Straight copper fins of rectangular profile, having thickness t = 1 mm, length L

Question

Straight copper fins of rectangular profile, having thickness t = 1 mm, length L = 10 mm and
thermal conductivity k = 380 W/mK are attached to a plane wall (surface area 2m x 2m) which is
maintained at a temperature of 230oC. The fins dissipate heat into the ambient air (which is at
30oC) with a convection heat transfer coefficient of 40 W/m2K). The fins are spaced 8 mm apart
(that is, 125 fins per metre). Assume negligible heat loss from the fin tips. Determine:
(a) The rate of heat loss per fin.
(b) The fin efficiency and effectiveness.
(c) The rate of heat loss from the plane wall.
(d) The overall surface efficiency.

Explanation / Answer

h = 40 w/m2-K
Afin = 2.w.l + w.t = 2x2x.01 + 2x.001 = .04+.002 = .042 m2

(a) rate of heat loss per fin = hAfin(Tb - To) = 40x.042x(230-30) = 336 W

(b) efficiency of fin = (l + t/2) {h/kt}
efficiency = (.01+.001){40/(380x.002)}0.5 =  0.0798 => 7.98%

fin effectiveness = Qfin / Qno fin

Qno fin = hAb(Tb-To) = 40x(2x.001)x(230-30)= 16 W

fin effectiveness = 336/16 = 21

(c) rate of heat loss from plane wall = h(Awall without fin)(Tb-To)
= 40x(2-250x.001)x2x(230-30) = 28000 W = 28 kW

(d) heat loss from all the fins = 336x250 = 84000 W = 84 kW

heat loss from wall without fins = 40x4x200 = 32000 W = 32 kW

total heat loss from the wall with fins = 28+84 = 112 kW

(actually not able to understand what this term actually signify. I have calculated all the data and you definitely can get this thing also from the values calcualted above)

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