simple question power, resistance, current, etc
in [Physics]
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From: none on 21 Feb 2010 14:31 On 2010-02-20 19:41:24 -0600, Darwin123 <drosen0000(a)yahoo.com> said: > On Feb 20, 7:51�pm, none <n...(a)none.com> wrote: >> On 2010-02-18 22:03:34 -0600, none <n...(a)none.com> said: >> >>> Is the "power" of a battery a constant? I assumed it should be, but >>> then I was thinking that if V=IR, say you have a battery with voltage > 9 >>> and circuits with resistance 3 or 4.5. The current should be 3 or 2, >>> respectively. Then, the power would be 27, or 18, if power = voltage > * >>> current, right? So does the power of a batter change depending on the >>> resistance of the circuit? Seems odd... >> >> OK, thanks for all of the replies! Very helpful, and now I do >> understand, of course the power changes, depending on the load. That's >> why a battery runs down at different rates when it drives a small light >> vs a big light, etc. >> >> Now, on to the real question behind this. We did a science project to >> determine the efficiency of a wind turbine, and hooked up an ammeter >> and voltmeter. It turned out that for many different windspeeds, if the >> resistance in the circuit was lower, then we got significantly higher >> Power being produced than if the resistance was higher. So we thought >> that meant that the wind turbine was more efficient at converting wind >> energy to electricity if the resistance was lower. But one of the >> judges told us that was not true �- �that the efficiency was actually >> the same. So who was right? >> >> thanks > I congratulate you and your class for the completion of such an > ambitious project. The answer will not detract in any way the > ambition, drive and curiosity shown by such a project. > I hypothesize that the judges are right based on an error you made > in the first post. You really wanted to ask a question about > generators. You made a statement that the power of a battery should be > constant. Batteries are always DC. However, your real question was > about a generator. Most practical generators are AC. If you took the > design from a book or website, it is very likely that you built an AC > generator. The judges may know something about AC generators that you > do not. > On the other hand, it is possible the opposite occurred. You > built a DC generator and the judges used an AC formula to calculate > efficiency. Then they may be wrong. However, the naivete with which > you asked about batteries makes that an unlikely, although plausible, > possibility. > The discussion continues on the assumption that the judges are > right. I conjecture that your problem comes from not understanding AC > circuits. A battery is implicitly a DC power supply. All answers that > we gave you are completely valid for a DC power supply. However, you > now tell us that the question was really about a generator. > Apparently, you haven't really learned the difference between a > generator and a battery. Therefore, I conjecture that your class built > an AC generator without realizing that it would work differently than > a battery. > To validate my hypothesis, I will ask a few questions > 1) How did you calculate the power? > I suspect you used the formula which is completely valid for > DC, > P=IV > where P is the power, I was the current, and V was the voltage. > However, this formula is unambiguous only for DC current. If the > current was AC, one has to specify at what times I and V were > measured. Which brings us to the next question. > 2) Was the voltage output of your wind generator AC, DC or some > combination? > It depends on exactly how the brushes were correct. If the > voltage is DC, the formula P=IV is valid and unambiguous. If the > output is completely AC, however, the formula for power is: > P=IV cosA > where P is the time averaged power (averaged over several cycles), I > is the rms current, V is the rms voltage, and A is the phase angle. > The phase angle, A, could vary with resistance. If your generator > produced AC power, it is possible that A increased with load > resistance. Although the rms current would increase, the angle A could > increase in such a way as to leave the time averaged power unchanged. > The power you calculated with the DC formula would show an increase, > but it is not a physically meaningful power. > I will close with the following question: > 3) How does an oscilloscope, in the ungrounded configuration, present > the voltage of the generator. > If it is a DC generator, the oscilloscope will show a high > background voltage with a small AC ripple. However, if it is an AC > generator, it will show a sinusoidal waveform oscillating about zero, > with maybe a small DC offset. The windmill was just a DC motor attached to a propeller. We thought that it was making a DC current because hooked up to our voltmeter one way gave a positive voltage and the other way gave a negative voltage. Also, it only lit up the LEDs when we connected it in one orientation. But we didn't try an oscilloscope. In terms of "efficiency of energy conversion" we weren't trying to determine the theoretical maximum efficiency, but simply blowing a constant stream of air at the propeller, and measuring how much power came out (by multiplying voltage x current). The numbers we were getting calculated out to higher power when the resistance was lower. Since the energy going in to the windmill was a constant (constant wind speed), but power coming out was higher, we just said that qualitatively, it appeared to convert more wind to electricity at lower resistance.
From: Darwin123 on 21 Feb 2010 21:00 On Feb 21, 2:31 pm, none <n...(a)none.com> wrote: > On 2010-02-20 19:41:24 -0600, Darwin123 <drosen0...(a)yahoo.com> said: > > > > > On Feb 20, 7:51 pm, none <n...(a)none.com> wrote: > >> On 2010-02-18 22:03:34 -0600, none <n...(a)none.com> said: > > Since the energy going in to the windmill was a constant (constant wind > speed), but power coming out was higher, we just said that > qualitatively, it appeared to convert more wind to electricity at lower > resistance. This assumption is also suspect. The input power into that windmill doesn't have to be the same even if the wind speed the same. The reason is that the torque and the rate of revolution can both change. It is possible that with the smaller resistance, the rotation rate of the windmill fans increased. When a small resistance load is hooked up, the rotation rate may have increased while the torque remained the same. This would result in an increase of input power. The windmill may have become more efficient in extracting energy from the wind, but keep the same theromodynamic efficiency.
From: none on 22 Feb 2010 00:33 >> Since the energy going in to the windmill was a constant (constant wind >> speed), but power coming out was higher, we just said that >> qualitatively, it appeared to convert more wind to electricity at lower >> resistance. > This assumption is also suspect. The input power into that > windmill doesn't have to be the same even if the wind speed the same. > The reason is that the torque and the rate of revolution can both > change. > It is possible that with the smaller resistance, the rotation > rate of the windmill fans increased. When a small resistance load is > hooked up, the rotation rate may have increased while the torque > remained the same. This would result in an increase of input power. > The windmill may have become more efficient in extracting energy from > the wind, but keep the same theromodynamic efficiency. Actually, this is exactly what we were thinking might happen. But then that means we were right. We are (only) interested, in this experiment, in the total process of capture of wind energy and generation of electricity.If the windmill is more efficient at extracting energy from the wind when the resistance is lower, then that is what matters to the electric company and to your energy bill.
From: Darwin123 on 22 Feb 2010 10:51 On Feb 22, 12:33 am, none <n...(a)none.com> wrote: > >> Since the energy going in to the windmill was a constant (constant wind > >> speed), but power coming out was higher, we just said that > >> qualitatively, it appeared to convert more wind to electricity at lower > >> resistance. > > This assumption is also suspect. The input power into that > > windmill doesn't have to be the same even if the wind speed the same. > > The reason is that the torque and the rate of revolution can both > > change. > > It is possible that with the smaller resistance, the rotation > > rate of the windmill fans increased. When a small resistance load is > > hooked up, the rotation rate may have increased while the torque > > remained the same. This would result in an increase of input power. > > The windmill may have become more efficient in extracting energy from > > the wind, but keep the same theromodynamic efficiency. > > Actually, this is exactly what we were thinking might happen. But then > that means we were right. We are (only) interested, in this experiment, > in the total process of capture of wind energy and generation of > electricity.If the windmill is more efficient at extracting energy from > the wind when the resistance is lower, then that is what matters to the > electric company and to your energy bill. I guess the immediate question is whether you and the judges have the same definition of efficiency. You probably wrote the word efficiency in your report without defining what efficiency is. The judges may have thought you meant the output electrical power divided by the input work. Then, they would be right. You are probably correct that your definition of efficiency is impacts the cost of electricity. However, the judges may not have been thinking that way. The word efficiency depends on context. In a heat engine, which you didn't build, efficiency would be work out divided by heat energy in. In the case of your windmill, a physicist would automatically think, "work out divided by work in." You have to look at the background. judges. If you used the efficiency in your report, they may think you used the same definition of efficiency that they used. The way to fix this may be to simply add a paragraph defining a different type of efficiency. You may want to invent a phrase like "wind efficiency," which you define as the electric power out divided by wind speed. You may try to resubmit your report with a correction. However, it may be too late to do this. It would be sad if you lost because of a simple misuse of the English language. I know you worked hard on the generator. Ambiguity is a big problem in technical writing. Your audience has to know precisely what words mean, especially with numerical calculations. The efficiency of extracting energy from the wind could be important with regard to ecological concerns. The place the unused wind energy goes, and the effect on the wind, could conceivably become a problem. To a scientist these days, it is not clear whether the inventor is concerned about cost or environmental concerns. Therefore, particular meaning of the word efficiency has to be specified. Recently, it was found that the noise windmills make has a bad effect on birds. The efficiency of the windmill in making noise could be an issue, here.
From: Jeroen Belleman on 22 Feb 2010 11:24
none wrote: > Now, on to the real question behind this. We did a science project to > determine the efficiency of a wind turbine, and hooked up an ammeter and > voltmeter. It turned out that for many different windspeeds, if the > resistance in the circuit was lower, then we got significantly higher > power being produced than if the resistance was higher. So we thought > that meant that the wind turbine was more efficient at converting wind > energy to electricity if the resistance was lower. But one of the judges > told us that was not true - that the efficiency was actually the same. You should have told us that right away. All that battery talk is beside the point. You seem to have missed an interesting observation: Evidently, with either infinite or zero resistance, the power delivered by your mill would be zero. There is an optimum somewhere in between. Where this optimum lies depends on the properties of your mill and generator, and probably on windspeed too, I haven't thought that through. As for efficiency, how exactly do you propose to measure it? Jeroen Belleman |