ChE150B PSET2

# ChE150B PSET2 - CHEMICAL ENGINEERING 150B PROBLEM SET#2...

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CHEMICAL ENGINEERING 150B PROBLEM SET #2 ARJAN SINGH PUNIANI Problem 1 We know the system is in a pseudo-steady-state. Let us assume that R A = 0, steady- state conditions, or ∂c a ∂t = 0, and unidirectional flux: dN A,z dz . Let us assert that y A, 2 0 and y A, 1 = P A P = 0 . 03737. We can immediately solve for an explicit relation between flux and height: N A,z = - cD AB dy A dz + y A ( N A,z + N B,z ) N A,z = - cD AB 1 - y A dy A dz Separating differentials: N A,z Z z 2 z 1 dz = - cD AB Z y A, 2 y A, 1 dy A (1 - y A ) N A,z ( z 2 - z 1 ) = cD AB ln 1 - y A, 2 1 - y A, 1 A mass balance of the system yields the time-dependence of the height: N A,z = ρ A,L M A dy A dz = cD AB zy B,lm ( y A, 1 - y A 2 ) Z t t =0 dt = ρ A,L ( y B,lm /M A ) cD AB ( y A, 1 - y A, 2 ) Z z ( t ) z (0) z dz Where y B,lm = y A, 1 - y A, 2 ln h 1 - y A, 2 1 - y A, 1 i = 0 . 03737 ln h 1 - 0 1 - 0 . 03737 i = 0 . 9794 To solve for c , c = n V = P RT = 1 atm 0 . 08206 × 298 = 0 . 0409 kgmol m 3 Since D AB = 8 . 44 × 10 - 6 m 2 /s: N A,z ( z 2 - z 1 ) = 1 . 31472 × 10 - 7 mol/cm-s Solving the time ODE above: z 2 ( t ) - z 2 ( t 0 ) = - 2 cD AB y A, 1 ρ A,L y B,lm /M A t = 2 0 . 0409 × 8 . 44(10 - 6 ) × 0 . 03737 0 . 866 × 0 . 9794 92 . 2 × 10 6 t t = ρ A,L M A 1 1 . 472 × 10 - 8 [(Δ z ) 2 - 0 . 01 m 2 ] Since at t = 0, z 2 ( t 0 ) = 0 . 01 m 2 . Thus: t = 0 . 866 92 . 2 1 1 . 314 × 10 - 7 [(Δ z ) 2 - 100 cm 2 ] 1

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2 ARJAN SINGH PUNIANI For an increasing Δ z from 10.0 cm to 13.0 cm, we can compute the time, and, simultane- ously, the flux: Δ z t (in seconds) N A,z 10.0 0 1 . 314 × 10 - 9 10.5 348 861 1 . 252 × 10 - 9 11.0 714 739 1 . 195 × 10 - 9 11.5 1 . 1 × 10 6 1 . 14 × 10 - 9 12.0 1 . 5 × 10 6 1 . 0895 × 10 - 9 12.5 2 . 0 × 10 6 1 . 052 × 10 - 9 13.0 2 . 4 × 10 6 0 . 999 × 10 - 10 Table 1. Tabular of Values We expect the toluene to drain in 670 hours .
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