Table 1: The solution of a moderately nonlinear heat transfer problem based on q = hΔT and ΔT = f{q}.

Based on q = hΔT.       Based on ΔT = f{q}.
U = 1/(1/h1{ΔT1} + t/k + 1/h2{ΔT2) ΔTtotal = ΔT1{q} + ΔTwall{q} + ΔT2{q} (1)
ΔTtotal = 320 – 200 = 120 ΔTtotal = 320 – 200 = 120 (2)
h1 = .40(ΔT1).33 ΔT1 = 1.99q.75 (3)
twall/kwall = .05 ΔTwall = .05q (4)
h2 = .80(ΔT2).50 ΔT2 = 1.16q.667 (5)
U = 1/(1/.4(ΔT1).33 +.05 + 1/.8(ΔT2).50) 120 = 1.99q.75 + .05q + 1.16q.667 (6)

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