Let’s say I have an insulated box in a room that is precisely maintained at 72 ºF. I place an electric heater inside the box and apply 40 watts. After it equilibrates, I measure the inside temperature: it’s 120 ºF.
I remove the heater and place an electric fan inside the box. I apply 40 watts to the fan. Will the temperature be 120 ºF once equilibrium is reached?
I suspect it would be 120 ºF in either case. Or would the temperature w/ the fan be somewhat lower because there is more air flow along the inside wall?
I don’t think your problem is well-posed enough to answer. It depends on where you measure the temperature. Since heat is flowing out of the heater to other parts of the inside of the box, the heater itself could be hotter than 120 degrees, while the inside surface of the box is lower than 120 degrees, and the thermometer in between registers exactly 120 degrees.
In other words, if you put one thermometer on the heater, one on the wall, and one between the two, the three might register very different temperatures. With a fan instead of a heater, the three could register almost the same temperature.
To maybe answer your question, I think the thermometer on the wall should give the same reading, assuming the entire inside surface in each case is at one temperature.
As ZenBeam pointed out, it’s a little bit unclear as to how the experiment proceeds or what you’re trying to figure out.
If the box is insulated from the room, the room temperature of 72 doesn’t matter. Unless you want to take into account heat transfer via imperfect insulation. To do that, one would have to assume some things like type of insulation, box dimensions, etc.
Are we assuming as well, that the removal of a heater/addition of a fan does not involve any heat transfer?
Do we take into account heating produced by the fan, or is it assumed to be perfectly efficient?
Note that watts are a measure of power; you need to specify the time the device is on or what exactly is meant by equilibrium (as ZenBeam alluded to, the heater will be hotter than the air as long as it is on, unless you meant that its maximum temperature is 120 F ). If the fan is inefficient, then are the 40 watts electrically applied or a measure of output?
I don’t know if this is the problem you wanted, but I’ll set it up as follows :
Both the fan and the heater are already inside a perfectly insulated box. There is no heat transfer along the wires to the appliances. There are temp. sensors in the center of the box, in the heater, and in the fan.
The (40-Watt) heater is turned on for 5 minutes. It is switched off. When all three sensors give the same reading, it is 120 F.
The fan is switched on. When all three sensors give the same result, it is still 120 F. No heat transfer is possible in or out of the box, so the fan will not change it.
With a perfectly insulated box, no heat could leave - - so it would remain 120 F. Since this is not possible, the box would eventually reach thermodynamic equilibrium with the room.
Convection does draw away heat, but in your perfectly insulated box, the moving air is the same temp as the thermometer, so no heat is transferred.
If you keep the heater on, the temperature will never equilibrate if the room is perfectly insulated - it will just keep getting hotter. Perhaps you meant to say you’d shut it off after it got to 120F?.
If you replace the heater with a fan of similar wattage, there’s really no difference. That’s the rate at which you’ll be dumping energy into the air, in whatever form it takes - turbulence, heating from local air compression on the fan blades, electrical resistance in the motor, etc. The wattage will go into the room but it won’t come back out.
I assumed he meant imperfectly insulated. In this case, he achieves equilibrium when the amount of heat getting out through the insulation is 40 watts in each case.
The reason I asked this question is because someone wants me to “measure the performance of an insulated box,” particularly “how well it insulates.”
One idea I had was the following:
Place the box in a temperature-controlled environment (tightly controlled at 72 ºF).
Place a small electric heater inside the box.
Apply constant power to the heater (using a PID controller that monitors I & V).
Allow the box to come to equilibrium (2 or 3 hours).
Measure the temperature inside the box (using a small thermocouple or whatever).
I figure that, theoretically, the temperature inside the box will be constant at equilibrium. More importantly, I would think that the temperature inside the box will be a function of how well the box is insulated, i.e. the better the insulation, the higher the temperature.
However, there is an obvious problem, as some of you have already pointed out: where do you measure the temperature inside the box? After all, there is bound to be a large temperature gradient within the box.
This is where the fan comes in: What if I supply constant electrical power to a heater and a fan? If the circulation was adequate, I suspect the temperature gradient would be minimized and I could get away with measuring the temperature at just one point. Using a fan and heater (controlled to a precise power), I would think the performance of the insulated box could be determined simply by measuring the temperature within the box (high temp = good insulation, low temp = bad insulation).
Thinking from the point of view of what I would do in this case, assuming I had a thermocouple so I could measure the temperature without opening the box, I’d probably fill a bottle with hot water, tape the thermocouple to the side of it, put it in the box, and measure the temperature every 15 minutes (or less often if the temperature is dropping slowly). Lot’s of measurements means any measurement errors tend to average out.
If I was feeling industrious, I’d do the same using ice-cold water (but no ice).
Plotting the curves from different boxes on the same graph should clearly show which insulated better. The first few measurements might include transients from opening the box, but after a bit you should get clean data. You should look at the shape of the plots, not just who had a higher temperature at 30 minutes, say.
That’s a good point: tying the temperature probe to a thermal mass will help “average out” the readings (at the expense of increasing the time-to-equilibrium, of course).
So I assume you think this technique will work, i.e. all else being equal, a higher equilibrium temperature indicates a better-insulated box?
It seems to me that an easier way to measure the performance of the box is to heat the inside of the box up to some definate temp, then turn off the heat source and measure the rate at which the temp inside the box decreases. There are formulas for calculating the insulation rating based on this criteria. I don’t have them handy right now, but you could look them up in any good Thermodynamics text.
Crafter_Man, I meant to drop the heating element completely. Just see how quickly the temperature of the thermal mass approaches room temperature in each box you want to test, starting at the same temperature in each case.