { 1 \over 0.951} \right) \times 1.854 \times 0.509 \times 308 \cong & 299.9 K$

For the ``critical'' points "a" and "b" are the points that the shock doesn't occur and yet the flow achieve Mach equal 1 at the throat. In that case we don't have to go through that shock transition. Yet we have to pay attention that there two possible back pressures that can ``achieve'' it or target. The area ratio for both cases, is $A / A^{*} = 3$ In the subsonic branch (either using equation or the isentropic Table or GDC-Potto as

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{} \times P_0}$*
0.19745	0.99226	0.98077	3.0000	0.97318	2.9195
2.6374	0.41820	0.11310	3.0000	0.04730	0.14190

$\displaystyle P_{exit} = \left({ P_{exit} \over P_{0}} \right) P_0 = 0.99226 \times 4 \cong$

$\displaystyle 3.97 [Bar]$

For the supersonic sonic branch

$\displaystyle P_{exit} = \left({ P_{exit} \over P_{0}} \right) P_0 = 0.41820 \times 4 \cong$

$\displaystyle 1.6728 [Bar]$

It should be noted that the flow rate is constant and maximum for any point beyond the point "a" even if the shock is exist. The flow rate is expressed as following

$\displaystyle \dot{m} = \rho^{*} A^{*} U = \overbrace{ P^{*} \over R T^{*}}^{\rho^{*}} A \overbrace{c M}^{M=1}$

$\displaystyle = {\left(\overbrace{ { P^{*} \over P_0} P_0}^{P^{*}} \right) \ove... ...over R \left( {T^{*} \over T_0} T_0 \right) } A {\sqrt{kR{T^{*}\over T_0} T_0}}$

The temperature and pressure at the throat are:

$\displaystyle T^{*} = {\left( {T^{*}\over T_0} \right) T_0} = 0.833\times 308 = 256.7 K$

The temperature at the throat reads

$\displaystyle P^{*} = { \left( {P^{*}\over P_0} \right) P_0} = 0.5283\times 4 = 2.113[Bar]$

The speed of sound is

$\displaystyle c = \sqrt{1.4\times 287 \times 256.7} = 321.12 [m/sec]$

And the mass flow rate reads

$\displaystyle \dot{m} = { 4 10^{5} \over 287 \times 256.7} 3\times 10^{-4} \times 321.12 = 0.13 [kg/sec]$

It is interesting to note that in this case the choking condition is obtained ( ) when the back pressure just reduced to less than 5% than original pressure (the pressure in the tank). While the pressure to achieve full supersonic flow through the nozzle the pressure has to be below the 42% the original value. Thus, over 50% of the range of pressure a shock occores some where in the nozzle. In fact in many industrial applications, these kind situations exist. In these applications a small pressure difference can produce a shock wave and a chock flow.

For more practical example^6.1 from industrial application point of view.

$\begin{examl} In the data from the above example \eqref{variableShock:ex:tank} where would be shock's location when the back pressure is $2[Bar]$? \end{examl}$

Solution

A solution procedure similar to what done in previous example (6.1) can be used here. The solution process starts at the nozzle's exit and progress to the entrance.

The conditions at the tank are again the stagnation conditions. Thus, the exit pressure is between point ``a'' to point ``b''. It follows that there must exist a shock in the nozzle. Mathematically, there are two main possibles ways to obtain the solution. In the first method, the previous example information used and expanded. In fact, it requires some iterations by ``smart'' guessing the different shock locations. The area (location) that the previous example did not ``produce'' the ``right'' solution (the exit pressure was . Here, the needed pressure is only which means that the next guess for the shock location should be with a larger areaThe second (recommended) method is noticing that the flow is adiabatic and the mass flow rate is constant which means that the ratio of the $P_0 \times A^{*} = \left. P_{y0} \times A^{*} \right\vert _{@ y}$ (upstream conditions are known, see also equation (4.71)).

$\displaystyle { P_{exit} A_{exit} \over {P_x}_0 \times {A_x}^{*}} = { P_{exit}... ... {P_y}_0 \times {A_y}^{*}} = {2 \times 9 \over 4 \times 3} = 1.5 [unit less!]$

With the knowledge of the ratio ${P A \over P_0 A^{*}}$ which was calculated and determines the exit Mach number. Utilizing the Table (4.2) or the GDC-Potto provides the following table is obtained

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$	$\mathbf{F \over F^{*}}$
0.38034	0.97188	0.93118	1.6575	0.90500	1.5000	0.75158

With these values the relationship between the stagnation pressures of the shock are obtainable e.g. the exit Mach number, , is known. The exit total pressure can be obtained (if needed). More importantly the pressure ratio exit is known. The ratio of the ratio of stagnation pressure obtained by

$\displaystyle {{P_0}_y \over {P_0}_x} = \overbrace{\left({ {P_0}_y \over P_{ex... ...{exit} \over {P_0}_x} \right) = {1 \over 0.905 } \times { 2 \over 4} = 0.5525$

Looking up in the Table (4.2) or utilizing the GDC-Potto provides

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.3709	0.52628	2.0128	3.1755	6.3914	0.55250

With the information of Mach number (either or ) the area where the shock (location) occurs can be found. First, utilizing the isentropic Table (4.2).

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$
2.3709	0.47076	0.15205	2.3396	0.07158	0.16747

Approaching the shock location from the upstream (entrance) yields

$\displaystyle A = {A \over A{*} } A^{*} = 2.3396 \times 3 \cong 7.0188 [cm^2]$

Note, as ``simple'' check this value is larger than the value in the previous example.

Subsections

Next: Nozzle efficiency Up: gasDynamics Previous: Tables of Normal Shocks, Index

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