A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 3 Part 3 doc

§10.3 Radiant heat exchange between two finite black bodies 547Figure 10.11 Heat transfer from a disc heater to its radiation shield.. §10.4 Heat transfer among gray bodies 549Electrical

§10.3 Radiant heat exchange between two finite black bodies 539

Of course, Qnet1–2 = −Qnet 2–1 It follows that

A1F1–2σ

T14− T4 2

= −A2F2–1σ

T24− T4 1

or

This result, called view factor reciprocity, is very useful in calculations.

Example 10.1

A jet of liquid metal at 2000◦C pours from a crucible It is 3 mm in

di-ameter A long cylindrical radiation shield, 5 cm diameter, surrounds

the jet through an angle of 330◦, but there is a 30◦ slit in it The jet

and the shield radiate as black bodies They sit in a room at 30◦C, and

the shield has a temperature of 700◦C Calculate the net heat transfer:

from the jet to the room through the slit; from the jet to the shield;

and from the inside of the shield to the room

Solution. By inspection, we see that Fjet–room = 30/360 = 0.08333

and Fjet–shield = 330/360 = 0.9167 Thus,

Qnetjet–room= AjetFjet–roomσ

The heat absorbed by the shield leaves it by radiation and convection

to the room (A balance of these effects can be used to calculate the

shield temperature given here.)

To find the radiation from the inside of the shield to the room, we

need Fshield–room Since any radiation passing out of the slit goes to the

540 Radiative heat transfer §10.3

room, we can find this view factor equating view factors to the room

with view factors to the slit The slit’s area is Aslit= π(0.05)30/360 =

0.01309 m2/m length Hence, using our reciprocity and summationrules, eqns (10.12) and (10.15),

Hence, for heat transfer from the inside of the shield only,

Qnetshield–room= AshieldFshield–roomσ

Calculation of the black-body view factor, F1–2 Consider two elements,

dA1 and dA2, of larger black bodies (1) and (2), as shown in Fig 10.8

Body (1) and body (2) are each isothermal Since element dA2 subtends

a solid angle dω1, we use eqn (10.6) to write

dQ1 to 2= (i1dω1)(cos β1dA1)

But from eqn (10.7b),

i1= σ T

4 1

π

Note that because black bodies radiate diffusely, i1 does not vary withangle; and because these bodies are isothermal, it does not vary withposition The element of solid angle is given by

dω1= cos β2dA2

s2

§10.3 Radiant heat exchange between two finite black bodies 541

Figure 10.8 Radiant exchange between two black elements

that are part of the bodies (1) and (2)

where s is the distance from (1) to (2) and cos β2 enters because dA2 is

not necessarily normal to s Thus,

dQ1 to 2= σ T

4 1

The view factors F1–2 and F2–1are immediately obtainable from eqn

(10.16) If we compare this result with Qnet1–2 = A1F1–2σ (T14− T4

542 Radiative heat transfer §10.3

From the inherent symmetry of the problem, we can also write

in-We list some typical expressions for view factors in Tables10.2 and

10.3 Table 10.2 gives calculated values of F1–2 for two-dimensionalbodies—various configurations of cylinders and strips that approach in-finite length Table10.3gives F1–2 for some three-dimensional configu-rations

Many view factors have been evaluated numerically and presented

in graphical form for easy reference Figure10.9, for example, includesgraphs for configurations 1, 2, and 3 from Table10.3 The reader shouldstudy these results and be sure that the trends they show make sense

Is it clear, for example, that F1–2
→ constant, which is < 1 in each case,

as the abscissa becomes large? Can you locate the configuration on theright-hand side of Fig.10.6in Fig.10.9? And so forth

Figure10.10shows view factors for another kind of configuration—one in which one area is very small in comparison with the other one.Many solutions like this exist because they are a bit less difficult to cal-culate, and they can often be very useful in practice

Example 10.2

A heater (h) as shown in Fig. 10.11 radiates to the partially conical

shield (s) that surrounds it If the heater and shield are black,

calcu-late the net heat transfer from the heater to the shield

Solution. First imagine a plane (i) laid across the open top of the

shield:

F h −s + F h −i = 1

But F h −i can be obtained from Fig 10.9 or case 3 of Table 10.3,

Table 10.2 View factors for a variety of two-dimensional

con-figurations (infinite in extent normal to the paper)

2

−



h w

Table 10.3 View factors for some three-dimensional configurations

W2

H2(1 + H2+ W2) (1 + H2)(H2+ W2)

Figure 10.10 The view factor for three very small surfaces

“looking at” three large surfaces (A1 A2).

546

§10.3 Radiant heat exchange between two finite black bodies 547

Figure 10.11 Heat transfer from a disc heater to its radiation shield.

for R1 = r1/h = 5/20 = 0.25 and R2 = r2/h = 10/20 = 0.5 The

result is F h−i = 0.192 Then

Suppose that the shield in Example10.2were heating the region where

the heater is presently located What would F s −hbe?

Solution. From eqn (10.15) we have

548 Radiative heat transfer §10.3

view factor in terms of the unknown F1–2 and other known view tors:

fac-(2A)F (1 +3)–(4+2) = AF1–4+ AF1–2+ AF3–4+ AF3–2

2F (1 +3)–(4+2) = 2F1–4+ 2F1–2

F1–2= F (1 +3)–(4+2) − F1–4

And F (1 +3)–(4+2) can be read from Fig.10.9 (at φ = 90, w/ = 1/2,

and h/ = 1/2) as 0.245 and F1–4as 0.20 Thus,

F1–2= (0.245 − 0.20) = 0.045

Figure 10.12 Radiation between two

offset perpendicular squares

§10.4 Heat transfer among gray bodies 549

Electrical analogy for gray body heat exchange

An electric circuit analogy for heat exchange among diffuse gray bodies

was developed by Oppenheim [10.6] in 1956 It begins with the definition

of two new quantities:

H (W/m2) ≡ irradiance =

flux of energy that irradiates thesurface

and

B (W/m2) ≡ radiosity =

total flux of radiative energyaway from the surfaceThe radiosity can be expressed as the sum of the irradiated energy that

is reflected by the surface and the radiation emitted by it Thus,

We can immediately write the net heat flux leaving any particular

sur-face as the difference between B and H for that sursur-face Then, with the

help of eqn (10.18), we get

Equation (10.21) is a form of Ohm’s law, which tells us that (e b − B) can

be viewed as a driving potential for transferring heat away from a surface

through an effective surface resistance, (1 − ε)/εA.

Now consider heat transfer from one infinite gray plate to another

parallel to it Radiant energy flows past an imaginary surface, parallel

to the first infinite plate and quite close to it, as shown as a dotted line

550 Radiative heat transfer §10.4

Figure 10.13 The electrical circuit analogy for radiation

be-tween two gray infinite plates

in Fig.10.13 If the gray plate is diffuse, its radiation has the same metrical distribution as that from a black body, and it will travel to otherobjects in the same way that black body radiation would Therefore, wecan treat the radiation leaving the imaginary surface — the radiosity, that

geo-is — as though it were black body radiation travelling to an imaginarysurface above the other plate Thus, by analogy to eqn (10.13),

Qnet1–2 = A1F1–2(B1− B2) = B1− B2

1

A1F1–2

where the final fraction shows that this is also a form of Ohm’s law:

the radiosity difference (B1− B2), can be said to drive heat through the

geometrical resistance, 1/A1F1–2, that describes the field of view betweenthe two surfaces

When two gray surfaces exchange heat by thermal radiation, we have

a surface resistance for each surface and a geometric resistance due totheir configuration The electrical circuit shown in Fig.10.13 expresses

the analogy and gives us means for calculating Qnet 1–2 from Ohm’s law

Recalling that e b = σ T4, we obtain



1− ε εA

§10.4 Heat transfer among gray bodies 551

(Fig.10.6), and, with qnet1–2= Qnet 1–2/A1, we find

for infinite parallel plates Notice, too, that if the plates are both black

(ε1= ε2= 1), then both surface resistances are zero and

F1–2= 1 = F1–2

which, of course, is what we would have expected

Example 10.5 One gray body enclosed by another

Evaluate the heat transfer and the transfer factor for one gray body

enclosed by another, as shown in Fig.10.14

Solution. The electrical circuit analogy is exactly the same as that

shown in Fig.10.13, and F1–2is still unity Therefore, with eqn (10.23),

Qnet1–2 = A1qnet1–2 = σ

T14− T4 2

Figure 10.14 Heat transfer between an

enclosed body and the body surroundingit

552 Radiative heat transfer §10.4

The transfer factor may again be identified by comparison to eqn (10.2):

Qnet1–2= A1

11

(10.27)

This calculation assumes that body (1) does not view itself

Example 10.6 Transfer factor reciprocity

DeriveF2–1for the enclosed bodies shown in Fig.10.14

Solution.

Qnet1–2= −Qnet2–1

A1F1–2σ

T14− T4 2

= −A2F2–1σ

T24− T4 1

from which we obtain the reciprocity relationship for transfer factors:

Example 10.7 Small gray object in a large environment

DeriveF1–2 for a small gray object (1) in a large isothermal ment (2), the result that was given as eqn (1.35)

environ-Solution. We may use eqn (10.27) with A1/A2 1:

Note that the same result is obtained for any value of A1/A2 if the

enclosure is black (ε2= 1) A large enclosure does not reflect much

ra-diation back to the small object, and therefore becomes like a perfectabsorber of the small object’s radiation — a black body

§10.4 Heat transfer among gray bodies 553

Additional two-body exchange problems

Radiation shields A radiation shield is a surface, usually of high

re-flectance, that is placed between a high-temperature source and its cooler

environment Earlier examples in this chapter and in Chapter1show how

such a surface can reduce heat exchange Let us now examine the role

of reflectance (or emittance: ε = 1 − ρ) in the performance of a radiation

shield

Consider a gray body (1) surrounded by another gray body (2), as

discussed in Example 10.5 Suppose now that a thin sheet of reflective

material is placed between bodies (1) and (2) as a radiation shield The

sheet will reflect radiation arriving from body (1) back toward body (1);

likewise, owing to its low emittance, it will radiate little energy to body

(2) The radiation from body (1) to the inside of the shield and from the

outside of the shield to body (2) are each two-body exchange problems,

coupled by the shield temperature We may put the various radiation

resistances in series to find (see Problem10.46)

T14− T4 2

assuming F1–s= Fs–2 = 1 Note that the radiation shield reduces Qnet1–2

more if its emittance is smaller, i.e., if it is highly reflective

Specular surfaces The electrical circuit analogy that we have developed

is for diffuse surfaces If the surface reflection or emission has

direc-tional characteristics, different methods of analysis must be used [10.2]

One important special case deserves to be mentioned If the two gray

surfaces in Fig.10.14 are diffuse emitters but are perfectly specular

re-flectors — that is, if they each have only mirror-like reflections — then

the transfer factor becomes

F1–2= 1 1

ε1 + ε1

2 − 1

 for specularlyreflecting bodies (10.32)

This result is interestingly identical to eqn (10.25) for parallel plates

Since parallel plates are a special case of the situation in Fig 10.14, it

follows that eqn (10.25) is true for either specular or diffuse reflection

554 Radiative heat transfer §10.4

Example 10.8

A physics experiment uses liquid nitrogen as a coolant Saturatedliquid nitrogen at 80 K flows through 6.35 mm O.D stainless steel

line (ε l = 0.2) inside a vacuum chamber The chamber walls are at

T c = 230 K and are at some distance from the line Determine the

heat gain of the line per unit length If a second stainless steel tube,12.7 mm in diameter, is placed around the line to act as radiationshield, to what rate is the heat gain reduced? Find the temperature

of the shield

Solution. The nitrogen coolant will hold the surface of the line atessentially 80 K, since the thermal resistances of the tube wall and theinternal convection or boiling process are small Without the shield,

we can model the line as a small object in a large enclosure, as inExample10.7:

Qgain= (πD l )ε l σ

T c4− T4

l

= π(0.00635)(0.2)(5.67 × 10 −8 )(2304− 804) = 0.624 W/m

With the shield, eqn (10.31) applies Assuming that the chamber area

is large compared to the shielded line (A c A l),

The radiation shield would cut the heat gain by 47%

The temperature of the shield, T s, may be found using the heatloss and considering the heat flow from the chamber to the shield,with the shield now acting as a small object in a large enclosure:

§10.4 Heat transfer among gray bodies 555

The electrical circuit analogy when more than two gray bodies

are involved in heat exchange

Let us first consider a three-body transaction, as pictured in at the

bot-tom and left-hand sides of Fig 10.15 The triangular circuit for three

bodies is not so easy to analyze as the in-line circuits obtained in

two-body problems The basic approach is to apply energy conservation at

each radiosity node in the circuit, setting the net heat transfer from any

one of the surfaces (which we designate as i)

If the temperatures T1, T2, and T3 are known (so that e b1, e b2, e b3 are

known), these equations can be solved simultaneously for the three

un-knowns, B1, B2, and B3 After they are solved, one can compute the net

heat transfer to or from any body (i) from either of eqns (10.33)

Thus far, we have considered only cases in which the surface

temper-ature is known for each body involved in the heat exchange process Let

us consider two other possibilities

556 Radiative heat transfer §10.4

Figure 10.15 The electrical circuit analogy for radiation

among three gray surfaces

An insulated wall If a wall is adiabatic, Qnet = 0 at that wall For

example, if wall (3) in Fig.10.15 is insulated, then eqn (10.33b) shows

that e b3 = B3 We can eliminate one leg of the circuit, as shown on theright-hand side of Fig.10.15; likewise, the left-hand side of eqn (10.34c)equals zero This means that all radiation absorbed by an adiabatic wall

is immediately reemitted Such walls are sometimes called “refractorysurfaces” in discussing thermal radiation

The circuit for an insulated wall can be treated as a series-parallelcircuit, since all the heat from body (1) flows to body (2), even if it does

so by travelling first to body (3) Then

§10.4 Heat transfer among gray bodies 557

A specified wall heat flux The heat flux leaving a surface may be known,

if, say, it is an electrically powered radiant heater In this case, the

left-hand side of one of eqns (10.34) can be replaced with the surface’s known

Qnet, via eqn (10.33b)

For the adiabatic wall case just considered, if surface (1) had a

speci-fied heat flux, then eqn (10.35) could be solved for e b1and the unknown

temperature T1

Example 10.9

Two very long strips 1 m wide and 2.40 m apart face each other, as

shown in Fig.10.16 (a) Find Qnet1–2 (W/m) if the surroundings are

black and at 250 K (b) Find Qnet1–2 (W/m) if they are connected by

an insulated diffuse reflector between the edges on both sides Also

evaluate the temperature of the reflector in part (b)

Solution. From Table 10.2, case 1, we find F1–2 = 0.2 = F2–1 In

addition, F2–3 = 1 − F2–1 = 0.8, irrespective of whether surface (3)

represents the surroundings or the insulated shield

In case (a), the two nodal equations (10.34a) and (10.34b) become

1451− B1

2.333 = B1− B2

1/0.2 + B1− B3

1/0.8 459.3 − B2

1 = B2− B1

1/0.2 + B2− B3

1/0.8

Equation (10.34c) cannot be used directly for black surroundings,

since ε3 = 1 and the surface resistance in the left-hand side

denom-inator would be zero But the numerator is also zero in this case,

since e b3 = B3 for black surroundings And since we now know

s2