Combustion Efficiency: 0%
A combustion efficiency calculator is a tool used to determine how efficiently fuel is burned in a combustion system, such as a boiler, furnace, or engine. It calculates the percentage of energy from the fuel that is effectively converted into useful heat or power, while accounting for losses like unburned fuel and exhaust gases. This calculator helps in optimizing fuel usage, reducing emissions, and improving overall energy efficiency.
Formula of Combustion Efficiency Calculator
The formula for combustion efficiency is:
Combustion_efficiency = [(Input_energy – Losses) / Input_energy] * 100
Where:
- Combustion_efficiency is the efficiency of the combustion process, expressed as a percentage.
- Input_energy is the total energy content of the fuel.
- Losses include energy lost due to unburned fuel and heat carried away by exhaust gases.
Detailed Breakdown
- Losses Due to Exhaust Gas
Losses_exhaust = m_gas * c_gas * (T_gas – T_ref)
Where:- Losses_exhaust is the heat carried away by flue gases.
- m_gas is the mass flow rate of exhaust gas.
- c_gas is the specific heat capacity of the gas.
- T_gas is the exhaust gas temperature.
- T_ref is the reference temperature.
- Losses Due to Unburned Fuel
Losses_unburned = (Unburned_fuel_fraction) * Input_energy
Where:- Unburned_fuel_fraction is the fraction of unburned fuel.
- Input_energy is the energy content of the fuel.
- Input Energy
Input_energy = m_fuel * LHV
Where:- m_fuel is the mass of the fuel burned.
- LHV is the lower heating value of the fuel.
Combined Formula
Combustion_efficiency = [(m_fuel * LHV – (m_gas * c_gas * (T_gas – T_ref) + Unburned_fuel_fraction * m_fuel * LHV)) / (m_fuel * LHV)] * 100
Dependent Variables
- Mass of Exhaust Gas: m_gas = m_fuel * Air_fuel_ratio
- Unburned Fuel Fraction: Unburned_fuel_fraction = Mass_unburned_fuel / m_fuel
Useful Conversion Table
| Parameter | Unit | Typical Values/Notes |
|---|---|---|
| Specific Heat Capacity (c_gas) | J/(kg·K) or BTU/(lb·°F) | Depends on gas composition, typically 1.005 J/(kg·K) for air |
| Air-to-Fuel Ratio | Dimensionless | Varies by fuel, e.g., ~14.7:1 for gasoline |
| Lower Heating Value (LHV) | J/kg or BTU/lb | ~43 MJ/kg for natural gas, ~42 MJ/kg for gasoline |
| Exhaust Gas Temperature (T_gas) | °C or °F | Typical range: 150°C–250°C for boilers |
| Reference Temperature (T_ref) | °C or °F | Typically 25°C or 77°F |
Example of Combustion Efficiency Calculator
A boiler burns 10 kg of natural gas with a lower heating value (LHV) of 43 MJ/kg. The air-fuel ratio is 15:1. The exhaust gas temperature (T_gas) is 200°C, and the specific heat capacity of the exhaust gas is 1.005 J/(kg·K). The reference temperature (T_ref) is 25°C. Assume that 2% of the fuel remains unburned.
- Calculate m_gas:
m_gas = m_fuel * Air_fuel_ratio = 10 kg * 15 = 150 kg - Calculate losses due to exhaust gas:
Losses_exhaust = m_gas * c_gas * (T_gas – T_ref)
Losses_exhaust = 150 kg * 1.005 J/(kg·K) * (200 – 25) K = 26,362.5 J - Calculate losses due to unburned fuel:
Losses_unburned = Unburned_fuel_fraction * Input_energy
Losses_unburned = 0.02 * (10 kg * 43 MJ/kg) = 8.6 MJ - Calculate input energy:
Input_energy = m_fuel * LHV = 10 kg * 43 MJ/kg = 430 MJ - Compute combustion efficiency:
Combustion_efficiency = [(Input_energy – (Losses_exhaust + Losses_unburned)) / Input_energy] * 100
Combustion_efficiency = [(430 MJ – (26.3625 MJ + 8.6 MJ)) / 430 MJ] * 100 = 90.36%
Most Common FAQs
A good combustion efficiency is typically above 85%, with highly optimized systems achieving over 90%.
Efficiency can be improved by reducing excess air, insulating equipment to minimize heat loss, and maintaining optimal fuel-to-air ratios.
Unburned fuel leads to wasted energy, higher costs, and increased emissions. Addressing this issue can significantly improve efficiency and reduce environmental impact.