Airation rates for composting can be estimated using three different methods:

1. Temperature difference method

This is most commonly used for ASP (Aerated Static Pile). The aim is to keep the temperature of the compost pile constant and below 65°C. The calculation is based on the amount of air needed to evaporate the moisture from the compost pile. It is loosely based on the temperature generated by the microbes, but the actual calculation is based on mathematical modeling of water evaporation at a given temperature.

Comment:

• Excellent point about the temperature difference method being more about moisture management than microbial activity. This method often leads to over-aeration in the thermophilic phase, which could be counterproductive for the goals.
• It does not account for the fact that actively cooling the pile with large amounts of air will increase the rate of microbial respiration and thus will speed up the composting process. We want to avoid this, because we are aiming for a longer composting period than with common ASP systems. Our goal is not to create compost from the feedstock as fast as possible, but to maintain high temperatures for a specified period of time while creating high quality compost.
• Excessive aeration can also lead to nitrogen loss through ammonia volatilization, which would reduce the fertilizer value of your final product.

2. Carbon dioxide or Oxygen consumption method

This method is based on the amount of carbon dioxide generated or oxygen consumed by the microbes. The calculation is based on laboratory measurements of carbon dioxide production or oxygen consumption. Then the amount of air needed to replace the oxygen consumed is calculated. This is the most accurate method, but it is also the most complex. It does not account for the fact that the air exchange rate is also affected by the temperature of the air.

Comment:

• This method requires continuous monitoring equipment (like O₂/CO₂ sensors) which adds complexity and cost. The stoichiometry of microbial respiration (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy) is the basis here, but real-world microbial communities are more complex.
• It does not account for the fact that the oxygen consumption is affected by the availability of oxygen in the feedstock.
• The method assumes uniform oxygen distribution, which isn’t the case in practice. You might want to mention that this is particularly relevant for feedstocks with poor structure or high bulk density.
• Again our goal is not to create the most optimal conditions for microbial respiration (e.g., high airflow), but to create high quality compost while maintaining high temperatures for a specified period of time. So we aim for ‘just enough’ airflow to maintain thermophilic conditions.

3. Free air exchange method

This method is based on the amount of Free Air Space (FAS) that is available in the Feedstock. The calculation is not based on the temperature of the air, the humidity of the air or the pressure of the air. It is simply based on the number of air exchanges per hour. This is the simplest method, but it is also the least accurate.

Comment:

• The number of air exchanges per hour is often based on practical experience.
• FAS typically needs to be >30% for adequate aeration, and that this method is more of a rule-of-thumb than a precise calculation. It’s useful for initial design but should be validated with other methods.

It doesn’t give us an actual calculation method to find a good air exchange rate. It is however useful to use as a minimum requirement for the air exchange rate and to compare our calculation to other experiences from practice.

TEAPOTS approach:

First we calculate the total amount of energy that can be produced by microbial respiration during full consumption of the feedstock. Then we will make assumptions about how much of this energy we want to become available during each phase of composting. We will cross check these values with laboratory measurements of microbial respiration in different phases of composting. We will use this to calculate the amount of air that is needed to maintain the microbial respiration at a certain level (method 2). Then we will cross check this with commonly used air flows used in ASP systems (method 1) and with common numbers of air exchanges per hour (method 3). With this approach we calculate an airflow that is fitting to our goals: maintaining high temperature during the 6-12 month period of composting while producing as much excess energy as possible.

In steps:

Step 1. Calculate the total amount of energy that can be produced by microbial respiration during full consumption of the feedstock.

Step 2. Make assumptions about how much of this energy we want to become available during each phase of composting.

Step 3. Cross check these values with laboratory measurements of microbial respiration in different phases of composting.

Step 4. Use this to calculate the amount of air that is needed to maintain the microbial respiration at a certain level (method 2).

Step 5. Cross check this with commonly used air flows used in ASP systems (method 1) and with common numbers of air exchanges per hour (method 3).

Step 6. With this approach we calculate an airflow that is fitting to our goals: maintaining high temperature during the 6 month period of composting while producing as much excess energy as possible.

Comment: This is a robust, multi-method approach that addresses the limitations of each individual method. The energy-based approach is particularly innovative for your goal of heat recovery. You might want to include:

1. The heat production potential of different feedstocks (e.g., food waste ≈ 18-22 MJ/kg VS, yard waste ≈ 12-15 MJ/kg VS)
2. That only about 30-50% of this energy is typically recoverable as heat
3. The need to balance heat recovery with maintaining proper composting conditions

So, what do we need for this approach?

• laboratory measurements of microbial respiration in different phases of composting, for each feedstock
• splitting the total amount of energy that can be produced by microbial respiration during full consumption of the feedstock into different phases (thermophilic, mesophilic, maturation)
• an assumption about how much of the total amount of energy we want to become available during each phase
• an assumption about how long we want to maintain each phase
• Heat loss calculations for the specific reactor design
• The heat capacity of the feedstocks
• Data on how aeration rate affects both microbial activity and heat transfer
• Consideration of thermal mass effects in the system
• Potential for heat recovery from exhaust gases (latent and sensible heat)

AeroX Heat Exchanger Configuration Analysis

Introduction

This document contains a detailed analysis of various configurations of AeroX 1.6 heat exchangers for cooling air from 50°C to various target temperatures. The analysis focuses on determining the required number of units for different cooling scenarios.

Baseline Parameters

Basic Data

• Inlet air: 50°C, 100% RH (87.7 g/kg)
• Total air flow rate: 100 m³/h (distributed across units)
• Water flow rate per unit: 500 l/h (0.1389 kg/s) or 1000 l/h (0.2778 kg/s)
• Inlet water temperature: 10°C
• AeroX 1.6 specifications:
  • Surface area per unit: 17.71 m²
  • U-value: 14.6 W/m²K
  • Max. air flow rate per unit: 1500 m³/h
• Air density at 50°C: 1.11 kg/m³
• Total mass air flow: 100 m³/h × 1.11 kg/m³ = 0.0308 kg/s
• Latent heat of evaporation of water (h_v): 2257 kJ/kg
• Specific heat capacity of air (c_p_air): 1.005 kJ/kgK
• Specific heat capacity of water (c_p_water): 4.18 kJ/kgK

Detailed Heat Transfer Analysis

Theoretical Background

Heat transfer in the heat exchanger is determined by:

Q = h × A × ΔT_lm

Where:

• Q = heat flow (W)
• h = heat transfer coefficient (W/m²K) – dependent on air velocity
• A = heat transfer surface area (17.71 m² per unit)
• ΔT_lm = log mean temperature difference (K)

Determination of Heat Transfer Coefficient (h)

For tube bundles, the following relationship applies between Nusselt (Nu), Reynolds (Re) and Prandtl (Pr) numbers:

Nu = C × Re^m × Pr^n

Where:

• Nu = h × D/k (Nusselt number)
• Re = (ρ × v × D)/μ (Reynolds number)
• Pr = (μ × cp)/k (Prandtl number)

For air at 50°C:

• ρ = 1.11 kg/m³ (density)
• μ = 1.96 × 10^-5 kg/ms (dynamic viscosity)
• k = 0.0275 W/mK (thermal conductivity)
• Pr = 0.7 (Prandtl number)
• D = 0.00908 m (outer tube diameter)

Velocity and Flow Rate Calculations

Geometry Analysis:

1. Frontal area (perpendicular to air flow):
   • Width: 1,600 mm (tube length)
   • Height: 147 mm
   • Total frontal area: 1.6 m × 0.147 m = 0.2352 m²
2. Free flow area (50% of frontal area):
   • Assumption: 50% of area is air (50% tubes)
   • Effective flow area: 0.5 × 0.2352 m² = 0.1176 m²
3. Air flow path:
   • Number of tube rows: 21 (as stated in documentation)
   • Total air flow depth: 345 mm
   • Average free path length between tubes: 345 mm / 21 = 16.43 mm
4. Hydraulic diameter calculation:
   • For channel between tubes: Dh ≈ 2 × free space = 2 × 9.08 mm = 18.16 mm
   • This is the characteristic length for Reynolds calculation
5. Velocity calculation:
   • Volume flow rate (V̇) = 100 m³/h = 0.0278 m³/s (total for all units)
   • Velocity (v) = V̇ / (A_eff × n_units) Where n_units is the number of parallel units
6. Reynolds number:
   • Re = (ρ × v × Dh) / μ
   • At 50°C: ρ = 1.11 kg/m³, μ = 1.96 × 10^-5 kg/ms
   • Re = (1.11 × v × 0.01816) / 1.96×10^-5 = 1,029 × v

Calculation per Configuration (total 100 m³/h):

Units	Flow/Unit (m³/h)	Velocity (m/s)	Re	Flow Type
1	100	0.24	247	Laminar
2	50	0.12	124	Laminar
3	33.3	0.08	82	Laminar
4	25	0.06	62	Laminar
5	20	0.05	51	Laminar
6	16.7	0.04	41	Laminar

Note: Velocities are based on 50% free flow through the frontal area

Heat Transfer Calculations

Assumptions and Formulas:

1. Heat transfer:
   • Q = h × A × ΔT
   • Where h = heat transfer coefficient (W/m²K)
   • A = heat transfer surface area (17.71 m² per unit)
   • ΔT = temperature difference (25K)
2. Heat transfer coefficient (h):
   • For laminar flow (Re < 2300): Nu = 0.664 × Re^0.5 × Pr^0.33
   • Where Pr = 0.7 (Prandtl number for air)
   • h = (Nu × k) / Dh
   • With k = 0.0275 W/mK (thermal conductivity of air at 50°C)
3. Effective heat transfer surface area:
   • Total tube length per unit: 620.8 m
   • Outer tube diameter: 9.08 mm
   • Effective area: π × D × L = π × 0.00908 × 620.8 = 17.71 m²

Calculation per Configuration (total 100 m³/h):

Units	Flow/Unit (m³/h)	Vel. (m/s)	Re	Nu	h (W/m²K)	Q/Unit (W)	Total Q (W)
1	100	0.24	247	10.4	15.7	6,947	6,947
2	50	0.12	124	7.4	11.2	4,957	9,914
3	33.3	0.08	82	6.0	9.1	4,029	12,086
4	25	0.06	62	5.2	7.9	3,496	13,983
5	20	0.05	51	4.7	7.1	3,143	15,717
6	16.7	0.04	41	4.2	6.4	2,833	16,998

Note: Calculated for a total air flow rate of 100 m³/h, distributed across the specified number of units

Scenario 3: Detailed Analysis with 3 Parallel Units

Parameters:

• Inlet air: 55°C, 100% RH (87.7 g/kg)
• Outlet air: 20°C, 50% RH (7.3 g/kg)
• Water flow rate: 500 l/h total (approximately 167 l/h per unit)
• Inlet water temperature: 10°C
• Number of units: 3 parallel
• Heat transfer surface area per unit: 17.71 m²
• Effective flow area per unit: 0.1176 m²
• Condensation per kg dry air: 87.7 – 7.3 = 80.4 g/kg

Calculations per Total Air Flow Rate:

1. Total Air Flow Rate 25 m³/h

• Air flow per unit: 25 / 3 = 8.33 m³/h = 0.00231 m³/s
• Air velocity: 0.00231 / 0.1176 = 0.0196 m/s
• Reynolds number (Re): (0.0196 × 0.01816) / (1.8×10⁻⁵) = 19.8 (laminar flow)
• Nusselt number (Nu): 0.664 × √19.8 × 0.7^(1/3) = 2.63
• Heat transfer coefficient (h): (2.63 × 0.0275) / 0.01816 = 3.98 W/m²K
• Heat transfer per unit: 3.98 × 17.71 × 25 = 1,762 W
• Total power (3 units): 5,286 W

Water heating:

• Total water flow rate: 500 l/h = 0.1389 kg/s
• ΔT water: 5,286 / (4.18 × 0.1389) = 9.1°C
• Outlet water: 10 + 9.1 = 19.1°C

Condensation:

• Air mass flow rate: 25 × 1.07 = 26.75 kg/h = 0.00743 kg/s
• Condensate: 80.4 × 0.00743 = 0.597 g/s = 2.15 kg/h
• Latent heat (2,257 kJ/kg): 2.15 × 2.257 = 4.85 MJ/h = 1.35 kW

2. Total Air Flow Rate 50 m³/h

• Air flow per unit: 50 / 3 = 16.67 m³/h = 0.00463 m³/s
• Air velocity: 0.00463 / 0.1176 = 0.0394 m/s
• Reynolds number (Re): (0.0394 × 0.01816) / (1.8×10⁻⁵) = 39.7
• Nusselt number (Nu): 0.664 × √39.7 × 0.7^(1/3) = 3.72
• Heat transfer coefficient (h): (3.72 × 0.0275) / 0.01816 = 5.63 W/m²K
• Heat transfer per unit: 5.63 × 17.71 × 25 = 2,492 W
• Total power (3 units): 7,476 W

Water heating:

• ΔT water: 7,476 / (4.18 × 0.1389) = 12.9°C
• Outlet water: 22.9°C

Condensation:

• Air mass flow rate: 50 × 1.07 = 53.5 kg/h = 0.0149 kg/s
• Condensate: 80.4 × 0.0149 = 1.20 g/s = 4.31 kg/h
• Latent heat: 4.31 × 2.257 = 9.73 MJ/h = 2.70 kW

3. Total Air Flow Rate 100 m³/h

• Air flow per unit: 100 / 3 = 33.33 m³/h = 0.00926 m³/s
• Air velocity: 0.00926 / 0.1176 = 0.0787 m/s
• Reynolds number (Re): (0.0787 × 0.01816) / (1.8×10⁻⁵) = 79.4
• Nusselt number (Nu): 0.664 × √79.4 × 0.7^(1/3) = 5.26
• Heat transfer coefficient (h): (5.26 × 0.0275) / 0.01816 = 7.96 W/m²K
• Heat transfer per unit: 7.96 × 17.71 × 25 = 3,524 W
• Total power (3 units): 10,572 W

Water heating:

• ΔT water: 10,572 / (4.18 × 0.1389) = 18.2°C
• Outlet water: 28.2°C

Condensation:

• Air mass flow rate: 100 × 1.07 = 107 kg/h = 0.0297 kg/s
• Condensate: 80.4 × 0.0297 = 2.39 g/s = 8.60 kg/h
• Latent heat: 8.60 × 2.257 = 19.41 MJ/h = 5.39 kW

Summary of Results

Total Air Flow (m³/h)	Power (kW)	Water ΔT (°C)	Outlet Water (°C)	Condensate (kg/h)	Latent Power (kW)
25	5.29	9.1	19.1	2.15	1.35
50	7.48	12.9	22.9	4.31	2.70
100	10.57	18.2	28.2	8.60	5.39

Key Insights:

• Power: Power increases less than linearly with flow rate due to the decreasing heat transfer coefficient at lower flow rates.
• Water temperature: Even at the highest flow rate (100 m³/h), the outlet water temperature remains below 30°C, which is manageable.
• Condensation: Condensate production varies from 2.15 to 8.60 kg/h depending on air flow rate.
• Efficiency: At lower flow rates, efficiency decreases due to lower Reynolds numbers, resulting in lower heat transfer coefficients.

Performance Analysis per Scenario

Key Assumptions:

1. Total air flow rate: 100 m³/h (55°C, 100% RH)
2. Water flow rate: 500–1000 l/h at 10°C
3. Temperature difference (ΔT): 25K between air and water
4. Air flow: Laminar flow (Re < 2300)
5. Heat transfer surface area: 17.71 m² per unit

Performance Characteristics:

• 1 Unit:
  • Highest heat transfer per unit (6.95 kW)
  • Lowest total power (6.95 kW)
  • Most efficient configuration
• 2–3 Units:
  • Better distribution of air flow
  • Total power 9.9 – 12.1 kW
  • Good balance between power and efficiency
• 4–6 Units:
  • Highest total power (14.0 – 17.0 kW)
  • Lower efficiency per unit
  • More complex installation and higher costs

Considerations:

• At lower air flow rates, heat transfer decreases due to lower Reynolds numbers
• Water flow rate has limited influence on heat transfer (not the limiting factor in this configuration)
• Pressure drop across the heat exchanger decreases with an increasing number of units

Scenario Analysis

Scenario 1: Cooling to 40°C (100% RH)

• Outlet air: 40°C, 100% RH (51.1 g/kg)
• Condensation: 87.7 – 51.1 = 36.6 g/kg
• Required power: 2.86 kW
• Water heating:
  • 500 l/h: ΔT = 4.9°C → Outlet water = 14.9°C
  • 1000 l/h: ΔT = 2.5°C → Outlet water = 12.5°C
• Required configuration: 1 unit (6.95 kW)
• Effective overcapacity: 143% (6.95/2.86)
• Recommendation: 1 unit is more than sufficient

Scenario 2: Cooling to 30°C (75% RH)

• Outlet air: 30°C, 75% RH (20.1 g/kg)
• Condensation: 87.7 – 20.1 = 67.6 g/kg
• Required power: 5.33 kW
• Water heating:
  • 500 l/h: ΔT = 9.2°C → Outlet water = 19.2°C
  • 1000 l/h: ΔT = 4.6°C → Outlet water = 14.6°C
• Required configuration: 1 unit (6.95 kW)
• Effective overcapacity: 30% (6.95/5.33)
• Recommendation: 1 unit is sufficient

Scenario 3: Cooling to 20°C (50% RH)

• Outlet air: 20°C, 50% RH (7.3 g/kg)
• Condensation: 87.7 – 7.3 = 80.4 g/kg
• Required power: 6.53 kW
• Water heating:
  • 500 l/h: ΔT = 11.2°C → Outlet water = 21.2°C
  • 1000 l/h: ΔT = 5.6°C → Outlet water = 15.6°C
• Required configuration: 1 unit (6.95 kW)
• Effective overcapacity: 6% (6.95/6.53)
• Recommendation: 1 unit is just sufficient; consider 2 units for margin

Detailed Analysis: 3 Parallel Units

Parameters:

• Inlet air: 55°C, 100% RH (87.7 g/kg)
• Outlet air: 20°C, 50% RH (7.3 g/kg)
• Water flow rate: 500 l/h total (approximately 167 l/h per unit)
• Inlet water temperature: 10°C
• Number of units: 3 parallel
• Heat transfer surface area per unit: 17.71 m²
• Effective flow area per unit: 0.1176 m²

Calculations per Air Flow Rate:

1. Air Flow Rate 25 m³/h per Unit (Total 75 m³/h)

• Air velocity: 0.059 m/s
• Reynolds number (Re): 59.5 (laminar flow)
• Nusselt number (Nu): 4.53
• Heat transfer coefficient (h): 6.86 W/m²K
• Heat transfer per unit: 3,037 W
• Total power (3 units): 9,111 W
• Water heating:
  • ΔT water: 15.7°C
  • Outlet water: 25.7°C
• Condensation:
  • Condensate: 6.45 kg/h
  • Latent heat: 4.04 kW

2. Air Flow Rate 50 m³/h per Unit (Total 150 m³/h)

• Air velocity: 0.118 m/s
• Reynolds number (Re): 119
• Nusselt number (Nu): 6.41
• Heat transfer coefficient (h): 9.70 W/m²K
• Heat transfer per unit: 4,295 W
• Total power (3 units): 12,885 W
• Water heating:
  • ΔT water: 22.2°C
  • Outlet water: 32.2°C
• Condensation:
  • Condensate: 12.92 kg/h
  • Latent heat: 8.10 kW

3. Air Flow Rate 100 m³/h per Unit (Total 300 m³/h)

• Air velocity: 0.236 m/s
• Reynolds number (Re): 238
• Nusselt number (Nu): 9.06
• Heat transfer coefficient (h): 13.72 W/m²K
• Heat transfer per unit: 6,075 W
• Total power (3 units): 18,225 W
• Water heating:
  • ΔT water: 31.4°C
  • Outlet water: 41.4°C
• Condensation:
  • Condensate: 25.81 kg/h
  • Latent heat: 16.18 kW

Summary of Results

Air Flow (m³/h)	Water ΔT (°C)	Outlet Water (°C)	Total Power (kW)	Condensate (kg/h)	Latent Power (kW)
75 (3×25)	15.7	25.7	9.11	6.45	4.04
150 (3×50)	22.2	32.2	12.89	12.92	8.10
300 (3×100)	31.4	41.4	18.23	25.81	16.18

Conclusions:

1. Power: Total power increases almost linearly with air flow rate.
2. Water temperature: At higher air flow rates, the outlet water temperature may become too high (>40°C). Consider a higher water flow rate or more units.
3. Condensation: The amount of condensate is substantial (up to 25.8 kg/h at 300 m³/h), requiring adequate drainage.
4. Practical application: For continuous operation, a water temperature of 10°C is difficult to maintain. Consider a higher inlet temperature or a larger heat exchanger.