homework_20.1_sol

homework_20.1_sol - Homework #2 Solution 12.1) [a] 6...

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Unformatted text preview: Homework #2 Solution 12.1) [a] 6 marbles 6 marbles = = 0.1( marbles / sec ) 1 min 60 sec 0.005 kg marble e 0.1 = 5 104 ( kg/ sec ) marble sec A= [b] [c] D2 =< 4 4 ( 0.025m ) 2 4.909 10-4 m 2 marble -4 2 number flux = 0.1 4.909 10 m sec ( ) marble 203.72 2 m sec < 5 [d] mass flux =' 10 12.2) 4 kg -4 2 4.909 10 m sec ( ) kg 1.02 2 m sec k = 1.21 ( T -T ) J Because, Q = A in out = 105 k , h sec J J we have Tin - Tout =105 3100 m 2 1.21 0.05 m / 1.33K sec m K sec so, Tin = 301.33K 12.3) W J = 1.21 ; m K sec m K A = ( 50 m +< 30 m ) 2 10 m +B m 30 m = 3100 m 2 ; 50 V A P = At x 8 A2 P x= 8 Q 2 2 2.5 10-4 m ( 2.346 - 1) Pa = -4 -12 3 8 8.6 10 Pa sec 1 10 m ( ( ) ) 2.4m 12.4) Increase. 12.11) d J fluid < 2300 d Q d 4Q < 2300 < 2300 A d2 Laminar flow: 4 178m3 / sec 1000 kg/ m3 4Q d > = km 115 2300 2300 8.6 10-4 Pa sec ( ( ) ( ) ) d actural = 1.2d V A P = At x 8 128 Qx P = 2d 4 = 138 km 128 -4 Pa sec 8.6 10 7.48 10-8 Pa ( 2 ( 138 km ) ) ( 178m 4 3 / sec ) 43.5 km Not practical. ...
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This note was uploaded on 04/26/2009 for the course ENGR 112 taught by Professor V during the Spring '08 term at Texas A&M.

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