Given,

K = 380W/mK (Thermal conductivity)

L = 600mm = 0.6m (Length of fin)

d = 5mm = 0.005m (Diameter of fin)

ta = 20℃ = 293K (Surrounding temperature)

h = 20W/m2K (Convective heat transfer coefficient)

ts = 150℃ = 423K (Fin base temperature)

Find:

(1) Q (Heat transfer rate)

(2) ηfin (Fin efficiency)

Solution:

This is a case of fin with finite length and tip uninsulated

Heat transfer rate

Q = h.P.K.Ac(ts - ta)[tanh (m.L) + hK.m 1 + hK.m.tanh (m.L) ] …(1)

where,

m = h.PK.Ac = 20 × π × 0.005380 × π4 × 0.0052

m = 6.488

∴from (1)

Q = 20 × π × 0.005 × 380 × π4 × 0.0052 × (423 - 293) × [tanh (6.488 × 0.6) + 2 0380 × 6.488 1 + 20380 × 6.488 × tanh (6.488 × 0.6) ]

Q = 6.288 W

Find efficiency

ηfin = QactQ through fin maintained at ts

ηfin = Qacth.As.(ts - ta) = 6.28820 × π × 0.005 × 0.6 × (423 - 293)

ηfin = 0.2566 = 25.66%