(d) Explain the concept of heating value of a fuel and the difference between lo
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Question
(d) Explain the concept of heating value of a fuel and the difference between lower and higher heating values. (e) For two identical piston-cylinder assembly A and B, initially with the same amount of fuel and air mixture, same initial temperature and pressure. The piston in A is fixed such that combustion occurs under constant volume, while piston in B can move freely such that combustion occurs under constant pressure. Assume both adiabatic, which one has a higher combustion temperature? Write down 1 law for both systems to explain. (f) (bonus 5pts) Given 1 mole of CH/air mixture at stoichiometric condition, what is the mass of extra air needed for mixing such that we can prepare a CH/air mixture with equivalence ratio of 0.7? Assume the mean molecular weight of air is 29 g/mol, and the mole fraction of O, is 0.21 in airExplanation / Answer
d).
The lower heating value (also known as net calorific value) of a fuel is defined as the amount of heat released by combusting a specified quantity of fuel (initially at 25°C) and returning the temperature of the combustion products to 25°C, which assumes the latent heat of vaporization of water in the reaction products is not recovered.
The higher heating value (also known gross calorific value or gross energy) of a fuel is defined as the amount of heat released by a specified quantity of fuel (initially at 25°C) once it is combusted and the products have returned to a temperature of 25°C, which takes into account the latent heat of vaporization of water in the combustion products.
e) Adiabetic flame temperature is more in constant volume than constant pressure procees. You can compare it from Otto cycle and diesel cycles.
du=dq-pdv
for constant volume dv=0, so du=dq hence temperature rises more, du=dq;
for constant pressure pdv work is there so internal energy is reduced and therefore temperature is also reduced.
f) equivalence ratio = actual air fuel ratio/stoichiometric air fuel ratio
stoichiometric air fuel ratio = mass of air to completely burn 1 mole of fuel/ mass of one mole of fuel
(A/F)st. = 17.5
so,
(A/F)act. = 0.7* 17.5 = 12.28;
so, mass of air = 12.28* 16.042 =197.06;
which is less than air at stoichiometry(280) so we do not have to add extra air we have to remove 280-197= 83 g.
equivalence less that one leads to rich mixture that means air supplied to fuell is less than air supply at stoichiometric condition.
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