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

$${ T_2 \over T^{*} } = {T_1 \over T^{*} } {T_{2} \over T_{1} }\\ = 0.304 \times {400 \over 300} = 0.4053$$

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

$$P_2 = P_1 { P^{*} \over P_1} {P_2 \over P^{*}} = 3 \times {1 \over 2.2069}\times 2.1341 = 2.901[Bar]$$

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, $P_{0_2}$ can be calculated as the following:

$$P_{0_2} = P_1 \overbrace{P_{0_1} \over P_1}^{isentropic} { {P_0}^... ... \times {1\over 0.95745} \times {1 \over 1.2177} \times 1.1992 = 3.08572[Bar]$$

The heat released (heat transferred) can be calculated from obtaining the stagnation temperature from both sides. The stagnation temperature at the entrance, $T_{0_1}$

$$T_{0_1} = T_1 \overbrace{T_{0_1} \over T_1}^{isentropic} = 300 / 0.98765 = 303.75 [K]$$

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

$$T_{0_2} = T_2 \overbrace{T_{0_2} \over T_2}^{isentropic} = 400 / 0.98765 = 407.12[K] %\celsius$$

The heat released becomes

$${Q \over \dot{m} } = C_p \left( T_{0_2} - T_{0_1} \right) 1 \time... ...004 \times (407.12 - 303.75 ) = 103.78 \left[ kJ \over sec kg \celsius\right]$$

The maximum temperature occurs at the point the Mach number reach $1/\sqrt{k}$ 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:

$$T_{max} = T_{1} { T^{*} \over T_1} { T_{max} \over T^{*} } {300 \over 0.3044 } \times 1.0286 = 1013.7[K]$$

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

$$T_{0_{max}} = T_{max} * { T_{0_{max}} \over T_{max}} = { 1013.7 \over 0.875 } = 1158.51[K]$$

The maximum heat can be calculated as

$${Q \over \dot{m}} = C_p \left( T_{0_{max}} - T_{0_1} \right) = 1 ... ... 1.004 \times ( 1158.51 - 303.75 ) = 858.18 \left[ kJ \over kg sec K \right]$$

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.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$	$\mathbf{F \over F^{*}}$
1.0000	0.83333	0.63394	1.0000	0.52828	0.52828	0.52828

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The stagnation temperature

$$T_{0_2} = T_2 { T_{0_2} \over T_2} = {1000 \over 0.83333} = 1200.0[K]$$

The entrance temperature is

$${T_{0_1} \over T_{0_2} } = 1 - {Q /\dot{m} \over T_{0_2} C_P} = 1200 - {600 \over 1200 \times 1.004} \cong 0.5016$$

It must be noted that $T_{0_2} = {T_0}^{*}$ . Therefore with ${T_{0_1} \over {T_0}^{*}} = 0.5016$ either by using Table (10.1) or by Potto-GDC the following is obtained

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T^{*}}$	$\mathbf{T_0 \over {T_0}^{*}}$	$\mathbf{P \over P^{*}}$	$\mathbf{P_0 \over {P_0}^{*}}$	$\mathbf{\rho^{*} \over \rho}$
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

$$T_1 = T^{*}{T_1 \over T^{*}} = 1000 \times 0.58463 = 584.6 [K]$$

The difference between the supersonic branch to subsonic branch

Solution

To achieve maximum heat transfer the exit Mach number has to be one, $M_2 = 1$ .

$${Q \over \dot{m} } = C_p \left( T_{0_2} - T_{0_1} \right) = C_p {T_0}^{*} \left( 1 - {T_{0_1} \over {T_0}^{*}} \right)$$

The table for $M=3$ as follows

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T^{*}}$	$\mathbf{T_0 \over {T_0}^{*}}$	$\mathbf{P \over P^{*}}$	$\mathbf{P_0 \over {P_0}^{*}}$	$\mathbf{\rho^{*} \over \rho}$
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

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.0000	0.47519	2.6790	3.8571	10.3333	0.32834

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With Mach number of $M=0.47519$ the maximum heat transfer requires information for Rayleigh flow as the following

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T^{*}}$	$\mathbf{T_0 \over {T_0}^{*}}$	$\mathbf{P \over P^{*}}$	$\mathbf{P_0 \over {P_0}^{*}}$	$\mathbf{\rho^{*} \over \rho}$
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.

$${\left.{Q \over \dot{m} }\right\vert _{subsonic} \over \left.{Q ... ...ver {T_0}^{*}} \right)_{supersonic} } = { 1 - 0.65398 \over 1 - 0.65398 } = 1$$

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