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256) On measuring electric energy

Ludwik Kowalski (9/15/05)
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043

1) What can be more simple than measuring electric energy delivered to a resistor at constant current? Measure the current, measure the voltage, multiply amps by volts (to get power in watts) and multiply by time in seconds. This gives energy in joules. The same is true of common alternating current, for example, at 60 Hz. Most voltmeters and ammeters show so-called effective values and that is what is needed to calculate the power in joules correctly. Electric power meters, calibrated in watts, can also be used in d.c. and a.c. circuits. Furthermore, electric energy meters can be used. Nearly every house has an energy meter; cost of electricity is usually calculated on the basis of such instruments. (They display energy in kWh; 1 kWh=3.600.00 J).

2) But what about a case in which an electric current is highly irregular? I am referring to an experiment described in unit #252. Two days ago I received a message from a German engineer, XX, who read that unit. He suspects that at least some researchers did not measure electric energy properly. He wrote: “

“. . . Some comments in #254 shows that I'm not the only one which sees the measurement of electrical energy consumption as one of the keys to this experiment. On the other side there seems to be some believe that if one take two multimeters and take the average I and U one could easily measure the electrical power of the experiment. At that point I want to stress that doing so could easily generate errors of nearly any magnitude. I will give you and example. Take the case of a pulsed waveform. Let us assume a pulse width of 1/10s and a pulse voltage of 1V. The current which is forced by the pulse of 1V should be 1A . The pulse should be repeated every second. So how can I calculate the energy used in 100s? It is really simple. Is it?

Method ONE (I'm sure the correct one): The energy of one pulse is 1A*1V*1/10s = 1/10Ws. In 100 second we have 100 pulses so we have a consumption of 100*1/10Ws=10Ws= 10J.

Method TWO (I'm sure the erroneous one): Let us calculate the energy taking the values of our two multimeters. The voltmeter in our case shows us a value of 1/10V and the ampere meter shows us and value of 1/10A. If we now calculate the energy simply by using U*I*t we get 1/10V*1/10A*100s = 1Ws= 1J. We have on error of factor 10 (exactly the inverse of the 1/10 duty factor of our case).

Now you can think what's the point of this discussion? The point is that I have found some experiments I assume exact this kind of error has happened. One of them is in an experiment from Kanarev you mentioned in #172. Here is the link to the description At the end of the paper the calculation of consumed energy is discussed and Kanarev insist on the calculation like method two. Because duty factor in that experiment is small he get huge excess power. Also some people from SITIS (?? obviously some guys he cooperate with) made him aware of the wrong calculations he argued that his method is correct. I have to admit that I'm a little bit surprised at this. Everybody can make errors but a professor should realize such an error at least after he was point the error. The other one I found was on the homepage of Naudin you mentioned also in 172 and 174 (I want to note that I like his homepage) in his experiments with MAHG. There I assume the same error happens. He calculates the energy consumption of his pulsed system with values reading from his two RMS multimeters. At the least he should have been skeptical after looking on his own results The COP factor scales nicely with the inverse of the duty factor. It is exactly the error one get if using method TWO.

I write this message not to accuse somebody to spread false claims or do a bad job I do not even totally exclude the chance that my thoughts in this email are based on an error of myself also I strongly belief that I'm right. I write this long message only to make interested people aware of the fact that using the values of two multimeters (even if they are true RMS multimeters) to calculate the energy consumption of an experiment could result in huge errors. The size of this error depends strongly on the waveform used during the experiments. Please do not hesitate to comment my email even if the comments should be critical.

Replying to the above I wrote that arcing and sparking produces highly irregular current. This makes energy measurement even more difficult that for the pulsed waveforms described by XX. Fortunately a simple calorimetric method can often be used to calibrate the system. Here is an illustration. Produce an irregular current by inserting a vibrating contact (like in a Tesla coil, or in a door bell) into a circuit containing your resistive heater. Immerse this heater into a calorimeter and turn the power on, for example, for 300 seconds. Suppose the heat capacity of your calorimeter (including the cup and the heater) is 500 cal/C = 2093 J/C. Suppose the temperature goes up by 30 C in 5 minutes. Then the average power (rate of heating) is 2093*30/300 = 209.3 J/s. Your electric power meter is reliable if the average power is close to 209.3 W. The calorimeter tells us that 209.3*300 = 62,790 J of thermal energy was delivered and that should agree with what is measured electrically.

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