1

2.7 Radiation2.7.1 Definitions and laws Heat transfer by conduction and convection required the existence of a material medium, either a solid or a fluid. However, it is not required in heat transfer by radiation. Radiation can travel through an empty space at the speed of light in the form of an electromagnetic wave. As shown in the electromagnetic spectrum in Fig. 2.7.1, thermal radiation covers the range of wavelength from 0.1~100 μm.

2.7.1.1 Absorptivity

Thermal radiation impinging on the surface of an opaque solid is either absorbed or reflected. The absorptivity is defined as the fraction of the incident radiation that is absorbed:

2

)(

)(

i

a

q

q

[2.7.1] Where q(a) is the energy absorbed per unit area per unit time and q(i) is the energy impinging per unit area per unit time.

Let us define q(a) and q

(i) such that q(a)dand q(i)d represent respectively the absorbed and incident energies per unit area per unit time in the wavelength range to+d. The monochromatic absorptivity is defined as :

)(

)(

)(

)(

i

a

i

a

q

q

dq

dq

[2.7.2]

For any real body αλ< 1 and depend on . A graybody is a hypothetical one for which αλ < 1 but independent of and temperature. The limiting case of αλ = 1 for all . and temperature is known as a blackbody. In other words, a blackbody absorbs all the incident radiation.

2.7.1.2 Emissivity

The emissivity of a surface is defined as )(

)(

eb

e

q

q [2.7.3]

Where q(e) and qb(e) are the energies emitted per unit area per unit time by a

real body and a blackbody, respectively. Table 2.7-1 lists the emissivities of some Material. These averaged value have been used widely even though the emissivity actually depends on the wavelength and the angle of emission.

3

Let us define q(e) and qb

(e) such that q(e)dand qb

(e)d represent respectively the energies emitted per unit area per unit time in the wavelength range to+dof a real body and blackbody. The monochromatic emissivity is defined as :

)(

)(

)(

)(

eb

e

eb

e

q

q

dq

dq

[2.7.4]

=1 for blackbody and <1 for a real body.

4

2.7.1.3 Kirchhoff’s law

Consider the body enclosed in the cavity shown in Fig. 2.7.2. Suppose the body is at the same temperature as the wall of the cavity, that is, the two are at thermodynamic equilibrium with each other. Since there is no net heat transfer from the cavity wall to Thebody, the energy emitted by the body must be equal to the energy absorbed.

AqAq ie )()( [2.7.5]

Where q(e) is the energy emitted per unit area of the body per unit time, A the surfacearea of the body, q(i) the energy impinging per unit area of the body per unit time, and the absorptivity of the body. If the body is replaced by a blackbody, Eq.[2.7.5] becomes

AqAq ieb

)()( [2.7.6]

Dividing Eq. [2.7.5] by Eq.[2.7.6], we get )(

)(

eb

e

q

q [2.7.7]

Substituting Eq. [2.7.7] into Eq.[2.7.3], we have [2.7.8]

5

2.7.1.4 Plank’s distribution law

Plank derived the following equation for the energy emitted by a blackbody as afunction of the wavelength and temperature:

[2.7.10]

Where T = temperature in K

c = speed of light h = Plank’s constant k = Boltzmann’s constant

2 5( ) 2

exp( / ) 1eb

c hq

ch k T

This is Kirchhoff’s law, which states that for a system in thermodynamic equilibrium the emissivity and absorptivity are the same. Following a similar procedure, we can show that

[2.7.9]

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2.7.1.6 The solid angle

2.7.1.5 Stefan-Boltzmann law

The Plank distribution law can be integrated over wavelengths from zero to infinity to determine the total emissive energy of a blackbody:

[2.7.11]

or 4)( Tq e

b [2.7.12]

Where (Stefan-Boltzmann constant) = 5.676 x 10-8 Wm-2K-4.

Consider a hemisphere of radius r surrounding a differential area dA1at the center. Fig. 2.7-4a shows only one-quarter of the hemisphere. On the hemisphere

[2.7.13]

5 4 4( ) ( )

2 30

2

15e eb b

k Tq q d

c h

2 ( )( )

( sin )( )

dA ab cd

r d rd

7

The solid angle that intersects dA2 on the hemisphere, shown in Fig.2.7-4b, is defined as:

ddr

dAd sin

22 [2.7.14]

The energy leaving dA1 in the direction given by the angle θis IbdA1cosθ, where Ib is the blackbody intensity. The radiation arriving at some area dA2 at a distance r from A1 would be

Where dA2 is constructed normal to the radius vector. The dA2/r2 represents the solid angle subtended by the area dA2 which can be expressed as

so that2 / 2( )

1 1 20 0

2 / 2

1 10 0

sin cos

sin cos

eb b

b b

r d rdq dA I dA

r

I dA d d I dA

2 ( sin )( )dA r d rd

21 2

( cos )b

dAI dA

r

J. P. Holman, “Heat Transfer”, 1977, p.285

r1

r2dθ

r1dθr2dθ

8

This yields the following equation:

[2.7-18]

and

2.7.2 Radiation between blackbodies

Consider two black surface A1 and A2 shown in Fig. 2.7-6. Define the view factors:

F12= fraction of energy leaving surface 1 that reaches surface 2F21= fraction of energy leaving surface 2

that reaches surface 1

( )eb

b

qI

( )eb bq I

9

As such, the energy leaving surface 1 that reaches surface 2 per unit time is

Similarly, the energy leaving surface 2 that reaches surface 1 per unit time is

Therefore, the net energy exchange rate is as follows:

If both surfaces are at the same temperature T, there is no net energy exchangeand Q12=0. Then substituting Eq.[2.7-12] into Eq. [2.7-22], we have

so that

Which is called the reciprocity relationship. Substituting Eqs. [2.7-12] and [2.7-24]into [2.7-22], we get

[2.7-20]

[2.7-21]

[2.7-22]

[2.7-23]

[2.7-24]

[2.7-25]

( )

1 2 1 1 12

e

bQ q A F

( )

2 1 2 2 21

e

bQ q A F

( ) ( )

12 1 2 2 1 1 1 12 2 2 21

e e

b bQ Q Q q AF q A F

412 1 12 2 21 0Q T AF A F

1 12 2 21A F A F

4 4 4 412 1 12 1 2 2 21 1 2Q AF T T A F T T

10

Let us now proceed to determine the view factors F12 and F21 between the two black surfaces. From Eq. [2.7-19]

[2.7-26]

where dΩ12 is the solid angle subtended by dA2 as seen from dA1. Let r12 be the distance between the centers of dA1 and dA2. Consider the hemisphere of radius r12 and centered at dA1. The projection of dA2 on the hemisphere is cos2dA2. Therefore, from Eq.[2.7-14]

[2.7-27]

Substituting Eq.[2.7-27] into Eq. [2.7-26]

And so

Similarly it can be shown that

[2.7-28]

[2.7-29]

[2.7-30]

( )1

1 2 1 1 12cosebqdQ dAd

2 212 2

12

cos dAd

r

( ) 1 2 1 21 2 1 2

12

cos coseb

dAdAdQ q

r

( ) 1 21 2 1 1 22

12

cos cosebQ q dAdA

r

( ) 1 22 1 2 1 22

12

cos cosebQ q dAdA

r

11

Substituting Eq. [2.7-29] and Eq. [2.7-30] respectively into Eqs. [2.7-20] and[2.7-21], we get Eqs. [2.7-31] and [2.7-32]

[2.7-31]

[2.7-32]

and

The integration in Eq. [2.7-31] and [2.7-32] is often difficult and needs to be done numerically. Analytical equations are available for a number of special cases. Some examples are given below:

1 212 1 22

1 12

cos cos1F dAdA

A r

1 221 1 22

2 12

cos cos1F dAdA

A r

12

For large (infinite) parallel plates, long (infinite) concentric cylinders and concentric spheres F12 = 1, as shown in Fig.2.7-7.

13

For the two identical, parallel directly opposed rectangles shown in Fig. 2.7-8.

where X and Y are defined in the figure.[2.7-33]

1 22 2

2 1 2 1 1 112 2 2 2 2

1 12ln 1 tan 1 tan tan tan

1 1 1

X Y X YF X Y Y X X X Y Y

XY X Y Y X

14

For the two parallel concentric circular disks shown in Fig. 2.7-9,

[2.7-34]

where X, R1, and R2 are defined in the figure.

For the sphere and the disk shown in Fig. 2.7-10,

[2.7-35]

2

2 212

1

14

2

RF X X

R

12 22

1 11

2 1F

R

15

Example

Consider the radiation from the small area dA1 to the flat disk A2, as shown in the Fig. The element of area dA2 is chosen as the circular ring of radius x. Calculate FdA1A2.

16

2.7.3 Radiation between graybodies

In radiation heat transfer between blackbodies, all the radiant energy that strikes a surface is absorbed. In radiation heat transfer between nonblackbodies, the energy striking a surface will not all be absorbed; part will be reflected. Since the energy emitted and the energy reflected by a nonblack surface both contribute to the total energy leaving the surface (J), we can write

[2.7-36]

Where J is called radiosity, is the total energy leaving a surface per unit area per unit time and , called the reflectivity, is the fraction of the incident energy reflected.

From the definition for J, we see that

J - q(i) = net energy leaving a surface per unit area per unit timeand q(i) - J= net energy received at a surface per unit area per unit time

For an opaque material the incident energy is either reflected or absorbed:

+ =1 [2.7-37]

( ) ( ) ( ) ( )e i e ibJ q q q q

17

If Kirchhoff’s law can be applied, that is, = according to Eq. [2.7-8], Eq.[2.7-37] becomes + = 1 [2.7-38]

From Eqs. [2.7-36] and [2.7-38]

[2.7-39]

Let us consider radiation heat transfer between two gray surface A1 and A2. Since the net energy transfer from A1 to A2 (Q12) equals either the net energy leaving A1 or the net energy received at A2, we can write, with the help of Eq. [2.7-39]:

and

[2.7-40]

[2.7-41]

( ) ( ) ( )( ) (1 )

1 1 (1 1)

e e ei b b bJ q J J q q J

J q J

1( )12( )

( )1 1 112 1 1 1 1 1 1

1 1

1( )

1 1

ei

ebb

Qq J

Q A J q A q JA

2( )12

( ) ( )2 2 212 2 2 2 2 2 2

2 2

1( )

1 1

e

i ebb

QJ q

Q A q J A J qA

( ) ( )

( ) ( )( )( )

1

e i

e eei b b

J q q

J q J qJ qq

18

Similar to Eqs. [2.7-20] and [2.7-21] for two black surfaces, we can write the following expressions for two gray surfaces in terms of the view factors:

[2.7-42]

[2.7-43]

121121 FAJQ

212212 FAJQ

and

Since A1F12 =A2F21 according to Eq. [2.7-24]

)( 21121122112 JJFAQQQ [2.7-44]

From Eqs. [2.7-40], [2.7-44], and [2.7-41]

[2.7-45]

[2.7-46]

[2.7-47]

( )1

1 1 121

1 1e

bq J QA

1 2 121 12

1J J Q

A F

( )2

2 2 122

1 1e

bJ q QA

For blackbody( )

1 2 1 1 12ebQ q A F [2.7-20]

19

Adding these equations, we have

[2.7-48]

[2.7-49]

[2.7-50]

Substituting Eq. [2.7-12] ( ) into Eq. [2.7-48], we have

1 2( ) ( )1 2 12

1 1 12 2

1 1 1 11e eb bq q Q

A A F A

4 4 4 412 1 12 1 2 2 21 1 2Q AF T T A F T T

1 2

1 12 2 21 1 1 12 2

1 2 1 2

1 11 1 1 1

1 1 1 1 1 11 1 1 1

A A A F

A A

F F

( ) 4ebq T

4 4

4 4 4 41 21 2 12 12 1 2

12

(T T ) 1(T T ) or (T T ) Q Q

Q

For blackbody

Compared to Eq.[2.7-48a], therefore,

[2.7-48a]

1 21 2

1 1 1, therefore, A A

A A 12 21

12 21

F FF F

For A1=A2

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Example 2.7-3 Radiation shields

Given: Hot surface T1, 1, cold surface T2, 2, radiation shield T3, 3

Surfaces 1 & 2: two infinite parallel gray surface with the area of A

Find: The radiation heat transfer between two parallel gray surfaces

Since A1=A2=A3, and the surfaces are infinite and parallel F12=F32=1 (by definition, F13 = fraction of energy leaving surface 1 and reaches surface 2)

From [2.7-50] 1 3

13 1

1 3 1 3

1 1 1 11 1

1

1 1 1 1 1 11 1 1 1

A A A A

A A

F

Substituting above two equations into Eq.[2.7-49], we get

32 3 2

1 1 1 11

A A

F

21

4 4 4 41 3 3 2

13 32

1 3 3 2

( ) ( ) and

1 1 1 11 1

A T T A T TQ Q

Since Q13=Q32=Q132, we have

4 4 4 4132 1321 3 3 2

1 3 3 2

1 1 1 11 and 1

Q QT T T T

A A

Adding the two equations, we have

4 4 1321 2

1 3 3 2

1 1 1 11 1

QT T

A

and 4 4

1 2

132

1 3 3 2

1 1 1 11 1

A T TQ

22

4 41 2

12

1 2

( )

1 11

A T TQ

The ratio of radiation heat transfer with a shield to that without one is

1 2132

12

1 3 3 2

1 11

1 1 1 11 1

Q

Q

If 1=2=3=

Without a radiation shield, from Eqs [2.7-49] and [2.7-50]

132

12

21

12 2 21 1

Q

Q

23

Example: Two very large parallel planes with emissivities 0.3 and 0.8 exchange heat. Find the percentage reduction in heat transfer when a polished-aluminum radiation shield (=0.04) is placed between them.

Sol: The heat transfer without the shield is given by

4 412 1 2

1 2

( )

1 11

Q T T

A

The radiation network for the problem with the shield is placed, based on Eq. [2.7-50], is given

1 2

1 2

1 12 2 21 1 1 12 2

1 1

1 1 1

A A A A F A

F F

'b1 1 3 3 3 2 b2E J J J J EbE

3 31 2

1 13 3 3 32 2

1- 1-1- 1-1 1

F F

1

1

3

3

2

2

1 1 0.30.333

0.3

1 1 0.0424

0.04

1 1 0.80.25

0.8

The total resistance is 2.333+2 (24) + 2(1) + 0.25 =52.5834 4

4 412 1 21 2

( )=0.01712 ( )

52.583

Q T TT T

A

The heat transfer is reduced by 93.6%

E’b1 E’b2E’b3

24

Summary

1. Emission and absorption of a black body and a gray body

( ) 41

ebq T

(black body)

T1

(black body)

T2, A2

( ) ( )

( ) ( )

e eemission b

a iabsorption

q q

q q

2. Radiation between black bodies (Q12: The net energy exchange rate )

(black body)

T1, A1

(gray body)

T1

4 4 4 412 1 12 1 2 2 21 1 2( ) ( )Q AF T T A F T T

4 41 212 1 2

12 12

( ) =

1 1b bE EQ T T

AF F

b1 2E bE

12

1

F

and

and

25

Summary

(gray body)

T2, A2

3. Radiation between gray bodies

(gray body)

T1, A1

4 4 4 412 1 12 1 2 2 21 1 2( ) ( )Q AF T T A F T T

1 2

1 12 2

1 12 2 21 1 1 1

1 111 1

+ + F

AF A F A A A

1 1 2 2E J J E

1 2

1 12 2

1- 1-1

F

( ) ( )

2(J:total energy leaving a surface/time.cm )

e iJ q q

( )( )

( ) 2

1( 1)

( :net energy leaving a surface/time.cm )

ei b

i

q JJ q

J q

b1 b2E E