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Unformatted text preview: 0.002 Secondorder linear ordinary differential equation c r 2 r c r 73 030. cr
2 Alternate forms: Differential equation solution: cr r 7.481260885311541 1080 870 d1 H36 516. r d1 1 F1 18 258.; 1 2 c0.01 0.002 Plots of sample individual solution: ; r2 1.997379776974881 1080 867 H36 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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