Chapter 12 Solutions - View Factor Assignment PDF

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View Factor Assignment...

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12-8 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. 12-6,

L3 1   0.5  W 2 F  0.24  31 L1 1   0.5 W 2 

W=2m L2 = 1 m

L1 = 1 m

and

A2

(2)

A1

(1)

L3 = 1 m

A3

(3)

L3 1   0.5 W 2   F3 (1 2) 0.29 L1  L2 2   1 W 2  We note that A1 = A3. Then the reciprocity and superposition rules gives A1 F13  A3 F31  F13  F31  0.24 F 3

Finally,

(1 2)

F 31 F 32 

0.29 0.24 F 32  F 32 0.05

A2  A3   F23  F32 0.05

12-13 The view factors from the base of a cube to each of the other five surfaces are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis Noting that L1 / w  L2 / w 1 , from Fig. 12-6 we read

(2)

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

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

(1) 12-28 The base, top, and side surfaces of a furnace of cylindrical shape are black, and are maintained at uniform temperatures. The net rate of radiation heat transfer to or from the top surface is to be determined. Assumptions 1 Steady operating conditions exist 2 The surfaces are T1 = 700 K black. 3 Convection heat transfer is not considered. 1 = 1 Properties The emissivity of all surfaces are  = 1 since they are black. r1 = 2 m Analysis We consider the top surface to be surface 1, the base surface to be surface 2 and the side surfaces to be surface 3. The cylindrical furnace can be considered to be three-surface enclosure. We assume that h =2 m T3 = 500 K steady-state conditions exist. Since all surfaces are black, the radiosities are equal to the emissive power of surfaces, and the net rate of radiation 3 = 1 heat transfer from the top surface can be determined from 4 4 4 Q  A1 F12  (T1  T 2 )  A1F13 (T1  T 3 4 )

and

A1  r   (2 m) 12 .57 m 2

2

2

T2 = 1200 K

2 = 1

r2 = 2 m

The view factor from the base to the top surface of the cylinder is F12 0.38 (From Figure 12-44). The view factor from the base to the side surfaces is determined by applying the summation rule to be F11  F12  F13 1   F13 1  F12 1  0.38 0.62 4 4 4 4 Q  A1 F12 (T1  T2 )  A1 F13 ( T1  T3 )

Substituting,

 (12.57 m 2 )(0.38)(5.6710 -8 W/m 2 .K 4 )(700 K 4 - 500 K 4 )  (12.57 m2 )(0.62)(5.6710-8 W/m 2 .K4 )(700 K4 -1200 K 4 )

 7.62 10 5 W -762 kW Discussion The negative sign indicates that net heat transfer is to the top surface. D2 = 0.5 m T2 = 500 K

12-30 Two very long concentric cylinders are maintained at uniform temperatures. The net rate of radiation heat transfer between the two cylinders 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 1 = 1 and 2 = 0.7. Analysis The net rate of radiation heat transfer between the two cylinders per unit length of the cylinders is determined from

D1 = 0.2 m T1 = 950 K 1 = 1

2 = 0.7

Vacuum

A ( T1 4  Q 12  1 1 1  2  1 2

T2 4 )

[ (0.2 m)(1 m)](5.67 10  8 W/m 2 K 4 )[(950 K) 4  (500 K )4 ]  1 1  0.7  2     1 0.7  5 

 r1    r   2  22,870 W 22.87 kW

12-43 Two perpendicular rectangular surfaces with a common edge are maintained at specified temperatures. The net rate of radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are 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 the horizontal rectangle and the surroundings are  = 0.75 and  = 0.85, respectively. Analysis We consider the horizontal rectangle to be surface 1, the vertical rectangle to be surface 2 and the surroundings to be surface 3. This system can be considered to be a three-surface enclosure. The view factor from surface 1 to surface 2 is determined from

L1 0.8   0.5 W 1.6 F 0.27  L2 1.2  12  0.75 W 1.6 

T2 = 550 K 2 = 1

W = 1.6 m

(Fig. 12-6)

The surface areas are

A1  (0.8 m )(1.6 m ) 1.28 m2 A2  (1.2 m)(1.6 m ) 1.92 m 2

L2 = 1.2 m L1 = 0.8 m

A2

(2)

A1

(1) T1 =400 K 1 =0.75

(3) T3 = 290 K 3 = 0.85

A3  2 

1.2 0.8  2

0.82  1.2 2 1.6 3.268 m 2

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 A1 F12  A2 F21   (1. 28)(0.27 ) (1.92 ) F21   F21 0.18

(reciprocity rule)

F 11  F 12  F 13 1  0  0.27  F 13 1  F 13 0.73

(summation rule)

F 21  F22  F23 1   0.18 0  F23 1   F23 0.82

(summation rule)

A1 F13  A3 F31

 (1.28)(0.73) (3.268) F31   F31 0.29

(reciprocity rule)

A2 F23  A3 F32   (1.92)(0.82) (3.268) F32   F32 0.48 (reciprocity rule) We now apply Eq. 12-35 to each surface to determine the radiosities.

1  1  F12 ( J 1  J 2 )  F13 ( J 1  J 3 ) 1 1  0.75  0.27( J 1  J 2 )  0.73( J 1  J 3 ) (5.67 10  8 W/m 2 .K 4 )(400 K ) 4 J 1  0.75

T1 4 J 1 

Surface 1:

T 2 4  J 2   (5.6710 8 W/m 2 .K 4 )(550 K ) 4  J 2 1  3  F31 (J 3  J 1 )  F32 (J 3  J 2 ) T3 4  J 3  3

Surface 2:

Surface 3:

8 2 4 4 (5.6710  W/m .K )(290 K )  J 3 

1  0.85  0.29(J 1  J 2 )  0.48( J 1  J 3 ) 0.85

Solving the above equations, we find

J1 1587 W/m2 , J2 5188 W/m2 , J3 811.5 W/m 2 Then the net rate of radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are determined to be   A F ( J  J ) (1.28 m2 )(0.27)(1587 5188)W/m 2 1245 W Q  Q 21

12

1 12

1

2

  A F ( J  J )  (1.28 m2 )(0.73)(1587  811.5)W/m 2 725 W Q 13 1 13 1 3...