How to Calculate Combustion Efficiency
The mere term "combustion efficiency" can be misleading since a combustion process where all of the fuel is burned stoichiometrically is 100-percent efficient. It means all of the fuel's heat of combustion has been realized. In actuality, it is the efficiency of the processes that harness this heat of combustion that require the greatest consideration in using fuel resources efficiently. Most efficiency improvement efforts are aimed at assuring complete combustion, and then efficiently using the heat produced. Calculating combustion efficiencies in boilers and internal combustion engines are good examples of this two-part process.
Steam Power Boilers
Define the boiler combustion application. In this example, a small steam boiler burns #6 fuel oil to produce 20,000 pounds per hour of saturated steam at 150-pounds-per-square-inch pressure. During a given hour, the boiler uses 145 gallons of the oil to produce the 20,000 pounds/hour of the steam, with an excess oxygen reading of 2-1/2 percent. With this information, you can calculate the boiler's overall combustion efficiency.
Calculate the overall Btu from the oil's combustion. The British Thermal Unit value of #6 fuel oil is 141,000 Btu/gal. Since the 2-1/2 percent excess oxygen in the stack gas lies in the stoichiometrically correct realm of combustion of the oil in the combustion chamber, then virtually all of the heating value of the fuel oil is being realized. Substituting the values, 141,000 Btu/gal x 148-gal burned = 20,868,000 Btu heat from the combustion.
Calculate the amount of heat used to make the saturated 150-psi-steam from saturated liquid. The steam table chart shows that 150-psi-saturated steam has a Btu/lb value of 1,192.4 for saturated vapor and 322 for saturated liquid, which means the boiler adds 870.4 Btu/lb of steam produced. Multiplying the 20,000 lbs of steam x 870.4 Btu/lb yields 17,795,328 Btu added.
Calculate the boiler combustion efficiency based on the fuel used to produce steam by dividing 17,795,328 Btu/20,868,000 Btu = 0.8528. Boiler efficiency during that hour is 85.28 percent, which is within design parameters.
Internal Combustion Engine Generators
Define the internal combustion engine application. This example considers an emergency generator that produces 12,000 watts of electrical power during a 1-hour time period. The gasoline engine driving the generator burns 2.2-gallons of regular gasoline during the hour. This is sufficient information to calculate the overall combustion efficiency of a gasoline engine driving a generator.
Calculate the amount of heat in Btu from the efficient combustion of 2.2 gallons of gasoline. The average Btu value for regular gasoline is about 115,000 Btu/gallon. Therefore, 2.2 gallons x 115,000-Btu/gal = 253,000 Btu from burning the gasoline.
Convert the 12 kW hours of electricity produced to Btu/hr of energy produced by the engine. Taking 12 kW hr x 3,413 Btu/kW hr = 40,956 Btu.
Calculate the overall efficiency of the generator set in producing the electricity from 2.2 gallons of gasoline. 40,956 Btu produced/253,000 Btu from combustion = 0.16188 or an efficiency of 16.19 percent. This is less than one-fifth the boiler efficiency, which is why most public utility electricity is produced by boiler-powered steam generators and not gasoline engines.
Things You Will Need
- Steam Tables (Saturated Steam Data)
- Always keep more fuel than estimated for emergency generators because very few emergency scenarios present ideal conditions.
- Combustion efficiency is predicated on enough air for safe combustion and the least amount of carbon monoxide in the operating environment.