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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 wall of 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 this temperature 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.

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[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-stream velocity 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 to work 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 of one 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 air is 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 = 1.726x10-5 m2/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 Reynolds number of 500,000 we have made a good guess! Now we can complete our solution by finding the velocity that gives us this Reynolds number.

[pic]5.70 m/s

7-67C In flow across tube banks, why is the Reynolds number based on the maximum velocity instead of the uniform approach velocity?

The uniform approach velocity is not as important to the flow phenomena across tube banks as the local velocity between tubes. Thus the local maximum velocity is a better indicator of the heat transfer effects and is used in the correlation equations.

8-22 Cooling water available at 10°C is used to condense steam at 30°C in the condenser of a power plant at a rate of 0.15 kg/s by circulating the cooling water through a bank of 5-m-long 1.2-cm-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of 4 m/s, and leaves at a temperature of 24°C. The tubes are nearly isothermal at 30°C. Determine the average heat transfer coefficient between the water and the tubes, and the number of tubes needed to achieve the indicated heat transfer rate in the condenser.

Here we are using the data given in the problem to determine the heat transfer coefficient that is responsible for the heat transfer performance represented by those data.

To start, we find the following properties for water at the average temperature of (10oC + 24oC)/2 = 17.5oC from Table A-9: ρ = 998.7 kg/m3, cp = 4183.8 J/kg(oC, and latent heat of vaporization, hfg = 2431 kJ/kg. The mass flow rate of water in one tube can be found as follows.

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We can apply the first law of thermodynamics to determine the heat addition rate required to increase the temperature from the inlet temperature of 10oC to the outlet temperature of 24oC in a single tube.

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The log-mean temperature difference, which is used to calculate the heat transfer, is found below.

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The surface area, As, available for heat transfer is the wall area of the cylindrical tube.

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We now have all the data required to compute the average heat transfer coefficient.

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The number of tubes required are found by determining the total heat transfer required to condense 0.15 kg/s of steam with the latent heat found above: hfg = 2431 kJ/kg. The total heat transfer required is

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Since each tube has a heat transfer of 2.656x104 W, the number of tubes required is simply the ratio of the two heat transfers.

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Thus, 14 tubes are required.

8-29C Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will (a) double, (b) more than double, (c) less than double, (d) reduce by half, or (e) remain constant. (This problem was not assigned and was included here when the instructor misread 39 as 29.)

For fully developed flow in a pipe the pressure drop is proportional to the length. Thus if entrance effects are negligible, the pressure drop in the entire pipe is proportional to length. So, if the pipe length is doubled, the pressure drop will double.

8-39 Determine the convection heat transfer coefficient for the flow of (a) air and (b) water at a velocity of 2 m/s in an 8-cm-diameter and 7-m-long tube when the tube is subjected to uniform heat flux from all surfaces. Use fluid properties at 25oC.

We have to repeat the same calculations for both air and water. Starting with air we find the following properties for air at 25oC from Table A-15: k = 0.02551 W/m(oC, ν = 1.562x10-5 m2/s, and Pr = 0.7296. We start by computing the Reynolds number.

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This corresponds to turbulent flow. Using the rule of thumb that the entry length is 10 diameters for turbulent flow gives an entry length of 10(0..08 m) = 0.8 m, which is only (0.8 m)/(7 m) = 11.4% of the total tube length. For the accuracy of the correlation equations, we can assume that the entire length is fully developed turbulent flow. (Using the equation that the entry length to diameter ratio is 1.359Re1/4 given a slight larger entry length of 13.6 diameters.)

We can use the Gnielinski equation for the Nusselt number

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Where we find the friction factor from the first Petukhov equation

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Substituting this friction factor and the Reynolds and Prandtl numbers found above into the Nusselt number equation gives

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We can now find the average heat transfer coefficient for air.

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Repeating the calculations for water, we start by finding the properties of water at 25oC from Table A-9: : k = 0.607 W/m(oC, μ = 0.891x10-3 kg/m(s, ρ = 997 kg/m3, and Pr = 6.14; we first use properties to find the Reynolds number.

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Again, this is turbulent and the entry length of 10 diameters is small enough so that we can neglect entry effects. We then repeat the calculations of friction factor, Nusselt number, and heat transfer coefficient as done for air.

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We see that the heat transfer coefficient in water is over 700 times the heat transfer coefficient for air. This is due to a combination of effects. The viscosity and density of water give a higher Reynolds number for the same flow velocity and diameter. This, in combination with a higher Prandtl number results in a higher Nusselt number. The effect of the larger Nusselt number is magnified by the higher thermal conductivity of water.

8-41 An 8-m long, uninsulated square duct of cross section 0.2 m by 0.2 m and relative roughness 10-8 passes through the attic space of a house. Hot air enters the duct at 1 atm and 80oC at a volume flow rate of 0.15 m3/s. The duct surface is nearly isothermal at 60oC. Determine the rate of heat loss from the duct to the attic space and the pressure difference between the inlet and outlet sections of the duct.

If we assume that the air will not cool very much in the duct length we can use the inlet temperature of 80oC to find the properties of air from Table A-15: k = 0.02953 W/m▪oC, ν = 2.097x10-5 m2/s, ρ = 0.9994 kg/m3, cp = 1008 J/kg(oC, and Pr = 0.7296. We start by computing the mean velocity of air from the volume flow rate and the cross sectional area of a2 = (0.2 m)2= 0.04 m2. (Note a denotes the width of one side of the square duct; L denotes the 10 m length of the duct.)

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We now compute the Reynolds number based on the hydraulic diameter.

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This corresponds to turbulent flow. Using the rule of thumb that the entry length is 10 diameters for turbulent flow gives an entry length of 10(0.2 m) = 2 m, which is only (2m)/(8 m) = 25% of the total tube length. For the accuracy of the correlation equations, we can assume that the entire length is fully developed turbulent flow.

For this problem, the relative roughness, ε/Dh = (10-8 m)/(0.2 m) = 5x10-8.

We can use the Gnielinski equation for the Nusselt number with the friction factor found from the Colebrook equation. We are justified in using the Gnielinski formula, rather than a special formula for roughness, for this duct because the roughness is so small. The Colebrook equation is shown below.

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Since this is an implicit equation, we can use the Haaland equation to estimate f

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Substituting this estimate for f into the Colebrook equation gives our final answer for f.

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Before completing the heat transfer calculation, we can use this friction factor to find the pressure drop.

[pic]6.34 Pa

Substituting the friction factor and the Reynolds and Prandtl numbers found above into the Gnielinski equation for the Nusselt number gives

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We can now find the average heat transfer coefficient.

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Next, we find the exit temperature of the air by computing the number of transfer units, NTU, and substituting that NTU value into the equation for exponential behavior of the fluid temperature..

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[pic]72.0oC

We can find the heat transfer by using h and the log-mean temperature difference or by using the first law. Here we use the first law because it is simpler.

[pic]= –1212 W

The negative sign for the heat transfer indicates that the heat is transferred from the fluid to the wall.

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