Posted: 9/26/2012 8:14:09 PM EDT
Okay, its been awhile since I took thermo (Like, the better part of a decade ), and I've encountered a problem which I initially thought was going to be simpler than a google search is turning up. In the back of my mind, I thought there was a simpler,way to do it, but then again, I may be thinking of something else.Anyway, here is the problem: Lets say I have a container that I want to heat it to 160*F higher than room temperature. I'm not sure exactly what insulation materials this container (other than that they are rated for much higher temperatures than 160*F) is made out of, so I don't feel comfortable assuming the values. So, I put a 75w light bulb in the container, and over the course of a day, found out that this bulb heated up the container 14*F with respect to the room surrounding it. In my mind, there should be a way for me to calculate the "insulation ability" (can't even recall the correct term) of of this container based on my measured data rather than trying to estimate it based on the geometry of the container and material properties. What am I missing here? |
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You obviously have change in temp (ΔT), and change in time (Δt). To calculate the efficiency of your setup, you'll need to know the energy generated, or input, into your system. In this case, the heat created by the light bulb. 75W is it's electrical usage, if it's an incandescent only 10%-15% of this is actually used for lighting while the rest is lost as heat produced.
You'll need also the volume of your container, and can probably assume some things to get close. Density of air @ 30C is 1.2 kg/m^3, specific heat capacity is 1.00kj/kgC, etc. Comparing what you know to have happened to what the ideal insulative case is, you can find the efficiency of your container. Disclaimer: I'm an engineering student taking heat transfer this semester. Seeing as how I'm only 5 weeks into the semester, I could very well be off base. |
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Quoted:
You obviously have change in temp (ΔT), and change in time (Δt). To calculate the efficiency of your setup, you'll need to know the energy generated, or input, into your system. In this case, the heat created by the light bulb. 75W is it's electrical usage, if it's an incandescent only 10%-15% of this is actually used for lighting while the rest is lost as heat produced. You'll need also the volume of your container, and can probably assume some things to get close. Density of air @ 30C is 1.2 kg/m^3, specific heat capacity is 1.00kj/kgC, etc. Comparing what you know to have happened to what the ideal insulative case is, you can find the efficiency of your container. Disclaimer: I'm an engineering student taking heat transfer this semester. Seeing as how I'm only 5 weeks into the semester, I could very well be off base. 75W is it's electrical usage, if it's an incandescent only 10%-15% of this is actually used for lighting while the rest is lost as heat produced.
If the light stays in the can it becomes heat also. The input is the full 75W. All you need is equilibrium temp inside and outside and you can calculate the heat moving across the boundary. The is used to determine the U value (overall heat transfer coefficient). Density of air @ 30C is 1.2 kg/m^3, specific heat capacity is 1.00kj/kgC, etc.
These are the some of the values you would need to determine how quickly the article would heat up. None of this matters if temperature equilibrium is present. Heat out = heat in at that point. You could compute U (or its reciprocal R) units of W/(m2 °C) or Btu/(hr-sq ft °F). Disclaimer: I'm an engineering student taking heat transfer this semester. Seeing as how I'm only 5 weeks into the semester, I could very well be off base. You have a long way to go. |
The key assumption here is steady state. Let's run with that because transient analysis is more complicated, and I don't want to have to bill you.  Our main equation for steady state heat transfer that we are concerned with is: Q= U*A*(delT) Q is the heat transfer, in our case it's Watts A is the area; here we should actually use the logarithmic mean area, but we'll skip that as you see later (therefore units don't matter) delT is the difference in temperature between the inside and outside of the container, here in units of degF (but again doesn't matter for this case) U is the overall heat transfer coefficient The inverse of U*A is called the thermal resistance. You already acknowledged its existence in recognizing the container's "insulation ability." It is most important to recognize that as long as you aren't using a container that changes shape or other physical property, it's staying the same. Soooo....if that's staying the same, we understand that the ratio of delT/Q is the same for all steady state conditions.  Define variables at Light Bulb test (denoted with LB) and desired (denoted with D): Q_LB = 75 W delT_LB = 14 degF delT_D = 160 degF Therefore, Q_D = (160/14)*75 = 857 W Certainly, very simplifying assumptions are made here. Therefore, it's up to you do determine which way you want to go with that number to determine a conservative estimate. Now what's the biggest assumption? That your container reached steady state at delT = 14 in one day. That'd be the first thing I measured to validate. If it is actually transient information, skip the heartache and wait for it to get steady.
(I bet it's steady, though.) |
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You can calculate till you are blue in the face, but since the insulation value varies with the temperature drop across them you are wasting your time.
Just build it and try. Adding more insulation is not hard if you cannot get the temperature high enough (especially if you plan ahead...). |
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Quoted: Thanks!The key assumption here is steady state. Let's run with that because transient analysis is more complicated, and I don't want to have to bill you. Our main equation for steady state heat transfer that we are concerned with is: Q= U*A*(delT) Q is the heat transfer, in our case it's Watts A is the area; here we should actually use the logarithmic mean area, but we'll skip that as you see later (therefore units don't matter) delT is the difference in temperature between the inside and outside of the container, here in units of degF (but again doesn't matter for this case) U is the overall heat transfer coefficient The inverse of U*A is called the thermal resistance. You already acknowledged its existence in recognizing the container's "insulation ability." It is most important to recognize that as long as you aren't using a container that changes shape or other physical property, it's staying the same. Soooo....if that's staying the same, we understand that the ratio of delT/Q is the same for all steady state conditions. Define variables at Light Bulb test (denoted with LB) and desired (denoted with D): Q_LB = 75 W delT_LB = 14 degF delT_D = 160 degF Therefore, Q_D = (160/14)*75 = 857 W Certainly, very simplifying assumptions are made here. Therefore, it's up to you do determine which way you want to go with that number to determine a conservative estimate. Now what's the biggest assumption? That your container reached steady state at delT = 14 in one day. That'd be the first thing I measured to validate. If it is actually transient information, skip the heartache and wait for it to get steady. (I bet it's steady, though.)I had forgotten about this thread actually. I'm pretty sure that it had reached steady state because I took multiple readings over that 24+ hour period, and it clearly was reaching an asymptote somewhere in the 13.XX degree F difference range. The equation you quoted was exactly the one that I I was thinking of, but for whatever reason Google search failed me (or should I say, I failed at google searching .) As others have mentioned I know that this involves allot of assumptions, and takes into account only steady state. Obviously we are going to want allot more power so if the door is opened and a bunch of heat escapes it will be able to heat back up within a reasonable time frame. I basically wanted to get a baseline to say "we definitely want to be significantly higher than this." I kept coming up with more and more complicated equations, and online insulation calculators (one of which yielded an estimated power need of 1200W, though I had to assume so many things I didn't feel comfortable with it). A heating and electric guy I knew guessed 1000W. A vendor was going to sell us a drop in unit and calculated 1500W, but unfortunately we were not allowed to go with that vendor due to electrical certification issues...which would have cost more than the unit itself.  At least we know we are in the ballpark before applying our factor of safety. Thanks for all the responses! |
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Quoted: A heating and electric guy I knew guessed 1000W Barely\y more than the average blow dryer. Recovery time will be VERY long (if it IS tightly sealed). A thermostat and some decent size electric heaters would likely work better. You only use power when needed. That's what I thought originally, cheap and easy, but the environmental health and safety people said we'd have to get the wiring certified, and it would end up costing allot more. (Its like they don't trust us or something. )We're going to end up with some sort of prepacked unit with a decent temperature controller. |
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Quoted:
Quoted:
A heating and electric guy I knew guessed 1000W
Barely\y more than the average blow dryer. Recovery time will be VERY long (if it IS tightly sealed). A thermostat and some decent size electric heaters would likely work better. You only use power when needed. That's what I thought originally, cheap and easy, but the environmental health and safety people said we'd have to get the wiring certified, and it would end up costing allot more. (Its like they don't trust us or something. )
We're going to end up with some sort of prepacked unit with a decent temperature controller. If it plugs into a 120 V receptacle you will not get much more tan about 1,500 W no matter what. |