Exercise 3.2.1: Methane and ethane combustion.A mixture of fuels, ethane (C2H6) and methane (CH4), is burned in excess air. The fuel stream enters the furnace at 25°C and 1 atm,while a stream of air enters the furnace at 200°C and 1 atm. At steady state, the flue gas stream leaves the furnace at 800°C and 1 atmand 257 mol/s. The composition of the flue gas is 7.39 mol% 02, 73.2 mol % N2, 12.3 mol% H20, 5.43 mol% CO2, and 1.67 mol% CO.(a)Draw and label a process flow diagram for this process. Number each stream.(b)Calculate the component molar flow rates (mol/s) of the streams entering to the reactor.(c)Calculate the percent excess oxygen.(d)The total molar flow rate exiting the reactor increases by 15%. Will the percent excess of oxygen increase, decrease, or stay thesame?

Note: as per the guideline we are authorized to answer three subparts at a time, since you have not mentioned which part you are looking for, so we are answering the first three subparts, please repost your question separately for the remaining subpart. Given value: The composition of flue gas The temperature of fuel stream = 25 0C Temperature of air stream = 200 0C Temperature of flue gas stream = 800 0C The molar flow rate of flue gas stream = 257 mol/s Composition of flue gas stream is, O2 = 7.39 mol % N2 = 73.2 mol % H2O = 12.3 mol % CO2 = 5.43 mol % CO = 1.67 mol %(a) The process flow diagram is shown below: (b) The flow rate of feed stream is calculated as follows: The molar flow rate of flue gas stream = 257 mol/s Composition of flue gas stream is, O2 = 7.39 mol % N2 = 73.2 mol % H2O = 12.3 mol % CO2 = 5.43 mol % CO = 1.67 mol % Therefore the molar flow rate of the species present in the flue gas becomes. O2= 18.99 mol/s N2= 188.124 mol/s H2O = 31.61 mol/s CO2 = 13.95 mol/s CO = 4.29 mol/s Let he feed consists F1mol/s of CH4 and F2mol/s of C2H6 and F3 mol/s of air The molar flow rate of nitrogen will remain same in the feed and flue gas stream. Therefore the molar flow rate of nitrogen in the feed becomes, N2 fed = 188.124 mol/s The composition of nitrogen in the air = 0.79 Therefore the molar flow rate of air in the feed becomes, F3 = 188.124 mols0.79=238.13 mols Therefore the oxygen fed to the reactor becomes, O2fed = 238.13mols ×0.21 = 50 molsApply the carbon atom balance F1 × 1 + F2 × 2 - 1 ×13.95 - 1 ×4.29 = 0F1 +2 F2 - 13.95 -4.29 = 0 F1 +2 F2 =18.24 ......... (1) Apply the hydrogen atom balance F1 × 4 + F2 ×6 - 2 ×31.61 = 04F1 +6 F2 = 2 ×31.61 F1 +1.5 F2 =15.82 .......... (2) Solve equation (1) and (2) F1 = 8.56 molsF2 = 4.84 mols% F1 = 8.568.56 +4.84 =63.88 %% F2 = 4.848.56 +4.84 =36.12 %(C) For% excess of air assume the complete combustion of fuels. Therefore the theoretical O2 is calculated as follows: For complete combustion of 1 mol of CH4, 2 mol of oxygen are required and for complete combustion of 1 mol of C2H6, 7/2 mol of O2 are required. Therefore the theoretical O2 becomes O2theoretical = 2 × 8.56 + 72 × 4.84 =34.06 molsTherefore the % excess O2 becomes, % excess O2 = O2fed -O2theoreticalO2theoretical ×100= 50-34.0634.06 ×100 = 46.8%