An experimental device that produces excess heat is passively cooled. The additi

Question

An experimental device that produces excess heat is passively cooled. The addition of pin fins to the casing of this device is being considered to augment the rate of cooling. Consider a copper pin fin 0.25 cm in diameter that protrudes from a wall at 95 degree C into ambient air at 25 degree C. The heat transfer is mainly by natural convection with h = 10 W/m^2-K. Calculate the heat loss, assuming that the fin is 'infinitely long" the fin is 2.5 cm long and the coefficient at the end is the same as around the circumference. How long would the fm have to be for the infinitely long solution to be correct within 5%?

Explanation / Answer

For the given system,

copper pin fins are used to augment the rate of cooling.

Diameter of fin (D) = 0.25cm = 0.0025m

wall temperature (T0) = 950C

Air temperature (T) = 250C

for copper fin, thermal conductivity (k) = 401 w/m.K

For natural convection, heat transfer coefficient (h) = 10 w/m2.k

Cross sectional area of fin (A) = D2/4 = 4.9*10-6 m2

Perimeter of fin (p) = *D = 7.85*10-3 m

a)

i) For infinitely long fin,

we take temperature of fin tip equal to (T)

Heat loss(Q) = (hPkA)0.5 (T - T)

(Q) = (10 * 7.85*10-3 * 401 * 4.9*10-6 )0.5 (95-25)

Q = 0.37 W

ii) for fin of lenght (L) 2.5 cm = 0.025 m

through convective heat transfer,

The heat loss (Q) = (hPkA)0.5 (T - T) (sinh mL + (h/mk) cosh mL) / (cosh mL + (h/mk) sinh mL)

where m = (hP/kA)0.5 = (10*7.85*10-3 /401*4.9*10-6 )0.5 = 6.32

mL = 6.32*0.025 = 0.158

h/mk = 10/6.32*401 = 3.9*10-3

(Q) =  (hPkA)0.5 (T - T) (sinh mL + (h/mk) cosh mL) / (cosh mL + (h/mk) sinh mL)

= [(10 * 7.85*10-3 * 401 * 4.9*10-6 )0.5][(95-25)][(sinh 0.158 + 3.9*10-3 cosh 0.158 )/(cosh 0.158 + 3.9*10-3 sinh 0.158)]

= 0.37 * 0.9136

Q = 0.338 W

b) For infinitely long fin solution to be correct within 5%,the lenght should be

(Q for lenght L) = 0.95 * (Q for infinite long fin)

(hPkA)0.5 (T - T) (sinh mL + (h/mk) cosh mL) / (cosh mL + (h/mk) sinh mL) = 0.95 *  (hPkA)0.5 (T - T)

(sinh mL + (h/mk) cosh mL) / (cosh mL + (h/mk) sinh mL) = 0.95

We have calculated

m = (hP/kA)0.5 = (10*7.85*10-3 /401*4.9*10-6 )0.5 = 6.32

h/mk = 10/6.32*401 = 3.9*10-3

sinh mL +  3.9*10-3 cosh mL = 0.95 (cosh mL + 3.9*10-3 sinh mL)

sinh mL +  3.9*10-3 cosh mL = 0.95 cosh mL +  3.7*10-3 sinh mL

0.9963 sinh mL = 0.9461 cosh mL

tanh mL = 0.9461/0.9963 = 0.9496

mL = tanh -1 0.9496

= 0.5 ln[(1+0.9496)/(1-0.9496)]

= 1.83

where m = 6.32

L = 1.83 / 6.32

L= 0.289 m

The fin should be 0.289m long

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