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 11.23

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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[1]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[1]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[2]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[2]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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

 Isentropic 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

 Isentropic 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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[5]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[2]/div/table[5]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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

 Isentropic 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 11.24

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.

 Isentropic 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

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 11.25

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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[1]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[1]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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 $$M_z$$ $$M_y$$ $$\dfrac{T_y}{T_x}$$ $$\dfrac{\rho_y}{\rho_x}$$ $$\dfrac{P_y}{P_x}$$ $$\dfrac{{P_0}_y }{ {P_0}_x}$$ 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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[3]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[3]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 0.33138 0.47519 0.40469 2.0802 1.1857 0.22844

or for subsonic branch

 Rayleigh Flow Input: $$M$$ k = 1.4 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[4]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[6]/div/table[4]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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

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 11.26

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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[1]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[1]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[2]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[2]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 1.000 1.000 1.000 1.000 1.000 1.000

And the isentropic relationships for Mach 0.3 are

 Isentropic 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}}

ParseError: EOF expected (click for details)
Callstack:
at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/p[10]/span, line 1, column 4

{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 $$M$$ $$\dfrac{T}{T^{\star}}$$ $$\dfrac{T_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[4]/tbody/tr[2]/td[3]/span, line 1, column 4 $$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.9:_Rayleigh_Flow/11.9.4:_Examples_For_Rayleigh_Flow), /content/body/div[8]/div/table[4]/tbody/tr[2]/td[5]/span, line 1, column 4 $$ $$\dfrac{\rho}{\rho^{\star}}$$ 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.

Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.