College

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|College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer | |
| |Spring 2007 Number 17629 Instructor: Larry Caretto |
April 11 Homework Solutions

7-16 During a cold winter day, wind at 55 km/h is blowing parallel to a 4-m-high and 10-m-long wallof a house. If the air outside is at 5°C and the surface temperature of the wall is 12°C, determine the rate of heat loss from that wall by convection. What would your answer be if the wind velocity was doubled?

Here we have convection over a flat surface with a length of 10 m and a width of 4 m. The mean temperature is (5oC + 12oC)/2 = 8.5oC. We find the following properties of air at thistemperature from Table A-15: k = 0.02428 W/m(oC, ν = 1.413x10-5 m2/s, and Pr = 0.7340.

We first find the Reynolds number.

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This means that the flow becomes turbulent over the plate and we can use the Nusselt number equation for combined laminar and turbulent flow.

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We then use this Nusselt number to find the heat transfer coefficient and the heat transfer.

[pic][pic]9.08 kW

If the wind velocity were doubled, the Reynolds number would be doubled and we would repeat the calculations above, starting with this revised Reynolds number..

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[pic]16.21 kW

7-32 A transformer that is 10 cm long, 6.2 cm wide, and 5 cm high is to be cooled by attaching a 10-cm by 6.2-cm wide polished aluminum heat sink (emissivity = 0.03) to its top surface.The heat sink has seven fins, which are 5 mm high , 2 mm thick, and 10 cm long. A fan blows air at 25°C parallel to the passages between the fins. The heat sink is to dissipate 12 W of heat and the base temperature of the heat sink is not to exceed 60°C. Assuming the fins and the base plate to be nearly isothermal and the radiation heat transfer to be negligible, determine the minimum free-streamvelocity the fan needs to supply to avoid overheating.

With the simplifying assumptions given in this problem, we really have a problem of 7 flat plates with heat transfer from two sides. The spaces at the base of the fins also form flat plates. So we find the heat transfer from one flat plate with a total area of all the fins plus the unifnned surface at the base of the fins.

We have towork this problem in reverse. We are essentially given the required heat transfer – the 12W that the transformer has to dissipate – and data on the fin area, and the temperature difference. We can therefore compute the heat transfer coefficient and work backwards to find the air velocity.

The flat plates in this case have a length of 10 cm and a width of 0.5 cm. The total area from one side ofone fin is (0.005 m)(0.1 m) = 0.0005 m2. There are 14 such sides giving an area of 0.007 m2. In addition, the total area at the top of the transformer is (0.1 m)(0.062 m) = 0.0062 m2. The bases of the seven fins take up an area of 7(0.1 m)(0.002 m) = 0.0014 m2. Thus the unfinned area on the base is the difference in these areas: 0.0062 m2 – 0.0014 m2. = 0.0048 m2. The total area for the airis the sum of the fin area and the unfinned area: 0.007 m2 + 0.0048 m2 = 0.0118 m2. From these area data and the other given data we can compute the heat transfer coefficient as follows.

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We need air properties to continue. The mean temperature is (60oC + 25oC)/2 = 42.5oC. We find the following properties of air at this temperature from Table A-15: k = 0.02681 W/m(oC, ν = 1.726x10-5m2/s, and Pr = 0.7248. The Nusselt number based on the common length of 0.1 m for all the surfaces that we are treating as flat plates, can now be computed as follows

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Now we have to decide which equation to use for Nu: the laminar or turbulent equation. Since the length is short we choose the laminar equation for the first attempt.

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Since this is less than the critical...