91影视

First, let's find the emissive power of the small disk and the enclosure. The emissive power can be calculated using the Stefan-Boltzmann law, which is defined as: \(E = \epsilon \sigma T^4\) where \(E\) is the emissive power (W/m虏), \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}\)), and \(T\) is the temperature (K). We are given the emissivities and temperatures for the disk and enclosure. We can find their emissive powers using these values. For the disk: \(\epsilon_d = 0.7\) \(T_d = 900 \mathrm{~K}\) For the enclosure: \(\epsilon_e = 0.85\) \(T_e = 300 \mathrm{~K}\) Now, let's calculate their emissive powers using the Stefan-Boltzmann law: \(E_d = \epsilon_d \sigma T_d^4\) \(E_e = \epsilon_e \sigma T_e^4\)

Next, we need to find the view factor between the disk and the aperture. The view factor is a measure of the geometrical relationship between two surfaces and determines how much radiation emitted by one surface is intercepted by the other. For this problem, we will use the circular disk to hemispherical aperture view factor formula, which is defined as: \(F_{d \rightarrow a} = \frac{1}{2} \left ( 1 - \frac{D}{\sqrt{D^2 + h^2}} \right )\) where \(F_{d \rightarrow a}\) is the view factor between the disk and the aperture, \(D\) is the diameter ratio between the aperture and the hemisphere, and \(h\) is the height-to-diameter ratio for the disk and hemisphere. In our case, we are given that the diameter of the disk is \(5 \mathrm{~mm}\), the diameter of the aperture is \(2 \mathrm{~mm}\), and the radius of the hemisphere is \(100 \mathrm{~mm}\). First, let's convert these values into meters: \(D_d = 5 \mathrm{~m} = 0.005 \mathrm{~m}\) \(D_a = 2 \mathrm{~m} = 0.002 \mathrm{~m}\) Now, calculate the diameter ratio (\(D\)) and the height-to-diameter ratio (\(h\)): \(D = \frac{D_a}{D_d + D_a} = \frac{0.002}{0.005 + 0.002}\) \(h = \frac{2 \times 100 \mathrm{~mm}}{5 \mathrm{~mm}} = 40\) Finally, we can calculate the view factor between the disk and the aperture: \(F_{d \rightarrow a} = \frac{1}{2} \left ( 1 - \frac{D}{\sqrt{D^2 + h^2}} \right )\)

The radiative exchange between the disk and the aperture (\(q_{d \rightarrow a}\)) is given by the following expression: \(q_{d \rightarrow a} = F_{d \rightarrow a} \times E_d - F_{a \rightarrow d} \times E_e\) where \(F_{a \rightarrow d}\) is the view factor from the aperture to the disk. Note that due to the reciprocity rule, the following relation holds for the view factors: \(F_{a \rightarrow d} = \frac{A_d}{A_a} F_{d \rightarrow a}\) where \(A_d\) is the area of the disk, and \(A_a\) is the area of the aperture. We can calculate the areas using the given diameters: \(A_d = \pi \left ( \frac{D_d}{2} \right )^2\) \(A_a = \pi \left ( \frac{D_a}{2} \right )^2\) Now, we can find the radiative exchange between the disk and aperture: \(q_{d \rightarrow a} = F_{d \rightarrow a} \times E_d - F_{a \rightarrow d} \times E_e\)

Finally, we need to find the radiant power leaving the aperture. This can be obtained by multiplying the radiative exchange (\(q_{d \rightarrow a}\)) by the area of the aperture: \(P_a = q_{d \rightarrow a} \times A_a\) Now, we have calculated the radiant power leaving the aperture.