Respuesta :
higher heating value = 31199.344 kJ/kg
Volatile matter(%) = 89.5 - fixed carbon (%)
i) mass air/fuel ratio
basis of 1 kg of coal
Species
Weight Percentage (%)
mass (g) No. of moles(mol)
Carbon C 70 0.700*1000 = 700 700/12 = 58.3333
Hydrogen H2 6 0.060*1000 = 60 60/2 = 30
Oxygen O2 8 0.080*1000 = 80 80/32 = 2.5
Sulfur S 3.3 0.033*1000 = 33 33/32 = 1.01325
Nitrogen N2 12.7 0.127*1000 = 127 127/28 = 4.5358
Species reaction oxygen required(mol)
Carbon C C+02 + CO 58.3333*1 = 58.3333
Hydrogen H2 H2 +502 + 2H20 30*0.5 = 15
Sulfur S S + O2 + SO , 1.01325*1 = 1.1325
Total oxygen required = 58.3333 + 15 + 1.1325 = 74.4658 mol
Oxygen available = 2.5 mol
Theoretical oxygen required =74.4658 - 2.5 = 71.9658 mol
Percentage excess air = 25%
Percentage \, excess\, air=\frac{O_{2}\, supplied-O_{2}\, theoretial}{O_{2}\, theoretial} \times 100
25 = O2 supplied – 71.9658 71.9658 x 100
O_{2}\, supplied = 89.9573 \, mol
air supplied = \frac{O_{2}\, supplied}{0.21} = \frac{89.9573}{0.21}=428.3681\, mol
molecular mass of air = 29 g/mol
mass of air supplied = 428.3681*29 = 12422.67 g = 12.42kg
the mass ratio of air to fuel = 12.42/1 = 12.42
(ii) The higher heating value(HHV)
HHV(kJ/kg) = 33823\, \, \textbf{C} +144249\, \left (\boldsymbol{\textbf{} H}-\frac{\textbf{O}}{8} \right )+9418\, \, \textbf{S}
where C, H, O, S are the mass fractions of elements from the ultimate analysis
HHV(kJ/kg) = 33823 \times 0.70 +144249 \left (0.06-\frac{0.08}{8} \right )+9418 \times 0.033
KJ HHV(kJ/kg) = 31199.344 kg
higher heating value = 31199.344 kJ/kg
(iii) Percentage of volatile matter
Fixed carbon (%) + Volatile matter (%) = 100 - moisture (%) - ash (%)
Fixed carbon (%) + Volatile matter (%) = 100 - 4 - 6.5 = 89.5
Volatile matter(%) = 89.5 - fixed carbon (%)
here Fixed carbon % is not given. So more information is needed to find Volatile matter (%). so we are not able to find the volatile matter %.
To learn more about the heating value and volatile matter the link is given below:
https://brainly.com/question/13039778?
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