Mec3609 ht radiation

ME C3609 ME C3609 AN INTRODUCTION TO RADIATION AN INTRODUCTION TO RADIATION

HEAT TRANSFERHEAT TRANSFERby by

M.N.A. HawladerM.N.A. Hawlader

RADIATION HEAT TRANSFER

Lecturer: Dr M.N.A. Hawlader Contents:

Fundamental Concepts Laws of blackbody radiation Intensity and shape factor Radiation exchange between

blackbody and gray surfaces Radiation shield

RADIATION HEAT TRANSFER

References

1.Heat Transfer by J.P. Holman, Seventh Edition, McGraw Hill, Singapore 1992

2. Principles of Heat Transfer by Frank Kreith and M.S. Bohn, Fourth Edition, Harper and Row, Singapore 1986.

3. Fundamentals of Heat Transfer by F.P. Incropera and D.P. Dewitt, John Wiley and Sons, Singapore 1985

RADIATION HEAT TRANSFERFundamental Concepts

Conduction and convection: require a medium to transport energy

(a)Conduction +

Convection

Fluid

(b) Conduction/convection

Fig.1 Heat transfer in a (a) fin and (b) pipe flow

Radiation : no carrier or medium is required

Radiation : no carrier or medium is required

Medium: Participating & Non-participating

Hotbody

vacuum

Enclosure

Q

q

q < Q

ParticipatingMedium

Q

Q

Non-participating

For gas and semi-transparent materials, radiation interaction takes place within the volume of the medium.

For opaque materials,absorption, reflectionand emission takes placeat the surface


Irradiation: radiation incident on a surface, W/m2

incident

transmitted

reflected

Absorptance = Gabsorbed/Gincident

= Reflectance = Greflected/Gincident

= Transmittance = Gtrasmitted/Gincident

=

Radiosity: radiation leaving a surface due to reflection and emission

diffuse

specular

radiosity

Absorptance, Reflectance and Transmittance

For semi transparent material + + = 1 If the properties are averaged over the entire spectrum + + = 1 For opaque medium – + = 1 + = 1


Nature of radiation

X-rays UltravioletThermal Microwave

Gama rays Radiation

10-5 10-4 10-3 10-2 10-1 1 10 102 103 104

Figure: Spectrum of electromagnetic radiation


Thermal RadiationSpectral Range for thermal radiation: 0.2 µm 100 µm

0.01 0.36 µm uv

0.36 0.76 µm visible (0.55 – 0.56µm) most sensitive λ > 0.76 µm Infrared

Thermal radiation and light: Light is the visible part of thermal radiation


Radiation is emitted at different wavelength and different direction

Spectral distribution Directional distribution

E

wavelength(a) (b)

E = Monochromatic radiation emission

Figure Radiation emitted by a surface. (a) Spectral distribution. (b) Directional distribution


Plane and Solid angles

r r

d d dl dAn

Solid Angle = ratio of the element, Plane angle = ratio of the element dAn and the square of arc length and of the radius, r radius, r = dAn / r

2 = dl/r


Radiation Intensity

N

X

Z (a) (b)

Figure (a) Directional nature of radiation, (b) spherical co-ordinate system

dAn

dA1

Intensity I of radiationat any is defined asthe rate at which theradiant energy is emitted in ( direction per unit dAn

per unit solid angle,per unit wavelength.W/(m2.sr. m)


Blackbody Radiation absorbs all radiation irrespective of

wavelength and direction; no surface can emit more energy than a

blackbody for a prescribed temp. and wavelength;

the blackbody is a diffuse emitter.

G= E b

I e= I bI i

Isothermal blackbody enclosure


The Planck DistributionRadiation intensity for blackbody emission is given by

I,b (,T) = 1/exp

25

2

kThc

hc

o

o

h = universal Planck constant = 6.6256 x 10 -34 (J.s)k = Boltzman constant = 1.3805 x 10 -23 (J/K)co = speed of light = 2.988 x108 (m/s)T = absolute temp. of the blackbody (K)


Planck Distribution EquationSince the blackbody is a diffuse emitter,

E,b (,T) = I,b (,T)

= 1/exp25

1

TC

C

Where,C1 = 2hco

2 = 3.742 x 108 (W.m4/m2)C2 = (hco/k) = 1.439 x 104 (m.K)

Example

Emissive power of a blackbody is 1kW/(m2.m) at a wavelength of 4 m. Find the temperature of the body.

Solution:

Given- λ = 4 m

Ebλ(T) = 1kW /(m2.m)

Find – T, the temperature of the blackbody.

E,b (,T) = I,b (,T)

= 1/exp25

1

TC

C

Where,C1 = 2hco

2 = 3.742 x 108 (W.m4/m2)C2 = (hco/k) = 1.439 x 104 (m.K)

]1)}4/(10439.1[exp{4

10743.31000

45

8

Tx

x

KTT

x

609

53.366ln4

10439.1 4

Exercise

There is a small circular opening of 40 mm in diameter in a large spherical cylinder whose inner surface is maintained at 527oC. Find the rate of emission of radiation through this opening.

Outline of solution

Emissive power at the given temp

Eb(T) =σT4

=(5.67x10-8)(800)4

Q = A Eb(T) , Area of the aperutre


Wien’s Displacement Law

The black body spectral distribution is characterised by maximum and the wavelength associated with this maximumdepends upon temperature. Differentiating the followingequation w.r.t

E,b (,T) = 1/exp25

1

TCC

and setting it to zero, gives

maxT = C3 = 2897.6, m.K


The Stefan- Boltzman Law

Total emissive power

E = dE

0

=

0 1/exp2

51

TCC

d

= T4 = Stefan Boltzman constant = 5.67 x 10-8 (W/m2.K4)


Band Emission

Eb(T)

0 1

Eb(T)dF 0- =

0

,

0

,

dE

dE

b

b

= f(T)

F1-2 =

0

,

2

0

1

0

,,

dE

dEdE

b

bb

= f(T)

= F (0-2) - F(0-1)

Fraction of radiation emitted withina wavelength band:

1 2

Radiation Example

Example 3Consider a large isothermal enclosure that is maintained at a uniform temperature of 2000 K.1. Calculate the emissive power of the radiation that

emerges from a small aperture on the enclosure surface.

2. What is the wavelength 1 below which 10% of the emission is concentrated. What is the wavelength above which 10% of the emission is concentrated.

3. Determine the maximum emissive power and the wavelength at which this emission occurs.

4. If a small object is placed within the enclosure, what is the irradiation incident on the object

Example 3 (contd)

Assumption: aperture area is very small compared to the surface area of the enclosure

Enclosure T = 2000 K

E, b (T)

10%

10%

1

2

Example 3-Solution1. E = Eb (T) = T4 = 5.67 x 10-8

42Km

W (2000)4 (K)4

= 9.07 x 105 W/m2

2. 1.0F10 from table 1

1T 2200 mK 1 = 1.1 m

9.0F20 2T 9382 mK

2 = 4.69 m

Example 3-Solution(contd)

3. From Wien’s Displacement Law maxT = 2898 mK

For T = 2000, max = 1.45 m

Maximum emissive power:

1T/Cexp

CT,E

max2max5

1maxB,

=

1Km2898

Km10439.1expm45.1

m/mW10742.34

55

248

= 4.10 x 105 W/m2 m

Example 3-Solution(contd)

4. The answer is same as (1)

E = Eb (T) = T4 = 5.67 x 10-8

42Km

W (2000)4 (K)4

= 9.07 x 105 W/m2

Example 4A diffuse surface at a temperature of 1600 K has the spectral, hemispherical emissivity shown below:

1. Determine total, hemispherical emissivity 2. Calculate the total emissive power 3. At what wavelength will the spectral emissive power

be a maximum?

0.8

0.4

2 4 6

()

, m

0 0

2

1

Example 4-solutionAssumption: Surface is a diffuse emitter. Equation (37)

(T) = )(

),(),(0

,

TE

dTET

b

b

(37)

= b

b

E

dE2

0

,1

+ b

b

E

dE5

2

,2

= 1 F(0 - 2 m) + 2 F(2 - 5 m)

or = 1 F(0 - 2 m) + 2[F(0-5 m) – F(0 – 2 m)]

Example 4-solution(contd)

1T = 2 (m) x 1600 (K) = 3200 (mK); F0 – 2 = 0.318 2T = 5 (m) x 1600 (K) = 80, 000 (mK); F0 – 5 = 0.856 = 0.4 x 0.318 + 0.8 (0.856 – 0.318) = 0.558


2. E = Eb = T4 = 0.558 x 5.67 x 10-8 (W/m2 K4) 16004 (K4) = 207 kW/m2 3. max = 2898

1600 = 1.81 m

Since is different for different , we have to calculate emissive power for both

Exercise

The filament of tungsten bulb is heated to a temperature of 2227oC. Find the fraction of the enrgy in the visible range. The visible range of the spectrum may be considered as

0.4 m ≤ λ ≤ 0.7 m

Example Example 7A flat plate collector with no cover plate has a selective absorber surface emittance of 0.1 and a solar absorptanceof 0.95. At a given time of the day, the absorber surfacetemperature is Ts = 120 oC when the solar irradiation is750 W/m2, the effective sky temperature is -10 oC and theambient air temperature is 30 oC. Assume that the convectiveheat transfer coefficient for the calm day condition can beestimated from the following equation is: h = 0.22 (Ts -T )1/3 W/m2.KCalculate the useful heat removal rate from the collector for these conditions. What is the efficiency of the collectorfor these conditions?

Example Example 7 - contd

Tsky = -10 oC

G= 750 W/m2

qu

T = 30 oC

=0.1, =0.95

Example 7 - contd

Assumption: a) Steady-state conditions

b) Bottom of collector well insulated

c) Absorber surface is diffuse

1. Energy balance on the absorber Ein – Eout = 0 Or sGs + sGsky – qcon – E – qu = 0 Gsky = Tsky

4

Sky radiation is concentrated in approximately the same spectral region as that of the surface emission and as such it may be reasonable to assume. sky = 0.1

qcon = h (Ts – T) = 0.22 (Ts - T)43

and E = Ts4

Hence, qu = sGs + sky Gsky* - qcon – E*

= 0.95 (750 W/m2) + 0.1 x 5.67 x 10-8 (W/m2K4) x (393)4K4- 0.22(393-303)1.33 – 0.1x5.67x108

(W/m2K4)x3934K4 qu = 712.5 (W/m2) + 27.127 (W/m2) – 87.4 (W/m2) –

135.25 (W/m2) = 516.9 W/m2

2. Collector efficiency, = u

s

q 516.90.689

G 750= =

qu = sGs - (Ts4 – Tsky

4) – h (Ts - T)

= sGs - (Ts4 – Tsky

4) - 0.22 (Ts - T)43

263

Radiation Exchange between Surfaces

Assumption:surfaces are separated by non-participating medium which neither emits, absorb or scatter radiation.

a) black surfaces,b) diffuse-gray surfaces, and c) enclosures.

The view factor: The view factor Fij is defined as the fractionof the radiation leaving surface i, which is intercepted by surface j. ::

Ai, Ti

Aj, Tj

RReciprocity relation: Fij Ai = Fji Aj :

Summation Rule: J=1 N Fij = 1

11 2 F11 + F12 =1

A1F12=A2F21

N

The view factor: Example

The view factor F1-3 between the base and the top surface of the cylinder shown in Figure can be found from charts. Develop expressions for the view factors F1-2 and

F2-1 between the base the lateral cylindrical surface in terms of F1-3.

R

H

3

1

2

Solution

From the summation rule F1-1 + F1-2 + F1-3 = 1 Since F1-1 = 0, the above equation reduces to the following

form F1-2= 1-F1-3

From the reciprocity relation, A1F1-2 =A2F2-1

Hence, F2-1= (A1/A2) F1-2

(Find the values of the shape factors, if R = 40 mm and H = 100 mm)

Blackbody Radiation Exchange

Blackbody Radiation Exchange

Consider radiation exchange between two black surfaces – qi. j = (AiJi) Fij (74) where qi j = rate at which radiation leaves surface i and is intercepted by surface j For a black surface, radiosity = emissive power .

. . qi j = Ai Ebi Fij (75) Similarly, qj i = Aj Ebj Fji (76) .

. . net radiative exchange between the two surfaces, qij = qi j – qj i (77) = Ai Ebi Fij - Aj Ebj Fji

= Ai Fij (Ti4 – Tj

4) (78) (since Ai Fji = Aj Fji)

Example 10A furnace cavity, which is in the form of a cylinder of diameter D = 75 mm and length L = 150 mm, is open at one end to surroundings that are at a temperature of 27oC. The sides and bottom may be approximated as blackbodies, heated electrically, and are maintained at temperatures of T1 = 1350oC and T2 = 1650oC ,respectively. The sides and bottom are considered to be well insulated.

How much power is required to maintain the furnace at the prescribed conditions?

Insulation

Heater wire

Side, T1

Bottom, T2

L

Example 10 - contd

Solution: Assumption: 1. Interior surfaces behave as blackbodies 2. Negligible heat transfer by convection 3. Outer surface of the furnace is adiabatic

A2, T2 = 1923 K

T1 = 1623 K

Tsurr = 300 K = Ts

T3 = Ts A3

q

L = 0.15m

D = 0.075m

Power required = Heat losses from the furnace Heat loss is due to radiation (other losses have been neglected) from the hypothetical surface A3. The surroundings are large; the heat transfer from the furnace to the surroundings may be approximated by treating the surface A3 to be at T3 = Tsur . Heat balance, q = q13 + q23 = A1 F13 (T1

4 – T34) + A2 F23 (T2

4 – T34)

rj /L = 0.0375

0.25;0.15

= i

L 0.154,

r 0.0375= = From Fig 26

F23 = 0.06

From summation rule, F21 + F22 + F23 = 1 F21 = 1 - F23 = 0.94 since F22 = 0

From reciprocity theorem, A1F12 = A2F21 F12 = 2

211

AF

A

= [{0.25 π(0.075 m)2}/{ π(0.075mx0.15m)}]x0.94 = 0.118 From symmetry, F13 = F12

Therefore, q = π x 0.075 x 0.15 x 0.118 x 5.67 x 10 -8 [1623 4 -3004 ] +0.25 π (0.075)2 x 0.06 x 5.67 x 10-8 [19234 – 300 4] = 1639 + 205 W = 1844 W

Radiation exchange between diffuse-gray surfaces in an enclosure

Black surfaces are ideal surface, difficult to achieve in reality.

In real surfaces, complication arises due to multiple

reflection, with partial absorption occurring each time. Likely assumptions:

1. Isothermal surface characterized by uniform radiosity and irradiation.

2. opaque – diffuse surface 3. Non-participating medium.

Problem: To determine net radiative heat flux from each

surface. (Ti associated with each surface is known.)

Radiation exchange between diffuse-gray surfaces in an enclosure (contd)

Fig. 29 Radiation exchange in an enclosure consisting of gray surfaces and non-participating medium

Ji qi Gi

Tj, Aj, j T1, A1, 1

Ti, Ai, i

Net radiation exchange at a surface Qi = Ai (Ji – Gi) (79)

qi is the net rate at which radiation leaves surface i.

Radiosity: Ji = Ei + iGi (80) qi = Ai (Ei – iGi) (81) since, ρi = 1 - i for an opaque surface = 1 - i for opaque, diffuse gray surface .

. . Ji = i Ebi + (1 - i) Gi (82)

or Gi= i i bi

i

J F

1 - e-

Ji,Ai Gi,Ai

Ai

Gi,Ai

iGi,Ai

iGi,Ai

Ei,Ai

qi

Substituting in (79)

ii

iibiii

JEAQ

/)1(

= bi i

i i i

E J

(1 ) / A

-- (83)

where, Ebi – Ji = driving potential

i

i i

1

A

- = surface radiative resistance

i

ii

i

A1

Ji

qi

Ebi

qi

Fig. 29 Radiation exchange in an enclosure consisting of gray surfaces and non-participating medium

Ji qi Gi

Tj, Aj, j T1, A1, 1

Ti, Ai, i

Radiation exchange between surface

For an enclosure, the irradiation at surface I,

N

jjjjiii JAFGA

1

= j

N

jiij JAF

1 (from reciprocity theorem)

. . . Gi = j

N

jij JF

1

Substituting into (79) gives

qi =

N

jjijii JFJA

1

using 11

N

jijF

qi =

N

jjiji

N

jiji JFJFA

11

hence, qi =

N

jijjiij

N

ji qJJFA

11

(84)

where qij =net radiation exchange

between i & j

Combining equations (83) & (84)

N

j iji

ji

iii

ibi

FA

JJ

A

JE

11)(/)1( (85)

Note:1. Equation (85) is useful when the surface temperature Ti (and hence Ebi ) is known.

1. Equation (84) is useful when the net radiation transfer rate, qi , is known.

qi1

qi2

qi3

qi Ebi

(AiFi1)-1

Ji

(AiFi2)-1

(AiFi3)-1

(AiFin)-1

i

i i

1

A

J1

J2

J3

Jn

qin

The two-surface Enclosure: Since there are only two surfaces,

Net radiation transfer from surface 1, q1 = net radiation transfer to surface 2, - q2

q1 = - q2 = q12

q1 = bi 1

1

1 1

E J1

A

--

; -q2 = 2 b2

2 2

J E1

A

--

q12 = 1 2

1 12

J J1

A F

-

2

2 2

1

A

q1

(A1F12)-1

1

1 1

1

A

J1

J2

q12 q1

q1

Eb1

Eb2

-q2 q12

A1, T1, 1

A2,T2,

Eb1 - J1 = q1

11

1

A = q12

11

11

A

J1 - J2 = q12

121

1

FA

J2 - Eb2 = -q2

22

21

A = q12

22

21

A

Adding Eb1 - Eb2 = q12 ]111

[22

2

12111

1

AFAA

= 42

41 TT

22

2

12111

1

42

41

12 111)(

AFAA

TTq

(86)

Eqn (86) may also be written as:

2

1

2

2

121

1

42

411

12111

)(

A

A

F

TTAq

Concentric sphere Concentric Finite Infinite parallel Cylinders plane

21 1

1222 2

A r; F 1

A r= =

1 112

2 2

A r; F 1

A r= = A1 = A2 = A; F12 = 1

r1 r2

Radiation Shields

Radiation shield, constructed from low emittance materials can be used to reduce heat transfer between two surfaces.

.

q1

Eb1

Eb2 (A1F13)

-1 1

1 1

1

A

J1

J2 J31

31

3 31

1

A

Eb3

32

3 32

1

A

2

2 2

1

A

J32

(A3F32)-1

q1 -q2

q13 q32

A3T3

A1, 1, T1 A2, 2, T2

q12

32 31

Total Resistance Rtotal = R1+R2+R3+R4+R5+R6

Where, R1= (1-ε1)/A1ε1

R2= (A1F13)-1

R3= (1-ε31)/A3ε31

R4= (1-ε32)/A3ε32

R5= (A3F32)-1

R6= (1-ε2)/A2ε2

For large parallel plates

32

32

31

31

21

42

411

12 1111)(

TTAq

when ε31 and ε32 are small, the resistance become very large.

Special case:

ε1 = ε2 = ε31 = ε32

One radiation shield reduces the radiation heat transfer by 50%

Example 12

A cryogenic fluid flows through a long tube of diameter D1

= 20 mm, the outer surface of which is diffuse-gray with ε1

= 0.03 and T1 = 77 K. This tube is concentric with a larger tube of diameter D2 = 50 mm, the inner surface being diffuse-gray with ε2 = 0.05 and T2 = 300 K. The space between the surfaces is evacuated. 1. Calculate the heat gain by cryogenic fluid per unit length

of the tube. 2. If a thin radiation shield of diameter D = 35 mm and

emittance, ε3 = 0.02 (both sides) is inserted midway between the inner and outer surfaces, calculate the change in percent of heat gain per unit length of the tube.

Example 12-solution

Solution

Radiation Shield D3 = 35 mm 3 = 0.02

D2 = 50 mm T2 = 300 K 2 = 0.05

D1 = 20 mm T1 = 77 K 1 = 0.02

(A1F12)-1

1

1 1

1

A

J1

J2 Eb1 Eb2

2

2 2

1

A

Network: (without shield)

Network: (with shield)

R1 R2 R3 R4 R5 R6

Eb1 J1 J31 Eb3 J32 J2 Eb2

Example 12-solution(contd) Total Resistance Rtotal = R1+R2+R3+R4+R5+R6

Where, R1= (1-ε1)/A1ε1

R2= (A1F13)-1

R3= (1-ε31)/A3ε31

R4= (1-ε32)/A3ε32

R5= (A3F32)-1

R6= (1-ε2)/A2ε2


1. Heat gain by cryogenic fluid, q

21

2

2

1

42

411

/11

))((

DD

TTLDq

Heat gain per unit length of the tube

Lqq /' 21

2

2

1

42

411

/11

))((

DD

TTD


Lqq /' 21

2

2

1

42

411

/11

))((

DD

TTD

=

)50/20(05.0

05.01

02.0

1)(30077)02.0()./(1067.5 444428

KKmWx

= 28.7

0.498 W/m57.6

=-


2. When a radiation shield is placed in between the two concentric tubes, heat gained by the cryogenic given by

tottot

bb

R

TT

R

EEq

42

4121

where,

LDLDFLDLDFLDRtot

22

2

332331

31

11311

1 11}

1{2

11

= ( )1779.9 15.9 891.3 9.1 121.0

L+ + + +

= )/1(,1817 2mL

Example 12-solution(contd)Hence :

q' = q

L =

1817

300771067.5 448 x

= -0.25, 2m

W

Percentage reduction in heat gain

100498.0

25.0498.0100

'

''xx

q

qq

wo

wwo

w = with shield wo = without shield = 49.8 %

Mec3609 ht radiation

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