A flow-through combustion chamber consists of 15 -cm diameter long tubes immersed in water. Compressed air is routed to the tube, and fuel is sprayed into the compressed air. The combustion gases consist of 70 percent \(\mathrm{N}_{2}\), 9 percent \(\mathrm{H}_{2} \mathrm{O}, 15\) percent \(\mathrm{O}_{2}\), and 6 percent \(\mathrm{CO}_{2}\), and are maintained at \(1 \mathrm{~atm}\) and \(1500 \mathrm{~K}\). The tube surfaces are near black, with an emissivity of \(0.9\). If the tubes are to be maintained at a temperature of \(600 \mathrm{~K}\), determine the rate of heat transfer from combustion gases to tube wall by radiation per \(m\) length of tube.

The Stefan-Boltzmann law states that the net radiative heat transfer between two black bodies is given by: $$Q = \sigma A(T_{1}^{4} - T_{2}^{4})$$ where \(Q\) is the heat transfer rate, \(\sigma\) is the Stefan-Boltzmann constant, which is equal to \(5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}\), \(A\) is the surface area of the tube, \(T_{1}\) is the temperature of the combustion gases, and \(T_{2}\) is the temperature of the tube wall. However, we are given that the tubes have an emissivity of \(0.9\), which means we need to modify the Stefan-Boltzmann law for the given tubes. The modified formula is: $$Q = \varepsilon \sigma A(T_{1}^{4} - T_{2}^{4})$$ where \(\varepsilon\) is the emissivity of the tube surfaces.

To calculate the surface area of the tube per meter length, we use the formula for the lateral surface area of a cylinder: $$A = 2 \pi r L$$ where \(r\) is the radius of the tube and \(L\) is the length of the tube. Given the diameter of the tube is \(15 \thinspace\mathrm{cm}\), we first need to convert it to meters and then find the radius: $$D=\frac{15}{100} =0.15\mathrm{~m}$$ $$r=\frac{D}{2}=0.075\mathrm{~m}$$ Let the length of the tube be \(1\mathrm{ ~m}\), since we need to calculate the heat transfer per meter length of the tube: $$L = 1\mathrm{ ~m}$$ Now we can find the surface area \(A\) per meter length of the tube: $$A = 2 \pi (0.075)(1) = 0.15 \pi \mathrm{~m^2}$$

Now, we can plug in the given values and find the heat transfer rate, Q, using the modified Stefan-Boltzmann law: $$Q = \varepsilon \sigma A(T_{1}^{4} - T_{2}^{4})$$ where \(\varepsilon=0.9\), \(\sigma = 5.67 \times 10^{-8}\mathrm{W/(m^2 K^4)}\), \(A=0.15 \pi \mathrm{~m^2}\), \(T_{1}=1500\mathrm{ ~K}\), and \(T_{2}= 600\mathrm{ ~K}\). Plugging in the values, we get: $$Q = 0.9 \times 5.67 \times 10^{-8} \times 0.15 \pi (1500^{4} - 600^{4})\thinspace\mathrm{W}$$ Calculating the above expression, we get: $$Q \approx 15,096\thinspace\mathrm{W}$$ Thus, the rate of heat transfer from combustion gases to the tube wall by radiation per meter length of the tube is approximately \(15,096\thinspace\mathrm{W}\).