Radiation Shield Temperature Between Parallel Plates
This solution determines the steady-state temperature of a radiation shield positioned between two large parallel plates based on their temperatures and emissivities.
Problem Overview
- Plate 1 Temperature: $T_1 = 900$ K
- Plate 2 Temperature: $T_2 = 300$ K
- Plate Emissivities: $\epsilon_1 = \epsilon_2 = \epsilon$
- Shield Properties: Low emissivity ($\epsilon_{sh}$), negligible conductive resistance, parallel placement.
- Objective: Calculate the shield's steady-state temperature, $T_{sh}$.
Heat Transfer Principles
The core principle is the energy balance at the shield. In steady state, with negligible conduction, the net radiation absorbed by the shield equals the net radiation emitted. This implies the net heat flux ($q$) leaving Plate 1 towards the shield is equal to the net heat flux leaving the shield towards Plate 2.
Heat Flux Equations
The net heat flux per unit area ($q$) between surfaces is calculated using the appropriate thermal resistance:
Heat flux from Plate 1 to the shield:
$ q = \frac{\sigma (T_1^4 - T_{sh}^4)}{\frac{1}{\epsilon_1} + \frac{1}{\epsilon_{sh}} - 1} $Heat flux from the shield to Plate 2:
$ q = \frac{\sigma (T_{sh}^4 - T_2^4)}{\frac{1}{\epsilon_{sh}} + \frac{1}{\epsilon_2} - 1} $Where $\sigma$ is the Stefan-Boltzmann constant.
Deriving Shield Temperature Formula
Given that the emissivities of the plates are equal ($\epsilon_1 = \epsilon_2 = \epsilon$), the denominators of the heat flux equations become identical:
$ R_{denom} = \frac{1}{\epsilon} + \frac{1}{\epsilon_{sh}} - 1 $By equating the two expressions for $q$:
$ \frac{\sigma (T_1^4 - T_{sh}^4)}{R_{denom}} = \frac{\sigma (T_{sh}^4 - T_2^4)}{R_{denom}} $After canceling common terms ($\sigma$ and $R_{denom}$), we get:
$ T_1^4 - T_{sh}^4 = T_{sh}^4 - T_2^4 $Rearranging to solve for $T_{sh}$:
$ T_1^4 + T_2^4 = 2 T_{sh}^4 $ $ T_{sh}^4 = \frac{T_1^4 + T_2^4}{2} $ $ T_{sh} = \left( \frac{T_1^4 + T_2^4}{2} \right)^{1/4} $This formula gives the shield temperature when plate emissivities are equal. The shield's low emissivity ensures it effectively reflects radiation, but its specific value cancels out in this derivation due to equal plate emissivities.
Temperature Calculation
Substitute the provided temperatures into the derived formula:
$ T_{sh} = \left( \frac{(900 \text{ K})^4 + (300 \text{ K})^4}{2} \right)^{1/4} $Calculate the fourth powers:
$ 900^4 = 6561 \times 10^8 \text{ K}^4 $ $ 300^4 = 81 \times 10^8 \text{ K}^4 $Substitute these values back:
$ T_{sh} = \left( \frac{6561 \times 10^8 + 81 \times 10^8}{2} \right)^{1/4} \text{ K} $ $ T_{sh} = \left( \frac{6642 \times 10^8}{2} \right)^{1/4} \text{ K} $ $ T_{sh} = \left( 3321 \times 10^8 \right)^{1/4} \text{ K} $Simplify the expression:
$ T_{sh} = (3321)^{1/4} \times (10^8)^{1/4} \text{ K} $ $ T_{sh} = (3321)^{1/4} \times 10^2 \text{ K} $Calculating the fourth root:
$ (3321)^{1/4} \approx 7.59 $Final calculation:
$ T_{sh} \approx 7.59 \times 100 \text{ K} $ $ T_{sh} \approx 759 \text{ K} $Conclusion
The calculated steady-state temperature of the radiation shield is 759 K.
Important Questions from Radiation
Stefan Boltzmann's constant is expressed in the unit-
A body whose absorptivity does not vary with temperature and wavelength of the incident ray is known as
The process of heat transfer from a hot body to a cold body in a straight line, without affecting the intervening medium, is known as ______.
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- Radiosity is defined as _______.
