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Initial data
U n = 220 V - rated voltage
2 p = 4 - four-pole motor
R n = 55 kW - rated power
n n = 550 rpm - nominal rotation frequency
I n = 282 A - rated armature current
r I + r dp = 0.0356 Ohm - resistance of the armature winding and additional poles
N = 234 - number of active armature conductors
2a = 2 - number of parallel armature branches
Ф n = 47.5 mVb - nominal magnetic flux of the pole
k = pN / 2a = 2 * 234/2 = 234 - design factor of the engine
kFn = E / u = (Un.-In. (Rya. + Rd.p.)) / u = 3.65 (Wb.)
u n = 2pn n / 60 = 57.57 (rad / s.)
SCH(I)
u = 0, I = 6179.78 (A.)
I = 0, n = 60.27 (rad / s.)
SCH(M)
u (M) = Uн - M (Rя. + Rd.) / (kFn)
u = 0, M = 22 (kN / m)
M = 0, n = 60.27 (rad / s.)
2. Determine the value of the additional resistance that must be introduced into the armature circuit to reduce the speed to u = 0.4 nat rated armature current of the motorI= I n... Build an electromechanical characteristic at which the engine will operate at a reduced speed
Scheme of rheostat regulation of an independent excitation motor:
n = 0.4 n = 23.03 (rad / s)
u = (Un. - In (Rya. + Rd.p. + Rd)) / kFn
kFn * u = Un. - In (Rя. + Rd.p. + Rd)
In (Rya. + Rd.p. + Rd) = Un - kFn * u
Rd = (Un - kFn * n) / In - (Rа. + Rd.p) = (220-84.06) / 282-0.0356 = 0.4465 (Ohm) - additional resistance
Plotting an electromechanical characteristic - SCH(I)
u (I) = (Un. - I (Rya. + Rd.p. + Rd)) / kFn
u = 0, I = 456.43 (A)
I = 0, n = 60.27 (rad / s.)
motor armature brake electromechanical
3. Determine the additional braking resistance that limits the armature current to two times the nominal value I=2 In during the transition from the nominal mode to the generator mode:
a) braking by opposition
From the formula: u (I) = (E - I R) / kFn we find Rtotal:
Rtotal = (wn. (KF) n. - (-Un.)) / - 2In = (57.57 * 3.65 + 220) / (2 * 282) = 0.7626 (Ohm.)
Rd = Rtot - (Rp. + Rd.p) = 0.727 (Ohm)
We take, in the calculations, resistance modulo.
Plotting an electromechanical characteristic - SCH(I)
u (I) = (E - I R) / kFn
u = 0, I = -288.5 (A.)
I = 0, n = -60.27 (rad / s.)
Plotting a mechanical characteristic - SCH(M)
u (M) = E - M * R / (kF)
u = 0, M = -1.05 (kN / m)
M = 0, n = -60.27 (rad / s.)
b) dynamic braking
Since during dynamic braking the anchor chains of the machine are disconnected from the network, the voltage in the expression should be equated to zero U n, then the equation will take the form:
M = - I n F = -13.4 N / m
u = M * Rtot / (kFn) 2
Rtot = wn * (kFn) 2 / M = 57.57 * 3.65 2 / 13.4 = 57.24 (Ohm)
Rd = Rtot - (Ry. + Rd.p) = 57.2 (Ohm)
Plotting an electromechanical characteristic - SCH(I)
u (I) = (E - I R) / kFn
u = 0, I = -3.8 (A.)
I = 0, n = 60.27 (rad / s.)
Plotting a mechanical characteristic - SCH(M)
u (M) = E - M * R / (kFn)
u = 0, M = -14.03 (kN / m)
M = 0, n = 60.27 (rad / s.)
F = 0.8Fn = 0.8 * 47.5 = 38 (mVb)
kF = 2.92 (Wb.)
Plotting an electromechanical characteristic - SCH(I)
u (I) = (Uн. - I (Rа. + Rd.)) / kФ
u = 0, I = 6179.78 (A.)
I = 0, u = 75.34 (rad / s.)
Plotting a mechanical characteristic - SCH(M)
u (M) = Uн - M (Rа. + Rd.) / kФ
u = 0, M = 18 (kN / m)
M = 0, n = 75.34 (rad / s.)
Plotting an electromechanical characteristic - SCH(I)
u (I) = (U. - I (Rya. + Rd.)) / kFn
u = 0, I = 1853.93 (A.)
I = 0, u = 18.08 (rad / s.)
Plotting a mechanical characteristic - SCH(M)
u (M) = U - M (Rn. + Rd.) / (kFn)
u = 0, M = 6.77 (kN / m)
M = 0, n = 18.08 (rad / s.)
6. Determine the engine speed during regenerative lowering of the load, if the engine torque isM = 1.5Mn
M = 1.5Mn = 1.5 * 13.4 = 20.1 (N / m)
u (M) = Uн - M (Rя. + Rd.p.) / (kFn) = 60 (rad / s)
n = 60 * n / (2 * p) = 574 (rpm)
Connection diagram for starting resistors
The values of the switching currents I 1 and I 2 are selected based on the technology requirements for the electric drive and the switching capacity of the motor.
l = I 1 / I 2 = R 1 / (Rя + Rdp) = 2 - the ratio of switching currents
R 1 = l * (Rя + Rdp) = 0.0712 (Ohm)
r 1 = R 1 - (Rя + Rdp) = 0.0356 (Ohm)
R 2 = R 1 * l = 0.1424 (Ohm)
r 2 = R 2 - R 1 = 0.1068 (Ohm)
R 3 = R 2 * l = 0.2848 (Ohm)
r 3 = R 3 - R 2 = 0.178 (Ohm)
Building a start-up diagram
u (I) = (Un. - I (Rya. + Rd.)) / kFn
u 0 = 0, I 1 (R 3) = 772.47 (A)
u 1 (I 1) = (Un. - I 1 R 2) / kFn = 30.14 (rad / s)
u 2 (I 1) = (Un. - I 1 R 1) / kFn = 45.21 (rad / s)
u 3 (I 1) = (Un. - I 1 (Rя + Rdp)) / kFn = 52.72 (rad / s)
I = 0, n = 60.27 (rad / s.)
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Department: "Electrical Equipment for Ships and Power Engineering"
Course work
on the topic:
"Calculation of the electric drive of the lifting mechanism"
Kaliningrad 2004
Initial data for calculations ……………………………………………
Construction of a simplified load diagram of the mechanism
Construction of a simplified engine load diagram ………….
2.3 Calculation of the static power on the motor shaft ……………………… ...
2.4 Construction of a simplified load diagram of the engine ………… ..
2.5 Calculation of the required engine power according to the simplified load
diagram ……………………………………………………………… ...
3. Construction of mechanical and electromechanical characteristics …… ..
3.1 Calculation and construction of mechanical characteristics …………………… ...
3.2 Calculation and construction of electromechanical characteristics …………… ..
4. Construction of a load diagram ……………………………………… ..
4.1 Lifting the rated load …………………………………………… ..
4.2 Braking release of load ………………………………………………… ...
4.3 Lifting the idle hook ………………………………………………… ..
4.4 Power release of the power hook ……………………………………………
5. Checking the selected engine to ensure the specified
winch performance …………………………………………… ...
6. Checking the selected motor for heating …………………………………
7. Power circuit of a frequency converter with a voltage inverter …… ..
8. List of used literature ………………………………………… ..
Initial data for calculations
Continuation of table 1 grabbing device G x.g, kg freight drum D, m diagrams t i, s Continuation of table 1 Continuation of table 1 |
||||||||||||||||||||||||||||
| Landing speed υ` s, m / s | Name executive mechanism | System management | Current type |
|||||||||||||||||||||||||
| | Asynchronous engine | Converter frequency with voltage inverter | Network alternating current 380V |
|||||||||||||||||||||||||
Table -1- Initial data for calculations
2. Construction of a simplified load diagram of the mechanism
and preselection of engine power
2.1 Construction of a simplified motor load diagram
The duration of inclusion is calculated by the formula:
(1)
where
(2)
Engine running time when lifting a load:
Engine running time when lowering the load:
(5)
Engine running time when lifting the idle hook:
(6)
Engine running time when lowering the idle hook:
Here, the idle hook lowering speed is equal to the idle hook lifting speed
Cumulative time of engine on state:
Determine the duration of the engine on
2.2 Calculation of the static power at the output shaft of the mechanism.
Static power on the output shaft when lifting a load:
(8)
Static power on the output shaft when lowering the load:
Static power on the output shaft when landing a load:
(10)
Static power on the output shaft when lifting the idle hook:
(11)
Static power on the output shaft when lowering the idle hook:
2.3 Calculation of the static power on the motor shaft.
Static power on the motor shaft when lifting a load:
(13)
Static power on the motor shaft when lowering the load:
(14)
Static power on the motor shaft when landing a load:
Static power on the motor shaft when lifting the idle hook:
Here η x.z = 0.2
Static power on the engine shaft when lowering the idle hook:
2.4 Construction of a simplified motor load diagram. 
Figure 1 - Simplified motor load diagram
2.5 Calculation of the required motor power according to a simplified load diagram
WITH 
the average square power is calculated by the formula:
(18)
where β i is a coefficient that takes into account the deterioration of heat transfer and is calculated for all working sections by the formula:
(19)
Here β 0 is a coefficient that takes into account the deterioration of heat transfer with a stationary rotor
For motors of open and protected design β 0 = 0.25 ÷ 0.35
For motors with closed blown design β 0 = 0.3 ÷ 0.55
For engines closed without blowing β 0 = 0.7 ÷ 0.78
For motors with forced ventilation β 0 = 1
We accept β 0 = 0.4 and υ nom = m / s
When lifting a load:
(20)
When lowering the load to one meter:
(21)
When landing the cargo:
(22)
When lifting the idle hook:
(23)
When lowering the idle hook:
(24)
Table 2 - Summary table of data for calculating the root mean square
power
| Plot | P with | t p, s | υ, m / s | υ n | β |
| 1 | |||||
| 2 | |||||
| 2 landing | |||||
| 3 | |||||
| 4 |
Let's write an expression for calculating the rms power of the engine:
=
We find the rated power of the engine by the formula: 
(26)
where k s = 1.2 - safety factor
PV nom = 40% - nominal duration of inclusion
According to the reference book, we select a brand engine that has the following characteristics:
Rated power P n = kW
Nominal slip s n =%
Rotation speed n = rpm
Nominal stator current I nom = A
Nominal efficiency η n =%
Rated power factor cosφ n = 
Moment of inertia J = kg m 2
Number of pole pairs p =
3. Construction of mechanical and electromechanical characteristics.
3.1 Calculation and construction of mechanical characteristics.
Rated angular rotation speed:
(26)
N
(27)
nominal moment:
Determine the critical slip for the motor mode:
where
overload capacity λ =
(29)
The critical moment of rotation is found from expression 29:
According to the Kloss equation, we find M dv:
(31)
Let's write the expression for the angular velocity:
(32)
where ω 0 = 157 s –1
Using formulas 31, 32, we compose a calculation table:
Table 3 - Data for constructing mechanical characteristics.
| | | | | | | | | |
|
| ω, s -1 | | | | | | | | |
|
| M, N m | | | | | | | | | |
3.2 Calculation and construction of electromechanical characteristics.
No-load current:
(33)
where
(34)
The current, the value of which is due to the parameters of slip and moment on the shaft:
(35)
Using formulas 33, 34, 35, we will compose a calculation table:
Table 4 - Data for constructing the electromechanical characteristics.
| | | | | | | | | |
||
| M, N m | | | | | | | | | |
|
| I 1, A | | | | | | | | | | |
Figure 2 - Mechanical and electromechanical characteristics of asynchronous
motor type at 2p =.
4.Building a load diagram
4.1 Lifting the rated load.
(36)
Ratio:
(37)
Motor shaft torque:
Acceleration time:
(39)
where the angular velocity ω 1 is determined from the mechanical characteristics of the engine and corresponds to the moment M 1st.
The selected motor type is equipped with disc brake type c M t = Nm
Constant losses in the electric motor:
(40)
Braking torque due to constant losses in the electric motor:
(41)
Total braking torque:
Stopping time of the lifted load when the engine is turned off:
(43)
The steady-state lifting speed of the rated load:
(44)
Lifting time in steady state:
The current consumed by the motor, within the limits of permissible loads, is proportional to the torque on the shaft and can be found by the formula:
4.2 Braking release of the load.
Moment on the motor shaft when lowering the rated load:
Since within the limits of permissible loads, the mechanical characteristic for the generator and motor modes can be represented by a single line, the regenerative braking speed is determined by the formula:
(49)
where the angular velocity ω 2 is determined from the mechanical characteristics of the engine and corresponds to the moment M 2st.
If the current of the braking mode I 2 is taken equal to the current of the motor operating with the moment M 2st, then:
Acceleration time when lowering the load with the engine running:
(51)
Braking torque when the engine is disconnected from the mains:
Stopping time of the lowering load:
Lowering speed:
(54)
Distance covered by the load during acceleration and deceleration:
(55)
Time of lowering the load at steady state:
(56)
Lifting the idle hook.
Moment on the motor shaft when lifting the idle hook:
(57)
According to the mechanical characteristics, the motor speed ω 3 = rad / s corresponds to the moment M 3st = Nm
Motor current:
(58)
The moment of inertia of the electric drive reduced to the motor shaft:
(59)
Acceleration time when lifting the idle hook:
(60)
Braking torque when the engine is turned off at the end of the hook lift:
Hook stopping time:
(62)
Idle hook lifting speed:
(63)
(64)
Time of steady-state movement when lifting the idle hook:
Power release of the power hook.
The moment on the shaft of the electric motor when lowering the idle hook:
(66)
The motor speed ω = rad / s corresponds to the moment М 4st = Nm
and consumed current:
(67)
Acceleration time when lowering the idle hook:
(68)
Braking torque when the engine is switched off:
(69)
Lowered hook stop time:
(70)
Idle hook lowering speed:
The distance covered by the hook during acceleration and deceleration:
(72)
Time of steady motion when lowering the idle hook:
(73)
The calculated data of engine operation are summarized in table 5.
Table 5 - Estimated data of engine operation.
| Working hours | Current, A | Time, s |
| Lifting the rated load: acceleration ………………………………………… steady state ……………………… braking…………………………………… Horizontal movement of cargo ……………. Brake release of the load: acceleration ………………………………………… steady state ……………………… braking…………………………………… Unslinging the cargo ............................................................. Lifting the idle hook: acceleration ………………………………………… steady state ……………………… braking…………………………………… Horizontal movement of the hook …………… ... Idle hook pull: acceleration ………………………………………… steady state ……………………… braking…………………………………… Slinging of cargo ………………………………… | t 01 = t 02 = t 03 = t 04 = |
5. Checking the selected engine for provision
the given productivity of the winch.
Total cycle time:
Cycles per hour:
6. Checking the selected motor for heating.
Estimated duration of inclusion:
(76)
Equivalent current for intermittent operation,
corresponding to the calculated duty cycle% (assuming the current smoothly decreasing
from starting to working, we take its average value for calculation,
especially since the time of the transient process is negligible): 
Equivalent current in intermittent duty, converted to standard duty cycle% of the selected motor, according to the equation:
(78)
Thus, I ε n = A
8. Bibliography.
Chekunov K. A. “Ship electric drives for electric propulsion of ships”. - L .:
2. The theory of the electric drive. guidelines for term paper for
students of full-time and part-time faculties of higher educational institutions in
specialty 1809 "Electrical equipment and automation of ships."
Kaliningrad 1990
3. Chilikin MG “General course of electric drive” .- M .: Energy 1981.
7. Power circuit of a frequency converter with a voltage inverter.
A converter with a voltage inverter includes the following main power units (Figure 3): a controlled HC rectifier with an LC filter; voltage inverter - AI with groups of gates for forward DC and reverse OT current, cut-off diodes and switching capacitors; slave inverter VI with LC filter. The choke windings of the UV and VI filters are made on a common core and are included in the arms of the valve bridges, while also performing the functions of current limitation. In the converter, the amplitude method of regulating the output voltage by means of an SW is carried out, and the AI is made according to a scheme with one-stage phase-to-phase switching and a device for recharging capacitors from a separate source (not shown in the diagram). The driven inverter VI provides the regenerative braking mode of the electric drive. When constructing the converter, the joint control of the HC and VI was adopted. Therefore, in order to limit the equalizing currents, the control system must provide a higher DC voltage of the HV than that of the HC. In addition, the control system must provide a given law of voltage and frequency control of the converter.
Let us explain the formation of the output voltage curve. If initially thyristors 1 and 2 were in the conducting state, then when thyristor 3 opens, the capacitor charge is applied to thyristor 1, and it closes. Thyristors 3 and 2 turn out to be conductive. Under the action of the EMF of self-induction and phase A, diodes 11 and 16 open, since the potential difference between the beginning of phases A and B turns out to be the greatest. If the duration of the turn-on of the reverse diodes, determined by the self-induction of the load phase, is less than the duration of the operating interval, diodes 11 and 16 are closed.
A capacitor is connected to the DC link in parallel with the inverter, which limits the voltage ripple that occurs when the thyristors of the inverter are switched. As a result, the DC link has a resistance for the AC component of the current, and the input and output voltages of the inverter at constant load parameters are connected by a constant coefficient.
The inverter arms are bi-directional. To ensure this, thyristors are used in the arms of the inverter, shunted by oppositely connected diodes.
0Faculty of Electrical Power
Department of Automated Electric Drive and Electromechanics
COURSE PROJECT
in the discipline "Theory of electric drive"
Calculation of the electric drive of the freight elevator
Explanatory note
Introduction ………………………………………………………… ... ………………
1 Calculation of the electric drive of the freight elevator ………………………………………
1.1 Kinematic diagram of the working machine, its description and technical data …………………………………………………………………………… ...…
1.2 Calculation of static moments ………………………………………… ... ……
1.3 Calculation of the load diagram ………………………………………………
1.4 Preliminary calculation of the power of the electric motor and its selection ………
1.5 Calculation of reduced static moments …………………………… ...…
1.6 Construction of the load diagram of the electric motor ……………………
1.7 Preliminary check of the electric drive for heating and performance …………………………………………………………………….
1.8 Selection of the electric drive system and its structural diagram …………………
1.9 Calculation and construction of natural mechanical and electromechanical characteristics of the selected engine …………………………………………………
1.9.1 Calculation and construction of natural characteristics of a DC motor of independent excitation …………………………………… .. ……
1.10 Calculation and construction of artificial characteristics ………………………
1.10.1 Calculation and construction of a starting diagram of an engine with a linear mechanical characteristic graphically ………………………. …… ..
1.10.2 Construction of braking characteristics …………………………… ... ……
1.11 Calculation of transient modes of an electric drive …………………………… ..
1.11.1 Calculation of mechanical transient processes of an electric drive with absolutely rigid mechanical connections ………………………………………
1.11.2 Calculation of the mechanical transient process of an electric drive in the presence of an elastic mechanical connection …………………………………………… ...…
1.11.3 Calculation of the electromechanical transient process of an electric drive with absolutely rigid mechanical connections …………………………………… ..…
1.12 Calculation and construction of an updated engine load diagram
1.13 Checking the electric drive for a given capacity, heating and overload capacity of the electric motor ………………………………… ..…
1.14 Schematic diagram electrical part of the electric drive
Conclusion ………………………………………………………………..………
Bibliography……………………………………………………………..…
Introduction
The way to get the energy you need to perform mechanical work in production processes, at all stages of the history of human society, he exerted a decisive influence on the development of the productive forces. The creation of new, more advanced engines, the transition to new types of drives for working machines were major historical milestones in the development of machine production. Replacement of motors that realize the energy of the falling water, steam engine, served as a powerful impetus to the development of production in the last century - the century of steam. Our 20th century. Received the name of the century of electricity primarily because the main source of mechanical energy has become a more perfect Electrical engine and the main type of drive of working machines is an electric drive.
The individual automated electric drive is now widely used in all spheres of life and activities of society - from the sphere of industrial production to the sphere of everyday life. Thanks to the features discussed above, the improvement of the technical performance of electric drives in all areas of application is the basis of technical progress.
The breadth of application determines an exceptionally large range of power of electric drives (from fractions of a watt to tens of thousands of kilowatts) and a significant variety of designs. Industrial plants of unique performance - rolling mills in the metallurgical industry, mine hoists and excavators in the mining industry, powerful construction and assembly cranes, long high-speed conveyor plants, powerful metal-cutting machines and many others - are equipped with electric drives with a capacity of hundreds and thousands of kilowatts. ... Converting devices of such electric drives are direct current generators, thyristor and transistor converters with direct current output, thyristor frequency converters of the corresponding power. They provide a wide range of flow control options electrical energy entering the engine in order to control the movement of the electric drive and the technological process of the driven mechanism. Their control devices, as a rule, are based on the use of microelectronics and in many cases include control computers.
1 Calculation of the electric drive of the freight elevator
1.1 Kinematic diagram of the working machine, its description and technical data
1 - electric motor,
2 - brake pulley,
3 - reducer,
4 - traction pulley,
5 - counterweight,
6 - cargo cage,
7 - lower platform,
8 - upper platform.
Figure 1 - Kinematic diagram of the elevator
The freight elevator lifts the load placed in the freight cage from the lower platform to the upper one. The empty cage is lowered down.
The cycle of operation of the freight elevator includes the loading time, the time for lifting the stand at a speed V p, the time for unloading and the time for lowering the stand at a speed V in> V p
Table 1 - Initial data
|
Designation |
Indicator name |
Dimension |
|
|
Stand weight |
|||
|
Carrying capacity |
|||
|
Counterweight mass |
|||
|
Traction sheave diameter |
|||
|
Trunnion diameter |
|||
|
Coeff., Sliding friction in bearings |
|||
|
Linear rigidity of the mechanism |
|||
|
Cage lifting height |
|||
|
Travel speed with load |
|||
|
Travel speed without load |
|||
|
Permissible acceleration |
|||
|
Cycles per hour |
|||
|
Total operating time, no more |
According to the assignment, when calculating the mechanism, it is necessary to take a DC motor with independent excitement.
1.2 Calculation of static moments
The moment of static resistance of the freight elevator consists of the moment of gravity and the moment of friction in the bearings of the traction sheave and the friction of the freight cage and counterweight in the guides of the shaft.
The moment of gravity is determined by the formula:
where D is the diameter of the traction sheave, m;
m res - the resulting mass that is raised or lowered by the electric drive of the elevator, kg.
The resulting mass is determined by the ratio of the masses of the load, cage and counterweight and can be calculated using the formula:
m res = m k + m g - m n = 1500 + 750-1800 = 450 kg
The frictional moment in the bearings of the traction sheave can be determined by the expression:
The frictional moment of the load cage and the counterweight in the guides of the shaft is almost impossible to mathematically accurately determine, since the magnitude of this resistance depends on many factors that cannot be taken into account. Therefore, the magnitude of the frictional moment of the stand and counterweight in the guides is taken into account by the efficiency of the mechanism, which is determined by the design assignment.
Thus, the total moment of static resistance of the freight elevator is determined by the expression:
if the engine is operating in motor mode, and by the expression:
if the engine is operating in braking (generator) mode.
1.3 Calculation of the load diagram of the working machine
In order to roughly estimate the required for this mechanism engine power, it is necessary to determine in one way or another the power or moment of the production mechanism in different areas of its operation and the speed of movement of the working body of the mechanism in these areas. In other words, it is necessary to build a load diagram of the production mechanism.
The mechanism, operating in intermittent mode, in each cycle makes a forward stroke with full load and reverse at idle or light load. Figure 2.1 shows the load diagram of the mechanism with the limitation of the permissible acceleration of the working body of the mechanism.
Figure 2 - Load diagram of a mechanism with limited acceleration
The load diagram shows:
-, - static moments during forward and reverse moves;
-, - dynamic moments during forward and reverse moves;
-, - starting moments during forward and reverse moves;
-, - braking moments during forward and reverse strokes;
-, - speeds of forward and reverse moves;
-, - times of start-up, braking and steady motion during forward motion;
-, - times of start, deceleration and steady motion during the return stroke.
For the given speeds V c 1, V c 2, the length of movement L, and the permissible acceleration a, t p1, t p2, t t1, t t2, t y1, t y2 are calculated.
Start and deceleration times:
The path traversed by the working body of the machine during the start (braking):
The path traversed by the working body of the machine during the steady motion:
Time of steady motion:
The operating time of the mechanism for forward and reverse moves:
Dynamic moments of the working machine
where D is the diameter of a rotating element of a working machine that converts rotary motion into translational motion, m,
J pm1, J pm1 - moments of inertia of the working machine during forward and reverse moves.
The total moment of the working body of the mechanism, in dynamic mode (start, braking) during forward and reverse moves, are determined by the expressions:
1.4 Preliminary calculation of the power of the electric motor and its selection
Thus, as a result of calculations according to the above formulas, the coordinates of the load diagrams receive specific values that make it possible to calculate the root-mean-square value of the torque per cycle of operation.
For a load diagram, with acceleration limitation:
The actual relative operating time is determined from the expressions:
where t c is the duration of the work cycle, s,
Z is the number of starts per hour.
Having the value of the rms moment of the production mechanism per cycle, the approximate required engine power can be determined by the ratio:
where V cn is the speed of the working body of the mechanism V c 2,
PVN - the nominal value of the on-time closest to the actual PV N,
K is a coefficient that takes into account the magnitude and duration of the dynamic loads of the electric drive, as well as losses in mechanical attachments and in the electric motor. For our case, K = 1.2.
Now the engine is selected that is suitable for the operating conditions.
Engine parameters:
Crane metallurgical engine direct current, UН = 220 V, duty cycle = 25%.
Table 2 - Engine data
Determine the gear ratio of the gearbox:
where w N is the rated speed of the selected motor.
The gearbox can be selected according to the reference book, taking into account a certain gear ratio, rated power and engine speed, as well as the operating mode of the mechanism for which this gearbox is intended.
Such a choice of a gearbox is very primitive and suitable only for mechanisms such as a winch. In reality, the gearbox is designed for a specific working mechanism and is an integral part of it, limited to both the electric motor and the working body. Therefore, if the choice of gearbox is not particularly limited in the design assignment.
1.5 Calculation of the reduced static moments, moments of inertia and the coefficient of stiffness of the system electric motor - working machine
In order to be able to calculate the static and dynamic characteristics of the electric drive, it is necessary to bring all static and dynamic loads to the motor shaft. In this case, not only the gear ratio of the gearbox must be taken into account, but also the losses in the gearbox, as well as constant losses in the engine.
Engine idling losses (constant losses) can be determined by taking them to be equal to variable losses in the nominal operating mode:
where η n is the rated efficiency of the engine.
If the value of η n is not given in the catalog, it can be determined by the expression:
Moment of constant engine loss
Thus, the static moments of the electric motor - working machine system reduced to the motor shaft at each work site are calculated using the formulas:
if the motor in steady state is running in motor mode.
The total moment of inertia of the electric motor - working machine system, reduced to the electric motor shaft, consists, as it were, of two components:
a) the moment of inertia of the rotor (armature) of the engine and associated elements of the electric drive, rotating at the same speed as the engine,
b) the total moment of inertia of the moving executive bodies of the working machine and the associated moving masses involved in the technological process of this working mechanism, reduced to the motor shaft.
Thus, the total moment of inertia reduced to the motor shaft, with direct and reverse moves determined by the expressions:
where J d is the moment of inertia of the armature (rotor) of the engine,
a - coefficient taking into account the presence on the high-speed shaft of other elements of the electric drive, such as clutches, brake pulley, etc.
For the mechanism presented in the assignment for course design, the coefficient is a = 1.5.
J prrm1, J prrm2 - the total moment of inertia of the moving executive bodies reduced to the motor shaft, and the associated masses of the working machine during forward and reverse strokes:
In order to get an idea of the effect of elastic mechanical bonds on the transient processes of the electric motor - working machine system, the task presents the torsional stiffness C k.
The stiffness of the elastic mechanical connection C CR reduced to the engine shaft is determined through the value of the torsional stiffness:
1.6 Construction of a load diagram of an electric motor
To build a load diagram of an electric motor, it is necessary to determine the values of the dynamic torques required for starting and braking, as well as the values of the starting and braking moments of the motor.
For our load diagram of the acceleration limited mechanism, the values of these moments are determined by the following expressions.
Starting and braking torques for the case when the engine in steady state operates in motor mode is determined by the formula:
For building performance characteristics value of speed w c 1 is required. The speed w c2 is equal to the rated speed of the electric motor.
Figure 3 - Approximate load diagram of the electric motor
1.7 Preliminary check of the electric motor for heating and performance
A preliminary motor heat test can be performed using the motor load diagram using the equivalent torque method. In this case, this method does not give a significant error, since both DC motor and motor alternating current will work in the designed electric drive on the linear part of the mechanical characteristics, which gives reason with a high degree of probability to consider the motor torque proportional to the motor current.
The equivalent moment is determined by the expression:
Permissible moment of a preselected motor operating at PV f:
Correctness condition preselection engine:
For our case
which satisfies the conditions for choosing an electric motor.
1.8 The choice of the electric drive system and its structural diagram
The projected electric drive together with a given production mechanism forms a single electromechanical system. The electrical part of this system consists of an electromechanical converter of DC or AC power and a control system (energy and information). The mechanical part of the electromechanical system includes all associated moving masses of the drive and mechanism.
As the main representation of the mechanical part, we take the calculated mechanical system (Figure 4), a frequent case of which, while neglecting the elasticity of mechanical links, is a rigid reduced mechanical link.
Figure 4 - Two-mass design mechanical system
Here J 1 and J 2 are the moments of inertia of two masses of the electric drive, reduced to the motor shaft, connected by elastic connection,
w1, w2 - the speed of rotation of these masses,
c12 is the stiffness of the elastic mechanical bond.
As a result of the analysis of the electromechanical properties of various motors, it was found that, under certain conditions, the mechanical characteristics of these motors are described by identical equations. Therefore, under these conditions, the basic electromechanical properties of motors are also similar, which makes it possible to describe the dynamics of electromechanical systems by the same equations.
The above is true for motors with independent excitation, motors with successive arousal and mixed excitation with linearization of their mechanical characteristics in the vicinity of the point of static equilibrium and for an induction motor with a phase rotor with linearization of the working section of its mechanical characteristics.
Thus, applying the same notation for three types of motors, we obtain a system of differential equations describing the dynamics of a linearized electromechanical system:
where M s (1) and M s (2) are the parts of the total load of the electric drive applied to the first and second masses,
M 12 is the moment of elastic interaction between the moving masses of the system,
β is the modulus of static stiffness of the mechanical characteristic,
T e is the electromagnetic time constant of the electromechanical converter.
The block diagram corresponding to the system of equations is shown in Figure 5.
Figure 5 - Block diagram of the electromechanical system
Parameters w0, Te, β are determined for each type of engine by their own expressions.
The system of differential equations and the structural diagram correctly reflects the basic laws inherent in real nonlinear electromechanical systems in the modes of permissible deviations from the static state.
1.9 Calculation and construction of natural mechanical and electromechanical characteristics of the selected electric motor
The equation of natural electromechanical and mechanical characteristics of this engine is as follows:
where U is the voltage at the motor armature,
I - motor armature current,
M is the moment developed by the engine,
R iΣ - total resistance of the anchor circuit of the engine:
where R i is the resistance of the armature winding,
R dp - winding resistance of additional poles,
R ko - resistance of the compensation winding,
Ф is the magnetic flux of the motor.
K - design factor.
From the expressions given above it can be seen that the characteristics of the engine are linear under the condition Ф = const and can be plotted using two points. These points select the ideal idle point and the nominal mode point. The rest of the quantities are determined:
Figure 6 - Natural characteristic of the engine
1.10 Calculation and construction of artificial characteristics of the electric motor
The artificial characteristics of the engine in this course project include the rheostat characteristic for obtaining a reduced speed when the engine is running at full load, as well as rheostatic characteristics that provide the specified conditions for starting and braking.
1.10.1 Calculation and construction of a starting diagram of an engine with a linear mechanical characteristic graphically
Construction begins with the construction of a natural mechanical characteristic. Next, you need to calculate the maximum torque developed by the engine.
where λ is the overload capacity of the motor.
To construct the operating characteristic, we use the values of w 1 and M c1, the point of ideal idling.
When reaching the natural characteristic, there is an inrush current that goes beyond M 1 and M 2. To start from operating characteristic, it is necessary to keep the current start pattern. Since when starting up to the operating and natural characteristic, one stage is required and there is no need for additional stages.
M 1 and M 2 are taken equal:
Figure 7 - Starting characteristic of the engine
According to the figure, starting resistances are calculated using the following formulas:
The starting sequence is shown in the figure as symbols.
1.10.2 Calculation and construction of the engine operating characteristic with a linear mechanical characteristic.
The operating characteristic of a DC motor with independent excitation is built on two points: the point of ideal idling and the point of the operating mode, the coordinates of which were determined earlier:
Figure 8 - Engine performance
Depending on how the operating characteristic is located relative to the starting diagram of the motor, one or another correction of either the starting diagram or the trajectory of starting the engine under load Mc1 up to the speed wc1 is necessary.
Figure 9 - Engine performance
1.10.3 Construction of braking characteristics
The terms of reference determined the maximum allowable acceleration in transient processes, then the initial values for the construction of braking characteristics are the values of the average, constant in magnitude, braking moments defined in clause 6. Since, in their determination, the acceleration was considered constant, the braking moments during braking with different load and at different initial speeds can differ significantly from each other, moreover, up or down. In theory, even their equality is possible:
Therefore, both braking characteristics must be plotted.
The drawing should take into account that the rheostat characteristics of braking by the Opposition must be constructed in such a way that the area between the characteristics and the coordinate axes is approximately equal in one case:
and in another case:
Often the values of the braking torques are much less than the peak moment M 1, at which the starting resistances are determined. In this case, it is necessary to construct the natural characteristic of the motor for the reverse direction of rotation and determine the values of the braking resistances according to the expressions according to the figure:
1.11 Calculation of transient modes of an electric drive
In this course project must be calculated transients of starting and braking with different loads. As a result, the time dependences of the moment, speed and angle of rotation should be obtained.
The results of calculating transient processes will be used to construct load diagrams of the electric drive and to check the motor for heating, overload capacity and specified performance.
1.11.1 Calculation of mechanical transient processes of an electric drive with absolutely rigid mechanical connections
When introducing the mechanical part electric drive rigid a mechanical link and neglect of electromagnetic inertia, an electric drive with a linear mechanical characteristic is an aperiodic link, with a time constant T m.
The transient equations for this case are written as follows:
where M s is the engine torque in a steady state,
w c - motor speed in steady state,
M start - the moment at the beginning of the transient process,
W start - engine speed at the beginning of the transient.
T m - electromechanical time constant.
The electromechanical time constant is calculated according to the following formula, for each stage:
For braking performance:
The operating time on the characteristic, during transient processes, is determined by the following formula:
To reach the natural characteristic, we consider:
To reach the operating characteristic:
For braking performance:
The time of the transient processes during start-up and braking is determined as the sum of the times at each stage.
To reach the natural characteristic:
To reach the operating characteristic:
The operating time on the natural characteristic is theoretically equal to infinity, respectively, it was considered as (3-4) Tm.
Thus, all the data for calculating the transient processes were obtained.
1.11.2 Calculation of the mechanical transient process of an electric drive in the presence of an elastic mechanical connection
To calculate this transient, it is necessary to know the acceleration and the frequency of free oscillations of the system.
The solution to the equation has the form:
In an absolutely rigid system, the load of the gears during the start-up process is equal to:
Due to elastic vibrations, the load increases and is determined by the expression:
Figure 13 - Elastic load fluctuations
1.11.3 Calculation of the electromechanical transient process of an electric drive with absolutely rigid mechanical connections
To calculate this transient, it is necessary that the following values be calculated:
If the ratio of the time constants is less than four, then we use the following formulas to calculate:
Figure 14 - Transient process W (t)
Figure 15 - Transient process M (t)
1.12 Calculation and construction of an updated load diagram of an electric motor
The refined load diagram of the motor should be built taking into account the starting and braking modes of the motor in the cycle.
Simultaneously with the calculation of the load diagram of the motor, it is necessary to calculate the value of the rms torque at each section of the transient process.
RMS torque characterizes the heating of the motor when the motors operate on the linear part of their characteristics, where the torque is proportional to the current.
To determine the rms values of the torque or current, the real curve of the transient process is approximated by straight sections.
The values of the root-mean-square moments at each section of the approximation are determined by the expression:
where M start i is the initial value of the moment in the considered section,
M kon i - the final value of the moment in the considered area.
For our load diagram, six rms moments need to be determined.
For movement on a natural characteristic:
For driving on the working characteristic:
1.13 Checking the electric drive for a given capacity, heating and overload capacity
Checking for a given performance of the mechanism is to check whether the calculated operating time is within the specified technical specifications t p.
where t pp is the estimated operating time of the electric drive,
t p1 and t p2 are the times of the first and second starts,
t т1 and t т2 are the times of the first and second braking,
t у1 and t у2 - times of steady-state modes during operation with high and low load,
t p2, t p1, t t2, t t12 - are taken from the calculation of transient processes,
Checking the selected motor for heating in this course project should be performed using the equivalent torque method.
The permissible motor torque in repeated - short-term mode is determined by the expression:
1.14 Principled electrical circuit power section of the electric drive
The power section is presented in the graphic section.
Description of the power circuit of the electric motor
The control of the electric drive consists, firstly, in connecting the motor windings to the mains at start-up and disconnecting at stopping, and secondly, the gradual switching of the starting resistor stages by the relay-contactor equipment as the motor accelerates.
Removing the steps of the starting resistor in the rotor circuit is possible in several ways: as a function of speed, as a function of current and as a function of time. In this project, the motor is started as a function of time.
Conclusion
In this coursework, the electric drive of the overhead crane trolley was calculated. The selected motor does not quite satisfy the conditions, since the torque developed by the motor is greater than required for the given mechanism, therefore, it is necessary to choose a motor with a lower torque. Since the list of offered engines is not complete, we leave this engine as amended.
Also, in order to use the operating characteristic for starting in both directions, we allowed a slightly larger current surge when going to natural characteristic. But this is permissible, since a change in the starting scheme would lead to the need to introduce additional resistance.
Bibliography
1.Klyuchev, V.I. Electric drive theory / V.I. Key. - M .: Energoatomizdat, 1998.- 704s.
2.Chilikin, M.G. General course of electric drive / M.G. Chilikin. - M .: Energoatomizdat, 1981.-576 p.
3.Veshenevsky, S.N. Characteristics of motors in an electric drive / S.N. Veshenevsky. - M .: Energiya, 1977 .-- 432 p.
4.Andreev, V.P. Electric drive fundamentals / V.P. Andreev, Yu.A. Sabinin. - Gosenergoizdat, 1963 .-- 772 p.
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1. Analysis and description of the "electric drive - working machine" system
1.1 Quantification of the state vector or tachogram of the required motion process
1.2 Quantifying the moments and forces of resistance
1.3 Drawing up a design diagram of the mechanical part of the electric drive
1.4 Construction of the load diagram and the mechanical characteristics of the working machine
2. Analysis and description of the "electric drive - network" and "electric drive - operator" systems
3. Choice of fundamental decisions
3.1 Construction of the mechanical part of the electric drive
3.2 Selecting the type of drive (motor)
3.3 Choosing a method for adjusting coordinates
3.4 Evaluation and comparison of selected options
4. Calculation of the power drive
4.1 Calculation of parameters and selection of the motor
4.2 Calculation of parameters and selection of the power converter
5. Calculation of static mechanical and electromechanical characteristics of the motor and drive
6. Calculation of transient processes in an electric drive for a cycle of work
7. Checking the correctness of the power calculation and the final choice of the engine
1. Analysis and description of the "electric drive - working machine" system
1.1 Quantification of the state vector or tachogram of the required motion process
The speed of the electric drive during dressing and at idle speed, which is selected from the speed range from 1.45 m / s to 2.4 m / s.
According to the description of the technological process [1], you can build a tachogram of the required movement process. At the request of the process, the cycle of work occurs at a constant speed. Start-up and changeover to another speed are not included in the work cycle. The tachogram is shown in Figure 1.
Figure 1- Tachogram of the workflow
Let us determine the value of the minimum angular speed of the engine based on the tachogram and the conditions of the task:
(1)
where i is the gear ratio of the gearbox;
v 1 - the minimum speed of movement of the sheet;
R is the radius of the working and supporting rollers.
Maximum angular speed of the engine:
(2)
where v 2 is the maximum speed of movement of the sheet.
Consider two cases:
1) driving the maximum length of the sheet at the minimum speed;
2) running the minimum length of the sheet at maximum speed.
First case.
Rolling time:
(3)
where L max is the maximum sheet length.
According to the condition of setting the PV of the mechanism - 75%. Let's define the cycle time:
(4)
Idle time:
Second case.
(6)
We will choose an engine with a design mode S1 since there are no pauses during the drive cycle.
1.2 Quantifying the moments and forces of resistance
Knowing the total total moment at maximum load, referred to the work rolls, it is possible to determine the static moment reduced to the shaft:
(7)
where is the efficiency of the mechanism (considered unchanged).
The idling torque, given to the engine shaft, is set and is equal to:
The torque on the motor shaft during dressing is determined by the formula:
1.3 Drawing up a design diagram of the mechanical part of the electric drive
For a theoretical study, we replace the real mechanical part of the electric drive (Figure 2) with a dynamically equivalent reduced design scheme, consisting of concentrated inertial elements interconnected by elastic links, and having the same energy reserve as the real original drive system. The parameters of the equivalent reduced design scheme are the total reduced moments of inertia of the masses formed by the reduced masses, the bonds between which are assumed to be rigid, and the equivalent reduced stiffnesses of elastic mechanical bonds.
Figure 2 - Kinematic diagram of the mechanism
The electric drive consists of the following kinematic elements:
1 - electric motor;
2 - reducer;
3 - six cage;
4 - universal spindles;
5 - working stand.
The moment of inertia of the couplings between the engine and the gearbox is 16 kg * m 2, the moment of inertia of the couplings between the gearbox and the gear stand is 40.2 kg * m 2, of one spindle - 0.003 kg * m 2. The moment of inertia of the gearbox, reduced to the motor shaft, is equal to 30% of J dv.
The number of spindles -17, the number of working rollers -17, support - 15.
The mechanical part of the electric drive of the sheet-straightening mill is a three-mass system consisting of rotors (armatures) of motors with half-couplings on the shafts - J1, a gearbox with half-couplings on its input and output shafts - J2 and the working body of the machine, also with half-couplings on the input shaft - J3. The elastic links of this system are the stiffnesses of the C 12 and C 23 couplings.
Let's calculate the parameters of the resulting circuit.
First mass moment of inertia:
where J pm1 is the moment of inertia of the half-couplings on the motor shafts.
The moment of inertia of the gearbox with half couplings on its input and output shafts (taking into account that the moment of inertia of the gearbox, reduced to the motor shaft, is equal to 30% of J dv) is:
where J pm2 is the moment of inertia of the half-coupling on the output shaft of the gearbox.
The moment of inertia of the working member of the drive with half-couplings on the input shaft, reduced to the motor shaft, is calculated by the following expression:
(11)
where J roll is the total moment of inertia of the working and support rollers;
J shp is the moment of inertia of the spindles;
J pm - moment of inertia of half-couplings;
i - gear ratio of the gearbox.
Let's define the moment of inertia of the roller:
where L is the length of the roller, m;
D - roller diameter, m;
Density of the material (= 7.66 * 10 3 kg / m 3).
Given the number of working and support rollers, we get:
Spindle moment of inertia:
Then the moment of inertia of the working body will be equal to:
Rigidity of the coupling between the gearbox and the gear cage, referred to the motor shaft:
.(15)
Considering that when the elastic elements are connected in parallel, the stiffnesses add up, we find the stiffness of the C 12 and C 23 couplings, which are the elastic links of the three-mass system:
where С м1 - stiffness coupling sleeve between motor and gearbox.
Calculation of transients in a three-mass system is complicated, therefore we transform the system into a two-mass one.
 |
Let's calculate the parameters of the circuit. Equivalent stiffness of a two-mass design scheme:
The transition and justification for the transition to a single-mass settlement scheme will be given below.
1.4 Construction of the load diagram and the mechanical characteristics of the working machine
The load diagram of the mechanism is the dependence of the torque given to the motor shaft as a function of time per cycle.
The working cycle is the alternation of the drive during the movement of the sheet and idle work the machine before the start of the next cycle. We build a simplified load diagram of a working machine, which is built according to the static loads calculated for each section of the work cycle, that is, without taking into account dynamic loads. Dynamic loads are not included in the work cycle as the machine runs at a constant speed.
A simplified load diagram is as follows:
At the idle speed interval, the torque is equal to the idle torque;
In the dressing interval, the torque is equal to the sum of the moments of static on the axis of the work rolls, reduced to the motor shaft and idling.
The load diagram is shown in Figure 5.
Figure 5 - Load diagram of the mechanism
The mechanical characteristic of the working machine is the dependence of the reduced static moment on the speed of the motor shaft. According to the assignment, this dependence is close to parabolic.
The mechanical characteristics of the working machine are shown in Figure 6.
Figure 6 - Mechanical characteristics of the working machine
2. Analysis and description of the "electric drive-network" and "electric drive-operator" systems
The electric drive of the sheet-straightening mill is powered by a 3-phase alternating current network with a frequency of 50 Hz and a voltage of 380V.
The standards provide for and allow for a voltage variation of ± 10% and a frequency of ± 2.5% (GOST 13109-87). This phenomenon is caused, among other things, by the presence of other powerful consumers of energy in the conditions of the workshop, plant. This significantly affects the operation of engines, imposes additional requirements on the organization of their work.
Using the QF1 circuit breaker, we connect the voltage to the frequency converter.
By pressing the START button, the drive is switched on, then the drive works in automatic mode, no operator is required for constant control of the drive operation.
3. Choice of fundamental decisions
3.1 Construction of the mechanical part of the electric drive
The kinematic diagram of the main electric drive of the sheet-straightening mill is shown in Figure 2. The main operation - straightening, is performed using rotating rolls located in the working stand. The upper work roll moves in a vertical plane, and the axis of the lower roll is always in a fixed position.
The transmission mechanisms in the rolling mill consist of a gearbox, a gear stand, working spindles and couplings.
The gearbox is designed to enable the use of a motor with a relatively high rated speed at low rolling speeds and thereby reduce the size and cost of the engine and the entire installation as a whole.
The spindles are used to transfer rotation to the rolls from the gear stand. The need for their application lies in the fact that with a change in the position of the upper roll, the distance between this roll and the gear stand, as well as the angle between the shaft of the gear stand and the spindle, also change.
Couplings are used to connect the gear cage and the motor to the gearbox.
3.2 Selecting the type of drive (motor)
The basis for choosing the type of motor is the technical specifications for the design of the drive of the sheet-straightening mill:
Long-term operation;
Infinitely variable speed control within a given range.
The following drives meet the above conditions:
1 Frequency converter - asynchronous motor;
2 Controlled rectifier - DC motor;
3 Cascade circuit;
4 Generator - engine.
3.3 Choosing a method for adjusting coordinates
When choosing a method for controlling coordinates (speed), it is necessary to take into account the energy aspect of choosing a control method. This means that the minimum size of the motor and its full use for heating takes place when the way of speed control according to the indication of the permissible load corresponds to the dependence of the load on the speed.
Since the mechanical characteristic of the mechanism is a viscous load, it is advisable to use a speed control method with a constant power, i.e. regulation with P = const. In the case of using this method, the engine is provided with the best thermal conditions.
In the system, a frequency converter (AIN PWM) is an asynchronous motor, the required speed is obtained by changing the frequency and generating the voltage on the stator (volt-frequency control) or by regulating the frequency and forming the vector of the main flux linkage of the machine (vector control).
In systems controlled rectifier - DC motor and generator - motor, the required speed is obtained by changing the armature supply voltage.
In a cascade scheme, speed control is carried out by introducing an additional EMF into the rotor circuit of the machine.
3.4 Evaluation and comparison of selected options
The generator - engine system is outdated, therefore, when comparing the selected options, it will not be taken into account.
Carrying out rigorous technical and economic calculations is not possible due to the lack of the required initial data, therefore, to evaluate and compare the selected options, we will use an approximate method - the “method of expert assessments”. Comparison of solution options is made with respect to n characteristics of the system that are important from the point of view of the design goal by comparing certain values of the corresponding quality indicators q i. Quality indicators serve to quantitatively characterize the degree of fulfillment of the requirements of the assignment for the design of the electric drive, as well as other requirements of the working machine.
The evaluation of electric drives will be carried out according to the following quality indicators:
1 - regulation range;
2 - efficiency of the electric drive;
3 - power factor;
4 - weight and size indicators;
5 - the cost of the electric drive;
6 - reliability of the electric drive;
7 - resource of work;
8 - operating costs;
9 - control accuracy;
Let us evaluate the fulfillment of the requirements for the i-th characteristic of the system according to the following criterion:
5 - the requirements for the i-th characteristic of the system are fulfilled very well;
qi = 4 - the requirements for the i-th characteristic of the system are fulfilled well;
3 - the requirements for the i-th characteristic of the system are satisfied satisfactorily;
2 - the requirements for the i-th characteristic of the system are fulfilled unsatisfactorily.
The FC - IM and UV - DPT systems with speed feedback provide a very large control range, therefore the requirements for the control range are fulfilled very well. In a cascade circuit, the range is limited by the power of the converter, i.e. when the range is increased, the inverter power becomes greater than the motor power, so the requirements for the control range are satisfied satisfactorily.
The efficiency of power drives is high enough that the drive efficiency requirements are met very well.
All drives meet the power factor requirements well.
The mass and dimensions of the drive are determined by the mass and dimensions of the motor and converter. Modern drives FCh - IM and UV - DPT have very good weight and size indicators, therefore the requirements for the weight and size indicators of the drive are met very well, and the cascade circuit has slightly worse weight and size indicators, therefore the requirements for the weight and size indicators of the drive are met well.
The cost requirement in HC - DC drives and the cascade circuit is fulfilled very well, and in the FC - IM drive it is somewhat worse due to the fact that the cost of the PC - IM is slightly higher than the cost of the HC - DC and the cascade circuit.
An asynchronous squirrel-cage motor does not have a collector unit and brush contacts, therefore, the requirements for reliability and service life are met very well. In a cascade circuit, the engine does not have a collector unit, but has a brush contact, therefore, the requirements for reliability and service life are met well. The DC motor has a collector assembly, therefore, the reliability requirements are not met satisfactorily, and with proper maintenance of the collector, the service life requirements are satisfied satisfactorily.
The drive of the PCh - IM does not require operating costs, therefore the requirements for operating costs performed very well. In a cascade circuit, periodic checking of the brush contacts is necessary, so the operating cost requirements are met well. In the UV-DPT drive, a more frequent inspection of the collector unit is required, as well as periodic cleaning of the brushes, therefore, the requirements for operating costs are satisfied satisfactorily.
In the UV - DPT drive, the control accuracy requirements are met very well. In the drive of the PCh-IM, the requirements for the control accuracy are well fulfilled. In the cascade scheme, the requirements for the control accuracy are satisfied satisfactorily.
The choice of the option as the best one depends on how equal the characteristics of the system are, i.e. you need to evaluate their significance. For this, the weight coefficients λ i are introduced, which can be determined as follows:
5 - i-th characteristic the system is critical to the design goal;
4 - - “- very large, but not decisive;
li = 3 - - “- important;
2 - - “- it is desirable to take into account;
1 - - “- immaterial for the purpose of development.
The task of an electric drive is to perform useful work with a minimum of losses, therefore, the efficiency of an electric drive is of decisive importance.
The consumption of reactive power from the network is normalized (for exceeding the norm, the enterprise has to pay a fine), therefore the power factor is of decisive importance.
Since a sheet-straightening mill is a continuous-action unit and its unforced downtime leads to huge losses, therefore, reliability and service life are of decisive importance.
According to the assignment, the drive must provide a relatively small control range, therefore this quality indicator is not very large and decisive and can be described as important.
Cost matters a lot. However, as you know, cost is closely related to quality; therefore, such an indicator as cost is of great, but not decisive, importance.
Usually at metallurgical enterprises there are premises sufficient to accommodate the mill, therefore, the mass and size parameters of the mill are not very large and of decisive importance. However, with an increase in the mass of the mill, its cost also increases, therefore this indicator can be described as important.
The evaluation chart is shown in Figure 7.
Figure 7 - Evaluation diagram (quality indicators: 1 - control range; 2 - efficiency of the electric drive; 3 - power factor; 4 - weight and size indicators; 5 - cost of the electric drive; 6 - reliability of the electric drive; 7 - service life; 8 - operating costs; 9 - regulation accuracy)
The choice of the best solution is made by determining the weighted sum, ( the best way has a large amount) according to the formula:
where is the quality indicator;
Weight coefficient;
Weighted amount.
Let's define the weighted sums:
As a result, we get that the following drive has the maximum weighted sum: frequency converter - asynchronous motor.
Therefore, this drive is subject to further calculation.
4. Calculation of the power drive
4.1 Calculation of parameters and selection of the motor
The design operating mode of the engine is long-term with variable load, since there are no pauses during engine operation, and the load changes in jumps (Figure 5).
Since the necessary initial data for calculating the motor power by the methods of average losses, there is no equivalent current, therefore we will use a less accurate method - the method of equivalent torque, assuming that constant losses, motor resistances do not change during operation, and also that the torque developed by the motor, proportional to the current.
According to the load diagram and the mechanical characteristics of the working machine, the equivalent torque is equal to:
(21)
where 
- coefficient of deterioration of cooling of the machine when working at speed;
Cooling deterioration coefficient during pauses, depending on engine ventilation (for closed self-ventilated motors = 0,45 -0,55)
Range of regulation when working at speed.
The additional load created by the dynamic moment will be taken into account by the safety factor.
Let's calculate the equivalent torque without taking into account the coefficient of deterioration of the cooling of the machine when operating at a speed different from the nominal one for two limiting operating modes of the drive:
1) drive the maximum length of the sheet at the minimum speed:
;
2) running the minimum length of the sheet at the maximum speed:
Let's take the moment the largest of the two given cases:
.
According to the instructions of the project, it is required to ensure operation in the speed range, therefore, the engine speed:
rpm; (22)
rpm; (23)
The minimum engine speed is n dv = 500 rpm, it is less than required. Therefore, we will regulate the drive in the 1st zone.
Using a variable frequency drive, we will be able to provide the required speed.
Let's estimate the required engine power:
The criteria for choosing an engine are as follows:
When choosing, it is necessary to choose an engine with in order to more fully use the engine in terms of power.
However, the industry produces engines (standard series 4A) with a power of more than 197.3 kW (200 kW) only for revolutions above 1000 rpm (104.6 rad / s) and above, and with an increase in power, the rated speed of the motors increases.
Also, with an increase in the rated speed of the motor, the rated torque decreases, according to the formula
whence it follows that in order for the engine not to overheat during operation, it is necessary to overestimate the engine power.
Thus, it is necessary to select an engine with a capacity 
and 
rpm However, there is no standard motor (4A series) with such parameters.
Due to the impossibility of performing a high-power drive with one motor, we will build an electric drive consisting of two machines. The interconnected electric drive in high-power installations allows to reduce the load of each drive and thereby facilitate transmission to the working body, reduce the total moment of inertia of the motor rotors.
Thus, we select engines (series 4A) with identical parameters from the reference book (therefore, further calculations will be performed for one engine):
4A355M12U3 (IP44),
R n = 110kW - rated power,
n = 500 rpm - synchronous rotation frequency,
s n = 0.02 - nominal slip,
Nominal efficiency,
- the moment of inertia of the rotor,
The multiplicity of the critical moment,
Multiplicity of starting torque,
Oe .; o.e.; o.e.; o.e.; o.e. - parameters of the equivalent circuit in p.u.
The rated speed of the motor is:
Motor rated torque:
(28)
In order for the motor not to overheat, it is necessary that the torque permissible for heating the motor (equal to the rated torque of the motor) be greater than or equal to the torque equivalent:
(29)
Thus, the selected motor is heated.
We check the correct choice of the motor in terms of overload capacity and starting conditions.
The drive starts up at idle speed, then:
(30)
Overload capacity:
(31)
where U = 0.9U n - we take into account a possible decrease in the supply voltage by 10%.
4.2 Calculation of parameters and selection of the power converter
It is required to select a frequency converter with the following characteristics:
Converter type - AIN PWM;
Control law - P = const;
Power supply: ~ 3 380V 50Hz;
Converter power - P = 75 kW.
We select the Omron 3G3FV A4750 CUE converter. Highly dynamic with great control depth. Starting torque up to 150% from 3 Hz. It features a vector control mode, the ability to work with full torque in the region of zero frequencies and improved dynamic characteristics: has the function of automatically detecting the parameters of the electric motor. 7 digital inputs (6 of them are programmable), 3 analog inputs (1 programmable) (0-10V or 4-20mA). 2 analog outputs for monitoring frequency or current. 2 programmable relay outputs (up to 1A). 2 opto-isolated outputs Built-in RS232 / RS485 / 422 + PID + Energy saving + neuro-Fuzzy + crane characteristics.
Table 1 - Characteristics of the converter
| Parameter |
Meaning |
| Power, kWt) |
|
| Input voltage (V) |
|
| Input Frequency (Hz) |
|
| Allowable voltage fluctuation |
from -15% to + 10% |
| Frequency range (Hz) |
|
| Output frequency resolution (Hz) |
|
| Engine management |
volt-frequency / vector with feedback |
| Carrier frequency (kHz) |
|
| Communication capabilities |
Modbus; Compo Bus / D (Device Net); Profibus DP Sysmac Bus; Interbus |
| Analog output (0-10V) |
|
| Number of fixed speeds |
|
| Analog speed reference |
|
| Acceleration / deceleration time |
from 0.01 to 6000 sec. |
| Degree of protection |
The frequency converter provides complete protection of the converter and the motor against overcurrent, overheating, earth leakage, and phase failure.
5. Calculation of static mechanical and electromechanical characteristics of the motor and drive
The mechanical characteristic is calculated by the formula:
(32)
where is the phase voltage across the stator;
Active resistance of the stator phase, Ohm;
Active resistance of the rotor phase, reduced to the stator circuit, Ohm;
Inductive resistance of the stator phase, Ohm;
Inductive resistance of the rotor phase, reduced to the stator circuit, Ohm;
s - slip;
Ideal idle speed (magnetic field).
The resistances of the stator phases and the reduced resistances of the rotor phases are calculated according to the reference data.
Basic resistance value:
(33)
where as the basic values of voltage and current we take the nominal values of the phase voltage and stator current:
Let us construct a natural mechanical characteristic according to the formula (41) using the mathematical package Mathcad, taking into account that, substituting, postponing the moment M along the x-axis, and the motor speed along the y-axis.
The natural mechanical characteristics of the engine are shown in Figure 8.
Figure 8 - Natural mechanical characteristic of the engine
Let's calculate the electromechanical characteristics of the engine.
As the base current value, we take the nominal value of the rotor current reduced to the stator circuit.
The dependence of the reduced rotor current on slip is determined by the formula:
(36)
The dependence of the stator current on slip is determined by the formula:
(37)
where is the relative rotor current;
The maximum value of the relative rotor current;
Relative magnetizing current;
Rated stator current.
Maximum value of relative rotor current:
(38)
where is the critical slip;
.(39)
Relative magnetizing current:
(40)
Rotor current:
(41)
We construct the natural electromechanical characteristic of the rotor current and the electromechanical characteristic of the stator current, using the mathematical package Mathcad, substituting, plotting the current I along the x-axis, and the motor speed along the y-axis.
Natural EMR of the engine are shown in Figure 9.
Figure 9 - Natural electromechanical characteristics of the engine
Since a PI regulator is used for speed regulation (it will be shown below), which gives zero static error, therefore, the mechanical characteristic of the drive will be absolutely rigid.
Figure 10 - Mechanical characteristics of the drive
6. Calculation of transient processes in an electric drive for a cycle of work
To obtain simpler transfer functions of regulators, it is necessary to go from a two-mass design scheme to a single-mass design scheme.
Justification for the transition to a single-mass settlement scheme:
Only motor variable feedbacks are used;
Natural vibration frequency:
Transition condition:.
As shown below, the T of the drive is 0.0258, then 
... Then the transition condition is fulfilled () and, therefore, one can go to a single-mass design scheme.
The total moment of inertia of a single-mass design scheme will be equal to:
A single-mass design scheme is shown in Figure 11.
Figure 11 - Single-mass design scheme
When regulating, the dependence of the torque permissible for heating the motor on the speed should repeat the dependence of the static torque on the speed.
To control the drive, we will use a two-loop automatic control system with volt / frequency control with sequential link correction, with an internal torque control loop and an external speed control loop.
With volt / frequency control, two control channels are organized: a supply frequency control channel and a voltage control channel. The speed is stabilized by adjusting the voltage as a function of frequency and as a function of load.
Consider a frequency control channel.
By expanding the equations of the dynamic mechanical characteristics in a series and linearizing the obtained equations in the vicinity of the point M = 0, s = 0, we obtain a linearized model of an induction motor, which is valid for.
Due to the fact that it is difficult to measure the motor torque in an asynchronous electric drive, disturbance control is used instead of torque control by deviation. Because the disturbing effect for the torque control loop is speed, then we will introduce a positive feedback on the speed, with the transmission coefficient.
We will regulate the engine speed by the deviation by introducing a negative speed feedback.
The block diagram of the frequency control channel is shown in Figure 12.
Figure 12 - Block diagram of the frequency control channel
Consider a torque control loop.
For static mode:
Zero error will be provided if:
.(44)
Maximum motor torque:
With volt / frequency control with:
(46)(47)
Electromagnetic time constant:
(48)
Mechanical stiffness:
(49)
The frequency transfer coefficient of the converter is determined by the ratio of the maximum signal at the output of the converter to the maximum signal at the output of the torque regulator:
.(50)
The maximum value of the limiting torque is equal to the critical moment of the natural characteristics of the engine:
From equation (45) we find K pm:
The torque regulator is represented as a P - regulator.
Gain Limit feedback providing torque control with zero error:
(53)
To calculate the velocity contour, we represent the torque contour in the form of a link:
By designating 
, we obtain the transfer function of the optimized torque control loop:
(55)
The block diagram of the speed control loop is shown in Figure 13.
Figure 13 - Block diagram of the speed control loop
The negative speed feedback sensor gain is calculated as the ratio maximum speed for the corresponding task voltage:
(56)
The small uncompensated time constant of the speed control loop is the electromagnetic motor constant, i.e. accept 
.
The large compensated time constant of the speed control loop is the mechanical constant of the motor.
To obtain a zero error in statics and forcing transient processes in dynamics, the speed controller must be a PI controller.
Let's tune the speed controller to the symmetrical optimum.
Desired transfer function of the speed loop set to symmetrical optimum:
.(57)
Transfer function of the controlled object:
(58)
Dividing the desired speed loop transfer function by the control object transfer function, we obtain the speed controller transfer function:
;
.
In order to remove overshoot, according to the command, it is necessary to put a filter with a time constant and the following transfer function at the input of the speed loop:
(61)
Calculation of transients is performed in the Matlab package.
In the model, we will use a single-mass conservative design drive scheme.
The drive model is shown in Figure 14.
Figure 14 - Drive model
The graphs of transient processes - the torque of the electromagnetic motor and the speed of the first mass, reduced to the motor shaft - are presented in Figures 15, 16.
Figure 15 - Graph of the transient process of the speed of the first mass
Figure 16 - Graph of the transient process of the electromagnetic moment
As a result of the simulation, it was obtained that the overshoot of the speed is:
7. Checking the correctness of the power calculation and the final choice of the engine
We will check the correctness of the power calculation using the method of average losses.
The total nominal motor losses are:
Variable rated motor losses are:
Then the constant losses will be equal:
Average losses for a cycle of work are equal:
(65)
where are losses in i-th moment time,
Cooling deterioration rate when operating at speed
T c = 6.9 s - cycle time.
Losses at the i-th moment of time can be determined from the following expression:
,(66)
where 
,
Engine load degree.
.(66’)
Substituting (66 ') in (65) we get:
(67)
Using expressions (67), we find the average losses per operating cycle.
To find the average losses by formula (67), we use the drive model.
First, we square the torque of the electromagnetic motor. Then we divide the resulting value by the square of the nominal torque and add. Then we integrate the resulting value and multiply by, we get the value of the average losses per cycle.
The model for finding the average losses per cycle is shown in Figure 17.
Figure 17 - Model for finding the average losses per cycle of work
As a result of modeling, it was obtained that the average losses per operating cycle are equal to:
.
Then the engine load factor is:
(68)
Thus, the engine is loaded at 80% (70%<80%<100%), следовательно, оставляем выбранный двигатель.
The choice of the electric motor and elements of the control system of the automated drive, which provides the desired range of rotation speed control at a given load diagram. Drawing up a schematic diagram and calculating static characteristics.
Saratov State Technical University
Department of AEU
Coursework on Electric Drive
"Calculation of the electric drive"
Saratov - 2008
1. Choosing an electric motor
2. Calculation of the parameters of the transformer
3. Choice of valves
4. Calculation of the parameters of the anchor chain
5. Calculation of the parameters of the control system
5.1 For the upper limit of the range
5.2 For the lower end of the range
6. Calculation of cutoff parameters
7. Construction of static characteristics
Conclusion
Application
1. Select the electric motor and the elements of the automated drive control system, which, at a given load diagram, provides a range of rotation speed control D = 75 with a relative error = 15%. When starting the engine and overloading, the torque must be kept within the range from M1cr = 85 Nm to M2cr = 115 Nm. Rated angular speed n = 1950 rpm.
2. Make a schematic diagram of the drive.
1. Choosing an electric motor
Let's calculate the equivalent moment using the load diagram:
Let's calculate the engine power:
Based on the power of the engine and the rated angular speed, we select the PBST-63 electric motor with the nominal parameters:
Un = 220 V; Pн = 11 kW; In = 54 A; nн = 2200 rpm; wя = 117; Rя = 0.046 Ohm; Rd = 0.0186 Ohm; ww = 2200; Rv = 248 Ohm.
Let's calculate the actual torque and engine parameters:
2. Calculation of transformer parameters
Secondary voltage and transformer power:
kc = 1.11-coefficient of the scheme
kz = 1,1-safety factor, taking into account the possible voltage drop
kR = 1.05 is a safety factor that takes into account the voltage drop in the valves and the switching of the current in the valves.
ki = 1,1-safety factor, taking into account the deviation of the current shape in the valves from the rectangular km = 1.92-scheme factor
Based on the voltage of the secondary circuit and power, we select the transformer TT-25 with nominal parameters: Str = 25 kW; U2 = 416 ± 73 V; I2ph = 38 A;
uк = 10%; iхх = 15%. Let's calculate the resistance of the transformer:
3. Choice of valves
Taking into account the speed control range, we select a single-phase electric drive control system. Average valve current:. Valve rated current:. kz = 2,2-safety factor, m = 2-factor, depending on the rectification circuit. Highest reverse voltage applied to the valve:
Valve rated voltage:
We select valves T60-8.
4. Calculation of the parameters of the anchor chain
The largest permissible value of the variable component of the rectified current:
Required armature inductance:
The total inductance of the motor and transformer is less than required, therefore, a smoothing choke with inductance must be included in the armature circuit:
Choke active resistance:
Active resistance of the armature circuit:
5. Combingt control system parameters
For the upper limit of the range
What corresponds to the adjustment angle According to the dependence, we determine the change in the EMF and the adjustment angle:
which in percentage terms:
Lower range limit:
Which corresponds to the adjustment angle
According to the dependence, we determine the change in the EMF and the angle of regulation:
In this case, the transmission coefficient of the converter is equal to:
The transmission coefficient of the SPFU is determined according to Fig. 2 Applications:
Overall system open-loop gain:
Largest open-state static error:
which in percentage terms:
Largest static error when closed:
Consequently, at the lower limit of the control range, the relative error is greater than the permissible one. To reduce the static error, we introduce an intermediate amplifier into the control system. Determine the required transmission ratio of the entire system in the open state:
Therefore, the transfer coefficient of the intermediate amplifier must be at least:
6. Calculation of cutoff parameters
As a Zener diode V1, we take a Zener diode D 818 (stabilization voltage Ust1 = 9 V Uy max = 11 V).
Current cutoff transfer ratio:
Stabilization voltage of the Zener diode V2:
The functional diagram of the electric drive is shown in Fig. 1 Appendices.
An integrated amplifier-limiter with zener diodes in the feedback circuit was used as an amplifier.
7. Plotting static characteristics
The limiting voltage is found from the static characteristics of the SPPC (Fig. 2 Appendix.):
Conclusion
In the course of calculating the course work, the methodology for calculating the parameters of the main components of an electric drive, such as an electric motor, a transformer, a pulse-phase control system and a thyristor converter, was studied. The static characteristic of the electric drive was calculated and built, which gives an idea of the speed of the drive with a change in the armature current of the electric motor, a load diagram that gives an idea of the load that the drive is experiencing during operation. Also, the schematic and functional diagrams were drawn up, giving an idea of the electrical elements included in the control system of the electric drive. Thus, a whole complex of calculations and constructions was implemented, which develops the student's knowledge and ability to calculate the electric drive, as a whole, as well as its main parts.
Application
Fig.1 Functional diagram of the electric drive.
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