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Fanno
Chapter 11 Gd (continue)
11.9 Rayleigh Flow
Rayleigh flow is a model describing a frictionless flow with heat transfer
through a pipe of constant cross sectional area.
In practice, Rayleigh flow isn't a really good model to describe real situations.
Yet, Rayleigh flow is practical and useful concept in a obtaining trends and
limits such as the density and pressure change due to external cooling or heating.
As opposed to the two previous models, the heat transfer can be in two
directions not like the friction (there is no negative friction).
This fact creates a different situation as compared to the previous two models.
This model can be applied to cases where the heat transfer is significant
and the friction can be ignored.
Flow of steam in steam boiler is good examle where Rayleigh flow can be used.
11.9.1 Introduction
Fig. 11.39 The control volume of Rayleigh Flow.
The third simple model for 1–dimensional flow with a constant heat transfer for frictionless flow.
This flow is referred to in the literature as Rayleigh Flow (see historical notes).
This flow is another extreme case in which the friction effects
are neglected because their relative magnitute is significantly smaller than the heat transfer effects.
While the isothermal flow model has heat transfer and friction,
the main assumption was that relative length is enables significant
heat transfer to occur between the surroundings and tube.
In contrast, the heat transfer in Rayleigh flow occurs between
unknown temperature and the tube and the heat flux is maintained constant.
As before, a simple model is built around the assumption of constant
properties (poorer prediction to case where chemical reaction take a place).
This model is used to roughly predict the conditions which occur
mostly in situations involving chemical reaction.
In analysis of the flow, one has to be aware that properties do change
significantly for a large range of temperatures.
Yet, for smaller range of temperatures and lengths the calculations are more accurate.
Nevertheless, the main characteristics of the flow such as a choking condition etc.~are encapsulated
in this model.
The basic physics of the flow revolves around the fact that the gas is highly compressible.
The density changes through the heat transfer (temperature change).
Contrary to Fanno flow in which the resistance always oppose
the flow direction, Rayleigh flow, also, the cooling can be applied.
The flow acceleration changes the direction when the cooling is applied.
11.9.2 Governing Equations
The energy balance on the control volume reads
\begin{align}
Q = C_p\, \left({T_0}_2  {T_0}_1 \right)
\label{ray:eq:energy}
\end{align}
The momentum balance reads
\begin{align}
A\, ( P_1  P_2 ) = \dot{m} \, ( V_2  V_1)
\label{ray:eq:momentum}
\end{align}
The mass conservation reads
\begin{align}
\rho_1 U_1 A = \rho_2 U_2 A = \dot{m}
\label{ray:eq:mass}
\end{align}
Equation of state
\begin{align}
{P_1 \over \rho_1 \,T_1} =
{P_2 \over \rho_2 \,T_2}
\label{ray:eq:state}
\end{align}
There are four equations with four unknowns, if the upstream conditions
are known (or downstream conditions are known).
Thus, a solution can be obtained.
One can notice that equations qref{ray:eq:momentum}, qref{ray:eq:mass} and qref{ray:eq:state}
are similar to the equations that were solved for the shock wave.
Thus, results in the same as before qref{gd:shock:eq:pressureRatio}
Pressure Ratio
\begin{align}
\label{ray:eq:Pratio}
\dfrac{P_2 }{ P_1} =
\dfrac {1 + k\,{M_1}^{2} }{ 1 + k\,{M_2}^{2}}
\end{align}
The equation of state qref{ray:eq:state} can further assist in obtaining the
temperature ratio as
\begin{align}
\dfrac{T_2 }{ T_1} = \dfrac{P_2 }{ P_1} \dfrac{\rho_1 }{ \rho_2}
\label{ray:eq:Tratio}
\end{align}
The density ratio can be expressed in terms of mass conservation as
\begin{align}
\dfrac{\rho_1 }{ \rho_2} = \dfrac{U_2 }{ U_1} =
\dfrac{ \dfrac{U_2 }{ \sqrt{k\,R\,T_2} } \sqrt{k\,R\,T_2} }
{ \dfrac{U_1 }{ \sqrt{k\,R\,T_1} } \sqrt{k\,R\,T_1} }
= \dfrac{M_2 }{ M_1} \sqrt{ T_2 \over T_1}
\label{ray:eq:rhoR1}
\end{align}
or in simple terms as
Density Ratio
\begin{align}
\label{ray:eq:rhoR}
\dfrac{\rho_1 }{ \rho_2} = \dfrac{U_2 }{ U_1} =
\dfrac{M_2 }{ M_1} \sqrt{\dfrac{ T_2 }{ T_1}}
\end{align}
or substituting equations qref{ray:eq:Pratio} and qref{ray:eq:rhoR}
into equation qref{ray:eq:Tratio} yields
\begin{align}
{T_2 \over T_1} =
{1 + k\,{M_1}^{2} \over 1 + k\,{M_2}^{2}}\,
{M_2 \over M_1} \sqrt{ T_2 \over T_1}
\label{ray:eq:t2t1a}
\end{align}
Transferring the temperature ratio to the left hand side and squaring the
results gives
Temperature Ratio
\begin{align}
\label{ray:eq:t2t1b}
\dfrac{T_2 }{T_1} =
\left[ \dfrac{1 + k\,{M_1}^{2} }{ 1 + k\,{M_2}^{2}} \right]^{2}\,
\left(\dfrac{M_2 }{ M_1}\right)^{2}
\end{align}
Fig. 11.40 The temperature entropy diagram for Rayleigh line.
The Rayleigh line exhibits two possible maximums one for $dT/ds = 0 $ and for $ds /dT =0$.
The second maximum can be expressed as $dT/ds = \infty$.
The second law is used to find the expression for the derivative.
\begin{align}
{s_1 s_2 \over C_p} = \ln {T_2 \over T_1}
 \dfrac{k 1 }{ k}\, \ln {P_2 \over P_1}
\label{ray:eq:2ndLaw}
\end{align}
\begin{align}
\label{ray:eq:sndRlawEx}
\dfrac{s_1 s_2 }{ C_p} =
2\, \ln \left[
{({1 + k\,{M_1}^{2}) \over (1 + k\,{M_2}^{2} ) } {M_2 \over M_1}}
\right] +
{k 1 \over k} \ln \left[
{1 + k\,{M21}^{2} \over 1 + k\,{M_1}^{2} } \right]
\qquad \,
\end{align}
Let the initial condition $M_1$, and $s_1$ be constant and the variable parameters are $M_2$, and $s_2$.
A derivative of equation qref{ray:eq:sndRlawEx} results in
\begin{align}
\dfrac{1 }{ C_p } \dfrac{ds }{ dM} =
\dfrac{2\, ( 1  M^{2} ) }{ M\, (1 + k\,M^{2} )}
\label{ray:eq:sndRlawExDrivative}
\end{align}
Taking the derivative of equation qref{ray:eq:sndRlawExDrivative}
and letting the variable parameters be $T_2$, and $M_2$ results in
\begin{align}
{dT \over dM} =
constant \times { 1  k\,M^{2} \over \left( 1 + k\,M^2\right)^{3} }
\label{ray:eq:dTdM}
\end{align}
Combining equations qref{ray:eq:sndRlawExDrivative} and
qref{ray:eq:dTdM} by eliminating $dM$ results in
\begin{align}
{dT \over ds} = constant \times
{M (1  kM^2 ) \over ( 1 M^2) ( 1 + kM^2 )^2 }
\label{ray:eq:dTds}
\end{align}
On T–s diagram a family of curves can be drawn for a given constant.
Yet for every curve, several observations can be generalized.
The derivative is equal to zero when $1  kM^2 = 0$ or
$M = 1 /\sqrt{k}$ or when $M \rightarrow 0$.
The derivative is equal to infinity, $dT/ds = \infty$ when $M = 1$.
From thermodynamics, increase of heating results in increase of entropy.
And cooling results in reduction of entropy.
Hence, when cooling is applied to a tube the velocity decreases and when
heating is applied the velocity increases.
At peculiar point of $M = 1/\sqrt{k}$ when additional
heat is applied the temperature decreases.
The derivative is negative, $dT/ds < 0$, yet note this point is not the choking point.
The choking occurs only when $M= 1$ because it violates the second law.
The transition to supersonic flow occurs when the area changes,
somewhat similarly to Fanno flow.
Yet, choking can be explained by the fact that increase of energy must be
accompanied by increase of entropy.
But the entropy of supersonic flow is lower (see Figure 11.40) and
therefore it is not possible (the maximum entropy at $M=1$.).
It is convenient to refer to the value of $M=1$.
The equation qref{ray:eq:Pratio} can be written between choking
point and any point on the curve.
Pressure Ratio
\begin{align}
\label{ray:eq:Pratioa}
\dfrac{P^{*} }{ P_1} = {1 + k\,{M_1}^{2} \over 1 + k}
\end{align}
The temperature ratio is
Pressure Ratio
\begin{align}
\label {ray:eq:Tratioa}
{T^{*} \over T_1} = {1 \over M^2}
\left( {1 + k{M_1}^{2} \over 1 + k} \right)^{2}
\end{align}
The stagnation temperature can be expressed as
\begin{align}
{{T_0}_1 \over {T_0}^{*}} =
{T_1 \left( 1 + \dfrac{k 1 }{ 2} {M_1}^{2} \right)
\over T^{*} \left( \dfrac{1 + k } {2} \right)}
\label{ray:eq:T0ratio2}
\end{align}
or explicitly
Stagnation Temperature Ratio
\begin{align}
\label{ray:eq:T0ratio}
{{T_0}_1 \over {T_0}^{*}} =
\dfrac{ 2\, ( 1 + k )\, {M_1}^{2} }{ (1 + k\,M^{2})^2}
\left( 1 + {k 1 \over 2} {M_1} ^2 \right)
\end{align}
The stagnation pressure ratio reads
\begin{align}
\dfrac{{P_0}_1 }{ {P_0}^{*}} =
{P_1 \left( 1 + \dfrac{k 1 }{ 2}\, {M_1}^{2} \right)
\over P^{*} \left( {1 + k } \over 2 \right)}
\label{ray:eq:P0ratio2}
\end{align}
or explicitly
Stagnation Pressure Ratio
\begin{align}
\label{ray:eq:P0ratio}
\dfrac{{P_0}_1 }{ {P_0}^{*}} =
{\left({ 1 + k \over 1 + k\,{M_1}^2}\right)}
\left( { 1 + k\,{M_1}^2 \over {(1 + k) \over 2}} \right)^{\dfrac{k }{ k 1}}
\end{align}
11.9.3 Rayleigh Flow Tables and Figures
Several observations can be made in regards to the stagnation temperature.
The maximum temperature is not at Mach equal to one.
Yet the maximum enetropy accurs at Mach equal to one.
$M$ 
0.03 
0.04 
0.05 
0.06 
0.07 
0.08 
0.09 
0.10 
0.20 
0.25 
0.30 
0.35 
0.40 
0.45 
0.50 
0.55 
0.60 
0.65 
0.70 
0.75 
0.80 
0.85 
0.90 
0.95 
1.0 
1.1 
1.2 
1.3 
1.4 
1.5 
1.6 
1.7 
1.8 
1.9 
2.0 
2.1 
2.2 
2.3 
2.4 
2.5 
2.6 
2.7 
2.8 
2.9 
3.0 
3.5 
4.0 
4.5 
5.0 
5.5 
6.0 
6.5 
7.0 
7.5 
8.0 
8.5 
9.0 
9.5 
10.0 
20.0 
25.0 
30.0 
35.0 
40.0 
45.0 
50.0 
55.0 
60.0 
65.0 
70.0 
$\dfrac{T}{T^{\star}}$ 
0.00517 
0.00917 
0.014300 
0.020529 
0.027841 
0.036212 
0.045616 
0.056020 
0.20661 
0.30440 
0.40887 
0.51413 
0.61515 
0.70804 
0.79012 
0.85987 
0.91670 
0.96081 
0.99290 
1.014 
1.025 
1.029 
1.025 
1.015 
1.00 
0.96031 
0.91185 
0.85917 
0.80539 
0.75250 
0.70174 
0.65377 
0.60894 
0.56734 
0.52893 
0.49356 
0.46106 
0.43122 
0.40384 
0.37870 
0.35561 
0.33439 
0.31486 
0.29687 
0.28028 
0.21419 
0.16831 
0.13540 
0.11111 
0.092719 
0.078487 
0.067263 
0.058264 
0.050943 
0.044910 
0.039883 
0.035650 
0.032053 
0.028972 
0.00732 
0.00469 
0.00326 
0.00240 
0.00184 
0.00145 
0.00117 
0.000971 
0.000816 
0.000695 
0.000600 
$\dfrac{T_0}{{T_0}^{\star}}$ 
0.00431 
0.00765 
0.011922 
0.017119 
0.023223 
0.030215 
0.038075 
0.046777 
0.17355 
0.25684 
0.34686 
0.43894 
0.52903 
0.61393 
0.69136 
0.75991 
0.81892 
0.86833 
0.90850 
0.94009 
0.96395 
0.98097 
0.99207 
0.99814 
1.00 
0.99392 
0.97872 
0.95798 
0.93425 
0.90928 
0.88419 
0.85971 
0.83628 
0.81414 
0.79339 
0.77406 
0.75613 
0.73954 
0.72421 
0.71006 
0.69700 
0.68494 
0.67380 
0.66350 
0.65398 
0.61580 
0.58909 
0.56982 
0.55556 
0.54473 
0.53633 
0.52970 
0.52438 
0.52004 
0.51647 
0.51349 
0.51098 
0.50885 
0.50702 
0.49415 
0.49259 
0.49174 
0.49122 
0.49089 
0.49066 
0.49050 
0.49037 
0.49028 
0.49021 
0.49015 
$\dfrac{P}{{P}^{\star}}$ 
2.397 
2.395 
2.392 
2.388 
2.384 
2.379 
2.373 
2.367 
2.273 
2.207 
2.131 
2.049 
1.961 
1.870 
1.778 
1.686 
1.596 
1.508 
1.423 
1.343 
1.266 
1.193 
1.125 
1.060 
1.00 
0.89087 
0.79576 
0.71301 
0.64103 
0.57831 
0.52356 
0.47562 
0.43353 
0.39643 
0.36364 
0.33454 
0.30864 
0.28551 
0.26478 
0.24615 
0.22936 
0.21417 
0.20040 
0.18788 
0.17647 
0.13223 
0.10256 
0.081772 
0.066667 
0.055363 
0.046693 
0.039900 
0.034483 
0.030094 
0.026490 
0.023495 
0.020979 
0.018846 
0.017021 
0.00428 
0.00274 
0.00190 
0.00140 
0.00107 
0.000846 
0.000686 
0.000567 
0.000476 
0.000406 
0.000350 
$\dfrac{P_0}{{P_0}^{\star}}$ 
1.267 
1.266 
1.266 
1.265 
1.264 
1.262 
1.261 
1.259 
1.235 
1.218 
1.199 
1.178 
1.157 
1.135 
1.114 
1.094 
1.075 
1.058 
1.043 
1.030 
1.019 
1.011 
1.005 
1.001 
1.00 
1.005 
1.019 
1.044 
1.078 
1.122 
1.176 
1.240 
1.316 
1.403 
1.503 
1.616 
1.743 
1.886 
2.045 
2.222 
2.418 
2.634 
2.873 
3.136 
3.424 
5.328 
8.227 
12.50 
18.63 
27.21 
38.95 
54.68 
75.41 
1.0E+2 
1.4E+2 
1.8E+2 
2.3E+2 
3.0E+2 
3.8E+2 
1.1E+4 
3.2E+4 
8.0E+4 
1.7E+5 
3.4E+5 
6.0E+5 
1.0E+6 
1.6E+6 
2.5E+6 
3.8E+6 
5.5E+6 
$\dfrac{{\rho}^{\star}}{\rho}$ 
0.00216 
0.00383 
0.00598 
0.00860 
0.011680 
0.015224 
0.019222 
0.023669 
0.090909 
0.13793 
0.19183 
0.25096 
0.31373 
0.37865 
0.44444 
0.51001 
0.57447 
0.63713 
0.69751 
0.75524 
0.81013 
0.86204 
0.91097 
0.95693 
1.000 
1.078 
1.146 
1.205 
1.256 
1.301 
1.340 
1.375 
1.405 
1.431 
1.455 
1.475 
1.494 
1.510 
1.525 
1.538 
1.550 
1.561 
1.571 
1.580 
1.588 
1.620 
1.641 
1.656 
1.667 
1.675 
1.681 
1.686 
1.690 
1.693 
1.695 
1.698 
1.699 
1.701 
1.702 
1.711 
1.712 
1.713 
1.713 
1.714 
1.714 
1.714 
1.714 
1.714 
1.714 
1.714 
The data is presented in Figure 11.41.
Fig. 11.41 The basic functions of Rayleigh Flow (k=1.4).
11.9.4 Examples For Rayleigh Flow
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 maximum flow rate.
Air enters a pipe with pressure of $3[Bar]$ and temperature of $27^{\circ}C$ at Mach number of $M=0.25$.
Due to internal combustion heat was released and the exit temperature was found to be $127^{\circ}C$.
Calculate the exit Mach number, the exit pressure, the total exit pressure, and heat released and transferred to the air.
At what amount of energy the exit temperature will start to decrease?
Assume $C_P = 1.004 \left[ \dfrac{kJ }{ kg \, ^{\circ}C} \right]$
Solution
The entrance Mach number and the exit temperature are given and from Table
or from Potto–GDC 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 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.2500 
0.30440 
0.25684 
2.2069 
1.2177 
0.13793 
and
\begin{align*}
\dfrac{ T_2 }{ T^{*} } = \dfrac{T_1 }{ T^{*} } \,\dfrac{T_{2} }{ T_{1} }
= 0.304 \times \dfrac{400 }{ 300} = 0.4053
\end{align*}
Rayleigh Flow 
Input: $\dfrac{T}{T^{\star}}$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.29831 
0.40530 
0.34376 
2.1341 
1.1992 
0.18991 
The exit Mach number is known, the exit pressure can be calculated as
\begin{align*}
P_2 = P_1 \,\dfrac{ P^{*} }{ P_1} \,\dfrac{P_2 }{ P^{*}} =
3 \times \dfrac{1 }{ 2.2069}\times 2.1341 = 2.901[Bar]
\end{align*}
For the entrance, the stagnation values are
Insentropic Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T_{0}}$ 
$\dfrac{\rho}{\rho_{0}}$ 
$\dfrac{A}{A_{0}}$ 
$\dfrac{P}{P_{0}}$ 
$\dfrac{A\,P}{A^{\star}\,P_0}$ 
$\dfrac{F}{F^{\star}}$ 
0.2500 
0.98765 
0.96942 
2.4027 
0.95745 
2.3005 
1.0424 
The total exit pressure, $P_{0_2}$ can be calculated as the following:
\begin{align*}
P_{0_2} = P_1 \overbrace{P_{0_1} \over P_1}^{isentropic}
{ {P_0}^{*} \over P_{0_1}} \,{P_{0_2} \over {P_{0}}^{*}}
= 3 \times {1\over 0.95745} \times {1 \over 1.2177} \times 1.1992
= 3.08572[Bar]
\end{align*}
The heat released (heat transferred) can be calculated from obtaining
the stagnation temperature from both sides.
The stagnation temperature at the entrance, $T_{0_1} $
\begin{align*}
T_{0_1} = T_1 \overbrace{T_{0_1} \over T_1}^{isentropic}
= 300 / 0.98765 = 303.75 [K]
\end{align*}
The isentropic conditions at the exit are
Insentropic Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T_{0}}$ 
$\dfrac{\rho}{\rho_{0}}$ 
$\dfrac{A}{A_{0}}$ 
$\dfrac{P}{P_{0}}$ 
$\dfrac{A\,P}{A^{\star}\,P_0}$ 
$\dfrac{F}{F^{\star}}$ 
0.29831 
0.98251 
0.95686 
2.0454 
0.94012 
1.9229 
0.90103 
The exit stagnation temperature is
\begin{align*}
T_{0_2} = T_2 \overbrace{\dfrac{T_{0_2} }{ T_2}}^{isentropic}
= 400 / 0.98765 = 407.12[K]
\end{align*}
The heat released becomes
\begin{align*}
\dfrac{Q }{ \dot{m} } = C_p \left( T_{0_2}  T_{0_1} \right)
1 \times 1.004 \times (407.12  303.75 ) =
103.78 \left[ \dfrac{kJ }{ sec\, kg \, ^{\circ}C }\right]
\end{align*}
The maximum temperature occurs at the point where the Mach number
reaches $1/\sqrt{k}$ and at this point the Rayleigh relationship are:
Rayleigh Flow 
Input: $\dfrac{T}{T^{\star}}$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.84515 
1.0286 
0.97959 
1.2000 
1.0116 
0.85714 
The maximum heat before the temperature can be calculated as following:
\begin{align*}
T_{max} = T_{1} \dfrac{ T^{*} }{ T_1} \dfrac{ T_{max} }{ T^{*} } \sim
\dfrac{300 }{ 0.3044 } \times 1.0286 = 1013.7[K]
\end{align*}
The isentropic relationships at the maximum energy are
Insentropic Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T_{0}}$ 
$\dfrac{\rho}{\rho_{0}}$ 
$\dfrac{A}{A_{0}}$ 
$\dfrac{P}{P_{0}}$ 
$\dfrac{A\,P}{A^{\star}\,P_0}$ 
$\dfrac{F}{F^{\star}}$ 
0.84515 
0.87500 
0.71618 
1.0221 
0.62666 
0.64051 
0.53376 
The stagnation temperature for this point is
\begin{align*}
T_{0_{max}} = T_{max} \, \dfrac{ T_{0_{max}} }{ T_{max}} =
\dfrac{ 1013.7 }{ 0.875 } = 1158.51[K]
\end{align*}
The maximum heat can be calculated as
\begin{align*}
\dfrac{Q }{ \dot{m}} = C_p \left( T_{0_{max}}  T_{0_1} \right)
= 1 \times 1.004 \times ( 1158.51  303.75 ) = 858.18
\left[ \dfrac{kJ }{ kg \,sec \, K } \right]
\end{align*}
Note that this point isn't the choking point.
After this point additional heat results in temperature reduction.
Heat is added to the air until the flow is choked in amount of 600 [kJ/kg].
The exit temperature is 1000 [K].
Calculate the entrance temperature and the entrance Mach number.
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 11.7 or from the PottoGDC the following ratios can be obtained.
Insentropic Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T_{0}}$ 
$\dfrac{\rho}{\rho_{0}}$ 
$\dfrac{A}{A_{0}}$ 
$\dfrac{P}{P_{0}}$ 
$\dfrac{A\,P}{A^{\star}\,P_0}$ 
$\dfrac{F}{F^{\star}}$ 
1.0000 
0.83333 
0.63394 
1.0000 
0.52828 
0.52828 
0.52828 
The stagnation temperature
\begin{align*}
T_{0_2} = T_2 \, \dfrac{ T_{0_2} }{ T_2} = \dfrac{1000 }{ 0.83333} = 1200.0[K]
\end{align*}
The entrance temperature is
\begin{align*}
{T_{0_1} \over T_{0_2} } =
1  {Q /\dot{m} \over T_{0_2} C_P} = 1200 
{600 \over 1200 \times 1.004}
\cong 0.5016
\end{align*}
It must be noted that $ T_{0_2} = {T_0}^{*}$.
Therefore with ${T_{0_1} \over {T_0}^{*}} = 0.5016 $ either by using Table qref{gd:tab:rayBasic}
or by Potto–GDC the following is obtained
Rayleigh Flow 
Input: $\dfrac{T}{T^{\star}}$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.34398 
0.50160 
0.42789 
2.0589 
1.1805 
0.24362 
Thus, entrance Mach number is 0.38454 and the entrance temperature can
be calculated as following
\begin{align*}
T_1 = T^{*}{T_1 \over T^{*}} = 1000 \times 0.58463 = 584.6 [K]
\end{align*}
The difference between the supersonic branch to subsonic branch
Air with Mach 3 enters a frictionless duct with heating.
What is the maximum heat that can be added so that there is no subsonic flow?
If a shock occurs immediately at the entrance, what is the maximum heat that can be added?
Solution
To achieve maximum heat transfer the exit Mach number has to be one,
$M_2 = 1$.
\begin{align*}
\dfrac{Q }{ \dot{m} } = C_p \, \left( T_{0_2}  T_{0_1} \right) =
C_p\, {T_0}^{\star} \left( 1  \dfrac{T_{0_1} }{ {T_0}^{*} } \right)
\end{align*}
The table for $M=3$ as follows
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
3.00 
0.28028 
0.65398 
0.17647 
3.4245 
1.5882 
The higher the entrance stagnation temperature the larger the heat amount that can be absorbed by the flow.
In subsonic branch the Mach number after the shock is
Shock Wave 
Input: $M_x$ 
k = 1.4 
$M_x$ 
$M_y$ 
$\dfrac{T_y}{T_x}$ 
$\dfrac{\rho_y}{\rho_x}$ 
$\dfrac{P_y}{P_x}$ 
$\dfrac{{P_0}_y }{ {P_0}_x} $ 
3.000 
0.47519 
2.6790 
3.8571 
10.3333 
0.32834 
With Mach number of $M=0.47519$ the maximum heat transfer requires information for Rayleigh
flow as the following
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.33138 
0.47519 
0.40469 
2.0802 
1.1857 
0.22844 
or for subsonic branch
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.47519 
0.75086 
0.65398 
1.8235 
0.1244 
0.41176 
It also must be noticed that
stagnation temperature remains constant across shock wave.
\begin{align*}
\dfrac{\left.\dfrac{Q }{ \dot{m} }\right_{subsonic} }
{ \left.\dfrac{Q }{ \dot{m} }\right_{supersonic}} =
\dfrac{\left( \dfrac{ 1  {T_{0_1} } }{ {T_0}^{*}} \right)_{subsonic} }
{\left( 1  \dfrac{T_{0_1} }{ {T_0}^{*}} \right)_{supersonic} } =
\dfrac{ 1  0.65398 }{ 1  0.65398 } = 1
\end{align*}
It is not surprising for the shock wave to be found in the Rayleigh flow.
One of the reason that Rayleigh flow model was invented is to
be analyzed the flow in a combustion chamber.
Consider a flow of air in conduct with a fuel injected into the flow
as shown in Figure 11.42. Calculate
Fig. 11.42 Schematic of the combustion chamber.
what the maximum fuel–air ratio.
Calculate the exit condition for half the fuel–air ratio.
Assume that the mixture properties are of air.
Assume that the combustion heat is 25,000[KJ/kg fuel] for the average
temperature range for this mixture.
Neglect the fuel mass addition and assume that all the fuel is
burned (neglect the complications of the increase of the entropy
if accrue).}
Solution
Under these assumptions, the maximum fuel air ratio is obtained when the flow is choked.
The entranced condition can be obtained using PottoGDC as following
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.3000 
0.40887 
0.34686 
2.1314 
1.1985 
0.19183 
The choking condition are obtained using also by PottoGDC as
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
1.000 
1.000 
1.000 
1.000 
1.000 
1.000 
And the isentropic relationships for Mach 0.3 are
Insentropic Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T_{0}}$ 
$\dfrac{\rho}{\rho_{0}}$ 
$\dfrac{A}{A_{0}}$ 
$\dfrac{P}{P_{0}}$ 
$\dfrac{A\,P}{A^{\star}\,P_0}$ 
$\dfrac{F}{F^{\star}}$ 
0.30000 
0.98232 
0.95638 
2.0351 
0.93947 
1.9119 
0.89699 
The maximum fuelair can be obtained by finding the heat per
unit mass.
\begin{align*}
\dfrac{\dot{Q}}{\dot{m}} = \dfrac{Q}{m} =
C_p \left( T_{02}  T_{01} \right) =
C_p T_1 \left( 1  \dfrac{ T_{01}}{T^*} \right)\
\end{align*}
\begin{align*}
\dfrac{\dot{Q}}{\dot{m}} = 1.04 \times 350 / 0.98232 \times
( 1  0.34686) \sim 242.022 [ kJ/kg]
\end{align*}
The fuel–air mass ratio has to be
\begin{align*}
\dfrac{ m_{fuel}} {m_{air}} = \dfrac{\hbox{needed heat}}
{\hbox{combustion heat}} =
\dfrac{242.022} {25,000} \sim 0.0097 [kg\,\,fuel/kg\,\, air]
\end{align*}
If only half of the fuel is supplied then
the exit temperature is
\begin{align*}
T_{02} = \dfrac{Q}{m C_p} + T_{01} =
\dfrac{ 0.5\times 242.022 } { 1.04} + 350/0.98232
\sim 472.656 [K]
\end{align*}
The exit Mach number can be determined from the exit stagnation
temperature as following:
\begin{align*}
\dfrac{T_2}{T^*} = \dfrac{T_{01}}{{T_0}^*}
\dfrac{T_{02}}{T_{01}}
\end{align*}
The last temperature ratio can be calculated from the value
of the temperatures
\begin{align*}
\dfrac{T_2}{T^*} = 0.34686 \times \dfrac {472.656}{350/0.98232}
\end{align*}
The Mach number can be obtained from a Rayleigh table or using PottoGDC
Rayleigh Flow 
Input: $M$ 
k = 1.4 
$M$ 
$\dfrac{T}{T^{\star}}$ 
$\dfrac{T_0}{{T_0}^{\star}}$ 
$\dfrac{P}{P^{\star}}$ 
$\dfrac{P_0}{{P_0}^{\star}}$ 
$\dfrac{\rho}{\rho^{\star}}$ 
0.33217 
0.47685 
0.40614 
2.0789 
1.1854 
0.22938 
It should be noted that this example is only to demonstrate how
to carry the calculations.
