Arrian Forbush Ryan Heap

Introduction We wanted to experiment on the last problem from exam #1 and figure out how accurate the calculations can be. Also, for those times when you put off buying your part of the picnic until the last minute, it is essential to know how much time you have. We want to calculate the time it takes for a honeydew melon to cool to the optimum chilled temperature for consumption. (Watermelon just ain’t in season) 2

The Setup T i =68°F T f =53°F T ∞ =40°F h assumed =5(W/m 2 ∙K) r hd = m (assuming a sphere and averaging circumferences) Since the Bi number is greater than 0.1, we cannot use lumped capacitance. We use the approximate solutions method instead. 3

The Solution 4 For sphere, using eq. 5.50a Since, sin(x)/x=1 when x–›0 (r*=0)

Results 5 T Fh5 Ti Fr Tinf Fr*0 Tavg F vf kf Cp4185 α E-07 bi C ξ0.806 θ* Fo = t = seconds hrs

Conclusions and Recommendations We were surprised that the melon cooled faster than our calculations showed. We realized that the fridge temperature varied with the cycling. This would change the affect of the heat transfer. Also, the center of a honey dew is not solid, so the temperature that we measured was 1.5” from the center. Other calculations suggest that the center and the position 1.5” from the center did not differ much in time. We thought that the properties of honeydew would be closer to water like a watermelon, but apparently it is not close enough. Notice that this experiment happened in the winter time where room temperature is about 10 degrees cooler than during summer. The cooling of the honey dew won’t take a long time, so as long as you give yourself at least 16 hours, the honey dew will be the same temperature as the fridge. 6

Appendix From the Thermophysical Properties Calculator (T ave =60°F): Cp=4185 k= v= Table 5.1: Bi=0.2273, interpolation yeilds, C 1 =1.067 ξ=