Appendix E: Useful Conversion Factors

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Key terms and units of measure important to radiometry are presented in Table E.1. Converting units of measure for other selected quantities can be accomplished with the information presented in Table E.2. The multipliers needed to convert from the unit in the first column to the unit in the second column are given in the third column of Table E.2.

E.1 The Many Meanings of WM^–2

Solar irradiance, the amount of solar radiation received by a surface (i.e., radiant energy density) for the minute, hour, day, month, or year is often reported in units of average Wm^–2 or averaged radiant power per unit area for a given time period. The energy per unit area per unit time (Wm^–2) multiplied by the time interval becomes the radiant energy density because energy is power multiplied by time. For example, one Colorado town received an average of 5.6 kWhm^–2 of solar radiation per day. This value is also reported as an average of 233 Wm^–2 over the day.

233 \frac{W \cdot d a y}{m^{2}} =233 \frac{W \cdot 24 \cdot h r}{m^{2}} =5592 \frac{W \cdot h r}{m^{2}} =5 .6 \frac{k W \cdot h r}{m^{2}} (E.1)

$233 \frac{W \cdot d a y}{m^{2}} =233 \frac{W \cdot 24 \cdot h r}{m^{2}} =5592 \frac{W \cdot h r}{m^{2}} =5 .6 \frac{k W \cdot h r}{m^{2}} (E.1)$

Another example of quantifying energy and power is to assume the measured solar irradiance (power density) was 1,000 Wm^–2 over a 5-minute period. The equivalent energy density is 83 Whm^–2.

1000 \frac{W \cdot 5 \min}{m^{2}} =1000 \frac{W \cdot 5 \cdot \min \cdot \frac{1 h r}{60 \min}}{m^{2}} =83 \frac{W \cdot h r}{m^{2}} (E.2)

$1000 \frac{W \cdot 5 \min}{m^{2}} =1000 \frac{W \cdot 5 \cdot \min \cdot \frac{1 h r}{60 \min}}{m^{2}} =83 \frac{W \cdot h r}{m^{2}} (E.2)$

The general public is interested in kilowatt-hours (kWh) that can be produced by a photovoltaic (PV) system. Consider that the average U.S. household consumes between 900 and 1,000 kilowatt-hours of electricity each month. In the previous example, a town in Colorado received an annual average of 5.6 kilowatt-hours of solar energy per meter squared per day on a surface that is tilted to the south by an angle equal to the site latitude (the angle made with the horizontal is 40° for a site at 40°). In an average 30-day month each square meter of a tilted surface will receive about 168 kilowatt-hours of energy; therefore, 6 square meters would be needed to intercept a little over 1,000 kilowatt-hours of energy. However, if the PV panels receiving this energy are about 16% efficient at converting sunlight into electricity, then 36 square meters of PV panels are required to meet the average monthly electrical load of the house. This is roughly a square area that is 6 meters (∼20 feet) on a side.

TABLE E.1
Radiometric Terminology and Units of Measurement

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TABLE E.2
Useful Conversion Factors

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Questions

Given that an average 5.6 kWhm^–2 is incident on a surface and solar panels are 16% efficient, what is the area of photovoltaic panels needed to supply an average of 1200 kWh to a house? What is the answer in meters? Feet?
What is the difference between energy and power?

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Table of Contents for Appendix E: Useful Conversion Factors

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