Drying Energy Requirement

From water removal to energy

For thermal drying, multiply the water mass to evaporate by the specific useful heat demand. Divide useful heat by thermal efficiency to obtain energy input. Dividing by duration gives average thermal power. For example, 100 kg of water at an assumed useful demand of 3600 kJ/kg and 50 percent efficiency requires 200 kWh. Over two hours, average power is 100 kW.

Choosing inputs

These numbers illustrate the calculation only. Specific heat demand depends on pressure, temperature, bound water and product heating. Entering measured total consumption as useful heat and dividing by efficiency again counts losses twice. Mechanically separated water is not evaporated.

Keeping the energy boundary consistent

Specific useful heat describes heat needed by the process per unit of evaporated water. It is not the same as total energy consumption measured by a fuel or electricity meter. Define the boundary for input energy, useful heat and recovery before choosing efficiency. Consumption measured under other conditions may not transfer directly even to the same dryer.

For a hand check, assume 40 kg of water, 1800 kJ/kg of useful heat, 80 percent efficiency and 2 hours. Useful heat is 40 × 1800 = 72000 kJ. Since one kWh is 3600 kJ, this equals 20 kWh. Energy input is 20/0.8 = 25 kWh and average thermal power is 25/2 = 12.5 kW. The specific heat and efficiency are illustrative inputs, not material constants or design recommendations.

With unchanged water mass, extending duration does not reduce energy input in this model. It only reduces the average power obtained by dividing energy by time. If duration affects actual losses or efficiency, reassess those inputs separately.

Do not select heater capacity directly from average power. Startup heating, changing drying rates and peak loads are not resolved. Add fan and auxiliary motor electricity separately. This preliminary estimate also does not establish allowable product temperature, drying time or fire safety.