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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

Rayleigh Flow Control Volume

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}

 Rayleigh Line

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.
Rayleigh Flow Figure

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.

Example 11.23

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.

Example 11.24

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 Potto-GDC 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

Example 11.25

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.

Example 11.26

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

Combunstion Chamber

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 Potto-GDC 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 Potto-GDC 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 fuel-air 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 Potto-GDC

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.


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