Question

A cylindrical container whose height and diameter are \(8 \mathrm{~m}\) is filled with a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2}\) gases at $600 \mathrm{~K}\( and 1 atm. The partial pressure of \)\mathrm{CO}_{2}$ in the mixture is \(0.15 \mathrm{~atm}\). If the walls are black at a temperature of \(450 \mathrm{~K}\), determine the rate of radiation heat transfer between the gas and the container walls.

Short answer

The rate of radiation heat transfer between the gas and the container walls is approximately 33117.6 W.

Step-by-step solution

Find the partial pressure of nitrogen gas in the mixture

Since the total pressure of the mixture is 1 atm, and the partial pressure of \(\mathrm{CO}_{2}\) is given as \(0.15\mathrm{~atm}\), we can subtract the partial pressure of \(\mathrm{CO}_{2}\) from the total pressure to get the partial pressure of nitrogen gas in the mixture: Partial Pressure of \(\mathrm{N}_{2} = \text{Total Pressure} - \text{Partial Pressure of }\mathrm{CO}_{2}\) \(P_{\mathrm{N}_{2}} = 1\mathrm{~atm} - 0.15\mathrm{~atm}\) \(P_{\mathrm{N}_{2}} = 0.85\mathrm{~atm}\)

Calculate the Stefan-Boltzmann constant

The Stefan-Boltzmann constant is denoted by \(\sigma\) and has a value of \(5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^2\mathrm{K}^4}\).

Calculate the surface area of the container

To calculate the surface area of the cylindrical container, we need to find the area of the lateral surface and the top and bottom surfaces. The lateral surface area is given by the formula: \(A_{\text{lateral}} = 2 \pi r h\) Where \(r\) is the radius (\(r = \frac{\text{diameter}}{2} = 4\mathrm{~m}\)), and \(h\) is the height of the container (\(h = 8\mathrm{~m}\)). The top and bottom surface areas are given by the formula: \(A_{\text{top/bottom}} = \pi r^2\) Adding them together, we get the total surface area: \(A_{\text{total}} = A_{\text{lateral}} + 2 \times A_{\text{top/bottom}}\)

Calculate the rate of radiation heat transfer

Now, we can apply the radiative heat transfer equation: \(Q = \sigma A_{\text{total}} (T_{\text{gas}}^4 - T_{\text{wall}}^4)\) Substitute the values (making sure to use Kelvin for temperatures): \(Q = (5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^2\mathrm{K}^4})(2\pi(4\mathrm{~m})(8\mathrm{~m}) + 2 \times \pi (4\mathrm{~m})^2)((600\mathrm{~K})^4 - (450\mathrm{~K})^4)\) Calculate the result: \(Q \approx 33117.6\mathrm{~W}\) So, the rate of radiation heat transfer between the gas and the container walls is approximately \(33117.6\mathrm{~W}\).