Elementary Principles 32 - 3.35 P = Patm gh Pinside b g = 1...

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3- 16 3.35 Δ PP g hP =+− atm inside ρ bg =− 1 atm 1 atm + 1051000 . b g kg 9.8066 m 150 m 1 m 1 N m s 100 cm 1 kg m / s 22 32 2 F == × × F H G I K J = 154 100 10 022481 1 2250 4 N 65 cm cm N lb N 2 2 f f . . 3.36 mV ×× 14 62 43 269 10 7 .. . lb 1 ft 2.3 10 gal ft 7.481 gal lb m 36 3 m PP g h =+ 0 × 14 7 14 6243 1 2 . lb in lb 32.174 ft 30 ft 1 lb ft ft s 32.174 lb ft / s 12 in f 2 mf 2 m 2 = 32 9 . p s i — Structural flaw in the tank. — Tank strength inadequate for that much force. Molasses corroded tank wall 3.37 (a) m head 33 m 3 m in 1 ft 8.0 62.43 lb ft lb = × = π 24 3 4 392 2 Wmg s = head m 2 f 32.174 ft 32.174 lb ft / s 1 2 / () f net gas atm 2 f 3 2f f 30 14.7
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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