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On Mon, 14 Apr 2008 09:10:20 -0500
Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. My view is that any source must both absorb and deliver power at some none zero impedance. As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source. While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. Charging a battery with zero internal resistance is a good example. Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator. Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. Some would call that a world of science fiction. -- 73, Roger, W7WKB |
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On Apr 14, 10:10*am, Cecil Moore wrote:
Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. Your equation above completely ignores reflections which is the subject of the thread. That it does. We get to that later. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. They are quite different. I am quite will to explore (and am doing so) the concept of imputed energy flow in the reflected wave. It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. You should tread back through the posts, the question was answered. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Well that is the question, isn't it? It could be in the source. Or, if it can not be determined where the energy in Pr goes, then the only other answer is that Pr does not represent an energy flow. Think Sherlock: "when the impossible has been eliminated the residuum, however improbable, must contain the truth." Even my ten year old grandson can solve that problem and he's no future rocket scientist. Ah yes, but was he presented with ALL the options? ...Keith |
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Roger Sparks wrote:
While it can be argued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. I thought I had already addressed that topic when I added the one-wavelength of transmission line to the example in between the source and source resistance. But here's an example that may allow better tracking of the energy flow. Let's modify my Part 1, Fig. 1-1 to add a 50 ohm circulator and load to the ground leg of the source. Everything else remains the same. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? Does this example answer your questions? -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. You can probably answer your own question by figuring out how the light energy in an interference pattern gets from the dark ring to the bright ring some distance away. Do you think it happens faster than the speed of light? It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. I am satisfied with the destructive/constructive interference explanation. That you have come up short in tracking those component energies is not unexpected given your prejudices. If you told me your car disappeared from existence because you cannot find it, I wouldn't believe you either. You should tread back through the posts, the question was answered. "It depends" was no answer - that was just mealy-mouthing. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. The Vs source has zero resistance rendering dissipation impossible. I repeat: There are only two sources of power *DISSIPATION* in the entire system, the load resistor and the source resistor. The source can certainly throttle back its output when there is destructive interference in the source resistor and increase its output when there is constructive interference in the source resistor, but it CANNOT dissipate any power. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. Nope, you are confused. The source can adjust its output but the source cannot *DISSIPATE* power. As I said, the only sources of power *DISSIPATION* are the two resistors. No amount of obfuscation is going to change that. Perhaps adding a circulator to my Part 1, Fig. 1-1 will allow you to see things in a clearer light. Of course, using light would be even better. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? -- 73, Cecil http://www.w5dxp.com |
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On Apr 14, 12:06*pm, Roger Sparks wrote:
On Mon, 14 Apr 2008 09:10:20 -0500 Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil *http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. *My view is that any source must both absorb and deliver power at some none zero impedance. * I am not sure why you desire a non-zero impedance. The usual definition of an ideal voltage source is that it provides or sinks what ever current is needed to hold the desired output voltage. When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. When it is sinking current, it is absorbing energy. No statement is made about where this energy is going. A non-zero impedance is not required to make any of the above behaviour work. If you include a non-zero impedance, then you have a more real world source which can often be modeled using the Thevenin equivalent circuit; an ideal voltage source (zero impedance) in series with a resistor representing the impedance of the real world source. As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source. This is true. While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers *power into the source. *Charging a battery with zero internal resistance is a good example. *Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator. Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. *Some would call that a world of science fiction. ...Keith |
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On Apr 14, 1:36*pm, Cecil Moore wrote:
Keith Dysart wrote: Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. You can probably answer your own question by figuring out how the light energy in an interference pattern gets from the dark ring to the bright ring some distance away. Do you think it happens faster than the speed of light? What is your explanation for this phenomenon? It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. I am satisfied with the destructive/constructive interference explanation. That you have come up short in tracking those component energies is not unexpected given your prejudices. I suppose. If you are happy with energy equations that don't balance. If you told me your car disappeared from existence because you cannot find it, I wouldn't believe you either. You should tread back through the posts, the question was answered. "It depends" was no answer - that was just mealy-mouthing. Except, that it does depend. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. The Vs source has zero resistance rendering dissipation impossible. I repeat: There are only two sources of power *DISSIPATION* in the entire system, the load resistor and the source resistor. You have to read more carefully. I did not use the word dissipation. This is because there are ways other than dissipation to remove energy from the system. Just as the source provides energy and we do not care where it comes from, it can remove energy and we do not care where it goes. Examine Ps(t). You will find that for some of the time energy is being absorbed by the source. This occurs when the sign of Ps(t) is negative. The source can certainly throttle back its output when there is destructive interference in the source resistor and increase its output when there is constructive interference in the source resistor, but it CANNOT dissipate any power. Since we don't know the internals of the source, we do not know if it is dissipating or not. But we do know that when the sign of Ps(t) is negative, the source is absorbing energy from the system, exactly analagous to it providing energy when the sign of Ps(t) is positive. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. Nope, you are confused. The source can adjust its output but the source cannot *DISSIPATE* power. As I said, the only sources of power *DISSIPATION* are the two resistors. No amount of obfuscation is going to change that. You really should rethink this a bit. When current flows into a voltage source, the voltage source is absorbing energy. Perhaps adding a circulator to my Part 1, Fig. 1-1 will allow you to see things in a clearer light. Of course, using light would be even better. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL * * * *\ / * * * * 3 * * * * | * * *50 ohms * * * * | * * * *GND How much power is dissipated in the circulator resistor? This circulator, while often used, does not in any way add clarity. Changing the circuit changes the results, especially when you add a circulator which alters rather dramatically the energy flows. How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? This depends on the design of the generator and the length of the line. With a shorted line that is 90 degrees long, and a generator constructed using the circuit of Fig 1-1, the source supplies no energy to maintain an imputed forward power of 50 watts on the line. ...Keith |
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Keith Dysart wrote:
When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. The question is: Is that energy being created or dissipated as needed according to your omnipotent whims? -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
What is your explanation for this phenomenon? You first! :-) I suppose. If you are happy with energy equations that don't balance. My energy equations balance perfectly. Yours are the energy equations that don't balance in violation of the conservation of energy principle. "It depends" was no answer - that was just mealy-mouthing. Except, that it does depend. Sounds more like religion than anything else. You have to read more carefully. I did not use the word dissipation. You have to read more carefully. I used the word "dissipation" and you disagreed with me. Please read it again to verify that fact. Since we don't know the internals of the source, we do not know if it is dissipating or not. Sorry, the source is, by definition, lossless. All of the source dissipation is lumped in the source resistance drawn separately on the diagram as Rs. You really should rethink this a bit. When current flows into a voltage source, the voltage source is absorbing energy. But the source is NOT *DISSIPATING* energy because Rs is not inside the source. Rs is clearly drawn outside the lossless source so the source generates *ZERO* heat. I am amazed at the lengths to which you will go to try to obfuscate the discussion. Changing the circuit changes the results, ... When you get more serious about the technical facts of physics than you are about saving face, please get back to us, but please, not before. Everyone is tired of your delusions of grandeur where the laws of physics obey your every whim. -- 73, Cecil http://www.w5dxp.com |
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On Mon, 14 Apr 2008 16:54:47 GMT
Cecil Moore wrote: Roger Sparks wrote: While it can be argued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. I thought I had already addressed that topic when I added the one-wavelength of transmission line to the example in between the source and source resistance. I thought the addition of a one wavelength transmission line did not address the issue, and only added more reflections. We still need a reason to assume that a voltage source should *not* absorb power. But here's an example that may allow better tracking of the energy flow. Let's modify my Part 1, Fig. 1-1 to add a 50 ohm circulator and load to the ground leg of the source. Everything else remains the same. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? Does this example answer your questions? -- 73, Cecil http://www.w5dxp.com No, I'm sorry but no. I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor). I think that we must allow our voltage source to have that same real property. I do understand that when we allow the source to receive power, then we need to address source impedance. If we assign a single impedance, then we expect reflections from the source. The simple solution that I propose is to add a source property of absorbing all reflections. This can be accomplished in the real world by making the transmission so long that reflections never return from the source over any reasonable time, or by making the tranmission line sufficiently lossy to absorb reflections. Your example uses the first method. Does the idea of source receiving power run counter to what you were planning to write in Parts 2 and 3? I am trying to understand why you have such great reluctance to accept that the source could receive power for part of a cycle, especially when it could easily bring the instantaneous power and energy calculations into balance. -- 73, Roger, W7WKB |
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On Mon, 14 Apr 2008 13:40:20 -0700 (PDT)
Keith Dysart wrote: On Apr 14, 12:06*pm, Roger Sparks wrote: On Mon, 14 Apr 2008 09:10:20 -0500 Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil *http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. *My view is that any source must both absorb and deliver power at some none zero impedance. * I am not sure why you desire a non-zero impedance. The usual definition of an ideal voltage source is that it provides or sinks what ever current is needed to hold the desired output voltage. When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. When it is sinking current, it is absorbing energy. No statement is made about where this energy is going. A non-zero impedance is not required to make any of the above behaviour work. My thought needed more developement. When the source delivers power, we readily accept that the impedance of delivery will be the impedance of the attached circuit. We make the same assumption when a reflection is returned to the source. If we make the assumption that the source has the same impedance as the refection, then no reflection from the source is expected. So yes, I agree with your observation. If you include a non-zero impedance, then you have a more real world source which can often be modeled using the Thevenin equivalent circuit; an ideal voltage source (zero impedance) in series with a resistor representing the impedance of the real world source. -- 73, Roger, W7WKB |
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