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Infiltration/Ventilation

Infiltration/Ventilation. HVAC 7a CNST 305 Environmental Systems 1 Dr. Berryman. Infiltration. Types of heat losses?. Types of heat gains?. Calculating Infiltration. Crack method Air change method Effective leakage-area method. Sensible BH = 1.08 x  T x CFM Latent

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Infiltration/Ventilation

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  1. Infiltration/Ventilation HVAC 7a CNST 305 Environmental Systems 1 Dr. Berryman

  2. Infiltration Types of heat losses? Types of heat gains?

  3. Calculating Infiltration • Crack method • Air change method • Effective leakage-area method Sensible BH = 1.08 x T x CFM Latent BH = 0.68 x G x CFM Equations w/ Air Constants

  4. Air Crack Method Divide by 60 min for CFM Lower pressure differentials mean tighter construction

  5. Crack Example (winter) BHsens = 1.08 x T x CFM • Given: • 3o 5o Double Hung Window • No added humidification • IA conditions: DB=75oF • OA conditions: DB=-5oF • Medium Construction • How many BTUH are lost due to frame wall infiltration? 1.08 x 80oF x (16LF)(21 CF/60)= EXAMPLE 484 BH

  6. BHsens = 1.08 x T x CF 60 Air Change • Converts equation to cubic feet • Equals one room change • Designer will use 0.3 to 2.0 room changes per hour • Occupancy • Climatic condition (i.e. winter vs. summer) • Construction (tight or loose) • Least accurate of the three methods

  7. BHsens = 1.08 x T x CF 60 (.35) 1.08 x 80oF x 12000 CF 60 Air Change Example • Given: • 20’ x 60’ x 10’ room • Winter - 0.35 RCH • No added humidification • IA conditions: DB=75oF • OA conditions: DB=-5oF • How many BTUH lost due to infiltration? EXAMPLE BHsens= 6,048

  8. Ventilation FA 200 CFM SA 1600 CFM RA + pressure 1400 CFM

  9. Sensible BH = 1.08 x T x CFM Latent BH = 0.68 x G x CFM Equations w/ Air Constants Calculating Ventilation • CFM per Person • Use ASHARE chart • Room Air Change per Hour • 0.5 – 2.0 RCH

  10. Determining Ventilation Requirements • Determinates • Building Application • # Occupants

  11. Ventilation - CFM per Person • Given: • 2000 SF Coin Operated Laundry • Winter design • No added humidification • IA conditions: DB=75oF • OA conditions: DB=-5oF • How many BTUH lost due to ventilation? BHsens = 1.08 x T x CFM EXAMPLE (1.08)(80oF)(15cfm)(40 people) BHsens= 51,840

  12. BHsens = 1.08 x T x CF 60 1.08 x 80oF x 20,000 CF 60 (2.0 RCH) Ventilation - Room Change • Given: • 2000 SF Coin Operated Laundry - 10’ walls • Winter design 2.0 RCH • No added humidification • IA conditions: DB=75oF • OA conditions: DB=-5oF How many BTUH lost due to ventilation? BHsens= 57,600

  13. Summer Conditions- CFM per Person Given: • 2000 SF Coin Operated Laundry • 15 CFM/person summer • IA conditions: DB=75oF 40%RH • OA conditions: DB= 95oF WB=80oF How many BTUH gained due to ventilation? BHsens = 1.08 x T x CFM (1.08)(20oF)(15cfm)(40 people) BHsens= 12,960 BHlat= 0.68 x G x CFM (0.68)(80g)(15cfm)(40 people) BHlat= 32,640 BHsummer = 45,600

  14. In Class: Summer Conditions- Room Change Given: • 2000 SF Coin Operated Laundry 10’ walls • 1.5 RCH • IA conditions: DB=75oF 40%RH • OA conditions: DB= 95oF WB=80oF How many BTUH gained due to ventilation?

  15. roof lights people infiltration glass solar equipment glass conduction exterior wall floor Next Time • Heat Gains • Chapter 2.6 – 2.8 (Toa)

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