unit iv radiation

8/10/2019 Unit IV Radiation

1/83


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3/83

Spectrum of Electromagnetic Radiation

104 103

102

101

100

10-1

10-2

10-3

1011 1012 1013 1014 1015 1016 1017

m)

Thermal Radiation(100m to 0.1m)

Solar (3-0.1m)Radiation

Visible Light(0.78-0.38)

(freq) Infrared(1000-0.7m)


4/83

Radiation

Radiation emissions propagate in the form ofwaves. Since waves propagate through somemedium, this theory assumes that Universe

is filled with a hypothetical medium ETHER.

Waves travel with the speed of light

1. Wave Theory or Maxwells Classical Theory

Every wave possesses certain amount ofenergy, a part of which is transferred onbeing impinged by some object in its routeof travel


5/83

Radiation

Radiation emissions are in the form of seriesof entities known as quanta.

Each quanta possesses certain amount ofenergy, which is proportional to its frequencyof emission.

2. Quantum Theory or Plancks Theory

Quanta moves with the speed of light andreleases its energy on being impinged bysome object in its route of travel


6/83


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Some Definitions

Black Body:

A body which absorbs all incident energy anddoes not transmit and reflects at all, is called

Black Body. It is also the highest emitter ofradiation

1;1;0;0

Examples:Surface coated with lamp black,milk, ice, water, white paper etc


8/83

Some Definitions

White Body:A body which reflects the entire radiationfalling on it, is called White Body

1;0;0;0

Gray Body:

The body having same value of emissivity atall wavelengths , which is equal to average

emissivity, is known as Grey body.

Generally, all engg metals are grey bodies,for which , when in thermal equilibrium


9/83

Some Definitions

Monochromatic Emissive Power (q):

It is the rate, at which radiant flux is emitted

with a specific wave length at certain temp; itis dependent emissive power

It is the rate, at which the radiant flux isemitted from the surface at certain temp

Emissive Power (q):


10/83

Some Definitions

Monochromatic Emissivity ( ):

It is the ratio of monochromatic emissivepower of a surface to that of black bodywhen both are at same temp for same givenwavelength

bq

q

bq

q

Emissivity ():

It is the ratio of emissive power of a surfaceto that of black body when both at same temp


11/83

Some Definitions

Irradiation (G):

It is the net energy incident/falling on thesurface (need not necessarily be absorbed)

Radiosity (J):

It is the net energy leaving the surface.It consists of the radiant energy emitted andenergy reflected out of the incident radiation

from the surface.J= 1qb + (1-1)G


12/83

Plancks Law

Plancks law is based on Quantum theory andit gives the relationship among monochromaticEmissive powerof black body, the absoluteTempof the surface and corresponding

Wavelengthand is given as:

014387.0

&10596.0

;/

1.

2

2

16

1

2

5

1

2

C

xCwhere

mW

e

Cq

TCb


13/83

Plancks LawPlot shows the following:

qat certain temp firstincreases with , attainssome max value andthen decreases

For specific wavelength,qof black surfaceincreases with temp

Most of the thermal radiations lie in wavelengthregion from 0.3 to 10 m

Wavelength m), at which peak qobtained,decreases with increase in temp


14/83

Wiens Displacement Law

Wiens Law gives therelationship betweenthe wavelengths (m) ,at which peak (q)monochromatic emissive

power is obtained and theabsolute temp andgiven as:

m.T=0.0029 mK

Plot and above relation show that the value ofwavelength, at which peak/max monochromaticemissive power is obtained, decreases (displaces/shifts)with increase in surface temperature of the black body.


15/83

Derivation of Wiens Law

As per Plancks law,

1

2

25

1

TCb

e

Cq

xT

Cx

T

CPutting 22

1

2

55

5

2

1

x

b

eTx

C

CqngSubstituti

5

2

155

1 1...2

C

eTxCqOr

x

b

This eqn expressesqof black body asa function of x


16/83


For obtaining the wavelength (m) for specified temp,

at which max qoccurs, we have to differentiatethis equation wrt xand equate it to zero.

0

1...2

52

155

1

C

eTxC

dx

d x

01..2 155

2

5

1 x

ex

dx

d

C

TCOr

01 15 xexdx

dOr


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01 15

x

exdx

d

0.1.1.5.1 2541 xxx eexxe

0

1.

15 2

54

x

x

xe

exe

x

0

1

.5

1

4

x

x

x

e

ex

e

x

05501

.55

xe

e

exeOr x

x

xx


18/83


055 xehavenowWe x

This eqn is satisfied by putting x=4.96

TCxHence

296.4,

0029.0

96.4

014387.0 Tm

mKTTherefore m 0029.0,


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Stefan Boltzmanns Law

Kirchhofs Law

Emissive power of a black body is directlyproportional to fourth power of its absolutetemperature:

428

44

/1067.5

;

KmWxwhere

TqorTq bb

When a surface is in thermal equilibrium withits surroundings, the emissivity of the surfaceis equal to its absoptivity

isThat


20/83

Solid Angle

Solid angle subtended bysurface A2at surface dA1(elementary surface)is numerically equal to thearea on a surface of sphere

with unit radius and centreat elementary area, whichis cut by conical surfacehaving its base as perimeterof A2and vertex at dA1

Solid angle is measured in Steradians (Sr) anddenoted by symbol

2

2

r

Ad


21/83

Solid Angle Between Two Elementary Areas

Solid Angle subtendedby elementary area dA2at dA1 can be given as:

222

r

CosdAd

Similarly, solid anglesubtended by area dA1at dA2can be given as:

2

11

r

CosdAd


22/83

Intensity of RadiationIntensity of radiation emitted by

a surface is equal to the radiantenergy passing in a specifieddirection per unit solid angle

Intensity of radiation varies in different directionsand is max in the direction normal to the surface

Lambert Cosine Law:Intensity of radiation in any direction is proportionalto the Cosine of the angle made by that directionwith the normal.

That is, I=InCos; where Inis the intensity (max) inthe normal direction and is the angle made by thatdirection with the normal

4T

IIqpoweremissiveTotal nn

II


23/83

Shape Factor/Geometric Factor

Shape factor is defined as the fraction ofenergy emitted by one surface and directlyintercepted by the other.

1 2

22121

1

121:

A A r

dAdACosCosA

FFactorShape

Shape Factor depends upon:1. Shape and size of surfaces2. Orientation of surfaces wrt each other3. Distance between the surfaces


24/83

Relations/ Theorems of Shape Factors

1. Reciprocal Relation: F12.A1=F21.A2

2. Enclosure Relation: If nno of surfaces form anenclosure, then: F11+F12+F13+..+F1n=1

F21+F22+F23+..+F2n=1

Fn1+Fn2+Fn3+..+Fnn=13. Summation Relation:

Shape Factor F12between twosurfaces A1and A2is equal

to the sum of shape factorsF13& F14, if the two areas A3& A4together make up area A2

413121141312 ,; FFFHoweverFFF


25/83

Relations/ Theorems of Shape Factors

4. Shape factor depends on geometry andorientation of surfaces and it does notchange with temp.

5. Shape Factor wrt itself (F11, F22, F33)means radiation emitted by a portion of asurface falling on the other portion of itselfdirectly

Example : Shape Factor for concave surface

Shape factor for convex or Flat surface wrtitself is zero.


26/83

Radiation Heat Exchange BetweenTwo Parallel Plates

T11

T22

Consider two grey opaque parallel platesmaintained at temperatures T1& T2havingemissivities 1& 2respectively

For grey bodies, absoptivity = emissivity


27/83

Radiation Heat Exchange Between Two Parallel Plates

T11

T22 q2

1q2

2(1-1)q2

(1-1)q2

(1-1)(1-2)q2

Consider radiant flux q2emitted by surface 2.

Out of q2, a fraction 1q2will be absorbed by surface 1and rest (q21q2) will be reflected towards surface 2

Out of this, 2(1-1)q2will be absorbed by surface 2

and balance (1-1)(1-2)q2will be reflected to 1


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T11

T22 q2

1q2

2(1-1)q2

(1-1)q2

(1-1)(1-2)q2

1(1-1)(1-2)q2

(1-1)2(1-2)q2

2(1-1)2(1-2)q2

(1-1)2(1-2)

2q2

1(1-1)2(1-2)

2q2

(1-1)3(1-2)

2q2

2(1-1)3(1-2)

2q2

(1-1)3(1-2)

3q2

1(1-1)3(1-2)

3q2

This process of absorption and reflection goes onindefinitely, the quantities involved being

successively smaller.

Out of this, 1(1-1)(1-2)q2will be absorbed by

surface 1 and balance (1-1)2(1-2)q2will be reflectedback to 2


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Thus, total radiant flux absorbed by surface 1out of

q2emitted by surface 2will be:

.............11

1111

2

3

2

3

11

2

2

2

2

11221121

q

qqq


30/83

Radiation Heat Exchange Between TwoParallel Plates

2112

111

q

2121

12

q

.....1111111 323

1

2

2

2

12112 q

Similarly, considering radiation flux q1emitted by

surface 1and fraction out of which absorbed bysurface 2can be given as:

2121

21

q

2121

12

11

q


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Assuming T1> T2, net radiant flux absorbed by 2:

2121

12

2121

2112

qqq

4

222

4

111 & TqTqSince

2121

4

221

4

11212

TTq

111

21

42

41

12

TTq

111

21

42

41

12

TTAQOr


32/83

Electrical Analogy of RadiationShape/Space Resistance:

Heat flow between two black surfaces at temps T1&T2can be written as:

121

4

2

4

14

2

4

111212 1

FA

TTTTAFQ

212121

4

2

4

1

11tanRe

,

FAFAcesisEquivalentAnd

TTDiffPotentialEquivalentHere

Due to finite dimensions of the surfaces, 100%ofemitted radiation from surface 1does not fall on

surface 2, hence some part of emitted energy go tosurroundings, thus lost. This loss is conceptuallyexplained to be caused due to resistance offered byfiniteness of dimensions of surfaces and theirorientation. Hence, it is called Shape/Space Resistance


33/83

Electrical Analogy of Radiation

Surface Resistance:

Black body emits max possible radiation and itsemissivity is taken as 1( the datum). However,Grey bodies emit less due to surface properties;and hence their emissivities are taken as less than 1

(in comparison).Therefore, emission of radiation from grey bodies isalways less than that of black body. This lesseremission is conceptually assumed to be caused dueto a resistance offered by surface of the body as it

depends on surface property; the emissivity. Thisresistance is called Surface Resistance and given as:

)2(1

&)1(1

22

2

11

1 surfaceofA

surfacetheofA

Rsurface


34/83

Radiation Heat Exchange Between TwoParallel Plates (By Other Method)

1 2

1

T1

2

T2

22

2

12111

1

4

2

4

112 111

AFAA

TTQ

Since F12=1& A1=A2=A;

11

111

21

4

2

4

112

TT

A

Q

111

21

4

2

4

112

TTq


35/83

R1R2

T11

Radiation Heat Exchange Between TwoConcentric Infinitely Long Grey Cylinders

22

2

12111

1

4

2

4

112

111

AFAA

TTQ

F12=1as inner cylinder is completely enclosed by 2


36/83

Radiation Heat Exchange Between TwoConcentric Infinitely Long Grey Cylinders

22

2

12111

1

4

2

4

112 111

AFAA

TTQ

Putting F12=1,

We have:

11

1111

22

1

11

4

2

4

112

A

A

A

TTQ

111

22

1

1

42

411

12

A

A

TTAQ

A1=2R1L

A2=2R2L


37/83

Radiation Heat Exchange Between Two Surfaces

1

11

22

1

1

4

2

4

1112

A

A

TTA

Q

This expression is very

useful as it can be appliedto so many situations:

1. For heat exchange between two concentric spheres;Only diff will be : A

1=4R

1

2& A2=4R

2

2

2. For eccentric cylinders and spheres

4. For convex/Flat surface completely enclosed byother body as F12=1and F21=A1/A2

If enclosure (A2) is very large, A1/A20;Hence, Q = 1A1(T1

4 - T24)

3. For heat exchange between two parallel platesas A1 = A2 = A


38/83

Radiation ShieldIn order to reduce the radiation heat transfer

rate between two surfaces, a third surface isinserted between them. This surface isknown as Radiation Shield.

Requirements of Shield (Surface):- Highly reflecting- Lowest emissivity (also absorptivity)- Lowest thickness (thinnest)

Applications in more effective thermos flasks,for reducing error in temp measurement bythermocouples etc


39/83

Radiation Shield 1T11

2T22

3T33

Heat Flow Rate

assuming T1>T2:

111

21

4

2

4

112

TTq

Now, a shield having both side emissivity 3isplaced between the surfaces 1& 2.

On achieving steady state, the shield will

attain steady temp T3between T1& T2.Since T3remains steady, that means whateverradiation, the shield is receiving from surface 1,the same it is giving out to surface 2.


40/83

Radiation Shield

111

111,

23

4

2

4

3

31

4

3

4

13213

TTTTqqHence

yandxngSubstituti 111

1

11

2331

y

T

x

T

x

T

y

T

y

TT

x

TThaveWe

4

2

4

1

4

3

4

3

4

2

4

3

4

3

4

1,

yx

xTyTTxTyTyTxTOr

4

2

4

14

3

4

2

4

1

4

3

4

3


41/83

Radiation Shield

yxx

xTyTyTxT

x

yx

xTyTT

q

ressionqinTngSubstituti

4

2

4

1

4

1

4

1

4

2

4

14

1

13

13

4

3 ;exp

yxTT

yxx

TTxq

4

2

4

1

4

2

4

113

..


42/83

Radiation Shield

On simplification:

12

111

321

42

41

13

TTq

Since 3will be very small, hence denominatorof q13will be very large, therefore, there shallbe large reduction of q12to q13.

1

111

11;&

2331

4

2

4

113

TTqyxngSubstituti


43/83

Radiation Shield

If 1=

2

3= ;

12

12

12

111

4

2

4

1

4

2

4

113

TTTTq

shieldonewithtwiceusedismeansThis

1

2,

shieldsnwith

n

TTqHence

1

21

,

4

2

4

113


44/83

Radiation Shield

shieldsnwith

n

TTqHence

12

1

,

4

2

4

113

21112

;

12

4

2

4

113

qTTq

shieldONEWith

Hence, q13now becomes half of q12


45/83

Radiation Shield

Home Assignment:

.&

&int&

2&1/

,

1

2

1

11Pr

21

2121

3

33

1

22

1

1

4

2

4

1113

AAareashaving

TTtempsatainedmaesemissiviti

havingSphheresCylindersTWObetween

placedisemissivityhavingshieldawhen

A

A

A

A

TTAQthatveo


46/83

Q1:Effective temp of a body having an areaof 0.12m2is 527C. Calculate the following:

a) Rate of radiation energy emissionb) Intensity of normal radiationc) Wavelength of max monochromatic

emissive power

Solution:

a) Total emission of radiation Q=A T4

WxxxQ 9.278627352712.01067.5 48

srmWxxTq

I

RadiationNormalofIntensityb

bn ./5.73922735271067.5

)

2484


47/83

c) Wavelength of max monochromatic

emissive power:

From Wiens Displacement Law;

mmx

T

mKT

m

m

m

625.310625.3

273527

0029.00029.00029.0

6

Answer


48/83

Q2: A sphere of radius 5cmis concentric with anothersphere. Find the radius of the outer sphere so that

shape factor of outer sphere wrt inner sphere is 0.6.

1

2

R1

R2

Solution:

A1=4r12

A2=4r22 F21=0.6

Since sphere 1is completely enclosedby sphere 2, hence F12=1

We know that F12.A1=F21.A2

222

46.005.041;,

RxxhavewevaluesngSubstituti

AnswercmR 45.62


49/83

Q3: Find F12.

5

8

4 3

1

3

4 2

5

6141612 FFF 41

1

461

1

6 FA

AF

A

A

43451

46365

1

6 FFA

AFF

A

A

L1=y/x

L2=z/x

65.0

20

13&35.0

20

7: 2165

W

L

W

LF

From graph: F65=0.32


50/83

Solution (Contd):

4.020

8&35.0

20

7: 2163 W

L

W

LF


65.02013&2.0

204: 2145

WL

WLF



51/83

Solution (Contd):

4.020

8

&2.020

4

: 21

43 W

L

W

L

F


43451

46365

1

6

12 FF

A

AFF

A

AF

06.033.036.0205

20426.032.0

205

20712

x

x

x

xF


52/83

Q 3: Find out heat transfer rate due to radiationbetween two infinitely long parallel planes. One plane

has emissivity of 0.4and is maintained at 200C.Other plane has emissivity of 0.2and is maintained at30C. If a radiation shield ( =0.5) is introducedbetween the two planes, find percentage reduction inheat transfer rate and steady state temp of the shield.

Solution:

111

21

42

41

12

TTq

2

448

12

/363

12.0

1

4.0

1273302732001067.5

mW

xq


53/83

Solution (Contd):

;insertedisshieldWhen

12111

321

4

2

4

113

TTq

2448

13 /4.248

15.0

21

2.0

1

4.0

1273302732001067.5 mWxq

%57.31100363

4.248363

10012

1312

x

xq

qq

reductionPercentage


54/83

Solution (Contd):

111

111

23

4

2

4

3

31

4

3

4

13213

TTTTqq

Under Steady StateConditions, we have:

12.0

1

5.0

1

27330

15.0

1

4.0

1

273200 44

3

4

3

4TT

or

KT 67.4313


55/83

Q4: Cryogenic fluid flows through annular space ofinner tube dia of 30mm and outer tube of 90mm dia.

Surface emissivitiesof inner and outer tubes are 0.2and 0.5, while respective temps are 100and 300K.Find Heat gain rate by the fluid per meter length.Also, percentage reduction in heat gain, if a radiationshield of tubular shape having 45mm diameter and

emissivitiesof 0.1on inner surface and 0.05on outersurface is introduced between the two tubes.

Solution:

1

23

111

22

1

1

4

2

4

1112

A

A

TTAQ

W

xx

xx

xxxx8

15.0

1

109.0

103.0

2.0

1

300100103.01067.5 448


56/83

Solution (Contd):

With shield:

11111132313

1

22

1

1

4

2

4

1113

AA

AA

TTAQ

W

xxxx

xxxx

xxxx

732.1

105.01

1.01

1045.0103.01

5.01

109.0103.0

2.01

300100103.01067.5 448

10012

1312

xQ

QQreductionPercentage

ANSWERx %38.781008

732.18


57/83

Q5: A pipe carrying steam having an outside dia of20cmpasses through a large room and is exposed to

air at temp of 30C. Pipe surface temp is 200C. Findthe heat loss per meter length of pipe both byconvection and radiation taking emissivity of the pipesurface as 0.8.

Use following relations:Nu=0.53(Gr.Pr)0.25for horizontal pipe

Air properties:

Temp C K (W/mK) x106

(m2/s)

Pr

30 0.0267 18.60 0.701

115 0.0330 24.93 0.687

200 0.0393 26.00 0.680


58/83

Solution:

For Radiation:

11122

1

1

4

2

4

11

AA

TTAQr

0)()(

2

112

A

APipeARoomASince

4

2

4

111 TTAQr

mW

xxxxx

/6.1185

2733027320012.01067.58.0 448


59/83

Solution (Contd):

For Convection:CTmean

115

2

30200

7

26

3

2

3

1053.51093.24273115

2.030200181.9x

x

xxxTLgG r

67 1038687.01053.5Pr. xxxGr

k

hDGrNuselectingHence

25.0Pr.53.0,

KmW

xxh

2

25.06

/87.62.0

103853.0033.0

mWxxxTTAhQ wc /8.7333020012.087.6

mWQQQTotal rc /4.19196.11858.733


60/83

Q7: Find shape factor for the following wrt itself:a) Cylindrical cavity of dia Dand depth H

b) Conical hole of dia Dand depth Hc) Hemispherical hole of dia D

Solution: a) We know F21=1

As F12A1= F21A2 F12 = A2/A1

We know that F11 + F12= 1as surfaces 1and 2form enclosure

1

21211 11

A

AFFHence

2

2

2

1

4

4

DAAnd

DDHANow

DH

H

DDH

DF

4

4

4

412

2

11


61/83


62/83

Solution:

c) F11+F12=1as surfaces 1& 2

form enclosureHence F11=1- F12

221112 AFAFthatknowalsoWe

1211

221

1

212 FasA

AFA

AFHence

1

211 1

A

AF 2

2

2

2

14

&22

24

DAD

D

ANow

5.05.015.0

2

4112

2

1

2 FHenceD

D

A

A


63/83

Q7: Two concentric spheres 210mmand 300mmdiameters with the space between evacuated are to

be used to store liquid air (-153C) in a room at 27C.The surfaces of the spheres are flushed with aluminum=0.03) and latent heat of vaporization of liquid airis 209.35 kJ/kg. Calculate the rate of evaporation ofliquid air per hour. ( Ans. 0.0217 kg/h)

Solution:

1

2

Liquid O2:21 asgivenissurfaceouterto

surfaceinnerfromFlowHeat

111

22

1

1

42

411

12

A

A

TTAQ


64/83

Solution (Contd):

WQ

xx

Q

26.1

103.0

1

150.04

105.04

03.0

1

273272731531067.5105.04

12

2

2

4482

12

)(1 Oxygensurfacebyreceived

beingheatindicatessignNegative

hkgx

xQm

nevaporatioofRate

/0217.0100035.209

360026.1:


65/83

Q8: The filament of a 75Wlight bulb may beconsidered a black body radiating in to black enclosureat 70C. The filament dia is 0.1mm and length is 5cm.Considering radiation, determine the filament temp.(Ans. 2756C)

Solution:

111

.

22

1

1

4

2

4

1112

A

A

TTAQ

Since small A1is enclosed in large A2, hence A1/A2 0

441834

2

4

11112

273701067.5105.0101.075

Txxxxxx

TTAQ

)2756(30291 CKT


66/83

Q9: Determine the heat loss rate by radiation from asteel tube of outside dia 70mmand 3mlong at a tempof 227C, if the tube is located within a square brickconduit of 0.3mside and at 27C. Take emissivity ofsteel=0.79and that for brick=0.93. (Ans. 1589.7W)

Solution:

111

22

1

1

4

2

4

1112

AA

TTAQ

AnswerW

xx

xx

xxxxQ

7.1589

193.0

1

33.04

)307.0(

79.0

1

273272732271067.5)307.0( 448

12


67/83

Q10: Three hollow thin walled cylinders having dia10cm, 20cmand 30cmare arranged concentrically.

The temps of the innermost and outermost cylindricalsurfaces are 100Kand 300Krespectively. Assumingvacuum in annular spaces, find the steady state tempattained by the surface having dia of 20cm. Takeemissivities of all surfaces as 0.05.

(Ans. 269K)

Solution:

111

111

33

2

2

43

422

23

22

1

1

42

411

12

A

A

TTAQ

A

A

TTAQ

Under steady state:

12

3


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Solution (Contd):

A1DL=3.14x0.1x1=0.314

A2=3.14x0.2x1=0.628 A3=3.14x0.3x1=0.942

1

11

1

11:

33

2

2

4

3

4

22

22

1

1

4

2

4

11

A

A

TTA

A

A

TTAhaveWe

105.0

1

942.0

628.0

05.0

1

300628.0

105.0

1

628.0

314.0

05.0

1

100314.0 44

2

4

2

4TT

nsweKT 2692


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Gas RadiationGases in many cases are transparent to radiation

When they absorb and emit radiation, they usuallydo so only in certain narrow wavelength bands.

Some gases such as N2, O2and other non-polargases are essentially transparent to radiation and

they do not emit radiation

While polar gases like CO2, H2O and varioushydrocarbon gases absorb and emit radiation toan appreciable extent in narrow wavelength bands.

For solids and liquids, radiation occurs from thinlayer (1m to 1mm) of surface, hence it is surfacephenomenon. However, for gases it is not surfacebut volumetric phenomenon.


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Volumetric Absorption:

Let a monochromatic beam of radiation having an

Intensity Iiimpinges on the gas layer of thicknessdxas shown in Fig.

Decrease in intensity resultingfrom absorption in the layers is

proportional to the thickness oflayer and intensity of radiation atthat point

Thus;

Ii Io

X=0x dx

Ix

tcoefficienabsorptionticmonochromacallediskwhere

dxIkdI

;


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Volumetric Absorption:

Integrating this equation gives;

This is Beers Law and representsexponential decay of radiationintensity

Ii Io

X=0x dx

Ix

xxI

iI

dxkI

dI

0

.

xk

i

x eI

Ior


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Gas Radiation

If gas is non-reflecting, then;

we know that monochromatic transmissivity;

1

1

henceand

xke

xketyAbsorptiviTherefore

.1

xkeEmissivitysurfacegreyforAnd

.1,,


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Emissivity of CO2, H2O Vapor & Gas

Emissivity of a gas mixture is a function of total

pressure (P), partial pressure of a gas(p), gastemperature (T) and characteristic dimension of thesystem; also known as beam length

LTpPf ,,,

When the gas mixture is at 1 atm total pressure,the emissivity of CO2 and H2O vapors are given byfollowing empirical relations;

3

6.08.0

5.3

33.0

1005.3

100).(5.3

TLp

TLp

w

c

enclosureofareaSurface

mixtureGasofVolumexL

astakenisLengthBeamMean

casespracticalmostFor

6.3

,


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Heat Exchange between Gas Volume & its Enclosure

If the enclosure surface is grey, the net heat transfer

to grey enclosure having emissivity greyis given by:

2

1

grey

black

grey

Q

Q


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77/83

Error In Temp Measurement By Thermocouple

For measurement of temp of a hot gas flowing

through a duct, a thermocouple (TC) is placed indirect contact with gas.

Tg

Tw

h

Tt Att

AwThermocouple receives heatfrom gas by convection andtries to attain gas temp.

As the temp of thermocouplejunction rises, it starts loosingenergy by radiation to ductwall , which is at a lower temp.

Thus, temp of thermocouple (TC) reduces. Hence,temp recorded by thermocouple (Tt) is always lessthan gas temp (Tg)


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When TC attains steady state

temp Tt, this means heatenergy being received by it byconvection from hot gas attemp Tgis equal to heat beinglost by it by radiation to duct

which is at Tw

111

..

44

ww

t

t

wtt

tgt

A

A

TTATTAh

tmeasuremeninerrorisTTwhere

TTTThhenceA

ASince

tg

wtttg

w

t

44

.,0


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Error (Tg - Tt) inmeasurement can be

reduced by;

1) Providing radiationshield around Thermocouple

2) Reducing emissivity of thermocouplejunction. For this, junction may be coatedwith some material having low emissivitylike Aluminum, Zinc, Chromium, etc

44

. wtttg TTTTh


80/83


81/83

Solution (Contd):

1

11

.

.

44

ww

t

t

wtt

tgt

A

A

TTA

TTAh

Since At/Aw0,

We have; 44.. wttttgt TTATTAh

448

27320273500

200

1067.56.0

xx

TT tg

KTT

tmeasurementempinErrorHence

tg 60

,


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Solution (Contd):

Assuming At


83/83

Solution (Contd):

Now taking up heat transfer from hot gases to TC byConvection and from TC to shield by radiation:

h.At.(Tg Tt) = .t.At.(Tt4 Ts

4)

Substituting:200 (833 Tt) = 5.67x10

-8x0.6x(Tt4 8154)

Tt=829K

Hence, error with shield now becomes:Tg Tt= 833 829 = 4 K

unit iv radiation

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