Vail Valley: How well do solar panels perform?
Vail, CO, Colorado
Matthew Charles, who co-authors these columns, recently showed me a spreadsheet in which he codifies the calculations needed to come up with the “COP,” or coefficient of performance of various types of heating and air conditioning systems. If you have ever investigated heat pump systems of any kind you’re probably familiar with the term.
If not, the COP is basically the ratio of energy delivered by a heat pump or other home heating system to the amount of energy it consumes. For example, an electric resistive heating element produces heat with 99 to 100 percent efficiency. A kilowatt of electricity passed across a resistive load like a baseboard electric heater produces a kilowatt of heat output which equals a COP of one. Most people express units of heat as British Thermal Units or Btu, of which it takes 3,412 to equal a kilowatt.
Geothermal or ground source heat pump technology uses the natural heat in the ground while air source heat pumps use heat in the air to supplement electrical energy with impressive results. A review of most ground or air source heat pumps will reveal a COP between 2 and 5, meaning that the heat pump will deliver between 2 to 5 kilowatts of capacity for each kilowatt it consumes to run its compressor. This makes a heat pump system a pretty good deal over time, since it can deliver its finished product – heating or cooling – using 20 to 30 percent of the energy a conventional resistive system can.
Matt was curious about how solar thermal systems would compare to heat pumps and ran his spreadsheet equations using solar energy as the source for Btu’s rather than ground or air temperatures. The results were surprising, even with the understanding that solar energy is an incredibly abundant natural resource.
The spreadsheet takes into consideration any part of the solar thermal system that uses an electrical component that is not powered directly by the sun. These components are typically circulation pumps and computerized controllers and, in larger systems, more pumps, controls and actuated valves.
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A basic solar thermal domestic hot water system comprised of two solar thermal collectors, a storage tank, one 30-watt pump and a controller using five watts while operating and two watts when standing by will produce in the neighborhood of 55,000 Btu per sunny day, or 16.11 kilowatt hours, for no charge from the sun. The pump and controller will use 312 watts (0.312 kilowatt hours) of grid electricity during the 8 hours of system operation. Divide the production by the input and the result is a COP of 51!
This means that for every unit of energy the system required to operate, 51 units of energy were delivered.
Certain types of solar thermal systems can be built without a grid-powered electric pump and computerized system controller and use a photovoltaic powered DC pump instead. These systems will have a small solar panel (usually 30 or 40 watts of power) mounted next to the solar thermal collectors to which the pump is directly attached.
The same panel used to power the DC pump also becomes the system controller due to the nature of how photovoltaic panels work ” when the sun is out, the panel produces electricity and runs the DC pump. When there is no sun, no electricity is made and the pump doesn’t run. It couldn’t be any easier. When no grid electricity is used, the COP for these types of systems is nearly infinite!
One thing to keep in mind is that these COP calculations do not include the embedded energy of the solar thermal components themselves. Because it requires electrical energy to manufacture the components, there is an energy debt that should be paid off by the energy the system produces before the debt-free energy is counted. We won’t worry too much about this because the heat pump manufacturers don’t account for embedded energy, either.
One thing becomes immediately clear when comparing solar thermal energy systems to heat pumps ” you’ll get much more bang for your buck when you put the sun to work for you.
Bill Sepmeier is chief technical officer and Matthew Charles is the design and sales manager for Grid Feeders in Eagle-Vail. Contact them at 970-688-4347.