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Heat Transfer - Engineering

Q1:

Out of 100 kcal/second of incident radiant energy on the surface of a thermally transparent body, 300 kcal/second is reflected back. If the transmissivity of the body is 0.25, the emissivity of the surface will be

A 0.35

B 0.45

C 0.55

D 0.85

ANS:B - 0.45

To find the emissivity of the surface, we can use the relationship between transmissivity (τ), reflectivity (ρ), and emissivity (ε): Transmissivity+Reflectivity+Emissivity=1Transmissivity+Reflectivity+Emissivity=1 Given that:

  • Incident radiant energy = 100 kcal/second
  • Reflected energy = 300 kcal/second
  • Transmissivity (τ) = 0.25
We can calculate the emissivity (ε) as follows: Reflectivity=Reflected energyIncident energy=300 kcal/second100 kcal/second=3Reflectivity=Incident energyReflected energy​=100kcal/second300kcal/second​=3 Emissivity=1−(Transmissivity+Reflectivity)Emissivity=1−(Transmissivity+Reflectivity) Emissivity=1−(0.25+3)Emissivity=1−(0.25+3) Emissivity=1−3.25Emissivity=1−3.25 Emissivity=−2.25Emissivity=−2.25 However, emissivity cannot be negative, so there seems to be an issue with the calculation. Let's correct it. The issue arises from the fact that the reflected energy is greater than the incident energy, which is not physically possible. The reflected energy cannot exceed the incident energy. This inconsistency suggests there may be a mistake in the given data or calculation. Could you please review the provided information to ensure its accuracy?