Dear Gigi, I do not read Italian as you suspected so I will need a good translation of what you have written in order to comment properly. I reviewed the link in Italian and it appears that you have run a test with the 5 mm pipe feed directly into a storage sink. Further within the report it looks like you substituted a 10 mm tube of the same length. This setup would not be a valid test of my theory since the flow rates will not be the same in both cases.
Who would doubt that far more water will be flowing within the 10 mm tube than within the 5 mm tube under this condition? The vastly greater amount of flow within the 10 mm tube will carry a correspondingly greater quantity of kinetic energy. If I recall, you intentionally used a short piece of pipe (.4 Meters) along with your 5 mm tubing in an effort to make the flow rate match that obtained by Mizuno with the 16 Meter long 10 mm tube. That is the goal that you must aim for if a fair test is to be conducted. The best way for you to perform a true test is to actually obtain a 16 Meter long 10 mm inside diameter pipe. If you make that substitution, I and most others will consider your experiment to be a reasonable replication attempt. There will be no further problems pertaining to the 5 mm pipe issue that we are discussing at this point. You can insulate your pipe coil well enough to allow you to make an accurate measurement of the heat being transported away from the pump motor and include all other sources of pump related heat that impact Mizuno's set up. I fail to understand why you are determined to use a test system that does not match the real world? If you match Mizuno's system, you will find it much easier for us to accept your test results. Dave -----Original Message----- From: Gigi DiMarco <[email protected]> To: vortex-l <[email protected]> Sent: Sat, Jan 10, 2015 4:18 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Dear Dave, you still insist on your calculation neglecting what I wrote to you in an earlier message regarding the fact that increasing the pipe the power goes to zero when calculated according to your mathematics. We have just published the new experiment with the theory and diagrams behind it. https://gsvit.wordpress.com/2015/01/10/ulteriori-misure-sulla-pompa-md-6k-n-utilizzata-da-tadahiko-mizuno/ Unfortunately it is only in Italian; you have to wait a bit to have the official English translation I'm not sure to finish it by tomorrow. However, google translate makes a good job. Feel free to make all your comments; I'd rather like on our site so that is very easy for us to reply. 2015-01-10 21:06 GMT+01:00 David Roberson <[email protected]>: Thanks Jed. If the water alone recovers 1.3 watts with average drive drive, and more resides within the vessel, then you are in great shape. If you have the chance, I would greatly appreciate it if you could ask Dr. Mizuno about the measured flow rate. My earlier calculation using 9 liters per minute clearly suggests that the skeptics made a major error by using the 5 mm pipe. As the calculations show, they will find that kinetic energy and thus power transport will be 16 times as much as seen had they used 10 mm pipe assuming the flow rate is constant. As you know I am discussing this aspect of their report and hope to resolve the issue soon. I am confident in my analysis. I have approached the problem from a couple of different directions and keep getting the same result. Dave -----Original Message----- From: Jed Rothwell <[email protected]> To: vortex-l <[email protected]> Sent: Sat, Jan 10, 2015 2:42 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised David Roberson <[email protected]> wrote: Jed, looking at figure 6, the Oct 21 data I calculate that the average power is 1.3888 watts. That is 20 watts * 500 seconds / 7200 seconds = 1.3888 watts. Yes, that is the answer I got, in Table 1. However, bear in mind that is for the water alone. Not for the reactor, which has a slightly larger thermal mass than the water, and much worse insulation. Estimating that, I get 3.4 W total, on average. Based on a very rough estimate of unaccounted for heat losses and Newton's law of cooling I guess the actual average power is about 7 W. In other words, the reactor metal plus the water are recovering about half of the heat. If Mizuno applies that amount of power continuously what would you expect the temperature to do? With 1.3 W input I expect to see nothing, as I said in the paper on p. 9. That is, in fact, what I saw when I did a similar test. There is too much noise, and the water recovers only about one-fourth of the heat, as I said. So I figure you would have to input ~7 W continuously to see this temperature rise. Mizuno hopes to do that kind of simulation but I do not know when. Actually, now that ambient fluctuations are reduced, you might see 1.3 W in the reactor. That would put ~0.5 W into the water I guess, about twice as much as the pump. It might raise the water temperature by ~1 deg C after an hour or two. It is hard to say. The only way to find out is to do a test and measure it. My gut feeling is that the temperature would increase along a constant slope once the transients are settled down. Well, it increases for a while, but at low power it then soon stops rising as the calorimeter goes from being adiabatic to isoperibolic. That takes 1.4 hours at ~0.2 W. I do not know how long it takes at 0.5 W or 3 W. At any power level it must eventually stop heating, when losses equal input power. Losses increase with the rising temperature, per Newton's law. Also, can you verify that the water flow rate is actually nominally 8 liters per minute? That's what Mizuno said. I suppose he measured it when dumping out the cooling water. He had to change out the Dewar reservoir a couple of times. I think that is what the pump spec. sheet says. There is hardly any resistance, and no grade, so I guess it should be close to maximum performance. - Jed
