Liquid methanol is burned with excess air (i.e. O_2 and N_2). The pure liquid me
ID: 948063 • Letter: L
Question
Liquid methanol is burned with excess air (i.e. O_2 and N_2). The pure liquid methanol enters the combustion chamber at rate of 5.45 L/h. Air enters the combustion chamber at 6.10 times 10^2 ft^3 / h. at a gauge pressure of 1.00 atm and a temperature of 170 degree F. Assume atmospheric pressure at this location is 1.00 atm. Methanol data: Formula: CH_3 OH; molecular weight = 32.04; Specific Gravity = 0.792 Determine the molar flow rate (mol) of the liquid methanol Determine the molar flow rate of oxygen in the incoming air stream (assume ideal behavior). Write the balanced equation for the complete combustion of methanol (CH_3OH) and calculate the percent excess (%XS) oxygen. If you did not gel answers for part (a) or (b), use 200 mol/h CH_3OH and 600 mol/h O_2 (these are not the correct values but will allow you to complete this part of the problem).Explanation / Answer
Density of methanol =791 kg/m3= 791g/L
Flow rate of liquid methanol= 5.45 L/h
Mass flow rate of liquid Methanol (CH3OH)= 5.45*791 gm=4310.95gm
Molecular weight of CH3OH= 12+3+16+1= 32
Molar flow rate of methanol= 4310.95/32=134.7172 moles/hr
2.Flow rate of air =6.1*100 =610ft3/hr, pressure = 1+1=2 atm and Temperature= 170F= 170+460R= 630R
From gas law, PV=nRT, n= moles of Air = PV/RT
R= Gas constant=0.7302ft3.atm/mole.R
Number of moles of air/hr , n= 2*610/(0.7302*630)=2.33moles/hr
Air contains 79%Nitrogen and 21% Oxygen , molar flow rate of oxygen =2.33*0.21=0.4893 moles/hr
b) The balanced reactino for combustion of CH3OH is
CH3OH+1.5O2--àCO2+2H2O
1 mole of CH3OH requires 1.5 moles of oxygen
200moles of CH3OH/hr requires 200*1.5 =300 moles/ hr of oxygen
Oxygen supplied =600 moles/Hr
Excess oxygen =600-300 =300 moles/hr
Percentage of excess oxygen =100*300/300= 100%
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