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Assuming you fix the temperature at the cell boundary, you can solve for the steady state temperature profile within the cell using (see any elementary heat transfer book) $$\nabla^2 T = -\frac{\dot q}{k}$$ where $k$ is the thermal conductivity on the inside of the cell and $\dot q$ is the heat generation rate per unit volume. The temperature only varies ...


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The thermal conductivity for water is k=0.56 W/m⋅K,all right,but this is for a lake,for exemple,when the heat transfer is between the air and the cold-water layer.The unit of measurement is W/mK ,not W/(m^2)K. So it's for the thickness ,not for the surface! If you want to calculate the heat production in the time unit ,H,start from: H=4π(r^2)ΔT α ,where ...


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An open cup of hot water loses a lot of its heat through evaporation. You can easily test this: prepare two cups of hot water, and then float a thin layer of cooking oil on top of one. Come back in a half hour: you'll find that the cup without oil is a lot cooler, and the water level has also significantly gone down. Edit: I did the experiment: I poured ...


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The fins don't have moisture so no difference from evaporative cooling there. Air with higher humidity has a higher capacitance. As humid air passes from the front the back it will adsorb the same amount of heat and temperature at the trailing edge will be lower than the dry air. So humid air will maintain a greater temperature difference which results ...


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It differs depending on how you formulate it. However, based on your experiments, "Body load" is more descriptive and realistic. Essentially you may want to add a diagonal body load (i.e. $xx,yy$ and $zz$).


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Sometimes it's just faster to measure. Bought 2 exactly same type of thermometer, left both in wrapping. They both said same temp. Walked home 12 minutes, both ended up at above 88. Put one inside, put other in shaded area near porch, under the "tree" that gives us all the cool shade, and the stairs, and otherwise surrounded by walls, in the shade. (also on ...



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