toda radiacion celgel

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12-28 The base, top, and side surfaces of a furnace of cylindrical shape are black, and aremaintained at uniform temperatures. The net rate of radiation heat transfer to or from the topsurface is to be determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are black. 3 Convection heat transfer is not

considered. Properties The emissivity of all surfaces are I = 1 since they are black.

Analysis We consider the top surface to be surface 1, thebase surface to be surface 2 and the side surfaces to besurface 3. The cylindrical furnace can be considered to bethree-surface enclosure. We assume that steady-stateconditions exist. Since all surfaces are black, the radiositiesare equal to the emissive power of surfaces, and the net rate of radiation heat transfer from the top surface can bedetermined from

)()(4

34

11314

24

1121 T T F AT T F AQ ! W W

and A r 12 22 12 57! ! !T T ( ) .m m2

The view factor from the base to the top surface of the cylinder is F 12 0 38! . (From Figure 12-44).

The view factor from the base to the side surfaces is determined by applying the summation rule tobe

F F F F F 11 12 13 13 121 1 1 0 38 0 62 ! � p� ! ! !. .

Substituting,

kW -762!v!

v

v!

W W !

W1062.7

)K 1200-K )(700.K W/m1067.5)(62.0)(m 57.12(

)K 500-K )(700.K W/m1067)(0.38)(5.m (12.57

)()(

5

44428-2

44428-2

43

41131

42

41121 T T F AT T F AQ

D iscussion The negative sign indicates that net heat transfer is to the top surface.

T 1 = 700

I1 = 1r 1 = 2 m

T 2 = 1200¡

I2 = 1r 2 = 2 m

T 3 = 500¢

I3 = 1

h =2 m



12-29 The base and the dome of a hemispherical furnace are maintained at uniform temperatures.The net rate of radiation heat transfer from the dome to the base surface is to be determined.

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

Analysis The view factor is first determined from

F

F F F

11

11 12 12

0

1 1

!

! p !

(flat surface)

(summation rule)

Noting that the dome is black, net rate of radiation heat transfer from dome to the base surface can be determinedfrom

kW 759.4!

v!

�v!

!!

W10594.7

])K 1000()K 400)[(K W/m1067.5)(1]( /4)m 5()[7.0(

)(

5

444282

42

411211221

T

W I T T F AQQ

The positive sign indicates that the net heat transfer is from the dome to the base surface, asexpected.

12-30 Two very long concentric cylinders aremaintained at uniform temperatures. The net rate of radiation heat transfer between the two cylinders isto be determined.

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

Properties The emissivities of surfaces are given tobe I1 = 1 and I2 = 0.7.

Analysis The net rate of radiation heat transferbetween the two cylinders per unit length of thecylinders is determined from

kW .8!!

¹º

¸©ª

¨

�£ T !

¹¹º

¸©©ª

¨

I

I

I

W!

W870,22

5

2

7.0

7.01

1

1

])¤

500(¤

) 950)[(¤

W/m1067.5](m) m)(1 2.0([

11

)( 44428

2

1

2

2

1

42

411

12

r

r

T T AQ

D2 = 0.5 mT 2 = 500

¥

I2 = 0.7

D1 = 0.2 mT 1 = 950

¦

I1 = 1

Vacuum

T 1 = 400§

I1 = 0.7

T 2 = 1000 ̈

I2 = 1

D = 5 m



12-31 A long cylindrical rod coated with a new material isplaced in an evacuated long cylindrical enclosure which ismaintained at a uniform temperature. The emissivity of thecoating on the rod is to be determined.

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

Properties The emissivity of the enclosure is given to beI2 = 0.95.

Analysis The emissivity of the coating on the rod isdetermined from

¹º

¸©ª

¨

�©

!

¹¹º

¸©©ª

¨

!

10

1

95.0

95.011

]

200

500)[

W/m1067.5](m) m)(1 01.0([W8

11

)(

1

44428

2

1

2

2

1

42

411

12

I

T

I

I

I

W

r

r

T T AQ

which gives

I1 = 0.074

12-32E The base and the dome of a long semicylindrical duct are maintained at uniformtemperatures. The net rate of radiation heat transfer from the dome to the base surface is to bedetermined.

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

Properties The emissivities of surfaces are given to be I1 = 0.5 and I2 = 0.9.

Analysis The view factor from the base to the dome is first determined from

F

F F F

11

11 12 12

0

1 1

!

! p !

(flat surface)

(summation rule)

The net rate of radiation heat transfer from dome to the basesurface can be determined from

Btu/h 101.311

v!

¼½

»¬« T

�v

!

I

I

I

I

W!!

)9.0(2

t) 1)(

t 15(

9.01

)1)(

t 15(

1

)5.0)(

t 15(

5.01

]R) 1800()R 550)[(R Btu/h.

t 101714.0(

111

)(

22

44428

22

2

12111

1

42

41

1221

AF AA

T T QQ

The positive sign indicates that the net heat transfer is from the dome to the base surface, asexpected.

D2 = 0.1 mT 2 = 200

I2 = 0.95

D1 = 0.01 mT 1 = 500

I1 = ?

Vacuum

T 1 = 550 R

I1 = 0.5

T 2 = 1800 R I2 = 0.9

D = 15 t



12-33 Two parallel disks whose back sides are insulated are black, and are maintained at auniform temperature. The net rate of radiation heat transfer from the disks to the environment is tobe determined.


Properties The emissivities of all surfaces are I = 1 since they are black.

Analysis Both disks possess same properties and theyare black. Noting that environment can also beconsidered to be blackbody, we can treat this geometryas a three surface enclosure. We consider the two disksto be surfaces 1 and 2 and the environment to besurface 3. Then from Figure 12-7, we read

F F

F

12 21

13

0 26

1 0 26 0 74

! !

! !

.

. . ( )

summation rule

The net rate of radiation heat transfer from the disksinto the environment then becomes

W55 5!

�v!

!

!!

]K 300K 700)[K W/m1067.5]()m 3.0()[74.0(2

)(2

2

444282

43

411133

1323133

T

W T T AF Q

QQQQ

12-34 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of the base surface is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 End effects are neglected.

Properties The emissivities of surfaces are given to be I1 = 0.8and I2 = 0.5.

Analysis This geometry can be treated as a two

surface enclosure since two surfaces have identicalproperties. We consider base surface to be surface 1and other two surface to be surface 2. Then the viewfactor between the two becomes F 12 1! . The

temperature of the base surface is determined from

K 3T

1!�p�

�v

!

!

)5.0()m 2(

5.01

)1)(m 1(

1

)8.0)(m 1(

8.01

]

500)[

W/m1067.5(W800

111

)(

222

441

428

22

2

12111

1

42

41

12

T

AF AA

T T Q

I

I

I

I

W

Note that .m 2 and m 1 22

21 !! AA

Disk 1, T 1 = 700

, I1 = 1

Disk 2, T 2 = 700

, I2 = 1

0.40 m

Environment

T 3 =300

I1 = 1

D = 0.6 m

q1 = 800 W/m2

I1 = 0.8

T 2 = 500

I2 = 0.5

b = 2 ! t



12-36 The floor and the ceiling of a cubical furnace are maintained at uniform temperatures. Thenet rate of radiation heat transfer between the floor and the ceiling is to be determined.


Properties The emissivities of all surfaces are I = 1 since they are black or reradiating.

Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the side surfaces tobe surface 3. The furnace can be considered to be three-surface enclosure with a radiation networkshown in the figure. We assume that steady-state conditions exist. Since the side surfaces arereradiating, there is no heat transfer through them, and the entire heat lost by the ceiling must begained by the floor. The view factor from the ceiling to the floor of the furnace is F 12 0 2! . . Then

the rate of heat loss from the ceiling can be determined from

1

231312

211

11

¹¹º

¸©©ª

¨

!

RRR

E E Q bb

where

24428422

24428411

W/m 5188)

"

550)(

"

.W/m 1067.5(

W/m 015,83)"

1100)("

.W/m 1067.5(

!v!!

!v!!

T E

T E

b

b

W

W

and

A A1 224 16! ! !( )m m2

RA F

R RA F

121 12

13 231 13

1 1

16 0 203125

1 1

16 0 80 078125

! ! !

! ! ! !

( )( . ).

( )( . ).

mm

mm

2

-2

2

-2

Substituting,

kW !v!

¹¹

º

¸

©©

ª

¨

!

W1047.7

)m 078125.0(2

1

m 3125.0

1

W/m)5188015,83( 5

1

2-2-

2

12Q

T 2 = 550#

I2 = 1

T 1 = 1100$

I1 = 1

Reradiating side

sur % acess

a = 4 m



12-37 Two concentric spheres are maintained at uniform temperatures. The net rate of radiationheat transfer between the two spheres and the convection heat transfer coefficient at the outersurface are to be determined.

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

Properties The emissivities of surfaces are given to be I1

= 0.1 and I2 = 0.8.

Analysis The net rate of radiation heat transfer betweenthe two spheres is

W669!

¹º

¸©ª

¨

�v!

¹¹

º

¸

©©

ª

¨

!

2

444282

22

21

2

2

1

42

411

12

m 4.0

m 15.0

7.0

7.01

5.0

1

]K 400K 700)[K W/m1067.5](m) 3.0([

11

)(

T

I

I

I

W

r

r

T T AQ

Radiation heat transfer rate from the outer sphere to the surrounding surfaces are

W685])K 27330()K 400)[(K W/m1067.5](m) 8.0()[1)(35.0(

)(

444282

4422

!�vT !

W I!

surr rad T T FAQ

The convection heat transfer rate at the outer surface of the cylinder is determined fromrequirement that heat transferred from the inner sphere to the outer sphere must be equal to theheat transfer from the outer surface of the outer sphere to the environment by convection andradiation. That is,

W9845685166912 !!! rad conv QQQ

Then the convection heat transfer coefficient becomes

? A CW/m5. 4

2r�!�p� T !

! g

hh

T T hAQconv

K) 303-K (400m) 8.0(W984 2

22.

D2 = 0.8 mT 2 = 400

&

I2 = 0.7

D1 = 0.3 m

T 1 = 700'

I1 = 0.5

T surr = 30rC

T g = 30rC

I = 0.35



12-38 A spherical tank filled with liquid nitrogen is kept in an evacuated cubic enclosure. The net rate of radiation heat transfer to the liquid nitrogen is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.


Analysis We take the sphere to be surface 1 and thesurrounding cubic enclosure to be surface 2. Noting that F 12 1! , for this two-surface enclosure, the net rate of

radiation heat transfer to liquid nitrogen can be determinedfrom

? A ? A

W 22!

¼¼½

»

¬¬«

�v!

¹¹º

¸©©ª

¨

!!

2

2

444282

2

1

2

2

1

42

411

1221

m) 6(3

m) 2(

8.0

8.01

1.0

1

K 240K 100K W/m1067.5m) 2(

11

T

T

I

I

I

W

A

A

T T AQQ

12-39 A spherical tank filled with liquid nitrogen is kept in an evacuated spherical enclosure. Thenet rate of radiation heat transfer to the liquid nitrogen is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.

Properties The emissivities of surfaces are given to beI1 = 0.1 and I2 = 0.8.

Analysis The net rate of radiation heat transfer to liquidnitrogen can be determined from

W227!

¹¹

º

¸

©©

ª

¨

�v

!

¹¹

º

¸

©©

ª

¨

!

2

2

444282

22

21

2

2

1

42

411

12

m) (1.5

m) 1(

8.0

8.01

1.0

1

](

100(

240)[(

W/m1067.5](m) 2([

11

)(

T

I

I

I

W

r

r

T T AQ

D2 = 3 m

T 2 = 240)

I2 = 0.8

D1 = 2 m

T 1 = 1000

I1 = 0.1

Vacuum

Liquid N2

Cube, a =3 m

T 2 = 2401

I2 = 0.8

D1 = 2 m

T 1 = 1002

I1 = 0.1

Vacuum

Liquid N2



12-41 A circular grill is considered. The bottom of the grill is covered with hot coal bricks, while thewire mesh on top of the grill is covered with steaks. The initial rate of radiation heat transfer fromcoal bricks to the steaks is to be determined for two cases.


Properties The emissivities are I = 1 for all surfaces since

they are black or reradiating.Analysis We consider the coal bricks to be surface 1, thesteaks to be surface 2 and the side surfaces to be surface 3.First we determine the view factor between the bricks and thesteaks (Table 12-1),

75.0m 0.20

m 15.0!!!!

L

r RR i

ji

7778.30.75

75.0111

2

2

2

2

!

!

!i

j

R

RS

2864.075.0

75.0

47778.37778.32

1

42

12/1

2

2

2/12

2

12 !±À

±

¿

¾

±°

±

¯

®

¼¼½

»

¬¬

«

¹º

¸

©ª

¨

!±À

±

¿

¾

±°

±

¯

®

¼¼

½

»

¬¬«

¹¹º

¸

©©ª

¨

!! i

j

ijR

R

S S F F

(It can also be determined from Fig. 12-7).

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

W1 74!

�v!

!

])3

278()3

1100)[(3

W/m1067.5](4/m) 3.0()[2864.0(

)(

444282

42

4111212

T

W T T AF Q

When the side opening is closed with aluminum foil, the entire heat lost by the coal bricks must begained by the stakes since there will be no heat transfer through a reradiating surface. The grill canbe considered to be three-surface enclosure. Then the rate of heat loss from the room can bedetermined from

1

231312

211

11

¹¹º

¸©©ª

¨

!

RRR

E E Q bb

whereW / m K K W / m

W / m K K W / m

4 2 2

4 2 2

E T

E T

b

b

1 18 4 4

2 28 4 4

67 10 1100 83 015

67 10 18 273 407

! ! v !

! ! v !

W

W

(5. . )( ) ,

(5. . )( )

and A A1 2

20 3

0 07069! ! !T ( . )

.m

4m

2

2-

2131

2313

2-

2121

12

m 82.19)2864.01)(m 07069.0(

11

m 39.49)2864.0)(m 07069.0(

11

!

!!!

!!!

F ARR

F AR

Substituting, W 3757!

¹¹º

¸©©ª

¨

!

1

2-2-

2

12

)m 82.19(2

1

m 39.49

1

W/m)407015,83(Q

Steaks, T 2 = 2784

, I2 = 1

Coal bricks, T 1 = 11005

, I1 = 1

0.20 m



12-42E A room is heated by lectric resistance heaters placed on the ceiling which is maintained at a uniform 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. 3 Convection heat transfer is not considered. 4 There is no heat loss through the side surfaces.

Properties The emissivities are I = 1 for the ceiling and I = 0.8 for the floor. The emissivity of insulated (or reradiating) surfaces is also 1.

Analysis The room can be considered to be three-surfaceenclosure with the ceiling surface 1, the floor surface 2 and theside surfaces surface 3. We assume steady-state conditions exist.Since the side surfaces are reradiating, there is no heat transferthrough them, and the entire heat lost by the ceiling must begained by the floor. Then the rate of heat loss from the roomthrough its floor can be determined from

2

1

231312

211

11R

RRR

E E Q bb

¹¹º

¸©©ª

¨

!

where

24428422

24428411

Btu/h.6

t 130)R 46065)(R .Btu/h.6

t 101714.0(

Btu/h.6

t 157)R 46090)(R .Btu/h.6

t 101714.0(

!v!W!

!v!W!

T E

T E

b

b

and

A A1 2212 144! ! !( )

7

t7

t 2

The view factor from the floor to the ceiling of the room is F 12 0 27! . (From Figure 12-42). The

view factor from the ceiling or the floor to the side surfaces is determined by applying thesummation rule to be

F F F F F 11 12 13 13 121 1 1 0 27 0 73 ! � p� ! ! !. .

since the ceiling is flat and thus F 11 0! . Then the radiation resistances which appear in the

equation above become

2-

2131

2313

2-

2121

12

2-

222

22

ft 009513.0)73.0)(ft 144(

11

ft 02572.0)27.0)(ft 144(

11

ft 00174.0)8.0)(ft 144(

8.011

!!!!

!!!

!

!

!

F ARR

F AR

AR

I

I

Substituting,

Bt / 2!

¹¹

º

¸©©

ª

¨

!

2-

1

2-2-

2

12

ft 00174.0)ft 009513.0(2

1

ft 02572.0

1

8

tu/h.ft )130157(Q

T 2 = 90rF

I2 = 0.8

Ceiling: 12 9 t v 12 9 t

T 1 = 90rF

I1 = 1

Insulated side

sur @ acess9@ t



12-43 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 thehorizontal surface and the surroundings are to be determined.


Properties The emissivities of the horizontal rectangle and the surroundings are I = 0.75 and I =

0.85, respectively.Analysis We consider the horizontal rectangle to be surface 1, the vertical rectangle to be surface2 and the surroundings to be surface 3. This system can be considered to be a three-surfaceenclosure. The view factor from surface 1 to surface 2 is determined from

27.0

75.06.1

2.1

5.06.1

8.0

122

1

!

±±À

±±¿

¾

!!

!!

F

W

L

W

L

(Fig. 12-6)

The surface areas are

22

21

m 92.1)m 6.1)(m 2.1(

m 28.1)m 6.1)(m 8.0(

!!

!!

A

A

2223 m 268.36.12.18.0

2

8.02.12 !vvv!A

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

18.0)92.1()27.0)(28.1( 2121212121 !�p� !�p� ! F F F AF A (reciprocity rule)

73.0127.001 1313131211 !�p� !�p� ! F F F F F (summation rule)

82.01018.01 2323232221 !�p� !�p� ! F F F F F (summation rule)



We now apply Eq. 9-52b to each surface to determine the radiosities.

Surface 1:

? A

? A)(73.0)(27.075.0

75.01)K 400)(K .W/m 1067.5(

)()(1

312114428

311321121

11

41

J J J J J

J J F J J F J T

!v

!

I

I W

Surface 2: 24428

24

2 )K 550)(K .W/m 1067.5( J J T !v�p� ! W

Surface 3:

? A

? A)(48.0)(29.085.0

85.01)

A

290)(A

.W/m 1067.5(

)()(1

312134428

23321331

3

33

43

J J J J J

J J F J J F J T

!v

!

I

I W

Solving the above equations, we find2

32

22

1 W/m 5.811 ,W/m 5188 ,W/m 1587 !!! J J J

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

W1245!!!! 22211211221 W/m)51881587)(27.0)(m 28.1()( J J F AQQ

W725!!! 223113113 W/m)5.8111587)(73.0)(m 28.1()( J J F AQ

W = 1.6 m

(2)L2 = 1.2 m

L1 = 0.8 mA1 (1)

A2 T 3 = 290

B

I3 = 0.85

T 2 = 550C

I2 = 1

T 1 =400D

I1 =0.75

(3)



12-44 Two long parallel cylinders are maintained at specified temperatures. The rates of radiationheat transfer between the cylinders and between the hot cylinder and the surroundings are to bedetermined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer is not considered.

Analysis We consider the hot cylinder to be surface 1, cold cylinder to be surface 2, and thesurroundings to be surface 3. Using the crossed-strings method, the view factor between twocylinders facing each other is determined to be

F s D s

D1 2

2 2

2

2 2

2 2 !

v

! § §Crossed strings Uncrossed strings

String on sur E

ace 1 ( / )T

or 099.0)16.0(

5.016.05.022 2222

21 !T

¹º¸

©ª¨

!T

¹º¸

©ª¨

!D

sDs

F

The view factor between the hot cylinder andthe surroundings is

901.0099.011 1213 !!! F F (summation rule)

The rate of radiation heat transfer betweenthe cylinders per meter length is

2m 2513.02/m) 1)(m 16.0(2/ !!! T T DLA

W.!rv!W ! 44428242

411212 K )275425)(C.W/m1067.5)(099.0)(m 2513.0()( T T AF Q

Note 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 permeter length of the cylinder is

21 m 5027.0m) 1)(m 16.0( !!! T T DLA

W629.!rv!W ! 44428243

4113113 K )300425)(C.W/m1067.5)(901.0)(m 5027.0()( T T F AQ

D

D (2)

(1)s

T 3 = 300F

I3 = 1T 2 = 275

G

I2 = 1

T 1 = 425H

I1 = 1

(3)



12-45 A long semi-cylindrical duct with specified temperature on the side surface is considered.The temperature of the base surface for a specified heat transfer rate is to be determined.


Properties The emissivity of the side surface is I = 0.4.

Analysis We consider the base surface to be surface 1, the side surface to be surface 2. Thissystem is a two-surface enclosure, and we consider a unit length of the duct. The surface areas andthe view factor are determined as

22

21

m 571.12/m) 1)(m 0.1(2/

m 0.1)m 0.1)(m 0.1(

!!!

!!

T T DLA

A

1101 12121211 !�p� !�p� ! F F F F (summation rule)

The temperature of the base surface is determined from

K 84.8!�p�

�v

!

I

I

W!

1

22

441

428

22

2

121

42

41

12

)4.0)(m 571.1(

4.01

)1)(m 0.1(

1

]I

) 650()[W/m1067.5(W1200

11

)(

T T K

AF A

T T Q

12-46 A hemisphere with specified base and dome temperatures and heat transfer rate isconsidered. The emissivity of the dome is to be determined.


Properties The emissivity of the base surface is I = 0.55.

Analysis We consider the base surface to be surface 1, the dome surface to be surface 2. This

system is a two-surface enclosure. The surface areas and the view factor are determined as

2222

2221

m 0628.02/)m 2.0(2/

m 0314.04/)m 2.0(4/

!!!

!!!

T T

T T

DA

DA

1101 12121211 !�p� !�p� ! F F F F (summation rule)

The emissivity of the dome is determined from

0.21!I�p�

I

I

�v

!

I

I

I

I

W!!

2

22

222

44428

22

2

12111

1

42

41

1221

)m 0628.0(

1

)1)(m 0314.0(

1

)55.0)(m 0314.0(

55.01

]P

) 600(P

) 400)[(P

W/m1067.5(W50

111

)(

AF AA

T T QQ

T 1 = ?

I1 = 1

T 2 = 650Q

I2 = 0.4

D = 1 m

T 1 = 400R

I1 = 0.55

T 2 = 600S

I2 = ?

D = 0.2 m



12-51 A thin aluminum sheet is placed between two very large parallel plates that are maintainedat uniform temperatures. The net rate of radiation heat transfer between the two plates is to bedetermined for the cases of with and without the shield.


Properties The emissivities of surfaces are given to be I1 = 0.5, I2 = 0.8, and I3 = 0.15.

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

2W/m1857!

¹º

¸©ª

¨¹

º

¸©ª

¨

�v!

¹¹

º

¸

©©

ª

¨¹

¹º

¸©©ª

¨

!

115.0

1

15.0

11

8.0

1

5.0

1

])K 650()K 900)[(K W/m1067.5(

111

111

)(

44428

2,31,321

42

41

shield one,12

I I I I

W T T Q

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

244428

21

42

41

shield ,12 W/m035,12

18.0

1

5.0

1

])K 650()K 900)[(K W/m1067.5(

111

)(!

¹º

¸©ª

¨

�v!

¹¹º

¸©©ª

¨

!

I I

W T T Q no

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

,

,

,

Q

Q

12

12

1857

12 035

one shield

no shield

W

W! $

1

6

T 2 = 650T

I2 = 0.8

T 1 = 900U

I1 = 0.5

Radiation shieldI3 = 0.15



12-53 Two very large plates are maintained at uniform temperatures. The number of thinaluminum sheets that will reduce the net rate of radiation heat transfer between the two plates toone-fifth is to be determined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.

Properties The emissivities of surfaces are given to be I1 = 0.2, I2 = 0.2, and I3 = 0.15.

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

2

44428

21

42

41

shield ,12

W/m3720

12.0

1

2.0

1

])K 800()K 1000)[(K W/m1067.5(

111

)(

!

¹º

¸©ª

¨

�v!

¹¹º

¸©©ª

¨

I

I

W !

T T Q no

The number of sheets that need to be inserted inorder to reduce the net rate of heat transfer betweenthe two plates to onefifth can be determined from

3$!�p�

¹º

¸©ª

¨¹

º

¸©ª

¨

�v!

¹¹

º

¸

©©

ª

¨

I

I¹¹

º

¸©©ª

¨

I

I

W !

92.2

115.0

1

15.0

11

2.0

1

2.0

1

])K 800()K 1000)[(K W/m1067.5( )W/m(3720

5

1

111

111

)(

shield

shield

444282

2,31,3shield

21

42

41

shields,12

N

N

N

T T Q

12-54 Five identical thin aluminum sheets are placed between two very large parallel plates which

are maintained at uniform temperatures. The net rate of radiation heat transfer between the twoplates 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. 3 Convection heat transfer is not considered. Properties The emissivities of surfaces are given tobe I1 = I2 = 0.1 and I3 = 0.1.Analysis Since the plates and the sheets have thesame emissivity value, the net rate of radiation heat transfer with 5 thin aluminum shield can bedetermined from

2 W/183!

¹º

¸©ª

¨

�v

!

¹¹

º

¸©©

ª

¨

!

!

11.0

1

1.0

1

])V

450()V

800)[(V

W/m1067.5(

15

1

111

)(

1

1

1

1

44428

21

42

41

shield no ,12shield 5,12

I I

W T T

N Q

N Q

The net rate of radiation heat transfer without the shield is

W10 8!v!!�p�

! W1836)1(1

1shield 5,12shield no ,12shield no ,12shield 5,12 QN QQ

N Q

T 2 = 800W

I2 = 0.2

T 1 = 1000X

I1 = 0.2

Radiation shields

I3 = 0.15

T 2 = 450Y

I2 = 0.1

T 1 = 800 ̀

I1 = 0.1

Radiation shields

I3 = 0.1



12-56E A radiation shield is placed between twoparallel 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 The surfaces are black. 3 Convection heat transfer is

not considered. Properties The emissivities of surfaces are given tobe I1 = I2 = 1 and I3 = 0.15.

Analysis From Fig. 12-44 we have 52.01332 !! F F .

Then 48.052.0134 !!F . The disk in the middle is

surrounded by black surfaces on both sides.Therefore, heat transfer between the top surface of the middle disk and its black surroundings canexpressed as

]})a

540([48.0])R 1200([(52.0){R Btu/h.b

t 101714.0()b

t 069.7(15.0

)]([)]([

443

443

4282

42

43323

41

433133

�v!

WIWI!

T T

T T F AT T F AQ

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

]})c

540([52.0])R 700([(48.0){R Btu/h.d

t 101714.0()d

t 069.7(15.0

)]([)]([

443

443

4282

45

43353

44

433433

�v!

WIWI!

T T

T T F AT T F AQ

Combining the equations above, the rate of heat transfer between the disks through the radiationshield (the middle disk) is determined to be

Btu/h 866!Q and T3 = 895 K

T 1 = 1200 R, I1 = 1

I3 = 0.15

T 2 = 700 R, I2 = 1

1e

t

1e

t

T g = 540f

I3 = 1



12-57 A radiation shield is placed between two large parallel plates which are maintained at uniform temperatures. The emissivity of the radiation shield is to be determined if the radiationheat transfer between the plates is reduced to 15% of that without the radiation shield.



Analysis First, the net rate of radiation heat transfer between the two large parallel plates per unit area without a shield is

244428

21

42

41

shield no,12 W/m4877

19.0

1

6.0

1

])K 400()K 650)[(K W/m1067.5(

111

)(!

�v!

!

I I

W T T Q

The radiation heat transfer in the case of one shield is

22

shield no,12shield one,12

W/m 6.731W/m 487715.0

15.0

!v!

v! QQ

Then the emissivity of the radiation shield becomes

¹¹º

¸©©ª

¨¹

º

¸©ª

¨

�v!

¹¹

º

¸

©©

ª

¨¹¹

º

¸©©ª

¨

!

12

19.0

1

6.0

1

])K 400()K 650)[(K W/m1067.5(W/m731.6

111

111

)(

3

444282

2,31,321

4

2

4

1shield one,12

I

I I I I

W T T Q

which gives 0.18!3I

T 2 = 400g

I2 = 0.9

T 1 = 650h

I1 = 0.6

Radiation shieldI3



12-59 A coaxial radiation shield is placed between two coaxial cylinders which are maintained at uniform temperatures. The net rate of radiation heat transfer between the two cylinders is to bedetermined and compared with that without the shield.


Properties The emissivities of surfaces are given to be I1 = 0.7, I2 = 0.4. and I3 = 0.2.

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

233shield

222outer pipe,

211inner pipe,

m 942.0)m 1)(m 3.0(

m 314.0)m 1)(m 1.0(

m 628.0)m 1)(m 2.0(

!!!!

!!!!

!!!!

T T

T T

T T

LDAA

LDAA

LDAA

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

W703!

�v

!

!

)4.0)(942.0(

4.01

)1)(628.0(

1

)2.0)(628.0(

2.012

)1)(314.0(

1

)7.0)(314.0(

7.01

])i

500()i

750)[(i

W/m1067.5(

111111

)(

44428

22

2

2,332,33

2,3

1,33

1,3

13111

1

42

41

shield one,12

I

I

I

I

I

I

I

I

W

AF AAAF AA

T T Q

If there was no shield,

W7465!

¹º

¸©ª

¨

�v

!

¹¹º

¸©©ª

¨

!

3.0

1.0

4.0

4.01

7.0

1

])p

500()p

750)[(p

W/m1067.5(

11

)(

44428

2

1

2

2

1

42

41

shield no,12

D

D

T T Q

I

I

I

W

Then their ratio becomes

,

,

Q

Q

12

12

703

7465

one shield

no shield

W

W! ! 0 .094

D2 = 0.3 mT 2 = 500

q

I2 = 0.4

D1 = 0.1 mT 1 = 750

r

I1 = 0.7

Radiation shield

D3 = 0.2 m

I3 = 0.2



12-66 The temperature, pressure, and composition of a gas mixture is given. The emissivity of themixture is to be determined.

Assumptions 1 All the gases in the mixture are ideal gases. 2 The emissivity determined is themean emissivity for radiation emitted to all surfaces of the cubical enclosure.

Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components,which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures

of CO2 and H2O are

atm 09.0atm) 1(09.0

atm 10.0atm) 1(10.0

2

2

H

CO

!!!

!!!

P yP

P yP

Ow

c

The mean beam length for a cube of side length 6 m for radiationemitted to all surfaces is, from Table 12-4,

L = 0.66(6 m) = 3.96 m

Then,

atms

t 1.57atmm 48.0m) atm)(3.96 09.0(

atms

t .301atmm 396.0m) atm)(3.96 10.0(

�!�!!

�!�!!

Lt

Lt

u

c

The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1000K and 1atm are, from Fig. 12-36,

17.0atm 1 , !I c and 26.0atm 1 , !Iw

Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity correction factor at T = Tg = 1000 K is, from Fig. 12-38,

039.0 474.0

10.009.0

09.0

87.257.130.1

!I(±À

±¿

¾

!

!

!!

cw

w

wc

P P

P

LP LP

Note that we obtained the average of the emissivity correction factors from the two figures for 800K and 1200 K. Then the effective emissivity of the combustion gases becomes

0.3 1!vv!I(II!I 039.026.0117.01atm 1 ,atm 1 , wwccg C C

Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm.

6 m

Combustion

gases

1000v



12-67 A mixture of CO2 and N2 gases at 600 K and a total pressure of 1 atm are contained in acylindrical container. The rate of radiation heat transfer between the gas and the container walls isto be determined.

Assumptions All the gases in the mixture are ideal gases.

Analysis The mean beam length is, from Table 12-4

L = 0.60D = 0.60(8 m) = 4.8 m

Then,

atmft .362atmm 72.0m) atm)(4.8 15.0( �!�!!LP c

The emissivity of CO2 corresponding to this value at the gastemperature of Tg = 600 K and 1 atm is, from Fig. 12-36,

24.0atm 1 , !I c

For a source temperature of Ts = 450 K, the absorptivity of the gas is again determined using theemissivity charts as follows:

atmw

t 1.77atmm 54.0x

600

x

450m) atm)(4.8 15.0( �!�!!

g

s

cT

T LP

The emissivity of CO2 corresponding to this value at a temperature of Ts = 450 K and 1atm are,from Fig. 12-36,

14.0atm 1 , !I c

The absorptivity of CO2 is determined from

17.0)14.0(y

450

y

600)1(

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I¹

¹

º

¸

©©

ª

¨!E c

s

g

ccT

T C

The surface area of the cylindrical surface is

222

m 6.3014

m) 8(2m) 8(m) 8(

42 !

T T !

T T !

DDH A

s

Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnacebecomes

W101. 15

v!

�v!

EIW!

])�

450(17.0)�

600(14.0)[�

W/m1067.5)(m 6.301(

)(

444282

44net sg g g s

T T AQ

8 m

8 m T g = 600�

T s = 450�



12-68 A mixture of H2O and N2 gases at 600 K and a total pressure of 1 atm are contained in acylindrical container. The rate of radiation heat transfer between the gas and the container walls isto be determined.



L = 0.60D = 0.60(8 m) = 4.8 m

Then,

atmft .362atmm 72.0m) atm)(4.8 15.0( �!�!!LP w

The emissivity of H2O corresponding to this value at the gastemperature of Tg = 600 K and 1 atm is, from Fig. 12-36,

36.0atm 1 !I w


atm�

t 1.77atmm 54.0�

600

�

450m) atm)(4.8 15.0( �!�!!

g

s

wT

T LP

The emissivity of H2O corresponding to this value at a temperature of Ts = 450 K and 1atm are,from Fig. 12-36,

34.0atm 1 , !Iw

The absorptivity of H2O is determined from

39.0)34.0(K 450

K 600)1(

45.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I¹

¹

º

¸©©

ª

¨!E w

s

g

wwT

T C

The surface area of the cylindrical surface is

222

m 6.3014

m) 8(2m) 8(m) 8(

42 !

T T !

T T !

DDH A

s

Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace

becomes

W105.2445v!

�v!

EIW !

])K 450(39.0)K 600(36.0)[K W/m1067.5)(m 6.301(

)(

444282

44net sg g g s

T T AQ

8 m

8 m T g = 600�

T s = 450�



12-69 A mixture of CO2 and N2 gases at 1200 K and a total pressure of 1 atm are contained in aspherical furnace. The net rate of radiation heat transfer between the gas mixture and furnacewalls is to be determined.



L = 0.65D = 0.65(2 m) = 1.3 m

The mole fraction is equal to pressure fraction. Then,

atmft .640atmm 195.0m) atm)(1.3 15.0( �!�!!LP c

The emissivity of CO2 corresponding to this value at the gastemperature of Tg = 1200 K and 1 atm is, from Fig. 12-36,

14.0atm 1 , !I c


atm�

t .320atmm 0975.0�

1200

�

600m) atm)(1.3 15.0( �!�!!

g

s

cT

T LP

The emissivity of CO2 corresponding to this value at a temperature of Ts = 600 K and 1atm are,from Fig. 12-36,

092.0atm 1 , !I c

The absorptivity of CO2 is determined from

144.0)092.0(�

600

�

1200)1(

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I¹

¹

º

¸

©©

ª

¨!E c

s

g

ccT

T C

The surface area of the sphere is

222 m 57.12m) 2( !T !T ! DAs

Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnacebecomes

W101. 365

v!

�v!

EIW!

])�

600(144.0)�

1200(14.0)[�

W/m1067.5)(m 57.12(

)(

444282

44net sg g g s

T T AQ

T g = 1200�

T s = 600�

2 m



12-70 The temperature, pressure, and composition of combustion gases flowing inside long tubesare given. The rate of heat transfer from combustion gases to tube wall is to be determined.


Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components,which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressuresof CO2 and H2O are

atm 09.0atm) 1(09.0

atm 06.0atm) 1(06.0

2

2

H

CO

!!!

!!!

P yP

P yP

Ow

c

The mean beam length for an infinite cicrcularcylinder is, from Table 12-4,

L = 0.95(0.15 m) = 0.1425 m

Then,

atm�

t 0.042atmm 0128.0m) 5atm)(0.142 09.0(

atm�

t .0280atmm 00855.0m) 5atm)(0.142 06.0(

�!�!!

�!�!!

LP

LP

w

c


034.0atm 1 , !I c and 016.0atm 1 , !I w


0.0 6.0

06.009.0

09.0

07.0042.0028.0

!I�

±À

±¿

¾

!

!

!!

cw

w

wc

P P

P

LP LP

Then the effective emissivity of the combustion gases becomes

0.050.0016.01034.01atm 1 ,atm 1 , !vv!I(II!I wwccg C C

Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For asource temperature of Ts = 600 K, the absorptivity of the gas is again determined using theemissivity charts as follows:

atm�

t 0.017atmm 00513.0�

1500

�

600m) 5atm)(0.142 09.0(

atm�

t 0.011atmm 00342.0�

1500

�

600m) 5atm)(0.142 06.0(

�!�!!

�!�!!

g

s

w

g

s

c

T

T LP

T

T LP

The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and1atm are, from Fig. 12-36,

031.0atm 1 , !I c and 027.0atm 1 , !Iw

Then the absorptivities of CO2 and H2O become

041.0)027.0(K 600

K 1500)1(

056.0)031.0(K 600K 1500)1(

45.0

atm 1 ,

45.0

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I

¹¹º

¸©©ª

¨!E

!¹º¸©

ª¨!I

¹¹º¸

©©ª¨!E

ws

g

ww

cs

g cc

T

T C

T T C

Also (E = (I, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts =600 K instead of Tg = 1500 K. There is no chart for 600 K in the figure, but we can read (I values

T s = 600�

D = 15 cm

Combustion

gases, 1 atm

T g = 1500�



at 400 K and 800 K, and take their average. At P w/(P w+ P c) = 0.6 and P cL +P wL = 0.07 we read (I = 0.0. Then the absorptivity of the combustion gases becomes

0.0970.0041.0056.0 !!E(EE!E wcg

The surface area of the pipe per m length of tube is

2m 4712.0m) 1(m) 15.0( !T !T ! DLAs

Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnacebecomes

W6427!

�v!

EIW !

])K 600(097.0)K 1500(05.0)[K W/m1067.5)(m 4712.0(

)(

444282

44net sg g g s

T T AQ



12-71 The temperature, pressure, and composition of combustion gases flowing inside long tubesare given. The rate of heat transfer from combustion gases to tube wall is to be determined.


Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components,which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressuresof CO2 and H2O are

atm 09.0atm) 1(09.0

atm 06.0atm) 1(06.0

2

2

H

CO

!!!

!!!

P yP

P yP

Ow

c

The mean beam length for an infinite cicrcularcylinder is, from Table 12-4,

L = 0.95(0.15 m) = 0.1425 m

Then,

atm�

t 0.042atmm 0128.0m) 5atm)(0.142 09.0(

atm�

t .0280atmm 00855.0m) 5atm)(0.142 06.0(

�!�!!

�!�!!

LP

LP

w

c


034.0atm 1 , !I c and 016.0atm 1 , !I w

These are base emissivity values at 1 atm, and they need to be corrected for the 3 atm totalpressure. Noting that (P w+P)/2 = (0.09+3)/2 = 1.545 atm, the pressure correction factors are,from Fig. 12-37,

Cc = 1.5 and Cw = 1.8


0.0 6.0

06.009.0

09.0

07.0042.0028.0

!I�

±À

±¿

¾

!

!

!!

cw

w

wc

P P

P

LP LP

Then the effective emissivity of the combustion gases becomes0.0800.0016.08.1034.05.1atm 1 ,atm 1 , !vv!I(II!I wwccg C C


atm�

t 0.017atmm 00513.0�

1500

�

600m) 5atm)(0.142 09.0(

atm�

t 0.011atmm 00342.0�

1500

�

600m) 5atm)(0.142 06.0(

�!�!!

�!�!!

g

s

w

g

s

c

T

T LP

T

T LP


031.0atm 1 , !I c and 027.0atm 1 , !I w


073.0)027.0(K 600

K 1500)8.1(

084.0)031.0(K 600

K 1500)5.1(

45.0

atm 1 ,

45.0

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I

¹¹

º

¸

©©

ª

¨!E

!¹º

¸©ª

¨!I

¹¹

º

¸

©©

ª

¨!E

ws

g

ww

cs

g

cc

T

T C

T

T C

T s = 600�

D = 15 cm

Combustion

gases, 3 atm

T g = 1500�



Also (E = (I, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts =600 K instead of Tg = 1500 K. There is no chart for 600 K in the figure, but we can read (I valuesat 400 K and 800 K, and take their average. At P w/(P w+ P c) = 0.6 and P cL +P wL = 0.07 we read (I = 0.0. Then the absorptivity of the combustion gases becomes

0.1570.0073.0084.0 !!E(EE!E wcg

The surface area of the pipe per m length of tube is2m 4712.0m) 1(m) 15.0( !T !T ! DLA

s

Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnacebecomes

W10,276!

�v!

EIW !

])K 600(157.0)K 1500(08.0)[K W/m1067.5)(m 4712.0(

)(

444282

44net sg g g s

T T AQ



12-72 The temperature, pressure, and composition of combustion gases flowing inside long tubesare given. The rate of heat transfer from combustion gases to tube wall is to be determined. Assumptions All the gases in the mixture are ideal gases.Analysis The mean beam length for an infinitecicrcular cylinder is, from Table 12-4,

L = 0.95(0.10 m) = 0.095 m

Then,

atm

t 0.056atmm 0171.0m) atm)(0.095 18.0(

atm

t .0370atmm 0114.0m) atm)(0.095 12.0(

�!�!!

�!�!!

LP

LP

w

c

The emissivities of CO2 and H2O corresponding to these valuesat the gas temperature of Tg = 800 K and 1atm are, from Fig.12-36,

055.0atm 1 , !I c and 050.0atm 1 , !Iw


0.0 6.0

12.018.0

18.0

093.0056.0037.0

!I(±À

±¿

¾

!

!

!!

cw

w

wc

P P

P

LP LP

Then the effective emissivity of the combustion gases becomes0.1050.0050.01055.01atm 1 ,atm 1 , !vv!I(II!I wwccg C C


atm

t 0.035atmm 01069.0

800

500m) atm)(0.095 18.0(

atm

t 0.023atmm 007125.0

800

500m) atm)(0.095 12.0(

�!�!!

�!�!!

g

s

w

g

s

c

T

T LP

T

T LP


042.0atm 1 , !I c and 050.0atm 1 , !Iw


062.0)050.0(K 500

K 800)1(

057.0)042.0(K 500

K 800)1(

45.0

atm 1 ,

45.0

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I

¹¹

º

¸

©©

ª

¨!E

!¹º

¸©ª

¨!I

¹¹º

¸

©©ª

¨!E

ws

g

ww

c

s

g

cc

T

T C

T

T C

Also (E = (I, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts =500 K instead of Tg = 800 K. There is no chart for 500 K in the figure, but we can read (I values at 400 K and 800 K, and interpolate. At P w/(P w+ P c) = 0.6 and P cL +P wL = 0.093 we read (I = 0.0.Then the absorptivity of the combustion gases becomes

0.1190.0062.0057.0 !!E(EE!E wcg

The surface area of the pipe is2m 885.1m) 6(m) 10.0( !T !T ! DLA

s

Then the net rate of radiation heat transfer from the combustion gases to the walls of the tubebecomes

T s = 500

D = 10 cm

Combustiongases, 1 atm

T g = 800



W3802!

�v!

EIW!

])

500(119.0)

800(105.0)[

W/m1067.5)(m 885.1(

)(

444282

44net sg g g s

T T AQ



12-73 The temperature, pressure, and composition of combustion gases flowing inside long tubesare given. The rate of heat transfer from combustion gases to tube wall is to be determined.Assumptions All the gases in the mixture are ideal gases.Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components,which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressuresof CO2 and H2O are

atm 10.0atm) 1(10.0atm 10.0atm) 1(10.0

2

2

H

CO

!!!!!!

P yP P yP

Ow

c

The mean beam length for this geometry is,from Table 12-4,

L = 3.6V/As = 1.8D = 1.8(0.20 m) = 0.36 mwhere D is the distance between the plates. Then,

atmj

t .1180atmm 036.0m) atm)(0.36 10.0( �!�!!! LP LP wc


080.0atm 1 , !I c and 055.0atm 1 , !Iw


0025.0 5.0

10.010.0

10.0

236.0118.0118.0

!I(±À

±¿

¾

!

!

!!

cw

w

wc

P P

P

LP LP

Then the effective emissivity of the combustion gases becomes0.13250025.0055.01080.01atm 1 ,atm 1 , !vv!I(II!I wwccg C C


atmk

t 0.059atmm 018.0l

1200

l

600m) atm)(0.36 10.0( �!�!!!

g

s

wg

s

cT

T LP

T

T LP

The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and

1atm are, from Fig. 12-36,065.0atm 1 , !I c and 067.0atm 1 , !Iw


092.0)067.0(K 600

K 1200)1(

102.0)065.0(K 600

K 1200)1(

45.0

atm 1 ,

45.0

65.0

atm 1 ,

65.0

!¹º

¸©ª

¨!I

¹¹º

¸©©ª

¨!E

!¹º

¸©ª

¨!I

¹¹

º

¸

©©

ª

¨!E

ws

g

ww

c

s

g

cc

T

T C

T

T C

Also (E = (I, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts =600 K instead of Tg = 1200 K. There is no chart for 600 K in the figure, but we can read (I valuesat 400 K and 800 K, and take their average. At P w/(P w+ P c) = 0.5 and P cL +P wL = 0.236 we read (I

= 0.00125. Then the absorptivity of the combustion gases becomes0.192800125.0092.0102.0 !!E

m

EE!E wcg

Then the net rate of radiation heat transfer from the gas to each plate per unit surface areabecomes

W101.424

v!

�v!

!

])n

600(1928.0)n

1200(1325.0)[n

W/m 1067.5)(m 1(

)(

444282

44net sg g g s

T T AQ EI W

T s = 600o

20 cm

Combustiongases, 1 atm

T g = 1200



12-81 A man wearing summer clothes feels comfortable in a room at 22rC. The room temperatureat which this man would feel thermally comfortable when unclothed is to be determined.

Assumptions 1 Steady conditions exist. 2 The latent heat loss from the person remains the same.3 The heat transfer coefficients remain the same. 4 The air in the room is still (there are no windsor running fans). 5 The surface areas of the clothed and unclothed person are the same.

Analysis At low air velocities, the convection heat transfer

coefficient for a standing man is given in Table 12-3 to be 4.0W/m2.rC. The radiation heat transfer coefficient at typical indoorconditions is 4.7 W/m2.rC. Therefore, the heat transfer coefficient for a standing person for combined convection and radiation is

h h hcombined conv rad2W / m . C! ! ! r4 0 4 7 8 7. . .

The thermal resistance of the clothing is given to be

Rcloth2 2clo m . C / W = 0.109 m . C / W! ! v r r0 7 0 7 0155. . .

Noting that the surface area of an average man is 1.8 m2, thesensible heat loss from this person when clothed is determined tobe

W 104

C.W/m 8.7

1+C/W.m 0.109

C)2033)(m 8.1(

1

)(

2

2

2

combined

cloth

ambientskinclothedsensible, !

rr

r!

!

hR

T T AQ

s

From heat transfer point of view, taking the clothes off isequivalent to removing the clothing insulation or setting Rcloth = 0.The heat transfer in this case can be expressed as

C.W/m 8.7

1

C)33)(m 8.1(

1

)(

2

ambient2

combined

ambientskinunclothedsensible,

r

r!

!

T

h

T T AQ s

To maintain thermal comfort after taking the clothes off, the skin temperature of the person and

the rate of heat transfer from him must remain the same. Then setting the equation above equal to

104 W gives

T ambient ! r26.4 C

Therefore, the air temperature needs to be raised from 22 to 26.4rC to ensure that the person willfeel comfortable in the room after he takes his clothes off. Note that the effect of clothing on latent heat is assumed to be negligible in the solution above. We also assumed the surface area of theclothed and unclothed person to be the same for simplicity, and these two effects should

counteract each other.

T room= 20°C

Clothed

person

T skin= 33°C

T room

T skin= 33°C

Un



12-82E An average person produces 0.50 lbm of moisture whiletaking a shower. The contribution of showers of a family of fourto the latent heat load of the air-conditioner per day is to bedetermined.

Assumptions All the water vapor from the shower is condensedby the air-conditioning system.

Properties The latent heat of vaporization of water is given to be1050 Btu/lbm.

Analysis The amount of moisture produced per day is

mvapor = (Moisture produced per person)(No. of persons)

= (0.5 lbm/person)(4 persons/day) = 2 lbm/day

Then the latent heat load due to showers becomes

Q latent = mvaporhfg = (2 lbm/day)(1050 Btu/lbm) = 2100 Btu/day

12-83 There are 100 chickens in a breeding room. The rate of total heat generation and the rateof moisture production in the room are to be determined.

Assumptions All the moisture from the chickens is condensed by the air-conditioning system.

Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The averagemetabolic rate of chicken during normal activity is 10.2W (3.78 W sensible and 6.42 W latent).

Analysis The total rate of heat generation of the chickens in the breeding room is

W1020=chickens) (100W/chicken) 2.10(

chickens) of (

o.totalgen,totalgen,

!

! qQ

The latent heat generated by the chicken and the rate of moistureproduction are

kW 0.642=

W642=chickens) )(100W/chicken42.6(

chickens) of (

o.latent gen,latent gen,

!

! qQ

g/s 0.264 kg/s 000264.0kJ/kg 2430

kJ/s 642.0

g

latent gen,

moisture !!!!h

Qm

Moisture

0.5 lbm

100 Chickens 10.2 W



12-84 Chilled air is to cool a room by removing the heat generated in a large insulated classroomby lights and students. The required flow rate of air that needs to be supplied to the room is to bedetermined.

Assumptions 1 The moisture produced by the bodies leave the room as vapor without anycondensing, and thus the classroom has no latent heat load. 2 Heat gain through the walls and theroof is negligible.

Properties The specific heat of air at room temperature is 1.00 kJ/kg�rC (Table A-15). The averagerate of metabolic heat generation by a person sitting or doing light work is 115 W (70 W sensible,and 45 W latent).

Analysis The rate of sensible heat generation by thepeople in the room and the total rate of sensibleinternal heat generation are

W 500,144000500,10

W 10,500=persons)150W/person)( 70(

people)of (

o.

lightingsensible gen,sensible total,

sensible gen,sensible gen,

!!

!

!

!

QQQ

qQ

Then the required mass flow rate of chilled air becomes

kg/s 1.45!rr�

!

!

C15)C)(25kJ/kg(1.0

kJ/s5.14

sensible total,air

T C

Qm

p

D iscussion The latent heat will be removed by the air-conditioning system as the moisturecondenses outside the cooling coils.

Chilled

air

Return

air

15°C 25°C

150 Students

Lights

4 kW



12-85 The average mean radiation temperature during a cold day drops to 18rC. The requiredrise in the indoor air temperature to maintain the same level of comfort in the same clothing is tobe determined.

Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skintemperature remains the same. 3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs remain constant.

Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer coefficient from the body in still air or air moving with a velocity under 0.2 m/s is hconv =3.1 W/m2�rC (Table 12-3).

Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss byconvection, radiation, and evaporation,

( ) Q Q Q Q Q Q Q Qbody, total sensible latent lungs conv rad latent lungs! !

Noting that heat transfer from the skin by evaporation and from the lungs remains constant, thesum of the convection and radiation heat transfer from the person must remain constant.

])27318()273[(95.0)(

)()(

])27322()273[(95.0)22(

)()(

44newair,

4newsurr,

4newair,newsensible,

44

4oldsurr,

4old air,oldsensible,

!

!

!

!

ssss

ssss

ssss

ssss

T AT T hA

T T AT T hAQ

T AT hA

T T AT T hAQ

W

W I

W

W I

Setting the two relations above equal to each other,canceling the surface area As, and simplifying gives

!

v v !

22 0 95 22 273 0 95 18 273

31 22 0 95 5 67 10 291 295 0

4 4

8 4 4

h hT

T

. ( ) . ( )

. ( ) . . ( )

W W air, new

air, new

Solving for the new air temperature gives

Tair, new = 29.0rC

Therefore, the air temperature must be raised to 29rC to

counteract the increase in heat transfer by radiation.

22°C

22°C



12-86 The average mean radiation temperature during a cold day drops to 12rC. The required risein the indoor air temperature to maintain the same level of comfort in the same clothing is to bedetermined.

Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skintemperature remains the same. 3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs remain constant.

Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer coefficient from the body in still air or air moving with a velocity under 0.2 m/s is hconv =3.1 W/m2�rC (Table 12-3).

Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss byconvection, radiation, and evaporation,

( ) Q Q Q Q Q Q Q Qbody, total sensible latent lungs conv rad latent lungs! !

Noting that heat transfer from the skin by evaporation and from the lungs remains constant, thesum of the convection and radiation heat transfer from the person must remain constant.

])27312()273[(95.0)(

)()(

])27322()273[(95.0)22(

)()(

44newair,

4newsurr,

4newair,newsensible,

44

4oldsurr,

4old air,oldsensible,

!

!

!

!

ssss

ssss

ssss

ssss

T AT T hA

T T AT T hAQ

T AT hA

T T AT T hAQ

W

W I

W

W I

Setting the two relations above equal to each other,canceling the surface area As, and simplifying gives

0)295285(1067.595.0)22(1.3

)27312(95.0)27322(95.022

448newair,

4newair,

4

!vv

!

T

hT h W W

Solving for the new air temperature gives

Tair, new = 39.0rC

Therefore, the air temperature must be raised to 39rC tocounteract the increase in heat transfer by radiation.

22°C

22°C

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