Solar Thermal Solar Energy Workshop Colorado School of Mines Presented by: Daimon Vilppu, President, Simply Efficient

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

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

Solar Ovens

Solar Pool Heating Collectors

Solar Hot Water and Heating

Flat Plate Collectors

Evacuated Tube Collectors

Viessman Vitosol 300, Cut-awayView Viessman Vitosol 300, Cut-away View

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

Schott Solar Power Plant

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 Surface Options: 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 Emittance (80°C) Absorber Envelope Material Borosilicate glass Envelope Anti-Reflective Coating Sol gel Transmittance Maximum Operating Temperature 550°F (288°C)

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.

Thermal systems use Diffuse and Direct Normal Light 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 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. A computer simulation (TRNSYS) uses that data to give us the amount of radiation on a tilted surface. –Denver, 45 degree, MJ/m 2 day, which is Mega Joules (10 6 ) per square meter for a day (1839 BTU/ft 2 day) –SRCC (Solar Rating and Certification Center) Joule is the SI standard unit of energy (equivalent to a BTU) Joule is the SI standard unit of energy (equivalent to a BTU) A Watt is Power in the SI units system. A Watt is Power in the SI units system. –Also know as a Joule/Sec (BTU/hr)

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

Energy Storage WATER

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

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)

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

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/m 2* day A typical flat plate panel can collect on a cool clear day about 34 MJ/m 2* day Energy needed = 74 MJ 74 MJ/ (24 MJ/m 2* day)= 3 m 2 A 4’x8’ Collector is approx. 3 m 2

The rest of the equation For a system in Denver, I would recommend two 4’x8’ panels. 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 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

Total Solar Power Available to the USA 0.40 x 1367 W/m 2 = W/m x 1367 W/m 2 = W/m 2 –Available to the Earths’ surface Available to the USA Available to the USA W/m 2 x 9x10 12 m 2 = Trillion KiloWatts W/m 2 x 9x10 12 m 2 = Trillion KiloWatts Daily, Assuming 5 hours of useful sunlight that would be 5x4.921 Trillion KW= Trillion KWatts/day Daily, Assuming 5 hours of useful sunlight that would be 5x4.921 Trillion KW= Trillion KWatts/day

USA use of Power, ,278 Trillion BTU’s/year Convert that to Daily Watts 1 BTU= 1055 Joules 365 days/year (100,278 Trillion BTU’s/year)x(1055 J/BTU)x(1 year/365 days)= 289,800 Trillion Watts/Day= Trillion KW/Day

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

Thanks! I will be happy to answer questions Daimon Vilppu www.simplyeff.com