From: Peter Gluck
OK, old friend I understand what you say, the energy of the pump is consumed, is money spent for making the generator to work. No connection with heat balance of the system- but goes to expenses. Right? Dear Peter, Yes, but we can take it further. As a student of ontology you are fully familiar with all of the logical arguments. Allow me to apply reductio ad absurdum to this situation. Let's say Rossi shows up with a reactor that puts out one megawatt of heat. It requires a large flow of water, which is coming from a local dam and goes into a sewer. This new reactor requires no electrical input at all !! The heat is measured by a thermal circuit that removes heat from the stainless steel reactor, and the new owners of this magical device use it to heat the factory. It remains warm all year without any electricity ! Let's say the device is opened up and found to contain nothing but flow constrictors - which convert water pressure into heat via friction - nothing else. Is Rossi entitled to claim that the megawatt of heat is "overunity" and therefore free energy ? He would be, if we followed this argument that power to a required pump is not input power. Jones wrote: Dear Peter I do not understand the problem. There are two systems involved: heat and electricity At the system level P-out is thermal and refers to net heat. The calorimetry determines P-out for heat. P-in for the system, not for the calorimetry, is determined by the sum of all the electrical inputs. The pump must be included as it is necessary. There is nothing in the calorimetry loop which is used to determine *system P-in*. Yes the heat loop itself may have it own designation for P-in and P-out, but that is not for the system; that is why I believe you could be conflating the two issues. Jones From: Peter Gluck Want I wanted to say- the pump is part of the cooling circuit to which the heat produced is transferred. Has nothing to do with the heat produced. peter On Wed, Apr 20, 2011 at 6:15 PM, Jed Rothwell <jedrothw...@gmail.com> wrote: Peter Gluck wrote: Fortunately the inlet temperature of water is measured and this includes or, if you wish excludes the effect of the pump/motor. But he effect is negligible- and not on the side of Pin- it is at Pout. No, not Pout. The heat from the pump shows up past Pout, at the place where the water stops moving. That would be either in your drain pipe, or -- if you recycle the water -- in the reservoir tank. As long as the water is moving at the same speed as it did when it left the pump, the energy has not yet converted to heat. - Jed -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
