Solutions to Exercise 4 – Fins

Solutions to Exercise Four – Fins 1. A 25-W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table 3–6. Select a heat sink that will allow the case temperature of the transistor not to exceed 55oC in the ambient air at 18oC. Using the thermal circuit Ohm’s law analogy as an inequality – we want to have a resistance that just matches the Ohm’s law value or a smaller one – gives the following result. ΔT 55o C − 18o C 1.48o C ≤ = R≤ 25 W W Q& Looking at Table 3-6 we see that the following heat sinks have the required resistance: HS 5030 (either horizontal or vertical) HS 6071 (vertical) and HS 6115 (either horizontal or vertical). Final selection would be based on other factors on cost, space requirements, and integration with other design components 2. Consider a stainless steel spoon (k = 8.7 Btu/h·ft·oF) partially immersed in boiling water at 200oF in a kitchen at 75oF. The handle of the spoon has a cross section of 0.08 in by 0.5 in, and extends 7 in into the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is 3 Btu/h·ft2·oF, determine the temperature difference across the exposed surface of the spoon handle. State your assumptions. We assume that the spoon is a fin with a constant cross section of (0.08 in)(0.5 in) = 0.04 in2 = 0.0002778 ft2 and a base temperature of 200oF. We assume negligible heat transfer from the contact between the spoon and the edge of the pot. The perimeter of the spoon is 2(0.08 in + 0.5 in) = 1.16 in = 0.09667 ft. If we use the equation for a finite length fin with an end correction for convection, we have Lc = L + Ac/p = 7 in + (0.04 in2)/(1.16 in) = 7.034 in. The equation for the temperature profile along a fin is given below. θ = θb cosh m( L − x) cosh mL ⇒ T − T∞ = (Tb − T∞ ) cosh m( Lc − x) cosh mLc with m= hp kA The parameter m can be computed from the data given in the problem and the values found above for perimeter and cross sectional area. m= hp = kA 3 Btu ⎛ h ⋅ ft ⋅ o F ⎞ 0.09667 ft 10.954 ⎜⎜ ⎟⎟ = 2 o 2 ft h ⋅ ft ⋅ F ⎝ 8.7 Btu ⎠ 0.0002778 ft The temperature at the end of the spoon (x = Lc) is found as follows. Exercise four solutions Tx = L = T∞ + (Tb − T∞ ) Page 2 ( cosh m( Lc − Lc ) = 75o F + 200 o F − 75o F cosh mLc ) 1 ⎛ 10.954 ⎞ (7.034 in ) ft ⎟⎟ cosh⎜⎜ 12 in ⎠ ⎝ ft = 75.4 o F So the temperature difference across the spoon, Tb – Tx=L = 200oF – 75.4oF = 124.6oF . 3. A hot surface at 100oC is to be cooled by attaching 3-cm-long, 0.25-cm-diameter aluminum pin fins (k = 237 W/m·oC) to it, with a center-to-center distance of 0.6 cm. The temperature of the surrounding medium is 30oC, and the heat transfer coefficient on the surfaces is 35 W/m2·oC. Determine the rate of heat transfer from the surface for a 1-m by 1-m section of the plate. Also determine the overall effectiveness of the fins. Here we have to consider the analysis for each fin then evaluate the overall surface consisting of fins and a remaining area without fins. For each fin the cross sectional area Ac = π(0.0025 m)2/4 = 4.909x10-6 m2 and the perimeter p = π(0.0025 m) = 0.007854 m. The corrected length, Lc = L + Ac/p = 0.03 m + (4.909x10-6 m2) / (0.007854 m) = 0.030625 m. With these data and the values for k and h given above we can compute the parameters m and mLc. m= 35 W ⎛⎜ m⋅o C hp = kAc m 2 ⋅o C ⎜⎝ 237 W ⎞ 0.007854 m 15.37 ⎟ = ⎟ 4.909 x10 − 6 m m ⎠ mLc = 15.37 (0.030625 m ) = 0.4708 m The heat transfer from one fin is given by the following equation. (4.909 x10 m ⋅ C 35 W 2 o Q& one −6 m2 fin = kAc hp (Tb − T∞ ) tanh mL = W ) 237 (0.007854 m )(100 C − 30 C )tanh(0.4708) = 0.5493 W m⋅ C o o o The fins with a diameter of 0.0025 m, separated at a distance 0.06 m between centers, have a distance from the start of one fin to the start of the next fin of 0.06 m. Over a length of 1 m there is room for 1/0.06 = 166 fins. Over the 1-m-square plate there will be a total of 1662 = 27556 fins. The plate area occupied by these fins is 27556(4.909x10-6 m2) = 0.1353 m2. The remaining unfinned area = 1 m2 – 0.1353 m2 = 0.8647 m2 will have a convective heat transfer with the same heat transfer coefficient and temperature difference. Thus, the total heat transfer is found as follows. Q& = Q& finned + Q& unfinned = N fin Q& one fin + hAunfjnned (Tb − T∞ ) 35 W = (27556 )(0.5493 W ) + 2 o 0.8647 m 2 100 o C − 30 o C = 172555 W m ⋅ C ( )( ) Q& = 17.26 kW Without fins the heat transfer would be with the same convection coefficient and temperature difference over an area of 1 m2. Q& no fins = hAno fins (Tb − T∞ ) = (1 m )(100 C − 30 C ) = 2450 W m ⋅ C 35 W 2 o 2 o The overall effectiveness is the ratio (17.26 kW) / (2.45 kW) = 7.04 . o