Question

A 5-m-diameter spherical furnace contains a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2}\) gases at \(1200 \mathrm{~K}\) and \(1 \mathrm{~atm}\). The mole fraction of \(\mathrm{CO}_{2}\) in the mixture is \(0.15\). If the furnace wall is black and its temperature is to be maintained at \(600 \mathrm{~K}\), determine the net rate of radiation heat transfer between the gas mixture and the furnace walls.

Short answer

Answer: The net rate of radiation heat transfer between the gas mixture and the furnace walls is approximately \(9.153\times 10^6\: W\).

Step-by-step solution

Determine the surface area of the furnace

Given the diameter of the sphere is 5 meters, we can find the surface area (A) of the furnace using the formula: \(A = 4\pi r^2\), where r is the radius. The radius is half of the diameter, so r =2.5 m. Then, \(A = 4\pi(2.5)^2 = 78.54\: m^2\)

Calculate the emissivity using mole fraction

We are given that the mole fraction of CO2 in the mixture is 0.15. To maintain the temperature of the furnace wall at 600K, the furnace wall must be black, meaning it has an emissivity (E) of 1.

Use Stefan-Boltzmann Law

To find the net rate of radiation heat transfer (Q) between the gas mixture and the furnace walls, we can use the Stefan-Boltzmann Law, which is given by: \(Q = E\cdot A\cdot\sigma\cdot(T_{1}^4 - T_{2}^4)\) Where: - E is the emissivity, which is 1 in our case - A is the surface area of the furnace, calculated in Step 1 as 78.54 m² - \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67\times 10^{-8} \: W/m^2K^4\) - \(T_{1}\) is the temperature of the gas mixture, 1200 K - \(T_{2}\) is the temperature of the furnace walls, 600 K

Calculate the net rate of radiation heat transfer

Substitute the values into Stefan-Boltzmann Law: \(Q = 1\cdot 78.54\cdot(5.67\times 10^{-8})\cdot(1200^4 - 600^4)\) \(Q = 78.54\cdot(5.67\times 10^{-8})\cdot (2.0736\times 10^{12})\) \(Q = 9.153\times 10^6\: W\) The net rate of radiation heat transfer between the gas mixture and the furnace walls is approximately \(9.153\times 10^6\: W\).