Chapter08 - CHAPTER 8 External Flows 8.1 B A B C A...

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162 CHAPTER 8 External Flows 8.1 8.2 Re . . . . = = = × × = × - - 5 5 1 51 10 20 3 78 10 5 5 VD D n m 8.3 8.4 8.5 ( C ) 8.6 ( C A B C A-B: favorable B-C: unfavorable A-C: favorable separated region inviscid flow boundary layer near surface inviscid flow viscous flow near sphere no separation separation wake separation separated region building inviscid flow boundary layer wake A-B: favorable B-C: unfavorable A-D: favorable C-D: undefined A B C D separated flow
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163 8.7 ( B ) 6 0. 8 0.008 R e 4880. 1.3 1 10 VD n - × = == × 8.8 5 5 5 1 22 10 8 12 000915 5 = = = × × = - V D V n n a) fps. / . . / . . b) V = × × = - 5 388 10 8 12 000291 5 . / . . fps. c) V = × × = - 5 1 6 10 8 12 0 012 4 . / . . fps. 8.9 Re . . . = = × × = × - D D n 20 1 51 10 13 25 10 5 5 a) . . . = × × = × 13 25 10 6 7 9 10 5 6 Separated flow. b) . . . . = × × = × 13 25 10 06 7 9 10 5 4 Separated flow. c) . . . = × × = 13 25 10 006 7950 5 Separated flow. 8.10 F pdA p A p r rdr p p D A = - = - = -  = back back front 0 2 0 0 0 1 1 2 2 1 2 1 4 1 2 ( ) p p p Bernoulli: p V p p + = = × × = 1 2 1 2 1 21 20 242 2 0 0 2 r . . Pa. = = F D 1 2 242 380 p ( ) N C F V A D D = = × × × × = 1 2 2 2 2 2 380 1 21 20 1 0 5 r p . . 8.11 F F F total bottom top 000 .3 .3 +10 000 N. = + = × × × × = 20 3 3 2700 . . F lift cos 10 N = = 2700 2659 o F drag 10 N = = 2700 469 sin o C F L L = = × × × × = 1 2 2 2 2 2659 1000 5 3 3 2 36 r . . . C F D D = = × × × × = 1 2 2 2 2 469 1000 5 3 3 0 417 r . . . 8.12 F p A Lw F p A u u u l l l o = = × = = × = 26 8000 2 5 4015 000 . cos F F F L u = - = l o o cos cos 5 10 21 950 F F F D u = - = l o o sin 5 10 1569
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164 C F V A Lw L L = = × × = 1 2 2 2 2 21 3119 750 0 25 r 950 . . C F D D = = × × = 1 2 2 2 2 1569 3119 750 0 0179 r . . 8.13 If C D = 1 0 . for a sphere, Re = 100 (see Fig. 8.8). × = V . , 1 100 n n V = 1000 . a) V F D = × × = = × × × × - 1000 1 46 10 0146 1 2 1 22 0146 05 1 0 5 2 2 . . . . . . m/ s. p = × - 3 25 10 7 . . N b) V F D = × × × = = × × × × × - 1000 1 46 10 015 1 22 0 798 1 2 015 1 22 798 05 1 0 5 2 2 . . . . (. . ) . . . m/s. p = × - 4 58 10 5 . . c) V F D = × × = = × × × × - 1000 1 31 10 00131 1 2 1000 00131 05 1 0 6 2 2 . . . . . p = × - 6 74 10 6 . . 8.14 a) Re . . . . = = × × = × = - VD C D n 6 5 1 5 10 2 10 0 45 5 5 from Fig. 8.8. = = × × × × × = F V AC D D 1 2 1 2 1 22 6 25 45 1 94 2 2 2 r p N. . . . . b) . . . . = × × = × = - 15 5 1 5 10 5 10 0 2 5 5 C D from Fig. 8.8. = = × × × × × = F V D D 1 2 1 2 1 22 15 25 2 5 4 2 2 2 r p N. . . . . 8.15 ( B ) Assume a large Reynolds number so that C D = 0.2. Then 2 22 1 1 8 0 1000 1.2 3 5 0. 2 4770 N. 2 2 3600 D F V AC rp ×  = = × × ×××=   8.16 ( D Assume a Reynolds number of 10 5 . Then C D = 1.2. 11 . 6 0 1.2 3 4 0 4 1.2 . 0.0041 m. D F V A C DD r = ∴ = ×× × ∴= 5 6 4 0 0.0041 R e 1.6 4 10 . 1.2. The assumption was OK. 10 D C n - × = = =× ∴=
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165 8.17 The velocities associated with the two Re's are V D 1 1 5 5 3 10 1 5 10 0445 101 = = × × × = - Re . . n m/s, V D 2 2 4 5 6 10 1 5 10 0445 20 = = × × × = - . . n m /s. The drag, between these two velocities, is reduced by a factor of 2.5 ( 29 ( 29 [ ] C C D D high low and = = 0 5 0 2 . . . Thus, between 20 m/s and 100 m/s the drag is reduced by a factor of 2.5. This would significantly lengthen the flight of the ball. 8.18 a) F V AC V C V C D D D D = = × × = 1 2 0 5 1 2 00238 2 12 4810 2 2 2 2 r p . . . / . . . : . = = × × = = = × - VD V V C V D n 4 12 1 6 10 2080 5 98 4 Try fps, Re = 2 10 5 Try fps, Re= 2.3 10 5 C V D = = × . : . 4 110 b) C V V D = = × ×  × = 0 2 1 2 00238 2 12 2 155 2 2 . : . . . 0.5 fps. p 8.19 4 2 1 2 1000 1 0 267 2 10 2 10 2 2 2 6 5 . . . . .
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This note was uploaded on 03/10/2008 for the course ME 312 taught by Professor Me during the Winter '08 term at Cal Poly Pomona.

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Chapter08 - CHAPTER 8 External Flows 8.1 B A B C A...

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