Chapter02

Mechanics of Fluids

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12 C HAPTER 2 Fluid Statics 2.1 Σ ∆ ∆ F ma p z p s y z a y y y y = - = : sin a r 2 Σ ∆ ∆ ∆ ∆ F ma p y p s y z a g y z z z z z = - = + cos a r r 2 2 Since s y cos a = and s z sin a = , we have p p y a y y - = r 2 ( 29 p p z a g z z - = + r 2 Let y 0 and z 0: p p p p y z - = - = 0 0 = = p p p y z . 2.2 p = g h. a) 9810 × 10 = 98 100 Pa or 98.1 kPa b) (0.8 × 9810) × 10 = 78 480 Pa or 78.5 kPa c) (13.6 × 9810) × 10 = 1 334 000 Pa or 1334 kPa d) (1.59 × 9810) × 10 = 155 980 Pa or 156.0 kPa e) (0.68 × 9810) × 10 = 66 710 Pa or 66.7 kPa 2.3 h = p/ g . a) h = 250 000/9810 = 25.5 m b) h = 250 000/(0.8 × 9810) = 31.9 m c) h = 250 000/(13.6 × 9810) = 1.874 m d) h = 250 000/(1.59 × 9810) = 16.0 m e) h = 250 000/(0.68 × 9810) = 37.5 m 2.4 ( C ) (13. 6 9810 ) (28. 5 0.0254 ) 96600 Pa Hg ph g ==×× ×= 2.5 S = 2 0 144 62. 4 20 p h g × = × = 2.31 . r = 1.94 × 2.31 = 4.48 slug/ft 3 . 2.6 a) p = g h = 0.76 × (13.6 × 9810) = 9810 h . h = 10.34 m . b) (13.6 × 9810) × 0.75 = 9810 h . h = 10.2 m . c) (13.6 × 9810) × 0.01 = 9810 h . h = 0.136 m or 13.6 cm . 2.7 a) p = g 1 h 1 + g 2 h 2 = 9810 × 0.2 + (13.6 × 9810) × 0.02 = 4630 Pa or 4.63 kPa . b) 9810 × 0.052 + 15 630 × 0.026 = 916 Pa or 0.916 kPa . c) 9016 × 3 + 9810 × 2 + (13.6 × 9810) × 0.1 = 60 010 Pa or 60.0 kPa . y z p z y p y z p s s α z y ρ g V

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13 2.8 p = r gh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi . 2.9 ( D ) 0 8400 0 1.0 0 9.8 1 400 0 44760 Pa p p gh r = - = -××= 2.10 inside 10 0 9.81 3 13.51 Pa .28 7 253 10 0 9.81 3 11.67 Pa .28 7 293 outsid eo o base i i pg p g hh RT p pg p g RT r r × = = = ×= × ∴∆ × = = = × = 1.84 Pa If no wind is present this p base would produce a small infiltration since the higher pressure outside would force outside air into the bottom region (through cracks). 2.11 p = r gdh where h = –z. From the given information S = 1.0 + h /100 since S (0) = 1 and S (10) = 1.1. By definition r = 1000 S, r water = 1000 kg/m 3 . Then dp = 1000 (1 + h /100) gdh. Integrate: dp h gdh p = + 1000 1 100 0 10 0 ( / ) p = × + × 1000 9 81 10 10 2 100 2 . ( ) = 103 000 Pa or 103 kPa Note: we could have used an average S : S avg = 1.05, so that r avg = 1050 kg/m 3 . 2.12 v = + + p p x i p y j p z k \$ \$ \$ = – r a i x \$ ra y j \$ z k \$ r gk \$ = – ( 29 r a i a j a k x y z \$ \$ \$ + + r gk \$ = – r v a r v g ∴∇ = - + v v v p a g r ( ) 2.13 / 00 [ ( ) /] gR atm p pT zT a a =- = 100 [(288 - 0.0065 × 300)/288] 9.81/.0065 × 287 = 96.49 kPa p p gh = - = - × × × r 100 100 287 288 9 81 300 1000 . . / = 96.44 kPa % error = 96 96 96 100 .44 .49 .49 - × = - 0.052% The density variation can be ignored over heights of 300 m or less.
14 2.14 / 0 0 0 gR at m atm Tz pp p T a a  - ∆=- =-   = 100 288 0065 20 288 1 9 81 0065 287 - × - × . . /. = - 0.237 Pa or - 0.000237 kPa This change is very small and can most often be ignored. 2.15 Eq. 1.5.11 gives 310,00 0 14 4. dp d r r ×= But, dp = r gdh . Therefore, 7 4.46 4 10 gd hd rr r × = or 27 32.2 4.46 4 10 d dh r r = × Integrate, using 0 r = 2.00 slug/ft 3 : 20 32.2 4.46 4 10 h d dh r r r = ∫∫ × . 11 2 r -- = 7.21 × 10 -7 h or 7 2 1 14.4 2 10 h r - = Now, 7 77 00 22 ln( 1 14.4 2 1 0) 1 14.4 2 1 0 14.4 2 10 hh gg p gd h d h r - = = = - × Assume r = const: 2. 0 32. 2 64.4 p g h r = =× a) For h = 1500 ft: p accurate = 96,700 psf and p estimate = 96,600 psf . 96,60 0 96,700 % error 10 0 0.103 % 96,700 - = × b) For h = 5000 ft: p accurate = 323,200 psf p = 322,000 psf. 322,00 0 323,200 % error 10 0 0.371 % 323,200 - = × c) For h = 15,000 ft: p accurate = 976,600 psf p = 966,000 psf. 966,00 0 976,600 % error 10 0 1.085 % 976,600 - = × 2.16 Use the result of Example 2.2: p = 101 e - gz/RT .

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Chapter02 - C HAPTER 2 Fluid Statics 2.1 yz ay 2 yz yz Fz =...

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