Question

Two concentric spheres with diameters of \(5 \mathrm{~cm}\) and $10 \mathrm{~cm}\( have their surface temperatures maintained at \)100^{\circ} \mathrm{C}\( and \)200^{\circ} \mathrm{C}$, respectively. The enclosure between the two concentric spherical surfaces is filled with nitrogen gas at \(\mathrm{atm}\). Determine the rate of heat transfer through the enclosure.

Short answer

Answer: The rate of heat transfer between the two concentric spheres filled with nitrogen gas is approximately 714.29 W.

Step-by-step solution

Identify the known variables and constants

We are given the following: - Inner sphere diameter: \(d_1 = 5 \mathrm{~cm}\) - Outer sphere diameter: \(d_2 = 10 \mathrm{~cm}\) - Inner sphere temperature: \(T_1 = 100^{\circ} \mathrm{C}\) - Outer sphere temperature: \(T_2 = 200^{\circ} \mathrm{C}\) - The gas is nitrogen at atmospheric pressure. We need to find the thermal conductivity of nitrogen gas, which is approximately \(k_{N_2} = 0.024 \mathrm{~W/m\cdot K}\) at room temperature and atmospheric pressure.

Calculate the radii of the spheres and the temperature difference

First, let's find the radii \(r_1\) and \(r_2\) of the inner and outer spheres by dividing the diameters by 2. \(r_1 = \frac{d_1}{2} = 2.5 \mathrm{~cm}\) \(r_2 = \frac{d_2}{2} = 5 \mathrm{~cm}\) For simplicity, we can convert the radii to meters. \(r_1 = 0.025 \mathrm{~m}\) \(r_2 = 0.05 \mathrm{~m}\) Now let's find the temperature difference between the spheres in Kelvin. \(\Delta T = T_2 - T_1 = 200 - 100 = 100 \mathrm{~K}\)

Calculate the thermal resistance of the gas-filled enclosure

The thermal resistance for concentric spheres is given by the following equation: \(R_{enclosure} = \frac{1}{4\pi k_{N_2}}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\) Plug in the values to calculate the thermal resistance: \(R_{enclosure} = \frac{1}{4\pi (0.024)}\left(\frac{1}{0.025} - \frac{1}{0.05}\right) \approx 0.14 \mathrm{~m \cdot K/W}\)

Determine the rate of heat transfer

Now that we have the thermal resistance, we can calculate the rate of heat transfer using the following equation: \(q = \frac{\Delta T}{R_{enclosure}}\) Plug in the values: \(q = \frac{100 \mathrm{~K}}{0.14 \mathrm{~m \cdot K/W}} \approx 714.29 \mathrm{~W}\) Therefore, the rate of heat transfer through the enclosure is approximately \(714.29 \mathrm{~W}\).