Most pumps do quite well at converting electrical energy into mechanical energy. When they do only 35% or 40% conversion they are called inefficient. I have not measured the efficiency of the Mizuno pump but only looked at the specifications issued by the Vendor for a similar pump of the MD-6 variety. At 8 liters per min flow it is quoted to take about 29 watts. I tend to believe the Vendors statements about the pump energy input-- anyone can check the specs online. This type of pump is popular in many countries including the USA. The amperage and voltage that results from a 60 hertz, single phase electrical input is identified in the specs to produce a flow of about 8 liters per minute. I think this is the electric power characteristic that Mizuno used to run the pump. 10.8 watts is considerably below the pumps specified need for power. I do not think it would operate at this low level.
>From my experience with pumps, most energy in circulating systems ends up as >heat. Modern day light water reactor plants and those not so modern plants >use circulating pumps to heat the whole reactor plant from cold conditions to >operating temperatures. There are no electric heaters used to heat the >coolant. I mention this, since it is the best way to heat up a big insulated >system with lots of metal and circulating water. I respectively disagree with Jed's conclusions. I await a independent confirmation test. Bob Cook ----- Original Message ----- From: Jed Rothwell To: [email protected] Sent: Saturday, January 10, 2015 8:18 PM Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Bob Cook made two large mistakes here. I wish he -- and others -- would The Iwaik pump, if running, would have added heat at about 29 watts per the pump specification. In my report, p. 24, I list the pump specifications. Mizuno measured the pump input power with the watt meter. It is 10.8 W, not 29 W. However, only a tiny fraction of this power is delivered to the water. Mizuno measured how much is delivered. It was only ~0.4 W. If you do not think so, explain why Fig. 19 is wrong. You can confirm that nearly all the electric power converts to heat at the pump motor. Touch a pump and you will feel the heat radiating. Many pumps have fans that blow the hot air out of the motor. With a good pump, the water is at the other end away from the motor, and very little heat transfers to it. This was more than enough to raise the temperature without any reactor heat source given the recorded decrease of 1.7 watts when nothing was running or reacting. Suppose this is true. Suppose it was 1.7 W and suppose that raises the temperature by 4 deg C. Pick any temperature rise you like: suppose it raises the temperature by 10 deg C, or 20 deg C. Here is the point, which I have made again and again: THE TEMPERATURE WAS ALREADY that much higher when the test began. The pump runs all the time. Using this method we measure from that starting baseline temperature up to the terminal temperature of the test. The pump heat -- however much there is -- is already included in the baseline. Therefore we never include it in excess heat. You need to answer these points if you want to have a serious discussion. - Jed
