Question

A flow-through combustion chamber consists of long, 15 -cm-diameter 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

Question: Determine the rate of heat transfer from the combustion gases to the tube wall by radiation per meter length of tube. Answer: The rate of heat transfer from combustion gases to the tube wall by radiation per 1-meter length of tube is approximately \(55814.05 \frac{\mathrm{W}}{\mathrm{m}}\).

Step-by-step solution

Identify relevant properties and constants

First, we will identify the relevant properties and constants needed for this problem. We are given: - Emissivity of tube wall (\(ε\)): 0.9 - Temperature of combustion gas (\(T_g\)): 1500 K - Temperature of tube wall (\(T_t\)): 600 K - Diameter of tube (\(D\)): 15 cm = 0.15 m Stefan-Boltzmann constant (\(σ\)): 5.67 x \(10^{-8} \frac{\mathrm{W}}{{\mathrm{m}^2} \cdot {\mathrm{K}^4}}\) Now let's proceed to calculate the radiative heat transfer rate.

Calculate area of radiation for a 1-meter length of tube

Since we want to find the heat transfer per meter length of the tube, we need to calculate the area of radiation for a 1-meter length tube. To calculate the area, we will use the following formula for the lateral surface area of a cylinder: \(A = π × D × L\) where L = 1 m (length of the tube) \(A = π × 0.15 × 1 = 0.47 m^2\)

Apply the radiative heat transfer equation to find the heat transfer rate

Now we will apply the radiative heat transfer equation which is given by: \(q = ε × A × σ × {(T_g^4 - T_t^4)}\) where: - q is the radiative heat transfer rate - ε is the emissivity of the tube wall - A is the area of radiation calculated in Step 2 - σ is the Stefan-Boltzmann constant - \(T_g\) is the temperature of the combustion gas - \(T_t\) is the temperature of the tube wall Plugging in the given values: \(q = 0.9 × 0.47 × 5.67 × 10^{-8} × {(1500^4 - 600^4)}\) Calculating the result: \(q ≈ 55814.05 \frac{\mathrm{W}}{\mathrm{m}}\)

Present the final answer

The rate of heat transfer from combustion gases to the tube wall by radiation per 1-meter length of tube is approximately \(55814.05 \frac{\mathrm{W}}{\mathrm{m}}\).