The maximum location of the shock

The main point in this discussion however, is to find the furthest shock location downstream. Figure (9.16) shows the possible $\Delta \left(4fL \over D \right)$ as function of retreat of the location of the shock wave from the maximum location. When the entrance Mach number is infinity, $M_1= \infty$ , if the shock location is at the maximum length, then shock at $M_x = 1$ results in $M_y=1$ .

The proposed procedure is based on Figure (9.16).

1. $\parbox{0.92\textwidth}{ Calculate the extra $\frac{4fL}{D}$ and su... ...l extra $\frac{4fL}{D}$ assuming shock at the left side (at the max length). }$

2. $\parbox{0.92\textwidth}{ Calculate the extra $\frac{4fL}{D}$ and su... ...al extra $\frac{4fL}{D}$ assuming shock at the right side (at the entrance). }$

3. According to the positive or negative utilizes your root finding procedure.

Figure: The maximum entrance Mach number as a function of 4fL/D

From numerical point of view, the Mach number equal infinity when left side assumes result in infinity length of possible extra (the whole flow in the tube is subsonic). To overcome this numerical problem it is suggested to start the calculation from $\epsilon$ distance from the right hand side.

Let denote

$$\Delta \left( \frac{4fL}{D} \right ) = \bar{\frac{4fL}{D}}_{actual} - \left.\frac{4fL}{D}\right\vert _{sup}$$ (9.51)

Note that $\left.\frac{4fL}{D}\right\vert _{sup}$ is smaller than $\left.\frac{4fL}{D}\right\vert _{max_{\infty}}$ . The requirement that has to be satisfied is that denote $\left.{\frac{4fL}{D}}\right\vert _{retreat}$ as difference between the maximum possible of length in which the supersonic flow is achieved and the actual length in which the flow is supersonic see Figure (9.15). The retreating length is expressed as subsonic but

$$\left. \frac{4fL}{D} \right\vert _{retreat} = \left.\frac{4fL}{D}\right\vert _{max_{\infty}} - \left.\frac{4fL}{D}\right\vert _{sup}$$ (9.52)

Figure (9.17) shows the entrance Mach number, $M_1$ reduces after the maximum length is exceeded.

Solution

The solution is obtained by an iterative process. The maximum $\left.\frac{4fL}{D}\right\vert _{max}$ for $k=1.4$ is 0.821508116. Hence, $\frac{4fL}{D}$ exceed the maximum length $\frac{4fL}{D}$ for this entrance Mach number. The maximum for $M_1 =8$ is $\frac{4fL}{D} = 0.76820$ , thus the extra tube is $\Delta \left( \frac{4fL}{D} \right) = 0.9 - 0.76820 = 0.1318$ . The left side is when the shock occurs at $\frac{4fL}{D} = 0.76820$ (flow chocked and no any additional $\frac{4fL}{D}$ ). Hence, the value of left side is $-0.1318$ . The right side is when the shock is at the entrance at which the extra $\frac{4fL}{D}$ is calculated for $M_x$ and $M_y$ is

Normal Shock		Input: Mx		k = 1.4
Mx	My	Ty/Tx	ρy/ρx	Py/Px	P0y/P0x
8	0.39289	13.3867	5.56522	74.5	0.00848783

With $(M_1)'$

Fanno Flow		Input: M			k = 1.4
M	fld	P/P*	P0/P0*	ρ/ρ*	U/U*	T/T*
0.39289	2.44172	2.74611	1.61362	2.35907	0.423896	1.16406

The extra $\Delta\left(\frac{4fL}{D}\right)$ is $2.442 - 0.1318 = 2.3102$ Now the solution is somewhere between the negative of left side to the positive of the right side.^9.17

In a summary of the actions is done by the following algorithm:

(a)

check if the $\frac{4fL}{D}$ exceeds the maximum ${\frac{4fL}{D}}_{max}$ for the supersonic flow. Accordingly continue.

(b)

Guess ${\frac{4fL}{D}}_{up}= \frac{4fL}{D} - \left.\frac{4fL}{D}\right\vert _{max}$

(c)

Calculate the Mach number corresponding to current guess of ${\frac{4fL}{D}}_{up}$ ,

(d)

Calculate the associate Mach number, $M_x$ with the Mach number, $M_y$ calculated previously,

(e)

Calculate $\frac{4fL}{D}$ for supersonic branch for the $M_x$

(f)

Calculate the ``new and improved'' ${\frac{4fL}{D}}_{up}$

(g)

Compute the ``new ${\frac{4fL}{D}}_{down} = {\frac{4fL}{D}} - {\frac{4fL}{D}}_{up}$

(h)

Check the new and improved $\left.\frac{4fL}{D}\right\vert _{down}$ against the old one. If it satisfactory stop or return to stage (b).

shock location are:

$\rule[-0.1in]{0.pt}{0.3 in} \mathbf{M_1}$	$\mathbf{M_2}$	$\mathbf{\left.{ 4fL \over D}\right\vert _{up}}$	$\mathbf{\left.{ 4fL \over D }\right\vert _{down}}$	$\mathbf{M_x}$	$\mathbf{M_y}$
8.0000	1.0000	0.57068	0.32932	1.6706	0.64830

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The iteration summary is also shown below

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{i}$	$\mathbf{\left.{ 4fL \over D}\right\vert _{up}}$	$\mathbf{\left.{ 4fL \over D }\right\vert _{down}}$	$\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{ 4fL \over D }}$
0	0.67426	0.22574	1.3838	0.74664	0.90000
1	0.62170	0.27830	1.5286	0.69119	0.90000
2	0.59506	0.30494	1.6021	0.66779	0.90000
3	0.58217	0.31783	1.6382	0.65728	0.90000
4	0.57605	0.32395	1.6554	0.65246	0.90000
5	0.57318	0.32682	1.6635	0.65023	0.90000
6	0.57184	0.32816	1.6673	0.64920	0.90000
7	0.57122	0.32878	1.6691	0.64872	0.90000
8	0.57093	0.32907	1.6699	0.64850	0.90000
9	0.57079	0.32921	1.6703	0.64839	0.90000
10	0.57073	0.32927	1.6705	0.64834	0.90000
11	0.57070	0.32930	1.6706	0.64832	0.90000
12	0.57069	0.32931	1.6706	0.64831	0.90000
13	0.57068	0.32932	1.6706	0.64831	0.90000
14	0.57068	0.32932	1.6706	0.64830	0.90000
15	0.57068	0.32932	1.6706	0.64830	0.90000
16	0.57068	0.32932	1.6706	0.64830	0.90000
17	0.57068	0.32932	1.6706	0.64830	0.90000

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This procedure is rapidly converted to the solution.