It appears as if Jed and Mizuno have the situation under control. I will be relieved when the variations in ambient are taken out of the picture since that will make analysis of the system much simpler.
You might be correct that I got the direction of the pseudo input wrong. It is so easy to get mixed up when you think about these types of systems. Allow me to present the logic that I used to derive the behavior observed when an ambient temperature step takes place. First, it is assumed that the ambient is steady and the temperature of the coolant has settled down and no longer is changing value with time. If power is leaking into the system from the outside such as by means of the pump network then the coolant must obtain a temperature above the static ambient. This is required in order for the heat to flow outwards from the test system into the outside world. In this case heat is flowing across the delta in temperature at a rate that is proportional to that delta. Since the coolant is hotter than the ambient, heat is flowing from the coolant to the ambient. If the ambient now undergoes a step upwards the difference between the static coolant and the new ambient is less than before. Since a lower delta is now measured, less heat flows outwards from the coolant to the ambient. Before the step, all of the heat associated with the pump power was flowing through the insulation and to the ambient environment. Now, once that step takes place, the delta become smaller and less heat flows outwards. The difference in heat flowing is directed into the thermal capacity of the coolant and test device. This then should cause the temperature of those components to rise. I believe this behavior would be the same as would be observed by a real signal adding its heat currents into the calorimeter. A check to this thought is established by considering where the ultimate coolant and device system temperature stabilizes. Eventually, that combined temperature rises until the same temperature delta as before the step is reached. If that happened to be 4 degrees before a +1 degree step, the coolant temperature should move in the same direction as the step beginning at the hypothetical 3 degrees of delta slowly upwards to 4 degrees where it stabilizes. Perhaps I am overlooking something in my mental model that someone will point out. Dave -----Original Message----- From: Jed Rothwell <jedrothw...@gmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Mon, Jan 12, 2015 6:55 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised David Roberson <dlrober...@aol.com> wrote: I agree completely with Jed as long as the ambient is kept at a constant temperature. When ambient changes a great deal over a short time, calorimetry becomes too complicated. You need to throw away those results. Or use them for a limited purpose. Mizuno has reduced fluctuations and I hope he soon eliminates the overnight temperature change. I consider the change in ambient as being the equivalent of an input power application who's value is proportional to the ratio of the rapid change in ambient degrees to the total change above the normal stable ambient. Do you mean when ambient temperature falls? This would look like input power . . . but only if you do not record the ambient temperature! As long as you see it is falling, you know this is not actually input power. The hard part is when you have actual power plus a falling ambient. It is difficult to separate them out. It is not worth the effort. Just toss out the data. If ambient temperature rises (and you don't notice) it looks like heat vanishing in a magic endothermic reaction. Now if the ambient rapidly changes by 1 degree I believe that this is exactly the same as a signal appearing that is 1 watt * (1 C/ 4 C) = .25 watts. Well, it is not the same because we have a record of the ambient temperature. As I said, suppose you take a glass of warm water and put it outside in January. The difference between ambient and the water suddenly increases by 20 deg C. That does not mean a heat source appeared out of nowhere. It means the water does not instantly cool off. With this system, if I have derived Newton's cooling coefficient right, 1 W going into the water should produce a 1.5 deg C temperature difference. - Jed
