What is higher calorific value (HCV) and Lower Calorific Value (LCV)?

What is higher calorific value (HCV) and Lower Calorific Value (LCV)?

Higher Calorific Value (HCV): When the steam in the exhaust gas is cooled to form water. The heat so liberated is adding up to the calorific value of the fuel. Hence, the calorific value in this case is known as higher calorific value (HCV) of the fuel. On the other hand, if the steam is let out along with the exhaust gas the calorific value of the fuel will be less and known as lower calorific value (LCV). For better understanding of analysis of the fuel just go through the other numerical problem given in this blog. 

The stoichiometric AFR by volume for combustion of CO is (a) 1.19

The stoichiometric AFR by volume for combustion of CO is

(a) 1.19, (b) 2.38, (c) 2.45 (d) 4.76

Solution: The reaction involved in the combustion of carbon mono-oxide (CO) is

2CO + O₂       à        2CO₂

                        2vol + 1vol    à     2vol of CO₂

So the combustion of 1vol of CO needs ½ volume of O₂

Now 21m³ of O₂ is present in 100 m³ of air

Therefore ½ m³ of O₂ will be present in ½*100/21=2.38 m³ of air

AFR = volume of air/ volume of fuel= 2.38/1=2.38

Hence the option (b) is correct.

Methane burns with stoichiometric quantity of air. The AFR ratio by weight is (a)13. 41

Methane burns with stoichiometric quantity of air. The AFR ratio by weight is

(a)13. 41, (b) 16.79, (c) 12.51, (d) 17.39

Solution: The reaction involved in the combustion of methane is

CH₄ + 2O₂ à CO₂ +2H₂O

Molar mass of methane is 16g

Molar mass of 2O₂ is 64g

It means, for combustion 16g methane needs 64g O₂ OR

1g of methane needs 64/16=4g of O₂

Therefore 1kg of methane needs 4kg of O₂

Now as we know 23kg of O₂ by mass is present in 100kg of air.

Therefore mass of air needed will be (100/23)*4=17.39kg of air

So AFR= Mass of air/Mass of fuel =17.39/1=17.39

So the option (d) is correct

If CH4 undergoes combustion with stoichiometric quantity of air, the AFR will be (a) 16.23

If CH4 undergoes combustion with stoichiometric quantity of air, the AFR will be

(a) 16.23, (b) 14.5, (c) 9.52, (d) 13.11

Determine AFR on the molar basis.

Solution:  The reaction involved in the combustion of methane is

CH₄ + 2O₂ à CO₂ +2H₂O

Here 1volume of methane needs 2 volume of O2

We know that in air there is 21% oxygen, which means

21m³ of oxygen is present in 100m³ of air

So 2m³ of oxygen present in (100/21)*2=9.52

AFR = mass of air/mass of fuel=9.52/1=9.52

What is Air-Fuel Ratio (AFR)? What is stoichiometric mixture?

What is Air-Fuel Ratio (AFR)? What is stoichiometric mixture?

Ans: Commonly abbreviated AFR stands for stands for Air-Fuel ratio which is the ratio of mass of air to mass of fuel present in the internal combustion engine. If the appropriate amount of air is provided to the engine so that the fuel burns completely then the ratio is known as stoichiometric mixture. AFR is very important as it is used in measurement for anti pollution and efficiency of engines.