From recycled TV Lens!
6-8 thousand projected for a production system
http://peswiki.com/index.php/Directo...l_Solar_Design
but
This is something people with basic skills can build themselves with a little determination and research!
Charles Shults' Fresnel Solar Design
A 60"x40" lens has an area of (60*0.0254)*(40*0.0254) = 1.55 m^2
Solar constant is 1340 W/m^2, but at the surface, due to scattering it's more like 800 W/m^2 on a sunny day.
Incoming power = (1.55 m^2) * (800 W/m^2) = 1240W, which is what he says (1.2 kW).
But absorption of this energy isn't perfectly efficient; the target has some reflectivity and also emits heat via thermal radiation.
So, actual available heat energy will be smaller than 1.2 kW, but I'll ignore that in case the lens is bigger than 40" tall.
I'll go with 1.2 kW of available heat energy.
Water from the thermal panels is at most at 180 F = 82C
Energy required to bring that water to 100C = (100C - 82C) * (4.186 kJ/kgC) = 75 kJ/kg
Latent heat of vaporization of water = 2260 kJ/kg
So, it takes (2260 + 75) = 2335 kJ per kg of water boiled.
Density of water is about 1 kg/liter
This means that the most water this system can boil is
Volume per unit time = power / energy per unit volume
= (1.2 kJ/s) / (2335 kJ per liter) = 0.514 ml/s (0.017 ounces per second, for the Americans)
That's a very small amount of water. Even if there were no thermal losses after the steam generator, there's not much steam power available.
Sorry.
It would be better to consider a Sterling engine, or maybe thermocouples, with that sort of power output.
A relevant example: boil a pot of water on a small stove burner. If you push down on the lid, the steam hisses out. Is it enough to run a useful turbine? No; but the power is about the same as the Fresnel system (bigger, actually).