Elementary Principles 474 - 11.3 (cont'd) 3000 2500 2000...

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11- 3 11.3 (cont’d) 11.4 a. Air initially in tank: N 0 492 0 0258 = ° ° = 10.0 ft R 1 lb - mole 532 R 359 ft STP lb - mole 3 3 bg . Air in tank after 15 s: PV NR T NRT NN P P ff f f 00 0 0 00258 0 2013 =⇒ == = . . lb- mole 114.7 psia 14.7 psia lb- mole Rate of addition: ± .. n = = 0 2013 0 0258 0 lb- mole air 15 s .0117 lb- mole air s b. Balance on air in tank: Accumulation = input dN dt = 00117 .l b - m o l e s s ; tN = = 0 0 0258 ,. l b - m o l e c. Integrate balance: dN n dt N t Nt 0 0258 0 00258 00117 . ± z z = + lb- mole air Check the solution in two ways:
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