Fin Heat Transfer Rate

Q˙f=ηfhAfΔTb\dot{Q}_f = \eta_f \, h A_f \, \Delta T_b

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Once you have the efficiency, the fin's duty is just Newton's law of cooling with a discount: full area, full base-to-fluid ΔT, multiplied by η_f. A fin at η_f = 0.92 with 0.12 m² of surface in an h = 50 W/(m²·K) airstream 60 K below its base gives 0.92 × 50 × 0.12 × 60 = 331 W. Note that the whole fin area counts — both faces, plus the edges if they matter — because the efficiency has already accounted for the fact that the far end is cooler than the base.

In a real finned coil this is one of two terms. The total surface duty is Q̇ = h(A_base + η_f A_fin)ΔT, where A_base is the exposed tube between the fins, and the bracketed quantity is what designers call the effective area. Everything about fin selection follows from wanting that bracket to be large per dollar and per pascal of air pressure drop. Trap: fin spacing. Packing fins tighter adds area but chokes the flow, dropping h and raising fan power; and in a wet or dusty duty, fins closer than about 2 mm will bridge with condensate or lint and the coil loses more capacity to blockage than the extra area ever bought.

Fin Heat Transfer Rate
Q˙f=ηfhAfΔTb\dot{Q}_f = \eta_f \, h A_f \, \Delta T_b
Where
  • Q˙f\dot{Q}_f= Fin heat transfer rate
  • ηf\eta_f= Fin efficiency
  • hh= Film coefficient
  • AfA_f= Fin surface area
  • ΔTb\Delta T_b= Base-to-fluid ΔT