Elementary Principles 160 - 5.72 a. For N 2 : Tc = 126.20 K...

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5- 48 5.72 a. For N T K R, P atm 2c o c :. . . == = 126 20 22716 335 After heater T R R P psia atm atm 14.7 psia z1 . 0 2 r o o r : . . . . . U V | | W | | ⇒= 609 7 268 600 1 12 ± . n= 150 SCFM 359 SCF / lb - mole lb - mole / min = 0418 ± ± .. . /min V= zRTn P lb - mole min psia lb - mole R R 600 psia . 3 o o 3 = = 102 0 418 10.73 ft 609 7 4 65 ft b. tank = 0.418 lb - mole min lb lb - mole 0.81 lb ft min h h day days week weeks m m 3 28 62 4 60 24 7 2 / ./ bg 4668 34 900 ft gal 3 , 5.73 a. For CO T K, P atm cc . = = 1330 34 5 Initially T K 133.0 K P 2514.7 psia 34.5 atm atm 14.7 psia z =1.02 r1 r1 Fig. 5.4-3 . U V | | W | | 300 226 1 50 n psia 1.02 L 300 K atm 14.7 psia mol K 0.08206 L atm mol 1 = = 2514 7 150 1 1022 . After 60h T K 133.0 K P 2258.7 psia 34.5 atm atm 14.7 psia z =1.02 r1 r1 Fig. 5.4-3 . U V | | W | | 300 1 45 n psia 1.02 L 300 K atm 14.7 psia mol K 0.08206 L atm mol 2 = = 2259 7 150 1 918 . ± . nn 60 h mol / h leak = 173 b. ny n y PV RT mol CO mol air atm 0.08206
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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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