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:
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:
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For radiation problems with a shield, use the fourth power law of temperature and average the radiation heat flux from the two surfaces.
Updated On: Jan 24, 2025
- 715 K
- 359 K
- 659 K
- 859 K
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The Correct Option is A
Solution and Explanation
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 \( 715 \, \text{K} \).
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