Cooling Coil Total Load from Enthalpy Difference
The total heat a cooling coil removes, sensible plus latent.
Example
You enter
- Airflow across the coil (cfm) 2000
- Entering-air enthalpy (Btu/lb) 31.48
- Leaving-air enthalpy (Btu/lb) 22.97
You get
- Total coil load 76,590 Btu/hr
- Tons 6.38 tons (dh = 8.51 Btu/lb)
Checked against published worked examples from: PDHonline (J. Primo).
Details, formula, and sources
ScopeQ = 4.5 x CFM x (h_ent - h_lvg) Btu/hr, the full sensible-plus-latent heat a coil removes, from the moist-air enthalpy drop. 2,000 CFM across a 31.48 -> 22.97 Btu/lb drop -> 76,590 Btu/hr (6.38 tons), well above the dry-bulb-only estimate. A design aid; the coil rating governs.
Formula or tableQ = 4.5 x CFM x (h_ent - h_lvg) Btu/hr; tons = Q / 12000.
Source and editionThe total-heat coil-load relation from the ASHRAE Handbook - Fundamentals, by name.
Free accessThe 4.5 x CFM x enthalpy-difference total-heat relation is a standard airside coil result. The coil rating governs.
What governsEstimate. AHJ and licensed professional govern.
API field namescfm, h_ent_btu, h_lvg_btu, q_btuh, tons
- Total-heat load Q = 4.5 x CFM x (h_ent - h_lvg); 4.5 = 60 x 0.075ASHRAE Fundamentals
- Enthalpy source entering/leaving enthalpies from moist-air-enthalpyASHRAE
- Tons conversion tons = Q / 12000; negative Q means heatingscope of this tile