At 260 Fahrenheit in the Sun between Mars and Earth and -280 F in the shade could 2 large tanks rotate slowly in and out of its own shade to transfer steam backwards and forwards past a turbine?
In principle, this setup could run with some tweaks (already addressed in other answers and comments). But the showstopper here is small radiator surface area, especially on the cold side. It wouldn't help you any to concentrate heat with a parabolic reflector on the sunny side, if the dark side lacked the capacity to radiate it away. The system would get hotter and hotter until it melted. So you would need a larger area at least on the dark side. You could make the radiators different sized, but AFAIU the symmetry is required, to make this invertable like a hourglass.
A similar, but improved setup would have two flat plates at right angles to each other (looking like the skewed kind of kayak paddle with large blades and a very short shaft). The sunlight would be edge on to one of the "blades", so both of its faces would radiate heat out into space. The other blade would be facing the sun, so one side of it would be illuminated: the hot reservoir, in thermodynamic terms. But the back face of this blade would also be radiating out into space. So essentially, with two identical "blades" you would have the cold side radiator with three times the area of the hot side acceptor, which would be about right. And you could rotate the whole thing a quarter turn on its "shaft" to irradiate the other blade and reverse the flow, just like you do in your model.
Except that the direction of flow would not be away from the sun, but sideways, back and forth in a cycle. From the insolated surface via the connecting "shaft" and turbine to both sides in succession of the "edge on" radiator, back through the "shaft" and turbine-driven pump to the shadowed surface behind the "face-on" blade, and back to the insolated surface in a cycle. A fully symmetric arrangement with four identical radiators, one of which is heated, and three are cooled.
They could, but there's no real need to. You can just focus sunlight on a static boiler that generates steam to run past a turbine, through a condenser, and back (in liquid form) into the boiler. Just like closed-cycle turbine engines on Earth.
That lets you limit thermal cycling of the components, which will increase lifespan, and better specialize the heater and radiator components, improving efficiency.
There will be limits on this.
The solar flux on Mars is about 43% that on Earth. The Mars day is 24.5 hours. So you need to have corresponding amounts of surface area. You need area to absorb the heat. Possibly you could set up some kind of system such as SEGS to collect the heat. This is a large collection of mirrors that concentrate heat on a central generating facility. This heats a synthetic oil, which then drives a steam turbine.
You need area to dissipate heat from the "out" side of whatever engine you run. The atmosphere on Mars is very thin, about 1% of Earth, so you have some trouble losing heat. Just being "in the shade" won't do. You need some way to carry the heat away. If you have to run some honking-big fan to push the thin air then you lose a lot of the energy from the generator. Without a fan you will just heat the air nearby and lose efficiency very quickly.
Except during one of Mars's wind storms, when the sand probably covers your collector with dust anyway. You will need some kind of mirror cleaning system.
Since there is extremely little water on Mars, one of the favorite methods of losing heat for Earth based power stations is impossible. You can't use a lake or river.
Possibly you could use some large radiator structure. Without water you probably need a heat radiator of similar size to the collector.
SEGS quotes a capacity factor of 19.2%. There is even less cloud cover on Mars than in California, so maybe that could be exceeded. There are also no birds on Mars, so you won't be finding fried eagles piled around your mirrors. So, scaling to SEGS, you might get into the range of 1000 kW hr per m^2 per Earth year.
No, it won't work.
What you have put in the picture is just a fancy version of a moka pot, therefore it's not going to work to produce energy.
If you want to set up a power generator using a Rankine cycle, you need to have a pump to compress the liquid water.
There are four processes in the Rankine cycle. The states are identified by numbers (in brown) in the T–s diagram.
- Process 1–2: The working fluid is pumped from low to high pressure. As the fluid is a liquid at this stage, the pump requires little input energy.
In other words Process 1-2 is [Isentropic compression in pump]
- Process 2–3: The high-pressure liquid enters a boiler, where it is heated at constant pressure by an external heat source to become a dry saturated vapour. The input energy required can be easily calculated graphically, using an enthalpy–entropy chart (h–s chart, or Mollier diagram), or numerically, using steam tables.
In other words Process 2-3 is [Constant pressure heat addition in boiler]
- Process 3–4: The dry saturated vapour expands through a turbine, generating power. This decreases the temperature and pressure of the vapour, and some condensation may occur. The output in this process can be easily calculated using the chart or tables noted above.
In other words Process 3-4 is [Isentropic expansion in turbine]
- Process 4–1: The wet vapour then enters a condenser, where it is condensed at a constant pressure to become a saturated liquid.
While compressing water is a trivial task (segment 1-2 in the picture below), the energetic outcome of using the resulting enthalpy from the steam (line 3-4) is what makes steam generators useful.
Without pump you are basically moving back and forth on line 4-1, extracting practically 0 useful work.