As you may expect, The solar thermal harvesting component is the central part of all domestic heating and the electric power production parts of our systems. This consists of a number of solar collectors, a thermal storage tank, circulation pumps, temperature sensors, the controller, and the plumbing to tie it into the home systems.

Here's small tutorial by example on how to calculate the collector sq footage, the flow rates and available power relationships.

  One thing about thermodynamics is that it has a HEAVY emphasis on calculus.  I’ve focused on discussions of the Carnot and Rankin and have tried to simplify and clarify where I can. Everything in [ square brackets ] is my inserted comments. In some instances descriptions are out of sequence. I’ve transcribed them in the order that I think would be a natural understandable order.

     η_max= maximum energy conversion efficiency
            =1 – (low temp. / high temp.) 
              condenser cool / boiler hot

Making the hot side even hotter increases efficiency.
Making the cold side even colder increases efficiency

Why does Carnot efficiency (η_max) have that limitation? It's simply the portion of thermal power available that can be harvested. It's useful. Don't get hung-up on it.

Carnot and Rankin cycles are noted as 2T systems. They MEAN two temperature systems – a high boiler temperature & a lower condenser temperature.

Here, I’ll include two textbook problems with solutions to illustrate the technique to determine the size of the solar thermal panels needed to gather a particular power.

Example 1 – A Solar Engine – from the book ‘Thermodynamics’

   It is proposed that a solar energy be used to warm a large “collector plate”; This energy would, in turn, be transferred as heat to a fluid within a heat engine, and the engine would reject energy as heat into the atmosphere. Experiments indicate that about 200 Btu/hr-Ft² of energy can be collected when the plate is operating at 190°F. Estimate the minimum collector area which would be required for a plant producing 1 kw of useful shaft power. [ Notice that this doesn’t specify the working fluid but the implication is that it is vapor to drive the ‘engine’ at 190°F.]

   We first estimate the maximum energy conversion efficiency of this system using the Carnot efficiency as the upper limit. The atmospheric temperature is assumed to be 70°F.

 η_max = 1 – (70°F + 460) / (190°F + 460) = 0.184  
 530°R / 650°R = .815
                             1 - .815 = .184 Carnot efficiency value
      18.4% efficient     

 [The ‘+ 460’ adjusts Fahrenheit to absolute zero –  called temperature Rankin – degrees Rankin - °R

   The efficiency of any real heat engine operating between the collector plate and atmospheric temperatures would be less than this, owing to inefficiencies in real devices. The minimum rate at which energy must be collected is related to the required power output and the maximum energy conversion efficiency.

 [ Ộ_min symbolizes the minimum rate at which energy must be collected ]
 [ ẁ symbolizes work done ]                                          
Ộ_min =  ẁ / η_max  =  1 kw / 0.184 (maximum conversion efficiency)
= 5.44 kw * [ 1 kw = 3,413 btu hr ]  = 18,600 btu hr

…so the minimum area required is : 

                        A_min = 18,600 btu / 200 btu ft²= 93 ft² of solar collector area, minimum.

 – say a 10’ x 10’ solar collector to get 1kw. 

   A real world system might be expected to need two or three times this area since the actual efficiency would probably be considerably less than 18.4%.

 Example 2 –  In the chapter on energy analysis of thermodynamic systems. 

   A small solar engine for desert water pumping uses steam as the working fluid. The hardware is shown in the figure below. 

                                                 2            Ó_e                3  220 degrees F 30 psi

    30 psi boost       ẁ_f    pump                                  ↓ turbine                        
                        work needed →۝                          ۞→  work produced  ẁt Work-Turbine
                       120 degrees F                                   ↓

                                                  1             Ó_c                 4   1 atmosphere

       Water enters the pump as a saturated liquid at 120°F and is pumped up to 30 psia by a small centrifugal pump. The boiler evaporates the water at 30 psia. The saturated vapor enters the turbine at this pressure. The steam leaves the turbine with 6% moisture at 220°F at 1 atmosphere and is subsequently condensed. The flow rate is 300 lbm / hr. [ lbm = pound mass ] The pump is driven by a ½ hp electric motor operating at full load.  

                    Between points 1 and 2 the working fluid is liquid at constant temperature. Pressure is increased.
                    Between points 3 and 4 gas at constant temperature. Pressure is decreased.
                    Between points 2 and 3 liquid boiled to gas at constant pressure. Temperature is increased.
                    Between points 1 and 4 gas condensed to liquid at constant pressure. Temp. is decreased.

Determine the net hp output of this plant,

Notice the English units given – remember 1 btu is 1°F per pound. The heat added is sensible temperature 220°F - 120°F = 100°F plus latent heat of vaporization 16 btu per pound to become steam. 300 * 16 = 4800 btu needed to swing the water temperature 100 degrees - per pound and vaporize it.

   The flow rate is 300 lbs/hr ( /8.4 pounds per gallon = 36 gallons per hour): 34,800 btu / hr

 34,800 btu / hr = ( / 3,413 btu hr = 10.2 kw ) ( / 745 watts / hp ) = 13.7 hp   [ẁt]

That minus the half hp [ẁ_f ] for the feed pump nets out at 13.2 hp will be available.

Determine the energy conversion efficiency [ net shaft-work output hp / energy transfer to the fluid in the boiler = 1 – (low temp. / high temp.)],

 η_max = 1 – (120°F + 460) / (220°F + 460) =
                                             580°R / 680°R             = .853 
                           1 - .853 = .147                      14.7% energy conversion efficiency

Estimate the number of square feet of solar collectors which would be required, assuming that the collectors can pick-up 250 btu / hr per square foot of exposed surface.

34,800 but hr / .147 = 236,734 btu hr needed.

At 250 btu hr per ft² this requires 946 ft² of solar collector area.

             – say an 8’ x 135’ solar collector to get 10 kw all the time the sun shines.

And when it doesn't, run off storage - either thermal or battery. Spend less than you make.

    Working fluids that experience phase change at temperatures lower than water are frequently called ‘organic’, being carbon based. Things such as Freons, propane, butane, methyl chloride and also nitrogen based ammonia.

    It would be useful to have a chart which would show pressure on one axis, heat content on another axis – drawing a line unique to each type of working fluid showing where it’s solid, liquid, and gas, That’s a Mollier diagram, a tool to get an idea of the range of boiler-condenser pressures and temperatures for a working fluid. These ‘Low Temperature Phase Change’ working fluids can be expensive, poisonous, or flammable so it’s important that the volume be minimized and containment is important but can be done - as is done in refrigerators and heat pumps. One must isolate the heat and cool by heat exchangers.