Fin Heat Transfer Rate

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 $\eta_f$. A fin at $\eta_f$ = 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 = $h(A_{\mathrm{base}} + \eta_f A_{\mathrm{fin}})\Delta T$, where $A_{\mathrm{base}}$ 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.