Chapter 13 Heat Transfer

8/10/2019 Chapter 13 Heat Transfer

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R DI TIVE TR NSFER

BETWEEN TWO OR MORE

SURF CES

Prepared byNurhaslina bt che radzi

FKK, UITM



• This chapter focus on radiative exchange between two or moresurfaces.

• This exchange depends strongly on the surface geometries andorientations as well as on their radiative properties andtemperatures.

• To compute radiation exchange between any two surfaces, theconcept of a view factor must first introduce.

RADIATIVE TRANSFER BETWEEN TWO ORMORE SURFACES



VIEW FACTOR RELATIONS

• The view factor Fij is defined as the fraction of the radiation leaving surface i that isintercepted by surface j

• For radiation exchange between two surfaces of areas Ai and Aj, the view factors arerelated by reciprocity relation :

A iFij = A jF ji

• An example might be the floor and ceiling of a room

the view factor, F12 :

F12 = fraction of energy

leaving A 1 reaching A 2 A 1F12 = A 2F21

• For an enclosure consisting of N surfaces, the view factor are related by summationrule :

∑ Fij = 1

• For all radiant leaving the inner surface (1) must reach the outer surface (2), the viewfactor are related by inspection:

F12 = 1

• For several common geometries, view factors may be determined using Table 13.1 and

13.2 and Figure 13.4 through 13.6.

N

j = 1

A 1

A 2



• Consider a simple, two surfaces enclosure involving the sphericalsurfaces

• To calculate radiation exchange in an enclosure of N surface a total

of N2 view factors is needed

N2 = 22 = 4 view factors

F11 , F12 , F21 , F22

• From figure, since all radiation leaving the inner surface (1) mustreach the outer surface (2), by inspection:

F12 = 1 or by summation rule

* concave surface, it see itself Fii ≠ 0

* convex/plane surface

Fii = 0

F11 + F12 = 1 F11 = 0

F12 = 1

A 1

A 2



• From the reciprocity relation,

A 1F12 = A 2F21

F21 = A 1 F12

A 2

F21 = A 1

A 2

• From the summation rule,F21 + F22 = 1

F22 = 1 – F21

= 1 – F21

= 1 – A 1/A 2

• Additive rule for view factor relation

Fij = ∑ Fik A jF(j)I = ∑ A k Fki A j = ∑ A k n

k = 1

n

k = 1

n

k = 1



Example - PROBLEM 13.1Determine F12 and F21 for the following configurations using thereciprocity theorem and other basic shape factor relations. Do not usetable or charts

a) Long duct (L)

b) Small sphere of area A 1 under a concentric hemisphere of area A 2 = 2A 1

c) Long duct. What is F22 for this case?

d) Long inclined plates (point B is directly above the center of A 1)



e) Sphere lying on infinite plane

f) Hemisphere – disk arrangement

g) Long open channel (L)

h) Long concentric cylinders (L)

A 1

A 2

D1

D2



Radiation Exchange Between Opaque, Diffuse,Gray Surfaces in an Enclosure

• Analyzing radiation exchange in an enclosure is assumed to beisothermal and to be characterized by a uniform radiosity and auniform irradiation

Net radiation Exchange at a Surface

• The term qi , which is the net rate at which radiation leaves surface i,represents the net effect of radiative interactions occuring at the

surface

• The net radiation may be expressed as

qi

= A i

(Ji

– Gi

) (1)

Ji A i Gi A i

qi



Ji = Radiosity

= Ei + ρiGi

• The net radiation may also be expressed as

qi = A i(Ei – αiGi) (2)

Ei = Emissive power

Gi = Absorbed irradiation

For an opaque surface , αi = 1 - ρi , αi = absorptivity

ρi = 1 - αi εi = emmisivity

= 1 - εi ρi = reflectivity

Ji = εiE bi + (1 – εi)Gi qi = A i Ji – Ji – εiE bi

1 - εi

qi = E bi- Ji , E bi – Ji = driving potential (3)

(1 – εi)/ εi A i (1 – εi) = surface radiative resistance

εi A i



for black surface , εi = 1

* (1 - εi ) = 0

εi A i

* Ji = E bi blackbody radiation exchange

Radiation Exchange Between Surfaces

• To determine surface radiosity, Ji , it is necessary to considerradiation exchange between the surfaces of the enclosure

• From reciprocity relation :

A iGi = ∑ A iFijJ j

qi = A i (Ji - ∑FijJ j) (4)

• From the summation rule,

qi = A i (∑Fij Ji - ∑FijJ j)

qi = ∑ A i Fij (J j – J j) = ∑qij (5)

J j – J j = driving potential

(A i Fij )-1 = space/geometrical resistance

N

j = 1 N

j = 1

N

j = 1

N

j = 1N

j = 1

N

j = 1



• Combining equations (3) and (5)

E bi – Ji = ∑ Ji – J j

(1 – ξ i)/ ξ i A i (A iFij)-1

qi = ∑ Ji – J j

(A iFij)-1

Blackbody Radiation Exchange

• Equation (5) reduces to

qi = ∑ A iFij σ (Ti4 – T j

4)

• When all surfaces of the enclosure are black, there is no reflectionand radiosity is composed solely of the emitted energy

N

j = 1

N

j = 1

N

j = 1



BLACKBODY RADIATION EXCHANGE

• Consider radiation exchange between two black surfaces of arbitraryshape

• Defining qi j as the rate at which radiation leaves surfaces i and isintercepted by surface j

qi j = (A i Ji)Fij

for black surface , Ji = E bi

qi j = A iFij E bi

q j i = A jF ji E bj

A i , Ti A j , T j

ni n j

Ji = E bi

J j = E bj



• The net radiative exchange between the two surface is

qij = qi j - q j i

= A iFij E bi - A jF ji E bj

from Stefan – Boltzman Law : E b = σT4 and

from Reciprocity Relation : A iFij = A jF ji

The net rate at which radiation leaves surface i and intercepted bysurface j may be defined as

qij = A iFij σ (Ti4 – T j

4)



RPOBLEM 13.7

Consider the right-circular cylinder of diameter D, length L andthe areas A 1 , A 2 and A 3 representing the base, inner and topsurfaces

a) Show that the view factor form F12= 2H[(1 + H2)1/2 – H] , where

H = L/D

b) Show that the view factor for the inner surface to itself has theform F22 = 1 + H – (1 + H2)1/2



PROBLEM 13.8

Consider the parallel rectangles shown schematically. Show that the view factor F12 can be expressed as

F12 = 1 [A (14) F(14)(23) - A 1F13 - A 4F42]

2A i

where all view factors on the right-hand side of the eqn. Can beevaluated from Figure 13.4 (see Table 13.2) for aligned parallelrectangles.



PROBLEM 13.10

The reciprocity relation, the summation rule and Equations 13.5to 13.7 can be used to develop view factor relations that allow forapplications of Figure 13.4 and/or Figure 13.6 to more complexconfigurations. Consider the view factor F14 for surfaces 1 and 4 ofthe following geometry. These surfaces are perpendicular but do

not share a common edge.

a) Obtain the following expression for the view factor F14 :

F14 = 1 [(A 1+A 2)F(12)(34) + A 2F23 – (A 1+A 2)F(12)3 – A 2F(2)(34)]

b) If L1 = L2 =L4 = w/2 and L3 = w , what is the value of F14



PROBLEM 13.11

Determine the shape factor, F12 for the rectangles shown

a) Perpendicular rectangles without a common edge

b) Parallel rectangles of unequal areas

13

2 4

6m

6m

6m3m

(x)

(z)

(y)

Figure 13.6



PROBLEM 13.30

Two plane coaxial disks are separated by a distance L = 0.20m. The

lower disks (A 1) is solid with a diameter Do = 0.80m and a temp. T1 = 300K. The upper disk (A 2) at temp. T2 = 1000K has the same outerdiameter but is ring shaped withan inner diameter Di = 0.40m.

Assuming the disks to be blackbody, calculate the net radiative heatexchange between them.

Chapter 13 Heat Transfer

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