Motores de Indução (Aula 02) - Ensaios no Motor Gaiola de Esquilo

Hey guys. Today we will begin testing the three-phase induction motor. And this here is the bench that we are going to use.

It is made up of the dynamometer, the control panel and our engine. This is the connection box, where the three three-phase power cables already enter. On the engine plate we can get some information that is very important, let's write it down.

Let's put here the main data on the engine plate that we are testing. It is a three-phase machine, nominal power of 750W, speed of 1730 RPM when it is performing nominal power, supply voltages 220V or 380V, remembering 220V is for a delta 380 connection via a star connection. Currents of 3.

08A when it is connected in delta, line value, and 1. 78A when it is connected in star. Efficiency of 78%, also in the condition where it will be performing the nominal power and cosine 0.

82 in this same condition. What does this bench allow us to do? Here we have a varivolt that allows gradual power supply to the induction motor, the control panel where we can regulate the CT and PT measurements.

These are the gauges and we also have the dynamometer load control system, so I can gradually increase and decrease the load that we are imposing on the induction motor. In fact, the dynamometer is the name given just to this scale here. This entire structure will be responsible for measuring the mechanical power transferred by the shaft to the engine.

Hours but for me to be able to measure a power being transferred, someone has to be performing it, someone has to be dissipating this power. This is the role of the Foucault brake. And how does this brake work?

The motor shaft is coupled to this aluminum disc. Here we have two coils and they make up a magnetic circuit. They are powered by direct current, controlled by our control panel, so that a flow circulates here that crosses both sides of this aluminum disc.

With this motor in motion, currents will be induced inside the aluminum, creating a counter force, dissipating power in the form of heat. But how do we average this power? Power is torque, on the shaft, multiplied by speed.

The speed of the machine can be obtained using a digital tachometer. We have a reflective tape here we put the tachometer, it measures the reflection pulses and we get the speed value. As for the torque, we obtain it through the dynamometer.

Note that the bearing that supports the structure is double made up of two moving parts: the internal one where the motor shaft passes and the external part is where the brake is fixed. What keeps a brake in this horizontal direction are these springs here and without them it becomes loose. As the brake is electrically excited, it holds the disc, imposing a force on the balance.

Multiplying the weight measured by our dynamometer by the gravitational acceleration we have the force that is being imposed at this point here. If we multiply this force through the lever arm to the central part of the shaft we obtain a touch, we measure the speed, multiplying the torque by the speed gives us the transferred power. The system here allows us to simulate several different load points on the engine, simply by changing the brake excitation.

But it is one of our tests, as we saw the introduction video, the objective here is to raise the equivalent circuit of the motor and manage to separate its losses. Let's start with the locked rotor test. Here we have a pin that allows the machine's rotor to be blocked, and it crosses the disc here and, now, the machine is unable to move.

Obviously, in this condition, if we turn on this machine at the nominal voltage, the motor will burn because the current will be a starting current, starting current is a data that we can have on the board, so it is in this specific motor that it is 6. 2 times the rated current. Certainly, the cable will not support it, it will melt and the machine will possibly catch fire.

That's why, our bench, we have the varivolt. The varivolt will allow us to put a voltage, such that the current does not exceed the nominal. In this case, we will only reach the nominal current and take measurements of current, voltage and power.

As these notes will be used to create the equivalent circuit of the machine, we will now include the phase values ​​of the measurements here. So let's move on to the locked rotor test: the current we measured was nominal 1. 78A, phase voltage 39.

8V and three-phase power 165. 5W. It is worth making some observations: the machine has the rotor stopped, rotor blocked.

So we don't have shaft speed, if we don't have shaft speed there is no power being transferred to this dynamometer structure. Even though we have torque being exerted here in this structure, if there is no speed, the power will be zero. Second is that, without having speed, we have no rotational losses.

There will be no friction on the fans, there will be no friction on the engine bearings. So we can remove this from our analysis. What's left?

There are magnetic losses and losses in the rotor and stator windings. The current is nominal, so obviously we have losses in the rotor and stator windings. But the voltage, as it is blocked, had to be greatly reduced so that the current was limited to the nominal, so our phase voltage is around 40V while the nominal is 220V.

In this case, we managed to considerably mitigate the magnetizing branch, so the magnetic losses were basically removed from our measurement, leaving only the losses in the rotor and stator windings. To begin with, let's remember how the induction motor equivalent circuit works. We have R1 and jX(1), which represent the resistance and reactance of the stator coils.

We have a second term that represents the resistance and reactance of the rotor, the term inversely proportional to the slip that represents the power transferred by the machine shaft. Remembering: slip can be obtained by subtracting the synchronous speed NS by the real speed of the machine divided by the synchronous speed. Finally, we have the magnetizing branch that represents the losses in the machine’s iron.

Bringing to the locked rotor test situation we have a situation where the slip is maximum, so the branch that represents the rotor in the machine will impose the lowest resistance to the passage of current, making the magnetizing branch practically negligible in this analysis. of circuits. We can redesign the equivalent circuit in a more simplified way, where we only have R1, jX1, R2', jX'2.

Through the measured values ​​of current, voltage and power, we can calculate the values ​​of these parameters by simplifying the problem by saying that R1 = R2' and jX1 = jX'2. Now we are going to do the empty test. To do this, we have to decouple the machine from here, otherwise the rotational loss measurements will be taken from the machine with the bearing system here from the brake.

We have to dismantle. Machine uncoupled so let's do an empty test. First, power it at nominal voltage.

Now we take all measurements of current, voltage, power and speed through our tachometer. As this is the empty period, we supply at nominal voltage, measured current was 1. 12A, three-phase power of 121.

6W and motor speed of 1,799 RPM. Let's analyze the equivalent circuit of the motor. Here, we analyze the issue of slippage, this is a 4-pole machine, so its synchronous speed is 1800 RPM, that is, slippage of only 1 RPM.

Very low. This means that, from an electrical circuit point of view, the branch that represents the rotor is practically excluded from the circuit analysis, because the impedance there will be very high. We can redesign our equivalent circuit disregarding the rotor branch.

From this, with the data we obtained from the measurements and, already knowing the values ​​of R1 and jX1, obtained in the previous test, we can calculate Rf and jXm. In the empty test, we measured three-phase power of 121. 6 watts.

Where is this being dispelled? The machine is operating at no load, it is not performing power under any load. The test is carried out at nominal voltage, so we certainly have losses in the core of the material, magnetic losses.

The rated current is 1. 78 amps and the measured current was 1. 12A, so we have significant stone in the stator windings at least.

As it is working uncoupled and has practically no resistance to movement, there is the possibility, in this test, of gradually reducing the voltage value, in order to mitigate the effects of losses in the stator windings and reducing the voltage, consequently, the We also reduce magnetic losses. The objective of this is to find a point where we can isolate only the rotational losses of the machine. Let's make a table measuring voltage, current, power and speed on the motor shaft.

Initially, we will draw a graph that analyzes the speed of the machine as a function of the decrease in voltage. Note that up to a third of the nominal voltage, the machine did not experience a considerable loss of speed. The nominal speed was only reached at 29V, just over 10% of the nominal.

At this point here we have a very considerable mitigation of losses in the machine's core. Let's now analyze the graph that relates the machine current as a function of the decrease in voltage. Note that the nominal speed point is known as the point where the current begins to rise with the decrease in voltage.

At this point, in particular, we have a current of just 280mA, which also brings a significant reduction in losses in the motor windings. Taking only the rotational ones, we will plot the electrical power supplied as a function of the decrease in the supply voltage. Exactly at the point where the speed is nominal, a three-phase power of 20W was obtained.

If the objective of the test was to obtain only an estimate of the rotational losses, this may be a reasonable approximation. It is important to highlight that this estimate is rough. Depending on the machine, voltage and current values ​​can be much more significant, in this case, there are methods to improve this loss estimate.

As you can see in theory, you can, for example, extrapolate the power curve as a function of the square of the voltage, seeking to remove the feathers in the core from the results. Furthermore, based on the current values ​​of the equivalent circuit, the losses in the windings for that operating point can be calculated. This way, it is possible to isolate only the rotational losses much better.

But this could be a good task. Reproduce this data and apply the methods to find isolated rotational losses, let's see how far it is from the 20 watts we found here. Finally, we will look at the engine’s performance curve.

It is coupled again, started to run, nominal voltage, everything perfect. And then, we put the load on our Foucault brake. For now I haven't turned on the brake yet, so, basically, the system is operating at no load, it's as if we had a second test at no load.

But now, in addition to the rotational losses of the motor, we are measuring the rotational losses of the brake bearing. Carrying out these electrical measurements again for the machine now coupled and operating at no load, for nominal voltage we have a current of 1. 13A, power of 186.

4W and speed of 1795 RPM. Here is a comparison, with the same test, but for an uncoupled machine. Note that, for the same nominal voltage, the current was practically the same.

The speed decreased by 4RPM, but the power went from 121. 6W to 186. 4W.

So, neglecting this 4 RPM variation, we can say that this difference in power refers to losses in the bearings and coupling. This subtraction will give a value of 64. 8 watts.

So, to calculate this engine performance curve, I have to keep in mind that the power being supplied to the machine shaft is the sum of the power being imposed by the brake and the power being expended on the brake bearing. Let's start loading. To create the machine's performance curve, we will create a table with the following measurements: three-phase power supplied, power dissipated in the load, already calculated by multiplying torque by speed as explained previously, and the speed itself on the motor shaft.

Based on the results obtained, we will insert two more columns. The first is the power transferred by the machine shaft, which is the sum of the power dissipated in the disc with the losses in the bearing and the brake coupling. Here we will use the obtained value of 64.

8 watts for all points, but it is worth remembering that it was obtained for a speed between 1795RPM~1799 RPM. So in this test, as we increase the load, the speed drops a little and this power should also decrease a little, but, proportionally, it is not so relevant because our lowest value here was 1,725 ​​RPM. We will accept this approach to the problem.

And the second column, the motor percentage efficiency, which we can obtain through the power supplied and the power delivered to the machine shaft. Note that, for nominal power, the corresponding value of efficiency is practically 78%, which coincides with the plate data. If we take one third of the nominal, based on the graph, we see that it is approximately 58%.

Note that, by oversizing the machine in a given application, it may be working with a performance much lower than the nominal. That's it folks. These were the rehearsals we had for today.

Thank you very much and until next time.