What this calculator does
A solar thermal system is not a PV array with hot water. Hot water is a heat demand, and a flat plate collects heat only while its absorber is hotter than the water leaving it. That means the collector usually has to satisfy two different questions at once, and the answer is whichever needs the larger plate.
How to use it
- Enter the daily hot water demand and the highest single hour you expect to serve.
- Enter the insolation for your design day and your site peak irradiation.
- Enter the collector’s optical efficiency and heat loss coefficient from its data sheet — this tool holds no catalogue and picks no model.
- Enter the mean absorber and water temperatures you actually run at; their difference drives the heat loss.
- Read the governing constraint: if the peak hour governs, a collector sized only on the daily total will still fail at 6 p.m.
Governing formula
q_u = η₀·G − U_L·(T̄_abs − T̄_fluid) Where:
-
q_u= Useful flux from the collector, W/m² -
η₀= Optical efficiency, net of optical losses -
G= Irradiation on the collector, W/m² -
U_L= Heat loss coefficient, W/m²·K -
T̄_abs, T̄_fluid= Mean absorber and mean fluid temperature, °C
Derived Equations:
A_energy = daily demand / daily yield per m² A_peak = peak-hour demand / useful power per m² Worked example — 12 kWh a day from 5 kWh/m² of insolation
The figures below were produced by running this page's own engine with the example inputs (Daily hot water demand (kWh/day), Design day insolation (kWh/m²/day), Peak solar irradiation (W/m²), Optical efficiency (–), Heat loss coefficient (W/m²·K), Mean absorber temperature (°C), Mean water temperature (°C), Ambient reference temperature (°C), Peak hourly demand (kWh)). If the calculator returns something different for the same numbers, that is a bug, not a rounding difference.
| Collector area required | 6.15 m² |
|---|---|
| What governs the size | daily energy |
| Useful energy per m² per day | 1.95 kWh/m²/day |
| Collector efficiency | 39 % |
| Useful power at peak sun | 525 W/m² |
| Area for daily energy alone | 6.15 m² |
| Area for the peak hour alone | 3.81 m² |
| Absorber-to-water difference | 15 K |
| What this means | The daily energy needs 6.15 m² and the peak hour 3.81 m², so the array is sized by energy. A collector large enough for the daily total still meets the peak hour here. |
- At 15 K above the water, the collector delivers 0.75 × 800 − 5 × 15 = 525 W/m² at peak.
- Over the day that is 1.95 kWh/m², about 39 % of the incident energy.
- The daily total needs 12 / 1.95 = 6.15 m².
- The peak hour needs 2 kWh / 525 W/m² = 3.81 m², so the daily energy governs here.
Assumptions & limitations
Engineering calculation. Flat-plate collector area from a heat balance, sized twice — once for daily energy and once for the peak hour — with whichever governs reported.
- The optical efficiency, heat loss coefficient and daily insolation are YOUR inputs. This tool holds no irradiance database, no climate file and no collector catalogue, and it selects no collector model.
- One collector temperature and one insolation value are used for the whole day. A real collector produces more at noon than at 8 a.m., so the daily yield is an average, not a profile.
- No stagnation temperature, freeze protection, pump power, expansion vessel or pipework heat loss is modelled. This sizes the collector plate only.
- The daily-energy and peak-power sizings use the same collector efficiency, so the switch between them moves with the demand shape you entered, not with a measured system.
This tool supports professional engineering work; it does not replace a sealed design. Confirm every result against the applicable standard and a licensed engineer before it is built.
References & standards
- ISO 9806 — flat-plate solar collector test and reporting, the basis for the η₀ and U_L inputs.
- ASHRAE Handbook — Solar Heat for Buildings, domestic hot water collector sizing.