HW3QuestionsAnswersME525SP2011

HW3QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework#3...

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1 ME 525 SP2011 Homework #3 Questions and Answers Due Thursday, February 17, 2011 Prof. Lucht (E-mail address: [email protected]) Question: can you put a .xls version of the spreadsheet on the website. i can not read .xlsx files from my computer thank you Answer: I have put .xls and .csv versions on the website. Question: On part a. I invoke the partial equilibrium assumption to get k_1f[H][O_2]=k_1r[O][OH] and k_2f[OH][H_2]=k_2r[H][H2O]. I calculate the rate coefficients using A,b,E_a(converting E_a to J/gmol) and calculate the unknown rate coefficients using K_eq via the given curve fit for gibbs free energies. When I use the given concentrations from the CHEMKIN results at the different times, I am unable to prove that the partial equilibrium holds for either time. Below are my values for the rate coefficients: k_1f=1.992e10 cm^3/gmol-s k_1r=1.203e13 k_2f=1.066e12 k_2r=1.115e9 Am I trying to prove the wrong assumption from above? Thanks again. Answer: Your rate coefficients look close but they do not match mine at either the 1.90 ms condition (T=935.1 K) or the 1.98 ms condition (T=1818.7 K). What temperature did you use in your calculations? Question: I have a problem understanding the meaning of the term "evaluate the applicability of the partial equilibrium assumption ". I found K1 and K2, I took down all the mole fractions at time 1.95ms, for 1.98ms I should interpolate right? How do I check the applicability of the partial eq. assumption? Answer: The partial equilibrium assumption holds when the forward reaction rate equals the backwards reactions rate. For reaction 1, partial equilibrium holds when k 1f [H][O2] = k 1r [O][OH] There is no need to interpolate, t = 1.98 ms = 1.98e-03 s is a table entry. Question: When I calculate [N]_ss via k_1[O][N_2]-k_2[N][O_2]-k_3[OH][N]=0, do I use the initial concentrations and conditions to calcualte [N]_ss. It seems like [N]_ss would be a constant and wouldn't change with time so it would be the same for t=0, 1.98 or 3ms. Thanks for any help.
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2 Answer: The term steady state is always confusing. The steady state concentration [N]_ss varies drastically with time in this flame. Steady state for a radical means that its concentration
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HW3QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework#3...

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