---
title: "Radiation between two parallel grey plates"
subject: "Heat Transfer"
chapter: 4
key: "semester_5/heat_transfer:radiation-exchange-parallel-grey-plates"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/radiation-exchange-parallel-grey-plates
site: BCT Engineering
---

# Radiation between two parallel grey plates

Heat Transfer · Chapter 4: Radiation Heat Transfer

- Asked in: year not given, 6 marks
- Video: [Radiation 08: Heat Transfer between two parallel plates](https://www.youtube.com/watch?v=1A8FjBkCd3s) — HeatQuiz: Digital Teaching of Heat & Mass Transfer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/radiation-exchange-parallel-grey-plates
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-4

## Question

Derive an expression for the net radiant heat exchange per unit area between two large parallel grey plates of emissivities e1 and e2 maintained at temperatures T1 and T2. Use the radiosity-irradiation (electrical network) approach.

## Short answer

Using radiosity network with surface resistances (1-ε)/εA and space resistance 1/(A F12), for large parallel plates F12=1, A1=A2=A. Net heat per unit area: q = σ(T1^4 - T2^4) / (1/ε1 + 1/ε2 - 1). For black plates, ε1=ε2=1 gives q = σ(T1^4 - T2^4).

## Answer

#### Setup
Two large parallel plates, so that all radiation leaving one reaches the other: $F_{12} = F_{21} = 1$. Plate 1 at $T_1$ ($\varepsilon_1$), plate 2 at $T_2$ ($\varepsilon_2$), $T_1 > T_2$, both opaque grey surfaces ($\rho = 1 - \varepsilon$).

Radiosity $J$ is the total radiation leaving a surface (emitted plus reflected). The net heat leaving a surface:

$$
Q = \frac{E_b - J}{(1-\varepsilon)/(\varepsilon A)}
$$

where $(1-\varepsilon)/\varepsilon A$ is the **surface resistance**. The exchange between two surfaces uses the **space resistance** $1/(A_1F_{12})$.

#### Network
```text
 Eb1   J1              J2    Eb2
  o--/\/\--o----/\/\----o--/\/\--o
  (1-e1)/   1/(A1F12)   (1-e2)/
   (e1 A1)              (e2 A2)
```
Total resistance (series):

$$
R = \frac{1-\varepsilon_1}{\varepsilon_1A} + \frac{1}{AF_{12}} + \frac{1-\varepsilon_2}{\varepsilon_2A}
$$

With $A_1 = A_2 = A$, $F_{12} = 1$:

$$
Q_{12} = \frac{E_{b1}-E_{b2}}{R} = \frac{\sigma\left(T_1^4 - T_2^4\right)A}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1}
$$

#### Result
Per unit area:

$$
\frac{Q_{12}}{A} = \frac{\sigma\left(T_1^4 - T_2^4\right)}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1} = \varepsilon_{eff}\,\sigma\left(T_1^4 - T_2^4\right)
$$

with $\varepsilon_{eff} = \left(\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1\right)^{-1}$. For black plates $\varepsilon_{eff} = 1$ and the result becomes $\sigma(T_1^4-T_2^4)$.

## Check yourself

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

1. Why is F12 = 1 taken for two large parallel plates?
   A. Because both plates are black
   B. Because the plates are at the same temperature
   C. Because the plates have equal emissivities
   D. Because all radiation leaving one plate reaches the other
   Answer: D. For infinitely large close plates, no radiation escapes to the surroundings, so the view factor is 1. (Key point: Shape factor for large plates)

2. What are the three resistances in series in the network between Eb1 and Eb2?
   A. (1-e1)/(e1 A), 1/(A F12), (1-e2)/(e2 A)
   B. e1 A, F12, e2 A
   C. 1/e1, 1/e2 and 1/sigma
   D. (1-e1)/A, F12/A, (1-e2)/A
   Answer: A. Two surface resistances (1-e)/(eA) and one space resistance 1/(A F12) add in series. (Key point: Network resistances)

3. The net exchange per unit area between the plates is
   A. sigma (T1^4 - T2^4) (e1 + e2 - 1)
   B. sigma (T1^4 - T2^4) / (e1 + e2)
   C. sigma (T1^4 - T2^4) / (1/e1 + 1/e2 - 1)
   D. sigma (T1 - T2) / (1/e1 + 1/e2 + 1)
   Answer: C. Summing the series resistances with A1 = A2 and F12 = 1 gives the denominator 1/e1 + 1/e2 - 1. (Key point: Final result)

4. What does the result reduce to for black plates (e1 = e2 = 1)?
   A. Zero heat exchange
   B. sigma (T1^4 - T2^4)
   C. 2 sigma (T1^4 - T2^4)
   D. sigma (T1^4 + T2^4)
   Answer: B. With e = 1 the denominator is 1, so the effective emissivity is 1. (Key point: Black plate limit)

## Related in Heat Transfer

- [Effect of radiation shield between plates](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aradiation-shield-between-parallel-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

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