Dave and Gigi-- In the overall evaluation of the energy balance, please note that the 6L pump uses .25amps x 115 VAC power = 28.75 watts. This is in accordance with a 2005 drawing of the pump from the vendor that I forwarded to Dave a day or so ago. (The actual pump used in the test may have had a different amperage/voltage rating, if it is a newer model, but I doubt it.)
As Gigi says some of this energy goes into the water as heat and some to the ambient. However as the ambient gets cooler a larger fraction goes to the ambient, because it becomes a better heat sink relative to the water pathway. This change may be small but should be considered in the modeling. The time constant for this change should be relative short and correspond to the decrease of the ambient temperature. This is because the heat capacity of the pump is small, and its pump body readily changes temperature in step with the ambient changes per my guess. Pump body measurements could confirm this assumption. I think this observation may be consistent with Jed's fancy of Newton's law of cooling, although it may not have been considered in his Adiabatic Calorimetric modeling of the Mizuno test. Bob ----- Original Message ----- From: David Roberson To: [email protected] Sent: Tuesday, January 13, 2015 11:49 AM Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Dear Gigi, I have begun to analyze your report and find something that does not seem logical according to my understanding of heat flow. On your figure A2 I see that you have overlaid your simulation results upon Jed's figure. The correspondence between the curves is remarkable and you should be commended for your work. The issue that I need to resolve is that the delta temperature between the Dewar and ambient is actually increasing during this time. Also, the delta for the reactor is becoming less with time as I was expecting. In order for the temperature delta to increase you would have to supply some form of heat power to that device. The model that you are using is extremely simple and certainly does not suggest that anything more complex would be happening. How do you explain that the delta is increasing? Is there some process that is supplying extra power into the Dewar once the pump is turned off? Regards, Dave -----Original Message----- From: Jed Rothwell <[email protected]> To: vortex-l <[email protected]> Sent: Tue, Jan 13, 2015 2:06 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Gigi DiMarco <[email protected]> wrote: The refrigerator example is quite evident, but is unfit to our situation, by various causes. The main one is that there you have an abrupt change of air temperature, while in the 18h test the air temperature is falling at a modest rate of 0,36 °C/h that is very simple to follow for the calorimeter. No, it isn't. That is why a gap opens between the room temperature and the calorimeter, and the gap persists until early morning. If from now on the losses are equal to the pump power, since you have Loss = K * deltaT and PumpPower = loss = constant and since K is valid over a broad range of deltaT you should have a constant deltaT. No, it isn't. See Newton's law of cooling. So going back to the plots in the missing file you considered only the first 1.5 hour only because just after the ambient temperature starts decreasing. What have it happened if the ambient did not change for 5-6 hours? Can you answer this question? Yes, I can. If ambient stays stable, the reactor and water temperature will remain stable at 0.6 deg C above room Where in the data do you see that 0.6 °C is the maximum? It goes no higher after 1.4 hours. You can see this in other data sets as well, such as early in the morning with this data set. Whenever ambient remains stable for a few hours or more, the reactor temperature always settles 0.6 deg C warmer. Please don't be contemptuous and dismissive; it is not the case. If someone does't understand calorimetry it is not me. You do not understand Newton's law of cooling and you cannot tell the difference between ambient cooling and heat generation in a cell. In my opinion, you are terribly confused and totally unqualified to do calorimetry. - Jed
