ChE150B PSET2

# R c c r c c 001 1 generated by wolframalpha

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Unformatted text preview: ￿ 0.002￿ Second￿order linear ordinary differential equation c ￿r￿ ￿ 2 r c ￿r￿ ￿ 73 030. c￿r￿ 2 Alternate forms: ￿￿ ￿ Differential equation solution: c￿r￿ ￿ ￿￿r ￿ 7.481260885311541 ￿ 1080 870 d1 H￿36 516. ￿r￿ ￿ d1 1 F1 18 258.; 1 2 c￿0.01￿ ￿ 0.002 Plots of sample individual solution: ; r2 ￿ 1.997379776974881 ￿ 1080 867 H￿36 516. ￿r￿ c￿ c r c c￿ ￿0.01￿ ￿ 1 Generated by Wolfram|Alpha (www.wolframalpha.com) on September 10, 2010 from Champaign, IL. © Wolfram Alpha LLC—A Wolfram Research Company Figure 3. WolframAlpha solution to the second-order ODE 1 Examining the sample individual solution, we can gain conﬁdence that the bioﬁlm’s concentration gradient proﬁle will share many similarities to the hypergeometric function, including: (1) A high concentration at the edge, corresponding to the diﬀusive ﬂux from the exterior volume and (2) A rapid falling-oﬀ with increasing length but with some maintenance of linearity due to a lower Peclet number (computation not shown). The actual solution, if the Hermite polynomials were solved and hypergeometric conditions obeyed would resemble the following manually-produced graph: 6 ARJAN SINGH PUNIANI Problem 3 Conversions: 0.02 mg/mm3 = 20 kg/m3 , 5 µg/cm3 = 0.005 kg/m3 and 1.2 mmol/(hr-g) = 3.33 ×10−7 mol/(g-s). Assume that 1. Steady-state. 2. Diﬀusion is uniaxial (in spherical coordinates: r, but neither θ nor φ). 3. Constant total c, DAB . 4. A is dilute in solution ⇒ yA 0. The general equation ·N+ Which becomes 1d 2 r NA,r − RA = 0 r2 dr Since NA,r = −cDAB Then DAB d r2 dr Integrating yields − r2 dcA dr − RA = 0 ⇒ d[r2 (dcA /dr)] = −RA r2 dr DAB dcA dyA = −DAB dr dr ∂c − RA = 0 ∂t r2...
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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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