ME476_HW4_Soln

ME476_HW4_Soln - HW #4 Problem 3.2 3.3 3.10 (a)-(c) And,...

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Double check N 2 CO 2 + O 2 + 1 = _______________________________________________________________________________ Problem 3.3 Dry exhaust gas composition given for propane, fine equivalence ratio CO 2 10.8% = O 2 4.5% = CO 0% = H 2 0% = Combustion Equation C 3 H 8 λ 5 ( ) O 2 3.773N 2 + ( 29 + 3CO 2 4H 2 O + 5 λ 1 - ( 29 O 2 + 5 3.773 λ N 2 + = n tot.dry 3 5 λ + 5 - 18.865 λ + = CO 2 10.8% = 3 n tot.dry = 3 3 5 λ + 5 - 18.865 λ + ( 29 = solve for lambda Given .108 3 3 5 λ + 5 - 18.865 λ + ( 29 = Find λ ( 29 1.248 = λ 1.248 = φ 1 λ = φ 0.801 = HW #4 Problem 3.2 3.3 3.10 (a)-(c) And, problem: #3. Ethanol is supplied to small SI engine at 3 g/s. The fuel is combusted with 100% excess air. Calculate the mass flow rate of air to the engine. _______________________________________________________________________________________ Problem 3.2 Calculate DRY exhaust gas composition of butane w/ equivalence ratio= 0.9 φ 0.9 = λ 1 φ = λ 1.111 = Combustion Equation
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This note was uploaded on 01/25/2011 for the course ME 476 taught by Professor Broch during the Spring '10 term at Nevada.

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ME476_HW4_Soln - HW #4 Problem 3.2 3.3 3.10 (a)-(c) And,...

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