How to determine the power and speed of an electric motor without disassembling it. How to calculate the rated current of a three-phase electric motor. The three-phase electric motor nameplate has been erased. How to find out the power

Hello, dear readers and guests of the Electrician's Notes website.

I decided to write an article about calculating the rated current for a three-phase electric motor.

This question is relevant and does not seem so complicated at first glance, but for some reason errors often arise in the calculations.

As an example for calculation, I will take a three-phase asynchronous motor AIR71A4 with a power of 0.55 (kW).

Here is its appearance and a tag with technical data.

If you plan to connect the motor to a three-phase 380 (V) network, then its windings need to be connected in a star configuration, i.e. on the terminal block it is necessary to connect the terminals V2, U2 and W2 to each other using special jumpers.

When connecting this motor to a three-phase network with a voltage of 220 (V), its windings must be connected in a triangle, i.e. install three jumpers: U1-W2, V1-U2 and W1-V2.

So let's get started.

Attention! The power on the engine nameplate is not electrical, but mechanical, i.e. useful mechanical power on the motor shaft. This is clearly stated in the current GOST R 52776-2007, clause 5.5.3:

Useful mechanical power is designated as P2.

Even more rarely, the label indicates horsepower (hp), but I have never seen this in my practice. For information: 1 (hp) = 745.7 (Watt).

But we are interested precisely in electrical power, i.e. power consumed by the motor from the network. Active electrical power is designated as P1 and it will always be greater than mechanical power P2, because it takes into account all engine losses.

1. Mechanical losses (Рmech.)

Mechanical losses include bearing friction and ventilation. Their value directly depends on engine speed, i.e. the higher the speed, the greater the mechanical losses.

For asynchronous three-phase motors with a wound rotor, losses between brushes and slip rings are also taken into account. You can learn more about the design of asynchronous motors.

2. Magnetic losses (Рmagn.)

Magnetic losses occur in the “iron” of the magnetic circuit. These include losses due to hysteresis and eddy currents during magnetization reversal of the core.

The magnitude of magnetic losses in the stator depends on the frequency of magnetization reversal of its core. The frequency is always constant and is 50 (Hz).

Magnetic losses in the rotor depend on the frequency of rotor magnetization reversal. This frequency is 2-4 (Hz) and directly depends on the amount of motor slip. But magnetic losses in the rotor are small, so they are most often not taken into account in calculations.

3. Electrical losses in the stator winding (Re1)

Electrical losses in the stator winding are caused by their heating from the currents passing through them. The greater the current, the more the motor is loaded, the greater the electrical losses - everything is logical.

4. Electrical losses in the rotor (Re2)

Electrical losses in the rotor are similar to those in the stator winding.

5. Other additional losses (Radd.)

Additional losses include higher harmonics of the magnetomotive force, pulsation of magnetic induction in the teeth, etc. These losses are very difficult to take into account, so they are usually taken as 0.5% of the consumed active power P1.

You all know that in an engine, electrical energy is converted into mechanical energy. To explain in a little more detail, when electrical active power P1 is supplied to the motor, some of it is spent on electrical losses in the stator winding and magnetic losses in the magnetic core. The residual electromagnetic power is then transferred to the rotor, where it is consumed by electrical losses in the rotor and converted into mechanical power. Part of the mechanical power is reduced due to mechanical and additional losses. As a result, the remaining mechanical power is the useful power P2 on the motor shaft.

All these losses are included in a single parameter - the efficiency of the engine, which is denoted by the symbol “η” and is determined by the formula:

By the way, the efficiency is approximately equal to 0.75-0.88 for engines with power up to 10 (kW) and 0.9-0.94 for engines over 10 (kW).

Let us once again turn to the data of the AIR71A4 engine discussed in this article.

Its nameplate contains the following information:

• engine type AIR71A4
• serial number No. ХХХХХ
• type of current - alternating
• number of phases - three-phase
• supply frequency 50 (Hz)
• winding connection diagram ∆/Y
• rated voltage 220/380 (V)
• rated current at delta 2.7 (A) / at star 1.6 (A)
• rated useful shaft power P2 = 0.55 (kW) = 550 (W)
• rotation speed 1360 (rpm)
• Efficiency 75% (η = 0.75)
• power factor cosφ = 0.71
• operating mode S1
• insulation class F
• protection class IP54
• name of the company and country of manufacture
• year of manufacture 2007

Calculation of motor rated current

First of all, it is necessary to find the electrical active power consumption P1 from the network using the formula:

P1 = P2/η = 550/0.75 = 733.33 (W)

Power values ​​are substituted into the formulas in watts, and voltage in volts. Efficiency (η) and power factor (cosφ) are dimensionless quantities.

But this is not enough, because we did not take into account the power factor (cosφ ) , but the motor is an active-inductive load, so to determine the total power consumption of the motor from the network, we use the formula:

S = P1/cosφ = 733.33/0.71 = 1032.85 (VA)

Let's find the rated current of the motor when connecting the windings in a star:

Inom = S/(1.73 U) = 1032.85/(1.73 380) = 1.57 (A)

Let's find the rated current of the motor when connecting the windings in a triangle:

Inom = S/(1.73 U) = 1032.85/(1.73 220) = 2.71 (A)

As you can see, the resulting values ​​are equal to the currents indicated on the motor tag.

To simplify, the above formulas can be combined into one general one. The result will be:

Inom = P2/(1.73 U cosφ η)

Therefore, in order to determine the rated current of the motor, it is necessary to substitute the mechanical power P2 taken from the tag into this formula, taking into account the efficiency and power factor (cosφ), which are indicated on the same tag or in the passport for the electric motor.

Let's recheck the formula.

Motor current when connecting the windings in a star:

Inom = P2/(1.73 U cosφ η) = 550/(1.73 380 0.71 0.75) = 1.57 (A)

Motor current when connecting the windings in a triangle:

Inom = P2/(1.73 U cosφ η) = 550/(1.73 220 0.71 0.75) = 2.71 (A)

I hope everything is clear.

Examples

I decided to give a few more examples with different types of engines and powers. Let's calculate their rated currents and compare them with the currents indicated on their labels.

As you can see, this motor can only be connected to a three-phase network with a voltage of 380 (V), because its windings are assembled into a star inside the motor, and only three ends are brought out into the terminal block, therefore:

Inom = P2/(1.73 U cosφ η) = 1500/(1.73 380 0.85 0.82) = 3.27 (A)

The resulting current of 3.27(A) corresponds to the rated current of 3.26(A) indicated on the tag.

This motor can be connected to a three-phase network with a voltage of both 380 (V) star and 220 (V) delta, because It has 6 ends connected to the terminal block:

Inom = P2/(1.73 U cosφ η) = 3000/(1.73 380 0.83 0.83) = 6.62 (A) - star

Inom = P2/(1.73 U cosφ η) = 3000/(1.73 220 0.83 0.83) = 11.44 (A) - triangle

The obtained current values ​​for different winding connection schemes correspond to the rated currents indicated on the label.

3. AIRS100A4 asynchronous motor with a power of 4.25 (kW)

Similar to the previous one.

Inom = P2/(1.73 U cosφ η) = 4250/(1.73 380 0.78 0.82) = 10.1 (A) - star

Inom = P2/(1.73 U cosφ η) = 4250/(1.73 220 0.78 0.82) = 17.45 (A) - triangle

The calculated current values ​​for different winding connection schemes correspond to the rated currents indicated on the motor nameplate.

This motor can only be connected to a three-phase network with a voltage of 6 (kV). The connection diagram of its windings is a star.

Inom = P2/(1.73 U cosφ η) = 630000/(1.73 6000 0.86 0.947) = 74.52 (A)

The rated current of 74.52(A) corresponds to the rated current of 74.5(A) listed on the tag.

Addition

The formulas presented above are of course good and according to them the calculation is more accurate, but among the common people there is a more simplified and approximate formula for calculating the rated current of the motor, which is most widespread among home craftsmen and craftsmen.

It's simple. Take the engine power in kilowatts indicated on the tag and multiply it by 2 - here is the finished result. Only this identity is relevant for 380 (V) engines assembled into a star. You can check and increase the power of the above engines. But personally, I insist that you use more accurate calculation methods.

P.S. But now, as we have already decided on the currents, we can begin to select a circuit breaker, fuses, thermal protection of the motor and contactors for its control. I will tell you about this in my next publications. In order not to miss the release of new articles, subscribe to the newsletter of the Electrician's Notes website. Until next time.

All electric motors are produced with plates on the housing, from which you can find out the main characteristics of the electric motor: its brand, rated operating current and power consumption, rotation speed, motor type, efficiency and cos(fi). This data is also indicated in the passport for the device.

Of all the parameters The most important factors for connection are: the power of the electric motor and the current consumed; this should not be confused with the starting current. It is these data that allow us to determine the sufficiency of power for the drive, the required cable cross-section to connect the motor, and select the appropriate circuit breaker and thermal relay for protection.

But it happens that there is no passport or plate and to determine these values ​​it will be necessary to make measurements. You will learn further in this article how to find out the power, operating current and reduce the starting current.

How to determine the power of an electric motor

The easiest way is to look at the plate and find the value in kilowatts. For example, in the picture it is 45 kW. Please note. that this value on the plate indicates the active power consumed from the electrical network. The total power will be equal to the sum of active and reactive power. Electric meters in a house or garage count only the consumption of active electricity, and reactive energy is recorded only in enterprises using special meters. The higher the cos(fi) of an electric motor, the lower the reactive energy component of total power will be. Do not confuse cos(fi) with efficiency. This indicator shows how much electricity is converted into useful mechanical work, and how much into useless heat. For example, an efficiency of 90 percent means that a tenth of the electricity consumed is spent on heat losses and friction in the bearings.

You should keep in mind. that the passport or plate indicates the rated power, which will be equal to this value only if the optimal load on the shaft is achieved. However, you should not overload the shaft for a number of reasons; it is better to choose a more powerful motor. At idle, the current will be much lower than the nominal value.

How to determine the rated power of an electric motor? On the Internet you will find many different formulas and calculations. For some, you need to measure the dimensions of the stator, for other formulas you will need to know the current value, efficiency and cos(fi). My advice is don't bother with all this. Practical measurements will still be better than these calculations. And you won’t need anything at all to carry them out.

How to determine the power of any electrical appliance in the house or garage? Of course, using an electricity meter. Before starting the measurement, unplug all electrical appliances from the sockets, lighting and everything connected to the electrical panel.

Further if you have an electronic meter like Mercury, everything is very simple, you just need to turn on the motor under load and drive for about 5 minutes. The electronic display should display the load value in kW currently connected to the meter.

If you have disk induction counter Keep in mind that he keeps records in kilowatt/hours. Write down the latest readings before starting measurements, turn on the engine strictly second per second for exactly 10 minutes, then after stopping, subtract the new readings from the previous ones and multiply kWh by 6. The result obtained will be the active power of this engine in Kilowatts; to convert to Watts, divide per 1000. I recommend reading the article: how to take electric meter readings.

If the engine is low power. then for higher accuracy you can count the disk revolutions. For example, in one minute it made 10 full revolutions, and the meter says 1200 revolutions = 1 kW/h. We multiply 10 by the number of minutes in an hour and get 600 revolutions per hour. Divide 1200 by 600 and we get 500 watts or 0.5 kW. The longer you measure, the more accurate the data will be. But the time must always be a multiple of a full minute. Then divide 60 by the number of minutes of measurement and multiply by the counted revolutions. After this, we divide the value of revolutions equal to one Kilowatt/hour for your electric meter model by the result obtained and obtain the required amount of power.

How to determine the current consumption of an electric motor

Knowing the power. You can easily calculate the amount of current consumed. For 3-phase motors connected in a star circuit at 380 Volts, it is necessary to multiply the power in kilowatts by 2. For example, with a power of 5 kilowatts, the current will be 10 Amperes. Again, keep in mind that the motor will take such current only under a load as close as possible to the nominal value. A semi-loaded electric motor, and even more so at idle, will consume significantly less current.

To determine the current in single-phase networks, it is necessary to divide the power by voltage. For example, when the engine is running, the voltage at its connection point is 230 Volts. This is important because after turning on the load, the voltage will most likely drop at the point where the electric motor is connected.

If, for example, the power of a 220 Volt motor was measured to be 1.5 kW or 1500 Watts. Divide 1500 by 230 Volts and we find that the operating current of the motor is approximately 6.5 Amperes.

Motor starting current

When starting any type of electric motor, a starting current occurs from 2 to 8 times the rated current in the operating mode of the electric motor. The magnitude of the starting current depends on the type of motor, rotation speed, connection diagram, presence of load on the shaft and other parameters.

The starting current occurs because at the moment of starting a very strong magnetic field is induced in the windings, which is necessary to move and spin the rotor. When the motor is turned on, the resistance of the windings is low, and therefore, according to Ohm's law, the current increases at a constant voltage in the circuit section. As the motor spins up, an EMF or inductive reactance appears in the windings and the current begins to decrease to the rated value.

These bursts of reactive energy negatively affect the operation of other electrical consumers connected to the same power supply line, which causes the occurrence of voltage surges or surges that are especially destructive for electronics.

Reduce starting current by half This is possible by using a thyristor unit specially designed for this purpose, or better yet, using a soft start device (SPD). The UPD with a lower starting current starts the motor one and a half times faster compared to thyristor starting.
Soft starters are suitable for both synchronous and asynchronous motors. UPZ are produced by enterprises in Ukraine and Russia.

To start a three-phase asynchronous motor Today frequency converters are also often used. Their widespread distribution is currently limited only by price. By changing the frequency of current and voltage, it is possible not only to make a smooth start, but also to regulate the speed of rotation of the rotor. There is no other way to regulate the rotation speed of an asynchronous motor by changing the frequency of the electric current. But you should know that the frequency converter creates interference in the electrical network, so use a surge protector to connect electronics and household appliances.

Using a soft start device and a frequency converter allows you not only to maintain the stability of the power supply for you and your neighbors connected to the same power supply line, but also to extend the service life of electric motors.

How do you find out the power if the current is no-load? Such experienced people will gather and drive away the blizzard. The engine is removed - there is no load on it. You turn it on and measure the idle current, but it is several times lower than the maximum - that is, the one that is written on the nameplate. And if you start loading, you will get anything, right up to the machine switching off, the wire burning out, or the engine burning out, it’s so easy for you - you measured the current with clamps and everything, like an electric stove or something, but I already wrote how to calculate the power of a three-phase current. Here's an example for you - a conveyor with an 18 kW engine has an idle current of 17 amperes, despite the fact that it turns the conveyor actually empty.

Yes, it’s true... They’ll recruit... First, let’s go over education. I have a specialty “Installation of electrical equipment of stations and substations”, a full course - 3 years of training in the specialty. Secondly, let’s be careful: I don’t say anywhere that you need to measure the current at idle, I’m talking about measuring it under load for what you plan to use the engine for. Thirdly, if you install capacitors according to the maximum current, which is indicated on the nameplate, then you will not get a circular field, you will get an oval field and the excess of this oval field will go into heating the engine. Fourthly, you won’t get any current. The engine is designed for a certain load and two options are possible: overload (but the engine does not stop, although it gets very hot) - here, in any case, whether you install capacitors or not, the varnish on the windings will burn and you will get an interturn short circuit and a load (not necessarily full ) - if the motor takes as much as it needs from a three-phase network, then with capacitors it must be given a very specific capacity, which is best selected according to the load, this way you can achieve a uniform circular electromagnetic field and reduce heating from incorrectly selected capacitors. My motors (2.2 kW) on the jointer operate from 60 uF working, on the circular saw there are two modes, if simple sawing is also 60 uF, and if I cut logs lengthwise I connect an additional 60 uF. So, during simple sawing, the engine practically does not heat up (I don’t take heating to operating temperature into account) and I can work on it all day without turning it off (just like on a jointer), but if I forget to turn off the additional 60 uF after half an hour I I “hear” the smell of the engine overheating, it is impossible to touch it with my hand. And let's take your example. In your case, it’s not completely idle, an empty conveyor is also a load, but if you judge by the maximum power, then you need to proceed from a current of 25-30 amperes per phase, and not 17. And the maximum power capacitors need 1200 uF, while for normal operation under given conditions (empty conveyor) requires only 370 (almost three times lower than the maximum. Moreover, unfortunately, I was not the one who wrote the article on connecting a three-phase motor to a single-phase network, and if I had written it, I would have indicated that the motor load with capacitor should not exceed 65-85% of the rated power indicated on the shield of the three-phase motor. And the formula for calculating capacitors looks like this: Cwork = X (Inom / U), where X is a number, depending on the connection diagram, Inom is the RATED CURRENT, not. the current indicated on the nameplate, and the current that flows at a given load. In a normal manual, it would look like this: start the engine with the planned load, measure the current in the network winding with a current clamp, substitute it into the formula and get the capacitance of the capacitor. And to be completely pedantic, cos F has not been canceled and it is also of great importance.

What single-phase network? I’m writing about three-phase 380V, Saratovets asks: “there is no sign on the engine. How to determine the power if it is known. that it was previously used in the drive of an industrial sewing machine at 380 V three-phase.” You write to measure the current with clamps and calculate how to calculate the power of a three-phase current, he already knows several posts there without you, only the efficiency needs to be taken into account. And your experiments with a circular electric field are the installation of a compensating capacitance.

Example: We have an Engine 4A 80846SU1 3ph 50Hz Star 3.6A 1.5kW 1400rpm Efficiency 77% cosPhi 0.83. And we calculate according to yours: 3 * 220 * 3.6 * 0.83 = 1972.08 W is probably a bit too much because we did not take into account the efficiency, multiply 0.77 and we get 1518.5 W - This looks more like the truth. The second formula is more accurate: 380*1.732*3.6*0.83*0.77=1514 W
But in fact, before measuring the current, you need to measure the effective voltage under load (with the motor connected) and then measure the current. (and then you get no-load current with the motor removed, and if you clamp the shaft, the maximum starting current lasts no more than 0.1 s) But without a nameplate you won’t know what the efficiency and cosine are. So, let’s determine using the barbaric method, divide the max start-up by 12 and get the maximum working)))

Well, for that matter, there is no point in using a barbaric method. It is known that at the moment of start-up the reactive load is practically zero, only the active one works, which means we measure the resistance and divide 220 volts by the resistance of one winding (if a triangle) or 380 by the winding resistance multiplied by two and get the starting current. In general, you are right, I looked at that post, maybe I was tired or something... I wrote the correct formulas, but didn’t think about the meaning of the question. In the way the question sounds, I don’t even know what to answer. Most likely, there is an option to get confused and try to calculate from the cross-section of the wire what is the optimal current for such a wire so that it does not melt, or rather the varnish on the wire does not melt and multiply by three, then multiply by the voltage of 220 volts and get an approximate value. Precisely approximate, because it is necessary to take into account the cosine and efficiency. In general, not a very meaningful idea.

Electricians often use the measuring method, i.e. measure the height of the axis of rotation and overall dimensions and the engine speed by eye, and then use the reference book to find the engine (if they can determine the type of engine by appearance).

Will the power consumption of the fan motor change depending on the air temperature? Let's say at -27°C the air density is 1.4 kg/m³, at 18 1.2 kg/m³. that is, the drop in mass of the measured air occurs 1.17 times. If our fan moves 20,000 m³/h, at -27°C it will be 28 tons/h and at +18° it will be 24 tons/h, will the engine power consumption change at the same time, and is there a graph of power consumption depending on the load on the shaft?

Alexey, hello. Yes, the power will change. The higher the air density, the heavier it is on the engine, the more it will consume. But as for the schedule, I can’t tell you anything. Either this needs to be done empirically and determine a schedule, or look for specialized literature.

Everything is clear - about currents, powers, etc. - I’m interested in something else: the power of a three-phase motor is 14 kW, judging by the conversations, the current consumption under a certain load will be equal to 28 amperes. How many amperes will flow through each phase? Divide 28 amperes into three phases and get 9.3 amperes? Or is this wrong?

Alexander, hello. The power of the electric motor consists of three phases. In order not to keep in mind the calculation formula using linear voltage and root, etc. for an approximate calculation, you can do it simpler, divide the power by three and by 220 and you get the current strength in one phase, therefore 4.7 kW per phase, and a current of 21 amperes per phase. This is at a voltage of 220 volts, at a voltage of 380 the current will be less.

And also - on the contactor it is written: 40A - 40 amperes - forty amperes for each contact, or is this the total current of all three contacts? If so, then again we divide forty amperes by three and get 13.3 amperes for each contact? Who will tell you the truth?

The most minimal losses are in the active component; the reactive component is always a loss. A boiler is purely active resistance (if it is electric on heating elements, and not on some clever devices containing the reactive part of electricity conversion. Just think where the highest efficiency is in analog devices (transformers) or digital (electronic). Electronics operate on constant voltage and current, plus semiconductor devices, which also do not have a reactive component, and as a result, low losses and high efficiency. Motors and generators will never approach (in the near future) the efficiency of electronic components. But in any case, any transformation is electronic. Whether it’s analog or not, there are ALWAYS losses. Somewhere there will be more, somewhere there will be losses. A simple example: you take a 50-liter flask, you understand that you won’t be able to carry it or you won’t be able to carry it in 200-liter bottles. ml (relatively) Now you pour the flask into bottles. No matter what cunning condition you come up with, you will still lose some of the moisture, which will simply evaporate while you pour the water. Then the same thing will happen when you pour the bottles into a large flask, some of the water will remain in small bottles. It would seem that a small part of the water, no more than 1-2%, BUT THERE WILL BE A LOSS OF THIS WATER. no matter what contraptions you come up with. And this is a simple example. More complex - a large thermos and small thermoses. Carrying one large thermos will result in less temperature loss than pouring boiling water into several thermoses and then draining it back. Here the losses will be 10-15%, etc. Draw your own conclusions.

Tell me how to determine the engine power. We can’t read the nameplate. 1966 single-phase with starting winding. Shaft 16 mm. The oscillation current is 1.8 amperes on the working winding. The current is 5 amperes on the starting winding. Engine without load. When switched on through the condenser in the 6-microstart winding, the engine starts and the current drops at the operating circuit to 1.3 amperes. I need at least a kilowatt for the unit, tell me who knows. Thank you.

Nikolay, hello. It's unlikely that anyone will help you. You can approximately calculate the cross-section of the wire. Well, or look for old documents and look out for the characteristics of your engine. According to modern reference books, it is very easy to make a mistake, because the dimensions can be two or three times larger than the dimensions of modern engines with the same power dimensions.

I noticed two errors in the article:
1) the electric motor plate indicates not the active electrical power, but the mechanical power on the shaft;
2) where “how to determine the current consumption of an electric motor”, multiplying the power by 2 we get the current for a triangle and not for a star (see photo of the plate)

Oleg, hello. Often articles are written by copywriters who, for one reason or another, cannot find a job, but have a computer and Internet access without problems. Consequently, as a rule, articles cannot be considered literate. There are many requirements for the text, and one of them is uniqueness, and in order to achieve this, you have to replace words with synonyms, so it turns out that there are many articles, but there is only one primary source and it may even be literate, but it is now difficult to find it on the Internet. This is why I exist here, to answer various misunderstandings and questions that arise from readers. And thank you for pointing out the mistakes. I take this into account when I refer people to certain articles.

Unfortunately, you are right. For ten years, my dissertation and scientific articles have been dismantled into quotations, and now by referring to my own robots, I risk becoming a plagiarist.

I think it's not all that scary???? It is not prohibited by law to refer to your own works, but selling an article can be problematic, since it must be, you see, “unique” and nothing else???? This means that you can either make your own website and not worry too much about uniqueness, the main thing is that there is definitely no plagiarism, or reinvent the wheel. But the Internet was intended to help people and find information :))))) But in reality, lately it has been a big headache. There are several worthwhile sites, and everything else is plagiarism in order to make money on affiliate links???? Such is the prose of life. But if we take into account that we used to live well without the Internet, essentially nothing will change if we use it in limited quantities. For example, I respond to comments on this site, sometimes I write articles for ordinary people, I go to several sites based on my interests and download movies and TV series (I rarely even go to social networks, and then only out of necessity to communicate with my family) and I don’t suffer at all?? ??

How to make an ammeter from a voltmeter. Maybe someone knows.

http://jelektro.ru

All electric motors have a plate on their housing indicating its electrical characteristics. We will talk about the main parameters of electric motors in this article.

Electric motor parameters: table

But sometimes the sign is missing or impossible to read. During operation, the engine is painted several times, often along with the nameplate. Therefore, it is necessary to determine its parameters by measurement.

Electric motor parameters No. 1: power

The rating data indicates the rated active power consumed from the network at the rated load on the shaft. To make measurements, you need to load the electric motor, testing it with a standard load (as part of the device it is intended to drive).

You can use an electric meter for measurements. To do this, you need to connect the electric motor as the only load on the meter for the time recorded by the stopwatch.

For convenience of calculations, the engine is connected for a time of 10 minutes. Before connection and after 10 minutes, readings are taken from the meter. The difference in readings in kWh, divided by 60/10 = 6, will be equal to the power of the electric motor in kilowatts.

Some electronic meters have the function of measuring instantaneous power, which simplifies the task. With the engine running, you need to go to the meter measurement menu and find the desired value in it.

Electric motor parameters No. 2: current consumption

To measure the current consumed by an electric motor, current clamp, measuring current in a circuit without breaking it.

Using multimeter() or ammeter you need to make sure in advance that the expected value of the measured parameter lies within the measurement range. The device is connected in series with an electric motor or with one of the windings of three phases. AND Don't forget about the starting current, before starting the device must be securely short-circuit so that it doesn't burn.

You can also use electronic counter with current measurement function.

If the power consumption is already known, the current can be calculated. For single phase motor:

For three phase:

If measurements are made without load, then it turns out no-load current. It is not possible to calculate the rated current, since the no-load current is not standardized and is 20-40% of the rated current. In this case, table data is used to calculate the no-load currents of three-phase asynchronous electric motors.

Electric motor parameters No. 3: type of winding connection

This is a very important parameter of a three-phase electric motor. All six terminals of the beginnings and ends of the windings are brought out into the engine bar. They can be connected either in a star or in a triangle.

Next to the symbols triangle/star The plate indicates the rated voltage – “220/380 V”. This means that when a three-phase current with a voltage of 380 V is connected to the network, the motor windings must be connected in a star. An error in the connection will result in motor failure.

The rated current is also indicated as a fraction. In the described case, the value specified in the denominator is required.

Motor starting current

At the moment of starting, the electric motor shaft is motionless. To unwind it, you need a force exceeding the nominal one. Therefore, the starting current exceeds the rated current. As the shaft unwinds, the current gradually decreases.

Inrush currents interfere with the operation of electrical equipment, causing sudden voltage dips. When starting powerful units, the starters of other electric motors may even disappear and the DRL lamps may go out.

To reduce the consequences of a launch, three methods are used.

1. Switching during acceleration of the electric motor circuit from star to triangle.
2. Use of electronic soft starters.
3. Usage frequency converters.

The operation of the motor is based on the principle of electromagnetic induction. The device consists of two parts. The fixed part is the stator for AC motors or the inductor for DC motors. The moving part is the rotor for AC motors or the armature for DC motors. Manufacturers produce motors of different technical characteristics and configurations, but the moving and fixed parts remain unchanged.

What is electric motor power

The power of an electric motor characterizes the rate of conversion of electrical energy; it is usually measured in watts. To understand how this works, we need two quantities: current and voltage. Current strength is the amount of current that passes through a cross section over a period of time; it is usually measured in amperes. Voltage is a quantity equal to the work done to move a charge between two points in a circuit; it is usually measured in volts.

In simple terms, current and voltage can be compared to water. Current strength is the speed at which water flows through pipes. The voltage can be seen in the example of two containers connected to each other by a tube. If you place one container higher than the other, water will flow out until the levels in both containers are equal. It is the difference in height that will be the tension. After you place the plug between the two containers, the flow of water (current) will stop, but the voltage will remain.

To calculate power, use the formula N = A/t, where:

N - power;

What about work;

Electric motor power calculation

Manufacturers indicate all technical parameters of electrical equipment. “Why do any calculations then?” you say. But the fact is that the declared power is not the actual power of the electric motor, but the maximum permissible power of the electric flow. So, if your appliance or tool says 1000 watts, for example, it's not what you think it is.

Three ways to determine the power of an electric motor

There are dozens of ways to calculate power. We will not talk about each of them, focusing only on the simplest and most accessible ones.

First way. Calculation using tables

For this calculation method you will need a ruler or caliper. Using them, measure the diameter of the shaft of your electric motor, the length of the motor (do not take into account the protruding parts of the shaft) and the distance to the axle. Using the obtained figures, you can determine the power of the electric motor from the tables of motor technical characteristics. Finding such tables is not difficult - they are publicly available on the Internet. Having opened the table, determine the series of the electric motor and, accordingly, its technical characteristics.

Second way. Calculation by meter

This method is considered the simplest; you do not need any additional equipment or calculations. Before you start measuring motor power, turn off all electrical appliances from the network. Turn on the electric motor under test and run it for 5-7 minutes. If your home has a modern meter, it will show the load in kilowatts.

Third way. Calculation by dimensions

For this method you will need a ruler or caliper. Measure the diameter of the core from the inside and the length (take into account the length of the ventilation holes). Determine the mains frequency and the synchronous shaft speed. Multiply the core diameter in centimeters by the synchronous shaft speed, multiply the resulting value by 3.14, divide by the network frequency multiplied by 120.

Sometimes you have to face the need to determine engine power in the absence of a tag. For example, the relevant documents were lost, and the inscriptions on the device itself are impossible to read (they often wear out over time).

Meter measurements

The simplest option is to check the readings of a household electricity meter. First turn off absolutely all equipment that operates from the network (including lights), since otherwise the results will not correspond to reality. Make sure the meter is not spinning or flashing. Then record the readings, then turn on the engine and let it run for ten minutes. After turning off the device, take the results again. The difference between the first and last readings must be multiplied by six. The resulting number will be the power of the electric motor.

Tables

If you carefully search for information on the Internet, you will probably be able to find tables from which you can find out the type of motor and its power. However, for this you may need a large number of parameters, which you often have to measure yourself. Among them: shaft diameter, dimensions of fasteners, rotation speed, motor length, distance to the axis, flange diameter (in the case of a flanged motor).

Calculation by parameters

If necessary, the power of the electric motor can be obtained using arithmetic calculations. Doing them using a calculator is not difficult for anyone. You will need three parameters:

• the radius of the shaft (denoted by the letter A), which can be measured using a caliper;
• the number of shaft revolutions per second (denoted by the letter B);
• indicator of the draft force of the motor (denoted by the letter C).

The power of the electric motor will be equal to the number obtained by the formula: A*6.28*B*C.

Engine power is one of its most important characteristics. Without knowing it, it is impossible to select a thermal relay and automatic circuit breaker with suitable parameters, or to determine the throughput and cross-section of suitable cables. Moreover, ignorance of the limit beyond which one cannot go during operation can lead to overloads and breakdowns.