Exam1SolnwPCME525SP2011

Exam1SolnwPCME525SP2011 - 1 Solution Key ME 525 Exam #1...

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1 Solution Key February 22, 2011 Prof. Lucht Room 262 Potter, 10:30 – 11:45 am EXAM INSTRUCTIONS Write your name on each work sheet. At the end of the exam be sure to turn in all your work pages, and staple any extra work sheets to the work sheets provided. One, three, and five blank pages follow problems 1, 2, and 3, respectively. Please use extra sheets rather than work on the back of a page. Please do not write outside the page borders on the exam sheets. This exam is closed book and closed notes. Equation sheets, thermodynamic tables, and a table of equilibrium constants are attached at the end the exam booklet. When working the problem, list all assumptions, and begin with the basic equations. Pay careful attention to units. On this exam, numerical calculations will be much easier if you work in concentration units of gmol/m 3 . When you us the thermodynamic property table, remember that 1 kJ/kmol = 1 J/gmol Problem #1 /25 Problem #2 /35 Problem #3 /40 Total /100 ME 525 Exam #1
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2 ME 525 Exam #1 22 Feb 2011 Solution Key 1. (25 points) A lean mixture of propane (C 3 H 8 ) and air at an equivalence ratio of 0.7  (142.8% theoretical air) enters the combustion zone of a power-generating gas turbine engine at 700 K, 10 atm, and the products of combustion exit the combustor at 1800 K, 10 atm. (a) What are the major products of combustion? Calculate the number of kmols of each of the major species of combustion produced for each kmol of C 3 H 8 . +10 (b) Carbon monoxide (CO) is not a major species of combustion for lean flames, it is present in very low concentration compared to the major species of combustion. Calculate the mole fraction and the concentration of CO in the products of combustion at 1800 K and 10 atm. +15  22 38 2 2 2 2 2 3 8 2 222 0 , 1 5 5 18.8 3 4 18.8 5 0.7 7.14 0.7 0.7 7.14 26.86 3 4 2.14 26.86 3 4 2.14 26.86 36.00 32 . 1 4 0.0833 x 0.0594 36 36 1800 stoich stoich tot CO O fCO aa CH O N CO HO N a a C H ON C O H O bn x gK     +3 +7 +2 2 2 0 , 0 , 269,164 1800 396,425 1800 0 fO J gmol J gmol J gmol
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3 ME 525 Exam #1 22 Feb 2011 Solution Key  2 2 2 2 22 1 0 2 1/2 0 00 0 0 ,, , 0 1 2 exp 1 1800 1800 1800 2 396,425 269,164 0 127,260 exp 8.314 1800 CO T p CO O u Tf C O f C O f O T p CO O CO x G P K xx P R T G g K g K g K J gmol J G gmol J gmol K J K gmol K           2 2 1/2 0 0 5 55 3 3 3 4930 10 1 0.0833 2.19 10 4930 0.0594 10 2.19 10 10.1325 10 8.314 1800 1.48 10 CO CO pO CO CO u x x Kx PP P atm P atm x J xP m CO RT J k gmol K gmol CO m   +5 +3 +3 +2
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4   2 2 2 2 22 1 1/2 2 3.693 4 2 0 2 1/2 0 0 4 5 1 2 10 2.028 10 10 1 2.028 10 0.0833 2.19 10 0.0594 10 CO O p CO pC O CO O CO CO Alternative solution using table of equilibrium constants : CO CO O xx P K xP Kx x P Pa t m P a t m x CO R      +5 +3 +3 55 3 3 3 2.19 10 10.1325 10 8.314 1800 1.48 10 u J m T J k gmol K gmol CO m   +2
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5 ME 525 Exam #1 22 Feb 2011 Solution Key 2. (35 points) A mixture of H 2 and Cl 2 is placed in a constant-volume reactor at a low pressure of 0.2 bars and at a low temperature of 200 K.
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Exam1SolnwPCME525SP2011 - 1 Solution Key ME 525 Exam #1...

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