Measuring kinetic energy with a DC motor - Page 1
Forum section: electronics / beginners

author: Floopy
author_rank: Regular Contributor
timestamp: March 23, 2022, 01:34:35 am
content: Hello,This is a little bit unusual, but I'm stumped and there are a lot of smarter people here than me. Pardon my lack of foresight and understanding.I'm working on a school project in which I'm trying to measure the kinetic energy stored in a flywheel through a DC motor. The idea was to use a DC motor to spin up a wheel to speeds up to 2500RPM and then shunt the motor with a resistor to dissipate the induced energy. I set it up like in so:The coil is used for detecting a magnet on the flywheel to measure the angular speed of the wheel. The idea was to use a PicScope to record the voltage drop over the resistor over time. Once I had that, I could find the power output and integrate it as a function of time. (using numerical integration methods of course)P=V^2/R -> P = V^2/R_resistor + V^2/R_motorI measured the resistance across the motor when it was off to get a general idea of what it should be. I got around 30for the motor and 15for the resistor.ResultsTesting provided consistent results, but they were consistently off of the predicted values.(too many sig figs in the picture, my bad)Assuming I got the numbers worked out correctly. Does my reasoning on this sound right? Is there something I'm overlooking that is causing an increase in error? I'm not an electrical engineer (mechanical) and there is probably something I obviously doing wrong as I don't believe there could be that many losses in a system.Also, here is the equation for the model E = 1/2 * I * omega ^2 ( I = 0.0178131 )-Thanks for taking a look,

author: twospoons
author_rank: Frequent Contributor
timestamp: March 23, 2022, 02:45:10 am
content: The other possibility is that you have a mistake in your predictions.

author: WattsThat
author_rank: Frequent Contributor
timestamp: March 23, 2022, 03:43:11 am
content: P=V^2/R -> P = V^2/R_resistor + V^2/R_motorShould be:P= V^2 / (R_resistor + R_motor)

author: Floopy
author_rank: Regular Contributor
timestamp: March 23, 2022, 04:20:36 am
content: Quote from: evb149 on March 23, 2022, 02:06:58 amGood point, it's my bad for not making it clear, I'm actually trying to show that the predicted and measured values are similar. So yes, I do agree, it could just be measured, but that's not exactly what I'm trying to demonstrate.Quote from: evb149 on March 23, 2022, 02:06:58 amI should have thought of that, I'll see if I need to retake another few samples.Quote from: evb149 on March 23, 2022, 02:06:58 amFunny enough, I realized this after acquiring all the data. Sometimes the solution to a problem is a lot simpler.Quote from: WattsThat on March 23, 2022, 03:43:11 amWhy is that so? I thought the resistors were taken into consideration as parallel? I did try to change it and I get smaller values.Quote from: twospoons on March 23, 2022, 02:45:10 amI'll check again, the engineering book I'm pulling this from can be pretty cryptic.Thank you all for the suggestions, I'll see on working through a few of them.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 23, 2022, 07:52:33 am
content: Quote from: WattsThat on March 23, 2022, 03:43:11 amThe formula is correct if the theoretical voltage is measured, but this voltage is not accessible. The measuremd voltage at the resistor isV_resistor = R_resistor / (R_resistor + R_motor) * VSo the total power, including the resistive loss in the motor is:P= V_resistor^2 * (R_resistor + R_motor) / R_resistor² = (R_resistor + R_motor) * I²The R*I² way to calculated the power is likely the easier one to understand.There are some additional losses in the usual DC motors. So expect the calculated power to still be lower than this. One could get an idea of the extra losses from the power needed to run the wheel at a constant speed. So it absolutely makes sense to have the relatively small external resistor and thus the relatively fast stopping of the flywheel.

author: Berni
author_rank: Super Contributor
timestamp: March 23, 2022, 08:24:36 am
content: I agree with KleinsteinThe voltage your oscilloscope is seeing is only the voltage across the resistor. For the 2nd "+ V^2/R_motor" part of your equation requires you to know that V that is the internal voltage drop inside the motors windings, you can't measure this voltage with simply an oscilloscope.For this reason it is better to use the 30 Ohm resistor as a shunt resistor to instead calculate current trough this resistor. This current is the same current as trough the motor so the whole thing can be treated as the energy dissipated by a 45 Ohm load at the given current.As others have said this number still won't precisely match because there are also magnetization losses in the motor, bearing friction etc.. But i am guessing the number should get pretty close, as in >90%

author: RoGeorge
author_rank: Super Contributor
timestamp: March 23, 2022, 12:52:47 pm
content: Never tried in practice, but I would expect to see significantly less energy dissipated in the resistor than it was stored in the flywheel.- 1. In an ideal setup (when internal resistance of the generator and the resistance of the load are equal) at best only 50% of power is dissipated in the external load resistor, the other 50% is dissipated in the generator's internal resistance. If the Rgenerator is different than Rload, even less power is transferred to the load.- 2. If the motor is small, I would expect a lot of energy to be dissipated internally as magnetic loss, and as friction in the bearing- 3. A dc motor is in fact an switched alternating machine, so the generator is in fact generating AC. That means, in theory, that there could be a phase shift (power factor) between I and V inside the motor. I have no hands on experience with electrical machines, so no idea how big of a percentage that could be in practice.For 3. I think the power factor should be possible to measure with an oscilloscope, if we attach a photo detector (or a mechanical switch) to the flywheel, so to synchronize the oscilloscope with the flywheel.- That would be equivalent with triggering the oscilloscope on the internal voltage of the motor, which is not the same with the voltage on Rload, so the I vs U phase inside the DC motor can be observed on the oscilloscope by looking at the V on the Rload and the phase shift from the flying wheel sync/trigger point.- By measuring/recording the RPM, the amplitude of the internal AC voltage inside the generator's coils can be also deducedKnowing R, L, Ui, I and the phase between U internal and I should be enough to calculate the active power disipated inside the generator. I think on top of that, it should be added the losses by the Eddy currents.

author: max_torque
author_rank: Super Contributor
timestamp: March 23, 2022, 05:11:57 pm
content: The motor has an efficiency with which it turns eletrical energy into useful work, and this is NOT 100%However, you can use the fact the flywheels energy doesn't change to estimate the efficiency!Run the flywheel at a series of constant speeds up to the speed you wish to spin it. At each speed measure the power input to the motor. This gives you the drag of the system including motor efficiencyRun the flywheel up to that speed, then immediately let it run back down to stopped. measure the energy you used to "charge" the system and the energy used to discharge the systemBecause a motor has roughly the same efficiency when it acts as a generator, the difference in energy measured, is the total drag during the spinning, plus TWICE the motor loss, because the motor lost energy when motoring and when absorbingSo, to make the numbers simple, imagine you put in 100 Wh and got out 50 Wh.You drag curve allows you to integrate all the drag energy, lets say 10 Wh (5Wh spining up, and 5Wh spinning down if accel and decel times were the same), that leaves 40 Wh for motor "round trip", efficiency loses meaning motor lost 20 Wh in each direction. This means your flywheel stores 75 Wh at the average speed of your test

author: twospoons
author_rank: Frequent Contributor
timestamp: March 23, 2022, 07:53:07 pm
content: Quote from: Kleinstein on March 23, 2022, 07:52:33 amUsing V is the wrong way to do this, unless Rload=Rmotor. You can see this is wrong if you take the extreme case where Rload=0. In that situation the measured V is also 0, and so your energy calculation also comes out at zero, which is clearly wrong. The flywheel still stops with Rload=0 because the energy is dissipated in the motor .The correct way to do this is to measure the current, which will be the same in the load and the motor as they are in series, then use P=I2*Rmotor +I2*Rload.Or if you can only measure voltages, I= V/Rload so P= V2/Rload+(V/Rload)2*Rmotor

author: Floopy
author_rank: Regular Contributor
timestamp: March 23, 2022, 08:17:58 pm
content: Thank you all, I will sift through this and see what I need to improve. I'm glad there are people that are better equipped at analyzing this than me.-Really appreciate it,

author: RoGeorge
author_rank: Super Contributor
timestamp: March 23, 2022, 09:26:36 pm
content: Found a paper with a mechanical to permanent magnet DC generator. In "2.2. Model of permanent magnet DC generator and buck converter" is a model of the generator, too. Good as an example and to know what to consider.https://www.researchgate.net/publication/275112342_Maximum_Efficiency_or_Power_Tracking_of_Stand-alone_Small_Scale_Compressed_Air_Energy_Storage_System/fulltext/554825930cf2b0cf7aceb574/Maximum-Efficiency-or-Power-Tracking-of-Stand-alone-Small-Scale-Compressed-Air-Energy-Storage-System.pdf?origin=publication_detailElectrical machines might be quite confusing when it's to distinguish what energy goes where. Because they work in a variable magnetic field, Kirckhhoff loop law doesn't hold (IMO - and I'm not a physicist so I might be wrong here - because of the non-conservative magnetic field, see the controversial "Kirchhoff is for the birds" of Prof Walter Lewin), therefor a new term (a correction) is needed to be introduced: the back EMF.Maybe some other demo might be better to prove the quantity of energy stored in that flywheel, maybe a mechanical only setup would be easier for the students to understand.

author: cncjerry
author_rank: Unknown
timestamp: March 23, 2022, 10:27:13 pm
content: What about the effective resistance in the flywheel bearings? Poutisn't going to equal Pinas the wheel is doing work, right?I remember the feeling I got realizing computers do no real work when I had to calculate the input power and cooling power, that which were for all purposes, equal.Here's another thought. If you measure the total power that it takes to bring the Flywheel up to a constant speed (integrate power against rpm over time), wouldn't that consider most of the variables including the resistance of the windings and measuring apparatus, bearing load, etc. except for heat as you would need to know the temperature at T0and Tconstantspeed and the mass of the motor... Maybe too complicated.And then you have the Chernobyl test, where they were trying to use the flywheel effect of the turbines to run the pumps. Saw how well that worked.

author: Floopy
author_rank: Regular Contributor
timestamp: March 24, 2022, 05:55:05 am
content: Quote from: max_torque on March 23, 2022, 05:11:57 pmQuote from: cncjerry on March 23, 2022, 10:27:13 pmIt actually worked! I changed the power equation along with estimating the losses by measuring the power input to the motor at specific speeds and curve fitting the data. I then subtracted those estimated losses from the measure power output at the given speed. My measured data has been reduced to at most a 10% error, and I'm getting some of the measured to be within 1% of predicted. Thank you so much, and it makes sense too. Also I ran a second run of samples, and I used the current instead of the voltage and it is easier. However, unfortunately I can't throw away the data I've already acquired (at least for this class).Also, thank you for the Chernobyl reference. I will definitely look into it.