Question

A spherical communication satellite with a diameter of \(2.5 \mathrm{~m}\) is orbiting the earth. The outer surface of the satellite in space has an emissivity of \(0.75\) and a solar absorptivity of \(0.10\), while solar radiation is incident on the spacecraft at a rate of $1000 \mathrm{~W} / \mathrm{m}^{2}$. If the satellite is made of material with an average thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the midpoint temperature is \(0^{\circ} \mathrm{C}\), determine the heat generation rate and the surface temperature of the satellite.

Short answer

The heat generation rate of the satellite is 1101.31 W, and the surface temperature is 17.58°C.

Step-by-step solution

Determine the surface area of the sphere

We can determine the surface area of the spherical satellite using its diameter, which is given as \(2.5 \mathrm{~m}\). The radius of the sphere is half the diameter, so \(r = \frac{2.5}{2} = 1.25 \mathrm{~m}\). The surface area of a sphere can be calculated as: \(A = 4 \pi r^2\) Now substitute the value of r: \(A = 4 \pi (1.25)^2 = 19.63 \mathrm{~m}^2\)

Calculate the absorbed solar radiation

The satellite absorbs solar radiation at a rate of \(1000 \mathrm{~W} / \mathrm{m}^2\), and its solar absorptivity is \(0.10\). To find the total absorbed solar radiation, we can multiply the satellite's surface area by the rate of solar radiation and the absorptivity: \(Q_{in} = A \cdot S \cdot \alpha\) where \(Q_{in}\) is the absorbed solar radiation, \(A\) is the surface area, \(S\) is the incident solar radiation, and \(\alpha\) is the solar absorptivity. Now substitute the values: \(Q_{in} = 19.63 \cdot 1000 \cdot 0.10 = 1963 \mathrm{~W}\)

Calculate the emitted thermal radiation

The satellite emits thermal radiation due to its emissivity, which is given as \(0.75\). The equation for the emitted thermal radiation is: \(Q_{out} = \epsilon \cdot A \cdot \sigma \cdot T^4\) where \(Q_{out}\) is the emitted thermal radiation, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4)\), and \(T\) is the temperature in Kelvin. First, we need to convert the midpoint temperature from Celsius to Kelvin: \(T = 0^{\circ} \mathrm{C} + 273.15 = 273.15 \mathrm{~K}\) Now substitute the values into the equation: \(Q_{out} = 0.75 \cdot 19.63 \cdot 5.67 \times 10^{-8} \cdot (273.15)^4 = 861.69 \mathrm{~W}\)

Determine the heat generation rate

The heat generation rate (\(Q_{gen}\)) can be determined by the difference between the absorbed solar radiation and the emitted thermal radiation: \(Q_{gen} = Q_{in} - Q_{out}\) Substitute the values: \(Q_{gen} = 1963 - 861.69 = 1101.31 \mathrm{~W}\)

Calculate the surface temperature

To find the surface temperature, we can use the following equation: \(T_{surface} = T + \frac{Q_{gen}}{4 \pi r k}\) where \(T_{surface}\) is the surface temperature, \(T\) is the satellite's midpoint temperature, \(r\) is the radius, and \(k\) is the average thermal conductivity. Now substitute the values: \(T_{surface} = 273.15 + \frac{1101.31}{4 \pi (1.25) 5} = 273.15 + 17.58 = 290.73 \mathrm{~K}\) Finally, convert the surface temperature from Kelvin to Celsius: \(T_{surface}^{\circ} \mathrm{C} = 290.73 - 273.15 = 17.58^{\circ} \mathrm{C}\) In conclusion, the heat generation rate of the satellite is \(1101.31 \mathrm{~W}\), and the surface temperature is \(17.58^{\circ} \mathrm{C}\).