Thermal Radiation - III- Radn. energy exchange between gray surfaces

Lectures on Heat Transfer –THERMAL RADIATION-III:Radiation energy exchange between gray surfaces

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,Mangalore,India

Preface:

• This file contains slides on THERMALRADIATION-III: Radiation energy exchange between gray surfaces.

• The slides were prepared while teaching Heat Transfer course to the M.Tech. Heat Transfer course to the M.Tech. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.

Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful to teachers, students, researchers and professionals working in this field.

• For students, it should be particularly useful to study, quickly review the subject, useful to study, quickly review the subject, and to prepare for the examinations.

• ��


References:• 1. M. Thirumaleshwar: Fundamentals of Heat &

Mass Transfer, Pearson Edu., 2006• https://books.google.co.in/books?id=b2238B-

AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false

• 2. Cengel Y. A. Heat Transfer: A Practical Approach, 2nd Ed. McGraw Hill Co., 2003

Aug. 2016 MT/SJEC/M.Tech. 4

Approach, 2nd Ed. McGraw Hill Co., 2003• 3. Cengel, Y. A. and Ghajar, A. J., Heat and

Mass Transfer - Fundamentals and Applications, 5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.

• 4. Incropera , Dewitt, Bergman, Lavine: Fundamentals of Heat and Mass Transfer, 6th

Ed., Wiley Intl.• 5. M. Thirumaleshwar: Software Solutions to • 5. M. Thirumaleshwar: Software Solutions to

Problems on Heat Transfer – Radiation Heat Transfer-Part-II, Bookboon, 2013

• http://bookboon.com/en/software-solutions-heat-transfer-radiation-ii-ebook


Thermal Radiation – III:Radiation energy exchange between gray surfaces :

Outline…

Radiation heat exchange between gray surfaces - electrical network method – two


surfaces - electrical network method – two zone enclosures – Problems - three zone enclosures – Problems - radiation shielding – Problems - radiation error in temperature measurement – Problems

Radiation energy exchange between gray surfaces:• Following assumptions are made to simplify the

solution:• All the surfaces of the enclosure are opaque (τ = 0),

diffuse and gray• Radiative properties such as ρ, ε and α are uniform

and independent of direction and frequency


• Radiative properties such as ρ, ε and α are uniform and independent of direction and frequency

• Irradiation and heat flux leaving each surface are uniform over the surface

• Each surface of the enclosure is isothermal, and• The enclosure is filled with a non-participating medium

(such as vacuum or air)

The electrical network method:• We shall discuss the ‘electrical network

method’, since it is simple to apply and gives a physical ‘feel’ of the problem.

• This method, introduced by Oppenheim in the 1950’s, is simple and direct; it emphasizes on the physics of the problem,


emphasizes on the physics of the problem, and is easy to apply.

• Before we introduce this method, let us define two new quantities, viz. irradiation and radiosity: (See Fig. 13.25)


• Irradiation, (G): is the total radiation incident upon a surface per unit time, per unit area (W/m2).

• Radiosity, (J): is the total radiation leaving a surface, with no regard for its origin (i.e., reflected plus emitted from the surface) per unit time, per unit area (W/m2).

• Now, from Fig.(13.25), it is clear that total radiation leaving the surface (i.e. Radiosity, J) is:

• For a gray, opaque (τ = 0) surface, we have:

J ρ G. ε E b.


have:

• Therefore,

ρ 1 α( ) 1 ε( ) ...from Kirchoff's law

J 1 ε( ) G. ε E b.

or, GJ ε E b

.

1 ε( )

• Now, net rate of radiation energy transfer from the surface is given by: (rate of radiation energy leaving the surface minus the rate of radiation energy incident on the surface), i.e. Q

AJ G

Q J ε E b.


• Therefore,

i.e. Q

AJ

J ε E b.

1 ε

Qε A.

1 εE b J.

i.e. QE b J

1 ε( )

A ε.

W.....(13.48)

• By analogy with Ohm’s law, we can think of Q in eqn.(13.48) as a current flowing through a potential difference (Eb-J), and the factor (1-ε)/A.ε as the resistance.

• Now, this resistance is the resistance to the flow of radiant heat due to the nature of the surface and is known as ‘surface resistance (R)’.


and is known as ‘surface resistance (R)’. • i.e.

• Surface resistance for a surface ‘i’ is shown schematically in Fig. (13.26,a).

R1 ε( )

A ε.....surface resistance

• For a black body emissivity ε = 1; so, the surface resistance is zero, and

• Also, many surfaces in numerous applications are adiabatic, i.e. well insulated, and net heat transfer

J i E bi ...for a black body....(13.49)


insulated, and net heat transfer through such a surface is zero, since in steady state, all the heat incident on such a surface is re-radiated. These are known as re-radiating surfaces.

• Walls of a furnace is the familiar example of a re-radiating surface. Obviously, for a re-radiating surface, Qi = 0, and from eqn. (13.48) we get:

• Note that the temperature of a re-radiating J i E bi σ T i

4. ....for a re-radiating surface....(13.50)


• Note that the temperature of a re-radiating surface can be calculated from the above eqn; further, note that this temperature is independent of the emissivity of the surface.

• Again, consider two diffuse, gray and opaque surfaces i and j, maintained at uniform temperatures Ti and Tj, exchanging heat with each other.

• Then, remembering the definitions of radiosity and view factor, we can write for the radiation leaving surface ‘i’ that strikes


the radiation leaving surface ‘i’ that strikes surface ‘j’:

• Similarly, for surface j, we have:

Q i A i F ij. J i

.

Q j A j F ji. J j

.

• Therefore, net heat interchange between surfaces i and j is:

Q ij A i F ij. J i

. A j F ji. J j

.

i.e. Q ij A i F ij. J i J j

. W..(13.51)..since A i F ij. A j F ji

. ...by reciprocity.

i.e. Q ijJ i J j

1W....(13.52)


• Again, by analogy with Ohm’s law, we can write eqn. (13.52) as:

1

A i F ij.

Q ijJ i J j

R ijW

where, R ij1

A i F ij.

...(13.53)

• Rij is known as ‘space resistance’ and it represents the resistance to radiative heat flow between the radiosity potentials of the two surfaces, due to their relative orientation and spacing.

• Space resistance is illustrated


• Space resistance is illustrated in Fig. (13.26,b).

• Note from eqn. (13.52) that if Ji > Jj, net heat transfer is from surface i to surface j; otherwise, the net heat transfer is from surface j to surface i.

• Thus, for each diffuse, gray, opaque surface, in radiant heat exchange with other surfaces of an enclosure, there are two resistances, viz. the surface resistance, Ri = (1- εi)/(Ai.εi), and a space resistance, Rij = 1/(Ai.Fij).


• For a N surface enclosure, net heat transfer from surface ‘i’ should be equal to the sum of net heat transfers from that surface to the remaining surfaces. i.e.

• This situation is shown in Fig. (13.27).

Q i1

N

j

Q ij= 1

N

j

A i F ij. J i J j

.

= 1

N

j

J i J j

R ij=

...W....(13.54)

i.e.E bi J i

R i 1

N

j

J i J j

R ij=

W...(13.55)

J


Fig. 13.27 Radiation heat transfer from surface ito other surfaces in a N-surface enclosure

Surface i

JiEbi

Qi

Ri

J1

J2

JN-1

JN

Ri1 Ri2

Ri(N-1)

RiN

• As can be seen from the above fig., rate of radiation ‘current’ flow to surface ‘i’ through its surface resistance must be equal to the sum of all the radiation current flows from surface ‘i’ to all other surfaces through the respective space resistances.

• In general, there two types of radiation problems:


problems:• first (and most common), when the surface

temperature Ti, and therefore, the emissive power Ebi is known;

• and, the second type is when the net radiation heat transfer at the surface i is known.

• Eqn. (13.55) is useful in solving the first type of problems, i.e. when the surface temperature is known; instead, if the net heat transfer rate at the surface is the known quantity, eqn. (13.52) is the applicable equation.


• Essentially, the problem is to solve for the radiosities J1, J2,….Jn.

• Electrical network method is convenient to use if the number of surfaces in an enclosure is limited to about five.

• If the number of surfaces is more than five, the direct approach is to apply eqn. (13.55) for each surface whose temperature is known, and eqn. (13.52) for each surface at which the net heat transfer rate is known, and solve the resulting set of N linear, algebraic equations for the N unknowns, viz. J1, J2,…Jn by standard


mathematical methods. • Once the radiosities are known, eqn. (13.48)

may be applied to determine either the heat transfer rate or the temperature, as the case may be.

Radiation heat exchange in two-zone enclosures:• Two-zone enclosure- simply means that the two

surfaces, together, make up the enclosure and ‘see’ only themselves and nothing else.

• Many, practically important geometries may be classified as two-zone enclosures, eg. a small body enclosed by a large body, a pipe passing


classified as two-zone enclosures, eg. a small body enclosed by a large body, a pipe passing through a large room, concentric spheres, concentric, long cylinders , long, parallel plane surfaces, etc.

• Fig. (13.28) shows a schematic of a typical two-zone enclosure and the associated radiation (or, thermal) network.

• Surfaces 1 and 2 forming the enclosure are diffuse, gray and opaque.

• Let their emissivities, temperatures and areas be (ε1, T1, A1) and (ε2, T2, A2) respectively.

• The radiation network is shown in Fig. 13.28.


in Fig. 13.28. • Each surface has one surface

resistance associated with it and there is one space resistance between the two radiosity potentials, and all the resistances are in series, as shown.

• The ‘heat current’ (Q12) in this circuit is calculated by dividing the ‘total potential’ (Eb1 – Eb2) by the ‘total resistance’ (R1 + R12 +R2). So, we write:

Q 12 Q 1 Q 2E b1 E b2

R 1 R 12 R 2

i.e. Q 12E b1 E b2

1 ε 1 1 1 ε 2


A 1 ε 1. A 1 F 12

. A 2 ε 2.

i.e. Q 12σ T 1

4 T 24.

1 ε 1

A 1 ε 1.

1

A 1 F 12.

1 ε 2

A 2 ε 2.

W......(13.56)

Eqn. (13.56) is an important equation, which gives net rate of heattransfer between two gray, diffuse, opaque surfaces which form anenclosure, i.e. which ‘see’ only each other and nothing else.

• Let us consider a few special cases of two-surface enclosure:

• Case (i): Radiant heat exchange between two black surfaces:

• For a black body, ε = 1, and J = Eb, as explained earlier. i.e. surface resistance [= (1 - ε)/(A.ε)] of a black body is zero.


[= (1 - ε)/(A.ε)] of a black body is zero. Then, the radiation network will consist of only a space resistance between the two radiosity potentials, as shown in Fig. (13.29):

• Then, from eqn. (13.56), we get:

Fig. 13.29 Radiation network for two blacksurfaces forming an enclosure

Eb1= J1

R12 = 1/(A1.F12 )

Q12Eb2= J2


Q 12σ T 1

4 T 24.

1

A 1 F 12.

i.e. Q 12 A 1 F 12. σ. T 1

4 T 24. W....for two black surfaces forming an

enclosure....(13.57)

• Next, we shall consider four cases of practical interest where the view factor between the inner surface 1 and the outer surface 2 (i.e. F12) is equal to 1:

• Case (ii): Radiant heat exchange for a small object in a large cavity:

• See Fig. (13.30,a). A practical example of a small object in a large cavity is the case of a steam pipe passing through a large plant room.


through a large plant room.• For this case, we have:

A 1

A 20

and, F 12 1

• And, eqn.(13.56) becomes:

• Case (iii): Radiant heat exchange between infinitely large parallel plates:

• See Fig. (13.30,b). In this case, A1 = A2 = A, say, and F12 = 1.

Q 12 A 1 σ. ε 1. T 1

4 T 24. ....for small object in a large cavity.....(13.58)


A, say, and F12 = 1. • Then, eqn. (13.56) becomes:

Q 12A σ. T 1

4 T 24.

1

ε 1

1

ε 21

....for infinitely large parallel plates.....(13.59)


Fig. 13.30. Important ‘two surface enclosures’.

• Case (iv): Radiant heat exchange between infinitely long concentric cylinders:

• See Fig. (13.30,c). In this case:

• Then, eqn. (13.56) becomes:

F 12 1

Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for infinitely long concentric cylinders.....(13.60)


• Remember that A1 refers to the inner (or enclosed) surface.• Eqn. (13.60) is known as ‘Christiansen’s equation’.

where

A 1

A 2

r 1

r 2

• Case (v): Radiant heat exchange between concentric spheres:

• See Fig. (13.30,d). In this case:• Then, eqn. (13.56) becomes:

F 12 1

Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for concentric spheres.....(13.61)


• Remember, again, that A1 refers to the inner (or enclosed) surface.

1 2 2

where

A 1

A 2

r 1

r 2

2

Problems on two-surface enclosures:

• Example 13.19 (M.U. 1991): A long pipe, 50 mm in diameter, passes through a room and is exposed to air at 20 deg. C. Pipe surface temperature is 93 deg.C. Emissivity of the surface is 0.6. Calculate the net radiant heat loss surface is 0.6. Calculate the net radiant heat loss per metre length of pipe.

• Solution:• The pipe is enclosed by the room; so, it is two-surface enclosure

problem. Further, area of the pipe is very small, compared to the area of the room. Therefore, this is a case of a small object surrounded by a large area, and we have:


Aug. 2016 36

This is a two-zone enclosure problem.Fig. shows the radiation network for this problem.

MT/SJEC/M.Tech.

We have:


Radiation heat exchange in three zone enclosures:• Fig. (13.32,a) shows an enclosure made of three

opaque, diffuse, gray surfaces. • Let the surfaces A1, A2, A3 be maintained at uniform

temperatures of T1, T2 and T3 respectively. Also, let the emissivities be ε1, ε2 and ε3 respectively.

• The radiation network for this system of three surface


• The radiation network for this system of three surface enclosure is shown in Fig. (13.32,b).

• While drawing the radiation network, the principle to be followed is quite simple: first, draw the surface resistance associated with each gray surface; then, connect the radiosity potentials between surfaces by the respective space resistances.

• It is considered that the temperature of each surface is known; i.e. emissive power Eb for each surface is known.

• Then, the problem reduces to determining the radiosities J1, J2 and J3.

• This is done by applying Kirchoff’s law of


• This is done by applying Kirchoff’s law of d.c. circuits to each node: i.e. sum of the currents (or, rate of heat transfers) entering into each node is zero.

• Doing this, we get the following three algebraic equations:

Node J1:E b1 J 1

R 1

J 2 J 1

R 12

J 3 J 1

R 130 .....(13.63,a)

Node J2:E b2 J 2

R 2

J 1 J 2

R 12

J 3 J 2

R 230 .....(13.63,b)

Node J :E b3 J 3 J 1 J 3 J 2 J 3

0 .....(13.63,c)


• Solving these three eqns. simultaneously, we get J1, J2and J3.

• Remember to write each eqn. such that current flows into the node; then, the magnitudes of the radiosities would adjust themselves when all the three equations are solved simultaneously.

Node J3: b3 3

R 3

1 3

R 13

2 3

R 230 .....(13.63,c)

• Once the magnitudes of the radiosities are known, expressions for net heat flows betweenthe surfaces are:

Q 12J 1 J 2

R 12

J 1 J 2

1

A 1 F 12.

....(13.64,a)


Q 13J 1 J 3

R 13

J 1 J 3

1

A 1 F 13.

....(13.64,b)

Q 23J 2 J 3

R 23

J 1 J 2

1

A 2 F 23.

....(13.64,c)

• And, net heat flow from each surface is:

Q 1E b1 J 1

R 1

E b1 J 1

1 ε 1

A 1 ε 1.

.....(13.65,a)

QE b2 J 2 E b2 J 2 .....(13.65,b)


Q 2b2 2

R 2

b2 2

1 ε 2

A 2 ε 2.

.....(13.65,b)

Q 3E b3 J 3

R 3

E b3 J 3

1 ε 3

A 3 ε 3.

.....(13.65,c)

• Eqn. set (13.64) is a set of general equations for three diffuse, opaque, gray surfaces.

• However, these equations will be modified depending upon any constraint that may be attached to any of the surfaces, i.e.

• say, if the surface is black or re-radiating:


• say, if the surface is black or re-radiating:Ji = Ebi = σ.Ti

4. • And, Qi = 0 for a re-radiating surface. • If Qi at any surface is specified instead of

the temperature (i,.e. Ebi), then, (Ebi – Ji)/Ri is replaced by Qi.

Few special cases of three-zone enclosures:• Case (i): Two black surfaces connected to a

third refractory surface:• This is a three-zone enclosure, with two of the

surfaces being black and the third surface being a re-radiating, insulated surface.


a re-radiating, insulated surface. • Typical example is a furnace whose bottom is

the ‘source’ and the top is the ‘sink’ and the two surfaces are connected by a refractory wall which acts as a re-radiating surface.

• In effect, the source and sink exchange heat through the re-radiating wall.

• However, in steady state, the re-radiating wall radiates as much heat as it receives, which means that net heat exchange through the re-radiating wall (= Q) is zero, i.e. Eb = J for the re-radiating wall.

• Therefore, once J (i.e. Eb) is calculated for the re-radiating surface, its steady state


bthe re-radiating surface, its steady state temperature can easily be calculated from: Eb = σ. T4.

Eb1= J1

E = J R

Eb2 = J2

R12

R12=1/(A1.F12 )

EbR= JR

R2R R1R

R1R=1/(A1.F1R )

R2R=1/(A2.F2R )

Fig.(a)


Fig. 13.33 Two black surfaces connected by a third re-radiatingsurface and its radiation network

Eb1 = J1 Eb2 = J2R12

R2R R1R JR

Fig.(b)

• The radiation network is drawn very easily by remembering the usual principles, i.e.

• for a black surface, the surface resistance is zero, i.e. Eb = J.

• For a re-radiating surface too, Eb = J, as already explained;

• Further, for a re-radiating surface, Q = 0. • Between two given surfaces, the radiosity


• Between two given surfaces, the radiositypotentials are connected by the respective space resistances, as shown.

• It may be observed that the system reduces to a series-parallel circuit of resistances as shown in Fig. (13.33,b).

• So, we write, for the total resistance of the circuit, Rtot:

1

R tot

1

R 12

1

R 1R R 2R

and, Q 12E b1 E b2

R totE b1 E b2

1

R 12

1

R 1R R 2R

.


• Here, Q12 is the net radiant heat transferred between surfaces 1 and 2. Similar expressions can be written for heat transfer between surfaces 2 and 3 (= Q23) and the heat transfer between surfaces 1 and 3 (= Q13).

i.e. Q 12 σ T 14 T 2

4. A 1 F 12. 1

1

A 1 F 1R.

1

A 2 F 2R.

. .....(13.66)

Case (ii): Two gray surfaces surrounded by a third re-radiating surface:


• In this case, there are two gray surfaces, and the third surface is an insulated, re-radiating surface.

• As already explained, the re-radiating surface radiates as much energy as it receives; therefore, net radiant heat transfer for that surface is zero, i.e. Q 3 0


surface is zero, i.e. Q 3 0

i.e.E b3 J 3

1 ε 3

A 3 ε 3.

0

i.e. E b3 J 3

• i.e. once the radiosity of the re-radiating surface is known, its temperature can easily be calculated, since Eb3 = σ.T3

4. • Further, note that T3 is independent of the

emissivity of surface 3.• Now, the radiation network reduces to a simple

series-parallel circuit of the relevant resistances.


series-parallel circuit of the relevant resistances.• Expression for heat flow rate is:

Q 1 Q 2E b1 E b2

R tot

where, Rtot is the total resistance, given by:

R tot R 11

1

R 12

1

R 13 R 23

R 2

i.e. R tot1 ε 1

A 1 ε 1.

1

A 1 F 12. 1

1 1

1 ε 2

A 2 ε 2.

....(13.67)


1

A 1 F 13.

1

A 2 F 23.

Radiation shielding:

• One or more ‘radiation shields’ are used to reduce radiant heat transfer between two given surfaces.

• Radiation shield is, simply a thin, high reflectivity


• Radiation shield is, simply a thin, high reflectivity surface placed in between the surfaces which exchange heat between themselves.

• Radiation shields may be made of aluminium foils, copper foils, or aluminized mylar sheets etc.

• Fig. (13.36,a) shows to large parallel plates, 1 and 2 exchanging heat between themselves;

• Let their areas, temperatures (in Kelvin) and emissivities be (A1, T1, ε1) and (A2, T2, ε2).

• Let a radiation shield 3, be placed between these plates. Plate 3 is thin and made of a material of high reflectivity.

• Let the emissivities of two sides of the radiation shield be ε and ε as shown.


shield be ε3-1 and ε3-2 as shown. • Radiation network for this system is shown in

Fig. (13.36,b). • This is drawn, as usual, remembering that each

gray surface has a ‘surface resistance’ associated with it, and the two radiosity potentials are connected by a ‘space resistance’.

• When there is no shield, the radiation heat transfer between plates 1 and 2 is already shown to be:

• With one shield placed between plates 1 and 2,

Q 12A σ. T 1

4 T 24.

1

ε 1

1

ε 21

....for infinitely large parallel plates.....(13.59)


• With one shield placed between plates 1 and 2, the radiation network will be as shown in Fig. (b) above.

• Note that now all the relevant resistances are in series.

• Net heat transfer between plates 1 and 2is given as:Q12-one shield = (Eb1-Eb2)/Rtot where Rtot is

the total resistance.• i.e.

E b1 E b2


Q 12_one_shieldE b1 E b2

1 ε 1

A 1 ε 1.

1

A 1 F 13.

1 ε 3_1

A 3 ε 3_1.

1 ε 3_2

A 3 ε 3_2.

1

A 3 F 32.

1 ε 2

A 2 ε 2.

....for two gray surfaces with one radiation shield placed in between.....(13.70)

• Now, for two large parallel plates, we note:

• Then, eqn. (13.70) simplifies to:

F 13 F 32 1 and, A 1 A 2 A 3 A

Q 12_one_shieldA σ. T 1

4 T 24.

1 11

1 11

....(13.71)


• Note that as compared to eqn. (13.59) for the case of no-shield, we have, with one shield, an additional term appearing in the denominator of eqn. (13.71).

1

ε 1

1

ε 21

1

ε 3_1

1

ε 3_21

• Therefore, if there are N radiation shields, we have, for net radiation heat transfer:

• If emissivities of all surfaces are equal, eqn. (13.72) becomes:

Q 12N_shields

A σ. T 14 T 2

4.

1

ε 1

1

ε 21

1

ε 3_1

1

ε 3_21 .....

1

ε N_1

1

ε N_21

...(13.72)


(13.72) becomes:

Q 12N_shields

A σ. T 14 T 2

4.

N 1( )1

ε1

ε1.

1

N 1( )Q 12no_shield

. ....(13.73)

• Note this important result, which implies that, when all emissivities are equal, presence of one radiation shield reduces the radiation heat transfer between the two surfaces to one-half,

• Two radiation shields reduce the heat transfer to one-third, 9 radiation shields reduce the heat transfer to one-tenth etc.

• For a more practical case of the two surfaces having ε ε


emissivities of ε1 and ε2, and all shields having the same emissivity of εs, eqn. (13.72) becomes:

Q 12N_shields

A σ. T 14 T 2

4.

1

ε 1

1

ε 21 N

2

ε s1.

....(13.74)

• To determine the equilibrium temperature of the radiation shield:

• Once Q12 is determined from eqn. (13.71), the temperature of the shield is easily found out by applying the condition that in steady state:

• We can use either of the conditions: Q12 = Q13 or Q12 = Q32. • Q13 or Q32 is determined by applying eqn. (13.59); i.e. we get:

Q 12 Q 13 Q 32 ...(13.75)

A σ. T 14 T 3

4.


• In both the above eqns., T3 is the only unknown, which can easily be determined.

Q 12 Q 13A σ. T 1 T 3

.

1

ε 1

1

ε 31

...(13.76,a)

or,

Q 12 Q 32A σ. T 3

4 T 24.

1

ε 3

1

ε 21

....(13.76,b)

For a cylindrical radiation shield placed in between two, long concentric cylinders:


• Consider the case of radiation heat transfer between two long, concentric cylinders.

• The radiation heat transfer between two long, concentric cylinders is already shown to be:

Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for infinitely long concentric cylinders.....(13.60)

where

A 1

A 2

r 1

r 2


• Now, let a cylindrical radiation shield, 3, be placed in between the inner cylinder (1) and the outer cylinder (2), as shown in Fig.

• The radiation network for this system is exactly the same as shown in Fig. (13.36,b):

• And, the radiation heat transfer between cylinders 1 and 2, when the shield is present, is given by:


2, when the shield is present, is given by:

• Now, for the cylindrical system, we have:

Q 12_one_shieldE b1 E b2

1 ε 1

A 1 ε 1.

1

A 1 F 13.

1 ε 3_1

A 3 ε 3_1.

1 ε 3_2

A 3 ε 3_2.

1

A 3 F 32.

1 ε 2

A 2 ε 2.

....for two gray surfaces with one radiation shield placed in between.....(13.70)

F 13 F 32 1

A 1 2 π. r 1. L.

A 2 2 π. r 2. L.

and, A 3 2 π. r 3. L.


• Then, eqn. (13.70) reduces to:

• In eqn. (13.77), we have: (A1/A2) = (r1/r2), and (A1/A3) = (r1/r3).

Q 12one_shield

A 1 σ. T 14 T 2

4.

1

ε 1

A 1

A 2

1

ε 21.

A 1

A 3

1

ε 3_1

1

ε 3_21.

...for concentric cylinders with one radiation shield...(13.77)

• Note that as compared to the relation for two concentric cylinders with no shield (i.e. eqn. 13.60), an additional term appears in the denominator of eqn. (13.77) (i.e. the third term) when one radiation shield is introduced;

• If there is a second radiation shield, say (4), then one more similar term will have to be added in


one more similar term will have to be added in the denominator to take care of the resistance of that shield.

• Equilibrium temperature of the shield is determined by applying the principle that, in steady state: Q 12 Q 13 Q 32

For a Spherical radiation shield placed in between two concentric spheres:

• The radiation network for this system is shown in Fig. (13.36,b).

• When there is no radiation shield, radiation heat transfer between surfaces 1 and 2 is given by eqn. (13.61), viz.


Q 12A 1 σ. T 1

4 T 24.

1

ε 1

A 1

A 2

1

ε 21.

....for concentric spheres.....(13.61)

where

A 1

A 2

r 1

r 2

2

• Again, when the radiation shield is present, the general relation for

radiation heat transfer between surfaces 1 and 2 is eqn. (13.70).

• Relation for radiant heat transfer between surfaces 1 and 2, is exactly as eqn. (13.77), i.e.

Q 12one_shield

A 1 σ. T 14 T 2

4.

1

ε 1

A 1

A 2

1

ε 21.

A 1

A 3

1

ε 3_1

1

ε 3_21.

...for concentric spheres with one radiation shield...(13.78)


• In eqn. (13.78), we have:

• Equilibrium temperature of the shield is determined by applying the principle that, in steady state:

A 1

A 2

r 1

r 2

2

and,A 1

A 3

r 1

r 3

2

Q 12 Q 13 Q 32

Radiation error in temperature measurement:• An important application of radiation shields is in

reducing the radiation error in temperature measurement.

• Consider the case of a hot fluid at a temperature Tf, flowing through a channel, whose walls are at a temperature T .


Tf, flowing through a channel, whose walls are at a temperature Tw.

• Let the convective heat transfer coefficient between the fluid and the thermometer bulb be h.

• To measure the temperature of the fluid, a thermometer (or a thermocouple) is introduced into the stream, as shown in Fig. (13.39 a).

Tf , h

(a) Thermometer without radiation shield

Tw

qradqconv

TcThermometer

TcThermometer

Tw


Fig. 13.39 Radiation shielding of thermometers

Tw

Tc

Ts

Tw

Tf , h

(b) Thermometer with radiation shield

• Let the reading shown by the thermometer be Tc.

• This reading, however, does not represent the true temperature of the fluid Tf, since the thermometer bulb will lose heat by radiation to the walls of the channel which are at a lower temperature Tw (which is usually the case).

• So, in steady state, the thermometer bulb will gain heat by convection from the flowing fluid


gain heat by convection from the flowing fluid and will lose heat by radiation to the walls, and as a result, the temperature Tc shown by the thermometer will be some value in between Tfand Tw .

• We wish to find out the true temperature of the fluid Tf , by knowing the thermometer reading Tc.

• Making an energy balance on the thermometer bulb, in steady state, we have:

• Without radiation shield:• qconv to the bulb = qrad from the bulb

i.e. h A c. T f T c

. ε c A c. σ. T c

4 T w4.

i.e. T f T cε c σ. T c

4 T w4.

.....(13.79)


where Ac = surface area of thermometer bulb, εc = emissivity of thermometer bulb surface

• Eqn. (13.79) gives the true temperature of the fluid Tf.• Second term on the RHS of eqn. (13.79) represents the

error in temperature measurement due to radiation effect

i.e. T f T c h.....(13.79)

• Radiation error can be minimized by:• (i) having low value of εc i.e. high reflectivity for the

bulb surface• (ii) high value for h, the convective heat transfer

coefficient• In practice, even if we start with a thermometer bulb

surface of high reflectivity, soon the emissivity value rises to about 0.8 or 0.9 due to deposit formation,


rises to about 0.8 or 0.9 due to deposit formation, corrosion or erosion of the bulb surface, etc.

• So, the most practical way to reduce the radiation errorin temperature measurement is to provide a cylindrical radiation shield around the thermometer bulb, as shown in Fig. (13.39,b) above.

• In steady state, the shield temperature (Ts) will stabilize somewhere in between the fluid temperature Tf and the wall temperature Tw.

• Then, in eqn. (13.79), Tw will be replaced by the effective shield temperature Ts.

• Energy balance on the thermometer bulb:• Heat transferred to the bulb from the fluid by convection =

Heat transferred from the bulb to the shield by radiation


Heat transferred from the bulb to the shield by radiation

• In eqn. (13.80), Fcs = view factor of thermometer bulb w.r.t the shield and is, generally equal to 1.

i.e. h A c. T f T c

.σ T c

4 T s4.

1 ε c

A c ε c.

1

A c F cs.

1 ε s

A s ε s.

....(13.80)

• Making an energy balance on the shield:heat transferred to shield from the fluid by convection + heat transferred to shield from bulb by radiation = heat transferred from shield to walls by radiation

i.e. 2 A s. h. T f T s

.σ T c

4 T s4.

1 ε c

A c ε c.

1

A c F cs.

1 ε s

A s ε s.

ε s A s. σ. T s

4 T w4. ....(13.81)


where, A s = area of shield on one side

ε s = emissivitty of shield surface

A c = area of bulb surface

ε c = emissivity of bulb surface

F cs = view factor of bulb w.r.t. shield

• In the first term of the above eqn. factor 2 appears since convective heat transfer to the shield occurs on both surfaces of the shield.

• Also, in writing the RHS, the inherent assumption is that:

F sw 1 ...view factor between the shield and the walls

and,


• Solving eqns. (13.80) and (13.81) simultaneously, we obtain the shield temperature Ts and the thermometer reading Tc, (if Tf is known), or Tf (if Tc is known).

A s

A w0 ...i.e. surface area of shield is negligible compared to the area of the channel

walls

Thermal Radiation - III- Radn. energy exchange between gray surfaces

Engineering

Transcript of Thermal Radiation - III- Radn. energy exchange between gray surfaces