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Lighting systems convert only a minority fraction of their electrical input into useful light output. Much of the rest is released directly as heat into the space. Therefore, any upgrade of the lighting system that reduces input wattage reduces the amount of heat that must be removed by the air cooling system. This results in air cooling energy savings during the operation of the building. In new construction, an energy-efficient lighting design can result in significant savings in the installed cost of cooling systems. A rule of thumb in the industry is that 1 kWh of air conditioning energy is saved for every 3 kWh of lighting energy. This, however, is often not accurate because it does not account for different climates. A retrofit in a building in Alaska, obviously, will not yield the same air conditioning energy savings benefit as in a building in Florida - in fact, in Alaska this heat is quite useful, and the retrofit could result in a much higher heating bill! In the northern regions, the cost of additional heating can cancel out the air cooling energy savings, but in many areas of the United States the air cooling savings, which will be 0-30% of the lighting energy savings, will exceed this additional heating cost. How to Calculate Air Cooling Energy Savings Robert Rundquist, PE is the president of R.A. Rundquist Associates of Northampton, MA and a professional engineer with nearly three decades' experience in heating, ventilation and air conditioning (HVAC) system design, energy analysis and energy calculation research. He offers a formula to assess a more accurate figure for air cooling energy savings that was derived from both independent research and research conducted for the American Society of Heating, Refrigeration and Air Conditioning Engineers (ASHRAE). It has been validated by DOE-2 computer runs and other methods. 1. Lighting energy consumption must be reduced by a specific amount that stays constant throughout the year. This is most predictable in a retrofit, but can also work for some controls and other applications where hours of operation are reduced. 2. Determine the fraction of the year of the cooling season. (Download the DOE's Strategy Guideline: Accurate Heating and Cooling Load Calculations to get typical cooling seasons in the US) 3. Determine the fraction of the daily load met by mechanical cooling. Basically, this question asks, how much of the lighting system's heat must be removed by the cooling system? Usually this is about 90%, with 10% dissipated. 4. Determine the air cooling system's coefficient of performance. Tests on cooling systems have shown that for every watt (W or kW) put into the system, 2.7 watts (W or kW) of cool air is produced. The actual figure can vary, however, due to a range of factors; the 2.7 figure can be used although it is best to use the actual system's coefficient of performance. 5. Calculate using the formula below: Fraction of Lighting Savings as Air Cooling Savings = Fraction of the Year of the Cooling Season x Lighting Load Met by Mechanical Cooling ÷ System's Coefficient of Performance 6. Example: Suppose we retrofit a system in Raleigh, North Carolina, which has a cooling season of 30 weeks, and remove 20,000kWh from the lighting load. Fraction of Lighting Savings as Air Cooling Savings = 30 ÷ 52 = 0.5769 or 0.58 x 0.9 ÷ 2.7 = 0.19 That means that for every 1 kWh of lighting saved, we save 0.19 kWh of air cooling energy. In our example, this means that we have removed 3,800kWh (20,000kWh x 0.19) of air cooling load. If the local utility charges an average commercial rate of $0.065 per kWh, then we have reduced energy costs by $1,300 per year for lighting and an additional $247 per year for air cooling. See also: Retrofit Economics
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