Re: [Vo]:Problems with energy calculation Rossi device
Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the chimney with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the condensed water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:Problems with energy calculation Rossi device
I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the chimney with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the condensed water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:Problems with energy calculation Rossi device
Ofcourse. degr/sec is a typo. But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30 degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W? Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 5:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the chimney with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the condensed water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:Problems with energy calculation Rossi device
Absolutely not. It's a flow. Water get in and get out. It's not a closed container of water. With 1 watt you can rise, when dynamic transients go away, an half degree! 1 watt = 1 J/s 1 J/s = 1.8 g/s * 4.2 J/(g*K) * deltaT deltaT = 0.1 kelvin (= 0.1 celsius) If you don't trust me, read the report. It's a simple calculation! From: P.J van Noorden Sent: Saturday, April 09, 2011 5:41 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Ofcourse. degr/sec is a typo. But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30 degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W? Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 5:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the chimney with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen
[Vo]:Problems with energy calculation Rossi device
Hello Mattia, Ofcourse, you are right. I made a mistake. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 6:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device Absolutely not. It's a flow. Water get in and get out. It's not a closed container of water. With 1 watt you can rise, when dynamic transients go away, an half degree! 1 watt = 1 J/s 1 J/s = 1.8 g/s * 4.2 J/(g*K) * deltaT deltaT = 0.1 kelvin (= 0.1 celsius) If you don't trust me, read the report. It's a simple calculation! From: P.J van Noorden Sent: Saturday, April 09, 2011 5:41 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Ofcourse. degr/sec is a typo. But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30 degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W? Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 5:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the chimney with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C
[Vo]:Problems with energy calculation Rossi device
Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the condensed water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:Problems with energy calculation Rossi device
In reply to P.J van Noorden's message of Thu, 7 Apr 2011 11:19:06 +0200: Hi, [snip] Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. If you just consider the water flow rate, then the graph makes no sense at all, since a constant flow rate with a constant power input of 300 W at the beginning should result in a flat line at about 58 ºC. However what we see is a rising line. The only conclusion I can come to is that you need to include the thermal mass of the device which is also being heated by the electric heater so that in the beginning the whole thing is heating up. The really interesting point is that the reaction appears to take off at precisely the point where the temperature of the device has reached the 58ºC which would be the constant water temperature point for heating by the electric heater alone (300 W can maintain a temp. differential of 40 ºC at that flow rate) . I can't but help find this strange, but perhaps there is a causal link between the two? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Problems with energy calculation Rossi device
When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the chimney of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the condensed water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
