Jed, is it possible to calculate the amount of power that is being added to the water by looking at the system? I assume that the water is not moving just prior to being accelerated to finally reach the speed that it is moving inside the pipe. That may allow you to calculate the kinetic energy that must be imparted to it which would be deposited into the far tank. The frictional losses within the pipe would have to be supplied by the pump as well. That heat would likely end up within the water instead of conducting through the hose surface.
A good first start would be to calculate the kinetic portion of the pump power. It does make me curious as to where they assume the 3 watts is being dissipated. I find it difficult to believe that this much power is always being delivered regardless of the load configuration. In a very short pipe the losses must be kinetic. Do you have information concerning the mass of water pumped per second and its velocity at exit? Energy is equal to 1/2 * mass * velocity squared. The power can then be determined by dividing by the time during which that energy is imparted. Give it a shot! Dave -----Original Message----- From: Jed Rothwell <[email protected]> To: vortex-l <[email protected]> Sent: Tue, Nov 18, 2014 5:45 pm Subject: [Vo]:Heat from the pump would not be a problem even if we could detect it Some people are still confused about the input power from the pump in Mizuno’s calorimetry. Let me point out two things about this: 1. While there has to be some heat from the pump, with this configuration, that heat is too small and too close to the noise to be detected with this equipment. This is obvious from the data I uploaded. 2. If the heat could be detected, it would still not be a problem. It would simply be included in the baseline. In calorimetry you sometimes see input power from the instruments themselves, or from something like a circulation fan in a Seebeck calorimeter. The pump runs under the same conditions at all times so power is stable and it would be easy to subtract. Let me discuss these two points in more detail. Some people seem to have difficulty grasping the notion that heat can be too small to measure with a given instrument. I suppose the heat from this pump is on the order of ~0.2 W. Based on other data I think ~0.5 W is barely detectable with this system. The pump heat cannot be measured because it is close to the noise from ambient temperature changes. With any calorimeter it is always more difficult to measure at the bottom of the scale down in the noise. You can measure the difference between 3.0 W and 3.2 W more easily than the difference between 0.0 W and 0.2 W. Mizuno left the pump running for a day to see whether he could detect heat from it. Looking at the water temperature for the day he did not see an elevation above ambient. No doubt there was one, but he could not see it. Ambient temperature changes swamped it. One minute the room is warmer than the water by 0.2°C. A few minutes later the room heater turns off and the reactor is soon warmer by 0.1°C. This is what I observed on October 23 when we did not conduct testing until afternoon and I left the Omega thermometer in the T1-T2 comparative mode. That means heat from the room is sloshing into and out of the water, albeit at a very low rate thanks to the insulation. Still, it is apparently doing that enough to hide the effects of the pump. Once the water is heated above ambient, the heat sloshes out only. After the heating and air-conditioning in Mizuno’s lab is upgraded, it may be possible to detect a slight average temperature rise above ambient caused by the pump. If that happens we can then subtract that difference from the temperature readings. That is what I mean by "included in the baseline." A low level of input power will cause a persistent average higher temperature compared to ambient. It will not cause the temperature to climb higher and higher indefinitely, until you can see it. The temperature instead reaches a peak where losses equal input. In other words, after a while the system functions as an isoperibolic calorimeter, not an adiabatic one. Because insulation is not perfect. That seems to confuse people. Let me go it over it again with an example. On October 21 the average power measured with the reactor metal and water is roughly 4.7 W. That is 1.4 W from the resistance heater pulses plus 3.3 W of anomalous power, ignoring losses. (If you want to estimate losses, which I figure are ~1 W, they should all be added to the anomalous power by this method.) The temperature rises throughout the day as you see in Fig. 7. In Fig. 9 we zoom in, and you can measure the water and wall temperature increase from hour 1.0 to hour 2.0. This increase is 0.3°C, which means the power during this time is ~3.5 W (ignoring losses). That was all anomalous power; by hour 1.0, the effect of the pulse is gone and temperatures are in equilibrium. If there had been no anomalous power, the curve at hour 1.0 would be monotonically dropping back to ambient, as it did in the evening and overnight after the anomalous heat went away. (The pump has to be contributing a little heat, and slowing down this decline, but obviously it does not contribute enough to overcome losses, because otherwise in the evening and overnight the temperature would not fall.) Suppose you remove the palladium wire and input a steady ~6 W with a resistance heater. After losses that would be about 4.7 W so the temperature will rise about the same way it does in Fig. 7. The curve will be fairly straight for 6 hours. However, the reactor walls and the cooling water will gradually become much warmer than ambient. Losses increase, per Newton’s law of cooling. Eventually, losses equal input power, and the temperature will rise no more. At that point we would have a poorly designed isoperibolic calorimeter with a ridiculously long settle time. It would take months to calibrate. As Hemminger and Hohne emphasize, with a properly designed isoperibolic calorimeter the heat losses are predictable and controlled. They are also fast enough to allow a calibration in a reasonable amount of time. Because the pump constantly does the same amount of work, and it is left running pretty much all the time, we know this terminal temperature remains the same. So if we reduce the noise enough to detect this temperature, we can simply subtract it. A piece of cake! If we still can't detect it with reduced noise, it is too small to worry about. (To be more accurate, we could estimate how much power the temperature indicates when nothing else is running, and then subtract that much power, rather than the temperature.) Let me add a note about this pump itself (Iwaki Co., Magnet Pump MD-6K-N). In the report I wrote: “the motor runs cool and it is well separated from the pump mechanism.” Small pumps are inefficient. This one is of excellent quality but at maximum power it consumes 18 W to produce 3 W of pump power. A badly designed pump will add a lot more heat to the fluid from the motor than from stirring. This one is designed to keep the motor heat away from the fluid. It is a cylinder, with the motor at one end and the pump at the other, and plastic housing in between. I put my hand on the motor side and felt it was warm. The pump side was room temperature. (I estimate the motor side was radiating about 3 W. If it had been the ~18 W it would be too hot to touch.) Mizuno has a cheaper pump with metal housing. He stopped using it because the whole thing seemed to be getting warm and he thought it might add heat to the cooling water. I asked Mizuno to check the power input to this pump with the WattChecker watt meter. This will tell us a little more about how much stirring power the pump produces. However, I expect the ratio of electric power to stirring power is not fixed at 6:1. It gets worse at the low end. - Jed
