Worked-out Examples for Shock Dynamics

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.

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

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

Shock Dynamics			Input: My′			k = 1.3
Open valve
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

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

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

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

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

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

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
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 $M_x$ . 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
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.

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

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This means that the required $M_x = 2.3574$ 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

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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

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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

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

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

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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

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

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

Solution

The stationary difference the two sides of the shock are:

$$\Delta U = {U_y}^{'} - {U_x}^{'} = c_y {U_y}^{'} - c_x {U_x}^{'}$$

$$= \sqrt{1.4\times 287\times 300 \strut} \left( 0.8 \times \overbrace{\sqrt{1.1519}}^ {\sqrt{{T_y} \over {T_x}}} - 0.5\right)$$

$$\sim 124.4 [m/sec]$$

Figure 5.19: Figure for Example (5.9)

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.

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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.

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The time for the shocks to collide is

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