HW6QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework...

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1 ME 525 SP2011 Homework #6 Questions and Answers Due Thursday, April 28, 2011 Prof. Lucht (E-mail address: [email protected]) Question #1) Do you plan on giving out a data/text file with the output from HPFLAME, or did you want us to go ahead try and run the code for this homework? #2) Do we simply use Roper's experimental correlation (1330*. ..) and change S^-1 according to the listed reactions? For (b) and (c) He and N2 are in equal mole fractions with H2 (fuel) and would act as inert species. Since the stoichiometric equations would have similar molar ratios of O2 to H2, would their flame lengths be equivalent? The book mentions that diluents tend to reduce the flame length, but because only mole fractions appear in the equations, wouldn't He and N2 have the same effects despite their difference in molecular weight? #3) I'm having trouble envisioning what passes through the control volumes. In lecture 26 you gave an equation for T(r) inside and outside the flame sheet, so logically we could take d/dR (T(r)) and find dT/dr at the requested locations. I was trying to use the CV approach, and came up with the relation that the heat conduction emanated from the surface does not travel into the droplet so it must all transfer into the flame sheet. From this relation I got that dT/dr @ r = r_f- = (r_s/r_f)^2 * dT/dr @ r = r_s Basically that the temp. gradient right before the flame sheet is the temperature gradient at the surface of the drop scaled by the ratio of the radii r_s and r_f. Something tells me this is wrong (mainly due to the difference of 1 radius term in the exponential of the numerator when taking d/dr (T(r)) from the equation in Lecture 26). Any help would be appreciated. Answer: (1) I have placed a data file with temperature and mole fractions as a function of equivalence ratio for HW6, Problem 1 (DataHW6P1.txt). (2) You are correct. I have also put at least partial answers to all the problems in a new pdf file of HW6 on the website. (3) Heat conduction at the control volume surface just outside the droplet, r=r_s^+, is from the gas phase to the droplet. This is what drives the evaporation of the droplet. If you take a control volume inside the surface of the droplet, r=r_s^-, however, the temperature gradient is zero because the droplet has a uniform temperature. There is a flux of liquid fuel through the surface at r=r_s^- and a flux of fuel vapor through the surface at r=r_s^+. Question: Example 14.1 illustrates how to find Rdiff for the burning of carbon at 1800 K. It shows how to find the binary diffusion coefficient (Dab is proportional to T^(1.5)). In the table in the appendix, the value for D for CO2,N2 is 1.63e-5 at T = 293 K. To find D at T = 1800 K, I
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2 would find it by: D(1800) = D(293)*(1800/293)^1.5, however the problem shows it as D(1800) = D(293)*(1800/393)^1.5. Why is 393 used instead of 293? Answer:
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This note was uploaded on 12/27/2011 for the course ME 525 taught by Professor Lucth during the Fall '11 term at Purdue.

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HW6QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework...

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