---
title: "Effect of radiation shield between plates"
subject: "Heat Transfer"
chapter: 4
key: "semester_5/heat_transfer:radiation-shield-between-parallel-plates"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/radiation-shield-between-parallel-plates
site: BCT Engineering
---

# Effect of radiation shield between plates

Heat Transfer · Chapter 4: Radiation Heat Transfer

- Asked in: year not given, 8 marks
- Video: [Radiation Shield](https://www.youtube.com/watch?v=IwCHpHDxGIQ) — Ekeeda
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/radiation-shield-between-parallel-plates
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-4

## Question

Two large parallel grey plates have emissivities 0.8 and 0.5 and are at 800 K and 500 K respectively. Calculate (a) the net radiant heat exchange per square metre, (b) the heat exchange if a thin radiation shield of emissivity 0.1 on both sides is placed between them, (c) the percentage reduction in heat transfer, and (d) the temperature of the shield. Use sigma = 5.67 x 10^-8 W/m^2 K^4.

## Short answer

Net radiant heat exchange without shield: 8747 W/m². With a radiation shield (ε=0.1 on both sides) placed between them, heat exchange reduces to 926 W/m², a reduction of 89.4%. The shield temperature is 701.5 K.

## Answer

For two parallel grey plates $q = \dfrac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}-1}$.

$$
\sigma(T_1^4-T_2^4) = 5.67\times10^{-8}\,(800^4 - 500^4) = 5.67\times10^{-8}(4.096\times10^{11}-6.25\times10^{10}) = 19\,681\ \text{W/m}^2
$$

#### (a) Without shield

$$
q_{12} = \frac{19\,681}{\dfrac{1}{0.8}+\dfrac{1}{0.5}-1} = \frac{19\,681}{2.25} = 8747\ \text{W/m}^2
$$

#### (b) With shield (emissivity $\varepsilon_3 = 0.1$ on both sides)
The shield splits the space into two gaps in series: plate 1 to shield, and shield to plate 2. The total resistance is the sum of the resistances of the two gaps:

$$
\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_3}-1\right)+\left(\frac{1}{\varepsilon_3}+\frac{1}{\varepsilon_2}-1\right)
= (1.25+10-1)+(10+2-1) = 10.25 + 11 = 21.25
$$

```text
 Plate 1    Shield    Plate 2
 800 K   |  T3  |    500 K
 e=0.8   | e=0.1|    e=0.5
```

$$
q_{with} = \frac{19\,681}{21.25} = 926\ \text{W/m}^2
$$

#### (c) Reduction

$$
\frac{8747 - 926}{8747}\times100 = 89.4\%
$$

#### (d) Shield temperature
Heat from plate 1 to shield equals $q_{with}$:

$$
q_{with} = \frac{\sigma(T_1^4-T_3^4)}{10.25}\;\Rightarrow\;T_3^4 = 4.096\times10^{11}-\frac{926\times10.25}{5.67\times10^{-8}} = 2.422\times10^{11}
$$

$$
T_3 = 701.5\ \text{K}
$$

Check with the other side: $T_3^4 - T_2^4 = 2.422\times10^{11}-6.25\times10^{10} = 1.797\times10^{11}$; $\sigma\times1.797\times10^{11}/11 = 926$ W/m$^2$. Correct.

**Answer:** (a) 8747 W/m$^2$; (b) 926 W/m$^2$; (c) 89.4 % reduction; (d) 701.5 K.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For plates with emissivities 0.8 and 0.5, the denominator in the exchange formula is
   A. 0.8 + 0.5 - 1 = 0.3
   B. 1/0.8 + 1/0.5 - 1 = 2.25
   C. 1/0.8 + 1/0.5 = 3.25
   D. 0.8 x 0.5 = 0.4
   Answer: B. The denominator is 1/e1 + 1/e2 - 1 = 1.25 + 2 - 1 = 2.25, giving 8747 W/m^2. (Key point: Two-plate formula)

2. How is the total resistance found when a shield is inserted between the plates?
   A. Resistances of the two gaps are added: (1/e1 + 1/e3 - 1) + (1/e3 + 1/e2 - 1)
   B. The shield's emissivity replaces e1
   C. The resistance is divided by two
   D. Only the shield emissivity matters: 1/e3
   Answer: A. The shield makes two gaps in series, each with its own exchange formula, giving 10.25 + 11 = 21.25. (Key point: Effect of shield on resistance)

3. Why does the shield reduce the heat transfer by about 89 percent?
   A. It lowers the plate temperatures
   B. It raises the emissivities of the plates
   C. Its low emissivity (0.1) greatly increases the total radiative resistance
   D. It absorbs all radiation
   Answer: C. Resistance rises from 2.25 to 21.25, so q falls from 8747 to 926 W/m^2. (Key point: Reduction in heat transfer)

4. How is the shield temperature found?
   A. Average of 800 K and 500 K
   B. Set heat from plate 1 to shield equal to the net flux q_with and solve for T3
   C. Equal to the temperature of the colder plate
   D. T3 = T1 x e3
   Answer: B. In steady state the same flux flows through both gaps; using q = sigma(T1^4 - T3^4)/10.25 gives T3 of about 701.5 K, which the other gap confirms. (Key point: Shield temperature)

## Related in Heat Transfer

- [Radiation between two parallel grey plates](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aradiation-exchange-parallel-grey-plates) — Heat Transfer, chapter 4
- [Radiation properties and black and grey bodies](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aabsorptivity-reflectivity-kirchhoff-black-grey) — Heat Transfer, chapter 4