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Power supply and electrical equipment for the metal-cutting machine shop. Calculation of workshop electrical loads Selecting the type of current and voltage value for the workshop network

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Essay

This course project for the course “Power supply industrial enterprises» consists of an explanatory note (49 pages); graphic part (2 sheets of A1 format); 28 tables; 3 drawings.

POWER TRANSFORMER, THERMAL PULSE, FUSE, STROBOSCOPIC EFFECT, BUS BAR, VACUUM SWITCH, SYNCHRONOUS MOTOR, SUPPORT INSULATOR.

Introduction

The purpose of this course project is to obtain new and consolidate existing knowledge, as well as to demonstrate creative abilities in the field of designing power supply for small workshops.

This course project (CP) is the final stage in studying the main course of the specialty “Power supply of industrial enterprises”.

During the design process, you will have to choose the configuration option for the workshop network at 0.4 kV. In the design version, it is necessary to determine short-circuit currents and select switching equipment, while ensuring that the power supply system has high technical and economic indicators and provides the appropriate degree of quality and the required degree of reliability of power supply to the designed facility.

Initial data for the course project

Figure number 1 (0.4 kV distribution network)

Option No. 2

Name of electrical receivers, their quantity and power

Name of electronic signature	Number on the plan	Power, kWt
Cylindrical grinding
Turning and turret
Vertical drilling
Semi-automatic lathe
Surface grinding
CNC lathe
Horizontal flow
Horizontal boring
Ventilation unit
Radial drilling
Centerless grinding
Lathe-screw-cutting
Sharpening and grinding
Heating furnace
Thermal oven
Electrothermal furnace
Ventilation unit
Point stationary
Butt welding
Welding seam rollers
Spot welding
Ventilation unit

1. Calculationthree-phase electrical loads in the 0.4 kV distribution network

Electrical loads are calculated using the calculation coefficient method. This method The calculation allows you to determine the electrical loads of electrical receivers with voltages up to 1000 V. Let us carry out the calculation for the electrical receiver of a “cylindrical grinding” machine.

Calculation algorithm

1) Rated power of the electrical receiver

2) Number of electrical receivers,

3) Using reference data, we will determine the values of utilization and power factors, as well as by;

4) Total power of a group of electrical receivers:

5) We determine the average active and reactive power of this group of electrical receivers:

6) Find the value of the quantity

We perform a similar calculation for all other types of electrical receivers, with the exception of the welding load. We summarize the obtained data in table No. 1

7) Let's calculate the effective number of power receivers:

8) Let's determine the weighted average utilization rate:

9) Determine the value of the calculated coefficient:

10) for the main busbar we have:

11) Define the values:

Taking into account lighting and welding loads:

We enter the obtained data into table No. 1.1

Ep name

Cylindrical grinding

Turning and turret

Vertical drilling

Semi-automatic lathe

Surface grinding

CNC lathe

Horizontal flow

Ventilation unit

Radial drilling

Centerless grinding

Lathe-screw-cutting

Sharpening and grinding

Heating furnace

Thermal oven

Electrothermal furnace

Ventilation unit

Horizontal boring

Lighting NG

Welding NG

Total for the workshop

Table 1.1 - Calculation of loads for selecting a workshop transformer and ShMA

2. Calculationweldingequivalent three-phase load

All resistance electric welding machines are single-phase with intermittent operation.

Electrical loads of resistance welding machines are calculated based on full power; the rms load is taken as the calculated heating load.

Table 2.1 - Initial data for calculating electrical loads of resistance welding machines

1. Load distribution over three pairs of phases (we start from the nominal values):

3. Determine the average power of each pair of phases:

6. The design power of all welding machines is determined by the two most loaded phase pairs:

7. We find the calculated active and reactive loads using the formulas:

3. Calculation of lighting load

Lighting is calculated based on the specific load per unit of production area:

Let's determine the workshop area:

where is the specific electrical load per unit of production area, kW/. Let us assume that the lighting is produced by fluorescent lamps with cos

The obtained values are entered into table No. 1

4. Crane load calculation

The crane has three motors: trolley, bridge, hoist.

Power ratios are 1:2:3. Crane power 50 kW

Trolley power:

Bridge power:

Lifting power:

Switching factors:

for trolley

for bridge

for lifting

Let's determine the engine power:

Let's determine the rated power of the crane:

The obtained values are entered into table No. 1.1

5. Selecting the number and power of a workshop transformertaking into account reactive power compensation

We use a single-transformer substation, because in the workshop there are power receivers that allow an interruption of power supply during the delivery of the warehouse reserve, i.e. for consumers of categories II and III, and they are also acceptable for small quantity(up to 20%) category I consumers.

Since mutual reservation is present, we will accept the load factor

The power selection of the KTP power transformer is made taking into account reactive power compensation.

The power of the transformer is determined by the active calculated load:

where is the number of transformers equal to 1;

Load factor equal to 0.8

taken from table No. 1

Select the transformer TM-1000/10-U1 with the following parameters: ;

Let us determine the reactive power that it is advisable to pass through a transformer into a network with a voltage of up to 1 kV:

The first component of the power of a capacitor bank in a network with voltages up to 1000 V:

The second component of the capacitor bank power, determined in order to optimally reduce losses in the transformer and reduce losses in the 10 kV network:

where - economic value = 0.25

We select standard compensating devices according to:

Let's determine the real load factor of the transformer taking into account the coefficient of efficiency:

Let's determine the losses in the transformer

Losses are determined using the following formulas:

6. Selection of trunk and distribution busbars

Selection of ShMA

We select the main busbar according to the design current. We select ShMA type ShMA-73 on.

Selection of SRA

Let's calculate the loads to select the welding joint. Let's compile a table of loads for calculating ShRA1,2 (tables No. 7.1-7.2)

The calculation algorithm is the same as for ShMA, but the calculated coefficient is found according to Table 1 (reference data) where Kr 1, reactive power is determined from the condition

for n: Qp = Qavg; Pр = Кр Рср

Based on the values of table No. for the calculated current. choose ShRA1 type ShRA-73 - 400

Based on the values of table No. for the calculated current. choose ShRA2 type ShRA-73 - 250

7. Selection of power points

Let's calculate the loads to select the joint venture. Let's compile a table of loads for calculating SP 1,2,3,4 (tables No. 7.3-7.6)

The calculation algorithm is the same as for the ShRA, the calculated coefficient is found according to Table 1 (reference data) where Kr 1, reactive power is determined from the condition

for n10: Qp =1.1 Qavg; Pр = Кр Рср

Let's check the strengthpoints on currents of outgoing lines

We select power points: No. 1. : ShRS1 - 54UZ for a rated cabinet current of 320 A with a number of outgoing lines of 8 and a rated current of fuses of 100 A, type PN2 - 100 (up to 100 A)

We select power points: No. 2: ShRS1 - 53UZ for a rated cabinet current of 250 A with a number of outgoing lines of 8 and a rated current of fuses of 60 A, type NPN - 60 (up to 63A)

Let's check the currents of the outgoing lines, take the most powerful receiver taking into account tg

(sharpening grinder) and determine its rated current:

We select the power point: No. 3: ShRS1 - 28 UZ for a rated cabinet current of 400 A with the number of outgoing lines 8 and rated fuse current: 2x60 + 4x100 + 2x250 A type PN2 - 100 (up to 100 A), NPN2-60 (up to 63 A) , PN2-250 (up to 250A)

Let's check the currents of the outgoing lines, take the most powerful receiver taking into account Ki (heating furnace) and determine its rated current:

We select the power point: No. 4: ShRS1 - 54UZ for a rated cabinet current of 320 A with a number of outgoing lines of 8 and a rated current of fuses of 100 A, type PN2 - 100 (up to 100 A)

Let's check the currents of the outgoing lines, take the most powerful receiver taking into account tg (Electrothermal furnace) and determine its rated current:

Selected power points are selected correctly

Table 7.1 - Calculation of SRA - 1.

Name of electronic signature

Cylindrical grinding

Turning and turret

Vertical drilling

Ventilation unit

Table 7.2 - Calculation of SRA - 2.

Name of electronic signature

Semi-automatic lathe

Surface grinding

CNC lathe

Horizontal flow

Horizontal-growing

Table 7.3 - Calculation of SP-1.

Name of electronic signature

Radial drilling

Centerless grinding

Lathe - screw-cutting

Table 7.4 - Calculation of SP-2.

Table 7.5 - Calculation of SP-3.

Name of electronic signature

Heating furnace

Thermal oven

Table 7.6 - Calculation of SP-4.

Name of electronic signature

Electrothermal furnace

Ventilation unit

Selection of power points of the welding department

Selection of power point No. 5

Let's create a load table (table No. 7.7)

Table 7.7 - Calculation of SP No. 5

Name of electronic signature
Point stationary
Spot welding

Calculation algorithm

2. Let’s determine the average loads of each machine:

Load factor of the i-th welding machine;

Switching factor of the i-th welding machine.

AB:

4. Determine the root mean square power of each welding machine:

AB, is determined by the formula:

We select power point No. 5: ShRS1 - 53UZ for a rated cabinet current of 320 A with a number of outgoing lines of 8 and a rated current of fuses of 60 A, type NPN2 - 60 (up to 63A)

Let's determine the rated current for one machine - point stationary with a maximum:

The power point is selected correctly

Selection of power point No. 6

Let's create a load table (table No. 7.8)

Table 7.8 - Calculation of SP No. 6

Calculation algorithm

1. Distribute the loads among three pairs of phases:

2. Let’s determine the average loads of each machine:

Load factor of the i-th welding machine;

Switching factor of the i-th welding machine.

3. Let's determine the average power of each pair of phases, for example, AB:

4. Determine the root mean square power of each welding machine:

5. RMS load of each phase pair, for example, AB, is determined by the formula:

6. The design power of all welding machines is determined by the 2 most loaded phase pairs:

7. Determine the estimated active and reactive and apparent power:

In addition to the welding load, two ventilation units are connected to SP-6, with We sum up the welding load and the load of the ventilation units.

We select power point No. 6: ShRS1 - 53UZ for a rated cabinet current of 320 A with a number of outgoing lines of 8 and a rated current of fuses of 60 A, type NPN2 - 60 (up to 63A)

Let's check the power point for currents in outgoing lines:

Let's determine the rated current for one machine - welding - butt with maximum:

The power point is selected correctly

8. Selecting cables and jumper cables

The cross-section of the workshop network cable cores is selected for heating with a long-term rated current according to the condition:

where is the calculated current, A;

long-term permissible current of a given cross-section, A.

rated power of the electrical receiver, kW;

rated power factor of the electrical receiver.

For asynchronous motors with a squirrel-cage rotor the following conditions must be met:

for furnaces and welding machines:

For the design current for welding machines we take the root mean square current:

Table 8.1 - Selection of cables for electric motors with short-circuit motors. the rotor is the drive.

Name of electronic signature
Cylindrical grinding
Turning and turret
Vertical drilling
Semi-automatic lathe
Surface grinding
CNC lathe
Horizontal flow
Horizontal boring
Ventilation unit
Radial drilling
Centerless grinding
Lathe-screw-cutting
Sharpening and grinding
Ventilation unit
Ventilation unit

Table 8.2 - Selection of cables for ED thermal separation

Table 8.3 - Selection of cables for the ED welding department

Table 8.4 - Selection of cables and cable jumpers between ShMA and ShRA,SP,

Busbar name
ShMA-ShRA - 1
ShMA-ShRA - 2
ShMA-SP - 1
ShMA-SP - 2
ShMA-SP - 3
ShMA-SP - 4
ShMA-SP - 5
ShMA-SP - 6

Let's check the cable for permissible voltage loss:

Let's check the cable for the cylindrical grinder:

rated current of the cable line, A;

cable line length, km;

linear active and reactive resistance of cables,

number of cables laid in parallel.

We enter the data in tables No. 8

Table 8.5 Checking cable lines for voltage loss.

Name of electronic signature
Cylindrical grinding
Turning and turret
Vertical drilling
Ventilation unit
Semi-automatic lathe
Surface grinder
CNC lathe
Horizontal flow
Horizontal-growing
Radial - drilling
Centerless grinding
Lathe - screw-cutting
Sharpening and grinding
Heating furnace
Thermal oven
Electrothermal furnace
Ventilation unit
Ventilation unit

All cables are tested.

Table 8.6 Checking cable lines from ShMA to SP welding department

Name of foreign wire

All cables are tested

Table 8.7 Checking the cable lines of the welding department for voltage loss.

Name of electronic signature
Point stationary
Spot welding
Butt welding
Welding suture roller

All cables are tested

9. Calculation of short circuit currents

We carry out the calculation for the two most electrically distant power receivers. This is a radial drilling machine (No. 45) connected to SP-1, and a ventilation unit (No. 42) connected to ShRA-1.

Figure No. 9.1 Single-line diagram for calculating short-circuit currents

Let's determine the parameters of the equivalent circuit

The resistance of straight cable lines is determined by the formula:

linear active and reactance of cable lines, respectively, .

length of cable lines, m.

number of parallel laid cables, pcs.

Zero sequence resistance of cable lines:

Table No. 9.1 Calculation of resistances of direct and zero sequence cable lines

Name of CL

Positive sequence resistance of trunk and distribution busbar:

Zero sequence resistance of the main and distribution busbar:

Table No. 9.2 Calculation of positive and zero sequence busbar resistances for various short circuit points

The transformer resistance is determined by the formula:

short circuit losses in the transformer, kW;

rated voltage on the secondary winding, kV;

rated power of the transformer, kVA;

transformer short circuit voltage, %.

From the reference book we find the resistance of circuit breakers and fuses:

for switches Electron E16V with

for switches VA 0436 with 400 A

for switches VA 0436 with 160 A

Contact resistance of busbar connections:

ShMA (K2, K3) 9 sections of 6 meters each

ShMA(K4,K5) 1.7 sections of 6 meters

ShRA (K4, K5) 18 sections of 3 meters each

Contact resistance of connecting cables (we take into account 2 contacts per cable):

Figure No. 9.2 Equivalent circuit for calculating short-circuit currents.

Calculation of single-phase and three-phase short circuit currents

The three-phase short circuit current is determined by the formula:

The single-phase short circuit current is determined by the formula:

average rated voltage of the network, V, where the short circuit occurred;

the total active and inductive resistances of the positive sequence equivalent circuit, respectively, relative to the short-circuit point, including the resistances of busbars, devices and transition resistances of contacts, starting from the neutral of the step-down transformer, mOhm;

the same, zero sequence.

The zero-sequence resistance of a transformer with a low voltage of up to 1 kV in the connection diagram of the Tr-11 windings is assumed to be equal to the positive sequence resistance.

We calculate the three-phase short circuit current at point K1.

We believe that the short circuit is at the beginning of the ShMA because it is necessary to calculate the maximum value of the short-circuit current

The total active resistance is:

The total reactance is equal to:

The three-phase short circuit current is equal to:

We calculate the single-phase short circuit current at point K1.

Determine the single-phase short circuit current. We find the resistances of the reverse (equal to the direct because there are no rotating machines) and zero sequence. It should be noted that the positive sequence resistance must take into account the active arc resistance. We take into account the influence of the active resistance of the arc on the short circuit by multiplying the calculated short-circuit current, found without taking into account the arc resistance at the short-circuit location on the correction factor K s, which depends on the resistance of the short-circuit circuit.

For all other points we find the short-circuit current without taking into account the arc.

We believe that the short circuit is at the end of the ShMA because it is necessary to calculate the minimum value of the short-circuit current.

Then, taking into account the arc resistance, we have a single-phase short circuit current.

For all other points we perform a similar calculation. We summarize the results in table No. 8.3

Table 9.3 Calculation of short-circuit currents

10. Calculation of starting and peak currents.

Calculation of starting currents

The inrush current is determined for receivers having an IM with a squirrel-cage rotor to check the fuse inserts.

The starting current of the receiver is determined by the formula:

Normal electric current, which is determined by the following formula:

Multiplicity of starting current, since there is no data, let’s assume: = 5

Table No. 10.1 Starting current values for receivers with IM

Name of electronic signature
Cylindrical grinding
Turning and turret
Vertical drilling
Semi-automatic lathe
Surface grinding
CNC lathe
Horizontal flow
Horizontal boring
Ventilation unit
Radial drilling
Centerless grinding
Lathe-screw-cutting
Sharpening and grinding
Ventilation unit
Ventilation unit

Peak current calculation

Determination of peak currents of trunk, distribution busbars and substations

To calculate the peak currents of trunk, distribution busbars and substations, use the following formula:

I p - calculated current ShMA, ShRA, SP, A;

I p.ma x - starting current of the highest power electric device connected to ShMA, ShRA, SP, A;

K and - utilization factor of the largest electric power unit, A;

I n. max - rated current of the highest power electric device.

Calculation of the peak current of the ShMA

Let's determine the rated current of the receiver with the highest power (in this case it is a CNC lathe with K and = 0.2):

Maximum rated current of the load node (LMA), taking into account reactive power compensation;

Calculation of peak current ShRA-1

The largest power receiver is a vertical drilling one with

Maximum rated current ShRA-1

Calculation of peak current ShRA-2

The largest power receiver is a CNC lathe with

Maximum rated current ShRA-2

Calculation of peak current SP-1

The largest power receiver is a radial drilling machine with

Maximum rated current SP-1

Calculation of peak current SP-2

The largest power receiver is a turret lathe with

Maximum rated current SP-2

Calculation of peak current SP-4

In addition to the ventilation unit, SP-4 powers electrothermal furnaces, the peak current of which practically does not differ from the nominal one, so we use the motor power of the ventilation unit with

Maximum rated current SP-4

Calculation of peak currents of resistance electric welding machines

Resistance electric welding machines are consumers with sharply variable operating modes and create peak loads with high frequency, as a result of which voltage fluctuations occur in the network.

The peak power of the machine at the time of welding is determined by the formula:

The calculated peak of any pair of phases, for example phase AB, is determined by the formula:

Where is the number of simultaneously working machines, determined from the probability curves

Number of machines connected to a given phase pair

When determining, the weighted average is calculated

The peak load for a linear wire is determined by the formula, corresponding to the peaks of two phase pairs, for example in phase B:

Where, is the peak load for a pair of phases AB and for a pair of phases BC

Peak Line Current:

Where is line voltage, kV

Calculation of peak current SP-5

Table 10.2 Calculation of SP No. 5

6. Let’s determine the peak power of the most loaded phase from the two most loaded pairs of phases, therefore the most loaded phase B:

Let's determine the peak current

Calculation of peak current SP-6

Table 10.3 Calculation of SP No. 6

Calculation algorithm

1. Distribute the loads among three pairs of phases:

2. Determine the peak power of each group of machines:

3. In each pair of phases we find the weighted average switching factor:

The curves determine the number of simultaneously operating machines m out of the total number n in each pair of phases:

5. In each pair of phases, select the machines with the highest peak power in accordance with the resulting number of simultaneously operating machines m, determine the total value of the peak power in each pair of phases:

6. Let’s determine the peak power of the most loaded phase based on the two most loaded phase pairs:

Let's determine the peak current

But in addition to the welding load, SP-6 powers two ventilation units, so we will determine the starting current of the IM ventilation units.

Motor power of the ventilation unit with

Maximum rated current SP-6

i.e., the starting current turned out to be less than the welding current, therefore, in the future we focus on the peak welding current.

11 . Protection of workshops electrical networks

In networks with voltages up to 1000 V, protection is provided by fuses and circuit breakers.

The fuse is designed to protect electrical installations from overloads and short-circuit currents. Its main characteristics are: rated current of the fuse link, rated current of the fuse, rated voltage of the fuse, rated disconnecting current of the fuse, protective (ampere-second) characteristic of the fuse.

Designations in the calculation:

Rated network voltage, kV;

Maximum short-circuit current networks, A;

Maximum rated current, A;

Motor starting current, A.

Long-term permissible current of the protected section of the network;

Minimum short-circuit current

Calculation algorithm

Let's look at the example of choosing a fuse for a cylindrical grinding machine (No. 1).

Select fuse type NPN - 60 s; ;

Since the fuse is selected for an individual receiver, the rated current is taken as the calculated current:

4) where 46.6 = 233 A;

Overload factor, which takes into account the excess of the motor current above the rated value in the starting mode, assumed to be 2.5 - for light starting conditions.

i.e. = 93.2 A - the selected fuse is not suitable. Let's choose a fuse of type PN-2 100 s = 50 kA; ; , Where

The fusing currents of the inserts must correspond to the multiples of permissible continuous currents (coordination with the cross-section):

Checking the fuse for:

6) - for sensitivity

7) - for breaking capacity

50 kA 5.01 kA, where = = 5.01 kA

Select fuse type PN-2 100: = 50 kA; ;

Using this algorithm, we select fuses and summarize the selection in table No. 11.1

Table No. 11.1 Selection of fuses for electric motors driven by an IM with a short-circuit rotor

Name of electronic signature

Cylindrical grinding

Turning and turret

Vertical drilling

Semi-automatic lathe

Surface grinding

CNC lathe

Horizontal flow

Horizontal boring

Ventilation unit

Radial drilling

Centerless grinding

Lathe-screw-cutting

Sharpening and grinding

Ventilation unit

Table 11.2 - Selection of fuses for ED thermal compartment

Table 11.3 - Selection of fuses for ED welding department

Name of electronic signature

Point stationary

Spot welding

Butt welding

Welding seam rollers

1 2 . Selection of circuit breakers

Let's write down the conditions for selecting circuit breakers:

where is the highest calculated load current;

Rated current of the circuit breaker release.

peak current of a group of electrical receivers, A

3) Detuning from long-term permissible currents:

For circuit breakers with only electromagnetic release (cut-off):

4) Detuning from minimum short circuit currents:

5) Breaking capacity test:

Let's look at the example of choosing a switch for ShMA (SF1).

Table No. 12.1 Selection of circuit breakers

Installation location

Calculation data

Passport details

Switch type

E25V: - ШМА

VA 04-36: - ShRA1

VA 04-36: - ShRA2

VA 04-36: - SP1

VA 04-36: - SP2

VA 04-36: - SP3

VA 04-36: - SP4

VA 04-36: - SP5

VA 04-36: - SP6

Listusedliterature

1. Burnazova L.V. Guidelines for completing the course project. Mariupol 2010

2. Block V.M. A manual for coursework and diploma design, second edition, revised and expanded. Moscow “Higher School” 1990.

3. Neklepaev B.N. Electrical part of power plants and substations. - M.: Energoatomizdat, 1986.

4. GOST 28249-93 Interstate standard “Short circuits in electrical installations up to 1000 V”.

5. Fedorov A.A., Starkova L.E. Tutorial for course and diploma design on power supply of industrial enterprises. Textbook for universities - M. "Energoatomizdat", 1986.

6. Gaisarov R.V. Selection of electrical equipment. Chelyabinsk 2002

7. Internet media

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To calculate the workshop load we use the method of ordered diagrams. This method is used for mass electrical receivers. It establishes a connection between the workload and the operating mode of electrical receivers based on a probabilistic scheme for generating a group load schedule.

General information about calculating electrical loads

The load of industrial enterprises or individual workshops usually consists of electrical receivers of various capacities. Therefore, all electrical receivers in the workshop are divided into groups of receivers of the same type of operation, with characteristic subgroups of electrical receivers with the same power utilization rates and power factors identified in each group.

When determining electrical loads, we use the maximum electrical load utilization factor method. This method establishes a connection between the design load and the operating modes of electrical receivers (ER) based on a certain probabilistic scheme for generating a group load graph. The method is used as the main one for mass ED.

The procedure for determining design loads:

All electrical receivers are divided into groups according to the value of the utilization factor K and, power factor cos, rated active power Рн. We determine the utilization factor and power factor from Table 4.10 2, and determine tg from the value of the power factor.

We count the number of ES in each group and for the object as a whole.

In each group, indicate the minimum and maximum powers at PV=100%, if PV<100%, то номинальная мощность определится по формуле:

where: P pass- EP power according to the passport, kW;

PV - duration of switching on.

The total power of all electric devices is calculated using the formula:

P n=P neither ; (2)

For each supply line, the power assembly indicator m is determined using the formula:

where: - rated power of the maximum consumer, kW;

Rated power of the minimum consumer, kW.

Average loads for the busiest shift of power electric drives of the same operating mode are determined by the formulas:

where: P cm- average active power of one or a group of receivers for the busiest shift, kW;

R nom- we take the rated power of electrical receivers according to Table 1, kW;

TO And- utilization factor, take according to table 4.10 2;

Q cm- average reactive power of one or a group of receivers for the busiest shift.

For several groups of electrical receivers we determine by the formula

We determine the average utilization rate of the EP group K using the formula:

The effective number of electrical receivers is determined by formulas based on the following relationships.

For n5, Кis 0.2, m3 and Р nom const ne is determined by the formula:

Formula 9 can also be used when none of the cases listed below are suitable for calculation.

For n >5, К is 0.2, m 3 and Р nom const we accept ne=n.

For n >5, K is 0.2, m< 3 и Р ном const принимаем nэn.

For n 5, K is 0.2, m 3 and P nom const ne is determined by the formula:

where: n* E is the relative value of the number of EPs, the value of which can be found in the table based on the dependence n* E = f(n*; P*).

Using formula 10, n* is found:

where: n 1 - the number of EPs in the group, the power of each of which exceeds the maximum power of the EPs of this group divided by 2.

P* is determined by the formula:

P nom- maximum unit power of the electric group, kW;

R nom1- the total rated power of a group of electrical receivers whose power exceeds the maximum power of a given group of electrical equipment divided by 2, kW.

The maximum active power is determined by the formula:

Where: TO m - maximum coefficient is determined according to table 3.2 5;

R nom - rated power of the electrical receiver.

Maximum reactive power is determined by the formula:

where: - maximum reactive power factor, at n E? 10 =1, at n E<10 -=1,1

The total maximum power is determined by the formula:

The maximum current is determined by the formula:

Distributing the load:

RP-1: EP No. 1,2,3,4,5,6,7;

RP-2: EP No. 17,18,19,21,22,23;

RP-3: EP No. 8,9,12,13,14,15;

RP-4: EP No. 23,24,25,26,29,30,31;

RP-5: EP No. 10,11,16,27,28;

Determination of the design load of the workshop

For example, consider determining the load on RP-1.

table 2

1) We determine the average load of the electric unit for the busiest shift using formulas (6), (7):

P cm.1 = 0.65 · 2 · 3 =3.9 kW; Q cm.1 = 0.75 · 3.9 = 2.92 kVAr;

P cm.2 = 0.35 · 2 · 76 · v0.65 =42.9 kW; Q cm.2 = 1.73·42.9=74.2 kVAr;

P cm.3 = 0.12 · 1 · 4.4 =0.53 kW; Q cm.3 = 2.29·0.53=1.21 kVAr;

P cm.4 = 0.2 1 3 = 0.6 kW; Q cm.4 = 1.17· 0.6= 0.7 kVAr;

P cm.5 = 0.1 1 115.5 v0.4 =7.3 kW; Q cm.5 = 1.73· 14.6 = 12.6 kVAr.

2) Define K and groups using formula (8):

3) The power assembly indicator according to formula (3) will be equal to:

4) Since n > 5, TO and > 0.2, m>3, then n e =n=7

5) The maximum coefficient is determined according to table 4.3 2. A more accurate value of Km is determined using the interpolation method:

6) Maximum active and reactive power is determined by formulas (13) and (14):

P max = 1.89 55.22 = 104.36 kW.

Because n E<10, то принимаем значение К" М = 1,1:

Q max = 1.1 91.67= 100.84 kVAr.

We find the total maximum power using formula 15:

The calculated current is determined by formula 16:

Similarly, we determine the calculated load for the remaining receivers and enter the calculation results in Table 2.

1) We divide all electrical equipment of the workshop into groups with the same operating modes and determine the total rated power of the workshop:

2) Determine the power assembly indicator:

3) Determine the total load of the workshop for the busiest shift:

4) Determine the load utilization factor of the workshop electrical equipment:

5) Since n > 5, TO and > 0.2, m> 3, then n e =31.

6) The maximum coefficient is determined according to table 4.3 2. A more accurate value of Km is determined using the interpolation method:

where: K and1 K and2, K m1, K m2 - boundary values of the coefficients K and and K m.

We determine the calculated active and reactive powers:

So, we take the value:

8)Full design power:

9) Rated current:

The results of all calculations are recorded in Table 2.

table 2

Coeff. maximum	Max. active power	Max.reagent- rated power Q MAX, kvar	Max. full power	Coeff. Use	Effect. number of EP n E

Workshop lighting calculation

According to research, in modern conditions the use of LED spotlights and industrial lamps in production workshops is very effective, since it meets all operating requirements. They are also an economical solution, as they allow you to reduce electricity costs by about 2.5 times. LED spotlights with a narrow luminous flux distribution pattern are especially effective. The most common and universal industrial lamps.

Industrial LED lamps have a number of undeniable advantages, which include:

* they provide high efficiency;

* are highly resistant to temperature changes;

* do not emit mercury vapor or other harmful substances;

* Have high moisture resistance and dust protection;

* can be used in difficult climatic conditions, where they provide instant switching on and stable operation;

* economical in terms of maintenance of electrical networks;

* easy to install;

* do not require special maintenance;

* have a long service life

When choosing light sources, you should take into account their advantages, disadvantages, and their cost-effectiveness.

Compared to incandescent lamps, fluorescent lamps have a more favorable emission spectrum, 4-5 times greater luminous efficiency, longer service life and significantly lower glare. However, fluorescent lamps require starting equipment; they create a pulsating light flux, do not light well at low temperatures, and are less reliable.

Let's determine the luminous flux necessary to create normal working lighting in the workshop. To calculate, we use the luminous flux utilization coefficient method.

Task lighting is the main type of lighting. It is intended to create normal vision conditions in a given room and is carried out, as a rule, with general lighting lamps.

Emergency lighting is used to continue work or evacuate people when the working lighting goes out. It must provide workplace illumination of at least 5% of that established for normal conditions. Workshop dimensions - 36 x 24 m.

For lighting we will use industrial LED lamps

GSSN-200, the parameters of which are specified in the application.

Let's calculate the lighting of the workshop:

The height of the room is 7 m. The height of the calculated surface above the floor is h p = 1.5 m. The calculated height can be determined by the formula:

H P = h p - h p - h c m.; (18)

H P = 7 - 1.5 -1 = 4.5 m;

To determine the distance between rows of lamps, we use the formula:

L = Н Р L wholesale, m.; (19)

where: L opt - lighting technically the most advantageous optimal relative distance between lamps, table. 2.1 [L.7]

L = 4.5 1.2 = 5.4 m;

L opt =0.8h1.2-deep

Then the number of rows of lamps can be determined by the formula:

where: B is the width of the design room, m.

Let's take the number of rows of lamps n p = 5.

We determine the actual distance between the rows using the formula:

where: L ST.V - distance from the outermost row of lamps to the wall, (m). We accept L ST.V = 2 m.

The number of lamps is determined as:

where Ф 1 is the flux of lamps in each lamp.

Coefficient z, characterizing illumination unevenness, for LED lamps z = 1.

To determine the utilization coefficient, the index of the room i is found and the reflection coefficients are presumably estimated: ceiling - p, walls - s, calculated surface or floor - p, (Table 2.13[L.7]) Determine. The index is found by the formula:

where: A is the length of the design room, m.

According to table 2.15 [L.7] we determine = 37%

We take the safety factor k equal to k = 1.5 (according to table 2.16 [L.7])

The area of the room is determined by the formula:

S = A B, m 2 (23)

S = 36 24 = 864 m2

The specified minimum illumination is determined from the table. 4-1 [L.3] for visual work of average accuracy, general illumination E = 200 lux.

For lighting we use GSSN-200 lamps with a luminous flux of 24,000 lm. Let's determine the number of lamps using formula 21:

Then the number of lamps in the row. We accept N St. row = 7 N St. = 35.

Let's find the distance between lamps in one row using the formula:

where: A is the length of the room without taking into account the thickness of the walls,

L A. ST - the distance from the first lamp in the row is determined by the formula:

The layout of lighting fixtures throughout the workshop is shown in Figure 3.

Active installed lighting power:

P mouth = N Р o.p., (27)

where: P o.p. - lamp power, 200 W;

P mouth..=35 200 = 7 kW

Reactive installed lighting power:

where: tg = 0.25 for LED lamps.

Let's determine the total lighting power:

Calculation of the total load of the workshop

Total design power of the workshop including lighting:

Estimated current of the workshop taking into account lighting:

INTRODUCTION

The purpose of the section “Power supply and electrical equipment of an industrial enterprise” of the final qualifying work is to systematize, expand and consolidate theoretical knowledge in electrical engineering, electrical machines, electric drives and power supply of industrial enterprises, as well as the acquisition of practical skills in solving problems necessary for a future specialist.

The power supply system of an industrial enterprise must ensure an uninterrupted supply of electricity to consumers while meeting the requirements for efficiency, reliability, safety, quality of electricity, availability of reserve, etc.

The selection of modern electrical equipment, development of control circuits, protection, automation, signaling of electrical receivers, development of a power supply circuit for the workshop and (or) the entire enterprise using progressive technical solutions are the tasks of the section “Power supply and electrical equipment of an industrial enterprise” of the final qualifying work.

The section “Power supply and electrical equipment of an industrial enterprise” of the final qualifying work includes consideration of the following issues:

5) select the number and type of workshop transformers 10/0.4 kV;

6) select switching equipment for the 0.4 kV network and 10 kV network;

7) calculate the costs of constructing a power supply network;

8) calculate the grounding loop of the transformer substation;

9) consider the use and operation of isolated bus systems.

The initial data for the electrical part of the final qualifying work are the production (energy) equipment and mechanisms necessary to ensure the technological processes specified in the technical specifications, as well as the area of the production premises of the workshop (enterprise), the parameters of installed electrical receivers, existing circuits of the power supply system, etc. Indicated automation object.

In the explanatory note of the final qualifying work, the electrical part is drawn up as a separate chapter. The volume and content of the graphic part are determined by the design assignment. The graphic part contains a power supply diagram for the enterprise (workshop).

Option 14

Calculation of the workshop power supply network

1.1 Initial data for design

The schematic plan of the enterprise is set on a scale of 1:1000

Table 1 specifies the rated powers of electrical receivers, utilization and start-up factors, power factors of the specified electrical receivers, and the lengths from electrical receivers to ShS-1.

Table 1 - Initial data for the first stage

№	Power receiver	N pcs.	Pnom kW	Ki	cos𝜑	Kp	PV%	L m
				0,16	0,61	5,35	-
	Slotting machine			0,14	0,43	6,40	-
	Overhead crane			0,1	0,5	6,79
	Lathe			0,4	0,75	5,58	-
	Exhauster		5,6	0,63	0,8		-
Average value	0,6

The design loads of the power cabinets of workshop No. 4, the weighted average utilization factor and the number of effective electrical receivers are specified. This information is presented in Table 2.

Table 2 - Initial data for the second stage

№	Closet	P kW	Q kVAR	cos𝜑	Nef	K.av.vzv
	ShS-2		36,62	0,88		0,6
	ShS-3		21,05	0,88		0,54
	ShS-4		51,82	0,88		0,4
	ShS-5		23,73	0,86		0,8
	ShS-6		30,60	0,87		0,7
	ShS-7		13,49	0,88		0,7
	ShS-8		58,74	0,86		0,86
Average value	0,87

The initial data are the calculated capacities of the remaining workshops at the specified enterprise, the length of the supply cable is 10 kV from the main production point to the distribution point. The data is shown in Table 3.

Table 3 - Initial data for the third stage

The plan of an industrial enterprise is shown in Figure 1.

Figure 1 - Industrial plant plan

Calculation of electrical loads of ShS-1 consumers

The first and main stage of designing a power supply system for an industrial enterprise is determining the calculated values of electrical loads. They are not a simple sum of the installed capacities of electrical receivers. This is due to the incomplete loading of some electronic devices, the non-simultaneity of their operation, the probabilistic random nature of turning on and off the electronic devices, etc.

The concept of “design load” follows from the definition of the design current, according to which all network elements and electrical equipment are selected.

The calculated current is the constant average current over a 30-minute time interval that leads to the same maximum heating of the conductor or causes the same thermal wear of the insulation as a real variable load.

Table 5 - Calculation of load ShS-1

Initial data	Calculation data
Naim EP	N pcs	Est. Power kW	Ki	Coefficient react	Wed.Change.Power	Ne	Kmax	Design power
1 EP	∑		cos𝜑	tg𝜑	Pcm kW	Qcm kvar	Ne	Kmax	Calculation kW	Qcalc kvar

Group A
Slitting machine				0,16	0,61	1,29	2,24	2,88	-	-	-	-
Slotting machine				0,14	0,43	2,09	1,96	4,09	-	-	-	-
Overhead crane				0,1	0,5	1,72		24,08	-	-	-	-
Lathe				0,4	0,75	0,88		10,56	-	-	-	-
Total				0,8	-	-	30,2	41,61		2,31	69,76	45,77
Group B
Exhauster		5,6	11,2	0,63	0,8	0,75	7,05	5,2	-	-	-	-
Total		5,6	11,2	-	-	-	7,05	5,2	-	-	7,05	5,2

Table 6

Parameter	cosφ	tgφ	Pm, kW	Q M, quar.	S M , kV*A
Total on NN without CG	0,83	0,68	495,81	287,02	572,89

The design power of the heat treatment unit is determined.

Q k.r = α R m (tgα – tgφ k) = 0.9“495.81“(0.68 – 0.29) = 174.02 kvar.

Cosφ k = 0.96 is accepted, then tanφ k = 0.29.

We find the transformer load after compensation and its load factor:

For installation, we select an automated capacitor unit type 2 AUKRM 0.4-100-20-4 UHL4

The current of the compensating device is found by the formula:

where 1.3 is the safety factor (30% of the nominal value);

Line voltage, 0.4 kV.

Since we have 2 bus sections with a sectional switch, the power of the heat exchanger for each section will be determined by the load of each section. In the first section, power cabinets 1,2,3,4 will be connected; in the second section 5,6,7,8 will be connected.

Table 7

where is the weighted average power factor of all loops;

Required power factor on transformer buses (not less than 0.95).

where k is the coefficient obtained from the table in accordance with the values of power factors and ;

Section 1 requires more reactive power compensation due to ShS-1, which has a low power factor.

Total amount of compensated reactive power in both sections

For two transformer substations rated power

transformer is determined by the condition of permissible overload of one

transformer by 40%, subject to emergency shutdown of another within 6

hours per day for 5 working days.

In this case, the rated power of the transformer TP-10/0.4

determined by the expression:

where k=1.4 permissible transformer overload coefficient;

n=2 – number of transformers at the substation.

From a number of standard rated powers we select two

transformer TMG-400/10.

Reference data for the transformer is given in Table 8.

Table 8 – Passport data of the TMG-400/10 transformer

Snom, KVA	Unom, kV	∆Рхх, kW	∆Ркз, kW	Ukz, %	Iхх,%	dimensions	Weight, kg
		0,8	5,5	4,5	2,1	1650x1080x1780

Losses of active and reactive power in transformers at TP:

where n is the number of installed transformers, pcs;

– no-load losses in the transformer, kW;

– losses due to short circuit in the transformer, kW;

– rated power of the transformer, kVA.

where Iх.х – transformer no-load current, %;

Us.c – short circuit voltage, %.

The total power of the electrical receivers of the workshop, taking into account losses in

transformer:

Since the calculated power of 370.11 kVA satisfies the selected

rated power of the transformer, then we select 2 transformers TMG-400/10. And after recalculation when choosing centralized compensation, we connect the capacitor bank to the 0.4 kV busbars of the workshop substation. And as can be seen from the calculation, in this case the transformers of the main step-down substation and the supply network are unloaded from reactive power. In this case, the use of the installed capacitor power is the highest.

Individual compensation is most often used at voltages up to 660 V. This type of compensation has a significant drawback - poor use of the installed power of the capacitor bank, since when the receiver is turned off, the compensating installation is also turned off.

In many factories, not all equipment is running at the same time; many machines are used for only a few hours a day. Therefore, individual compensation becomes a very expensive solution when there is a large amount of equipment and a correspondingly large number of installed capacitors. Most of these capacitors will not be used for a long period of time. Individual compensation is most effective when most of the reactive power is generated by a small number of loads that consume the most power over a sufficiently long period of time.

Centralized compensation is used where the load fluctuates (moves) between different consumers during the day. At the same time, reactive power consumption varies throughout the day, so the use of automatic capacitor units is preferable to unregulated ones.

Load recalculation

Column 13 records the maximum reactive load from power

ED node Qcalc, kVar:

since ne< 10, то

Total maximum active and reactive loads according to the design

to the unit as a whole for electric drives with variable and constant load schedules

are determined by adding the loads of ED groups according to the formulas:

The maximum full load of power electric drives Scalc.uch, kVA is determined:

The calculated current Icalc, A is determined:

We will calculate currents and total power before installing the heat exchanger and after installing the heat exchanger.

Table 9 - Summary sheet before and after installation of the heat exchanger on TP buses

№	S, kVA	cos𝜑	I, A
BEFORE	AFTER	BEFORE	AFTER	BEFORE	AFTER
ShS-1	92,18	77,68	0,6	0,96	140,05
ShS-2	75,47	67,65	0,88	0,96	114,66	102,78
ShS-3	44,31	39,97	0,88	0,96	67,32	60,72
ShS-4	109,09	98,4	0,88	0,96	165,74	149,5
ShS-5	46,5	41,43	0,86	0,96	70,64	62,94
ShS-6	62,06	55,68	0,87	0,96	94,29	84,59
ShS-7	28,4	25,62	0,88	0,96	43,14	38,92
ShS-8	111,69	102,54	0,86	0,96	169,69	155,79

As can be seen from the statement, the result is obvious, the installation of the CU allowed us to:

Table 10 - Change in reactive power in the AL after installing the KU at the TP

№	power, kWt			K	kvar
ShS-1	76,81	0,6	0,96	1,04	71,89
ShS-2		0,88	0,96	0,25	14,85
ShS-3		0,88	0,96	0,25	8,77
ShS-4		0,88	0,96	0,25	21,6
ShS-5		0,86	0,96	0,30	10,8
ShS-6		0,87	0,96	0,28	13,6
ShS-7		0,88	0,96	0,25	5,62
ShS-8		0,86	0,96	0,30	26,73
Total 174.02

Table 11 - Recalculation of ShS-1 load

Initial data	Calculation data
Naim EP	N pcs	Est. Power kW	Ki	Coefficient react	Wed.Change.Power	Ne	Kmax	Design power
1 EP	∑		cos𝜑	tg𝜑	Pcm kW	Qcm kvar	Ne	Kmax	Calculation kW	Qcalc kvar

Group A
Unlocked conveyor				0,16	0,96	0,29	2,24	0,64	-	-	-	-
Crane bridge.				0,14	0,96	0,29	1,96	0,56	-	-	-	-
Slotting Machine				0,1	0,96	0,29		4,06	-	-	-	-
Drilling machine				0,4	0,96	0,29		3,48	-	-	-	-
Total				0,8	-	-	30,2	8,74		2,31	69,75	9,61
Group B
Exhauster		5,6	11,2	0,63	0,96	0,29	7,05	2,04	-	-	-	-
Total		5,6	11,2	-	-	-	7,05	2,04	-	-	7,05	2,04

Calculation of EP peak loads

As a peak ED mode to check for voltage sag on

electrical receiver and selection of circuit breakers are considered

starting mode of the most powerful electric motor and the peak current is determined by

Ipeak cable line, supply transformer substation. Peak current for

ED group is found as the sum of the currents of the maximum operating current of the group without taking into account the current of the most powerful motor and the starting current of this motor according to the formula:

where Inomm is the rated current of the most powerful blood pressure, A;

Кп – multiplicity of the starting current of the most powerful IM.

The current of the most powerful motor among the ShS-1 electrical receivers is calculated. Longitudinal planing machine Pnom = 14 kW and after compensation cosφ = 0.96.

The peak current will be equal to:

Characteristics of the premises

The turning shop room is classified as dry, since the relative air humidity does not exceed 60% of clause 1.1.6 c. A turning shop is a very dusty facility, so the premises are classified as dusty; due to the production conditions, process dust is released in such quantities that it can settle on the wires and penetrate inside the machines - clause 1. 1.11 c. The premises are non-explosive, since substances that form explosive mixtures with air are not located or used in them. 1.3 in. In terms of fire hazard, the premises of the turning shop are classified as non-fire hazardous, since they do not contain the conditions given in Chapter. 1.4 in.

Selecting a brand of 0.4 kV cables

Based on an analysis of the cable laying and the characteristics of the workshop environment, a conclusion is made about the possibility of using the VVGng(a)-Ls-0.66 cable (copper conductor, insulation made of PVC plastic of reduced fire hazard, sheath made of PVC composition) to power ShS 1-8 and electrical receivers reduced flammability) Cables of this brand are intended for vertical, inclined and horizontal routes. Unarmored cables can be used in areas subject to vibration. Do not propagate combustion when laid in bundles

(standards GOST R IEC 332-2 category A). They are used in cable structures and premises. The permissible heating of the conductor in emergency mode should not exceed +80ºC with a duration of operation of no more than 8 hours per day and no more than 1000 hours over the service life.

Service life – 30 years.

Table 12 - Selection of cable lines from transformer substations to shs for workshop No. 4 before installation of the heat exchanger

Naim	Route KL	S kVA	I A	K1	K2	Id A	Iadd A	L m	R Ohm	X Ohm	Z Ohm	Brand	Scab mm²
KL3-1	TP-ShS1	92,18	140,05		0,8	175,06			6,36	1,96	6,65	VVGng(a)-Ls-0.66
KL3-2	TP-ShS2	75,47	114,66		0,8	143,32			1,85	0,42	1,89	VVGng(a)-Ls-0.66
KL3-3	TP-ShS3	44,31	67,32		0,8	84,15			48,84		49,2	VVGng(a)-Ls-0.66
KL3-4	TP-ShS4	109,09	165,74		0,8	207,17			7,6	3,15	8,22	VVGng(a)-Ls-0.66
KL3-5	TP-ShS5	46,5	70,64		0,8	87,63			38,48	4,73	38,76	VVGng(a)-Ls-0.66
KL3-6	TP-ShS6	62,06	94,29		0,8	117,86			4,81	1,1	4,93	VVGng(a)-Ls-0.66
KL3-7	TP-ShS7	28,4	43,13		0,8	53,92			62,64	5,13	62,84	VVGng(a)-Ls-0.66
KL3-8	TP-ShS8	111,69	169,69		0,8	211,48			10,92	4,53	11,82	VVGng(a)-Ls-0.66

Table 13 - Selection of cable lines from transformer substations to shs for workshop No. 4 after installing the control unit on the transformer substation busbars

Naim	Route KL	S kVA	I A	K1	K2	Id A	Iadd A	L m	R Ohm	X Ohm	Z Ohm	Brand	Scab mm²
KL3-1	TP-ShS1	77,68			0,8	147,5			8,88	2,04	9,11	VVGng(a)-Ls-0.66
KL3-2	TP-ShS2	67,65	102,78		0,8	128,47			1,85	0,42	1,89	VVGng(a)-Ls-0.66
KL3-3	TP-ShS3	39,97	60,72		0,8	75,9			48,84		49,2	VVGng(a)-Ls-0.66
KL3-4	TP-ShS4	98,4	149,5		0,8	186,87			7,6	3,15	8,22	VVGng(a)-Ls-0.66
KL3-5	TP-ShS5	41,43	63,94		0,8	78,67			38,48	4,73	38,76	VVGng(a)-Ls-0.66
KL3-6	TP-ShS6	55,68	84,59		0,8	105,7			6,89	1,14	6,98	VVGng(a)-Ls-0.66
KL3-7	TP-ShS7	25,62	38,92		0,8	48,65			99,36	5,34	99,5	VVGng(a)-Ls-0.66
KL3-8	TP-ShS8	102,54	155,79		0,8	194,73			10,92	4,53	11,82	VVGng(a)-Ls-0.66

KL2-10

TP-KU

93,81

4,24

0,7

4,29

VVGng(a)-Ls-0.66-4x35.

Table 14 - Selection of cable from ShS-1 to EP

Name	Route KL	P kW	I A	cos𝜑	Iadd A	L m	R Ohm	X Ohm	Z Ohm	Brand	Ssection mm²
KL1-1	From ShS-1 to EP1		22,15	0,96			29,6	0,46	29,6	VVGng(a)-Ls-0.66	2,5
KL1-2	From ShS-1 to EP2		22,15	0,96			44,4	0,69	44,4	VVGng(a)-Ls-0.66	2,5
KL1-3	From ShS-1 to EP3		55,39	0,96			14,72	0,79	14,74	VVGng(a)-Ls-0.66
KL1-4	From ShS-1 to EP4		47,47	0,96			11,04	0,59	11,05	VVGng(a)-Ls-0.66
KL1-5	From ShS-1 to EP5	5,6	8,86	0,96			62,5	0,63	62,5	VVGng(a)-Ls-0.66	1,5
KL1-6	From ShS-1 to EP6	5,6	8,86	0,96			62,5	0,63	62,5	VVGng(a)-Ls-0.66	1,5

Table 15 - Checking cable lines KL1 in normal mode

KL	A	A	IN	IN	dU V	IN
KL1-1	22,15	29,6	1,13	1,85	2,99
KL1-2	22,15	44,4	1,7	1,85	3,55
KL1-3	55,39	14,72	1,41	1,85	3,26
KL1-4	47,47	11,04	0,9	1,85	2,75
KL1-5	8,86	62,5	0,95	1,85	2,8
KL1-6	8,86	62,5	0,95	1,85	2,8

Table 16 – Checking cable lines KL2 in normal mode

Name	A	Z Ohm	IN	dU%
KL2-1		9,11	1,85	0,48
KL2-2	102,78	1,89	0,33	0,08
KL2-3	60,72	49,2	5,16	1,35
KL2-4	149,5	8,22	2,12	0,55
KL2-5	63,94	38,76	4,28	1,12
KL2-6	84,59	6,98	1,02	0,25
KL2-7	38,92	99,5	6,69	1,76
KL2-8	155,79	11,82	3,18	0,83

Powerful engine

Metal-cutting machines are designed for mechanical processing of metal workpieces with cutting tools.

The purpose of metal-cutting machines is to produce parts of a given shape and size with the required accuracy and quality of the machined surface. The machines process workpieces not only from metal, but also from other materials, so the term “metal-cutting machine” is conditional.

According to the type of work performed, metal-cutting machines are divided into groups, each of which is divided into types, united by common technological characteristics and design features.

Metal-cutting machines represent a whole class of equipment designed to produce metal blanks: boring machines, lathes, etc.

As an example, we will calculate and select the electrical equipment of a screw-cutting lathe model 16D20.

Lathes are designed for the manufacture and processing of parts in the shape of bodies of revolution. They are used for processing cylindrical, conical, shaped surfaces, trimming ends, as well as for drilling and reaming holes, threading and other operations.

2.1 Selecting the type of current and voltage value for the workshop network

For power electrical networks of industrial enterprises, three-phase alternating current is mainly used. Direct current is recommended to be used in cases where it is necessary under the conditions of the technological process (charging batteries, powering galvanic baths and magnetic tables), as well as for smooth control of the rotation speed

electric motors. If the need to use direct current is not caused by technical and economic calculations, then three-phase alternating current is used to power power electrical equipment.

When choosing voltage, you should take into account the power, number and location of electrical receivers, the possibility of their joint power supply, as well as technological features of production.

When choosing the voltage to power electrical receivers directly, you must pay attention to the following points:

1) The rated voltages used in industrial enterprises for power distribution are 10; 6; 0.66; 0.38; 0.22 kV;

2) It is recommended to use voltages higher than 1 kV at the lowest level of power distribution only if special electrical equipment operating at voltages higher than 1 kV is installed;

3) If motors of the required power are manufactured for several voltages, then the issue of voltage selection must be resolved through a technical and economic comparison of options;

4) If the use of voltages above 1 kV is not caused by technical necessity, options for using voltages of 380 and 660 V should be considered. The use of lower voltages to power power consumers is not economically justified;

6) Using a voltage of 660 V, electricity losses and consumption of non-ferrous metals are reduced, the range of operation of workshop substations is increased, the unit power of the transformers used is increased and, as a result, the number of substations is reduced, and the power supply circuit at the highest level of energy distribution is simplified. The disadvantages of the 660 V voltage are the impossibility of jointly powering the lighting network and power electrical receivers from common transformers, as well as the lack of low-power electric motors for a voltage of 660 V, since currently such electric motors are not produced by our industry;

7) In enterprises with a predominance of low-power electrical receivers, it is more profitable to use a voltage of 380/220 V (unless the feasibility of using a different voltage has been proven);

8) The voltage of DC networks is determined by the voltage of the powered electrical receivers, the power of converter installations, their distance from the center of electrical loads, as well as environmental conditions.

Electronic control and signaling circuits must be powered by a transformer.

For AC control circuits powered from a transformer, the following voltage values are recommended: 1) 24 or 48V, 50 and 60 Hz; 2) 110V, 50Hz or 115V, 60Hz; 3) 220V, 50Hz or 230V, 60Hz.

For DC control circuits, the recommended voltage is: 24, 48, 110, 220, 250V. It is permissible to use other low voltage values for electronic circuits and devices that are designed for such voltages. A ground fault in any control circuit must not cause the machine to turn on unexpectedly, cause the machine to move dangerously, or prevent the machine from shutting down.

The control circuit must be designed so that if the time limit has expired, both buttons must first be released and then pressed again to start the cycle.

It is recommended to connect the alarm circuit, which is not connected to the control circuit, to 24V AC or DC. In this case, lamps with voltages from 24V to 28V are used. If an individual transformer is used, then 6V or 24V lamps are used. In this case, the signaling circuit can be connected to the control circuit.

The use of fluorescent lamps for local lighting of lathes is prohibited. The most widely used are incandescent lamps with a voltage of 36V, connected through a step-down transformer. It is prohibited to use local lighting with a voltage higher than 36 V.

For a universal high-precision screw-cutting lathe, model 16D20, the most suitable parameters are:

Supply network: voltage 380V, current type - alternating, frequency 50 Hz;

Control circuit: voltage 110V, current type - alternating;

Local lighting: voltage 24 V.

Electrical loads determine the choice of the entire power supply system. To calculate them, the demand coefficient method and the diagram ordering method are used. The first method is usually used at the design stage, when the power of individual electrical receivers (ER) is unknown.

The diagram ordering method or the maximum coefficient method is fundamental in the development of technical and operational power supply projects. It allows you to determine the design load of any node of the power supply circuit based on the rated power of the electric power supply, taking into account their number and characteristics. According to this method, the calculated maximum load of the electric group is:

Group rated power R n is defined as the sum of the rated capacities of the electric power plant excluding reserve ones.

Usage rate TO and one or a group of electric power plants (Table 2.1) characterizes the use of active power and is the ratio of the average active power of one or a group of electric power plants for the busiest shift to the rated power.

Maximum coefficient TO m is the ratio of the calculated maximum active load power of the electric power group to the average load power for the busiest shift.

For a group of electrical equipment of one operating mode, the average active and reactive loads for the busiest shift are determined:

;
. (2.2)

Rated power P same type of electronic signatures

. (2.3)

Table 2.1

Design coefficients of electrical loads

Electrical receivers
Pumps, compressors
Industrial fans, blowers, smoke exhausters
Welding transformers: manual electric welding
automatic welding
Resistance furnaces
Incandescent lamps
Fluorescent lamps
Overhead cranes, beam cranes, hoists, elevators

For consumers with variable load (group A) the calculated active load R p (A) groups of electronic equipment of a department (section, workshop) are determined taking into account the maximum coefficient TO m and average compartment load:

, (2.4)

Where TO m (A) – determined depending on the effective number of EP n e and from the group utilization factor TO and for the busiest shift (Table 2.2).

Table 2.2

Maximum coefficients TO m for different utilization rates

depending on the n uh

	Meaning TO m at TO And

Weighted average utilization rate of group A ED department

, (2.5)

Where R n (A) – total rated active power of the electric group

;

R cm (A) – total shift-average active power of group A electric vehicles

The effective number of EPs of group A is found by the formula

, (2.6)

or in simplified terms.

The calculated reactive load of a group of electrical units with variable load for the department and for the workshop as a whole is determined taking into account the given number of electrical units:

at n e >10
, (2.7)

at n uh £10
. (2.8)

For consumers of group B with a constant load schedule ( TO m = 1) the load of the electric group is equal to the average load for the busiest shift. Estimated active and reactive powers of group B group EP of department:

;
. (2.9)

Such electric motors may include, for example, electric motors of water supply pumps, fans, unregulated smoke exhausters, compressors, blowers, unregulated resistance furnaces.

After determining the loads of the departments, the calculated load for the workshop is found:

,
, (2.10)

Where R cm j , Q cm j– active and reactive loads ED j-th department; m– number of branches.

Estimated active and reactive power of the workshop:

kW;
kV∙Ar. (2.11)

If there are single-phase electric motors in the workshop, distributed among the phases with an unevenness of £ 15%, they are taken into account as three-phase ones of the same total power. Otherwise, the calculated load of single-phase electric motors is assumed to be equal to triple the load of the most loaded phase.

When the number of single-phase electric motors is up to three, their conditional three-phase rated power is determined by:

a) when a single-phase electric motor is switched on to phase voltage with a three-phase system

Where S n– nameplate power; R n.f. – rated power of the maximum loaded phase;

b) when one ED is switched on to line voltage

. (2.13)

Maximum loads of single-phase electric motors when their number is more than three at the same TO and and cosj connected to phase or line voltage are determined:

;
. (2.14)

To determine the electrical loads of the workshop, a summary statement is drawn up (Table 2.3) with all calculated data filled in.

Table 2.3

Summary sheet of workshop electrical loads

Name of the characteristic group of EP

Number of electronic devices

Installed power of the electric unit, reduced to PV = 100%

Coefficient

use TO And

Average load for the busiest shift

Maximum rated power

one, kW

total, kW

R cm,

Q cm, kW

R m, kW

Q m, kV∙Ar

Lighting loads are calculated using an approximate method based on the specific power per illuminated area.

;
(2.15)

Where R udo – specific design capacity per 1 m 2 of production area of the department ( F);

TOс – lighting demand coefficient (Table 2.4).

Table 2.4

Calculated coefficient TO and, cosj, R ud0 and TO from individual workshops of industrial enterprises

	Name of workshops				R ud0,
	Compressor
	Pumping
	Boiler rooms
	Welding shop
	Electrical shop
	Assembly shops
	Mechanical
	Administrative premises

When using known values of the specific power of general uniform lighting, depending on the type of lamp and, based on their optimal location in the room, the power of one lamp is determined.

To illuminate the main workshops with a height of more than 6 m and in the presence of open spaces, gas-discharge lamps of the DRL type with cosj = 0.58 are used. For administrative and domestic premises, fluorescent lamps with cosj = 0.85 are used; for lighting small rooms, incandescent lamps with cosj = 1 are used.

The total design load of the workshop is determined by summing the design loads of power and lighting groups of electrical receivers

The transformer is selected based on the full design load, taking into account reactive power compensation.

Note : examples for determining electrical loads are presented in.