Question

The temperatures of two large parallel plates of equal emissivity are \( 900 \, \text{K} \) and \( 300 \, \text{K} \). A reflective radiation shield of low emissivity and negligible conductive resistance is placed parallelly between them. The steady-state temperature of the shield, in \( \text{K} \), is:

• A. 715 K
• B. 359 K
• C. 659 K
• D. 859 K

Hint:

For radiation problems with a shield, use the fourth power law of temperature and average the radiation heat flux from the two surfaces.

Answer 1

The Correct Option is A

Step 1: Radiation heat transfer with a shield. For a radiation shield between two plates at temperatures \( T_1 \) and \( T_2 \), the steady-state temperature \( T_s \) of the shield satisfies: \[ T_s^4 = \frac{T_1^4 + T_2^4}{2}. \] Step 2: Substitute the given values. Given: \[ T_1 = 900 \, \text{K}, \quad T_2 = 300 \, \text{K}. \] Substitute into the equation: \[ T_s^4 = \frac{900^4 + 300^4}{2}. \] Step 3: Calculate \( T_s^4 \). \[ T_s^4 = \frac{(900)^4 + (300)^4}{2} = \frac{(6.561 \times 10^{10}) + (8.1 \times 10^8)}{2}. \] Simplify: \[ T_s^4 = \frac{6.6421 \times 10^{10}}{2} = 3.32105 \times 10^{10}. \] Step 4: Solve for \( T_s \). \[ T_s = \sqrt[4]{3.32105 \times 10^{10}} \approx 715 \, \text{K}. \] Step 5: Conclusion. The steady-state temperature of the shield is \( 715 \, \text{K} \).