The initial temperature of the copper sphere is given at time zero You may

The initial temperature of the copper sphere is given

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C. The initial temperature of the copper sphere is given at time zero. You may assume that Bi < 0.1, so the lumped capacitance model can be used. You may assume k for copper is constant, and can look up properties at 300K. The heat transfer is unsteady state, and the temperature of the sphere is measured by a thermocouple inserted at the center of the sphere (r = 0). Note: You will need to use a spreadsheet to plot data, but be careful to make the plot fit an appropriate equation so that you can find h from a trendline fit to the graph. Spreadsheet has calculations, and is used to find h. Below are the calculations for the two AIR experiments. The blue data are for no fan, the red for with a fan. The equations give slopes which must match -hA/mc p Problem 7 Time (sec) Data Temp, C (no fan) Temp, C (fan) Temp, C (water) 0 185 185 185 2 180 176 160.1 4 175.5 168 141 6 171.2 160.5 125 8 167.1 153.6 110.8 10 163.2 147 97.5 12 159.4 140.8 85.2 14 155.7 134.4 75.1 16 152 128.3 66.6 18 149.1 122.4 60.8 20 146.4 116.8 55.5 22 143.8 111.6 51.6 24 141.3 106 48.4 26 138.9 101.3 45.1 28 136.6 97 42.8 30 134.4 93.1 40.9 no fan fan no fan fan no fan fan T-Tinf T-Tinf (T-Tinf)/(Ti-Tinf) (T-Tinf)/(Ti-Tinf) Time (sec) ln{(T-Tinf)/(Ti-Tinf)} ln{(T-Tinf)/(Ti-Tinf)} 160 160 1 1 0 0 0 155 151 0.96875 0.94375 2 -0.031748698 -0.057893978 150.5 143 0.940625 0.89375 4 -0.061210731 -0.112329185 146.2 135.5 0.91375 0.846875 6 -0.090198268 -0.166202175 142.1 128.6 0.888125 0.80375 8 -0.11864278 -0.218467003 138.2 122 0.86375 0.7625 10 -0.146471904 -0.271152771 134.4 115.8 0.84 0.72375 12 -0.174353387 -0.32330925 130.7 109.4 0.816875 0.68375 14 -0.202269195 -0.380162925 127 103.3 0.79375 0.645625 16 -0.230986729 -0.437536439 124.1 97.4 0.775625 0.60875 18 -0.254086123 -0.496347605 121.4 91.8 0.75875 0.57375 20 -0.276082937 -0.555561518 118.8 86.6 0.7425 0.54125 22 -0.297732408 -0.613874 116.3 81 0.726875 0.50625 24 -0.319000756 -0.680724661 113.9 76.3 0.711875 0.476875 26 -0.339852945 -0.740500877 111.6 72 0.6975 0.45 28 -0.360252765 -0.798507696 109.4 68.1 0.68375 0.425625 30 -0.380162925 -0.854196602 y = -0.0127x - 0.0147 R² = 0.9945 y = -0.0285x + 0.0082 R² = 0.9992 y = -0.0134x R² = 0.9902 y = -0.0281x R² = 0.9989 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0 10 20 30 40 ln{(T-Tinf)/(Ti-Tinf)} Time (sec) Series1 Series2 Linear (Series1) Linear (Series2) Linear (Series1) Linear (Series2)
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Both equations should have been forced to go through the origin, since the theoretical equation has no intercept when plotted this way. (Note: I converted the data into a linear plot; it is possible to leave with a ln or exp, as long as a different fit is used). A = 4 * pi * r 2 = 4 * pi * 0.0125^2 = 0.00196 m 2 m = 4/3 * pi * r 3 * density = 4/3 * pi * (0.0125)^3 * 8933 = 0.0731 kg density = 8933 kg/m 3 c p = 385 J/kg-K With no fan, hA/mc p = 0.0134 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K) h (no fan) = 192 W/m 2 K With the fan, hA/mc p = 0.0281 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K) h (fan)= 403 W/m 2 K Similarly, for water, a third line can be added to the graph: For the water, hA/mc p = 0.0801 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K) h (water)= 1150 W/m 2 K (higher value than air; as expected) y = -0.0127x - 0.0147 R² = 0.9945 y = -0.0134x R² = 0.9902 y = -0.0285x + 0.0082 R² = 0.9992 y = -0.0281x R² = 0.9989 y = -0.0801x R² = 0.9971 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 0 5 10 15 20 25 30 35 ln{(T-Tinf)/(Ti-Tinf)} Time (sec) Series1 Series2 Water Linear (Series1) Linear (Series1) Linear (Series2) Linear (Series2) Linear (Water)
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8. A large solid sphere with a 10 inch diameter initially has a uniform temperature of 67 o F until it is placed in an oven at 400 o F (for the purposes of this problem, you can assume that the sphere is surrounded by air on all sides and does not touch any surface). A. Draw a qualitative sketch of T(r) from –r to r for a yellow pine sphere for four time points, including t = 0, t = and two times in between. Assume Bi > 0.1. B. Draw a similar qualitative sketch of T(r) from –r to r for a 10 inch diameter aluminum sphere, assuming Bi < 0.1.
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