I'm going to simulate an automotive radiator . The method called $\varepsilon$- NTU , suggests following steps to calculate the heat rejection,$Q$, of a heat exchanger:
1-calculate $U$(overall heat transfer coefficient)
2- calculate $NTU$
3- calculate $\varepsilon$ and finally calculate $Q(W)$.
My question is about calculation of $U$ in automotive radiators. We can neglect fouling and wall resistances and write: $$\frac{1}{UA} = \frac{1}{\eta_aA_ah_a}+\frac{1}{A_wh_w}$$ $U$: Overall heat transfer coefficient.
$A$ : Air side or waterside area.
$A_a$ : Air side heat transfer area
$A_w$: Waterside heat transfer Area
$h$ : heat transfer coefficient
$\eta_a$ air side surface efficiency.
I want to use Shanoun & Webb formula to calculate the first term of the above equation(Air thermal resistance). They propose following equation for automotive radiators: $$(\eta Ah)_a=(\eta_f Ah)_{l} +(\eta_f Ah)_{S1}+(\eta_f Ah)_{S2}+h_eA_e$$ You can find regions named louvered(l),S1,S2 and e in following image: This is image of a fin between 2 flat tubes.
Question: Shanoun and Webb just enter areas relating to one fin in their equation. So when we want to calculate $UA$, we account total surface of water$A_w$ but just a small surface of air side$A_a$. It makes the answer wrong. How can we use this formula to calculate $UA$ ?

I should multiply the No. of fins to the right-hand-side statement. So we have: $$(\eta hA)_a=No.of fins \times( (\eta_fhA)_l+(\eta_fhA)_{S1}+(\eta_fhA)_{S2}+h_eA_e)$$