The boiler evaporation rate describes the capacity of a steam boiler to produce steam. It is quantified as the mass of steam generated per unit of time, typically expressed in kilograms per hour (kg/hr) or tons per hour (TPH). Accurate calculation of the evaporation rate is critical for sizing boilers, determining fuel requirements, and monitoring thermodynamic efficiency.
Key Parameters
To calculate the evaporation rate, several thermodynamic properties must be known:
Fuel Consumption: The amount of fuel burned per hour.
Calorific Value (GCV): The energy content of the fuel (e.g., Coal, Oil, Natural Gas).
Boiler Efficiency: The percentage of fuel energy successfully transferred to the water.
Enthalpy of Steam ($h_g$): Total heat content of the steam leaving the boiler, dependent on pressure and temperature.
Enthalpy of Feed Water ($h_f$): Heat content of the water entering the boiler, dependent on feed water temperature.
The Calculation Formulas
The logic behind the calculator uses the principle of heat balance. First, we determine the useful heat energy utilized by the boiler:
Useful Heat (Q) = Mass of Fuel × Calorific Value × Efficiency
Next, we calculate the energy required to convert 1 kg of feed water into steam at the operating pressure:
Heat Added per kg ($\Delta h$) = Enthalpy of Steam ($h_g$) – Enthalpy of Feed Water ($h_f$)
Finally, the Actual Evaporation Rate ($m_s$) is derived:
Actual Evaporation Rate = Useful Heat (Q) / Heat Added per kg ($\Delta h$)
Equivalent Evaporation (From and At 100°C)
Comparing boilers operating at different pressures and feed water temperatures can be misleading. To standardize capacity, engineers use Equivalent Evaporation. This metric converts the actual evaporation rate to a standard condition: evaporating water at 100°C into steam at 100°C at atmospheric pressure.
The latent heat of vaporization at 100°C is 2257 kJ/kg. The ratio of actual heat added to this standard latent heat is known as the Factor of Evaporation (Fe).
Equivalent Evaporation = Actual Evaporation × Factor of Evaporation
Example Calculation
Consider a boiler consuming 600 kg/hr of oil with a calorific value of 42,000 kJ/kg. The boiler efficiency is 85%.
The steam is generated at 10 bar ($h_g \approx 2778$ kJ/kg) and feed water is supplied at 85°C ($h_f \approx 355$ kJ/kg).