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HEAT GAIN

HEAT GAIN. Types Of Loads A. Sensible heat transmitted through: 1) Walls 2) Ceilings 3) Floors B. Latent Heat Loads 1) Humidity in air 2) To remove one pound of water from air requires 970 BTU’s C. Building Leakage 1) Ventilation 2) Infiltration.

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HEAT GAIN

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  1. HEAT GAIN Types Of Loads A. Sensible heat transmitted through: 1) Walls 2) Ceilings 3) Floors B. Latent Heat Loads 1) Humidity in air 2) To remove one pound of water from air requires 970 BTU’s C. Building Leakage 1) Ventilation 2) Infiltration

  2. D. Solar Load 1) Radiant from the sun through windows and doors E. Energy Devices 1) Electrical devices a. 3.414 BTU/H per watt 1. Lights 2. Motors 2) Gas Appliances a. Stoves b. Ovens

  3. 3) People a. 300 BTU/H per person, average 4) Duct Loss • Calculating Heat Gain A. Outdoor design temperature for Springfield / Hartford area is 96° F B. Indoor design temperature for cooling is 72° F and can be adjusted for the application C. Enter all values as is done for heat calculations D. Solar gain formula: 1) Surface area x sun factor (determined by it’s exposure)

  4. a. Depending on geographic location and compus direction of the sun on the surface of the glass b. Northern exposure will have little solar gain c. Southwestern exposure will have a high solar gain • Infiltration A. Cracks around windows and doors 1) Seepage • Ventilation A. Air that is intentionally brought into the structure

  5. Duct Loss A. Ducts running through an unconditioned space must be insulated to prevent picking up heat B. A/C ducts will sweat where the dew point is high (basements) C. Ducts run in attics will reduce the capacity of the system and the insulation will allow some heat transfer (leakage) D. Multipliers are used to increase the system size to compensate for the duct loss • Surface film factor has an effect and must be considered

  6. Cooling Calculations • Record the number of people and multiply by 300 and record people BTU/H load • Add cooling BTU/H including the average appliance BTU/H A. The appliance load may have to be calculated if there are more than 350 watts of appliances

  7. 3. Multiply the new subtotal by the moisture factor of 1.3 and record the total BTU/H gain 4. Add total BTU/H for each room for a total house BTU/H and record it 5. To find CFM requirements for each room A. Divide the room BTU/H by 12,000 the answer is tonnage requirements B. Multiply the answer by 400 (CFM per ton average for AC) the answer is your individual room CFM requirements

  8. 6. To find CFM requirements for whole house 7. Divide total BTU/H by 12,000 (BTU per ton) A. The answer is tonnage requirements 8. Enter tonnage requirements A. Multiply the answer by 400 (CFM per ton) average for AC B. The answer gives you CFM requirements 9. Enter the CFM requirement

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