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Chapter 18: Problem 48

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 \(T_{0}=27^{\circ} \mathrm{C}(300 \mathrm{~K})\). Suppose the engine generates 15 hp \((11 \mathrm{~kW})\) of power and, due to several deep surface fins, has a surface area of \(A=0.50 \mathrm{~m}^{2}\). A shiny engine has an emissivity \(e=0.050\), whereas an engine that is painted black has \(e=0.95 .\) Determine the equilibrium temperatures for the black engine and the shiny engine. (Assume that radiation is the only mode by which heat is dissipated from the engine.)

Short Answer

Expert verified
Answer: The approximate equilibrium temperatures for the black engine and shiny engine are 223.2 °C (496.2 K) and 566.6 °C (839.6 K), respectively.

Step by step solution

01

Write the Stefan-Boltzmann equation

The equation governing the power loss per unit area due to thermal radiation is: P/A = eσ(T^4 - T0^4)
02

Determine the power lost as a result of heat dissipation (P)

The problem states that the engine generates 15 hp (11 kW) of power. The power dissipated as heat is also equal to 15 hp (11 kW), as radiation is the only mode that heat is dissipated from the engine. P = 11 kW
03

Substitute the known values into the equation for both black and shiny engines

The equation becomes: 11 = e * 0.50 * 5.67 * 10^{-8} * (T^4 - 300^4) For the black engine, e = 0.95 and for the shiny engine, e = 0.050.
04

Solve the equation for the temperature of the black engine (T1) and the shiny engine (T2)

For the black engine: 11 = 0.95 * 0.50 * 5.67 * 10^{-8} * (T1^4 - 300^4) And for the shiny engine: 11 = 0.050 * 0.50 * 5.67 * 10^{-8} * (T2^4 - 300^4) Solving these equations for T1 and T2, we get: T1 ≈ 496.2 K (223.2 °C) for the black engine T2 ≈ 839.6 K (566.6 °C) for the shiny engine Hence, the equilibrium temperatures for the black engine and shiny engine are approximately 223.2 °C (496.2 K) and 566.6 °C (839.6 K), respectively.

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