This is a multi step problem, I have answer A B C and D and need help with E B.

Question

This is a multi step problem, I have answer A B C and D and need help with E

B. (2 pts) Write a balanced combustion reaction for natural gas composed of 90 mol% methane and 10 mol%
ethane. [Fractional moles are convenient for the subsequent calculations. Your reaction should have .90 mol
methane and .10 mol ethane on the left side. Do not include nitrogen.]

my answer: 0.9CH4 +0.2C2H6 +2.5O2 ----> 1.3O2 + 2.4O2

C. (2 pts) Now note that air (not pure oxygen) flows through the combustion chamber. Rewrite your combustion reaction from (b) to include all molecules present, whether or not they participate in the reaction. Assume air is 80 mol% N2 and 20 mol % O2. Assume no excess air is present.

my answer: 80% nitrogen means 4 parts N 20% oxygen means 1 part O inputting this into above reaction gives: 0.9CH4 + 0.2CH2H6 +2.5O2 + 10N2 ----> 2.4H2O +1.3CO2 + 10N2

D. (4 pts) Fuel and air are introduced into the combustion chamber at Ti = 25

Explanation / Answer

enthalpy of formation of compounds

water= -241.82 KJ/mol. ; CO2 = -393.5 KJ/mol. ; CH4 = -74.87 KJ/mol. : C2H6 = -84.68 KJ/mol.

using balanced equation of A:

energy content of natural gas mixture = enthalpy of formation of (products - reactants) =

(1.1*(-393.5) + 2.1*(-241.82) )- (0.9*(-74.87) + 0.1*(-84.48) ) = - 864.841 KJ/mol of natural gas.

energy content = 864.841 KJ/mol.

energy wasted in flue gas heating= 171.8 KJ

remaining energy = 864.841- 171.8 = 693.041 KJ

max .steady state efficiency of furnace = 693.041/864.841 = 0.80135

max % of efficiency = 80.135

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