Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does using thin, 2.2 m-by-4.1 m panels that have a working temperature of about 6 degrees Celsius.
How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible.
Hint: Don't forget that the panels have two sides!
I ended up with 3098.89 W but it's wrong...
Well, it seems like those panels are feeling a little chilly up there in space! I hope they brought their space sweaters! Now, let's get to calculating the power being radiated.
To start, we need to calculate the surface area of each panel. Since each panel is 2.2 m by 4.1 m, we can multiply those numbers together to get the area: 2.2 m * 4.1 m = 9.02 m².
Next, we need to convert the working temperature of 6 degrees Celsius to Kelvin. Adding 273.15 to 6 gives us a temperature of 279.15 K.
Now, we can use the Stefan-Boltzmann Law, which states that the power radiated from an object is proportional to its surface area and the fourth power of its temperature. The equation is:
Power = emissivity * Stefan-Boltzmann constant * surface area * (temperature ^ 4)
The emissivity of a surface determines how efficiently it radiates thermal energy. For simplicity, let's assume the emissivity of these panels is 1, which means they are perfect radiators.
The Stefan-Boltzmann constant is approximately 5.67 x 10^-8 W/(m²K⁴).
Plugging in the values, we get:
Power = 1 * (5.67 x 10^-8 W/(m²K⁴)) * 9.02 m² * (279.15 K ^ 4)
Calculating that, we find that each panel is radiating approximately 1,304 watts (rounded to the nearest whole watt).
So, each panel is releasing about 1,304 watts of thermal energy. That's enough to power a hairdryer, an electric grill, or even a small electric car! Just imagine the possibilities in space!
To determine the power radiated from each panel, we need to use the Stefan-Boltzmann Law. This law relates the power radiated by an object to its temperature. It states that the power radiated per unit area (P) is given by the equation:
P = σ * A * (T^4 - Ts^4)
Where:
P is the power radiated per unit area,
σ (sigma) is the Stefan-Boltzmann constant (approximately 5.67 × 10^-8 W/(m^2·K^4)),
A is the area of the panel (2.2 m * 4.1 m = 9.02 m^2),
T is the temperature of the panel (6 degrees Celsius = 279.15 Kelvin), and
Ts is the temperature of space (approximately 2.7 Kelvin).
Since the panel has two sides, we need to calculate the total power radiated by multiplying the power per unit area by the total area of both sides (2 * A).
So, the power radiated from each panel is given by:
P = σ * (2 * A) * (T^4 - Ts^4)
Now, let's calculate the power radiated from each panel:
P = 5.67 × 10^-8 * (2 * 9.02) * (279.15^4 - 2.7^4)
P ≈ 2.16 Watts
Therefore, each panel radiates approximately 2.16 Watts of thermal energy.
(number of sides on the panel)(e)(area)(sigma)(T^4)=P
(2)(1)(9.02)(5.67E-8)(279^4)=6197.78
You probably just put it into your calculator wrong, or forgot to multiply by 2.
Use the Stefan-Bolzmann law and assume an emissivity of 1.0
Radiated power = 2*A*sigma*T^4, where
T = 279 K.
'sigma' is the Stefan-Boltzmann constant, which you should look up.
A = 2.2*4.1 = 9.02 m^2