cessna7686 Posted July 20, 2005 Posted July 20, 2005 This might be a real simple one, not sure. I want to make a box (let's just say a cube) where 5 of the panels are the same width, and one panel is a different width. I also want to calculate the heat flow into the box which requires knowing the width. What's the best way to do this? I did (5*width A + 1*width B)/6. Is the average width the best method to use? Thanks
DQW Posted July 20, 2005 Posted July 20, 2005 You have not defined your problem sufficiently. Can you throw in a drawing of what you want ? If you want to make a cube, all its sides must be equal (by definition), so I don't understand what these widths are, and why one is different. And what quantity are you calculating by taking the average ? Yes, you calculated the average of 6 numbers but what does this average represent in your box ? Also, it is impossible to have a cuboid where 5 faces have a certain width and one face has a different width. But you could build a distorted box that comes "close" to this. And as for heat flow, surely you don't expect us to tell you how to calculate the heat flow into the box, when you've told us absolutely nothing about the heating mechanism or any physical conditions associated with the box and/or the heating scheme ?
cessna7686 Posted July 20, 2005 Author Posted July 20, 2005 Sorry, I meant thickness when I said width (I was thinking base, height, width). I can see what I can do about a drawing, but maybe now it won't be necessary. I'll try to go more in depth. The box is a cube, 5 sides are one thickness and one side is a different thickness. The equation for heat flow I'm using is: q=kA(T2-T1)/x k=thermal conductivity A=area T2=temperature outside box T1=temperature inside box x=thickness What I would like is to be able to model the amount of heat flowing into the whole box, not just through one side. Is it possible to generalize the thickness of the box? That's what I was trying to do when I took the average thickness. Is there a way to do this, or would I have to do it by calculating the rate for each side? I'm not asking for a numerical answer for heat flow, so I don't think the mechanisms are necessary (I'm not even sure what I'm going to use yet). Thanks for the help
DQW Posted July 20, 2005 Posted July 20, 2005 Okay, that's better. So 5 of the walls have thickness t1 and the 6th has thibkness t2. Correct ? As for the heat flow question, this is still not well-defined. Let me make a guess at one possible interpretation, and you can correct it from there. Now, I guess the box is sitting in an atmosphere of uniform temperature T2, that does not change with time. The initial temperature of the air inside the box is T0. Assuming the air inside the box is essentially isothermal, you want to find the heat flow into the box as a function of time, and hence the temperature of the inside as a function of time. There are no other sources or sinks (of heat). Am I even close ?
cessna7686 Posted July 20, 2005 Author Posted July 20, 2005 Exactly, ideally I want to know how much time it will take for the box to reach a certain temperature inside. Here are the values I'm using right now if it helps: Thickness of 5 of the walls: 20mm Thickness of 1 of the walls: 15mm Outside Temp: 296K Inside Temp: 77K Thermal conductivity of all walls: .00375 W/m*K Total surface area of the box: .00417 m^2 I hope that's all you need, let me know if more is necessary
DQW Posted July 20, 2005 Posted July 20, 2005 1. The specific heat capacity of the material of the walls 2. Is the box airtight ? 3. Is the inside just air or is the box filled with something ? 4. Are you ABSOLUTELY certain about the value of K (=0.00375W/m.K) for the walls ? That number looks way too small - much smaller than for wood (about 10 times) or plastic or even styrofoam (which is essentially air). Heck, it's a better insulator than air !
DQW Posted July 20, 2005 Posted July 20, 2005 And also, are you certain about the total surface area ? That makes the side of the cube about 25 mm. But the wall thickness is 20 mm, so that's not possible either. <gotta go now...will get back to this later, if need be>
cessna7686 Posted July 20, 2005 Author Posted July 20, 2005 The walls of the box are going to be vacuum insulation panels and I'm sure of the values for them. Thermal Conductivity: .00375 W/m*K Specific Heat: 800 J/kg*K I miscalculated the SA, but I'm going to change the dimensions anyway. The dimensions of the wall should be 10cm x 10cm x 2cm SA: .06 m^2 For now we're assuming the box will just have air, though we are exploring different gases that could be put in it. Oh, and it will be airtight. Thanks
DQW Posted July 21, 2005 Posted July 21, 2005 The walls of the box are going to be vacuum[/u'] insulation panels and I'm sure of the values for them. That explains the low conductivity value...but raises another problem. You'll now need the mass of the box as well. Or, more accurately, the mass of the vacuum insulation panels. Thermal Conductivity: .00375 W/m*KSpecific Heat: 800 J/kg*K I miscalculated the SA, but I'm going to change the dimensions anyway. The dimensions of the wall should be 10cm x 10cm x 2cm SA: .06 m^2 For now we're assuming the box will just have air, though we are exploring different gases that could be put in it. Oh, and it will be airtight. Thanks Final, question : what kind of accuracy do you want the answer to ? A rough, analytical approach will probably get you within 20% but if you want to do much better (<5% error), it may take more carefully setup PDEs or an FEM calculation.
cessna7686 Posted July 21, 2005 Author Posted July 21, 2005 Why is the mass of the panels necessary (I don't have a value right now, I can get a value tommorow)? Within 20% might work, but a smaller error 5-10% would be best. What's the best method?
DQW Posted July 22, 2005 Posted July 22, 2005 If the heat capacity of the panels is comparable to the heat capacity of the enclosed gas, then this (former) number will determine the time constant for temperature change. And of course, heat capacity =mC (proportional to mass).
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