According to this blog, if we use the figures from test 6 (maximum efficiency). The gasifier can produce a total of 256.85MJ in 50 minutes of active burning while consuming 10kg of chopped wood with 8% moisture content and 3kg of charcoal. We assume that for direct air heating the efficiency is 44% so we can use 113MJ throughout the 50 minutes.
Therefore, average usable thermal power is 113 MJ / (50·60 s) = 37.67 kW. To estimate how much larvae can be dried in that time and what airflow the dryer needs, I make the following assumptions:
_latent heat of evaporation of water ≈ 2,257 kJ/kg
_the larvae are heated from 25 °C to a drying air temperature of 50 °C
_the wet larvae’s effective specific heat is 3.8 kJ·kg⁻¹·°C⁻¹ (between 4.18 kJ·kg⁻¹·°C⁻¹ for water and 1.5–2.0 kJ·kg⁻¹·°C⁻¹ for the dry matter like proteins, lipids, chitin…)
_incoming ambient air is 25 °C at 50% relative humidity
_the drying air is allowed to reach 100% RH at 50°C before exhaust
Using those numbers, drying 1 kg of input larvae that starts at 50% moisture and finishes at 9.1% (450 g water removed) requires the following energy per kg:
- Latent energy to evaporate 0.45 kg water = 0.45 × 2,257 = 1,015.65 kJ.
- Sensible heating of 1 kg wet larvae (cp = 3.8 kJ·kg⁻¹·°C⁻¹) from 25 °C to 50 °C (ΔT = 25 °C) = 3.8 × 25 = 95 kJ.
- Psychrometrics: at 25 °C saturation ≈ 0.02298 kg H₂O/m³ (incoming at 50% RH = 0.01149 kg/m³); at 50 °C saturation ≈ 0.08288 kg H₂O/m³. So each m³ of 50 °C air can pick up ≈0.08288 − 0.01149 = 0.07139 kg water. That means the volumetric air requirement to remove 0.45 kg water is 0.45 / 0.07139 ≈ 6.303 m³ of drying air per kg of wet larvae.
At an approximate air density of 1.092 kg/m³ at 50 °C, the air mass per kg of larvae is 6.303 × 1.092 ≈ 6.883 kg. Heating that air by 25 °C with cp ≈ 1.005 kJ·kg⁻¹·°C⁻¹ uses about 6.883 × 1.005 × 25 ≈ 172.94 kJ per kg of larvae.
Summing the three terms gives a system energy demand of about 1,015.65 (latent) + 95 (sensible larvae) + 172.94 (air heating) = 1,283.59 kJ per kg (≈1.284 MJ/kg).
With 113 MJ usable over the 50-minute run the system can therefore process:
Processed wet mass ≈ 113,000 / 1,283.59 ≈ 88.03 kg of wet larvae in 50 minutes.
Dried product = 88.03 × 0.55 ≈ 48.42 kg.
Water evaporated ≈ 88.03 × 0.45 ≈ 39.62 kg.
• Volumetric airflow required = 6.303 m³/kg × 88.03 kg / (50*60) s ≈ 0.1850 m³/s ≈ 666 m³/h or 391.99 CFM (Cubic Feet per Minute)
A more streamlined yet still effective version of these estimations was done on this Excel spreadsheet, where we simplified the model by neglecting both the heating of the air and the larvae.
