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.
