Question

The emissivity of a grey surface is:

• A. Equal to 1
• B. Less than 1 and independent of wavelength
• C. Greater than 1
• D. Equal to the absorptivity at all wavelengths

Hint:

A grey surface is an idealization where spectral variation is ignored ($\epsilon_{\lambda} = \text{constant}$).
This allows us to perform radiation calculations using a single total hemispherical emissivity value ($\epsilon$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question focuses on the definition and characteristics of a grey surface in radiation heat transfer.

Step 2: Key Formula or Approach:
The monochromatic emissivity ($\epsilon_{\lambda}$) of a surface is defined as the ratio of its spectral emissive power to that of a perfect black body at the same temperature:
\[ \epsilon_{\lambda} = \frac{E_{\lambda}(\lambda, T)}{E_{b\lambda}(\lambda, T)} \]
For a perfect black body, $\epsilon = 1$. For any real body, $\epsilon < 1$.

Step 3: Detailed Explanation:

• A real surface emits radiation that varies in a complex manner with wavelength ($\lambda$), direction, and temperature.

• To simplify radiation analysis, engineering models utilize the concept of a "grey surface".

• A grey surface is an idealized material whose spectral radiation properties are constant over the entire wavelength spectrum:
\[ \epsilon_{\lambda} = \epsilon = \text{constant} \]

• This means its emissivity is independent of wavelength.

• Since no real surface can emit more radiation than a perfect black body, the emissivity of a grey surface must be strictly less than $1$ ($\epsilon < 1$).

• Option D relates to Kirchhoff's Law ($\alpha_{\lambda} = \epsilon_{\lambda}$), which is a general thermodynamic principle for any surface in thermal equilibrium, but the defining property of a grey surface specifically is that its properties do not vary with wavelength.

Step 4: Final Answer:
The emissivity of a grey surface is less than 1 and independent of wavelength.