Designing a Small-Scale Solar Energy System: Boiler and Turbine Considerations

In summary: The part I am afraid will use the most energy is the pump to introduce cold water into the boiler.The pump you may use you can find it by apply the formoula of the 1st law of thermodynamics.
  • #1
sngtdt
10
0
I am an amateur science/engineering enthusiast (AA electrical engineering), I would like to design/build a solar energy system consisting of a small tracking parabolic mirror heating a steam boiler to drive a small turbine-generator. The mirror I am thinking 3 foot diameter and I have some idea how to go about constructing it. The questions come in on the boiler design, pressure, safety, how to pump in the cold water feed, check valving, turbine construction, cooling the steam, and many phases of constructing same. Any info on any of the above would be appreciated.
 
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  • #2
I am also an amateur, so I can't answer much of that. What I would seriously recommend, however, is that you have your pressure vessel built by a professional shop, with an appropriate relief valve provided by them. Safety always has to come first.
 
  • #3
Well, a useful number that you will want to know is this- solar energy incident on the Earth at sea level is about 125 W/m^2. With a 3 ft diameter mirror, this means you will only have a maximum power of about 80 W to work with; that doesn't sound like a whole lot when you're wanting to drive a steam turbine and an automatic tracking system...
 
  • #4
Mech_Engineer said:
Well, a useful number that you will want to know is this- solar energy incident on the Earth at sea level is about 125 W/m^2.

Actually it's closer to 1000 W/m^2. You may be thinking of a photovoltaic cell, where the low efficiency results in an output power close to what you are saying.

See the paragraph under "Solar constant" here:
http://en.wikipedia.org/wiki/Solar_radiation
 
  • #5
Thanks for the info, I am aware the pressure vessel would need special design cosideration, as far as the area of the mirror, this would be more or less a model. If it worked, something larger could be made.
 
  • #6
Redbelly98 said:
Actually it's closer to 1000 W/m^2. You may be thinking of a photovoltaic cell, where the low efficiency results in an output power close to what you are saying.

See the paragraph under "Solar constant" here:
http://en.wikipedia.org/wiki/Solar_radiation

That would be the amount of power incident at the edge of Earth's atmosphere (still in space). By the time it gets through Earth's atmosphere to the ground, I think it is anywhere from 100-375 W/m^2 average. That's at sea level in North America. The number will rise as you go higher in the atmosphere or closer to the equator, and peak power will incease or decrease depending on time of day and weather conditions.

Wikipedia Solar Energy Article said:
...in North America the average insolation at ground level over an entire year (including nights and periods of cloudy weather) lies between 125 and 375 W/m² (3 to 9 kWh/m²/day).
http://simple.wikipedia.org/wiki/Solar_energy
 
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  • #7
Mech_Engineer said:
That would be the amount of power incident at the edge of Earth's atmosphere (still in space).

Well no, that number is 1300 to 1400 W/m^2. According to your link.

By the time it gets through Earth's atmosphere to the ground, I think it is anywhere from 100-375 W/m^2 average. That's at sea level in North America. The number will rise as you go higher in the atmosphere or closer to the equator, and peak power will incease or decrease depending on time of day and weather conditions.

http://simple.wikipedia.org/wiki/Solar_energy

Checked out your link. They are averaging in the time the surface is in total darkness, and cloudy weather, to get those numbers. They also factor in that a horizontal surface is not oriented at a perfect 90 degrees to the sun.

http://simple.wikipedia.org/wiki/Insolation

The 1000 W/m^2 figure is more applicable to the purpose at hand: boiling water during daylight hours, where one can aim the setup directly at the sun.
 
  • #8
Mech_Engineer said:
... that doesn't sound like a whole lot when you're wanting to drive a steam turbine and an automatic tracking system...

The part I am afraid will use the most energy is the pump to introduce cold water into the boiler
 
  • #9
The pump you may use you can find it by apply the formoula of the 1st low of thermodynamics.

Q= m . C .Dt

m= mass flow rate , c= water heat coefficient 4200, dt= temperature difference between the boile imput and output.

Q= heat gain you may have.

Note: the above formula is only valid for closed systems.
 
  • #10
Before you get to far along on a design, you should get a copy of the ASME Sections 1 and 8 for boiler and pressure vessel code requirements. Also, a copy of B31.1 Power piping would be needed. These are the American Society of Mechanical Engineers code books that govern the minimum design requirements. Also, make sure they are the 2007 editions and for Section 8 you should use Division 1 not Division 2.
 
  • #11
sngtd, I realize that what you're building is just a model, but you probably won't get much power at all with a 3-foot mirror, so the turbine generator isn't going to be able to be big.
 

1. What is solar energy design?

Solar energy design refers to the process of designing and implementing systems that harness the energy from the sun to generate electricity or heat. It involves the use of solar panels, inverters, batteries, and other components to convert sunlight into usable energy.

2. What are the benefits of using solar energy design?

There are several benefits of using solar energy design, including reducing carbon emissions, reducing reliance on fossil fuels, and lowering energy costs. It also provides a sustainable and renewable source of energy and can increase property value.

3. What factors should be considered in solar energy design?

When designing a solar energy system, factors such as location, climate, shading, and energy needs should be taken into account. The orientation and tilt of the solar panels, as well as the type and size of the system, are also important factors to consider.

4. How efficient are solar energy systems?

The efficiency of solar energy systems varies depending on various factors, such as the type of panel, location, and weather conditions. On average, solar panels have an efficiency of around 15-20%, meaning they can convert 15-20% of sunlight into usable energy.

5. Can solar energy systems be used in all areas?

Solar energy systems can be used in most areas, but the amount of energy they can generate may vary. Areas with high amounts of sunlight and minimal shading are ideal for solar energy systems. However, advancements in technology have made it possible to use solar energy in areas with less sunlight, but the efficiency may be lower.

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