Course work: Design of an asynchronous motor with a squirrel-cage rotor. Hello student h j - design height of the rotor yoke, m

FSBEI HPE "Yugorsky" state university»

Department of Energy

Karminskaya T.D., Kovalev V.Z., Bespalov A.V., Shcherbakov A.G.

ELECTRIC MACHINES

Tutorial

to perform course design on

discipline " Electrical machines»

for bachelors studying in

direction of training 03/13/02 “Electrical power engineering and electrical engineering”

Khanty-Mansiysk 2013

This tutorial describes the design methodology for an asynchronous motor with a squirrel-cage rotor, which is necessary to complete a course design assignment. During the course design, tasks such as choosing the main dimensions of the motor, calculating the parameters and magnetic system of the stator winding, calculating the parameters and magnetic system of the rotor winding, determining the parameters of the equivalent circuit and constructing the mechanical and operating characteristics of an asynchronous motor are solved.

The textbook is compiled in accordance with the work programs of the courses “Electrical machines” for students of the direction 13.03.02 “Electrical power engineering and electrical engineering”. It may be useful to students of other electrical and electromechanical fields and specialties, as well as specialists involved in research, design and operation of asynchronous machines for various purposes.

Introduction

Initial data for design

Options for design tasks

Chapter 1. Methodology for designing an asynchronous motor with a squirrel-cage rotor

1.1. Selection of main engine dimensions.

1.2. Calculation of stator winding parameters

1.3. Calculation of air gap parameters

1.4. Calculation of rotor winding parameters.

1.5. Magnetizing current calculation

1.6. Calculation of engine operating mode parameters

1.7. Calculation of active losses in the engine

1.8. Engine Performance Calculation

1.9. Calculation of starting characteristics.

Chapter 2. Use of computers for designing an asynchronous motor with a squirrel-cage rotor.

2.1. Description of the AD-KP program

2.2. An example of the application of the “AD-KP” program

Conclusion

APPLICATIONS

References

Introduction.

Asynchronous machine - brushless machine AC, in which the ratio of the rotor speed to the frequency of the current in the circuit to which the machine is connected depends on the loads. Like any electrical machine, an asynchronous machine has the property of reversibility, i.e. can operate in both motor and generator modes. However, in practice, the motor mode of operation of the machine is most widespread. Today, an asynchronous motor is the main engine of most mechanisms and machines. More than 60% of all generated electrical energy is consumed by electric machines, with asynchronous motors accounting for a significant share of this consumption (approximately 75%). Asynchronous motors have become quite widespread due to their following advantages: small overall dimensions, simplicity of design, high reliability, high efficiency, and relatively low cost. The disadvantages of an asynchronous motor include: difficulties in regulating the rotation speed, high starting currents, low power factor when the machine operates in a mode close to idle. The first and second of the shortcomings can be compensated by the use of frequency converters, the use of which has expanded the scope of application of asynchronous machines. Thanks to frequency converters, the asynchronous motor is being widely introduced into areas where other types of electrical machines, primarily DC machines, have traditionally been used.

Since existing asynchronous motors are characterized by a number of disadvantages, over time, new series of asynchronous motors are constantly being developed, which have higher technical and economic indicators compared to previous series of asynchronous motors, and better performance and mechanical characteristics in terms of quality. In addition, there is often a need for the development and modernization of special-purpose asynchronous motors. Such engines include:

submersible asynchronous motors (SEM) used to drive electric submersible pumps (ESP). A design feature of such engines is the limited size of the outer diameter, the dimensions of which are determined by the diameter of the pump-compressor pipe in which the engine is located. In addition, the engine is operated at fairly high temperatures, which leads to a decrease in its developed power. These circumstances require the development of a special design of asynchronous motors;

motors operating in conjunction with frequency converters that perform the functions of their regulation. Since frequency converters lead to the generation of a whole spectrum of harmonic components in the motor supply voltage curve, the presence of harmonic components leads to additional losses in the motor and a decrease in its efficiency below the rated one. The design of an asynchronous motor operating in conjunction with frequency converters should take this feature into account, and the presence of higher harmonics in the supply voltage curve should not lead to additional power losses.

The specified list of special-design asynchronous motors can be continued, and from here the following conclusions can be drawn:

there is a need to develop new series of asynchronous motors;

there is a need to master existing methods for designing asynchronous motors to solve the above problem;

There is a need to develop new methods for designing asynchronous motors, allowing, with less time spent on design, to develop a new series of asynchronous motors with better technical and economic indicators.

The purpose of completing the course design assignment is to develop an asynchronous motor with a squirrel-cage rotor having specified parameters, based on the existing and widely used in practice methodology for designing asynchronous motors.

Initial data for design.

The squirrel-cage induction motor being developed must have the following passport data:

Rated (phase) supply voltage U 1nf, V;

Mains supply frequency f 1, Hz;

Number of supply voltage phases m 1

Rated power P 2, kW;

Synchronous rotation speed n 1, rpm;

Nominal efficiency value η (not less), rel. units;

Nominal value of power factor cos(φ) (not less), rel. units;

Design;

Execution according to the method of protection from environmental influences;

During the course design, it is necessary to design an asynchronous motor with a squirrel-cage rotor having the specified passport data, and compare the main indicators of the resulting asynchronous motor with the indicators of a similar motor produced by industry (consider asynchronous motors of the AIR series, whose passport data are given in APPENDIX 1, as analogues)

Present the calculation results in the form of an explanatory note.

Draw a drawing of the developed asynchronous motor and present it in A1 format.

Note: this training manual for course design is made in the form of a workbook, which can serve as a model for preparing calculations in the form of an explanatory note. It also provides an example of calculating an asynchronous motor with a squirrel-cage rotor, having the following initial data:

n 1, rpm

no less

Cos(φ), p.u.

no less

Design – IM1001;

Execution according to the method of protection from environmental influences – IP44;

Options for design tasks.

Option number	Initial data for design
Option number		n 1, rpm	no less

For all job options, the following rating data of the designed engines have the same values:

Supply voltage (phase value) U 1ph, V – 220;

Supply voltage frequency f 1, Hz – 50;

Number of supply voltage phases m 1 – 3;

Design IM1001;

Execution according to the method of protection against environmental influences IP44;

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Introduction

A modern electric drive is a complex of devices and devices designed to control and regulate the physical and power parameters of an electric motor. The most common electric motor used in industry is the asynchronous motor. With the development of power electronics and the development of new powerful asynchronous motor control systems, electric drives based on asynchronous motors and frequency converters are best choice, to control various technological processes. An asynchronous electric drive has the best technical and economic indicators, and the development of new energy-saving motors makes it possible to create energy-efficient electric drive systems.

Asynchronous electric motor, an electrical asynchronous machine for converting electrical energy into mechanical energy. The operating principle of an asynchronous electric motor is based on the interaction of a rotating magnetic field that arises when a three-phase alternating current passes through the stator windings with the current induced by the stator field in the rotor windings. As a result, mechanical forces arise that force the rotor to rotate in the direction of rotation of the magnetic field, provided that the rotor rotation frequency n is less than the field rotation frequency n1. Thus, the rotor rotates asynchronously with respect to the field.

Purpose course work is the design of an asynchronous motor. By means of this design, we study the properties and characteristics of a given engine, and also study the features of these engines. This work is an integral part of the course of studying electric machines.

1. Motor magnetic circuit. Dimensions, configuration, material

1.1 Main dimensions

1. Height of the rotation axis of an asynchronous motor:

For Рн =75 kW, n1=750 rpm

h=280 mm, 2р=8.

2. Outer diameter of the core DН1 with a standard height of the axis of rotation h=280 mm. Under these conditions, DH1 = 520 mm.

3. To determine the internal diameter of the stator core D1, we use the dependence D1=f(DH1) given in Table 9-3. For DN1=520 mm;

D1=0.72 DН1 - 3;

D1=0.72 520-3 = 371.4 mm.

4. Let's find the average value of kН=f(P2) of asynchronous motors

For pH=75 kW; 2р=8;

5. For motors with squirrel-cage rotor protection IP44, preliminary values.

For pH=75 kW

6. For motors with squirrel-cage rotor protection IP44, we take the cos value according to Figure 9-3, and for 2p = 8

7. Design power P? for AC motors:

where is efficiency; cos - power factor at rated load;

8. Finding the linear load of stator winding A1

A1 =420 0.915 0.86=330.4 A/cm.

9. Finding the maximum value of magnetic induction in air gap B

B = 0.77 · 1.04 · 0.86 = 0.69 T.

10. To determine the length of the stator core, we will set the preliminary value of the winding coefficient kоь1, at 2р=8

11. Find the estimated length of the core l1

l1=366.7+125=426.7

12. The structural length of the stator core l1 is rounded to the nearest multiple of 5:

13. Coefficient

425 / 371,4 = 1,149

14. Find max R4=1.1

max = 1.46 - 0.00071 DN1;

max = 1.46 - 0.00071 520 = 1.091

max =1.091 1.1 = 1.2

1.2 Stator core

The core is assembled from separate stamped sheets of electrical steel 0.5 mm thick, with insulating coatings to reduce losses in the steel from eddy currents.

For steel 2312 we use insulating sheets with varnish.

Number of slots per pole and phase:

Based on the selected value q1, the number of stator core slots z1 is determined:

where m1 is the number of phases;

z1 = 8 3 3 = 72.

1.3 Rotor core

For a given height of the rotation axis, we select steel grade 2312.

The core is assembled from separate stamped sheets of electrical steel 0.5 mm thick.

For the core we use the same sheet insulation as for the stator - varnishing.

The filling factor of steel is taken to be equal to

We accept the size of the air gap between the stator and the rotor.

At h = 280 mm and 2р = 8;

Bevel of grooves ck (without bevel of grooves)

Outer diameter of the rotor core DН2:

DN2 = 371.4 - 2 0.8 = 369.8 mm.

For a rotation height h of 71 mm, the internal diameter of the rotor sheets D2:

D2 0.23 520 = 119.6 mm.

To improve cooling, reduce mass and dynamic moment of inertia of the rotor, round axial ventilation ducts are provided in the rotor cores with h250:

Rotor core length l2 at h>250 mm.

l2 = l1 + 5 = 425+5=430 mm.

Number of slots in the core for a motor with a squirrel-cage rotor at z1=72 and 2р=8

2. Stator winding

2.1 Parameters common to any winding

For our engine, we use a multi-section, two-layer concentric winding made of PETV wire (heat resistance class B), placed in rectangular semi-open slots.

Typically, the stator winding is made of six zones; each zone is equal to 60 electrical degrees. With a six-zone winding, the distribution coefficient kР1

kР1 = 0.5/(q1sin(b/20));

kР1 = 0.5/(3 sin(10)) = 0.95.

The shortening of step 1 is taken equal to

1 = 0.8, with 2p = 8.

We perform a two-layer winding with a shortened pitch yП1

yP1 = 1 z1 / 2p;

yP1 = 0.8 72 / 8 = 7.2.

Shortening factor ky1

ky1=sin(1 90)= sin(0.8 90)=0.95.

Winding coefficient kOB1

kOB1 = kР1 · ky1;

kOB1 = 0.95 · 0.95 = 0.9.

Preliminary value of magnetic flux F

Ф = В D1l1 10-6/p;

Ф = 0.689 371.4 42510-6/4 =0.027 Wb.

Preliminary number of turns in the phase winding?1

1 = knU1/(222 kOB1(f1/50) F);

1 = 0,96 380/(222 0,908 0.027) ?66.9.

We select the number of parallel branches of the stator winding a1 as one of the divisors of the number of poles a1 = 1.

Preliminary number of effective conductors in slot NP1

NП1 = 1а1(рq1);

NP1 = 155.3 1/(4 3) = 5.58

We accept the value of NP1 by rounding NP1 to the nearest integer value

Having chosen an integer, we specify the value 1

1 = NП1рq1а1;

1 = 4 4 3/1 = 72.

Magnetic flux value F

Ф = 0.023 66.5/64 = 0.028 Wb.

Induction value in air gap B

B = B? 1/ ? 1;

B = 0.8 66.9/72 = 0.689 T.

Preliminary value of rated phase current I1

I1 = Рн 103/(3U1cos);

I1 = 75,103/(3,380 0.93 0.84) = 84.216 A.

А1 = 10Nп1z1I1(D1a1);

A1 = 6 13 72 84.216/(3.14 371.4) = 311.8 A/cm.

Average value of magnetic induction in the back of stator BC1

At h = 280 mm, 2р = 8

BC1 = 1.5 T.

Tooth division along the inner diameter of the stator t1

t1 = p 371.4/72 =16.1 mm.

2.2 Stator winding with rectangular semi-closed slots

We accept the preliminary value of magnetic induction at the narrowest point of the stator tooth

31max = 1.8 T.

Tooth division of the stator at the narrowest point

Preliminary tooth width at narrowest point

Preliminary width of half-open and open groove in die

Slot width of half-open groove

Permissible width of effective conductor with turn insulation

b?eff =()/=3.665mm;

Number of effective conductors by slot height

Preliminary stator back height

Ф 106?(2 kc l1 Вc1);

0.027 106 ? (2 0.95 425 1.5) = 22.3 mm.

Preliminary groove height

= [ (D H1- D1)/ 2]- h c1;

= =[(520-371.4)/2]-22.3 =53 mm.

Permissible height of an effective conductor with turn insulation

Effective conductor area

Preliminary number of elementary conductors

Number of elementary conductors in one effective

Preliminary number of elementary conductors in one effective

Increase to 4

Dimensions of an elementary elementary conductor along the height of the groove

The final number of elementary conductors

Smaller and larger sizes of bare wire

Size according to groove height

Size according to the width of the groove in the stamp

Groove height

= [ (D H1- D1)/ 2]- h c1;

= =[(520-371.4)/2]-18.3 =56 mm.

Refined tooth width at the narrowest part

Refined magnetic induction in the narrowest part of the stator tooth

Current density in stator winding J1

J1 = I1(c S a1);

J1 = 84.216/(45.465 1) = 3.852 A/mm2.

А1J1 = 311·3.852 = 1197.9 А2/(cm mm2).

(A1J1)add = 2200·0.75·0.87=1435.5 A2/(cm mm2).

lв1 = (0.19+0.1p)bcp1 + 10;

lв1 = (0.19+0.1 3) 80.64+10= 79.4 mm.

Average tooth division of the stator tCP1

tCP1 = (D1 + hП1)/z1;

tCP1 = p(371.4 + 56)/72 = 18.6 mm.

Average width of stator winding coil bCP1

bCP1 = tCP1 UP1;

bCP1 = 18.6 7.2 = 133.6 mm.

Average length of the frontal part of the winding ll1

lл1 = 1.3=279.6 mm

Average winding length lcp1

lcp1 = 2 · (l1 + lл1) = 2 · (425 + 279.6) = 1409.2 mm.

Extension length of the frontal part of the winding lв1

3. Squirrel-cage rotor winding

asynchronous magnetic stator phase

Let's use a rotor winding with bottle slots, because h = 280 mm.

Groove height from fig. 9-12 is equal to hp2 = 40 mm.

Estimated height of the rotor back hc2 at 2р=8 and h = 280 mm

hc2 = 0.38 · Dн2 - hp2 - ?dk2;

hc2 = 0.38 · 369.8 - 40 - ? 40 = 73.8 mm.

Magnetic induction in the back of the Vs2 rotor

Вс2 = Ф · 106 / (2 · kc · l2 · hc2);

Vs2 = 0.028 106 / (2 0.95 430 73.8) = 0.464 T.

Tooth division along the outer diameter of the rotor t2

t2 = рDн2/z2 = р · 369.8/86 = 13.4 mm.

Magnetic induction in the teeth of the rotor Вз2.

Vz2 = 1.9 T.

Literature

1. Goldberg O.D., Gurin Y.S., Sviridenko I.S. Design of electrical machines. - M.: graduate School, 1984. - 431 p.

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MINISTRY OF EDUCATION AND SCIENCE

REPUBLIC OF KAZAKHSTAN

North Kazakhstan State University named after. M. Kozybaeva

Faculty of Energy and Mechanical Engineering

Department of Energy and Instrument Engineering

COURSE WORK

On the topic: “Design of an asynchronous motor with a squirrel-cage rotor”

discipline – “Electrical machines”

Completed by Kalantyrev

Scientific supervisor

Doctor of Technical Sciences, Prof. N.V. Shatkovskaya

Petropavlovsk 2010

Introduction

1. Selection of main sizes

2. Determination of the number of stator slots, turns in the winding phase, cross-section of the stator winding wire

3. Calculation of the dimensions of the stator tooth zone and air gap

4. Rotor calculation

5. Calculation of the magnetic circuit

6. Operating parameters

7. Calculation of losses

8. Calculation of performance characteristics

9. Thermal calculation

10. Calculation of performance characteristics according to pie chart

Appendix A

Conclusion

References

Introduction

Asynchronous motors are the main converters of electrical energy into mechanical energy and form the basis of the electric drive of most mechanisms. The 4A series covers a power rating range from 0.06 to 400 kW and has 17 axis heights from 50 to 355 mm.

In this course project the following engine is considered:

Execution according to the degree of protection: IP23;

Cooling method: IC0141.

Design according to installation method: IM1081 – according to the first digit – motor on feet, with bearing shields; by the second and third digits - from horizontal arrangement shaft and lower location of the paws; according to the fourth digit - with one cylindrical shaft end.

Climatic operating conditions: U3 – by letter – for moderate climates; by number - for placement in enclosed spaces with natural ventilation without artificially controlled climatic conditions, where fluctuations in temperature and air humidity, exposure to sand and dust, solar radiation are significantly less than in the open air of stone, concrete, wood and other unheated rooms.

1. Selection of main sizes

1.1 Determine the number of pole pairs:

(1.1)

Then the number of poles

1.2 Determine the height of the rotation axis graphically: according to Figure 9.18, b

, in accordance with, according to table 9.8, we determine the outer diameter corresponding to the axis of rotation.

1.3 Stator inner diameter

, we calculate using the formula: , (1.2) – coefficient determined according to table 9.9. lies in the interval: .

Let's choose a value

, Then

1.4 Define pole division

: (1.3)

1.5 Determine the design power

, W: , (1.4) – power on the motor shaft, W; – the ratio of the EMF of the stator winding to the rated voltage, which can be approximately determined from Figure 9.20. When and , .

Approximate values

and take it from the curves constructed from the data of the 4A series engines. Figure 9.21, c. At kW and , , a

1.6 Electromagnetic loads A and B d are determined graphically using the curves in Figure 9.23, b. At

kW and , , T.

1.7 Winding coefficient

. For two-layer windings at 2p>2, = 0.91–0.92 should be taken. Let's accept.

1.8 Let us determine the synchronous angular speed of the motor shaft W:

, (1.5) – synchronous rotation speed.

1.9 Calculate the length of the air gap

:
, (1.6) – field shape coefficient. .

1.10 The criterion for the correct choice of main dimensions D and

serves as the ratio, which must be within acceptable limits, Figure 9.25, b. . The l value lies within the recommended limits, which means the main dimensions are determined correctly.

2. Determination of the number of stator slots, turns in the winding phase and cross-section of the stator winding wire

2.1 Let’s determine the limit values: t 1 max and t 1 min Figure 9.26. At

And , , .

2.2 Number of stator slots:

, (2.1) (2.2)

Finally, the number of slots must be a multiple of the number of slots per pole and phase: q. Let's accept

, Then
, (2.3)

where m is the number of phases.

2.3 We finally determine the tooth division of the stator:

(2.4)

2.4 Preliminary stator winding current

(2.5)

2.5 Number of effective conductors in the slot (subject to

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Electrical machines

Course project

“Design of an asynchronous motor with a squirrel-cage rotor”

Technical specifications

Design an asynchronous three-phase motor with a squirrel cage rotor:

P = 15 kW, U = 220/380 V, 2р = 2;

n = 3000 rpm, = 90%, cos = 0.89, S NOM = 3%;

h=160 M p / M n = 1.8, M max / M n = 2.7, I p / I n = 7;

design IM1001;

IP44 protection;

cooling method IC0141;

climatic version and placement category U3;

insulation heat resistance class F.

operating mode S1

Determination of basic geometric dimensions

1. First select the height of the rotation axis according to Fig. 8.17, and (hereinafter all formulas, tables and figures from) h = 150 mm.

From the table 8.6 we take the nearest smaller value h = 132 mm and a = 0.225 m (D a is the outer diameter of the stator).

2. Determine the internal diameter of the stator:

D=K D D a =0.560.225=0.126 (m)

K D - proportionality coefficient, determined according to table. 8.7.

3. Pole division

m

where 2p is the number of pole pairs.

4. Determine the estimated power:

P = (P 2 k E)/(cos)

k E - the ratio of the EMF of the stator winding to the rated voltage, determined from Fig. 8.20, k E = 0.983

- Efficiency of an asynchronous motor, according to Fig. 8.21,a, = 0.89, cos = 0.91

P 2 - power on the motor shaft, W

P = (1510 3 0.983) / (0.890.91) = 18206 (W)

5. Determine electromagnetic loads (preliminarily) according to Fig. 8.22, b:

Linear load (ratio of the current of all turns of the winding to the circumference) A = 25.310 3 (A/m)

Induction in the air gap B= 0.73 (T)

6. We select the preliminary winding coefficient depending on the type of stator winding. For single-layer windings k O1 = 0.95 0.96.

Let's take k O1 = 0.96.

7. The estimated length of the air gap is determined by the formula:

= P / (k V D 2 k O 1 AB)

k B is the field shape coefficient, preliminarily taken equal to

kV = / () = 1.11

- synchronous angular speed of the motor shaft, rad/s, calculated by the formula

rad/s

where 1 is the supply frequency, Hz

= 18206 / (1.110.126 2 3140.9625.310 3 0.73) = 0.19 (m)

8. Check the relation = / . It should be within the range of 0.19–0.87, determined from Fig. 8.25:

= 0,19 / 0,198 = 0,96

The obtained value is higher than the recommended limits, therefore we accept the next largest from the standard series (Table 8.6) height of the axis of rotation h = 160 mm. We repeat the calculations according to paragraphs. 1-8:

D a = 0.272 (m) P = (1510 3 0.984) / (0.910.89) = 18224 (W)

D = 0.560.272 = 0.152 (m) A = 3410 3 (A/m)

= (3,140,152) / 2 = 0.239 (m) B = 0.738 (T)

= 18224 / (1.110.152 2 3140.963610 3 0.738) = 0.091 (m)

= 0,091 / 0,239 = 0,38

Calculation of the winding, slots and stator yoke

Definition Z 1 , 1 And sections wires windings stator

1. We determine the limiting values of tooth division 1 according to Fig. 6-15:

1 max = 18 (mm) 1 min = 13 (mm)

2. The limit values for the number of stator slots are determined by the following formulas

We accept 1 = 36, then q = Z 1 / (2pm), where m is the number of phases

q = 36 / (23) = 6

The winding is single-layer.

3. We finally determine the tooth division of the stator:

m = 1410 -3 m

4. Find the number of effective conductors in the groove (preliminarily, provided that there are no parallel branches in the winding (a = 1)):

u =

I 1H is the rated current of the stator winding, A, and is determined by the formula:

I 1H = P 2 / (mU 1H cos) = 1510 3 / (32200.890.91) = 28.06(A)

u= = 16

5. We accept a=2, then

u= au = 216 = 32

6. We get the final values:

number of turns in a winding phase

linear load

Vehicle

flow

Ф = (1) -1

k O1 - the final value of the winding coefficient, determined by the formula:

k О1 = k У k Р

k У - shortening coefficient, for a single-layer winding k У = 1

k P - distribution coefficient, determined from table. 3.16 for the first harmonic

k P = 0.957

Ф = = 0.01 (Wb)

air gap induction

Tl

Values A and B are within acceptable limits (Fig. 8.22b)

7. Current density in the stator winding (preliminary):

J 1 = (AJ 1)/ A= (18110 9)/ (33.810 3)= 5.3610 6 (A/m 2)

the product of the linear load and the current density is determined from Fig. 8.27, b.

Effective conductor cross-section (preliminary):

q EF = I 1 H / (aJ 1) = 28.06 / (25.1310 6) = 2.7310 -6 (m 2) = 2.73 (mm 2)

We accept n EL = 2, then

q EL = q EF / 2 = 2.73 / 2 = 1.365 (mm 2)

n EL - number of elementary conductors

q EL - section of an elementary conductor

We select the PETV winding wire (according to Table A3.1) with the following data:

nominal diameter of bare wire d EL = 1.32 mm

average diameter of insulated wire d IZ = 1.384 mm

cross-sectional area of bare wire q EL = 1.118 mm 2

cross-sectional area of the effective conductor q EF = 1.1182 = 2.236 (mm 2)

9. Current density in the stator winding (final)

Calculation sizes jagged zones stator And air gap

Groove stator - according to Fig. 1, a with a size ratio that ensures parallelism of the side edges of the teeth.

1. We accept in advance according to the table. 8.10:

the induction value in the stator teeth B Z1 = 1.9 (T) the induction value in the stator yoke B a = 1.6 (T), then the tooth width

b Z1 =

k C - coefficient of core filling with steel, according to table. 8.11 for oxidized steel sheets grade 2013 k C = 0.97

СТ1 - length of steel stator cores, for machines with 1.5 mm

ST1 = 0.091 (m)

b Z1 = = 6.410 -3 (m) = 6.4 (mm)

stator yoke height

2. The dimensions of the groove in the stamp are:

groove slot width b W = 4.0 (mm)

groove slot height h W = 1.0 (mm), = 45

groove height

h P = h a = =23.8 (mm) (25)

width of the bottom of the groove

b 2 = = = 14.5 (mm) (26)

width of the top of the groove

b 1 = = = 10.4 (mm) (27)

h 1 = h P - + = = 19.6 (mm) (28)

3. Clearance dimensions of the groove, taking into account assembly allowances:

for h = 160 250 (mm) b P = 0.2 (mm); h P = 0.2 (mm)

b 2 = b 2 - b P = 14.5 - 0.2 = 14.3 (mm) (29)

b 1 = b 1 - b P = 10.4 - 0.2 = 10.2 (mm) (30)

h 1 = h 1 - h P = 19.6 - 0.2 = 19.4 (mm) (31)

Cross-sectional area of the groove for placing conductors:

S P = S FROM S PR

cross-sectional area of gaskets S PR = 0

cross-sectional area of body insulation in the groove

S FROM = b FROM (2h P +b 1 +b 2)

b IZ - one-sided insulation thickness in the groove, according to table. 3.1 b IZ = 0.4 (mm)

S FROM = 0.4(223.8+14.5+10.4) = 29 (mm 2)

S P = 0.5(14.3+10.2)19.4 29 = 208.65 (mm 2)

4. Groove filling factor:

k Z = [(d IZ) 2 u n n EL ] / S P = (1.405 2 402)/ 208.65 = 0.757 (34)

The obtained value of k3 for mechanized winding installation is excessively high. The fill factor should be within the range of 0.70 - 0.72 (from Table 3-12). Let's reduce the fill factor by increasing the cross-sectional area of the groove.

Let's take B Z1 = 1.94 (T) and B a = 1.64 (T), which is acceptable, since these values exceed the recommended ones by only 2.5 - 3%.

5. Repeat the calculation according to paragraphs. 1-4.

b Z1 = = 0.0063(m)= 6.3(mm) b 2 = = 11.55 (mm)

h a = = 0.0353 (m) = 35.3 (mm) b 1 = = 8.46 (mm)

h P = = 24.7 (mm) h 1 = = 20.25 (mm)

b 2 = = 11.75 (mm)

b 1 = = 8.66 (mm)

h 1 = = 20.45 (mm)

S FROM = = 29.9 (mm 2)

S P = = 172.7 (mm 2)

k З = = 0.7088 0.71

The dimensions of the groove in the die are shown in Fig. 1, a.

Calculation of windings, slots and rotor yoke

1. Determine the air gap (according to Fig. 8.31): = 0.8 (mm)

2. Number of rotor slots (according to Table 8.16): Z 2 = 28

3. External diameter:

D 2 = D2 = 0.15220.810 -3 = 0.150 (m) (35)

4. Length of rotor magnetic circuit 2 = 1 = 0.091 (m)

5. Tooth division:

t 2 = (D 2)/ Z 2 = (3,140,150)/ 28 = 0.0168 (m) = 16.8 (mm) (36)

6. The internal diameter of the rotor is equal to the diameter of the shaft, since the core is directly mounted on the shaft:

D J = D B = k B D a = 0.230.272 = 0.0626 (m) 60 (mm) (37)

The value of the coefficient k B is taken from the table. 8.17: kV = 0.23

7. Preliminary current value in the rotor rod:

I 2 = k i I 1 i

k i is a coefficient that takes into account the influence of the magnetizing current and winding resistance on the I 1 / I 2 ratio. k i = 0.2+0.8cos = 0.93

i - current reduction coefficient:

i = (2m 1 1 k O 1) / Z 2 = (23960.957) / 28 = 19.7

I 2 = 0.9328.0619.7 = 514.1 (A)

8. Cross-sectional area of the rod:

q C = I 2 / J 2

J 2 - current density in the rotor rods; when filling the grooves with aluminum, it is selected within

J 2 = (2.53.5)10 6 (A/m 2)

q C = 514.1 / (3.510 6) = 146.910 -6 (m 2) = 146.9 (mm 2)

9. Rotor groove - as shown in Fig. 1. b. We design pear-shaped closed grooves with slot dimensions b W = 1.5 mm and h W = 0.7 mm. We choose the height of the jumper above the groove equal to h W = 1 mm.

Permissible tooth width

b Z2 = = = 7.010 -3 (m) = 7.0 (mm) (41)

B Z2 - induction in the rotor teeth, according to table. 8.10 V Z2 = 1.8 (T)

Groove dimensions

b 1 ===10.5 (mm)

b 2 = = = 5.54 (mm) (43)

h 1 = (b 1 - b 2)(Z 2 / (2)) = (10.5 - 5.54)(28/6.28) = 22.11 (mm) (44)

We take b 1 = 10.5 mm, b 2 = 5.5 mm, h 1 = 22.11 mm.

10. We specify the width of the rotor teeth

b Z2 = = 9.1 (mm)

b Z2 = = 3.14 9.1 (mm)

b Z2 = b Z2 9.1 (mm)

Full groove height:

h P 2 = h Ш + h Ш +0.5b 1 +h 1 +0.5b 2 = 1+0.7+0.510.5+22.11+0.55.5 = 31.81 (mm)

Rod cross section:

q C = (/8)(b 1 b 1 +b 2 b 2)+0.5(b 1 +b 2)h 1 =

(3.14/8)(10.5 2 +5.5 2)+0.5(10.5+5.5)22.11 = 195.2 (mm 2)

11. Current density in the rod:

J 2 = I 2 / q C = 514.1 / 195.210 -6 = 3.4910 6 (A/m 2)

12. Short circuit rings. Cross-sectional area:

qKL = IKL / JKL

JKL - current density in the closing rings:

JCL = 0.85J2 = 0.853.49106 = 2.97106 (A/m2) (51)

ICL - current in rings:

ICL = I2 /

= 2sin = 2sin = 0.224 (53)

ICL = 514.1 / 0.224 = 2295.1 (A)

qKL = 2295 / 2.97106 = 772.710-6 (m2) = 772.7 (mm2)

13. Dimensions of closing rings:

hKL = 1.25hP2 = 1.2531.8 = 38.2 (mm) (54)

bKL = qKL / hKL = 772.7 / 38.2 = 20.2 (mm) (55)

qKL = bKLhKL = 38.2 20.2 = 771.6 (mm2) (56)

DK. CP = D2 - hKL = 150 - 38.2 = 111.8 (mm) (57)

Magnetic circuit calculation

Steel magnetic core 2013; sheet thickness 0.5 mm.

1. Air gap magnetic voltage:

F= 1.5910 6 Bk, where (58)

k- air gap coefficient:

k= t 1 /(t 1 -)

= = = 2,5

k= = 1.17

F= 1.5910 6 0.7231.170.810 -3 = 893.25 (A)

2. Magnetic voltage of tooth zones:

stator

F Z1 = 2h Z1 H Z1

h Z1 - estimated height of the stator tooth, h Z1 = h P1 = 24.7 (mm)

H Z1 - value of the field strength in the stator teeth, according to table P1.7 at B Z1 = 1.94 (T) for steel 2013 H Z1 = 2430 (A/m)

F Z1 = 224.710 -3 2430 = 120 (A)

calculated induction in teeth:

B Z1 = = = 1.934 (T)

since B Z1 is 1.8 (T), it is necessary to take into account the flow branch into the groove and find the actual induction in the tooth B Z1.

Coefficient k PH height h ZX = 0.5h Z:

k PH =

b PH = 0.5(b 1 + b 2) = 0.5(8.66+11.75) = 12.6

k PH = = 2.06

B Z1 = B Z1 - 0 H Z1 k PH

We accept B Z1 = 1.94 (T), check the ratio of B Z1 and B Z1:

1,94 = 1,934 - 1,25610 -6 24302,06 = 1,93

rotor

F Z2 = 2h Z2 H Z2

h Z2 - design height of the rotor tooth:

h Z2 = h P2 - 0.1b 2 = 31.8 - 0.15.5 = 31.25 (mm)

H Z2 - the value of the field strength in the rotor teeth, according to table P1.7 at B Z2 = 1.8 (T) for steel 2013 H Z2 = 1520 (A/m)

F Z2 = 231.25 10 -3 1520 = 81.02 (A)

tooth induction

B Z2 = = = 1.799 (T) 1.8 (T)

3. Saturation coefficient of the tooth zone

k Z = 1+= 1+= 1.23

4. Yoke magnetic voltage:

stator

F a = L a H a

L a - length of the average magnetic line of the stator yoke, m:

L a = = = 0.376 (m)

H a - field strength, according to table P1.6 at B a = 1.64 (T) H a = 902 (A/m)

F a = 0.376902 = 339.2 (A)

B a =

h a - design height of the stator yoke, m:

h a = 0.5(D a - D) - h P 1 = 0.5(272 - 152) - 24.7 = 35.3 (mm)

B a = = 1.6407 (T) 1.64 (T)

rotor

F j = L j H j

L j is the length of the average magnetic flux line in the rotor yoke:

Lj = 2hj

h j - height of the rotor back:

h j = - h P2 = - 31.8 = 13.7 (mm)

L j = 213.7 10 -3 = 0.027 (m)

B j =

h j - design height of the rotor yoke, m:

h j = = = 40.5 (mm)

B j = = 1.28 (T)

H j - field strength, according to table P1.6 at B j = 1.28 (T) H j = 307 (A/m)

F j = 0.027307 = 8.29 (A)

5. Total magnetic voltage of the magnetic circuit per pair of poles:

F C = F + F Z1 + F Z2 + F a + F j = 893.25 + 120 + 81.02 + 339.2 + 8.29 = 1441.83 (A)

6. Magnetic circuit saturation coefficient:

k = F C / F = 1441.83/893.25 = 1.6

7. Magnetizing current:

I = = = 7.3 (A)

relative value

I = I / I 1H = 7.3 / 28.06 = 0.26

Calculation of parameters of an asynchronous machine for nominal mode

1. Active resistance of the stator winding phase:

r 1 = 115

115 - specific resistance of the winding material at the design temperature, Ohm. For insulation heat resistance class F, the design temperature is 115 degrees. For copper 115 = 10 -6 /41 Ohm.

L 1 - total length of effective conductors of the stator winding phase, m:

L 1 = CP1 1

CP1 - average length of stator winding turn, m:

CP1 = 2(P1 + L1)

P1 - length of the groove part, P1 = 1 = 0.091 (m)

L1 - frontal part of the coil

L1 = K L b CT +2V

K L - coefficient, the value of which is taken from table 8.21: K L = 1.2

B is the length of the straight part of the coil extending from the groove from the end of the core to the beginning of the bend of the frontal part, m. We take B = 0.01.

b CT - average coil width, m:

b CT = 1

1 - relative shortening of the stator winding pitch, 1 = 1

b CT = = 0.277 (m)

L1 = 1.20.277+20.01 = 0.352 (m)

CP1 = 2(0.091+0.352) = 0.882 (m)

L 1 = 0.88296 = 84.67 (m)

r 1 = = 0.308 (Ohm)

Extension length of the front part of the coil

OUT = K OUT b CT +B = 0.260.277+0.01= 0.08202 (m)= 82.02 (mm) (90)

According to table 8.21 K OUT = 0.26

Relative value

r 1 = r 1 = 0.308 = 0.05

2. Active resistance of the rotor winding phase:

r 2 = r C +

r C - rod resistance:

r C = 115

for cast aluminum rotor winding 115 = 10 -6 / 20.5 (Ohm).

r C = = 22.210 -6 (Ohm)

r CL - resistance of the section of the closing ring enclosed between two adjacent rods

r CL = 115 = = 1.0110 -6 (Ohm) (94)

r 2 = 22.210 -6 + = 47.110 -6 (Ohm)

We reduce r 2 to the number of turns of the stator winding:

r 2 = r 2 = 47.110 -6 = 0.170 (Ohm) (95)

Relative value:

r 2 = r 2 = 0.170 = 0.02168 0.022

3. Inductive resistance of the stator winding phase:

x 1 = 15.8(P1 + L1 + D1), where (96)

P1 - coefficient of magnetic conductivity of slot scattering:

P1 =

h 2 = h 1 - 2b IZ = 20.45 - 20.4 = 19.65 (mm)

b 1 = 8.66 (mm)

h K = 0.5(b 1 - b) = 0.5(8.66 - 4) = 2.33 (mm)

h 1 = 0 (conductors are secured with a groove cover)

k = 1 ; k = 1 ; = = 0.091 (m)

P1 = = 1.4

L1 - coefficient of magnetic conductivity of frontal scattering:

L1 = 0.34(L1 - 0.64) = 0.34(0.352 - 0.640.239) = 3.8

D1 - coefficient of magnetic conductivity of differential scattering

D1 =

= 2k SK k - k O1 2 (1+ SK 2)

k = 1

SK = 0, since there is no bevel of the grooves

k SC is determined from the curves in Fig. 8.51,d depending on t 2 /t 1 and SC

= = 1.34 ; SK = 0; k SC = 1.4

= 21,41 - 0,957 2 1,34 2 = 1,15

D1 = 1.15 = 1.43

x 1 = 15.8(1.4+3.8+1.43) = 0.731 (Ohm)

Relative value

x 1 = x 1 = 0.731 = 0.093

4. Inductive reactance of the rotor winding phase:

x 2 = 7.9 1 (P2 + L2 + D2 + SK)10 -6 (102)

P2 = k D +

h 0 = h 1 +0.4b 2 = 17.5+0.45.5 = 19.7 (mm)

k D = 1

P2 = = 3.08

L2 = = = 1.4

D2 =

= = = 1,004

since with closed slots Z 0

D2 = = 1.5

x 2 = 7.9500.091(3.08+1.4+1.5)10 -6 = 21510 -6 (Ohm)

We reduce x 2 to the number of stator turns:

x 2 = x 2 = = 0.778 (Ohm)

Relative value

x 2 = x 2 = 0.778 = 0.099 (108)

Power loss calculation

1. Main losses in steel:

P ST. OSN. = P 1.0/50 (k Yes B a 2 m a +k DZ B Z1 2 +m Z1)

P 1.0/50 - specific losses at an induction of 1 T and a magnetization reversal frequency of 50 Hz. According to the table 8.26 for steel 2013 P 1.0/50 = 2.5 (W/kg)

m a - mass of stator yoke steel, kg:

m a = (D a - h a)h a k C1 C =

= 3.14(0.272 - 0.0353)0.03530.0910.977.810 3 = 17.67 (kg)

C - specific gravity of steel; in calculations we take C = 7.810 3 (kg/m 3)

m Z1 - mass of stator teeth steel, kg:

m Z1 = h Z1 b Z1 SR. Z 1 CT 1 k C 1 C =

= 24.710 -3 6.310 -3 360.0910.977.810 3 = 3.14 (kg) (111)

k Yes and k ДZ are coefficients that take into account the effect on losses in steel of uneven flux distribution across sections of sections of the magnetic core and technological factors. Approximately we can take k Da = 1.6 and k DZ = 1.8.

PCT. OSN. = 2.51(1.61.64217.67+1.81.93423.14) = 242.9 (W)

2. Surface losses in the rotor:

PPOV2 = pPOV2(t2 - bSH2)Z2ST2

pSOV2 - specific surface losses:

pPOV2 = 0.5k02(B02t1103)2

B02 - amplitude of induction pulsation in the air gap above the crowns of the rotor teeth:

B02=02

02 depends on the ratio of the slot width of the stator slots to the air gap. 02 (at bШ1/ = 4/0.5 = 8 according to Fig. 8.53,b) = 0.375

k02 is a coefficient that takes into account the effect of surface treatment of the heads of the rotor teeth on specific losses. Let's take k02 =1.5

B02 = 0.3571.180.739 = 0.331 (T)

pPOV2 = 0.51.5(0.33114)2 = 568 (16.8 - 1.5)24 0.091 = 22.2 (W)

3. Pulsation losses in the rotor teeth:

PPUL2 = 0.11mZ2

BPUL2 - amplitude of induction pulsations in the middle section of the teeth:

BPUL2 = BZ2

mZ2 - mass of steel rotor teeth, kg:

mZ2 = Z2hZ2bZ2СТ2kC2C =

= 2826.6510-39.110-30.0910.977.8103 = 3.59 (kg) (117)

BPUL2 = = 0.103 (T)

PPUL2 = 0.11= 33.9 (W)

4. Amount of additional losses in steel:

PCT. ADD. = PPOV1+PPUL1+PPOV2+PPUL2 = 22.2 + 33.9 = 56.1 (W

5. Total losses in steel:

PCT. = PST. OSN. + PST. ADD. = 242.9 + 56.1 = 299 (W

6. Mechanical losses:

PMECH = KTDa4 = 0.2724 = 492.6 (W) (120)

For engines with 2р=2 KT =1.

7. Engine idling:

IX. X.

IХ.Х.а. =

PE1 H.H. = mI2r1 = 37.320.308 = 27.4 (W)

IХ.Х.а. = = 1.24 (A)

IX.H.R. I = 7.3 (A)

IX.X. = = 7.405 (A)

cos xx = IX.X.a / IX.X. = 1.24/4.98 = 0.25

asynchronous three-phase motor squirrel cage rotor

Performance calculation

1. Parameters:

r 12 = P ST. OSN. /(mI 2) = 242.9/(37.3 2) = 3.48 (Ohm)

x 12 = U 1H /I - x 1 = 220/7.3 - 1.09 = 44.55 (Ohm)

c 1 = 1+x 1 / x 12 = 1+0.731/44.55 = 1.024 (Ohm)

= = =

= arctan 0.0067 = 0.38 (23) 1 o

Active component of the synchronous no-load current:

I 0a = (P ST. BASIC +3I 2 r 1) / (3U 1H) = = 0.41 (A)

a = c 1 2 = 1.024 2 = 1.048

b = 0

a = c 1 r 1 = 1.0240.308 = 0.402 (Ohm)

b = c 1 (x 1 +c 1 x 2) = 1.024(0.731+1.0241.12) = 2.51 (Ohm)

Losses that do not change when slip changes:

P ST. +P FUR = 299+492.6 = 791.6 (W)

Calculation formulas	Dimension	Slip S




Z = (R 2 +X 2) 0.5



I 1a = I 0a +I 2 cos 2
I 1p = I 0p +I 2 sin 2
I 1 = (I 1a 2 +I 1p 2) 0.5

P 1 = 3U 1 I 1a 10 -3
P E 1 = 3I 1 2 r 1 10 -3
P E 2 = 3I 2 2 r 2 10 -3
P ADD = 0.005P 1
P=P ST +P MECH +P E1 +P E2 +P ADD

Table 1. Induction motor performance characteristics

P2NOM = 15 kW; I0p = I = 7.3 A; PCT. +PMECH. = 791.6 W

U1NOM = 220/380 V; r1 =0.308 Ohm; r2 = 0.170 Ohm

2р=2 ; I0a = 0.41 A; c1 = 1.024; a = 1.048; b = 0 ;

a = 0.402 (Ohm); b = 2.51 (Ohm)

2. Calculate performance characteristics for sliding

S = 0.005;0.01;0.015

0.02;0.025;0.03;0.035, preliminary assuming that SNOM r2 = 0.03

The calculation results are summarized in table. 1. After constructing the performance characteristics (Fig. 2), we clarify the value of the nominal slip: SН = 0.034.

Rated data of the designed motor:

P2NOM = 15 kW cos NOM = 0.891

U1NOM = 220/380 V NOM = 0.858

I1NOM =28.5 A

Calculation of starting characteristics

Calculation currents With taking into account influence changes parameters under influence effect repression current (without accounting influence nasy tion from fields scattering)

Detailed the calculation is given for S = 1. The calculation data for the remaining points are summarized in table. 2.

1. Active resistance of the rotor winding taking into account the influence of the current displacement effect:

= 2h C = 63.61h C = 63.610.0255= 1.62 (130)

calc = 115 o C; 115 = 10 -6 /20.5 (Ohm); b C /b P =1; 1 = 50 Hz

h C = h P - (h W + h W) = 27.2 - (0.7+1) = 25.5 (mm)

- “reduced height” of the rod

according to fig. 8.57 for = 1.62 we find = 0.43

h r = = = 0.0178 (m)= 17.8 (mm)

since (0.510.5) 17.8 (17.5+0.510.5):

q r =

h r - depth of current penetration into the rod

q r - cross-sectional area limited by height h r

b r = = 6.91 (mm)

q r = = 152.5 (mm 2)

k r = q C /q r = 195.2 / 152.5 = 1.28 (135)

K R = = 1.13

r C = r C = 22.210 -6 (Ohm)

r 2 = 47.110 -6 (Ohm)

Reduced rotor resistance taking into account the influence of current displacement effect:

r 2 = K R r 2 = 1.130.235 = 0.265 (Ohm)

2. Inductive reactance of the rotor winding taking into account the influence of the current displacement effect:

for = 1.62 = kD = 0.86

KX = (P2 +L2 +D2)/(P2 +L2 +D2)

P2 = P2 - P2

P2 = P2(1- kD) = =

= = 0,13

P2 = 3.08 - 0.13 = 2.95

KX = = 0.98

x2 = KXx2 = 0.980.778 = 0.762 (Ohm)

3. Starting parameters:

Mutual induction reactance

x 12P = k x 12 = 1.644.55 = 80.19 (Ohm) (142)

with 1P = 1+x 1 / x 12P = 1+1.1/80.19 = 1.013 (143)

4. Calculation of currents taking into account the influence of the current displacement effect:

R P = r 1 +c 1 P r 2 /s = 0.308+1.0130.265 = 0.661 (Ohm)

Calculation formulas	Dimension	Slip S

63.61h C S 0.5



K R =1+(r C /r 2)(k r - 1)





R P = r 1 +c 1 P r 2 /s
X P = x 1 +c 1P x 2
I 2 = U 1 / (R P 2 +X P 2) 0.5
I 1 = I 2 (R P 2 + +(X P +x 12 P) 2) 0.5 /(c 1 P x 12 P)

Table 2. Calculation of currents in the starting mode of an asynchronous motor with a squirrel-cage rotor, taking into account the influence of the current displacement effect

P2NOM = 15 kW; U1 = 220/380 V; 2р=2 ; I1NOM = 28.5 A;

r2 = 0.170 Ohm; x12P = 80.19 Ohm; s1P = 1.013; SNOM = 0.034

XП = x1 + s1Пх2 = 0.731+1.0130.762 = 1.5 (Ohm)

I2 = U1 / (RP2+HP2)0.5= 220/(0.6612+1.52)0.5= 137.9 (A)

I1 = I2 (RP2+(HP+x12P)2)0.5/ (c1Px12P)=

=137.9(0.6612+(1.5+80.19)2)0.5/(1.01380.19)= 140.8 (A)

Calculation launchers characteristics With taking into account influence effect repression current And saturation from fields scattering

Calculation carry out for characteristic points corresponding to S=1; 0.8; 0.5;

0.2; 0.1, in this case we use the values of currents and resistances for the same slips, taking into account the influence of current displacement.

The calculation data are summarized in table. 3. Detailed calculation is given for S=1.

1. Inductive resistance of the windings. We accept k US =1.35:

Average MMF of the winding, related to one slot of the stator winding:

F P. SR. = = = 3916.4 (A)

C N = = 1.043

Fictitious leakage flux induction in the air gap:

B Ф =(F P. SR. /(1.6С N))10 -6 =(3916.410 -6)/(1.60.810 -3 1.043)=5.27(T)

for B Ф = 5.27 (T) we find k = 0.47

Magnetic conductivity coefficient of slot leakage of the stator winding taking into account the influence of saturation:

сЭ1 = (t1 - bШ1)(1 - к) = (14 - 4)(1 - 0.47) = 6.36

P1 US. =((hШ1 +0.58hK)/bШ1)(сЭ1/(сЭ1+1,5bШ1))

hK = (b1 - bШ1)/2 = (10.5 - 4)/2 = 3.25 (153)

P1 US. =

P1 US. = P1 - P1 US. = 1.4 - 0.37 = 1.03

Magnetic conductivity coefficient of differential leakage of the stator winding taking into account the influence of saturation:

D1 US. = D1k = 1.430.47 = 0.672

Inductive resistance of the stator winding phase taking into account the influence of saturation:

x1 US. = (x11 US)/ 1 = = 0.607 (Ohm)

Magnetic conductivity coefficient of slot leakage of the rotor winding taking into account the influence of saturation and current displacement:

P2. US. = (hШ2/bШ2)/(cЭ2/(сЭ2+bШ2))

sE2 = (t2 - bSh2)(1 - k) = (16.8 - 1.5)(1 - 0.47) =10.6

hШ2 = hШ +hШ = 1+0.7 = 1.7 (mm)

P2. US. =

P2. US. = P2 - P2. US. = 2.95 - 0.99 = 1.96

Magnetic conductivity coefficient of rotor differential leakage taking into account the influence of saturation:

D2. US. = D2k = 1.50.47 = 0.705

Reduced inductive reactance of the rotor winding phase taking into account the influence of current displacement and saturation effects:

x2 US = (x22 US)/ 2 = = 0.529 (Ohm)

s1P. US. = 1+ (x1 NAS./x12 P) = 1+(0.85/80.19) = 1.011

Calculation formulas	Dimension	Slip S



BФ =(FP.SR.10-6) / (1.6CN)

сЭ1 = (t1 - bШ1)(1 - к)
P1 US. = P1 - P1 US.
D1 US. = to D1
x1 US. = x11 US. / 1
c1P. US. = 1+x1 US. / x12P
сЭ2 = (t2 - bШ2)(1 - к)
P2 US. = P2 - P2 US.
D2 US. = to D2
x2 US. = x22 US. /2
RP. US. = r1+c1П. US. r2/s
XP.US=x1US.+s1P.US.x2US
I2US=U1/(RP.US2+HP.US2)0.5
I1 US=I2 US (RP.NAS2+(HP.NAS+ x12P) 2) 0.5/(c1P.NASx12P)
kUS. = I1 US. /I1
I1 = I1 US. /I1 NOM
M = (I2US/I2NOM)2КR(sHOM/s)

Table 3. Calculation of the starting characteristics of an asynchronous motor with a squirrel-cage rotor, taking into account the effect of current displacement and saturation from stray fields

P2NOM = 15 kW; U1 = 220/380 V; 2р=2 ; I1NOM = 28.06 A;

I2NOM = 27.9 A; x1 = 0.731 Ohm; x2 = 0.778 Ohm; r1 = 0.308 Ohm;

r2 = 0.170 Ohm; x12P = 80.19 Ohm; СN = 1.043; SNOM = 0.034

2. Calculation of currents and moments

RP. US. = r1+c1П. US. r2/s = 0.393+1.0110.265 = 0.661 (Ohm) (165)

XP.NAS.=x1NAS.+s1P.NAS.x2NAS. = 1.385 (Ohm) (166)

I2NAS.=U1/(RP.NAS2+HP.NAS2)0.5= 220/(0.6612+1.3852)0.5= 187.6 (A)

I1 US. = I2NAS.= = 190.8 (A) (168)

IP = = 6.8

M = = = 1.75

kUS. = I1 US. /I1 = 190.8/140.8 = 1.355

kUS. differs from the accepted one by US. = 1.35 by less than 3%.

To calculate other characteristic points, we set kNAS. , reduced depending on the current I1. We accept when:

s = 0.8 kUS. = 1.3

s = 0.5 kUS. = 1.2

s = 0.2 kUS. = 1.1

s = 0.1 kUS. = 1.05

The calculation data are summarized in table. 3, and the starting characteristics are presented in Fig. 3.

3. The critical slip is determined after calculating all points of the starting characteristics (Table 3) using the average resistance values x1 of the NAS. and x2 US. , corresponding to slips s = 0.2 0.1:

sKR = r2 / (x1 NAS / c1P NAS + x2 NAS) = 0.265/(1.085/1.0135+1.225) = 0.12

The designed asynchronous motor meets the requirements of GOST both in terms of energy indicators (and cos) and starting characteristics.

Thermal calculation

1. The temperature of the inner surface of the stator core exceeds the air temperature inside the engine:

pov1 =

PE. P1 - electrical losses in the slot part of the stator winding

PE. P1= kPE1= = 221.5 (W)

PE1 = 1026 W (from table 1 at s = sNOM)

k = 1.07 (for windings with insulation class F)

K = 0.22 (according to table 8.33)

1 - heat transfer coefficient from the surface; 1 = 152 (W/m 2 C)

pov1 =

2. Temperature difference in the insulation of the slot part of the stator winding:

from. n1 =

P P1 = 2h PC +b 1 +b 2 = 220.45+8.66+11.75 = 66.2 (mm) = 0.0662 (m)

EKV - average equivalent thermal conductivity of groove insulation, for heat resistance class F EKV = 0.16 W/(mS)

EKV - average value of the thermal conductivity coefficient, according to Fig. 8.72 at

d/d IZ = 1.32/1.405 = 0.94 EKV = 1.3 W/(m 2 C)

from. n1 = = 3.87 (C)

3. Temperature difference across the thickness of the insulation of the frontal parts:

from. l1=

PE. L1 - el. losses in the frontal part of the stator winding

PE. L1 = kPE1= = 876 (W)

PL1 = PP1 = 0.0662 (m)

bIZ. L1 MAX =0.05

from. l1= = 1.02 (C)

4. The temperature of the outer surface of the frontal parts exceeds the air temperature inside the engine:

pov l1 = = 16.19 (C)

5. Average temperature rise of the stator winding over the air temperature inside the engine

1 = =

= = 24.7 (C)

6. The air temperature inside the engine exceeds the ambient temperature

B =

P B - the sum of losses released into the air inside the engine:

P B = P - (1 - K)(P E. P1 +P ST. BASIC) - 0.9P MEC

P is the sum of all losses in the engine at rated mode:

P = P +(k - 1)(PE1+PE2) = 2255+(1.07 - 1)(1026+550) = 2365 (W)

PV = 2365 - (1 - 0.22)(221.5+242.9) - 0.9492.6 = 1559 (W)

SCOR - equivalent cooling surface of the housing:

SCOR = (Da+8PR)(+2OUT1)

PR - conditional perimeter of the cross section of the ribs of the engine housing, for h = 160 mm PR = 0.32.

B is the average value of the air heating coefficient, according to Fig. 8.70, b

B = 20 W/m2C.

SCOR = (3.140.272+80.32)(0.091+282.0210-3) = 0.96 (m2)

B = 1559/(0.9620) = 73.6 (C)

7. Average temperature rise of the stator winding over the ambient temperature:

1 = 1 +B = 24.7+73.6 = 98.3 (C)

8. Checking engine cooling conditions:

Air flow required for cooling

B =

km = = 9.43

For engines with 2р=2 m= 3.3

B = = 0.27 (m3/s)

Air flow provided by outdoor fan

B = = 0.36 (m3/s)

The heating of engine parts is within acceptable limits.

The fan provides required consumption air.

Conclusion

The designed engine meets those supplied in terms of reference requirements.

List of used literature

1. I.P. Kopylov “Design of electrical machines” M.: “Energoatomizdat”, 1993. Part 1,2.

2. I.P. Kopylov “Design of electrical machines” M.: “Energy”, 1980.

3. A.I. Woldek “Electric machines” L.: “Energy”, 1978.

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Course work: Design of an asynchronous motor with a squirrel-cage rotor. Hello student h j - design height of the rotor yoke, m

Chapter 1. Methodology for designing an asynchronous motor with a squirrel-cage rotor

Chapter 2. Use of computers for designing an asynchronous motor with a squirrel-cage rotor.

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Electrical machines

Course project

“Design of an asynchronous motor with a squirrel-cage rotor”

Technical specifications

Design an asynchronous three-phase motor with a squirrel cage rotor:

P = 15 kW, U = 220/380 V, 2р = 2;

n = 3000 rpm, = 90%, cos = 0.89, S NOM = 3%;

h=160 M p / M n = 1.8, M max / M n = 2.7, I p / I n = 7;

design IM1001;

IP44 protection;

cooling method IC0141;

climatic version and placement category U3;

insulation heat resistance class F.

operating mode S1

Determination of basic geometric dimensions

1. First select the height of the rotation axis according to Fig. 8.17, and (hereinafter all formulas, tables and figures from) h = 150 mm.

From the table 8.6 we take the nearest smaller value h = 132 mm and a = 0.225 m (D a is the outer diameter of the stator).

2. Determine the internal diameter of the stator:

D=K D D a =0.560.225=0.126 (m)

K D - proportionality coefficient, determined according to table. 8.7.

3. Pole division

m

where 2p is the number of pole pairs.

4. Determine the estimated power:

P = (P 2 k E)/(cos)

k E - the ratio of the EMF of the stator winding to the rated voltage, determined from Fig. 8.20, k E = 0.983

- Efficiency of an asynchronous motor, according to Fig. 8.21,a, = 0.89, cos = 0.91

P 2 - power on the motor shaft, W

P = (1510 3 0.983) / (0.890.91) = 18206 (W)

5. Determine electromagnetic loads (preliminarily) according to Fig. 8.22, b:

Linear load (ratio of the current of all turns of the winding to the circumference) A = 25.310 3 (A/m)

Induction in the air gap B= 0.73 (T)

6. We select the preliminary winding coefficient depending on the type of stator winding. For single-layer windings k O1 = 0.95 0.96.

Let's take k O1 = 0.96.

7. The estimated length of the air gap is determined by the formula:

= P / (k V D 2 k O 1 AB)

k B is the field shape coefficient, preliminarily taken equal to

kV = / () = 1.11

- synchronous angular speed of the motor shaft, rad/s, calculated by the formula

rad/s

where 1 is the supply frequency, Hz

= 18206 / (1.110.126 2 3140.9625.310 3 0.73) = 0.19 (m)

8. Check the relation = / . It should be within the range of 0.19–0.87, determined from Fig. 8.25:

= 0,19 / 0,198 = 0,96

The obtained value is higher than the recommended limits, therefore we accept the next largest from the standard series (Table 8.6) height of the axis of rotation h = 160 mm. We repeat the calculations according to paragraphs. 1-8:

D a = 0.272 (m) P = (1510 3 0.984) / (0.910.89) = 18224 (W)

D = 0.560.272 = 0.152 (m) A = 3410 3 (A/m)

= (3,140,152) / 2 = 0.239 (m) B = 0.738 (T)

= 18224 / (1.110.152 2 3140.963610 3 0.738) = 0.091 (m)

= 0,091 / 0,239 = 0,38

Calculation of the winding, slots and stator yoke

Definition Z 1 , 1 And sections wires windings stator

1. We determine the limiting values ​​of tooth division 1 according to Fig. 6-15:

1 max = 18 (mm) 1 min = 13 (mm)

2. The limit values ​​for the number of stator slots are determined by the following formulas

We accept 1 = 36, then q = Z 1 / (2pm), where m is the number of phases

q = 36 / (23) = 6

The winding is single-layer.

3. We finally determine the tooth division of the stator:

m = 1410 -3 m

4. Find the number of effective conductors in the groove (preliminarily, provided that there are no parallel branches in the winding (a = 1)):

u =

I 1H is the rated current of the stator winding, A, and is determined by the formula:

I 1H = P 2 / (mU 1H cos) = 1510 3 / (32200.890.91) = 28.06(A)

u= = 16

5. We accept a=2, then

u= au = 216 = 32

6. We get the final values:

number of turns in a winding phase

linear load

Vehicle

1. We determine the limiting values of tooth division 1 according to Fig. 6-15:

2. The limit values for the number of stator slots are determined by the following formulas

Values A and B are within acceptable limits (Fig. 8.22b)

cross-sectional area of bare wire q EL = 1.118 mm 2

cross-sectional area of the effective conductor q EF = 1.1182 = 2.236 (mm 2)

Cross-sectional area of the groove for placing conductors:

cross-sectional area of gaskets S PR = 0

cross-sectional area of body insulation in the groove

The obtained value of k3 for mechanized winding installation is excessively high. The fill factor should be within the range of 0.70 - 0.72 (from Table 3-12). Let's reduce the fill factor by increasing the cross-sectional area of the groove.

Let's take B Z1 = 1.94 (T) and B a = 1.64 (T), which is acceptable, since these values exceed the recommended ones by only 2.5 - 3%.