ChE150B PSET2

# Png figure 2 graph relating the toluene level to time

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Unformatted text preview: drain in 670 hours . as a Function of Time.png Figure 1. Graph relating Molar Flux to Time CHEMICAL ENGINEERING 150B PROBLEM SET #2 3 Level as a Function of Time.png Figure 2. Graph relating the Toluene Level to Time Problem 2 (a). Use spherical coordinates: r, θ, φ 4 ARJAN SINGH PUNIANI Some reasonable assumptions include: 1. Axisymmetric concentration gradient ﬂow (No dependencies on θ or φ). 2. Steady-state ∂c ∂t =0 . 3. Constant total c and DAB . 4. Assume that diﬀusion forces dominate convective forces. (b). Start with ∂c − RA = 0 ⇒ · NA,r = RA ∂t The gradient reduces to the radial term ﬂux change only, from Eqn. (25-29): NA + 1 d(r2 NA,r ) cA = −RA,max 2 r dr (KA + cA ) Where NA,r = −cDAB yA + yA (NA + NB ) Substituting to above: d −DAB r12 dr r2 dcA dr cA = −RA,max (KA +cA ) DAB d cA dr By the no-penetration principle, the ﬂux across the boundary of the inert pellet is zero. By the coordinates above, this translates to: d dr cA r=0.005 m =0 CHEMICAL ENGINEERING 150B PROBLEM SET #2 5 Similarly, at the boundary between the bioﬁlm and the eﬀectively inﬁnite volume it is situated in: cA (0.01) = 0.002 mol/m3 (c). Since: 1d r2 dr r2 dcA dt = RA,max cA 1d ⇒2 DAB KA r dr r2 dcA dr − 0.7307 × 105 cA = 0 Which we should input, in the simpliﬁed form, to WolframAlpha as c (r) + 2rc (r) − 73030c(r) = 0, c(0.01) = 0.002, c (0.005) = 0 c'' ￿r￿ ￿ 2r c' ￿r￿ ￿ 0.7303￿10^5 c￿r￿ ￿ 0, c￿0.01￿ ￿ 0.002 Computation for P2 Cropped.pdf Input interpretation: ODE classification: ￿c￿￿ ￿r￿ ￿ 2 r c￿ ￿r￿ ￿ 0.7303 ￿ 105 c￿r￿ ￿ 0, c￿0.01￿...
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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 University of California, Berkeley.

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