Solar Water Heating

Size and rate a flat-plate solar thermal collector with the Hottel-Whillier equation. Returns the useful energy gain Qu, instantaneous efficiency, daily hot-water energy, required collector area, and solar fraction, with an efficiency-line chart.


Hottel-Whillier · Duffie & Beckman · ISO 9806

Mode

Collector & Conditions

W/m²
0–1
W/m²·K
°C
°C

Hot-Water Demand & Sizing

L/day
°C
°C
MJ/m²·d
0–1

Results

1096W
Useful gain Qu
548W/m²
Useful flux
60.9%
Efficiency
37.67MJ/d
Daily demand
10.46kWh/d
Daily demand
4.65
Required area
16.20MJ/d
Delivered (this area)
43%
Solar fraction

Efficiency vs (Ti−Ta)/I

00.040.080.15(Ti−Ta)/I (m²·K/W)00.20.40.68

About Solar Water Heating Calculator

The solar water heating calculator models a flat-plate solar thermal collector with the Hottel-Whillier-Bliss equation, the standard relation used in solar engineering. The useful energy gain is Qu = A·FR·[I·(τα) − UL·(Ti − Ta)], where A is the collector area, FR the heat-removal factor, I the plane-of-array irradiance, (τα) the transmittance-absorptance product, UL the overall loss coefficient, Ti the inlet fluid temperature, and Ta the ambient temperature. Dividing by A·I gives the instantaneous efficiency, the familiar straight "Bliss line" that drops as the fluid runs hotter relative to the irradiance.

In performance mode you enter the collector area and operating conditions to get the useful gain and efficiency. In sizing mode you enter a daily hot-water demand — a volume and a temperature rise — and the tool returns the collector area required to meet it from the daily solar irradiation and a system efficiency, plus the solar fraction covered by any installed area you specify. The daily hot-water energy is E = V·ρ·cp·(T_hot − T_cold), and the required area is A = E / (H_daily · η_system).

How It Works

  1. Pick performance or sizing mode and enter the irradiance I, the collector test coefficients FR(τα) and FR·UL, and the inlet and ambient temperatures.
  2. The tool computes the useful flux I·FR(τα) − FR·UL·(Ti−Ta) and multiplies by the area to get Qu (clamped to zero below the no-loss point), and the efficiency η = FR(τα) − FR·UL·(Ti−Ta)/I.
  3. For the demand it computes the daily energy E = V·ρ·cp·(T_hot − T_cold) with water ρ = 1000 kg/m³ and cp = 4186 J/kg·K, reported in MJ and kWh.
  4. In sizing mode it returns the required area A = E / (H_daily · η_system); with an installed area it reports the daily delivered energy and the solar fraction (delivered ÷ demand).

Worked Example

A 2 m² flat-plate collector with FR(τα) = 0.68 and FR·UL = 3.2 W/m²·K runs at I = 900 W/m², inlet 40 °C, ambient 20 °C. The useful flux is 900×0.68 − 3.2×(40−20) = 612 − 64 = 548 W/m², so Qu = 2×548 = 1096 W and the efficiency is 548/900 = 0.609, i.e. 60.9%. For a household using 200 L/day heated from 15 °C to 60 °C, the daily energy is 0.2×1000×4186×45 = 37.674 MJ (≈10.47 kWh). With 18 MJ/m²·day of irradiation and a 45% system efficiency, the required area is 37.674 / (18×0.45) = 4.65 m².

Formulas

Useful energy gain (Hottel-Whillier)
Qu = A * FR * [ I*(tau*alpha) - UL*(Ti - Ta) ]
Instantaneous efficiency (Bliss line)
eta = Qu/(A*I) = FR*(tau*alpha) - FR*UL*(Ti - Ta)/I
Daily hot-water energy
E = V * rho * cp * (T_hot - T_cold)
Required collector area
A_req = E / (H_daily * eta_system)
Solar fraction
f = delivered / demand = (H_daily * eta_system * A) / E

Standards & References

  • Duffie & Beckman, "Solar Engineering of Thermal Processes" (Hottel-Whillier equation)
  • ISO 9806 (solar thermal collectors — test methods)
  • Bliss collector efficiency line (intercept FR(τα), slope FR·UL)
  • ASHRAE 93 (methods of testing solar collectors)

Frequently Asked Questions

What are FR(τα) and FR·UL?

They are the two coefficients from a collector efficiency test. FR(τα) is the optical intercept — the efficiency when the fluid is at ambient temperature — typically 0.6–0.8 for a good flat plate. FR·UL is the loss slope (W/m²·K) describing how fast efficiency falls as the fluid runs hotter than ambient.

Why does the efficiency fall as the inlet temperature rises?

Heat losses scale with the temperature difference (Ti − Ta). The Bliss line η = FR(τα) − FR·UL·(Ti − Ta)/I drops linearly with the reduced temperature (Ti − Ta)/I, so the same collector is far more efficient pre-heating cold water than boosting already-hot water.

Why is the useful gain clamped at zero?

When (Ti − Ta) is large and irradiance is low, the loss term can exceed the optical gain, giving a negative Qu. Physically the controller stops circulating so the collector neither gains nor sheds heat to the loop; the calculator reports zero useful gain at and below that no-loss (stagnation-approach) point.

What system efficiency should I use for sizing?

The overall system efficiency rolls up collector efficiency, piping and tank losses, and array shading/orientation. Typical annual values are about 0.35–0.55 for domestic solar water heaters; 0.45 is a reasonable first estimate. Lower it for cloudy climates or long runs.