An air-cooled motorcycle engine loses a significant amount of heat through thermal radiation according to the Stefan-Boltzmann equation. Assume that the ambient temperature is T0 = 22.3°C (295.45 K). Suppose the engine generates 13.5 hp (10.07 kW) of power and, due to several deep surface fins, has a surface area of A = 0.530 m2. A shiny engine has an emissivity e = 0.0490, whereas an engine that is painted black has e = 0.950.
a) Determine the equilibrium temperature (in K) for the shiny engine. (Assume that radiation is the only mode by which heat is dissipated from the engine.)
b) And determine the equilibrium temperature (in K) for the black engine.
To solve this problem, we will use the Stefan-Boltzmann equation, which relates the power radiated by an object to its surface area, temperature, and emissivity. The equation is given by:
P = εσAT^4
where P is the power radiated, ε is the emissivity of the surface, σ is the Stefan-Boltzmann constant (equal to 5.67 × 10^-8 W/(m^2·K^4)), A is the surface area, and T is the temperature in Kelvin.
a) To find the equilibrium temperature for the shiny engine, we need to determine the power radiated by the engine and set it equal to the power generated by the engine:
P_radiated = P_generated
Since radiation is the only mode of heat dissipation, P_generated is equal to the power generated by the engine:
P_generated = 10.07 kW = 10,070 W
Now, we can use the Stefan-Boltzmann equation to find the radiated power for the shiny engine:
P_radiated = εσAT^4
Substituting the given values:
10,070 W = (0.0490)(5.67 × 10^-8 W/(m^2·K^4))(0.530 m^2)(T_shiny)^4
Simplifying the equation, we get:
T_shiny = (10,070 W / (0.0490)(5.67 × 10^-8 W/(m^2·K^4))(0.530 m^2))^(1/4)
Using a calculator to evaluate the right-hand side of the equation, we find:
T_shiny ≈ 339.7 K
Therefore, the equilibrium temperature for the shiny engine is approximately 339.7 Kelvin.
b) To find the equilibrium temperature for the black engine, we follow the same procedure as before, but now we use the emissivity value for the black surface (e = 0.950):
P_radiated = εσAT^4
10,070 W = (0.950)(5.67 × 10^-8 W/(m^2·K^4))(0.530 m^2)(T_black)^4
Simplifying the equation, we get:
T_black = (10,070 W / (0.950)(5.67 × 10^-8 W/(m^2·K^4))(0.530 m^2))^(1/4)
Using a calculator to evaluate the right-hand side of the equation, we find:
T_black ≈ 319.3 K
Therefore, the equilibrium temperature for the black engine is approximately 319.3 Kelvin.