Hi,
Many years ago, I acquired a Freznel lens from a microfiche viewer. It is approximately 1mtr x 750mm. It focuses the sun to a 100mm square, and as can be imagined very hot when the sun is out.
I also acquired a 100mm liquid radiator.
I’ve been meaning to carry out experiments, comparing the power output of this set-up with a same area PV panel, but haven’t got round to it.
Can anyone assess what would be the results of such tests?
Camerart.
Do you mean assess? If so, please correct it.
I’m not sure what sort of assessment you want or would think likely?
I’d expect the Fresnel lens to be able to make some hot water; more than would be generated by the same area of PV, because I’d expect the losses from the radiator and whatever plumbing would be less than the losses due to the relatively low efficiency of whatever PV panel.
But you’d need to give a lot more information about the details of the experimental setup for anybody to be able to calculate some numbers, and I for one would need some more motivation 
There would need to be a quite complicated drive system to track the sun, and any inaccuracy there would seriously damage the efficiency.
Hi,
The focused heat will make steam, so try not to narrow ideas down to plumbing, as there may be a turbine involved.
The assessment would be a moment in time e,g, a calculation, so no moving parts yet.
A calculation would be: The area of the PV panel output at a certain sun/light level, compared with the area of the lens, which has concentrated the heat from low level to high level, which as you know can be more efficient.
C.
Hmm, I suggest you do your own calculations, with your own assumptions.
Hi D,
Ok, thanks.
C
For scale and simplicity PV is hard to beat. Aside from efficiency considerations (below), complexity and reliability of thermal (e.g. steam) systems will be a challenge. It’s hard to beat a system with no moving parts!
As a rule of thumb, for a good solar location, assume 1000 W solar radiation per sq meter at the surface around noon in summer (eg Sydney Australia). A good PV panel will convert 19-20% of that to electricity, so around 200 W per sqm. A fresnel lens - which is most likely plastic, might capture and concentrate 90% of that energy providing around 900 W thermal power for the same area. This then needs to be transferred to a fluid for example to heat water or for another use such as power generation. The efficiency of this process could be (very) low, depending on the heat exchanger design (material, pumping, radiative losses etc). Converting heat to electric power is thermodynamically inefficient unless you can achieve very high temperatures (even modern open-cycle gas turbines struggle to achieve better than 35%). At the sort of fluid temperatures you could achieve with a fresnel lens any heat engine would be woefully inefficient - say 10%. So the 900 W of thermal power might provide 90 W electric power - at much greater complexity and cost.
Modern solar thermal power stations use mirrors to concentrate sunlight, but even highly reflective parabolic troughs can only achieve fluid temperatures around 450 C (it’s also the maximum temperature that the oil can withstand) which limits the efficiency to about 20%. Using heliostats (almost flat mirrors) to focus sunlight to a tower can achieve much higher temperatures and therefore thermal efficiencies, but these come with other challenges such as the need for stable, strong materials to handle the temperatures.
Hi A,
An excellent explanation.
I looked into thermocouples and peltier effect, but again, very low efficiency of conversion.
I recall that salt is used instead of oil in high temperature sun focusing power stations. I doubt I could build one of those in my back garden:)
Thanks for saving me what could have been months of experiments.
C.