300°C while the other end extended from the surface at the temperature of mainta

Question

300°C while the other end extended from the surface at the temperature of maintained at 100°C. The fin exposed to ambient air at 40°C and heat transfer coefficient of 20W/m2K. Assume the fin made of copper (k380 W/m K) 02: - A circular cross sectional fin with 15 mn diameter and 300 mm long is a- Find the rate of conduction heat transfer to the fin from the surface at x0 b- The rate of heat transfers by the fin. c- The rate of heat transfer reaches to the other end at xL (14 points) T d. Fin efficiency KA Thermal performance of uniform cross-section fins (m2hp/ka, Tip boundary condition Adiabatic tip tanh (mL) Convection from tip cosh [m (L-x)+(h/mk) sinh h[TdL)-Tr--k1.- ||| Q_ (n-Tj).cos (mrt ((//mkesahmL sinh ml + (h/mk) cosh ml Fixed tip temperature TL)T(known) siunh (mL) cosh (mL)-(1 ) very long fin (L ) TUL) =Tj Tu)-T-(T-T) exp(x) ,

Explanation / Answer

Given,

Diameter of fin, d = 15mm = 0.015m

Length of fin, L = 300mm = 0.3m

Temperature at base of fin, Tb = 300°C

Temperature at tip of fin, Tl = 100°C

Ambient temperature, To = 40 °C

Boundary condition at fin base, o = Tb – T0 = 260°C

Boundary condition at fin tip, l = Tl – T0 = 60°C

Convection heat transfer coefficient, h = 20 W/m^2 – K

Thermal conductivity of copper, kcu = 380 W/m – K

Under steady conditions, Rate of heat transfer from the fin = Heat conduction to the fin at the base = Rate of heat transfer taking place at other end,L

For circular fin,

Perimeter, p = Pi * D = 0.047123890m

Cross sectional area of fin, Ac = (Pi/4)(0.015^2) = 0.000176715m^2

m = (4*h / k*d) = 3.7463 m^-1

mL = 1.12389

Rate of heat transfer from the fin,

Qf = (h*p*k*Ac) * b * [ (cosh mL - (l/ b) / sinh mL]

   = 69.9225W…… Ans (1), (2) & (3)

Maximum possible heat transfer from the fin,

Corrected fin length,Lc = L + (d/4) = 0.30375m

Surface area of fin = Afin = Pi * d * L = 0.01431m^2

Qmax = h * Afin * (Tb – To) = 74.432W

Efficiency of fin = (69.9225/74.432) * 100 % = 93.95% … Ans(4)

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