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Airflow Channels

5 min read

Air carries heat in and out of a rock or sand store, and because air holds only about 1.2 kJ per cubic metre per kelvin, moving 1 kW at a 30 K temperature rise takes 0.028 m³ of air every second. Natural draft through a 1 m column of warm air gives only about 2 Pa, so the channels and the bed must offer very little resistance: wide ducts, uniform stones, few bends, and a fan if you can drive one.

Part of Rock & Sand Heat Stores

Airflow channels are the ducts, pipes and open spaces that distribute air evenly through a heat store and carry it between the store and the room or heat source.

How Much Air Does A Kilowatt Need?

Air at room temperature has a density of about 1.2 kg/m³ and a specific heat of about 1.005 kJ/kg·K (standard values, recalled), so each cubic metre carries 1.2 × 1.005 = 1.21 kJ for every kelvin of temperature change.

Inputs: heat rate 1 kW; air temperature rise through the bed 30 K.

  1. Air needed = 1,000 W ÷ (1,206 J/m³·K × 30 K) = 0.0276 m³/s (0.028 rounded), about 28 L/s, or 99 m³/h.
  2. In a round duct of 100 mm diameter (area 0.00785 m²): speed = 0.0276 ÷ 0.00785 = 3.5 m/s.
  3. Through a bed 1 m² in cross-section: superficial speed = 0.0276 m/s, slow.

Natural Draft Is Weak

Warm air rises in a column. Inputs: column height 1 m; room air at 20°C (density 353 ÷ 293 = 1.204 kg/m³) and bed air at 80°C (353 ÷ 353 = 1.000 kg/m³), using ρ ≈ 353/T(K).

  • Δp = g × h × Δρ = 9.81 × 1 × (1.204 − 1.000) = 2.0 Pa.

The parent page notes hot air enters at the top of the bed and the heat rises. That drive of about 2 Pa per metre is small beside what a fan gives (tens of pascals) and must overcome the bed and the duct bends.

Resistance of the bed. The Ergun equation (a standard packed-bed relation, recalled): pressure drop per metre = 150 μ (1−ε)² v ÷ (ε³ d²) + 1.75 ρ (1−ε) v² ÷ (ε³ d).

Inputs: viscosity μ = 1.8 × 10⁻⁵ Pa·s; void fraction ε = 0.35; stone size d = 0.075 m; superficial speed v = 0.0276 m/s; ρ = 1.2 kg/m³.

  1. Viscous term = 150 × 1.8 × 10⁻⁵ × 0.65² × 0.0276 ÷ (0.35³ × 0.075²) = 0.13 Pa/m.
  2. Inertial term = 1.75 × 1.2 × 0.65 × 0.0276² ÷ (0.35³ × 0.075) = 0.32 Pa/m.
  3. Total = 0.45 Pa/m. For a bed 0.8 m deep: 0.36 Pa, well under the 2 Pa available.

So the bed itself passes the air easily; the bends, the inlet and the ducts will cost more. Use ducts of 150-300 mm rather than 100 mm for natural flow, keep every bend gentle, and avoid screens and dust.

Even Distribution

Air takes the path of least resistance, so a poor manifold feeds the bed unevenly. Let the pressure drop along the manifold be small compared with the drop across the bed, or put the air through a plenum space at top and bottom that is much wider than the ducts. The parent page asks for 30-40% void space between rocks, perforated pipes at top and bottom, and reversing the flow to extract heat: air enters at the bottom and leaves at the top.

What You Need

  • Ducts or open channels wider than you think
  • Perforated pipe or slotted boards at the top and bottom
  • A plenum at each end
  • Clean, uniform stones
  • A damper to stop flow when the bed is not in use, so it does not lose heat by natural draft

Method: Check The Airflow

Step 1: Hold a thread or a feather at the outlet and watch for steady flow when the bed is warm.

Step 2: Compare temperatures at several points across the bottom of the bed. Hot and cold patches mean uneven flow.

Step 3: Widen the narrowest duct first.

Common Mistakes

  • A 100 mm duct for natural draft. It chokes a store that wants 28 L/s per kilowatt.
  • Leaving the damper open. The bed loses heat all night by convection.
  • Fines in the bed.

Safety

Smoke and carbon monoxide travel the same channels as heat

A channel that carries clean hot air carries flue gas just as well. Never connect a stove exhaust directly to a duct into a living space; carbon monoxide is odourless and fatal. Keep combustible duct materials away from hot air at the inlet side, where it is hottest, and keep a damper that cannot jam open.

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