Capitulo 13 solucionario transferencia calor y masa cengel 4th ed PDF

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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for courseSolutions ManualforHeat and Mass Transfer: Fundamentals & ApplicationsFourth EditionYunus A. Cengel & Afshin J. GhajarMcGraw-Hill, 2011Chapter 13RADIAT...

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13-1

Solutions Manual for

Heat and Mass Transfer: Fundamentals & Applications Fourth Edition Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2011

Chapter 13 RADIATION HEAT TRANSFER

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13-2

View Factors

13-1C The view factor Fi → j represents the fraction of the radiation leaving surface i that strikes surface j directly. The view factor from a surface to itself is non-zero for concave surfaces.

13-2C The cross-string method is applicable to geometries which are very long in one direction relative to the other directions. By attaching strings between corners the Crossed-Strings Method is expressed as

Fi → j =

∑ Crossed strings − ∑ Uncrossed strings 2 × string on surface i

N

13-3C The summation rule for an enclosure and is expressed as

∑F

i→ j

= 1 where N is the number of surfaces of the

j= 1

enclosure. It states that the sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including to itself must be equal to unity. The superposition rule is stated as the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j, F1→( 2,3) = F1→ 2 + F1→ 3 .

13-4C The pair of view factors F i→ j and F j →i are related to each other by the reciprocity rule Ai Fij = A j F ji where Ai is the area of the surface i and Aj is the area of the surface j. Therefore, A1 F12 = A2 F21 ⎯ ⎯→ F12 =

A2 A1

F21

13-5 An enclosure consisting of five surfaces is considered. The number of view factors this geometry involves and the number of these view factors that can be determined by the application of the reciprocity and summation rules are to be determined. Analysis A five surface enclosure (N=5) involves N 2 = 5 2 = 25 view N ( N −1) 5(5 −1) = = 10 view factors factors and we need to determine 2 2 directly. The remaining 25-10 = 15 of the view factors can be determined by the application of the reciprocity and summation rules.

1 2 5 4

3

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13-3

13-6 An enclosure consisting of thirteen surfaces is considered. The number of view factors this geometry involves and the number of these view factors that can be determined by the application of the reciprocity and summation rules are to be determined. Analysis A thirteen surface enclosure (N = 13) involves N 2 = 13 2 = 169 view factors and we need to determine N ( N − 1) 13(13 − 1) = = 78 view factors directly. The 2 2 remaining 169 - 78 = 91 of the view factors can be determined by the application of the reciprocity and summation rules.

4 2

5

3

6

1

7 13 8 12

10

9

11

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13-4

13-7 A cylindrical enclosure is considered. (a) The expression for the view factor between the base and the side surface F13 in terms of K and (b) the value of the view factor F13 for L = D are to be determined. Assumptions 1 The surfaces are diffuse emitters and reflectors. Analysis (a) The surfaces are designated as follows: Base surface as A1, top surface as A2, and side surface as A3 Applying the summation rule to A1, we have (where F11 = 0 )

F11 + F12 + F13 = 1

or

F13 = 1 − F12

(1)

For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,

F12

⎧ ⎡ ⎛D 1⎪ = ⎨S − ⎢S 2 − 4 ⎜⎜ 2 2⎪ ⎢ ⎝ D1 ⎣ ⎩

S = 1+

1 + R22 2 R1

=2+

1 R

2

⎞ ⎟⎟ ⎠

2⎤

1/ 2

⎥ ⎥ ⎦

= 2+

⎫ ⎪ 1 2 1/ 2 ⎬ = [ S − (S − 4 ) ] ⎪ 2 ⎭ 4

( D / L)

2

= 2 + 4K 2

(2)

(3)

where

R1 = R 2 = R =

D 1 = 2 L 2K

Substituting Eq. (3) into Eq. (2), we get 1 {2 + 4K 2 − [( 2 + 4K 2 ) 2 − 4 ]1 / 2 } 2 1 = [2 + 4 K 2 − (16 K 4 + 16 K 2 )1 / 2 ] 2 1 = [2 + 4 K2 − 4 K( K2 +1)1 / 2 ] 2 = 1 + 2 K 2 − 2 K ( K 2 +1) 1 / 2

F12 =

Substituting the above expression for F12 into Eq. (1) yields the expression for F13:

F13 = 1 − [1 + 2K 2 − 2K ( K 2 + 1) 1 / 2] Hence, 2

F13 = 2 K ( K + 1)

1/ 2

−2K

2

(b) The value of the view factor F13 for L = D (i.e., K = 1) is F13 = 2 (1)(12 + 1)1 / 2 − 2(1) 2 = 2 2 − 2 = 0.828

Discussion If the cylinder has a length and diameter of L = 2D, then from the expression for F13 we have

F13 = 2(2)(22 + 1)1 / 2 − 2(2) 2 = 0.944

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13-5

13-8 The expression for the view factor F12 of two infinitely long parallel plates is to be determined using the Hottel’s crossed-strings method. Assumptions 1 The surfaces are diffuse emitters and reflectors. Analysis From the Hottel’s crossed-strings method, we have Fi → j =

Σ (Crossed strings ) − Σ (Uncrossed strings ) 2 × (String on surface i )

For uncrossed strings, we have L1 = L2 = ( w 2 + w 2 ) 1 / 2 = ( w 2 + w 2 )1 / 2 =

2w

For crossed strings, we have L3 = (w 2 + 4 w 2 )1 / 2 = 5w

and

L4 = w

Applying the Hottel’s crossed-strings method, we get F12 as F12 =

( L3 + L4 ) − ( L1 + L2 ) 2w

( 5 w + w) − ( 2 w + 2 w) 2w = 0.204 =

Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be considered to be two-dimensional and radiation interaction through the end surfaces is negligible.

13-9 The view factors between the rectangular surfaces shown in the figure are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis From Fig. 13-6,

W=4m

⎫ 1.2 = 0.3 ⎪ ⎪ W 4 ⎬F31 = 0.26 L1 1.2 = = 0.3 ⎪ ⎪⎭ W 4 L3

=

L2 = 1.2 m A 2

(2)

A1

(1)

L1 = 1.2 m

and L 3 1.2 ⎫ = = 0.3 ⎪⎪ W 4 ⎬ F3→ (1+2 ) = 0.32 L1 + L 2 2.4 = = 0.6⎪ ⎪⎭ W 4

L3 = 1.2 m

A3

(3)

We note that A1 = A3. Then the reciprocity and superposition rules gives A1 F13 = A3 F31 ⎯ ⎯→ F13 = F31 = 0.26

F3 →(1 +2) = F31 + F32 ⎯ ⎯→ Finally,

0.32 = 0.26+ F32 ⎯ ⎯→ F32 = 0.06

A2 = A3 ⎯ ⎯→ F23 = F32 = 0.06

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13-6

13-10 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base surface is to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis We designate the surfaces as follows:

Base surface by (1), (2)

top surface by (2), and side surface by (3). Then from Fig. 13-7 (3)

⎫ ⎪ ⎪ ⎬ F12 = F21 = 0.05 r2 r2 = = 0.25⎪ ⎪⎭ L 4r2 L 4r1 = =4 r1 r1

L=2D=4r (1) D=2r

summation rule : F11 + F12 + F13 = 1 0 + 0.05 + F13 = 1 ⎯ ⎯→ F13 = 0.95

πr1 πr A1 1 F13 = F13 = 1 2 F13 = (0.95) = 0.119 A3 2π r1 L 8 8π r1 2

reciprocity rule : A1 F13 = A 3 F31 ⎯ ⎯→ F31 =

2

Discussion This problem can be solved more accurately by using the view factor relation from Table 13-1 to be

r1 r = 1 = 0.25 L 4r1 r r R 2 = 2 = 2 = 0.25 L 4 r2

R1 =

S = 1+

F12

1 + R 22 R12

= 1+

1+ 0.252 0.25 2

⎧ ⎡ ⎛R ⎪ = ⎨S − ⎢S 2 − 4⎜⎜ 2 ⎢ ⎝ R1 ⎪ ⎣ ⎩ 1 2

⎞ ⎟ ⎟ ⎠

2

⎤ ⎥ ⎥ ⎦

= 18 0.5 ⎫

⎧ 2 0.5 ⎫ ⎡ 2 ⎪ 1⎪ ⎛1 ⎞ ⎤ ⎪ ⎬ = 2 ⎨18 − ⎢18 − 4 ⎜ ⎟ ⎥ ⎬ = 0.056 ⎢⎣ ⎝1 ⎠ ⎥⎦ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭

F13 = 1 − F12 = 1 − 0.056 = 0.944

πr πr A1 1 F 13 = 1 F13 = 1 2 F13 = (0.944) = 0.118 A3 2πr 1L 8 8πr1 2

reciprocity rule : A1 F13 = A3 F31 ⎯ ⎯→ F31 =

2

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13-7

13-11 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base is to be determined. Assumptions The surfaces are diffuse emitters and reflectors.

(2)

Analysis We number the surfaces as follows:

(1): circular base surface (1)

(2): dome surface Surface (1) is flat, and thus F 11 = 0.

D

Summation rule : F 11 + F 12 = 1→ F 12 = 1

πD 2 reciprocity rule : A 1F 12 = A 2F 21 ⎯ ⎯→ F 21 =

A1 A2

F12 =

A1 A2

(1) =

4 = 1 = 0.5 πD2 2 2

13-12 Three infinitely long cylinders are located parallel to each other. The view factor between the cylinder in the middle and the surroundings is to be determined. Assumptions The cylinder surfaces are diffuse emitters and reflectors.

(surr)

Analysis The view factor between two cylinder facing each other is, from Prob. 13-17,

D

2⎛⎜ s 2 + D 2 − s ⎞⎟ ⎠ F1− 2 = ⎝ πD

Noting that the radiation leaving cylinder 1 that does not strike the cylinder will strike the surroundings, and this is also the case for the other half of the cylinder, the view factor between the cylinder in the middle and the surroundings becomes

F 1−surr =1 −2 F 1− 2

D (2)

D (1)

s (2)

s

4⎛⎜ s 2 + D 2 − s ⎞⎟ ⎠ ⎝ =1 − D π

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13-8

13-13 The view factors from the base of a cube to each of the other five surfaces are to be determined.

(2)

Assumptions The surfaces are diffuse emitters and reflectors. Analysis Noting that L1 / D = L 2 / D = 1 , from Fig. 13-6 we read

(3), (4), (5), (6) side surfaces

F12 = 0.2

Because of symmetry, we have F12 = F13 = F14 = F15 = F16 = 0.2

(1)

13-14 The view factor from the conical side surface to a hole located at the center of the base of a conical enclosure is to be determined. Assumptions The conical side surface is diffuse emitter and reflector. Analysis We number different surfaces as

the hole located at the center of the base (1) the base of conical enclosure

(2)

conical side surface

(3)

h (3)

Surfaces 1 and 2 are flat, and they have no direct view of each other. Therefore, F11 = F22 = F12 = F21 = 0

(2)

summation rule : F11 + F12 + F13 = 1 ⎯ ⎯→ F13 = 1

reciprocity rule : A1 F13 = A3 F31 ⎯⎯→

πd 4

2

(1)=

(1) d

πDh 2

F31 ⎯⎯→ F31 =

2

d 2Dh

D

13-15 The four view factors associated with an enclosure formed by two very long concentric cylinders are to be determined. Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected. Analysis We number different surfaces as

(2)

the outer surface of the inner cylinder (1)

(1)

the inner surface of the outer cylinder (2) No radiation leaving surface 1 strikes itself and thus F11 = 0

D2

D1

All radiation leaving surface 1 strikes surface 2 and thus F12 = 1 reciprocity rule : A 1 F12 = A2 F21 ⎯ ⎯→ F21 =

A1 A2

F12 =

summation rule : F21 + F22 = 1 ⎯ ⎯→ F22 = 1 − F21 = 1 −

D πD1 h (1) = 1 D2 π D 2h D1 D2

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13-9

13-16 The view factors between the rectangular surfaces shown in the figure are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis We designate the different surfaces as follows:

4m

shaded part of perpendicular surface by (1), bottom part of perpendicular surface by (3),

1m

(1)

shaded part of horizontal surface by (2), and

1m

(3)

front part of horizontal surface by (4).

(2)

1m

(4)

1m

(a) From Fig.13-6 L2 2 L2 1 ⎫ ⎫ = = 0.5 ⎪ = = 0.25⎪ ⎪ ⎪ W 4 W 4 ⎬ F23 = 0.26 and ⎬F2→ (1+ 3) = 0.33 L1 1 L1 1 = = 0.25 ⎪ = = 0.25⎪ ⎪⎭ ⎪⎭ W 4 W 4

superposit ion rule : F 2→ (1+ 3) = F 21 + F 23 ⎯ ⎯→ F 21 = F 2→ (1+ 3) − F 23 = 0.33 − 0.26 = 0.07

reciprocity rule : A1 = A2 ⎯ ⎯→ A1 F12 = A2 F21 ⎯ ⎯→ F12 = F21 = 0.07

(b) From Fig.13-6, L2 1 ⎫ = = 0.25⎪ ⎪ W 4 ⎬ F(4 +2) →3 = 0.16 L1 2 = = 0.5 ⎪ ⎪⎭ W 4

and

L2 2 ⎫ = = 0.5 ⎪ ⎪ W 4 ⎬F ( 4+ 2)→ (1+ 3) = 0.24 L1 2 = = 0 . 5⎪ ⎪⎭ W 4

superposition rule : F(4 +2) →(1 +3) = F(4 +2 ) →1 + F( 4 +2) →3 ⎯ ⎯→ F(4 +2 ) →1 = 0.24 − 0.16 = 0.08

reciprocity rule : A( 4 +2) F(4 +2 ) →1 = A1 F1→( 4 + 2) ⎯ ⎯→ F1→ ( 4+ 2) =

A( 4 + 2 ) A1

F( 4 +2) →1 =

8 (0.08) = 0.16 4

4m 1m

(1)

superposition rule : F 1→ ( 4+ 2) = F 14 + F 12 ⎯ ⎯→ F14 = 0.16 − 0.08 = 0.08 since F12 = 0.07 (from part a). Note that F14 in part (b) is equivalent to F12 in part (a).

(3)

1m

(4) 1m 1m

(2)

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13-10

(c) We designate shaded part of top surface by (1),

3m (1) (3)

remaining part of top surface by (3), remaining part of bottom surface by (4), and shaded part of bottom surface by (2).

3m

1m 1m

From Fig.13-5, 3 ⎫ =1 ⎪ ⎪ D 3 ⎬ F(2 +4) →(1 +3) = 0.15 L1 2 = = 0.67⎪ ⎪⎭ D 3 L2

1m 1m

and

(2)

3 ⎫ =1 ⎪ ⎪ D 3 ⎬ F14 = 0.082 L1 1 = = 0.33⎪ ⎪⎭ D 3 L2

=

(4)

=

superposition rule : F( 2+ 4 )→ (1+ 3) = F( 2+ 4)→ 1 + F( 2+ 4)→ 3

symmetryrule : F( 2 +4) →1 = F( 2 +4) →3 Substituting symmetry rule gives F( 2 +4 ) →1 = F( 2 +4) →3 =

F( 2 +4) →(1 +3) 2

=

0.15 = 0.075 2

reciprocity rule : A1 F1→( 2+ 4) = A( 2 + 4) F( 2 +4 )→1 ⎯ ⎯→(2)F1→ ( 2+ 4 ) = (4)(0.075) ⎯ ⎯→ F 1→ ( 2+ 4 ) = 0.15 superposition rule : F1 →( 2 +4) = F12 + F14 ⎯⎯→ 0. 15 = F12 + 0. 082 ⎯⎯→ F12 = 0. 15 − 0.082 = 0.068

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13-11

13-17 The view factor between the two infinitely long parallel cylinders located a distance s apart from each other is to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis Using the crossed-strings method, the view factor between two cylinders facing each other for s/D > 3 is determined to be F1− 2 =

or

F1− 2

D

∑ Crossed strings − ∑ Uncrossed strings

D (2)

2 × String on surface 1

(1)

=

2 s 2 + D 2 − 2s 2( πD / 2)

=

2⎛⎜ s 2 + D 2 − s ⎞⎟ ⎠ ⎝ πD

s

13-18 View factors from the very long grooves shown in the figure to the surroundings are to be determined. Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected. Analysis (a) We designate the circular dome surface by (1) and the imaginary flat top surface by (2). Noting that (2) is flat, F22 = 0

D

summation rule : F21 + F22 = 1 ⎯ ⎯→ F21 = 1

(2)

D A 2 reciprocity rule : A1 F12 = A2 F21 ⎯ (1) = = 0.64 ⎯→ F12 = 2 F21 = πD π A1 2

(1)

(b) We designate the two identical surfaces of length b by (1) and (3), and the imaginary flat top surface by (2). Noting that (2) is flat, F22 = 0

a

summation rule : F21 + F22 + F23 = 1 ⎯ ⎯→ F21 = F23 = 0.5 (symmetry)

(2)

summation rule : F22 + F2 →(1 +3) = 1 ⎯ ⎯→ F2 →(1+3) = 1 b

reciprocity rule : A2 F2 →(1 +3) = A(1 +3) F(1 +3) →2 ⎯ ⎯→ F (1+ 3)→ 2 = F (1+ 3)→ surr =

A2 A(1 +3)

(1) =

(1)

b

a 2b

(c) We designate the bottom surface by (1), the side surfaces by (2) and (3), and the imaginary top surface by (4). Surface 4 is flat and is completely surrounded by other surfaces. Therefore, F44 = 0 and F4→ (1+ 2+ 3) = 1 . reciprocity rule : A 4 F4 →(1 +2 +3) = A(1 +2 +3) F(1 +2 +3) →4 ⎯ ⎯→ F(1+2 +3) →4 = F(1 +2 +3) → surr =

(3)

A4 A(1 +2 +3)

(1) =

a a + 2b

(4) b

b (2)

(3)

(1) a

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13-12

13-19 A circular cone is positioned on a common axis wi...