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```ME 452 - Heat Transfer
Lesson Outline
• View Factor
–
–
–
–
Reciprocity Relation
Summation Rule
Superposition Rule
Symmetry Rule
• Direct Method for
Problems
complex. Most of our analyses
will be greatly simplified.
– Gray body correction:
– Directional correction (View
Factor):
– Shielding correction, and/or
correction for non-vacuum
in atmosphere)
Q blackbody  As Ts4  T4 
4
4

Qgraybody  As Ts  T 
Q12  F12As T14  T24 
View Factor
• Thermal radiation follows the same
rules as other electromagnetic
• Directionality is important
– i.e. radio or TV antennae
• The View Factor (F) corrects for
directionality
– Note the view factor is different
depending upon which surface is
sending/receiving
Q A1  A2
1
cos 1 cos  2
F12 
  
dA1dA2
2

Q A1
A1 A2 A1
r
Q A2  A1
1
F21 


Q A2
A2
cos  2 cos 1
A1 A2 r 2 dA2dA1
Four Basic Rules of View Factors
•
•
•
•
Reciprocity Relation
Summation Rule
Superposition Rule
Symmetry Rule
Reciprocity Relation
• If the view factor of one
surface is known, the
view factor of the other
surface can be calculated
using the Reciprocity
Relation:
A1F12  A2 F21
Summation Rule
• Because radiation is a line of
sight phenomenon, the most
common application involves
enclosed spaces or enclosures
• Summation Rule:
– The sum of the view factors
from a surface to all other
surfaces within an enclosure
must equal one
N
F
j 1
i j
1
N  number of surfaces in enclosure
F  view factors
Superposition Rule
• Superposition Rule:
– The view factor from a surface “i”
to a surface “j” is equal to the
sum of view factors from surface
“i” to the parts of surface “j”
– This is useful when complex view
factors must be calculated
• In example to right:
– Known view factors:
» 1 to 2
» 1 to 2 and 3
– Allows calculation of view factor
1 to 3
Symmetry Rule
• Symmetry Rule:
– Two or more surfaces that
have symmetry with
another surface will all
have the same view factor
from that surface
– In example to right
• Surfaces 2,3,4 and 5 will
all have the same view
factor from surface 1
Four Basic Rules of View Factors
• Reciprocity Relation
A1F12  A2 F21
• Summation Rule
N
F
j 1
i j
1
• Superposition Rule
F12,3  F12  F13
• Symmetry Rule
Solved View Factors
• Most common view
factors have been
calculated and
tabulated
• A few common 3-D
be found
Graphical View Factors
• Less accurate, but less cumbersome are the
graphical representations of view factors
2D View Factors
• Just as with 3D view factors, many common 2D
configurations have been calculated and
tabulated
Crossed Strings Method
• The Crossed Strings
Method is used to
calculate view factors for
infinitely long surfaces (2
½ D)
Fi j
F12
crossed strings    uncrossed strings 


2  string on surface i 

L5  L6   L3  L4 

2 L1
Direct Vs indirect determination of View
factors
for N-sided enclosure We need N2 view factors
1
(example N=5)
2
N  5  25
2
2
5
3
From which, we need to directly find
N ( N  1) 5(5  1)

 10
2
2
The remaining 15 , we find them indirectly
using the previous rules
4
familiar at this point
• How to incorporate the view
factor between two surfaces:
Q12  F12As T14  T24 
• An enclosure may have multiple
surfaces, therefore:
4
4


Qi  Qi j   Fi jAi Ti  T j 
N
N
j 1
j 1
Graybody
• A blackbody is an idealization of the
perfect opaque diffuse emitter and
absorber
– Opaque - nontransparent
– Diffuse – independent of direction
• A graybody corrects for the amount
– Still assumes opaque and diffuse
• Radiosity (J) – The total amount of
including the emitted and reflected
• Radiosity is the total amount of
– Reflected from other sources
– Emitted from this surface
• A blackbody would absorb all
reflect none)
– Therefore the radiosity of a
blackbody is equal to the
emissive power
• A graybody radiosity is equal to
 emitted   reflected 
J i  
  

J i   iTi 4   iGi
• Because a graybody is opaque
(the transmittance equals zero)
absorptivity  reflectivity  transmissivity  1
therefore:
     1
opaque    0      1
gray    
 J i   iTi 4  1   i Gi
Electrical Analogy (Surface
Resistance)
•
Note that Surface Resistance is different
than the electrical analogy applied before
when we used Thermal Resistance
– Thermal resistance was used to calculate
the total heat transfer rate based on a
temperature difference
– Surface resistance is used to calculate the
•
I personally do not like this method
because of the ease of confusion
– Therefore, we will use what is called the
direct method or matrix method to solve
– The author discusses this method in detail
Problems
• Procedure:
– Set up a series of linear equations
– Solve for applicable unknowns
 Surface with known

Qi  Ai  Fi j J i  J j  

j 1
heat transfer rate Qi
 Surface with known
1  i N
4
Ti  J i 
Fi j J i  J j  

 i j 1
temperature Ti
N
Specific Concern – Radiation Effect on
Temperature Measurement
• Thermocouples are designed to read
via convection
• If there is a significant radiation
must be corrected:
T f  Tth 
 th Tth4  Tw4 
h
T f  actual fluid temperature
Tth  measured temperature value
Tw  wall temperature
h  convection heat transfer coefficien t
  emissivity of the thermocouple sensor
Medium
•
•
•
Air or vacuum can typically be ignored
With asymetric molecules such as H2O
or CO2 there can be significant
interaction
This can be very complex
– Gases require a volumetric
phenomenon (as opposed to the
surface analysis for opaque bodies)
– Gas emissivity, transmissivity and
absorptivity are all wavelength
dependent
•
We will not go into detail on this
subject
Lesson Outline
• View Factor
–
–
–
–
Reciprocity Relation
Summation Rule
Superposition Rule
Symmetry Rule
• Direct Method for
Problems
12–44 Two long parallel 16-cm-diameter cylinders are located
50 cm apart from each other. Both cylinders are black,
and are maintained at temperatures 425 K and 275 K. The
surroundings can be treated as a blackbody at 300 K. For a
1-m-long section of the cylinders, determine the rates of radiation
heat transfer between the cylinders and between the hot
cylinder and the surroundings.
F1 2
F1 2
Crossed strings   Uncrossed strings 2




2  String on surface 1
 

s2  D2  2s
2(D / 2)
2 s 2  D 2  s 2 0.52  0.16 2  0.5


 0.099
D
 (0.16)
F13  1  F12  1  0.099  0.901
A  DL / 2   (0.16 m)(1 m) / 2  0.2513 m 2
Q 12  AF12 (T1 4  T2 4 )  (0.2513 m 2 )(0.099 )(5.67 10 8 W/m 2 .C)( 425 4  275 4 )K 4  38.0 W
heat transfer between the hot
cylinder and the surroundings
per meter length
Q 13  A1 F13(T1 4  T3 4 )  (0.5027 m 2 )(0.901)(5.67 10 8 W/m 2 .C)( 425 4  300 4 )K 4  629.8 W
A1  DL   (0.16 m)(1 m)  0.5027 m 2
12–43 Consider two rectangular surfaces perpendicular to each other with a
common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and
the vertical surface is 1.2 m high. The horizontal surface has an emissivity of
0.75 and is maintained at 400 K. The vertical surface is black and is
maintained at 550 K. The back sides of the surfaces are insulated. The
surrounding surfaces are at 290 K, and can be considered to have an
emissivity of 0.85. Determine the net rate of radiation heat transfers between
the two surfaces, and between the horizontal surface and the surroundings.
L1 0.8


 0.5 

W 1.6
 F12  0.27
L2 1.2

 0.75 

W 1.6
A1  (0.8 m)(1.6 m)  1.28 m2
A2  (1.2 m)(1.6 m)  1.92 m2
A3  2 
1.2  0.8
 0.82  1.22  1.6  3.268 m2
2
A1F12  A2 F21 
(1.28 )(0.27 )  (1.92 ) F21 
 F21  0.18
F11  F12  F13  1 
 0  0.27  F13  1 
 F13  0.73
F21  F22  F23  1 
 0.18  0  F23  1 
 F23  0.82
A1 F13  A3 F31 
(1.28 )(0.73)  (3.268 ) F31 
 F31  0.29
A2 F23  A3F32 
(1.92 )(0.82 )  (3.268 ) F32 
 F32  0.48
J 1  1587 W/m2 , J 2  5188 W/m2 , J 3  811.5 W/m2
T1 4  J 1 
(5.67  10 8 W/m 2 .K 4 )( 400 K ) 4  J 1 
1 1
1
F12 ( J 1  J 2 )  F13 ( J 1  J 3 )
1  0.75
0.27 ( J 1  J 2 )  0.73( J 1  J 3 )
0.75
T2 4  J 2 
(5.67  10 8 W/m 2 .K 4 )(550 K ) 4  J 2
(5.67  10 8
1  3
F31 ( J 3  J 1 )  F32 ( J 3  J 2 )
3
1  0.85
0.29 ( J 1  J 2 )  0.48( J 1  J 3 )
W/m 2 .K 4 )( 290 K ) 4  J 3 
0.85
T3 4  J 3 
T1 4  J 1 
(5.67  10 8 W/m 2 .K 4 )( 400 K ) 4  J 1 
1 1
1
F12 ( J 1  J 2 )  F13 ( J 1  J 3 )
1  0.75
0.27 ( J 1  J 2 )  0.73( J 1  J 3 )
0.75
T2 4  J 2 
(5.67  10 8 W/m 2 .K 4 )(550 K ) 4  J 2
(5.67  10 8
1  3
F31 ( J 3  J 1 )  F32 ( J 3  J 2 )
3
1  0.85
0.29 ( J 1  J 2 )  0.48( J 1  J 3 )
W/m 2 .K 4 )( 290 K ) 4  J 3 
0.85
T3 4  J 3 
J 1  1587 W/m2 , J 2  5188 W/m2 , J 3  811.5 W/m2
Q 21  Q 12   A1 F12 ( J 1  J 2 )  (1.28 m 2 )(0.27)(1587  5188)W/m2  1245 W
Q 13  A1 F13 ( J 1  J 3 )  (1.28 m 2 )(0.73)(1587  811.5)W/m2  725 W
Q net  As ( gTg4   gTs4 )
 (1 m 2 )(5.67  10 8 W/m2  K 4 )[ 0.1325 (1200 K ) 4  0.1928 (600 K ) 4 ]
 1.42  10 4 W
HomeWork
13-24
13-37
13-41
13-43
13-44
13-32
13-41
131313-
```
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