A heat store’s retention is set by its time constant, proportional to the insulation thickness and to the ratio of volume to surface area, so retention improves only with the cube root of the size: a tenfold larger cube has a surface-to-volume ratio smaller by a factor of 2.15. With 1 m of straw-like insulation (k = 0.096 W/m·K) a 5 m³ water store loses 71% of its excess heat in 180 days, a 100 m³ store 37% and a 5,000 m³ store 12%, and heating one small house for a season needs about 100 m³ before losses.
Part of Seasonal Storage
Scale requirements are the volumes, insulation thicknesses and collector areas a seasonal heat store needs before it keeps enough heat to matter.
Retention Depends On Volume Over Surface
Model a cube of water of volume V and side a = V^(1/3), insulated all round by a layer of thickness t and conductivity k, in surroundings at a constant temperature. Heat capacity C = V × 4.18 MJ/m³·K; heat loss UA = k × S ÷ t, where S = 6a² is the surface area. The time constant is
τ = C ÷ UA = (4.18 × 10⁶ × t ÷ k) × (V ÷ S) = (4.18 × 10⁶ × t ÷ k) × (a ÷ 6),
in seconds, and the excess temperature decays as e^(−time ÷ τ). The model is the same one used for the pit page (Underground Pit Storage), and it ignores the ground warming around the store, which slows later loss, and any wet insulation, which speeds it.
Inputs: t = 1.0 m; k = 0.096 W/m·K (straw, corpus value from Natural Insulation Materials) and, for comparison, 0.05 (hay-like, Thermal Insulation Basics); time 180 days.
| V (m³) | a (m) | S (m²) | S/V | τ at k = 0.096 | Kept after 180 d | τ at k = 0.05 | Kept |
|---|---|---|---|---|---|---|---|
| 5 | 1.71 | 17.5 | 3.51 | 144 d | 29% | 276 d | 52% |
| 50 | 3.68 | 81.4 | 1.63 | 309 d | 56% | 594 d | 74% |
| 100 | 4.64 | 129 | 1.29 | 390 d | 63% | 749 d | 79% |
| 500 | 7.94 | 378 | 0.756 | 667 d | 76% | 1,280 d | 87% |
| 5,000 | 17.1 | 1,754 | 0.351 | 1,436 d | 88% | 2,758 d | 94% |
Worked line, V = 100 m³, k = 0.096: a = 100^(1/3) = 4.64 m; S = 6 × 4.64² = 129 m²; UA = 0.096 × 129 ÷ 1.0 = 12.4 W/K; C = 100 × 4.18 MJ/K = 418 MJ/K; τ = 418,000,000 ÷ 12.4 = 3.37 × 10⁷ s = 390 days; kept after 180 days = e^(−180 ÷ 390) = 0.63. So 37% is lost. With k = 0.05 the same 100 m³ store keeps 79% and loses 21%. By the same method, with k = 0.096, a 5 m³ store keeps 29% and so loses 71% (with k = 0.05 it keeps 52% and loses 48%), and a 5,000 m³ store keeps 88% and so loses 12%.
From V = 50 to V = 500 (ten times), S/V falls from 1.63 to 0.756, a factor of 2.15 = 10^(1/3). The surface-to-volume ratio of a cube is 6 ÷ a, and a grows only as the cube root of volume. That is why tens of cubic metres retain barely half of their heat through a winter, where thousands retain most. A sphere or a cylinder with height equal to diameter has less surface than a cube of the same volume, so it does a little better.
How Big Does A House’s Store Need To Be?
Inputs, assumed: a house needs an average 2 kW of heat for 120 days; the store is charged to 80°C and heat is usable down to 30°C (a 50 K swing); water 4.18 MJ/m³·K.
- Demand = 2 kW × 24 h × 120 days = 5,760 kWh = 20,736 MJ.
- Heat per cubic metre = 4.18 × 50 = 209 MJ = 58 kWh.
- Volume with no losses = 20,736 ÷ 209 = 99 m³.
- With 1 m of k = 0.05 insulation, 100 m³ keeps 79% (table), so about 99 ÷ 0.79 = 126 m³; with k = 0.096 it keeps 63%, so about 99 ÷ 0.63 = 157 m³. This is a rough first correction, because the loss really follows the store’s temperature as it falls.
Charging: a collector at an assumed 5 kWh/m² per day and 40% efficiency gathers 2.0 kWh/m² per day; over 100 summer days that is 200 kWh/m². To deliver 5,760 kWh at a retention of 79% takes 5,760 ÷ 0.79 = 7,291 kWh, so 7,291 ÷ 200 = 36 m² of collector (48 m² at 63% retention). That is several times the area of a hot-water heater (Sizing for Household Use). These are model figures with assumed inputs: the house demand, the sun and the efficiency all change with place.
The parent page’s 10-50 m³ for “meaningful seasonal storage” is the right scale for a supplement of a house’s heating, not for all of it; a store of 5 m³ keeps only 29% to 52% of its heat over 180 days even with 1 m of insulation.
What You Need
- Your house’s seasonal heat demand in kWh, from measurement over a winter
- The usable temperature swing of your system
- An insulation conductivity you can trust when wet and dry
- Your site’s summer sun, measured (Insolation & Angle)
Method: Size A Seasonal Store
Step 1: Compute demand in MJ; divide by (4.18 × swing) to get the loss-free volume.
Step 2: Choose the insulation thickness and compute τ for that volume; find the retention from e^(−t ÷ τ).
Step 3: Divide the volume by the retention for a first correction; repeat once.
Step 4: Compute collector area from sun × efficiency × days.
Step 5: If the numbers are too large, shrink the demand (insulate the house first) before enlarging the store.
Common Mistakes
- Building a small store and hoping. Retention follows the cube root of size.
- Quoting a loss figure without saying which insulation and which period.
- Planning for dry insulation in wet ground.
Safety
A store this large holds enough hot water to kill, and its collectors can reach steam
A hot water store of 100 m³ at 80°C holds 100 × 4.18 × 60 = 25,080 MJ above 20°C; a failure of a wall or lid releases a flood of scalding water, and a fall into an open one is fatal. Build covers that bear weight, vent the store, and site it away from dwellings. The collectors that charge it stagnate above boiling in summer, so vent them (Plumbing Connections).
What’s Next
- Learn next: District Heating — sharing one large store among many houses
- Combine with: Annualized Geo-Solar Heating — the ground as a store