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Solar Thermal Solar Energy Workshop Colorado School of Mines

Solar Thermal Solar Energy Workshop Colorado School of Mines. Presented by: Daimon Vilppu, President, Simply Efficient Daimon@Simplyeff.com. What it all boils down to. Radiation Conduction Convection. Or, as I like to say; Heat in/Heat out. Heat Transfer of the collectors.

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Solar Thermal Solar Energy Workshop Colorado School of Mines

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  1. Solar ThermalSolar Energy WorkshopColorado School of Mines Presented by: Daimon Vilppu, President, Simply Efficient Daimon@Simplyeff.com

  2. What it all boils down to • Radiation • Conduction • Convection Or, as I like to say; Heat in/Heat out

  3. Heat Transfer of the collectors • Allow sunlight in (transmittance) • Don’t allow sunlight out (emittance) • Absorb the sunlight (absorptance) • Transfer the sunlight energy to a fluid running through the collector (conductance and convection) • Lose as little heat as possible (convection)

  4. Solar Ovens

  5. Solar Pool Heating Collectors

  6. Solar Hot Water and Heating

  7. Flat Plate Collectors

  8. Evacuated Tube Collectors

  9. Evacuated Tube Collectors Viessman Vitosol 300, Cut-awayView

  10. High Temperature (Concentrating Systems), Industrial Power Generation, Hot Water, Process Heat, System Owned By Solucar.

  11. Schott Solar Power Plant

  12. Solucar Parabolic Trough Concentrating Collector • The physical characteristics of the concentrator modules are: Overall Module Size 7 ft. 6 in. x 20 ft.(2.3m x 6.1 m) • Concentrator Weight 178 lb ( 81 kg) • Concentrator Rim Angle72° • Materials of Construction: Aluminum Reflective SurfaceOptions: Aluminum acrylic Enhanced polished aluminumLightweight, low maintenance concentratorReceiver • The receiver specifications are: • Absorber Tube Outside Diameter 2.0 inch (5.08 cm) • Absorber Material SteelSelective Surface Blackened nickel • Absorptance  0.96 - 0.98 • Emittance (80°C)  0.15 - 0.25 • Absorber Envelope Material Borosilicate glass Envelope • Anti-Reflective Coating Sol gel • Transmittance  0.95 - 0.965 • Maximum Operating Temperature 550°F (288°C)

  13. Solar Radiation Is Radiant energy from the sun (electromagnetic radiation) produced by a nuclear fusion reaction. Half of this radiation is in the Visible spectrum, the other half is mostly in the near infra-red spectrum of light.

  14. Thermal systems use Diffuse and Direct Normal Light • The National Solar Radiation Database gives us TMY (typical meteorological year) radiation data for cities across the US • A computer simulation (TRNSYS) uses that data to give us the amount of radiation on a tilted surface. • Denver, 45 degree, 20.89 MJ/m2 day, which is Mega Joules (106) per square meter for a day (1839 BTU/ft2 day) • SRCC (Solar Rating and Certification Center) • Joule is the SI standard unit of energy (equivalent to a BTU) • A Watt is Power in the SI units system. • Also know as a Joule/Sec (BTU/hr)

  15. Solar Thermal Collectors Collect the suns radiation and transfer that to a fluid as it runs through them. • Water, Propylene Glycol, Oil, Air • High Temperature (Concentrating Systems), Industrial Power Generation • Medium Temperature (Evacuated Tubes) • Residential and Commercial • Low Temperature (Flat plates) • Unglazed (Pool Panels)

  16. Energy Storage WATER

  17. Why Water? • Specific Heat is the amount of energy required to raise 1 gram of a material by one degree Kelvin. • Water has a Constant Pressure specific heat of 4.183 J/g*K, Air= 1 J/g*K, Hydrogen= 14, Concrete= 0.88 • It is a measure of how much energy you can store in a mass of a material • Water is frequently used because it is relatively cheap, abundant and holds more energy than most materials

  18. How Much Water? Q=MCp(T2-T1) Q=energy Cp=Specific Heat T2=Final Temperature T1=Initial Temperature IF we had 2200 Kg of water ( approx. 120 gallons) And we wanted to raise its temperature 39 Kelvin (approx. 70 F)

  19. We would need how much energy? Q=(454 Kg)x(4.183 J/g*K)x(39K) Q= 74 MJ (Mega-Joules)

  20. How many flat plate collectors would we need to do this in one day? • A typical flat plate panel can collect on a cool clear day about 34 MJ/m2*day Energy needed = 74 MJ 74 MJ/ (24 MJ/m2*day)= 3 m2 A 4’x8’ Collector is approx. 3 m2

  21. The rest of the equation • For a system in Denver, I would recommend two 4’x8’ panels. • For a family of 3-4 • Solar Fraction, The amount of energy provided by the solar system divided by the amount of energy needed (or used). • Modeling shows a 51% solar fraction for one panel and an 81% solar fraction for two panels (using Retscreen, available free online), a 70% solar fraction is considered best. • What we haven’t considered • Thermal losses • In the pipes to and from the collector • In the heat exchange between the collectors and the solar storage tank • In the heat lost from the solar storage tank to the surroundings • In the energy used up to pump the fluid through the collectors

  22. Economics • That system would cost from $8-10,000 • Would save about $500/year of electricity Which gives a best case simple payback of 16-20 years for electricity. • This does not include; the tax federal tax credit, any increase in electricity, appreciation of the home or reduction of greenhouse gasses

  23. Thanks! I will be happy to answer questions Daimon Vilppu Daimon@Simplyeff.com 303-898-9951 www.simplyeff.com

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