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WRITTEN TEST ENERGY SYSTEMS 09 FEBRUARY 2021 Data Values from tables/Mollier charts Data from person code Person code: 12345678 1 2345678 X 36 Y 260 Problem 1 Question 1) Eq of reaction for stoichiometric combustion H2O mass fraction 0.45 ash mass fraction 0.02 C mass fraction 0.25 H mass fraction 0.05 O mass fraction 0.23 H2O mol weight kgH2O /Mol H2O 18.00 C mol weight kgC / Mol C 12.00 H mol weight kgH / Mol H 1.00 O mol weight kgO / Mol O 16.00 N mol weight kgN / Mol N 14.00 Mol H2O / Mol C 1.2000 Mol H / Mol C 2.4000 Mol O / Mol C 0.6900 Fuel equivalent composition = [CH 2.4O0.69 + 1.2 H2O] weight of volatile fraction kg volatile fraction / Mol C in fuel 47.040 1*12 + 2.4*1 + 0.69*16 + 1.2*18 weight of fuel, including ashes kg fuel / Mol C in fuel 48.000 (we ight of volatile fraction)/(1 - ash mass fraction) Eq of reaction = [CH2.4O0.69 + 1.2 H2O] + (1+ε)(1+ 2.4/2 - 0.69/2)(O2 + 3.76 N2) --> CO2 + (1.2 + 1.2) H2O + ε(1 + 1.2 - 0.345) O2 + (1+ε)(1 + 1.2 - 0.345) * 3.76 N2 Question 2) stove efficiency Excess air ε 0.433 Combustion efficiency % 98.0 Ash discharge temperature °C 180.0 Ash specific heat kJ/kg°C 1.30 Combustion products discharge temperature °C 225.0 Combustion products specific heat kJ/kg°C 1.15 Reference temperature °C 25.00 Log wood LHV MJ/kg fuel 10.00 CO2 H2O 12.00 43.20 kg in comb products / Mol C in fuel 1.0000 2.4000 y = kg / total kg 0.0333 0.1197 x = Mols / total Mols 0.0704 0.1690 Mols in products / Mol C in fuel Energy Systems Solution of exam give 09 Feb 2021 Problem 1 - page 1 O2 N2 Total combustion products kg of comb products / kg of fuel 7.51760 [Total kg of products / Mol C in fuel] / [kg fuel / Mol C in fuel]  ash = stove efficiency due to ash discharge % 0.0403 cp = stove efficiency due to discharge of comb products %…