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Next: Evacuating SemiRigid Chambers Up: Rayleigh Flow Previous: Rayleigh Flow Tables   Index

# Examples For Rayleigh Flow

Illustrative example

The typical questions that are raised in Rayleigh Flow are related to the maximum heat that can be transferred to gas (reaction heat) and to the flow rate.

Solution

The entrance Mach number and the exit temperature are given and from Table 10.1 or from the program the initial ratio can be calculated. From the initial values the ratio at the exit can be computed as the following.

Rayleigh Flow Input: M k = 1.4
M T/T* T0/T0* P/P* P0/P0* ρ*/ρ
0.25 0.3044 0.256837 2.2069 1.21767 0.137931

and

Rayleigh Flow Input: Tbar k = 1.4
M T/T* T0/T0* P/P* P0/P0* ρ*/ρ
0.298311 0.405301 0.343762 2.13412 1.19922 0.189915

The exit Mach number is known, the exit pressure can be calculated as

For the entrance the stagnation values are

Isentropic Flow Input: M k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
0.25 0.987654 0.969421 2.40271 0.957453 2.30048 1.04241

The total exit pressure, can be calculated as the following:

The heat released (heat transferred) can be calculated from obtaining the stagnation temperature from both sides. The stagnation temperature at the entrance,

The isentropic conditions at the exit are

Isentropic Flow Input: M k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
0.29831 0.982513 0.956855 2.04537 0.940123 1.9229 0.901028

The exit stagnation temperature is

The heat released becomes

The maximum temperature occurs at the point the Mach number reach and at this point the Rayleigh relationship are:

Rayleigh Flow Input: M k = 1.4
M T/T* T0/T0* P/P* P0/P0* ρ*/ρ
0.84515 1.02857 0.979591 1.20001 1.01162 0.857139

The maximum heat before the temperature can be calculated as follows:

Isentropic Flow Input: M k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
0.84515 0.875001 0.716179 1.02211 0.626657 0.640511 0.533757

0.84515& 0.87500& 0.71618& 1.0221& 0.62666& 0.64051& 0.53376

The stagnation temperature for this point is

The maximum heat can be calculated as

Note that this point isn't the choking point.

Solution

The solution involves finding the stagnation temperature at the exit and subtracting the heat (heat equation) to obtain the entrance stagnation temperature. From the Table (10.1) or from the Potto-GDC the following ratios can be obtained.

 1.0000 0.83333 0.63394 1.0000 0.52828 0.52828 0.52828
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The stagnation temperature

The entrance temperature is

It must be noted that . Therefore with either by using Table (10.1) or by Potto-GDC the following is obtained

 0.34398 0.50160 0.42789 2.0589 1.1805 0.24362
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Thus entrance Mach number is 0.38454 and the entrance temperature can be calculated as following

The difference between the supersonic branch to subsonic branch

Solution

To achieve maximum heat transfer the exit Mach number has to be one, .

The table for as follows

 3.0000 0.28028 0.65398 0.17647 3.4245 1.5882
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The higher the entrance stagnation temperature the larger the heat amount that can be observed by the flow. In subsonic branch the Mach number after the shock is

 3.0000 0.47519 2.6790 3.8571 10.3333 0.32834
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With Mach number of the maximum heat transfer requires information for Rayleigh flow as the following

 0.33138 0.47519 0.40469 2.0802 1.1857 0.22844
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It also must be noticed that stagnation temperature remains constant across the shock wave.

It is not surprising for the shock wave to be found in the Rayleigh flow.

Next: Evacuating SemiRigid Chambers Up: Rayleigh Flow Previous: Rayleigh Flow Tables   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21