ChE150B PSET2

Che150b pset2

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Unformatted text preview: 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 RA = 0, steadydNA,z state conditions, or ∂ca = 0, and unidirectional flux: dz . Let us assert that yA,2 0 ∂t PA and yA,1 = P = 0.03737. We can immediately solve for an explicit relation between flux and height: −cDAB dyA dyA + yA (NA,z + NB,z ) ⇒ NA,z = NA,z = −cDAB dz 1 − yA dz Separating differentials: z2 yA,2 NA,z z1 dz = −cDAB yA,1 1 − yA,2 dyA ⇒ NA,z (z2 − z1 ) = cDAB ln (1 − yA ) 1 − yA,1 t A mass balance of the system yields the time-dependence of the height: NA,z = Where yB,lm = To solve for c, P 1 atm kgmol n = = = 0.0409 V RT 0.08206 × 298 m3 −6 m2 /s: = 8.44 × 10 c= NA,z (z2 − z1 ) = 1.31472 × 10−7 mol/cm-s Solving the time ODE above: z 2 (t)−z 2 (t0 ) = −2 cDAB yA,1 ρA,L 0.0409 × 8.44(10−6 ) × 0.03737 1 t⇒t= t=2 [(∆z )2 −0.01m2 ] 0.9794 6 ρA,L yB,lm /MA MA 1.472 × 10−8 0.866 × 92.2 × 10 1 0.866 [(∆z )2 − 100 cm2 ] 92.2 1.314 × 10−7 1 ρA,L dyA cDAB = (yA,1 − yA2 ) ⇒ MA dz zyB,lm yA,1 − yA,2 ln 1−yA,2 1−yA,1 dt = t=0 ρA,L (yB,lm /MA ) cDAB (yA,1 − yA,2 ) = 0.9794 z (t) z dz z (0) = ln 0.03737 1−0 1−0.03737 Since DAB Since at t = 0, z 2 (t0 ) = 0.01 m2 . Thus: t= 2 ARJAN SINGH PUNIANI For an increasing ∆z from 10.0 cm to 13.0 cm, we can compute the time, and, simultaneously, the flux: ∆z t (in seconds) NA,z 10.0 0 1.314 × 10−9 10.5 348 861 1.252 × 10−9 11.0 714 739 1.195 × 10−9 6 11.5 1.1 × 10 1.14 × 10−9 6 12.0 1.5 × 10 1.0895 × 10−9 6 12.5 2.0 × 10 1.052 × 10−9 13.0 2.4 × 106 0.999 × 10−10 Table 1. Tabular of Values We expect the toluene to...
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This note was uploaded on 12/02/2010 for the course CHE 150B taught by Professor Bell during the Spring '10 term at Berkeley.

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