If a fuel mixture can be represented by the general formula CxH2x, prove that the stoichiometric gravimetric air/fuel ratio is 14.8:1.
The combustion reaction for a given fuel mixture will look as follows:
CxH2x + 1.5x(O2 + 3.76N2) → xCO2 + xH2O + 1.5x*3.76N2
The molar ratio is 1.5x:1, to find the gravimetric ratio, the masses of the air and fuel should be found. This can be done by multiplying molar ratio with molar weights of the fuel and air.
m(fuel) = 12*x + 2x*1 = 14x
m(air) = 1.5x(16*2 + 3.76*14*2) = 1.5x*137.28 = 205.92x
The gravimetric stoichiometric air/fuel ratio then is 205.92x:14x = 14.71:1
The dry exhaust gas analysis from an engine burning a hydrocarbon diesel fuel is as follows: CO2 = .121, O2=.037, N2 = .842. Determine the molar composition of the fuel, the gravimetric composition of the fuel, the equivalence ratio of the fuel/air mixture under these sampling conditions, the fuel/air ratio on a mass basis under these sampling conditions, and the stoichiometric air/fuel ratio for this fuel.
The chemical equation of fuel combustion is:
CxHy + (x + 0.25y + z)(O2 + 3.76N2) → xCO2 + 0.5yH2O + zO2 + (x + 0.25y + z)*3.76N2
Knowing the results of the exhaust gas analysis, the coefficients in the equation can be found. As the volume ratios correspond to the molar ratios, the following system of equations can be obtained from matching coefficients in the equation and exhaust gas analysis data:
x = 0.121; z = 0.037; (x + 0.25y + z)*3.76 = 0.842
0.121 + 0.25y + 0.037 = 0.224; y = 0.264
The molar ratio C:H of the fuel then is
x:y = 0.121:0.264 = 1:2.182
Having molar ratio, molar composition of the fuel can be found:
%m(C) = 1/(1 + 2.182) = 31.43%
%m(H) = 2.182/(1 + 2.182) = 68.57%
The gravimetric composition of the fuel can be derived from gravimetric ratio which can be found from molar ratio.
Gravimetric ratio x*M(C) : y*M(H) = 12x : y = 12 : 2.182
%w(C) = 12/(12 + 2.182) = 84.61%
%w(H) = 2.182/(12 + 2.182) = 15.39%
Equivalence ratio is defined as the actual fuel/air molar ratio divided by stoichiometric fuel/air ratio. The stoichiometric air is the actual air without excess air. As z represents excess air, for the given fuel equivalence ratio will be:
φ = [1 / (0.121 + 0.25*0.264 + 0.037)] / [1 / (0.121 + 0.25*0.264)] = 4.464 / 5.348 = 0.83
Fuel/air ratio on a mass basis:
F/A = (12x + y) : [(x + 0.25y + z)(16*2 + 3.76*2*14)] = (12*0.121 + 0.264) : [(0.121 + 0.25*0.264 + 0.037)(32 + 105.28)] = 1.716 : 30.751 = 1:17.92
Stoichiometric air/fuel ratio can be found similarly, with the exception that excess air (z) is not included in the calculations:
A/Fs = [(x + 0.25y)(16*2 + 3.76*2*14)] : (12x + y) = 25.671 : 1.716 = 15.01 : 1
Determine the adiabatic combustion temperature at constant pressure and at constant volume for the fuels Nitromethane (CH3NO2) and Methane (CH4), look at Heat of Reaction Activity attached. Graphically show that this corresponds to the product temperature where the heat of reaction is zero.
Exactly .655 grams of rape methyl ester (form of biodiesel fuel) is burned in a bomb calorimeter like the one shown in the video. It has an energy equivalent of 2425 cal/C. The initial temperature was 20.122 C and the final temperature was 22.755 C. The correction factor for the heat of combustion of the fuse wire was 20 cal. Based on these data, determine the gross heat of combustion (higher heating value) in MJ/kg. How does this compare with #2 diesel fuel (42 MJ/kg)?
First of all, the energy released in the combustion should be found:
Q = (T2 – T1) * C = (22.755 – 20.122) * 2425 = 6385.025 cal
The value should be corrected for the heat of combustion of the fuse wire:
Qc = Q – Qw = 6385.025 – 20 = 6365.025 cal
The value should be converted from cal to J:
Qc = 6365.025 cal * 4.184 J/cal = 26631.27 J = 0.02663 MJ
HHV = Qc / m = 0.02663 MJ / 0.655 g = 0.02663 MJ / 0.000655 kg = 40.66 MJ/kg
As it can be noted, the HHV of the tested rape methyl ester is slightly lower than the HHV for #2 diesel fuel. It is 3.2% lower.
