Compressible Flow Free Textbook

Next: Shock Tube Up: The Moving Shocks Previous: Partially Closed Valve Index

Worked-out Examples for Shock Dynamics

$\begin{examl} A shock is moving at a speed of 450 [m/sec] in a stagnated gas at ... ...perature behind the shock. Assume the specific heat ratio is k=1.3. \end{examl}$
Solution

It can be noticed that the gas behind the shock is moving while the gas ahead the shock is still. Thus it is the case of shock moving into still medium (suddenly open valve case). First, the Mach velocity ahead the shock has to calculated.

$\displaystyle {M_{y}}^{'} = { U \over \sqrt{kRT}} = {450 \over \sqrt{1.3 \times 287 \times 300}} \sim 1.296$

Utilizing POTTO-GDC or that Table (5.4) one can obtain the following table

Shock Dynamics			Input: My′			k = 1.3
Open valve			Input: My′			k = 1.3
Mx	Mx′	My	My′	Ty/Tx	Py/Px	P0y/P0x
2.41794	0	0.501926	1.296	1.80862	6.47856	0.496948

Using the above table, the temperature behind the shock is

$\displaystyle T_y = {T_y}^{'} = {T_y \over T_x} T_x = 1.809 \times 300 \sim 542.7 K$

In same manner it can be done for the pressure ratio as following

$\displaystyle P_y = {P_y}^{'} = {P_y \over P_x} P_x = 6.479 \times 1.0 \sim 6.479 [Bar]$

The velocity behind the shock wave is obtained (for confirmation)

$\displaystyle {U_y}^{'} = {M_y}^{'} c_y = 1.296 \times \sqrt{1.3 \times 287 \times 542.7} \sim 450 \left[m \over sec \right]$

$\begin{examl} Gas flows in a tube with a velocity of $450[m/sec]$. The static pr... ...cting shock. The specific heat ratio can be assumed to be $k=1.4$. \end{examl}$
Solution

The first thing which is needed to be done is to find the prime Mach number ${M_x}^{'}= 1.2961$ . Then, the prime properties can be found. At this stage the reflecting shock velocity is unknown.

Simply using the Potto-GDC provides for the temperature and velocity the following table:

Shock Dynamics			Input: Mx′			k = 1.4
Close valve			Input: Mx′			k = 1.4
Mx	Mx′	My	My′	Ty/Tx	Py/Px	P0y/P0x
2.04445	1.2961	0.569957	0	1.72395	4.70974	0.700101

Or if you insist on doing the steps yourself find the upstream prime Mach, ${M_x}^{'}$ to be 1.2961. Then using the Table (5.2) you can find the proper . If this detail is not sufficient enough then simply utilize the iteration procedure described earlier and obtain

Shock Dynamics			Input: Mx′		k = 1.4
Close valve			Input: Mx′		k = 1.4
i	Mx	My	TyTx	PyPx	Myp
0	2.2961	0.534878	1.94323	5.98409	0
1	2.04172	0.5704	1.72169	4.69671	0
2	2.04454	0.569942	1.72402	4.71016	0
3	2.04445	0.569957	1.72395	4.70972	0
4	2.04445	0.569957	1.72395	4.70974	0

The table was obtained by utilizing Potto-GDC with the iteration request.

$\begin{examl} What should be the prime Mach number (or the combination of the ve... ...uble the temperature when the valve is suddenly and totally closed? \end{examl}$
Solution

The ratio can be obtained from Table (5.3). It can also be obtained from the stationary normal shock wave table. Potto-GDC provides for this temperature ratio the following table

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.3574	0.52778	2.0000	3.1583	6.3166	0.55832

This means that the required and using this number in the moving shock table provides

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.3574	0.52778	0.78928	0.0	2.000	6.317	0.55830

$\begin{examl} A gas is flowing in a pipe with a Mach number of 0.4. Calculate th... ...mber is half (the new parameter that is added in the general case). \end{examl}$
Solution

Refer to the section (5.3.5) for the calculation procedure. Potto-GDC provide the solution of the above data

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.1220	0.89509	0.40000	0.20000	1.0789	1.3020	0.99813

If the information about the iterations are needed see following table.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{i}$	$\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{M_y}^{'}}$
0	1.4000	0.73971	1.2547	2.1200	0.20000
1	1.0045	0.99548	1.0030	1.0106	0.20000
2	1.1967	0.84424	1.1259	1.5041	0.20000
3	1.0836	0.92479	1.0545	1.2032	0.20000
4	1.1443	0.87903	1.0930	1.3609	0.20000
5	1.1099	0.90416	1.0712	1.2705	0.20000
6	1.1288	0.89009	1.0832	1.3199	0.20000
7	1.1182	0.89789	1.0765	1.2922	0.20000
8	1.1241	0.89354	1.0802	1.3075	0.20000
9	1.1208	0.89595	1.0782	1.2989	0.20000
10	1.1226	0.89461	1.0793	1.3037	0.20000
11	1.1216	0.89536	1.0787	1.3011	0.20000
12	1.1222	0.89494	1.0790	1.3025	0.20000
13	1.1219	0.89517	1.0788	1.3017	0.20000
14	1.1221	0.89504	1.0789	1.3022	0.20000
15	1.1220	0.89512	1.0789	1.3019	0.20000
16	1.1220	0.89508	1.0789	1.3020	0.20000
17	1.1220	0.89510	1.0789	1.3020	0.20000
18	1.1220	0.89509	1.0789	1.3020	0.20000
19	1.1220	0.89509	1.0789	1.3020	0.20000
20	1.1220	0.89509	1.0789	1.3020	0.20000
21	1.1220	0.89509	1.0789	1.3020	0.20000
22	1.1220	0.89509	1.0789	1.3020	0.20000

**Figure 5.18:** Schematic of a piston pushing air in a tube.

$\begin{examl} \parindent 0pt A piston is pushing air that flows in a tube with a... ...at there is no friction and the Fanno flow model is not applicable. \end{examl}$
Solution

The procedure described in the section. The solution is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.2380	0.81942	0.50000	0.80000	1.1519	1.6215	0.98860

The complete iteration is provided below

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{i}$	$\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$
0	1.5000	0.70109	1.3202	2.4583
1	1.2248	0.82716	1.1435	1.5834
2	1.2400	0.81829	1.1531	1.6273
3	1.2378	0.81958	1.1517	1.6207
4	1.2381	0.81940	1.1519	1.6217
5	1.2380	0.81943	1.1519	1.6215
6	1.2380	0.81942	1.1519	1.6216

The time it takes the shock to reach the end of the cylinder is

$\displaystyle t = {length \over \underbrace{U_s}_{c_x (M_x - {M_x}^{'})}} = {1 \over \sqrt{1.4\times 287\times 300} ( 1.2380 - 0.4)} = 0.0034 [sec]$

$\begin{examl} From the previous example \eqref{shock:ex:openValve} calculate the... ...ifference between initial piston velocity and final piston velocity. \end{examl}$
Solution

The stationary difference the two sides of the shock are:

$\displaystyle \Delta U =$	$\displaystyle {U_y}^{'} - {U_x}^{'} = c_y {U_y}^{'} - c_x {U_x}^{'}$
	$\displaystyle = \sqrt{1.4\times 287\times 300 \strut} \left( 0.8 \times \overbrace{\sqrt{1.1519}}^ {\sqrt{{T_y} \over {T_x}}} - 0.5\right)$
	$\displaystyle \sim 124.4 [m/sec]$

**Figure 5.19:** Figure for Example (5.9)

$\begin{examl} \par An engine is designed so that two pistons are moving toward e... ...[m]$. Calculate the time it will take for the two shocks to collide. \end{examl}$
Solution

This situation is open valve case where the prime information is given. The solution is given by equation (5.66) and it is the explicit analytical solution. For this case the following table easily be obtained from Potto-GDC for the left piston

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$	$\mathbf{{U_y}^{'} }$	$\mathbf{{c_x}}$
1.0715	0.93471	0.0	0.95890	1.047	1.173	0.99959	40.0	347.

While the velocity of the right piston is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$	$\mathbf{{U_y}^{'} }$	$\mathbf{{c_x}}$
1.1283	0.89048	0.0	0.93451	1.083	1.318	0.99785	70.0	347.

The time for the shocks to collide is

$\displaystyle t = {length \over {U_{sx}}_1 + {U_{sx}}_2 } = { 1 [m] \over ( 1.0715 + 1.1283) 347.} \sim 0.0013[sec]$

Next: Shock Tube Up: The Moving Shocks Previous: Partially Closed Valve Index

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

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