HW3SolnME525SP2011 - ME 525 Homework #3 Due Thursday,...

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1 ME 525 Homework #3 Due Thursday, February 17, 2011 Prof. Lucht (E-mail address: Lucht@purdue.edu) 1. The CHEMKIN code was used to perform a calculation for constant-volume combustion of a mixture of 20% H 2 , 8%O 2 , and 72% N 2 (on a molar basis) at an initial temperature of 900 K and pressure of 0.8 atm. A time step of 5.0e-06 sec (0.001 msec) was used and the total calculation time was 4.0e-03 sec (4 msec). The results of the calculation are contained in an Excel file on the class website (CHEMKINDataHW3P1StudentVersion.xlsx). (a) Using the results of the CHEMKIN calculation and the data from the H-O-N kinetic mechanism from Example 6.3 on pp. 195-196 in the text, evaluate the applicability of the partial equilibrium assumption for the following two reactions at times of 1.90 and 1.98 msec: 2 22 1 2 (1) (2) k HO OO H k OH H H H O    The following curve fit expressions for Gibbs free energies of formation may be useful:  2 02 , , , , 221,290 55.172 0.0010263 251,230 63.254 0.0007315 38,742 15.636 0.00042668 245,370 52.615 0.0010741 fH fO fOH fHO J gT T T gmol J T T gmol J T T gmol J T T gmol   (b) Use the three-step extended Zeldovich mechanism (Eqns. 21-23, from the H-O-N kinetic mechanism from Example 6.3, p. 196, Turns) to calculate d[NO]/dt for times of 1.98 and 3.0 msec. Neglect reverse reactions and assume that the N-atom is in steady state, but use actual computed values for [O], [O 2 ], [N 2 ], and [OH]. Comment on the relative effects of T and [O]. Note: If you wish, you can obtain the CHEMKIN code from me via email or by stopping by my office, ME 87. However, the results that you need for this problem are already contained in the Excel file. Solution Time (ms) T (K) P (atm) x H x O2 x O 1.90 9.35120E+02 8.29840E-01 1.37480E-03 7.76180E-02 1.17620E-04 1.98 1.8187E+03 1.54070E+00 3.7519E-02 6.8938E-03 5.0354E-03 3.00 2.33790E+03 1.91580E+00 2.28270E-03 9.15980E-05 5.48660E-05
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2 Time (ms) x OH x H2 x H2O x N2 P/R u T (gmol/cm 3 ) 1.90 7.02750E-05 1.94810E-01 4.78170E-03 7.21190E-01 1.083E-05 1.98 7.9454E-03 4.6014E-02 1.41110E-01 7.55480E-01 1.032E-05 3.00 1.53530E-03 4.32530E-02 1.71780E-01 7.81000E-01 9.987E-06 Partial equilibrium will hold for reactions 1 and 2 when:        11 2 22 2 2 ff r r r r Rk H ORk O O H R k O HH R k HHO  From the H-O-N mechanism in Turns: 3 14 0.5 1 91 . 3 2 3 . 3 3.61 10 3626 4.187 1.17 10 exp 8.314 1826 1.17 10 exp r f cm kT gmol s cal J gmol cal J T gmol K cm T Tg m o l s     The forward and reverse rates coefficients are related through the equilibrium constant:      2 12 1 1 1 2 0 00 0 0 0 1 , , , , exp fr r p eq T p T fO fOH fH u OO H kHO k kK HO G KG g g g g RT      2 2 0 0 0 0 2 , , , , exp f rf p eq T p T fHO u k OH H kk HHO K G g g g g
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3  2 02 , , , , 221,290 55.172 0.0010263 251,230 63.254 0.0007315 38,742 15.636 0.00042668 245,370 52.615 0.0010741 fH fO fOH fHO J gT T T gmol J T T gmol J T T gmol J T T gmol   Time (ms) T (K) P (atm) g 0 f,H g 0 f,O2 g 0 f,O 1.90 935.1 0.829840 168,800 0. 191,440 1.98 1818.7 1.54070 117,470 0. 133,770 3.00 2337.9 1.91580 86,690 0. 99,350 Time (ms) g 0 f,OH g 0 f,H2 g 0 f,H2O g 0 f,N2 1.90 24,490 0. -195,230 0.
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This note was uploaded on 04/17/2011 for the course ME 525 taught by Professor Lucth during the Spring '11 term at Purdue University-West Lafayette.

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HW3SolnME525SP2011 - ME 525 Homework #3 Due Thursday,...

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