ftfs chap22 p041

8/12/2019 FTFS Chap22 P041

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Chapter 12 Radiation Heat Transfer

22-41 A circular grill is considered. The bottom of the grill is covered with hot coal bricks, while the wiremesh on top of the grill is covered with steaks. The initial rate of radiation heat transfer from coal bricksto the steaks is to be determined for two cases.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.

Properties The emissivities are ε ! for all surfaces since theyare black or reradiating.

Analysis "e consider the coal bricks to be surface !, the steaks to be surface # and the side surfaces to be surface $. %irst wedetermine the view factor between the bricks and the steaks&Table ##'!(,

)*.+m+.#+m!*.+ ====

L

r R R i

ji

))).$+.)*

)*.+!!!

#

#

#

#

=+=+

+=i

j

R

RS

# -.+)*.+)*.+))).$))).$

#!

#!

#/!##

#/!##

!# =

−−=

−−==

i

jij R

RS S F F

&0t can also be determined from %ig. ##')(.

Then the initial rate of radiation heat transfer from the coal bricks to the stakes becomes

W1674=−⋅×=

−=− 1(2 #)&(2 !!++(3&2 "/m!+-).*1&/m($.+&(3# -.+&

(&##

#!!!#!#

π

σ T T A F Q

"hen the side opening is closed with aluminum foil, the entire heat lost by the coal bricks must be gained by the stakes since there will be no heat transfer through a reradiating surface. The grill can be considered

to be three'surface enclosure. Then the rate of heat loss from the room can be determined from

!

#$!$!#

#!!

!!−

++

−=

R R R

E E Q bb

where " / m 2 2 " / m

" / m 2 2 " / m

# #

# #

E T

E T

b

b

! !

# #

-) !+ !!++ $ +!*

-) !+ ! #)$ +)

= = × == = × + =

−

−σσ

&*. . (& ( ,

&*. . (& (

and A A! #

#+ $++)+-4= = =π& . (.

m m #

#'#

!$!#$!$

#'

#!#!!#

m#.!4(# -.+!(&m+)+-4.+&

!!

m$4.4(# -.+(&m+)+-4.+&

!!

=−

===

===

F A R R

F A R

Substituting,W3757=

+

−= −!

#'#'

#

!#

(m#.!4&#

!

m$4.4

!

"/m(+)+!*,$&Q

##'$!

Steaks, T # #) 2, ε

# !

Coal bricks, T ! !!++ 2, ε

! !

+.#+ m

8/12/2019 FTFS Chap22 P041



22-42E A room is heated by lectric resistance heaters placed on the ceiling which is maintained at auniform temperature. The rate of heat loss from the room through the floor is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered. 4 There is no heat loss through the side surfaces.

Properties The emissivities are ε ! for the ceiling and ε +. for the floor. The emissivity of insulated&or reradiating( surfaces is also !.

Analysis The room can be considered to be three'surface enclosurewith the ceiling surface !, the floor surface # and the side surfacessurface $. "e assume steady'state conditions exist. Since the sidesurfaces are reradiating, there is no heat transfer through them, andthe entire heat lost by the ceiling must be gained by the floor. Thenthe rate of heat loss from the room through its floor can bedetermined from

#

!

#$!$!#

#!!

!! R

R R R

E E Q bb

+

++

−= −

where

####

##!!

5tu/h.ft!$+(6 -+-*(&6 .5tu/h.ft!+!)!.+&

5tu/h.ft!*)(6 -+4+(&6 .5tu/h.ft!+!)!.+&

=+×=σ==+×=σ=

−

−

T E

T E

b

b

and

A A! ##!# != = =& (ft ft #

The view factor from the floor to the ceiling of the room is F !# + #)= . &%rom %igure ##'*(. The viewfactor from the ceiling or the floor to the side surfaces is determined by applying the summation rule to be

F F F F F !! !# !$ !$ !#! ! ! + #) + )$+ + = → = − = − =. .

since the ceiling is flat and thus F !! += . Then the radiation resistances which appear in the equationabove become

#'#

!$!#$!$

#'#

!#!!#

#'#

##

##

ft++4*!$.+()$.+(&ft!&

!!

ft+#*)#.+(#).+(&ft!&

!!

ft++!).+(.+(&ft!&

.+!!

====

===

=−=−=

F A R R

F A R

A R

ε ε

Substituting,

Btu/h2130=+

+−= −

#'!

#'#'

#

!#

ft++!).+(ft++4*!$.+&#

!

ft+#*)#.+

!5tu/h.ft(!$+!*)&Q

##'$#

T # 4+°%

ε# +.

Ceiling7 !# ft × !# ft

T ! 4+°%

ε! !

0nsulated sidesurfacess4 ft

8/12/2019 FTFS Chap22 P041



22-43 Two perpendicular rectangular surfaces with a common edge are maintained at specifiedtemperatures. The net rate of radiation heat transfers between the two surfaces and between the hori8ontalsurface and the surroundings are to be determined.


Properties The emissivities of the hori8ontal rectangle and the surroundings are ε +.)* and ε +. *,respectively.

Analysis "e consider the hori8ontal rectangle to be surface !, the vertical rectangle to be surface # andthe surroundings to be surface $. This system can be considered to be a three'surface enclosure. The viewfactor from surface ! to surface # is determined from

#).+

)*.+-.!

#.!

*.+-.!

,.+

!##

!

===

== F

W

L

W

L

&%ig. ##'-(

The surface areas are

##

#!

m4#.!(m-.!(&m#.!&

m#.!(m-.!(&m.+&

==

==

A

A

###$ m#-.$-.!#.!.+

#.+#.!

# =×++××= A

9ote that the surface area of the surroundings is determined assuming that surroundings forms flatsurfaces at all openings to form an enclosure. Then other view factors are determined to be

!.+(4#.!&(#).+(&#.!& #!#!#!#!#! = → = → = F F F A F A &reciprocity rule(

)$.+!#).++! !$!$!$!#!! = → =++ → =++ F F F F F &summation rule(

#.+!+!.+! #$#$#$###! = → =++ → =++ F F F F F &summation rule(

#4.+(#-.$&()$.+(&#.!& $!$!$!$!$! = → = → = F F F A F A &reciprocity rule(

.+(#-.$&(#.+(&4#.!& $#$#$#$#$# = → = → = F F F A F A &reciprocity rule("e now apply :q. 4'*#b to each surface to determine the radiosities.Surface !7

[ ]

[ ](&)$.+(&#).+)*.+

)*.+!(2 ++(&2 ."/m!+-).*&

(&(&!

$!#!!#

$!!$#!!#!

!!!

J J J J J

J J F J J F J T

−+−−+=×

−+−−+=

−

ε

ε σ

Surface #7 ##

## (2 **+(&2 ."/m!+-).*& J J T =× → = −σ

Surface $7

[ ][ ](&.+(&#4.+

*.+*.+!

(2 #4+(&2 ."/m!+-).*&

(&(&

!

$!#!$#

#$$#!$$!$

$

$$

J J J J J

J J F J J F J T

−+−−+=×−+−

−+=

−

ε

ε σ

Solving the above equations, we find#

$#

##

! "/m*.!! ,"/m*! ,"/m!* ) === J J J

Then the net rate of radiation heat transfers between the two surfaces and between the hori8ontal surfaceand the surroundings are determined to be

##'$$

W !.- m

(2) L# !.# m

L! +. m A

! (1)

A#

T $ #4+ 2

ε$ +. *

T # **+ 2

ε# !

T ! ++ 2

ε! +.)*

(3)

8/12/2019 FTFS Chap22 P041



W1245=−−=−−=−= ###!!#!!##! "/m(*!!* )(&#).+(&m#.!&(& J J F AQQ

W725=−=−= ##$!!$!!$ "/m(*.!!!* )(&)$.+(&m#.!&(& J J F AQ

##'$

8/12/2019 FTFS Chap22 P041



22-44 Two long parallel cylinders are maintained at specified temperatures. The rates of radiation heattransfer between the cylinders and between the hot cylinder and the surroundings are to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer isnot considered.

Analysis "e consider the hot cylinder to be surface !, cold cylinder to be surface #, and the surroundingsto be surface $. ;sing the crossed'strings method, the view factor between two cylinders facing each otheris determined to be

F s D s

D! #

# #

## #

# #− = −

× = + −∑ ∑Crossed strings ;ncrossed strings

String on surface ! & / (π

or +44.+(!-.+&

*.+!-.+*.+## ####

#! =π

−+

=π

−+

=− D

s D s F

The view factor between the hot cylinder and thesurroundings is

4+!.++44.+!! !#!$ =−=−= F F &summation rule(

The rate of radiation heat transfer between the

cylinders per meter length is#m#*!$.+#/m(!(&m!-.+&#/ === π π DL A

=−°×=−σ= − ###!!#!# 2 (#)*#*(&C."/m!+-).*(&+44.+(&m#*!$.+&(& T T AF Q

9ote that half of the surface area of the cylinder is used, which is the only area that faces the othercylinder. The rate of radiation heat transfer between the hot cylinder and the surroundings per meterlength of the cylinder is

#! m*+#).+m(!(&m!-.+& === π π DL A

629.8=−°×=−σ= − ##

$!!$!!$2 ($++#*(&C."/m!+-).*(&4+!.+(&m*+#).+&(& T T F AQ

##'$*

D

D(2)

(1) s

T $ $++ 2

ε$ !

T # #)* 2

ε# !

T ! #* 2 ε

! !

(3)

8/12/2019 FTFS Chap22 P041



22-45 A long semi'cylindrical duct with specified temperature on the side surface is considered. Thetemperature of the base surface for a specified heat transfer rate is to be determined.


Properties The emissivity of the side surface is ε +. .

Analysis "e consider the base surface to be surface !, the side surface to be surface #. This system is a

two'surface enclosure, and we consider a unit length of the duct. The surface areas and the view factor aredetermined as

##

#!

m*)!.!#/m(!(&m+.!&#/

m+.!(m+.!(&m+.!&

===

==

π π DL A

A

!!+! !#!#!#!! = → =+ → =+ F F F F &summation rule(

The temperature of the base surface is determined from

K684.8= → −+

−⋅×=ε

ε−+

−σ=

−!

##

!#

##

#

!#!

#!!#

(.+(&m*)!.!&

.+!

(!(&m+.!&

!12(-*+&(3"/m!+-).*&

"!#++

!!(&

T T K

A F A

T T Q

22-46 A hemisphere with specified base and dome temperatures and heat transfer rate is considered. Theemissivity of the dome is to be determined.


Properties The emissivity of the base surface is ε +.**.

Analysis "e consider the base surface to be surface !, the dome surface to be surface #. This system is atwo'surface enclosure. The surface areas and the view factor are determined as

####

###!

m+-#.+#/(m#.+&#/

m+$!.+/(m#.+&/

======

π π

π π

D A

D A

!!+! !#!#!#!! = → =+ → =+ F F F F &summation rule(

The emissivity of the dome is determined from

0.21=ε →

εε−++−

−⋅×−=

εε−++ε

ε−−σ

−=−=

−#

##

###

#

##

#

!#!!!

!

#!!##!

(m+-#.+&

!

(!(&m+$!.+&

!

(**.+(&m+$!.+&

**.+!12(-++&2(++(3&2 "/m!+-).*&

"*+

!!!(&

A F A A

T T QQ

##'$-

T ! <

ε! !

T # -*+ 2

ε# +.

D ! m

T ! ++ 2

ε! +.**

T # -++ 2

ε# <

D +.# m

8/12/2019 FTFS Chap22 P041



Radiation Shields and The Radiation Effe t

22-47C 6adiation heat transfer between two surfaces can be reduced greatly by inserting a thin, highreflectivity&low emissivity( sheet of material between the two surfaces. Such highly reflective thin plates orshells are known as radiation shields. =ultilayer radiation shields constructed of about #+ shields per cm.

thickness separated by evacuated space are commonly used in cryogenic and space applications tominimi8e heat transfer. 6adiation shields are also used in temperature measurements of fluids to reducethe error caused by the radiation effect.

22-48C The influence of radiation on heat transfer or temperature of a surface is called the radiationeffect. The radiation exchange between the sensor and the surroundings may cause the thermometer toindicate a different reading for the medium temperature. To minimi8e the radiation effect, the sensorshould be coated with a material of high reflectivity &low emissivity(.

22-49C A person who feels fine in a room at a specified temperature may feel chilly in another room atthe same temperature as a result of radiation effect if the walls of second room are at a considerably lowertemperature. %or example most people feel comfortable in a room at ## °C if the walls of the room are alsoroughly at that temperature. "hen the wall temperature drops to * °C for some reason, the interiortemperature of the room must be raised to at least #) °C to maintain the same level of comfort. Also,

people sitting near the windows of a room in winter will feel colder because of the radiation exchange between the person and the cold windows.

22-50 The rate of heat loss from a person by radiation in a large room whose walls are maintained at auniform temperature is to be determined for two cases.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are opaque, diffuse, and gray. 3 Convection heattransfer is not considered.

Properties The emissivity of the person is given to be ε! +.).

Analysis &a( 9oting that the view factor from the person tothe walls F !# != , the rate of heat loss from that person tothe walls at a large room which are at a temperature of $++2 is

W26.9=−⋅×=

−σε=− 1(2 $++&(2 $+$(3&2 "/m!+-).*(&m).!(&!(&*.+&

(&##

#!!!#!!# T T A F Q

&b( "hen the walls are at a temperature of # + 2,

W187=−⋅×=

−σε= − 1(2 # +&(2 $+$(3&2 "/m!+-).*(&m).!(&!(&*.+& (&

###!!!#!!# T T A F Q

##'$)

T #

Qrad

T ! $+°Cε! +. *

A !.) m #

6>>=

8/12/2019 FTFS Chap22 P041



22-51 A thin aluminum sheet is placed between two very large parallel plates that are maintained atuniform temperatures. The net rate of radiation heat transfer between the two plates is to be determinedfor the cases of with and without the shield.


Properties The emissivities of surfaces are given to be ε! +.*, ε# +. , and ε$ +.!*.

Analysis The net rate of radiation heat transfer with athin aluminum shield per unit area of the plates is

2 W/!18"7=

−++

−+

−⋅×=

−++

−+

−=

−

!!*.+!

!*.+!

!.+!

*.+!

1(2 -*+&(2 4++(3&2 "/m!+-).*&

!!!

!!!

(&

#

#,$!,$#!

#!shieldone,!#

ε ε ε ε

σ T T Q

The net rate of radiation heat transfer between the plates in the case of no shield is

##

#!

#!shield,!# "/m+$*,!#

!.+!

*.+!

1(2 -*+&(2 4++(3&2 "/m!+-).*&

!!!

(& =

−+

−⋅×=

−+

−=−

ε ε

σ T T Q no

Then the ratio of radiation heat transfer for the two cases becomes

,,

,

Q

Q!#

!#

! *)!# +$*

one shield

no shield

" "

= ≅ 1

6

##'$

T # -*+ 2

ε# +.

T ! 4++ 2

ε! +.*

6adiation shieldε

$ +.!*

8/12/2019 FTFS Chap22 P041


8/12/2019 FTFS Chap22 P041



0.05 0.1 0.15 0." 0."5500

1000

1500

"000

"500

3000

ε3

# 1 2 $ 1 s h i e l d

% W / !

2 &

##' +

8/12/2019 FTFS Chap22 P041



22-53 Two very large plates are maintained at uniform temperatures. The number of thin aluminumsheets that will reduce the net rate of radiation heat transfer between the two plates to one'fifth is to bedetermined.


Properties The emissivities of surfaces are given to be ε! +.#, ε# +.#, and ε$ +.!*. Analysis The net rate of radiation heat transfer between the plates in the case of no shield is

#

#

#!

#!shield,!#

"/m$)#+

!#.+

!#.+

!

1(2 ++&(2 !+++(3&2 "/m!+-).*&

!!!

(&

=

−+

−⋅×=

−

ε+

ε

−σ=

−

T T Q no

The number of sheets that need to be inserted in orderto reduce the net rate of heat transfer between the two

plates to onefifth can be determined from

3≅= →

−++

−+

−⋅×=

−ε+ε+

−ε+ε

−σ=

−4#.#

!!*.+!

!*.+!

!#.+

!#.+

!

1(2 ++&(2 !+++(3&2 "/m!+-).*& ("/m&$)#+

*!

!!!

!!!

(&

shield

shield

##

#,$!,$shield

#!

#!shields,!#

N

N

N

T T Q

22-54 %ive identical thin aluminum sheets are placed between two very large parallel plates which aremaintained at uniform temperatures. The net rate of radiation heat transfer between the two plates is to be

determined and compared with that without the shield. Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.

Properties The emissivities of surfaces are given to beε! ε# +.! and ε$ +.!.

Analysis Since the plates and the sheets have the sameemissivity value, the net rate of radiation heat transferwith * thin aluminum shield can be determined from

2 W/!183=

−+

−⋅×+=

−+

−+=+=

−

!!.+

!!.+

!1(2 *+&(2 ++(3&2 "/m!+-).*&

!*!

!!!

(&

!!

!!

#

#!

#!shieldno,!#shield*,!#

ε ε

σ T T

N Q

N Q

The net rate of radiation heat transfer without the shield is

W1098=×=+= → += "! $-(!&!

!shield*,!#shieldno,!#shieldno,!#shield*,!# Q N QQ

N Q

##' !

T # ++ 2

ε# +.#

T ! !+++ 2

ε! +.#

6adiation shieldsε$ +.!*

T # *+ 2

ε# +.!

T ! ++ 2

ε! +.!

6adiation shieldsε

$ +.!

8/12/2019 FTFS Chap22 P041



22-55

"GIVEN"N=5 "parameter to be varied"epsilon_3=0.1"epsilon_1=0.1 parameter to be varied"epsilon_ =epsilon_1T_1=$00 "[ !"T_ =+50 "[ !"si%ma=5.#&E'$ "[()m * ' *+ !, -te an' /olt mann onstant"

"2N2 4-I-"_dot_1 _s6ields=1)7N:1 8 _dot_1 _No-6ield_dot_1 _No-6ield=7si%ma87T_1*+' T_ *+ )71)epsilon_1:1)epsilon_ '1

" Q 12,shields [W/ 2!! **+# $--.)$ #)*

##+* ! $.$- !*).!) !$).*

!##.#4 !!+

!+ !++

1 Q 12,shields [W/ 2!+.! ! $.$

+.!* # #.+.# $ )

+.#* 4).-+.$ -! .)

+.$* )$ .4+. )+.

+. * !+!!+.* !!-!

+.** !$#!+.- ! 4$

+.-* !-))+.) ! )-

+.)* #+4++. #$##

+. * #*)*+.4 # *+

##' #

8/12/2019 FTFS Chap22 P041



1 " 3 + 5 # & $ 9 100

100

"00

300

+00

500

#00

'

# 1 2 $ s h i e l d s

% W / !

2 &

0.1 0." 0.3 0.+ 0.5 0.# 0.& 0.$ 0.9

0

500

1000

1500

"000

"500

3000

ε1

# 1 2 $ s h

i e l d s

% W / !

2 &

##' $

8/12/2019 FTFS Chap22 P041



22-56E A radiation shield is placed between two parallel disks which are maintained at uniformtemperatures. The net rate of radiation heat transferthrough the shields is to be determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are black. 3 Convection heat transfer is notconsidered.

Properties The emissivities of surfaces are given to beε! ε# ! and ε$ +.!*.

Analysis %rom %ig. ##') we have *#.+!$$# == F F .

Then .+*#.+!$ =−= F . The disk in themiddle is surrounded by black surfaces on both sides.Therefore, heat transfer between the top surface of themiddle disk and its black surroundings can expressedas

2 * +&3.+1(6 !#++&3&*#.+(@6 5tu/h.ft!+!)!.+&(ft+-4.)&!*.+

(1&3(1&3

$$##

#$$#$!$$!$$

−+−⋅×=−σε+−σε=

− T T

T T F AT T F AQ

Similarly, for the bottom surface of the middle disk, we have

(2 * +&3*#.+1(6 )++&3&.+(@6 5tu/h.ft!+!)!.+&(ft+-4.)&!*.+

(1&3(1&3

$$##

*$$*$$$$$

−+−⋅×=−σε+−σε=−

− T T

T T F AT T F AQ

Combining the equations above, the rate of heat transfer between the disks through the radiation shield&the middle disk( is determined to be

#t$/h866=Q and T $ 4* 2

##'

T ! !#++ 6, ε

! !

ε$ +.!*

T # )++ 6, ε

# !

! ft

! ft

T ∞ * + 2

ε$ !

8/12/2019 FTFS Chap22 P041



22-57 A radiation shield is placed between two large parallel plates which are maintained at uniformtemperatures. The emissivity of the radiation shield is to be determined if the radiation heat transfer

between the plates is reduced to !* of that without the radiation shield.


Properties The emissivities of surfaces are given to be ε! +.- and ε# +.4.

Analysis %irst, the net rate of radiation heat transfer between the two large parallel plates per unit areawithout a shield is

##

#!

#!shieldno,!# "/m))

!4.+

!-.+

!1(2 ++&(2 -*+(3&2 "/m!+-).*&

!!!

(& =−+

−⋅×=−+

−=−

ε ε

σ T T Q

The radiation heat transfer in the case of one shield is

##

shieldno,!#shieldone,!#

"/m-.)$!"/m))!*.+

!*.+

=×=×= QQ

Then the emissivity of the radiation shield becomes

−+

−+

−⋅×=

−++

−+

−=

−

!#

!4.+

!-.+

!

1(2 ++&(2 -*+(3&2 "/m!+-).*& "/m)$!.-

!!!

!!!

(&

$

##

#,$!,$#!

#!shieldone,!#

ε

ε ε ε ε

σ T T Q

which gives 0%18=$ε

##' *

T # ++ 2

ε# +.4

T ! -*+ 2

ε! +.-

6adiation shieldε

$

8/12/2019 FTFS Chap22 P041



22-58

"GIVEN"T_1=#50 "[ !"T_ =+00 "[ !"epsilon_1=0.#epsilon_ =0.9

";er ent<ed tion =$ 5 [>!, parameter to be varied"si%ma=5.#&E'$ "[()m * ' *+ !, -te an' /olt mann onstant"

"2N2 4-I-"_dot_1 _No-6ield=7si%ma87T_1*+' T_ *+ )71)epsilon_1:1)epsilon_ '1_dot_1 _1s6ield=7si% ma87T_1* +' T_ *+ )771)epsilon_1:1 )epsilon_ '

1 :71)epsilon_3:1)epsilon_3'1_dot_1 _1s6ield=71 ' ;er ent<ed tio n)10 0 8 _dot_1 _No-6ield

&er'e ted$'ti* [+!

3

+ +.4!*$* +. !

*+ +.)#** +.-$+-+ +.* **-* +. - 4)+ +.$ *)* +.$!*

+ +.# --* +.! +-

4+ +.!!)-4* +.+*)*!

+0 50 #0 &0 $0 90 1000

0."

0.+

0.#

0.$

1

(e) entRedu tion %*&

ε 3

##' -

8/12/2019 FTFS Chap22 P041



22-59 A coaxial radiation shield is placed between two coaxial cylinders which are maintained at uniformtemperatures. The net rate of radiation heat transfer between the two cylinders is to be determined andcompared with that without the shield.


Properties The emissivities of surfaces are given to be ε! +.), ε# +. . and ε$ +.#.

Analysis The surface areas of the cylinders and the shield per unit length are

#$$shield

###outer pipe,

#!!inner pipe,

m4 #.+(m!(&m$.+&

m$!.+(m!(&m!.+&

m-#.+(m!(&m#.+&

============

π π

π π

π π

L D A A

L D A A

L D A A

The net rate of radiation heat transfer between the two cylinders with a shield per unit length is

W703=

−++−++− −⋅×=

−++−+−++−−=

−

(.+(&4 #.+&.+!

(!(&-#.+&!

(#.+(&-#.+&#.+!

#(!(&$!.+&

!().+(&$!.+&

).+!1(2 *++&(2 )*+(3&2 "/m!+-).*&

!!!!!!

(&

#

##

#

#,$$#,$$

#,$

!,$$

!,$

!$!!!

!

#!shieldone,!#

ε ε

ε

ε

ε

ε

ε ε

σ

A F A A A F A A

T T Q

0f there was no shield,

W746"=

−+

−⋅×=

−+

−=

−

$.+

!.+

.+

.+!

).+

!

1(2 *++&(2 )*+(3&2 "/m!+-).*&

!!

(&

#

#

!

#

#

!

#!shieldno,!#

D

D

T T Q

ε

ε

ε

σ

Then their ratio becomes

,

,

Q

Q!#

!#

)+$) -*

one shield

no shield

" "

= = 0.094

##' )

D# +.$ m

T # *++ 2

ε# +.

D! +.! m

T ! )*+ 2

ε! +.)

6adiation shield D

$ +.# m

ε$ +.#

8/12/2019 FTFS Chap22 P041



22-60

"GIVEN"?_1=0.10 "[m!"?_ =0.30 "[m!, parameter to be varied"?_3=0. 0 "[m!"epsilon_1=0.&

epsilon_ =0.+epsilon_3=0. "parameter to be varied"T_1=&50 "[ !"T_ =500 "[ !"si%ma=5.#&E'$ "[()m * ' *+ !, -te an' /olt mann onstant"

"2N2 4-I-"= 1 "[m!, a nit len%t6 o t6e @linders is onsidered"

2_1=pi8 ?_18 2_ =pi8 ?_ 8 2_3=pi8 ?_38A_13=1A_3 =1

_dot_1 _1s6ield=7s i%m a87 T_1* +' T_ *+ )771' epsilon_1 )72_18epsilon_1 : 1)72_18A_13 :71' epsilon_3 )72_38epsilon_3 :7 1' epsilon_3 )72_38epsilon_3 : 1)72_38A_3 :71' epsilon_ )72_ 8epsilon_

2 [ ! Q 12,1 shield [W!+.#* -4#.

+.#)* -4 .-+.$ )+$.*

+.$#* )+).+.$* )!!.

+.$)* )! .)+. )!).*

+. #* )#+

+. * )##.$+. )* )# .$+.* )#-.!

3 Q 12,1 shield [W!+.+* #!!.!+.+) # ).+.+4 $-+.)+.!! #4.4+.!$ 4*.4+.!* ** .)+.!) -! .-+.!4 -)*.4+.#! )$+.-+.#$ ) $+.#* $$.!+.#) !.#+.#4 4#).+.$! 4)!.)+.$$ !+!+.$* !+**

##'

8/12/2019 FTFS Chap22 P041



0."5 0.3 0.35 0.+ 0.+5 0.5#90

#95

&00

&05

&10

&15

&"0

&"5

&30

+ 2 %!&

# 1 2 $ 1 s h i e l d

% W &

0.05 0.1 0.15 0." 0."5 0.3 0.35"00

300

+00

500

#00

&00

$00

900

1000

1100

ε3

# 1 2 $ 1 s h i e l d

% W &

##' 4

8/12/2019 FTFS Chap22 P041



Re,ie- ()o le!s

22-61 The temperature of air in a duct is measured by athermocouple. The radiation effect on the temperaturemeasurement is to be quantified, and the actual airtemperature is to be determined.

Assumptions The surfaces are opaque, diffuse, and gray.

Properties The emissivity of thermocouple is given to beε +.-.

Analysis The actual temperature of the air can bedetermined from

K1111=°⋅

−⋅×+=

−σε+=−

C"/m-+

1(2 *++&(2 *+(3&2 "/m!+-).*(&-.+&2 *+

(&

#

#h

T T T T wthth

th f

22-62 The temperature of hot gases in a duct is measured by a thermocouple. The actual temperature ofthe gas is to be determined, and compared with that without a radiation shield.

Assumptions The surfaces are opaque, diffuse, and gray.

Properties The emissivity of the thermocouple is given to be ε +.).

Analysis Assuming the area of the shield to be very close to the sensorof the thermometer, the radiation heat transfer from the sensor isdetermined from

#

#

#!

#!sensor fromrad,

"/m4.#*)

!!*.+!

#!).+

!

1(2 $ +&(2 *$+(3&2 "/m!+-).*&

!!

#!!

(&

=

−+

−

−⋅×=

−ε

+

−ε

−σ=

−

T T Q

Then the actual temperature of the gas can be determined from a heat transfer balance to be

K"32= → =−°⋅=−=

f f

th f

T T

T T h

qq

##

#

sensor fromconv,sensor toconv,

"/m4.#*)(*$+C&"/m!#+

"/m4.#*)(&

"ithout the shield the temperature of the gas would be

K"49.2=°⋅

−⋅×+=

−σε+=−

C"/m!#+

1(2 $ +&(2 *$+(3&2 "/m!+-).*(&).+&2 *$+

(&

#

#h

T T T T wthth

th f

##'*+

Air, T f

T w *++ 2

ThermocoupleT

th *+ 2

ε +.-

Air, T f

T w $ + 2

ThermocoupleT

th *$+ 2

ε! +.)

ε# +.!*

8/12/2019 FTFS Chap22 P041



22-63E A sealed electronic box is placed in a vacuum chamber. The highest temperature at which thesurrounding surfaces must be kept if this box is cooled by radiation alone is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered. 4 Beat transfer from the bottom surface of the box isnegligible.

Properties The emissivity of the outer surface of the box is ε +.4*.

Analysis The total surface area is#ft-).$(!!&(!#/!& =×+××= s A

Then the temperature of the surrounding surfaces is determined to be

43 °== → −⋅×=×

−=−

6 *+$

1(6 *4+(3&6 5tu/h.ft!+!)!.+(&m-).$(&4*.+&5tu/h(!#!.$!++&

(&##

surr

surr

surr s srad

T

T

T T AQ σ ε

22-64 A double'walled spherical tank is used to store iced water. The air space between the two walls isevacuated. The rate of heat transfer to the iced water and the amount of ice that melts a # 'h period are to

be determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are opaque, diffuse, and gray.

Properties The emissivities of both surfaces are given to be ε! ε# +.!*.

Analysis &a( Assuming the conduction resistance s of the wallsto be negligible, the rate of heat transfer to the iced water inthe tank is determined to be

m m # A D! ! # ## +! !# -4= = =π π& . ( .

W107.4=

−+

+−+⋅×=

εε−+ε

−σ=

−

#

##

#

#

!

#

#

!

!#!!#

+.#+!.#

!*.+!*.+!

!*.+!

1(2 #)$+&(2 #)$#+(3&2 "/m!+-).*(&m-4.!#&

!!

(&

D

D

T T AQ

&b( The amount of heat transfer during a # 'hour period iskC4#)*(s$-++#(&kC/s!+).+& =×=∆= t QQ

The amount of ice that melts during this period then becomes

kg27.8=== → =k /kg).$$$

k4#)*

if if h

QmmhQ

##'*!

D# #.+ m

T # #+°C

ε# +.!*

D! #.+! m

T ! +°C

ε! +.!*

Dacuum

'ed

.ater0 C

T surr

in!++ "ε +.4*

T s !$+ °%

!# in!# in

8/12/2019 FTFS Chap22 P041



22-65 Two concentric spheres which are maintained at uniform temperatures are separated by air at ! atm pressure. The rate of heat transfer between the two spheres by natural convection and radiation is to bedetermined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is anideal gas with constant properties.

Properties The emissivities of the surfaces are given to beε! ε# +.*. The properties of air at ! atm and the averagetemperature of & T ! ET #(/# &$*+E#)*(/# $!#.* 2 $4.* °C are &Table A'!*(

!'

#*

2 ++$#.+2 *.$!#

!)#*-.+Fr

/sm!+-4).!

C"/m.+#-*.+

==

=

×=

°=

−

β

υ

k

Analysis &a( 9oting that D i D ! and D o D # , the characteristic length is

m+.+*m(+.!*m#*.+&#!

(&#! =−=−= io! D D L

Then

##*

$'!#

#

$#! !*.)()#*-.+&

(/sm!+-4).!&

(m+*.+(&2 #)*$*+(&2 ++$#++.+(&m/s!.4&Fr

(& ×=×

−=−= −υ β ! LT T "

Ra

The effective thermal conductivity is

[ ] [ ]++*4++.+

m(#*.+&m(!*.+&m(#*.+&m(!*.+&

m+*.+

(&(& *)/*')/*'**/)*/)sph =+

=+

= −−oioi

!

D D D D

L F

[ ] "/m.!$!*(!+!*.)(&++*4+.+&)#*-.+-!.+

)#*-.+C("/m.+#-*.+&).+

(&Fr -!.+

Fr ).+

/!*/!

/!/!

eff

°=×

+°=

+= Ra F k k s#h

Then the rate of heat transfer between the spheres becomes

W23.3=−°=−

= 2 (#)*$*+&

(m+*.+&(m#*.+(&m!*.+&

(C"/m.!$!*.+&(&eff π π oi!

oi T T L

D Dk Q

&b( The rate of heat transfer by radiation is determined from

W32.3=

−+

−⋅×=

εε−+ε

−σ==π=π=

−

#

##

#

#

!

#

#

!

!#!!#

###!!

#*.+!*.+

4.+4.+!

4.+!

1(2 #)*&(2 $*+(3&2 "/m!+-).*(&m+)+).+&

!!

(&m+)+).+(m!*.+&

D D

T T AQ

D A

##'*#

D# #* cm

T # #)* 2 ε

# +.*

D! !* cm

T ! $*+ 2

ε! +.4

L! * c m

A06 ! atm

8/12/2019 FTFS Chap22 P041



22-66 A solar collector is considered. The absorber plate and the glass cover are maintained at uniformtemperatures, and are separated by air. The rate of heat loss from the absorber plate by natural convectionand radiation is to be determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are opaque, diffuse, and gray. 3 Air is an ideal gaswith constant properties.

Properties The emissivities of surfaces are given to be ε! +.4for glass and ε# +. for the absorber plate. The properties ofair at ! atm and the average temperature of & T ! ET #(/# & +E$#(/# *-°C are &Table A'!*(

!'

#*

2 ++$+ +.+2 (#)$*-&

!!)#!#.+Fr

/sm!+*).!

C"/m.+#))4.+

=+

==

=

×=

°=

−

f T

k

β

υ

Analysis %or θ = °+ , we have hori8ontal rectangular enclosure.The characteristic length in this case is the distance between the

two glasses L! L +.+$ m Then,

##*

$'!#

#

$#! !++ $.()#!#.+&

(/sm!+*).!&

(m+$.+(&2 $#+(&2 ++$+.+(&m/s!.4&Fr

(& ×=×

−=−= −υ β LT T "

Ra

#m*.m($&m(*.!& ==×= W $ A s

[ ] [

) ).$!

(#+cos&(!++ $.&

(#+cos&(!++ $.&

(#+.!sin&!)+!

(#+cos&(!++ $.&

!)+!.!!

!!

(cos6a&cos6a

(.!&sin!)+!

cos6a!)+

!.!! 9u

!-.!

$/!-.!

=

×+×

×−×

−+=

−+−−+=

+

++ θ θ θ

θ

W7"0=°−°=−=m+$.+

C($#+&(m*.(&) ).$(&C"/m.+#))4.+& ##!

LT T

kNuAQ s

9eglecting the end effects, the rate of heat transfer by radiation is determined from

W1289=−+

+−+⋅×=−+

−=−

!4.+

!.+!

1(2 #)$$#&(2 #)$+(3&2 "/m!+-).*(&m*.&

!!!

(& ##

#!

#!rad

ε ε

σ T T AQ s

Discussion The rates of heat loss by natural convection for the hori8ontal and vertical cases would be asfollows &9ote that the 6a number remains the same(7

$ori%onta& 7

!#.$!!

(!++ $.&

!++ $.

!)+!.!!!

!6a

6a!)+

!.!! 9u$/!$/!

=−×+×

−+=−+−+=++++

##'*$

Solar

radiation

θ #+° 0nsulation

Absorber plateT

! +°C

ε! +.

Glass cover,T

# $# °C

ε# +.4

!.* m

L $ cm

8/12/2019 FTFS Chap22 P041



W1017=°−°=−=m+$.+

C($#+&(m-(&!#.$(&C"/m.+#))4.+& ##!

LT T

kNuAQ s

'(rti!a& 7

++!.#m+$.+

m#()#!#.+&(!++ $.&#.+Fr #.+

$.++!#.+/!

$.++!#.+/! =

×=

=−−

L $

Ra Nu

W"34=°−°=−=m+$.+

C($#+&(m-(&++!.#(&C"/m.+#))4.+& ##!

LT T

kNuAQ s

##'*

8/12/2019 FTFS Chap22 P041



22-67E The circulating pump of a solar collector that consists of a hori8ontal tube and its glass coverfails. The equilibrium temperature of the tube is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Thetube and its cover are isothermal. 3 Air is an ideal gas. 4The surfaces are opaque, diffuse, and gray for infraredradiation. 5 The glass cover is transparent to solarradiation.

Properties The properties of air should be evaluated at theaverage temperature. 5ut we do not know the exittemperature of the air in the duct, and thus we cannotdetermine the bulk fluid and glass cover temperatures atthis point, and thus we cannot evaluate the averagetemperatures. Therefore, we will assume the glasstemperature to be * °%, and use properties at ananticipated average temperature of &)*E *(/# +H% &TableA'!*:(,

s

k

/ft!+!.-4)/hft -!!+.+

%ft5tu/h +! !.+#'# ×==

°⋅⋅=

υ

6 * +!!

)#4+.+Fr

ave==

=

T β

Analysis "e have a hori8ontal cylindrical enclosure filled with air at +.* atm pressure. The probleminvolves heat transfer from the aluminum tube to the glass cover and from the outer surface of the glasscover to the surrounding ambient air. "hen steady operation is reached, these two heat transfer rates mustequal the rate of heat gain. That is,

5tu/h$+gainsolarambient'glassglass'tube === QQQ &per foot of tube(

The heat transfer surface area of the glass cover is

#ft$+4.!ft(!(&ft!#/*&(& ==== π π W D A A o "&asso &per foot of tube(

To determine the 6ayleigh number, we need to know the surface temperature of the glass, which is notavailable. Therefore, solution will require a trial'and'error approach. Assuming the glass covertemperature to be *H%, the 6ayleigh number, the 9usselt number, the convection heat transfercoefficient, and the rate of natural convection heat transfer from the glass cover to the ambient air aredetermined to be

-

##

$#

#

$

!++4#.!()#4+.+&(/sft!+-)*.!&

(ft!#/*(&6 )**1&6(* +/&!(3ft/s#.$#&

Fr (&

6a

×=×

−=

−=

−

∞ν

β oo Do

DT T "

( )[ ] ( )[ ]4*.!

)#4+.+/**4.+!

(!++4#.!&$ ).+-.+

Fr /**4.+!

6a$ ).+-.+ 9u

#

#)/!-/4

-/!-#

#)/!-/4

!/-I

=

+×+=

++=

%ft5tu/h *$!*.+(4*.!&ft!#/*

%ft5tu/h+! !.+ 9u #

+°⋅⋅=°⋅⋅==

Dk

h o

5tu/h4-.-%()**(&ft$+4.!(&%ft5tu/h*$!*.+&(& ##

conv, =°−°⋅⋅=−= ∞T T AhQ oooo

Also,

##'**

D# * in

Aluminum tube D

! #.* in, T

!

ε! +.4

Air space+.* atm

Flastic cover,ε

# +.4, T

#

"ater

T ∞ )* °%

T sky

-+ °%

$+ 5tu/h.ft

8/12/2019 FTFS Chap22 P041



[ ]5tu/h*.$+

6(*$*&6(* *&(ft$+4.!(&6 ft5tu/h!+!)!.+(&4.+&

(&##

skyrad,

=−⋅⋅×=

−=−

T T AQ oooo σ ε

Then the total rate of heat loss from the glass cover becomes

5tu/h *.$)*.$++.)rad,conv,total, =+=+= ooo QQQ

which is more than $+ 5tu/h. Therefore, the assumed temperature of *H% for the glass cover is high.6epeating the calculations with lower temperatures &including the evaluation of properties(, the glasscover temperature corresponding to $+ 5tu/h is determined to be 81%5 .

The temperature of the aluminum tube is determined in a similar manner using the naturalconvection and radiation relations for two hori8ontal concentric cylinders. The characteristic length inthis case is the distance between the two cylinders, which is

ft!.#*/!#in#*.!#/(*.#*&#/(& ==−=−= io! D D LAlso,

#ft -* *.+ft(!(&ft!#/*.#&(& ==== π π W D A A itub(i &per foot of tube(

"e start the calculations by assuming the tube temperature to be !! .*H%, and thus an averagetemperature of & !.*E!! .*(/# !++ °% - + 6. ;sing properties at !++ °%,

##

$#

#

$

$$.!()#-.+&(/sft!++4.!&

(ft!#/#*.!(&6 *.!*.!!1&6(- +/&!(3ft/s#.$#&Fr

(&6a =

×−=−= −ν

β LT T " oi L


! --.+1ft(!#/*&ft(!#/*.#3&ft(&!.#*/!#

(1*.#/*3ln&

(&

(1/3ln&*$/*'$/*'$**/$*/$$cyc =

+=

+= −−

oi!

io

D D L

D D F

%ft5tu/h+$##).+

(!+$$.!! --.+&+.)#-+. -!

+.)#-%(ft5tu/h+!*#4.+&$ -.+

(6a&Fr -!.+

Fr $ -.+

/!

/!cyc

/!

eff

°⋅⋅=

××

+°⋅⋅=

+= L F k k

Then the rate of heat transfer between the cylinders by convection becomes

5tu/h.!+%(*.!*.!!&ln&*/#.*(

%(ft5tu/h+$##).+&#(&

(/ln&

# eff conv, =°−°⋅⋅=−=

π π

oiio

i T T D D

k Q

Also,

[ ]#

in*in*.#

4.+4.+!

4.+!

6(*.* !&6( *.*)&(ft-* *.+(&6 ft5tu/h!+!)!.+&

!!

(& ##

i

orad, =

−+

−⋅⋅×=

−+

−=−

o

i

o

o

iii

D

D

T T AQ

ε ε

ε

σ

Then the total rate of heat loss from the glass cover becomes

5tu/h.$*+.#*.!+rad,conv,total, =+=+= iii QQQ

which is more than $+ 5tu/h. Therefore, the assumed temperature of !! .*H% for the tube is high. 5ytrying other values, the tube temperature corresponding to $+ 5tu/h is determined to be 113%2 .Therefore, the tube will reach an equilibrium temperature of !!$.#H% when the pump fails.

##'*-

8/12/2019 FTFS Chap22 P041



22-68 A double'pane window consists of two sheets of glass separated by an air space. The rates of heattransfer through the window by natural convection and radiation are to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces areopaque, diffuse, and gray. 3 Air is an ideal gas with constant specificheats. 4 Beat transfer through the window is one'dimensional and theedge effects are negligible.

Properties The emissivities of glass surfaces are given to be ε! ε# +.4. The properties of air at +.$ atm and the average temperature of&T ! ET #(/# &!*E*(/# !+ °C are &Table A'!*(

!'

#**!

2 ++$*$.+2 (#)$!+&

!

)$$-.+Fr

/sm!+)*$./+.$!+#-.!$.+/

C"/m.+# $4.+

=+=

=×=×==

°=−−

β

υ υ atm

k

Analysis The characteristic length in this case is the distance between the glasses, m+*.+== L L!

##*$'!#

#$#! !+4!.!()$$-.+&

(/sm!+)*$.&(m+*.+&2 (*!*(&2 ++$*$.+(&m/s!.4&Fr (& ×=

×−=−= −υ

β LT T " Ra

*$4.!+*.+#

(!+4!.!&!4).+!4).+4/!

/!4/!

/! =

×=

=

−−

L $

Ra Nu

#m-(m$(&m#& == s A

Then the rate of heat transfer by natural convection becomes

W4".0=°−°=−=m+*.+

C(*!*&(m-(&*$4.!(&C"/m.+# $4.+& ##!

LT T

kNuAQ s!on)

The rate of heat transfer by radiation is determined from

( ) ( )

W2"2=

−++−+⋅×=

−+−=

−

!4.+

!4.+

!12 #)$*2 #)$!*(32 "/m!+-).*(&m-&

!!!

(&

###!

#!rad

ε ε

σ T T AQ s

Then the rate of total heat transfer becomes W297=+=+= #*#*rad !on)tota& QQQ

Discussion 9ote that heat transfer through the window is mostly by radiation.

##'*)

*°C!* °C L * cm

$ # m

QAir

8/12/2019 FTFS Chap22 P041



22-69 A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate ofheat loss from the water in the hose by natural convection and radiation is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is anideal gas with constant specific heats.

Properties The emissivities of surfaces are given to be ε! ε# +.4. The properties of air are at ! atmand the film temperature of & T sET ∞(/# & +E#*(/# $#.* °C are &Table A'!*(

!'

#*

2 ++$#)$.+2 (#)$*.$#&

!)#)*.+Fr

/sm!+-$#.!C"/m.+#-+).+

=+

=

=

×=

°=−

β

υ k

Analysis ;nder steady conditions, the heat transfer ratefrom the water in the hose equals to the rate of heat lossfrom the clear plastic tube to the surroundings by naturalconvection and radiation. The characteristic length in thiscase is the diameter of the plastic tube,

m+-.+# === D D L #&asti!! .

##*

$'!#

#

$# .#()#)*.+&

(/sm!+-$#.!&

(m+-.+&2 (#*+(&2 ++$#)$.+(&m/s!.4&Fr

(& =×

−=−= −∞

υ β DT T "

Ra s

( )[ ] ( )[ ]$+.!+

)# !.+/**4.+!

(!+#.#&$ ).+-.+

Fr /**4.+!

6a$ ).+-.+ 9u

#

#)/!-/4

-/!*#

#)/!-/4

!/-I =

+×+=

++=

#

##

#

#

m! *.+m(m(&!+-.+&

C."/m)*.($+.!+&m+-.+

C"/m.+#-+).+

====

°=°==

π π L D A A

Nu D

k h

#&asti!

Then the rate of heat transfer from the outer surface by natural convection becomes

W12.7=°−°=−= ∞ C(#*+(&m! *.+(&C."/m)*.&(& ### T T hAQ s!on)

The rate of heat transfer by radiation from the outer surface is determined from

W26.2=+−+⋅×=

−=− 12(#)$!*&(2 #)$+(3&2 "/m!+-).*(&m! *.+(&4+.+&

(&##

#rad sk* s T T AQ σ ε

%inally,

"4.$#.#-).!#, =+=&osstota& Q

Discussion 9ote that heat transfer is mostly by radiation.

##'*

D# - cm

Garden hose D

! # cm, T

!

ε! +.4

Air space

Flastic cover,ε

# +.4, T

# +°C

"ater

T ∞ #*°C

T sky

!* °C

8/12/2019 FTFS Chap22 P041



22-70 A solar collector consists of a hori8ontal copper tube enclosed in a concentric thin glass tube. Theannular space between the copper and the glass tubes is filled with air at ! atm. The rate of heat loss fromthe collector by natural convection and radiation is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is anideal gas with constant specific heats.

Properties The emissivities of surfaces are given to be ε! +. * for the tube surface and ε# +.4 for glass

cover. The properties of air at ! atm and the average temperature of & T ! ET #(/# &-+E +(/# *+°C are&Table A'!*(

!'

#*

2 ++$+4-.+2 (#)$*+&

!)##.+Fr

/sm!+)4.!

C"/m.+#)$*.+

=+

=

=

×=

°=

−

β

υ

k

Analysis The characteristic length in this case is

m+.+#m(+.+*'m+4.+&#!

(&#!

!# =−= D D L!

*+,!+()##.+&(/sm!+)4.!&

(m+#.+&2 (+-+(&2 ++$+4-.+(&m/s!.4&Fr

(&##*

$'!#

#

$#! =

×−=−= −υ

β ! LT T " Ra


[ ] [ ][ ]

!$+$.+m(+*.+&m(+4.+&m(+#.+&

(+*.+/+4.+ln&

(&

(/ln&*$/*'$/*'$**/$*/$$cyl =

+=

+= −−

oi!

io

D D L

D D F

[ ] "/m.+*$#!.+(*+,!+(&!$+$.+&)##.+-!.+

)##.+(C"/m.+#)$*.+&$ -.+

(6a&Fr -!.+

Fr $ -.+

/!/!

/!cyl

/!

eff

°=

+°=

+= F k k

Then the rate of heat transfer between the cylinders becomes

W11.4=°−°=−= C(+-+&(+*.+/+4.+ln&

(C"/m.+*$#!.+&#(&

(/ln&

# eff conv

π π

oiio

T T D D

k Q

&:q. !(

The rate of heat transfer by radiation is determined from

W13.4=

−+

+−+⋅×=

−

+

−=

−

4*

4.+4.+!

*.+!

1(2 #)$+&(2 #)$-+(3&2 "/m!+-).*(&m!*)!.+&

!!

(&

##

#!

##

!

#!!rad

D

D

T T AQ

ε

ε

ε

σ

%inally,

##'*4

D# 4 cm

Copper tube D

! * cm, T

! -+ °C

ε! +. *

Air space

Flastic cover,T

# +°C

ε# +.4

"ater

8/12/2019 FTFS Chap22 P041



".#.!$.!!, =+=&osstota& Q &per m length(

##'-+

8/12/2019 FTFS Chap22 P041



22-71 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transferrate at the bottom surface is considered. The temperature of the side surface and the net rates of heattransfer between the top and the bottom surfaces, and between the bottom and the side surfaces are to bedetermined.


Properties The emissivities of the top, bottom, and side surfaces are +.)+, +.*+, and +. +, respectively. Analysis "e consider the top surface to be surface !, the bottom

surface to be surface #, and the side surface to be surface $. Thissystem is a three'surface enclosure. The view factor from surface ! tosurface # is determined from

!).+

*.+#.!

-.+

#-.+

#.!

!# ===

== F

L

r

r

L

&%ig. ##')(

The surface areas are

#$

####!

m*#.(m#.!(&m#.!&

m!$!.!/(m#.!&/

===

====

π π

π π

DL A

D A A

Then other view factors are determined to be!).+#!!# == F F

$.+!!).++! !$!$!$!#!! = → =++ → =++ F F F F F &summation rule(, $.+!$#$ == F F

#!.+(*#.&($.+(&!$!.!& $!$!$!$!$! = → = → = F F F A F A &reciprocity rule(,

#!.+$!$# == F F "e now apply :q. ##'$* to each surfaceSurface !7

[ ]

[ ](&$.+(&!).+)+.+

)+.+!(2 *++(&2 ."/m!+-).*&

(&(&!

$!#!!#

$!!$#!!#!

!!!

J J J J J

J J F J J F J T

−+−−+=×

−+−−+=

−

ε

ε σ

Surface #7

[ ]

[ ](&$.+(&!).+*+.+

*+.+!(2 *++(&2 ."/m!+-).*&

(&(&!

$#!###

$##$!##!#

###

J J J J J

J J F J J F J T

−+−−+=×

−+−−+=

−

ε

ε σ

Surface $7

[ ]

[ ](&#!.+(&#!.++.+

+.+!(2 ."/m!+-).*&

(&(&!

$!#!$$#

#$$#!$$!$

$$$

J J J J J T

J J F J J F J T

−+−−+=×

−+−−+=

−

ε

ε σ

"e now apply :q. ##'$ to surface #

[ ] [ ](&$.+(&!).+(m!$!.!&(&(& $#!##

$##$!##!## J J J J J J F J J F AQ −+−=−+−=Solving the above four equations, we find

#$

##

#!$ "/m!4$ ,"/m$ ,"/m4) , ==== J J J T 631

The rate of heat transfer between the bottom and the top surface is W751%6=−=−= ##

!##!##! "/m(4)$(&!).+(&m!$!.!&(& J J F AQ

##'-!

T ! *++ 2

ε! +.)+

r ! +.- m

T $ <

ε$ +. +

h !.# m

T # -*+ 2

ε# +.*+

r # +.- m

8/12/2019 FTFS Chap22 P041



The rate of heat transfer between the bottom and the side surface is W644%0=−=−= ##

$##$##$ "/m(!4)$(&$.+(&m!$!.!&(& J J F AQ Discussion The sum of these two heat transfer rates are )*!.- E - !$4*.- ", which is practicallyequal to ! ++ " heat supply rate from surface #. This must be satisfied to maintain the surfaces at thespecified temperatures under steady operation. 9ote that the difference is due to round'off error.

##'-#

8/12/2019 FTFS Chap22 P041



22-72 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transferrate at the bottom surface is considered. The emissivity of the top surface and the net rates of heat transfer

between the top and the bottom surfaces, and between the bottom and the side surfaces are to bedetermined.


Properties The emissivity of the bottom surface is +.4+.

Analysis "e consider the top surface to be surface !, the base surface to be surface #, and the side surface to be surface $. This system is a three'surface enclosure. The view factor from the base to the top surface of thecube is F !# + #= . . The view factor from the base or the top to the sidesurfaces is determined by applying the summation rule to be

F F F F F !! !# !$ !$ !#! ! ! + # ++ + = → = − = − =. .

since the base surface is flat and thus F !! += . >ther view factors are

#+.+ ,+.+ ,#+.+ $#$!!$#$!##! ====== F F F F F F

"e now apply :q. 4'$* to each surface

Surface !7

[ ]

[ ](&+.+(&#+.+!

(2 )++(&2 ."/m!+-).*&

(&(&!

$!#!!

!!

#

$!!$#!!#!

!!!

J J J J J

J J F J J F J T

−+−−+=×

−+−−+=

−ε

ε

ε

ε σ

Surface #7

[ ]

[ ](&+.+(&#+.+

4+.+

4+.+!(2 4*+(&2 ."/m!+-).*&

(&(&!

$#!###

$##$!##!#

###

J J J J J

J J F J J F J T

−+−−+=×

−+−−+=

−

ε

ε σ

Surface $7$

#

$$

(2 *+(&2 ."/m!+-).*&

J

J T

=×=

−

σ

"e now apply :q. 4'$ to surface #

[ ] [ ](&+.+(&#+.+(m4&(&(& $#!##

$##$!##!## J J J J J J F J J F AQ −+−=−+−=Solving the above four equations, we find

#$

##

#!! "/m#$#* ,"/m4 *,! ,"/m)$-,!! , ==== J J J 0%44ε

The rate of heat transfer between the bottom and the top surface is##

#!m4(m$&

=== A A

W54%4=−=−= ##!##!##! "/m()$-,!!4 *,!(&#+.+(&m4&(& J J F AQ

The rate of heat transfer between the bottom and the side surface is##

!$ m$-(m4& === A A

W 285%6=−=−= ##$##$##$ "/m(#$#*4 *,!(&.+(&m4&(& J J F AQ

Discussion The sum of these two heat transfer rates are * . E # *.- $ + k", which is equal to $ + k"heat supply rate from surface #.

##'-$

T ! )++ 2

ε! <

T # 4*+ 2

ε# +.4+

T $ *+ 2

ε$ !

$ m

8/12/2019 FTFS Chap22 P041



22-73 A thin aluminum sheet is placed between two very large parallel plates that are maintained atuniform temperatures. The net rate of radiation heat transfer between the plates and the temperature of theradiation shield are to be determined.


Properties The emissivities of surfaces are given to be ε ! +. , ε # +.4, and ε $ +.!#.

Analysis The net rate of radiation heat transfer with athin aluminum shield per unit area of the plates is

2 W/748%9=

−++

−+

−⋅×=

−ε+ε+

−ε+ε

−σ=

−

!!#.+!

!#.+!

!4.+

!.+!

1(2 **+&(2 )*+(3&2 "/m!+-).*&

!!!

!!!

(&

#

#,$!,$#!

#!shieldone,!#

T T Q

The equilibrium temperature of the radiation shield is determined from

671%3= →

−+

−⋅×=

−ε+ε

−σ=

−$

$#

#

$!

$!!$

!!#.+!

.+!

1(2 )*+(3&2 "/m!+-).*& "/m4.)

!!!

(&

T T

T T Q

##'-

T # **+ 2

ε# +.4

T ! )*+ 2

ε! +.

6adiation shieldε

$ +.!#

8/12/2019 FTFS Chap22 P041



22-74 Two thin radiation shields are placed between two large parallel plates that are maintained atuniform temperatures. The net rate of radiation heat transfer between the plates with and without theshields, and the temperatures of radiation shields are to be determined.


Properties The emissivities of surfaces are given to be ε ! +.-, ε # +.), ε $ +.!+, and ε +.!*.

Analysis The net rate of radiation heat transfer withoutthe shields per unit area of the plates is

2 W/3288=

−+

−⋅×=

−ε

+ε

−σ=

−

!).+

!-.+

!1(2 $++&(2 -++(3&2 "/m!+-).*&

!

!!(&

#

#!

#!shieldno!#,

T T Q

The net rate of radiation heat transfer with two thinradiation shields per unit area of the plates is

2 W/206=

−++

−++

−+

−⋅×=

−ε+ε+

−ε+ε+

−ε+ε

−σ=

−

−

!!*.+!

!*.+!

!!+.+!

!+.+!

!).+

!-.+

!

1(2 $++&(2 -++(3&2 "/m!+-).*&

!!!

!!!

!!!

(&

#

$$#!

#!shieldstwo!#,

T T Q

The equilibrium temperatures of the radiation shields are determined from

549= →

−+

−⋅×=

−ε+ε

−σ=

−$

$#

#

$!

$!!$

!!+.+!

-.+!

1(2 -++(3&2 "/m!+-).*& "/m#+-

!!!

(&

T T

T T Q

429= →

−+

−⋅×=

−ε

+ε

−σ=

− ##

#

##

!).+

!!*.+!

1(2 $++&(32 "/m!+-).*& "/m#+-

!!!

(&

T T

T T Q

22 7" .. 22 77 +esi n and Essa ()o le!s

##'-*

T # $++ 2

ε# +.)

T ! -++ 2

ε! +.-

ε$ +.!+

ε +.!*

8/12/2019 FTFS Chap22 P041



ftfs chap22 p041

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