Question

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.

Short answer

Answer: The rate of heat transfer from combustion gases to the tube wall by radiation per meter length of the tube is approximately 15,096 W.

Step-by-step solution

State the Stefan-Boltzmann law

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.

Calculate the surface area A of the tube

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: $$A = 2 \pi (0.075)(1) = 0.15 \pi \mathrm{~m^2}$$

Calculate the heat transfer rate Q

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 \(15,096\thinspace\mathrm{W}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in the study of radiative heat transfer. It describes how much thermal radiation a body emits based on its temperature. The law is expressed with the formula \( Q = \sigma A (T_1^4 - T_2^4) \).

Where \( \sigma \) is known as the Stefan-Boltzmann constant, equal to \( 5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)} \). This constant applies universally to black bodies, which are idealized surfaces that absorb and emit radiation perfectly.

The temperatures \( T_1 \) and \( T_2 \) represent the absolute temperatures of the two bodies involved, in Kelvin. This law is traditionally used to calculate the net radiative heat transfer between two surfaces.

Emissivity

Emissivity is a key concept when calculating radiative heat transfer, as it modifies the Stefan-Boltzmann Law for real-world surfaces, which are seldom perfect black bodies.

Emissivity, represented by \( \varepsilon \), is a measure of a material's ability to emit thermal radiation. It is expressed as a ratio ranging from 0 to 1, where 1 signifies a perfect black body that fully absorbs and emits radiation. An emissivity of 0.9, for instance, indicates a material that is highly efficient at emitting radiation, but not perfect.

Given that practical surfaces often don't reach an emissivity of 1, incorporating this factor into the heat transfer formula is essential. For the combustion tubes in the exercise, using \( \varepsilon = 0.9 \) appropriately adjusts the formula to \( Q = \varepsilon \sigma A (T_1^4 - T_2^4) \), reflecting the realistic emission under practical conditions.

Heat Transfer Calculation

Calculating the rate of heat transfer between the combustion gases and the tube wall involves integrating surface properties and solving with the modified Stefan-Boltzmann equation.

The first step in this process is identifying and computing the surface area, \( A \), of the tube. For a cylinder, this translates to \( A = 2 \pi r L \), with \( r \) as the radius and \( L \) as the length. For a 15 cm diameter tube, the radius is calculated as 0.075 m, making the surface area for a 1 m length approximately \( 0.15 \pi \mathrm{~m^2} \).

With all values known, including the temperatures \( T_1 = 1500 \mathrm{~K} \) for the gases and \( T_2 = 600 \mathrm{~K} \) for the tube, insert these into the modified Stefan-Boltzmann law:

• \( \sigma = 5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)} \)
• \( \varepsilon = 0.9 \)
• \( A = 0.15 \pi \mathrm{~m^2} \)

Calculate \( Q \) to find the heat transfer rate which works out to approximately \( 15,096 \mathrm{W} \) over the meter length of tube.

Combustion Gases

In this exercise, combustion gases are a blend of different substances: \( 70\% \) \( \mathrm{N_2} \), \( 9\% \) \( \mathrm{H_2O} \), \( 15\% \) \( \mathrm{O_2} \), and \( 6\% \) \( \mathrm{CO_2} \). These components are typical in combustion processes, such as those occurring in engines and industrial burners.

The thermal properties of these gases, combined with their high temperature of \( 1500 \mathrm{~K} \), contribute to their ability to transfer heat via radiation. Each component may have different emissivity and specific heat capacities that affect the overall heat interactions.

Understanding these gaseous mixtures is essential when analyzing heat transfer in combustion chambers, as their composition impacts the efficiency and safety of thermal systems. The analysis involving such gases needs to consider their ability to absorb and emit radiation, affecting the rate at which they transfer energy to structures like the tube walls in this scenario.