According to Jed the pump is never turned off. So this is the real fact: there is no excess heat, only the pump. Plus the calorimeter external and internal time constant (capacity+resistance)
We can overlaid the experimental figures only by using the calorimeter parameter and an estimated pump power [in line with our measurements]. That's all Sorry about that. 2015-01-13 20:49 GMT+01:00 David Roberson <[email protected]>: > 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 > >
