Chapter 4 Radiation

BKF2422

Chapter 4

Principles of Steady-state Heat Transfer in Radiation

Topic Outcomes

It is expected that students will be able to: Utilize the basic equation of radiation for black and gray

bodies Solve problems related to combined radiation and

convection heat transfer mechanism Apply view factors to determine radiation heat transfer

rate for various geometries Analyze and solve problems fro radiation heat transfer

involves absorbing gases medium

electromagnetic radiation emitted by a body due to large

temperature difference.

transmitted through space & vacuum

medium not heated (medium heated by convection)

1. Thermal energy of a hot surface at T1 is converted into

energy of electromagnetic radiation waves

2. Waves travel through intervening space in a straight lines

& strike a cool object at T2

3. Electromagnetic waves are absorbed by the body and

converted to thermal energy or heat

ransmittedfraction t ivity transmiss

absorbedfraction ty absorptivi

reflectedn irradiatio offraction ty reflectivi

1

reflection

absorption

reflection

transmission

Opaque bodies, τ = 0 α + ρ = 1

bodyblack ofpower emissive total

surface a ofpower emissive total ,Emissivity

BE

E

1 body,Black

radiation emits also -

1 it scateringor

ing transmitt,reflecting without hs wavelengtallat directions all from

on)(irradiatiradiation incident all absorb that material-body Black

1 body,Gray

1an have materials All

Kirchhoff’s law:A body at the same temperature, T1

11

4111 TAq

42121 TAq

21

212

Tat

Tat 2body from 1body ofty absorptivi where

1T1

2T 2

1

2

q

:enclosure toradiation emitsbody Small

:Tat gssurroundin thefromenergy absorbsbody Small 2

4212

4111

42121

4111 TTATATAq

42

411 TTAq

21 Tat Where

:absorption ofheat Net

:Simplified

EXAMPLE 4.10-1 Page 303 Radiation to a Metal Tube

A small oxidized horizontal metal tube with an OD of 0.0254 M and being 0.61 m long with a surface temperature at 588 K is in a very large furnace enclosure with fire-brick walls and the surrounding air at 1088 K. The emissivity of the metal tube is 0.60 at 1088 K and 0.46 at 588K. Calculate the heat transfer to the tube by radiation.

K 5881 T m 0.0254

K 1088 air, 2 T

m 0.61

12qK 10882 T

6.0@

46.0@

21

11

T

T

2m )61.0)(0254.0( DLAi

W2130

1088588)10676.5)(6.0)(0254.0(

448

42

4112

TTAq

4

:bodyblack perfect a fromradiation by fer Heat trans

TAq

(K)body black theof re temperatu

./10676.5 =constant sBoltzmann' -Stefan =

body theof area surface

where

428

2

T

KmW

mA

4

:body)(gray body black -non a fromfer Heat trans

TAq

,When 2TT

radconv qqq :ferheat trans Total 211

211

where

TTAhq

TTAhq

rrad

cconv

2at

gssurroundin

T

convqradq

11 re temperatuand area , emissivity of surface TA

21

4

2

4

1

100100676.5

TT

TT

hr

21

4

2

4

1

1001001714.0

TT

TT

hr

21111

,When

TTAhTTAhqqq

TT

rcradconv

,1 when ,1 with unitsEnglish in 2-4.10 Figure

EXAMPLE 4.10-2 Page 305 Combined Convection Plus Radiation from a Tube

Recalculate Example 4.10-1 for combined radiation plus natural convection to the horizontal 0.0254-m tube.

.K W/m3.87

5881088100

588

100

1088

676.56.0

100100676.5

2

44

21

4

2

4

1

TT

TT

hr

.K W/m64.15

0254.0

588108832.1

32.1

2

4/1

4/1

D

Thc

W2507

)1088588)(0487.0)(3.8764.15(

211

TTAhhq rc

View Factors

There is net radiant exchange between surfaces.

A net flow of energy will occur from hotter surface to the colder surface

The shape, size, and orientation of two radiating surfaces determine the heat flow.

42

4112112

:2 plane to1 plane fromradiation Net

TTFAq

42

4121121

:1 plane to2 plane fromradiation Net

TTFAq

1 plane of area surface where 1 A

1 planes, parallel infiniteFor 2112 FF

1 body,black For 21

1

111

21

42

41112

TTAq

:2 plane to 1 plane from radiationNet

1 plane of emissivity Where 1

12112

1

FF

EXAMPLE 4.11-1 Radiation Between Parallel Planes

Two parallel gray planes which are very large he emmissivities of ε1 = 0.8 and ε2 = 0.7 and surface 1 is at 1100oF and surface 2 at 600oF. Use English units for the following: What is the net radiation from 1 to 2? If the surface are both black, what is the net radiation?

6.0@7.0

46.0@8.0

21

11

T

T

BodyGray (a)

2

448

21

42

41

12

21

42

41112

W/m15010

17.0

1

8.0

11

8.5885.86610676.5

111

1

111

1

TTA

q

TTAq

BodyBlack (b)

.K W/m25110

8.5885.86610676.5 2

448

42

41

1

12

42

41112

TTA

q

TTAq

The example shows the influence of surfaces with emissivities less than 1 on radiation.

This fact is used to reduce radiation loss or gain from a surface using planes as radiation shield

For example, two parallel surfaces of emissivity at T1 and T2, the exchange is

12

)()( 42

41012

TT

A

q

12

)(

1

1)( 42

4112

TT

NA

q N

Suppose we insert one or more radiation planes between the original surfaces,

General equation for view factor between black bodies

For finite size, some of the radiation from surface 1 does not strike surface 2, and vice versa.

Hence the net radiation interchange is less, since some is loss to the surroundings.

The fraction of radiation leaving surface 1 in all directions which is intercepted by surface 2 is called F12 and must be determined for each geometry by taking differential surface element and intergrating over the entire surface.

Two important quantities..

Solid angle, ω

dω1= dA1cosθ2/r2

Intensity of radiation, IB

IB = dq/(dA cos dω)

For general case for fraction of the total radiant heat that leaves a surface (black body) and arrives at the second surface, dq1→2 = ImdA cosθ1dω1

dq2→1 = ImdA cosθ2dω2

Finally,

q12= A1F12σ(T14 – T2

4) = A2F21σ(T14 – T2

4)

Where

A1F12 = A2F21

dω1= dA1cosθ2/r2

2 1

22121

112

coscos1A A r

AdA

AF

The reciprocity relationship of A1F12 = A2F21 can be applied to any two surfaces.

AiFij = AjFji

And if surface A1 sees a number of surfaces A2, A3,….

F11 + F12 + F13 + … = 1.0

EXAMPLE 4.11-2 View Factor from a plane to a Hemisphere

Determine the view factors between a plane A1 covered by a hemisphere A2 as shown in Fig. 4.11-5

212121 FAFA

1.0. factor view the,only see surface since 1221 FAA

2

1

2)0.1(

2

2

2

11221

R

R

A

AFF

0.1

,0010101 surfaceFor

2122

12111

FF

..F., FA

2

1

2

10.10.1

,for Solving

212122

22

FFF

F

EXAMPLE 4.11-3 Radiation Between Parallel Disks

In figure 4.11-6, a small disk of area A1 is parallel to a large disk of area A2 and A1 is centered directly below A2. The distance between the centers of the disks is R and the radius of A2 is a. Determine the view factor for radiant heat transfer from A1 to A2

a

A

A A

A A

dxxr

dxxrA

A

dxxr

dA

A

r

dAdA

AF

0 21

2

21

2

1

1

2121

1

22121

112

2

cos

2

cos

2

coscos1

coscos1

1

2 1

2 1

dxxdA 22

21

2/122 xR

R

1 cos

22

2

12 aR

aF

View factors when black bodies surfaces are connected by reradiating walls

Example: furnace

A larger fraction of the radiation from surface 1 is intercepted by 2. View factor =

For the case of two black bodies surfaces A1 & A2 connected with reradiating walls:

1 2

Refractory reradiating wall

12F

122121

21221

12121

21221

12 /21

/1

2 FAAAA

FAA

FAAA

FAAF

212121 FAFA 42

4111212 TTAFq

View factors when gray bodies surfaces are connected by reradiating walls

• For the case of two gray bodies surfaces A1 & A2 which cannot see themselves and connected with reradiating walls:

F12

q12 = F12 A1 σ (T14 – T2

4)

A1F12 = A2 F21

• For black or gray bodies, with no reradiating walls, again .

1

11

11

1

122

1

12 AA

F

1212 FF

View Factors F12 = fraction of the radiation leaving surface 1 that is

intercepted/reaching by surface 2.

2 1

22121

112

coscosA A r

dAdAAF

General equation:

212121: relationsy Reciprocit i FAFA

1... :ruleSummation ii 131211 FFF

0 :convex)or (flat itself seecannot 1 Surface iii 11 F

1A

2A

1 :2 surface seeonly can 1 Surface iv 12 F

32231132.12.1

2331332.133

23132.13

FAFAFA

FAFAFA

FFF

algebra Similar to v

1A

3A

2A

2.1A

EXAMPLE 4.11-5 Complex View Factor for Perpendicular Rectangles

Find the view factor F12 for the configuration shown in Fig. 4.11-9 of the rectangle with area A2 displaced from the common edge of rectangle A1 and perpendicular to A1. The temperature of A1 is T1 and that of A2 and A3 is T2

11 1

2

3 3

1A

2A

3A

1312)23(1

131121)23(11

42

41131

42

41121

42

41)23(11

FFF

FAFAFA

TTFATTFATTFA

13)23(112 FFF

+

1

2

1r

2r

?

surfaceblack areBoth

22 F

Example

2

2

122

2

2

12

2

21

22

21

2

121

212121

2122

2221

1211

1211

1

4

4

1

1

1 ,0

1

r

rF

r

r

r

r

r

r

A

AF

FAFA

FF

FF

FF

FF

sphereFor

Radiation in Absorbing Gas

• Most mono atomic & diatomic gases are transparent to thermal radiation. (He, H2, Ar, O2, N2).

Dipole gases and higher polyatomic gases emit and absorb radiant energy significantly. (CO2, CO, H2O, SO2, NH3).

Absorption of thermal radiant by gas depend on pressure and temperature.

i.e. partial pressure of gases, amount of absorption.

If the gas is heated, it radiates energy to surrounding.

Characteristic Mean Beam Length of Absorbing Gas

The absorption of energy by gas T1, P and characteristic length, L (mean beam length). Mean beam length, L specific geometry.

The mean beam length has been evaluated for various geometries in Table 4.11-1 (pg. 321).

For other shapes,

where,

L = m, V = m3 of gases A = surface area of enclosure m2.

A

VL 6.3

εG, αG and radiation of a gas

q = εG TG

4 - (rate of radiation emitted) from the gas.

q = α αG T14 - (rate of absorption of energy as it is

radiated back to the gas from midpoint of surface element.)

The net rate of radiant transfer between a gas at TG and a black surface of finite area A1 at T1.

41

4 TTAq GGG

Figure 4.11-10 gives the gas emissivity, εG of CO2 at a total pressure of the system = 1.0 atm abs using

at TG.

pG = partial pressure of CO2 in atm.

L = mean beam length in m (Table 4.11-1 ). αG of CO2 is determined also from Figure 4.11-10 but at T1.

(T1 =temperature of midpoint of surface element.)

But instead of using pG L, need to use:

• The value obtained from the chart is then multiplied by

correction factor .

GG T

TLp 1

65.0

1

TTG

LpG

For the case when wall of enclosure are not black. An approximation when the emissivity is greater than 0.7, effective emissivity, ε’ can be used

Hence the radiant transfer between a gas at TG and not black surface at T1

41

4' TTAq GGG

2

0.1'

Example 4.11.7 Page 321

A furnace is in the form of a cube 0.3 m on a side inside and these interior walls can be approximated as black surfaces. The gas inside at 1.0 atm total pressure at 1100K contains 10 mol% CO2 and the rest is O2 and N2. The small amount of water vapor present will be neglected. The walls of the furnace are maintained at 600K by external cooling. Calculate the total heat transfer to the walls neglecting heat transfer by convection.

Ans: 2.6 kW