Question 12.15: Rayleigh Flow in a Tubular Combustor A combustion chamber co...

Fuel is burned in a tubular combustion chamber with compressed air. The exit temperature, pressure, velocity, and Mach number are to be determined.
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid.
2 Combustion is complete, and it is treated as a heat addition process, with no change in the chemical composition of the flow. 3 The increase in mass flow rate due to fuel injection is disregarded.
Properties We take the properties of air to be k=1.4, c_{p}=1.005 kJ / kg \cdot K , and R = 0.287 kJ/kg · K.
Analysis The inlet density and mass flow rate of air are

\begin{aligned}\rho_{1} &=\frac{P_{1}}{R T_{1}}=\frac{480 kPa }{(0.287 kJ / kg \cdot K )(550 K )}=3.041 kg / m ^{3} \\\dot{m}_{\text {air }} &=\rho_{1} A_{1} V_{1}=\left(3.041 kg / m ^{3}\right)\left[\pi(0.15 m )^{2} / 4\right](80 m / s )=4.299 kg / s\end{aligned}

The mass flow rate of fuel and the rate of heat transfer are

\begin{gathered}\dot{m}_{\text {fuel }}=\frac{m_{\text {air }}}{ AF }=\frac{4.299 kg / s }{40}=0.1075 kg / s \\\dot{Q}=\dot{m}_{\text {fuel }} HV =(0.1075 kg / s )(42,000 kJ / kg )=4514 kW \\q=\frac{\dot{Q}}{\dot{m}_{\text {air }}}=\frac{4514 kJ / s }{4.299 kg / s }=1050 kJ / kg\end{gathered}

The stagnation temperature and Mach number at the inlet are

\begin{aligned}&T_{01}=T_{1}+\frac{V_{1}^{2}}{2 c_{p}}=550 K +\frac{(80 m / s )^{2}}{2(1.005 kJ / kg \cdot K )}\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}}\right)=553.2 K \\&c_{1}=\sqrt{k R T_{1}}=\sqrt{(1.4)(0.287 kJ / kg \cdot K )(550 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)}=470.1 m / s \\& Ma _{1}=\frac{V_{1}}{c_{1}}=\frac{80 m / s }{470.1 m / s }=0.1702\end{aligned}

The exit stagnation temperature is, from the energy equation q=c_{p}\left(T_{02}-T_{01}\right) ,

T_{02}=T_{01}+\frac{q}{c_{p}}=553.2 K +\frac{1050 kJ / kg }{1.005 kJ / kg \cdot K }=1598 K

The maximum value of stagnation temperature T_{0}^{*} occurs at Ma = 1, and its value can be determined from Table A–15 or from Eq. 12–65. At Ma _{1} = 0.1702 we read T_{0} / T_{0}^{*}=0.1291 . Therefore,

T_{0}^{*}=\frac{T_{01}}{0.1291}=\frac{553.2 K }{0.1291}=4284 K

\frac{T_{0}}{T_{0}^{*}}=\frac{(k+1) Ma ^{2}\left[2+(k-1) Ma ^{2}\right]}{\left(1+k Ma ^{2}\right)^{2}}                   (12–65)

The stagnation temperature ratio at the exit state and the Mach number corresponding to it are, from Table A–15

\frac{T_{02}}{T_{0}^{*}}=\frac{1598 K }{4284 K }=0.3730 \rightarrow Ma _{2}= 0 . 3 1 4 2

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A–15):

\begin{array}{llll}Ma _{1}=0.1702: & \frac{T_{1}}{T^{*}}=0.1541 & \frac{P_{1}}{P^{*}}=2.3065 & \frac{V_{1}}{V^{*}}=0.0668 \\Ma _{2}=0.3142: & \frac{T_{2}}{T^{*}}=0.4389 & \frac{P_{2}}{P^{*}}=2.1086 & \frac{V_{2}}{V^{*}}=0.2082\end{array}

Then the exit temperature, pressure, and velocity are determined to be

\begin{gathered}\frac{T_{2}}{T_{1}}=\frac{T_{2} / T^{*}}{T_{1} / T^{*}}=\frac{0.4389}{0.1541}=2.848 \rightarrow T_{2}=2.848 T_{1}=2.848(550 K )=1566 K \\\frac{P_{2}}{P_{1}}=\frac{P_{2} / P^{*}}{P_{1} / P^{*}}=\frac{2.1086}{2.3065}=0.9142 \rightarrow P_{2}=0.9142 P_{1}=0.9142(480 kPa )=439 kPa \\\frac{V_{2}}{V_{1}}=\frac{V_{2} / V^{*}}{V_{1} / V^{*}}=\frac{0.2082}{0.0668}=3.117 \rightarrow V_{2}=3.117 V_{1}=3.117(80 m / s )=249 m / s\end{gathered}

Discussion Note that the temperature and velocity increase and pressure decreases during this subsonic Rayleigh flow with heating, as expected. This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.