[Vo]:Pump power must be included
Peter, The Rossi effect is controlled in a narrow range by balancing heat removal and heat addition. It will not work reliably without constant heat removal. Therefore, power input related to the proper operation must be included as P-in. Jones From: Peter Gluck Dear Jones, just from curiosity, in what kind of P-in has to be included the pump's power and why? Peter We presume (hope) that the Swedes will not use a hose connected to plumbing where you get free water pressure, and will use a pump. The pump's power must be included in P-in. A liter/sec pump seems to require one horsepower or about .75 kW.
Re: [Vo]:Pump power must be included
Dear Jones, If the power has to be included, it has to be measured. But only a part of the energy consumed by the motor of the pump is used to make the water to moveand this produces a small heating of water due to friction, So the reverse is true- the power of the motor has to be subtracted from Pin. 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. Peter On Wed, Apr 20, 2011 at 5:34 PM, Jones Beene jone...@pacbell.net wrote: Peter, The “Rossi effect” is controlled in a narrow range by balancing heat removal and heat addition. It will not work reliably without constant heat removal. Therefore, power input related to the proper operation must be included as P-in. Jones *From:* Peter Gluck Dear Jones, just from curiosity, in what kind of P-in has to be included the pump's power and why? Peter We presume (hope) that the Swedes will not use a hose connected to plumbing where you get free water pressure, and will use a pump. The pump's power must be included in P-in. A liter/sec pump seems to require one horsepower or about .75 kW. -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:Pump power must be included
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
Re: [Vo]:Pump power must be included
Want I wanted to say- the pump is part of the cooling circuit to which the heat produced is transfered. 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
RE: [Vo]:Pump power must be included
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
Re: [Vo]:Pump power must be included
OK, old friend I understand what you sya, 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? Peter On Wed, Apr 20, 2011 at 6:42 PM, Jones Beene jone...@pacbell.net 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
RE: [Vo]:Pump power must be included
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
Re: [Vo]:Pump power must be included
Dear Jones, make some calculations please. for this case. OK? The word constrictors is terrific (Boa) Peter On Wed, Apr 20, 2011 at 7:18 PM, Jones Beene jone...@pacbell.net wrote: *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 -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
RE: [Vo]:Pump power must be included
a) The pump does add a tiny amount of heat to the water passing through it : the input temperature should be measured AFTER the pump. b) There is FRICTIONAL loss in a pipe http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm (Though in that calculation it's expressed as pressure drop). These aren't going to be anywhere near the measured power level. Hey! Another fake-let !!! c) The water could drive an internal turbine, generating heat directly (friction) or indirectly (generator connected to a resistor).
Re: [Vo]:Pump power must be included
yes, good point Jones, the system input power includes the power to operate the pump and the resistive heaters. Harry
Re: [Vo]:Pump power must be included
Alan J Fletcher wrote: a) The pump does add a tiny amount of heat to the water passing through it : the input temperature should be measured AFTER the pump. It is always measured after the pump. It would be rather difficult to measure it before the pump. Offhand, I can't imagine how you would arrange that. Perhaps in reservoir? Anyway, no one does it that way. However, when you do measure reservoir temperature, you find it is the same as the inlet to within 0.1°C. The pump does not add any measurable heat between the pump body and the inlet sensor. All of the heat it adds shows up with the water splashes to a halt. b) There is FRICTIONAL loss in a pipe Yes, that is the tiny amount of heat I referred to. - Jed
Re: [Vo]:Pump power must be included
hmm if water flow is required then could it be powered by the e-cat through convective heating. Harry From: Jones Beene jone...@pacbell.net To: vortex-l@eskimo.com Sent: Wed, April 20, 2011 10:34:59 AM Subject: [Vo]:Pump power must be included Peter, The “Rossi effect” is controlled in a narrow range by balancing heat removal and heat addition. It will not work reliably without constant heat removal. Therefore, power input related to the proper operation must be included as P-in. Jones
Re: [Vo]:Pump power must be included
Harry Veeder wrote: yes, good point Jones, the system input power includes the power to operate the pump and the resistive heaters. That is incorrect. Please review the messages I have posted. The input power does not include the pump any more than it includes the overhead lights or an oscilloscope. Heat generated before the inlet sensor or after the outlet sensor is not measured. - Jed
Re: [Vo]:Pump power must be included
Whether or not the water flow is powered by the pump or a waterfall, the kinetic energy of the flow may be a factor. A lack of water movement may explain why some PF type cells failed to perform in the past. They depended on the fickle nature of convection to spring to life. Harry yes, good point Jones, the system input power includes the power to operate the pump and the resistive heaters. That is incorrect. Please review the messages I have posted. The input power does not include the pump any more than it includes the overhead lights or an oscilloscope. Heat generated before the inlet sensor or after the outlet sensor is not measured. - Jed
Re: [Vo]:Pump power must be included
Harry Veeder wrote: Whether or not the water flow is powered by the pump or a waterfall, the kinetic energy of the flow may be a factor. No, it may not. That's out of the question. I have operated many flow calorimeters of all sizes and types, and there is absolutely no way you can detect the kinetic energy with this type of calorimeter. Furthermore, if you could, they would see it when they turn on the flow and prepare to do calorimetry, before they turn on the machine. The balance would be positive. It isn't. A lack of water movement may explain why some PF type cells failed to perform in the past. They depended on the fickle nature of convection to spring to life. Not a chance. - Jed
RE: EXTERNAL: Re: [Vo]:Pump power must be included
Harry, I agree the energy utilized should be subtracted from the output but how much of the pressure or flow rate is actually removed from the system? - the differential measurements are only for temp but you should also quantify the pressure/flow rate into the reactor and the pressure/flow rate out of the reactor if you want to determine if any energy was added or subtracted - otherwise the same pressure and flow are still potential - available for use downstream. I am sure the portion of pressure/flow removed from the system is only a small fraction. Maybe put the exiting water after measurement into same diameter pipe as the source and measure with an identical flow rate meter? Fran From: Harry Veeder [mailto:hlvee...@yahoo.com] Sent: Wednesday, April 20, 2011 3:11 PM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Pump power must be included Whether or not the water flow is powered by the pump or a waterfall, the kinetic energy of the flow may be a factor. A lack of water movement may explain why some PF type cells failed to perform in the past. They depended on the fickle nature of convection to spring to life. Harry yes, good point Jones, the system input power includes the power to operate the pump and the resistive heaters. That is incorrect. Please review the messages I have posted. The input power does not include the pump any more than it includes the overhead lights or an oscilloscope. Heat generated before the inlet sensor or after the outlet sensor is not measured. - Jed
Re: EXTERNAL: Re: [Vo]:Pump power must be included
Roarty, Francis X wrote: I agree the energy utilized should be subtracted from the output but how much of the pressure or flow rate is actually removed from the system? No measurable amount is removed. I guarantee that. -- the differential measurements are only for temp but you should also quantify the pressure/flow rate into the reactor and the pressure/flow rate out of the reactor if you want to determine if any energy was added or subtracted -- otherwise the same pressure and flow are still potential -- available for use downstream. I am sure the portion of pressure/flow removed from the system is only a small fraction. Maybe put the exiting wat er after measurement into same diameter pipe as the source and measure with an identical flow rate meter? Most flowmeters slow down the water and distort the numbers far more than the flow water through the pipes would. Especially rotary flowmeters take more energy out of the system than friction would. That is to say, they take out the energy to turn the rotor. Displacement ones also take energy. The ones that add a pulse of heat to detect it downstream add a tiny amount of energy. (But it is OUTSIDE THE SYSTEM!!! Doesn't count!) Anyway, there is no such thing as an identical flowmeter. Get two of the same make and model and you will find they each have a personality and a mind of their own. The variations between the flowmeters will be far greater than the energy losses from water friction. A couple of grains of sand in rotors or bearings will make a difference big enough to detect. I have put two or three flowmeters in line at about 100 ml/min. These were expensive ones. They produced different answers, for reasons beyond the scope of the discussion. The differences are too small to affect these conclusions but they are always present. Anyway, at 1 L/s you can put a dozen flowmeters in line above and below the inlet and it will not cause or detect any measurable change to the temperature or flow. Every one of them will register a different flow -- of that you can be sure -- but the variations will be caused by the instruments themselves and they will far exceed water friction. Ditto thermocouples, by the way. Install a dozen of them and you will get a dozen answers, with variations far larger than water friction can cause. They will have a different bias, speed, range, accuracy and precision. Gene and I installed multiple pairs of these as well, backed up with mercury thermometers. Although they misbehave and drift less than flowmeters do. Once you calibrate them, they are highly reliable in this range of temperatures. And contrary to assertions made here, any one of them can measure 5°C with confidence. Actually, they can measure 0.1°C, but real fluctuations in the water temperature are as large as that, for various reasons. Rest assured, Levi, Kullander and people like that know how flow meters and thermocouples work. Way more than I do. If were trying to measure the effect of friction from water flow in pipes, equipment such as the best $2000 flow meter and a top-notch RTD thermocouple would be ridiculously crude. That would be like trying to see a virus with a plastic magnifying glass. Someone like Rob Duncan has micro-calorimeters that could measure this friction easily, but they cannot measure a power level higher than a few milliwatts. Here is one with 0.002 micro-watt resolution. It does not say what the max power is, but it can't be much: http://www.stats-reports.com/webstat/dlcount.php?id=41061url=http://www.setaram.com/traitement/export_doc.php?doc=../files/documents/MICRODSC3-EVO.pdf - Jed
Re: [Vo]:Pump power must be included
In reply to Alan J Fletcher's message of Wed, 20 Apr 2011 09:37:23 -0700: Hi, [snip] If we assume 100 psi for the mains pressure, then a flow rate of 1 L /s equates to a total power of 724 W, assuming all the power in the water gets used. This would raise the temperature of that water by 0.173 ºC, so it would at most make a 4% difference, even if it were all included. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Pump power must be included
Jed Rothwell wrote: Harry Veeder wrote: Whether or not the water flow is powered by the pump or a waterfall, the kinetic energy of the flow may be a factor. No, it may not. That's out of the question. I have operated many flow calorimeters of all sizes and types, and there is absolutely no way you can detect the kinetic energy with this type of calorimeter. Furthermore, if you could, they would see it when they turn on the flow and prepare to do calorimetry, before they turn on the machine. The balance would be positive. It isn't. I agree that the flow of water adds an insignificant of heat to the water. I am speculating that a flow of water is part of the cold fusion recipe for making excess heat. A lack of water movement may explain why some PF type cells failed to perform in the past.They depended on the fickle nature of convection to spring to life. Not a chance. why do you say that? Harry
Re: EXTERNAL: Re: [Vo]:Pump power must be included
Sorry, I didn't mean to imply that the flow of water generates a significant amount of heat by friction. I was speculating that a flow of water is part of the recipe for making excess heat. If it is a part of recipe you have to include the power required to bring about that flow. Harry From: Roarty, Francis X francis.x.roa...@lmco.com To: vortex-l@eskimo.com vortex-l@eskimo.com Sent: Wed, April 20, 2011 3:34:37 PM Subject: RE: EXTERNAL: Re: [Vo]:Pump power must be included Harry, I agree the energy utilized should be subtracted from the output but how much of the pressure or flow rate is actually removed from the system? – the differential measurements are only for temp but you should also quantify the pressure/flow rate into the reactor and the pressure/flow rate out of the reactor if you want to determine if any energy was added or subtracted – otherwise the same pressure and flow are still potential – available for use downstream. I am sure the portion of pressure/flow removed from the system is only a small fraction. Maybe put the exiting wat er after measurement into same diameter pipe as the source and measure with an identical flow rate meter? Fran From:Harry Veeder [mailto:hlvee...@yahoo.com] Sent: Wednesday, April 20, 2011 3:11 PM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Pump power must be included Whether or not the water flow is powered by the pump or a waterfall, the kinetic energy of the flow may be a factor. A lack of water movement may explain why some PF type cells failed to perform in the past. They depended on the fickle nature of convection to spring to life. Harry yes, good point Jones, the system input power includes the power to operate the pump and the resistive heaters. That is incorrect. Please review the messages I have posted. The input power does not include the pump any more than it includes the overhead lights or an oscilloscope. Heat generated before the inlet sensor or after the outlet sensor is not measured. - Jed
