Testing a Solar Water Heater Efficiency Claim
A household solar water heater has a flat-plate collector area of 3.0 m². On a clear day the average solar irradiance on the collector plane is 700 W/m². In a test, 150 L of water is circulated through the collector and stored in an insulated tank. Over 4.5 hours the water temperature rises from 22 °C to 52 °C. The manufacturer claims a system efficiency of 60% under these conditions. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and the density of water is 1.0 kg/L. The pump uses 40 W of electrical power, which is not included in the manufacturer’s efficiency claim. The householder wants to know whether the claim is believable, and whether buying a second identical collector would double the temperature rise.
1. Calculate the useful thermal energy gained by the water and the solar energy incident on the collector. Hence find the system efficiency and judge the manufacturer’s claim.
Read the worked answer
Mass of water:
[ m=150\ \text{L}\times1.0\ \text{kg/L}=150\ \text{kg} ]
Useful heat:
[ Q=mc\Delta T=(150)(4200)(52-22) ]
[ Q=150\times4200\times30=1.89\times10^7\ \text{J}=18.9\ \text{MJ} ]
Incident solar energy:
[ E_{\text{solar}}=700\times3.0\times(4.5\times3600) ]
[ E_{\text{solar}}=700\times3.0\times16200=3.40\times10^7\ \text{J}=34.0\ \text{MJ} ]
Efficiency:
[ \eta=\frac{18.9}{34.0}=0.556\approx56% ]
The measured efficiency is about 56%, not 60%. The claim is slightly optimistic but not wildly so. If the pump’s 40 W is included, the electrical input over 4.5 hours is (40\times16200=0.648\ \text{MJ}), so the overall efficiency is about 54.5%. The manufacturer’s figure is therefore at the edge of what these measurements support; it is not proven by this test.
2. Explain why doubling the collector area would not double the temperature rise, and identify one measurement change that would make the test fairer.
Read the worked answer
Doubling the area roughly doubles the incident solar power, but heat losses also increase. The collector loses heat by conduction through insulation, convection to the air, and thermal radiation, and these losses depend on the collector temperature above ambient. A larger collector raises the water temperature faster, which increases the temperature difference and therefore increases losses. The tank also has a fixed heat-loss area, and the pump flow may not match the larger collector. The net useful gain is incident solar energy minus losses, so the temperature rise increases by less than a factor of two; in a closed system near boiling, it could increase hardly at all.
A fairer test would record the ambient temperature and wind speed, and measure the water temperature at several times rather than only at the start and end. This allows the average loss rate and the effect of changing conditions to be estimated. The pump energy should also be metered and reported separately from the solar gain.