Combustion Calculations and Stoichiometry — Lesson 5

This lesson covers the calculation of the amount of oxidizer required to burn 1 kg of fuel, assuming complete combustion. It explains the concept of stoichiometry and how to use it to solve problems related to combustion. The lesson provides a detailed walkthrough of three problems. The first problem involves calculating the product composition of a combustion process with a given equivalence ratio. The second problem involves determining the equivalence ratio, percent theoretical of the reactants, molecular weight of the products, and mole fraction of CO2 from a given volumetric analysis of a combustion process. The third problem involves determining the stoichiometric air-fuel ratio and product composition when a fuel mixture is burnt with 20% excess air.

Video Highlights

01:16 - Explanation of the chemical equation for combustion
09:11 - Calculation of equivalence ratio, percent theoretical of reactants, molecular weight of products, and mole fraction of CO2
19:45 - Calculation of stoichiometric air-fuel ratio and product composition with given fuel gas mixture

Key Takeaways

- Stoichiometry is a crucial concept in combustion calculations, helping to determine the amount of oxidizer required for complete combustion.
- The equivalence ratio is a measure of the actual fuel-air ratio to the stoichiometric fuel-air ratio.
- The product composition of a combustion process can be calculated using the given equivalence ratio and the stoichiometric equation.
- The stoichiometric air-fuel ratio can be determined from the molecular weight of the fuel and the number of moles of oxygen required for complete combustion.
- When a fuel mixture is burnt with excess air, the excess oxygen appears in the product, and the number of moles of nitrogen increases due to the excess air.