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Optimizing HVAC Performance: Understanding Latent Heat Removal

Setting out on the journey of HVAC engineering invites us to delve into the fascinating world of latent heat removal, where each calculation holds the key to optimizing system performance. Let's dive into a real-world scenario together to illuminate this essential process.


Latent Capacity Calculation - Insight Partners

Problem:


How much latent heat is this cooling coil removing?


Your air handling unit has the following conditions:

Airflow = 1,000 cfm

Coil EAT = 95F db / 78F wb

Coil LAT = 50F db / 49.5F wb





Answer: First, determine each condition's grains per pound of dry air.


EAT = 117.4 Grains / Lb of dry air

LAT = 51.6 Grains / Lb of dry air


Second, plug into this equation.

Q(L) = CFM x 0.68 x dw


  • Where dw = the difference in grains/lb of dry air

  • Q(L) = latent heat in BTU/hr


Conversion factor (at sea level, standard air conditions) = 0.68 =  60 min/hour x 0.075 lb/ft3 x 1076 Btu/lb x 1 lb/7000 grains


Q(L) = 1,000 x 0.68 x (117.4 - 51.6) = 44,744 BTU/hr


Answer = 44,744 BTU/hr of latent heat removal

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