What is a power network? How to write a paper on this topic? Help me ~ ~

This design is the preliminary technical design of Shandong Weihai Shangcheng International Power Distribution Project.

Shangcheng International is a large-scale shopping center built for the development of tourism in Weihai City, Shandong Province, which is divided into two adjacent buildings, the East Building and the West Building. Weihai, located at the easternmost tip of Shandong Peninsula, is the bridgehead for exchanges between China and South Korea. Its natural environment is beautiful, quiet and clean, and it is known as "the most livable city". Weihai is a tourist city with great development potential, so Shangcheng International is no less than a shopping center in a metropolis. Weihai's natural environment is very similar to Dalian's, belonging to a maritime climate, with moderate average air humidity and a lower temperature limit of -20? C or so, the high temperature limit does not exceed 40? C, non-corrosive, explosive and so on. There is no need to make special requirements in this respect in equipment selection.

Because Shangcheng International is located in the city and close to the load center, the introduction and distribution of power supply is very convenient. The high voltage value is 10KV, and the system capacity can be regarded as infinite. But at the same time, because it is located in the city, the distribution room can only be built indoors, and because it is a large shopping mall, it is more reasonable to build the substation in the basement for reasonable layout and customer safety. Indoor transformer should choose dry transformer; And it is necessary to introduce double incoming lines to ensure power supply.

According to statistics, Shangcheng International Power Distribution Project needs to supply power to the following equipment:

Fire shutter smoke exhaust fan 75KW, east elevator 36 KW, fan 64.5 KW, west elevator 36 KW, fire pump spray pump 100 KW, submersible pump 1- 18.5kW, submersible pump 2- 18.5kW, household pump 30 KW, and the calculation capacity is 6544.

A distribution project may have a variety of combination schemes, and the change of combination scheme will inevitably affect the change of investment cost and operation cost. Therefore, obtaining a reliable, safe, reasonable and economical scheme is the core content of design work. This design is aimed at Shangcheng international power distribution project, and the line of power distribution scheme should be relatively simple. It is preliminarily envisaged that the equipment will be introduced into the high-voltage room from the high-voltage network, sent to the substation after measurement, and sent to the low-voltage room for power distribution after substation.

2 Load calculation

2. 1 Significance and method of load calculation

Before the design of power transformation and distribution, the first task is to calculate the load. The so-called load calculation is to calculate the calculated load according to the installation capacity of electrical equipment provided by users, correctly estimate the power and electricity required by users, and then select the matching substation equipment. The accuracy of estimation affects the quality of user's power supply design. If the estimate is too high, it will increase the capacity of power supply equipment, complicate the power grid, waste non-ferrous metals and increase the initial investment and operation management workload. However, if the estimate is too low, electrical equipment will be used. After put into use, the lines and electrical equipment of the power supply system will overheat because they can't bear the actual reclosing current, which will accelerate the insulation aging speed, reduce the service life, increase the power consumption and affect the normal and reliable operation of the power supply system.

Therefore, before the distribution design of Shangcheng International Project, the first task is to calculate the load. Moreover, the load of substation is the basis for determining the transformer capacity, conductor cross-sectional area and measurement range of substation power supply system, and it is also an important data for relay protection setting.

The load calculation methods include coefficient method, binomial method, unit electricity consumption method and demand coefficient method. At present, the load calculation of general engineering mainly adopts demand coefficient method or utilization coefficient method, which is more convenient and fast. Demand coefficient method means that the actual load capacity of each equipment in the substation is always less than the rated total capacity of the equipment connected to it, and the ratio is called demand coefficient Kd. See table 2- 1[ 1] for the equipment demand coefficient involved in this design.

Name of electrical equipment group Kd cos

Dark color

Elevator category (because this building is a shopping mall) 0.8 ~ 0.9 0.5 1.73

Pump, exhaust fan, etc. 0.75 ~ 0.85 0.80.75

Lighting, air conditioning and other equipment sockets 0.9 ~ 1.0 0.9 0.48

Table 2- 1 Kd value of electrical equipment demand coefficient

The formula [1] for calculating load by demand coefficient method is as follows:

Pc？ n=KdnPNn (2- 1)

Qc？ n=Pcn tan n (2-2)

Where kdx-demand coefficient

Pc？ N- active power of equipment

Qc？ N- reactive power of equipment

-natural power factor angle

therefore

Pc =∑Pc？ One (2-3)

Qc=∑Qc？ One (2-4)

Sc= (2-5)

2.2 Load calculation

Fire shutter exhaust fan 75KW:

Pc？ 1 = KD 1pn 1 = 0.80 75 = 60KW

Qc？ 1 = PC 1 tan 1 = 60 0.75 = 45 KVAR

East Building Elevator 36 kW

Pc？ 2 = kd2pn2 = 0.836 = 28.8kW.

Qc？ 2 = pc2tan2 = 28.81.73 = 49.8kVA.

Fan 64.5 kW

Pc？ 3=Kd3PN3=0.75 64=48KW

Qc？ 3 = Pc3tan 3 = 0.75 48 = KVA

West Building Elevator 36 kW

Pc？ 4 = kd4pn4 = 0.836 = 28.8kW.

Qc？ 4 = pc4tan4 = 28.81.73 = 49.8kVA.

Fire pump spray pump 100 KW

Pc？ 5=Kd5PN5=0.75 100=75KW

Qc？ 5=Pc3tan 5=0.75 75=56.3KVAR

Submersible pump 1 —— 18.5kW

Pc？ 6 = kd6pn6 = 0.7518.5 =13.9kw.

Qc？ 6 = pc6 tan 6 = 0.75 13.9 = 10.4 kvar

Submersible pump 2-18.5kw.

Pc？ 7 = kd7pn7 = 0.7518.5 =13.9 kw.

Qc？ 7 = Pc7tan 7 = 0.75 13.9 = 10.4 kvar

Life pump 1 10 kw

Pc？ 8 = kd8pn 8 = 0.75 1 10 = 82.5 kw

Qc？ 8 = pc8 tan 8 = 0.75 82.5 = 6 1.9 kva

West building lighting socket 1000 KW

Pc？ 9=Kd9PN9=0.9 1000=900KW

Qc？ 9 = pc9tan9 = 0.48900 = 432kVA

East building lighting socket 700 KW

Pc？ 10 = KD 10pn 10 = 0.9 700 = 630 kw

Qc？ 10 = PC 10 tan 10 = 0.48 630 = 302.4 kvar

Reserved equipment branch100kw-calculated by ordinary lighting socket.

Pc？ 1 1 = KD 1 1pn 1 1 = 0.95 100 = 95KW

Qc？ 1 1 = PC 1 1 tan 1 1 = 0.48 95 = 45.6 kvar

55 KW—— DC screen capacitor energy storage

Pc？ 12 = KD 12pn 12 = 0.2 55 = 1 1KW

Active power:

Pc =∑Pc？ I

= 60+28.8+48+28.8+75+ 13.9+ 13.9+82.5+900+630+95+ 1 1

=1986.9kw

Reactive power:

Qc=∑Qc？ I

=45+49.8+48+49.8+56.3+ 10.4+ 10.4+6 1.9+432+302.4+45.6

=1111.6 kva

Total computing power

Sc=

= 2276.7 kW

Engineering natural power factor:

cos = = = 0.873

Because the natural power factor is lower than 0.9, reactive power compensation should be carried out. There are two ways to compensate: one is to use a synchronous camera; The other is to use electrostatic capacitance. At present, the company's distribution low-voltage cabinets mostly adopt the latter method, and the specific calculation is carried out when each transformer is selected below.

3 power distribution scheme

3. 1 Selection of transformer capacity and quantity

For the selection of transformer capacity and quantity, according to the requirements of load calculation and load grade (many equipment in large shopping malls, such as fire pumps and elevators, belong to secondary load), it is preliminarily envisaged to use three transformers to supply power to electrical equipment in groups.

Under normal working conditions, three transformers supply power at the same time, respectively supplying power to the following three groups of power supply equipment:

① Secondary load: East building elevator 36KW, submersible pump 1- 18.5kW, submersible pump 2- 18.5kW, fire pump spray pump 100KW, fire shutter smoke exhaust fan 75KW, west building elevator 36KW, fan 64.5KW, DC screen 650.

② Class III load: 700KW for the lighting socket in the east building.

③ Class III load: West Building 1000KW lighting socket.

According to the equipment load calculation in the previous chapter, the calculated load of each group can be obtained.

Group active power P(KW) reactive power Q(KVAR) calculation load S cos

1 456.9 377.2 592.5 0.77 1

2 630 302.4 700 0.9

3 900 432 1000 0.9

Table 3- 1 Statistics of three groups of equipment loads

The first group of equipment with higher load level leads to the standby branches on the other two groups of transformers, which are roughly grouped as follows:

(2) East building lighting socket 700 KW, East building elevator 36 KW, submersible pump 1- 18.5kW, submersible pump 2- 18.5kW, fire pump spray pump 100 KW.

(3) West building lighting socket 1000 KW, fire shutter smoke exhaust fan 75KW, west building elevator 36 KW and fan 64.5kw..

In order to improve the utilization rate of electric energy, reactive power compensation is needed for the system. Firstly, reactive power compensation is carried out for the first group of equipment; Then choose the capacity of the transformer.

I. Reactive Power Compensation

Considering the reactive power loss of the transformer, increase the power factor to 0.95, and then calculate the capacitor according to the following formula:

cos =0.77 1，cos '=0.95

tan =0.826，tan '=0.328

According to the formula:

QC=PZ (Tan-Tan') [2] (3- 1)

Where: QC- compensation capacity of electrostatic capacitor, (kvar);

Pz- active power calculation load of this group of equipment, (KW).

Alternative data:

QC=456.9(0.826-0.328)

=227.5 KVAR

Select 10 CLMD43/25KVAR capacitors, each with a capacity of 25KVAR. Connection mode: select triangle connection mode. As shown in figure 3- 1.

Therefore, the equipment capacity of the capacitor cabinet is 250KVAR.

After manual compensation:

Active power:

Pm = 456.9KW kw

Reactive power:

QM = QZ-QC = 377.2-250 =127.2kVA.

Apparent power:

Sm = = 474.2KVA kva

Power factor:

cos = =0.964

2. Selection of transformer

Only the loss of main transformer can be estimated before primary selection, and its calculation formula [2] is as follows

δPt = 0.02pm(3-2)

δQt = 0. 1Qm(3-3)

Where: Δ pt-the estimated value of the active part of transformer loss.

Δ Qt-the estimated value of reactive power loss of transformer.

PM-the calculated value of the maximum continuous active load;

QM-the calculated value of the maximum continuous reactive load of the bus after compensation.

Substitution data: δ Pt = 0.02 456.9 = 9. 138 kW.

δQt = 0. 1. 127.2 = 12.72 kvar

After considering the transformer loss, the total power is:

= Pm+δPt = 456.9+9. 138 = 466.0 kw

= Qm+δQt = 127.2+ 12.7 = 139.9 kvar

= =486.5KVA

The transformer capacity selected for the first group of equipment is

Sb≥ = (3-4)

Where:-the load guarantee coefficient in case of accident is generally 0.8 ~1[3];

COS '- power factor after considering artificial compensation and transformer loss.

Choose the transformer model as SG 10-? KVA/ 10kV/0.4kV- independent three-phase air ("dry" type) natural circulation cooling device, double windings, no excitation voltage regulation, conductor material copper [4].

In the same way, the computing power of the second and third groups of equipment in normal state is obtained. Related Data Sheet 3-2

Group 1 2 3

Reactive power compensation QC(KVAR) 250 250 360

The compensated active power pm (kw) is 456.9 630 900.

Compensating reactive power Qm(KVAR) 127.2 52.4 72

The compensated apparent power SM (KVA) is 474.2 632.2 902.9.

Compensating power factor Cos

0.964 0.996 0.997

Estimated active power (kW)

466.0 642.6 9 18

Estimated reactive power (KVAR)

139.9 57.64 79.2

Estimated apparent power (KVA)

486.5 645.2 92 1.4

Estimated power factor Cos'

0.958 0.996 0.996

Main capacity (kva)100010001250

Table 3-2 Computing Capacity of Three Groups of Equipment

Among them, the rated capacity of 1# transformer is much larger than the calculated capacity for two reasons: firstly, if 800KVA transformer is selected, three transformers with different specifications will appear in the same project, which is not conducive to future maintenance. Moreover, when the 2# or 3# transformer fails, the standby cannot be realized. Therefore, it is reasonable to choose 1000KVA transformer. 3# transformer is selected according to its calculation capacity, and it is not easy to reduce it for the sake of seeking common ground.

2# transformer shall provide backup for 36 KW east elevator, 1- 18.5 kW submersible pump, 2- 18.5 kW submersible pump and 100 KW fire pump, and check whether they meet the requirements.

P2 = 761.6kw.

Q2 = 429.3 kWh

= =874.5KVA

The compensation pow of that reactive power compensation cabinet is 250KVAR,

P2 = 761.6kw.

Q2'=429.3-250= 179.3KVAR

= =782.4KVA

Therefore, the capacity of 2# transformer can meet the requirements of safe standby.

The calculated loads of 1000 KW west building lighting socket, 75KW fire shutter smoke exhaust fan, 36 KW west building elevator and 64.5 KW fan are as follows: P3 =1036.8kw.

Q3 = 562.8KVAR KVA.

= = 1 179.7 kva

The compensation power of reactive power compensation cabinet is 360KVAR.

P3 =1036.8kW.

Q3' = 562.8-360 = 202.8

S3= = 1056.4KVA

Therefore, the capacity of 3# transformer can meet the requirements of safe standby.

According to the transformer selection manual, see 3-3[5] for the relevant parameter list.

3.2 Calculation of Transformer Loss

The loss calculation of transformer is based on the maximum loss. That is, when the transformer works at the maximum load, the load rate is:

= (3-5)

Among the three transformers, when 1# transformer fails, the load of 2# and 3# transformers reaches the maximum, and that of 3# transformers.

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