2.1 Selection of transformers

First, we determine the calculated loads of substations on 10 kV buses in the maximum (100%) mode. For the design substation No. 2 in normal mode (the ring jumper is off), the load is determined:

S =S +K

S =7+j4+1.1*1000

S _{max} = 8.049+j4.34

S _{H2} =

S _{H2} = 9.14 MVA

where: coefficient K takes into account power losses in the distribution network of 10 kV and it can be assumed that K = 1.05 – 1.10

The load on the 10 kV buses of substation No. 1 is set to S = 3+j1.

S _{H1} = = 3.16 MVA

For each substation, we select two transformers of the same type with on-load tap-changers. Given that one of them may fail, the power of each should be (0.6 … 0.7) S

S _{H1} u003d 3.16 * 0.7 u003d 2.26 kVA

S _{H2} u003d 9.14 * 0.7 u003d 6.53 kVA

We select the type of transformer according to the condition: S _{n.t.} ≥S _{calc.} We choose two transformers of the type ТМН-2500/110 and ТДН 10000/110 [7]

and check for emergency overload.

For PS No. 1

_{MAB} = _{*} 100= _{*} 100 = 126% ≤ 140% passes.

For PS No. 2

_{MAB} = _{*} 100= _{*} 100 = 91% ≤ 140% passes.

The data of the selected transformers are entered in the table.

Transformer type | S , MBA | Tense. HF HV MV LV | ΔP _{XX} , kW |
ΔP short kW | I , % | U ,% VN SN NN |

TDN-10000/110 | 0.9 | 10.5 – – | ||||

TMI-2500/110 | 2.5 | 5.5 | 1.5 | 10.5 – – |

We determine the parameters of the equivalent circuit of power transformers, knowing the main technical data of these transformers (nominal voltage of the windings of higher, medium, and lower voltage – U _{HV} ; U _{CH} ; U _{LV} , no-load losses – ΔP _{XX} and short circuit *–* ΔP short circuit, short circuit voltage U short _{circuit} and no-load current I _{XX} ), which we took from the reference materials and entered in the table.

With two transformers operating in parallel, the parameters of the equivalent equivalent circuit compared to the equivalent circuit for one transformer will be:

(2.7)

We determine the resistance of the windings of transformers by bringing them to the rated voltage of the higher voltage winding (HV). The active resistance of the equivalent circuit of two winding transformers, reduced to the rated voltage of the HV side, is expressed as follows:

R _{t} = Ohm, (2.3)

Where ΔР _{kz} – short circuit losses, kW;

U _{n} – the highest rated voltage, kV;

S _{n} – rated power, kVA.

For PS No. 2.

= = 3.84 ohm

For PS No. 1.

= = 21.3 ohms

Inductive reactance

X _{t} = Ohm, (2.4)

Where U _{r} – reactive voltage drop in the transformer windings,%.

In transformers with a power of 1000 kVA and above, it can be assumed that U _{r} % u003d U _{to} %, since in these transformers X _{t} u003e R _{t} .

For PS No. 2

= = 69.43 ohms

For PS No. 1

= = 254.1 ohms

We determine the no-load power of the transformer, i.e. load in its conductivities ΔS _{st}

ΔS _{st} = ΔP _{st} + j ΔQ _{st}

Where ΔР _{st} – active power losses in the steel of the transformer, which are taken equal to the no-load losses of the transformer, ΔР _{xx} .

ΔQ _{st} – power magnetizing steel;

ΔQ _{st} _{=} (2.6)

For PS No. 2

2 ΔР _{ST} u003d 2ΔР _{XX} u003d 2 * 14 u003d 0.028 MW

2ΔQ _{st} _{=} = 0.18 MVar

ΔS _{st} u003d 0.028 + j 0.18

For PS No. 1

2 ΔР _{ST} u003d 2ΔР _{XX} u003d 2 * 5.5 u003d 0.011 MW

2ΔQ _{st} _{=} = 0.075 MVar

ΔS _{st} u003d 0.011 + j 0.075

The flow distribution in the district network is calculated for the maximum (100%) and minimum (25%) normal load modes. In the maximum (100%) load mode are given from the condition. In this case, the load at the i – th point of the network in the minimum mode S will be

S = 0.25*S ,

Where S – maximum load at the i-th point. Then for PS No. 2

S _{MIN} _{2} u003d 0.25 * 8.049 + j4.34 u003d 2.012 + j1.085

And for substation No. 1, the minimum load will be

S _{MIN} _{1} u003d 0.25 * 3 + j1 u003d 0.75 + j 0.25

At the same time, we accept that cosφ = cosφ _{.} The time of using the maximum load at T ^{100} u003d T ^{25} is taken according to the condition for SS No. 1 – 3500, and for SS No. 2

– 4000.

To simplify the calculations and analysis of electrical networks, all substation loads on the low voltage sides S _{LV} are brought to the higher voltage side.

R _{t} /2 X _{t} /2 S _{max} /2

S _{pr} S _{2} S _{1}

4.02+j2.17

2∆S _{xx}

S _{1} R _{t} /2 X _{t} /2 S _{min} /2

2ΔР _{st} 2ΔQ _{w} 1.005+j0.54

Determination of the reduced powers in the maximum and minimum modes. For PS No. 2 in the maximum mode.

S _{1} _{max} = + ΔS = 4.02+ j2.17+ *(3.84+j69.43) = 4.026 + j2.38 (MVA)

S _{2} u003d S _{1max} * 2 u003d 8.052 + j4.76 (MVA)

S _{pr} _{.} _{max} u003d S _{2} + 2 ΔS _{xx} u003d 8.08 + j4.94 (MVA)

2 ΔS _{xx} = 0.028+ j0.18

For PS No. 2 in the minimum mode

S 1 _{min} = + ΔS = 1.005+ j0.54+ *(3.84+j69.43) = 1.006 + j0.55 (MVA)

S _{2} u003d S _{1min} * 2 u003d 2.012 + j1.1 (MVA)

S _{pr} _{.} _{m} _{in} u003d S _{2} + 2 ΔS _{xx} u003d 2.04 + j1.28 (MVA)

For PS No. 1 in maximum mode

S _{1} _{max} = + ΔS = 1.5+ j 0.5+ *(21.3+j254.1) = 1.504 + j0.55 (MVA)

S _{2} u003d S _{1max} * 2 u003d 3.008 + j1.1 (MVA)

S _{pr} _{.} _{max} u003d S _{2} + 2 ΔS _{xx} u003d 3.019 + j1.175 (MVA)

2 ΔS _{xx} = 0.011+ j 0.075

For PS No. 1 in the minimum mode

S 1 _{min} = + ΔS = 0.375+ j0.125+ *(21.3+j254.1) = 0.3753 + j0.1283 (MVA)

S _{2} u003d S _{1min} * 2 u003d 0.7506 + j0.256 (MVA)

S _{pr} _{.} _{m} _{in} u003d S _{2} + 2 ΔS _{xx} u003d 0.762 + j0.331 (MVA)

Calculation of the network in the maximum mode.

S 30 3 30 PS 2

2 8.08+j4.94

115 kV 4000

11.099+j6.115 10

one

3500 PS 1

3.019+j1.175

Determination of wire cross-sections according to the conditions of economic current density

Line C – 3

S _{H} = = 12.67 (MVA)

I _{max} = = 63A

F _{e} u003d 63 / 1.1 * 2 u003d 28.64 A / mm ^{2}

We choose AC 35 / 6.2 where I _{add} u003d 175 Au003e I _{max}

Passes by heating, no by the crown

Choose AC 70/11 I _{additional} = 265 A

Line 3 – 1

S _{H} = = 3.24 (MVA)

I _{max} = = 16.26 A

F _{e} u003d 16.26 / 1.1 * 2 u003d 7.39 A / mm ^{2}

Wire AC 16 / 2.7 passes through heating, but not through the crown.

We choose AC 70/11.

Line 3 – 2

S _{H} = = 9.47 (MVA)

I _{max} = = 47.8 A

F _{e} u003d 47.8 / 1.1 * 2 u003d 21.7 A / mm ^{2}

Wire AC 35 / 6.2 also does not pass through the crown.

We choose AC 70/11.

For double-circuit wire lines, we have a barrel D _{cf} = 3.5 m from [3]

We calculate the parameters of the wires in the sections and enter the results in the table

Plot | the wire | L, km | r _{o} , ohm/km |
x _{o} , ohm/km |
in _{o} *10 ^{6} , cm/km |
R, ohm | X, ohm | H*10 ^{6} cm |
Q _{in} , MVar |

C – 3 | AC 70/11 | 0.428 | 0.408 | 2.79 | 12.84 | 12.24 | 83.7 | 1.107 | |

3 – 1 | AC 70/11 | 0.428 | 0.408 | 2.79 | 4.28 | 4.08 | 27.9 | 0.37 | |

3 – 2 | AC 70/11 | 0.428 | 0.408 | 2.79 | 12.84 | 12.24 | 83.7 | 1.107 |

R=z _{0} LX=x _{0} LB=b _{0} L (2.1)

Where L is the length of the line, km.

The charging power of power lines Q _{in} it is permissible to determine the nominal voltage of the line U _{n} i.e.

Q _{in} = U In (2.2)

With U _{n} – in kV, V – in cm, dimension Q _{in} = in t.kvar.

Determination of power flow at the beginning of the line and at the end of each line.

R/2 X/2 R/2 X/2

With Q _{in1} 3 2

Q _{in3} Q _{in3}

Q _{v1} PS 2

Q _{v2}

8.08+j4.84

R/2

Q _{v2} X / 2

one

PS 1

3.019+j1.175

S _{ras1} = 3.019 + j1.175 – j0.37 = 3.019 + j0.805

S _{race} _{2} = 8.08 + j4.34- j1.107 = 8.08+ j3.83

R/2 X/2 S _{2} ^{i} R/2 X/2

S _{c} S _{1} ^{i} Q _{in} _{1} 3 S _{2}

Q _{in} _{3}

Q _{in} _{1}

Q _{in} _{2} S _{3} ^{i}

8.08+j3.83

R/2

X/2

3.019 + j0.805 S _{3}

S _{2} ^{i} u003d 8.08 + j3.83 + (6.42 + j6.12) = 8.12 + j3.87

S _{3} ^{i} = 3.019 +j0.805+ (2.14+j1.04) = 3.02+j0.806

S _{1} u003d S _{2} ^{i} + S _{3} ^{i} – jQ _{in} u003d 11.14 + j2.106

S _{1} ^{i} u003d 11.14 + j2.106 + (6.42+j6.12) = 11.2+j2.17

S _{c} u003d S _{1} ^{i} + (- jQ _{in} ) u003d 11.2 + j1.063

Determination of voltage at network points

U _{3} u003d 115 – = 114.32 kV

U _{2} u003d 114.32 – = 113.66 kV

U _{1} u003d 114.32 – = 114.26 kV

Network calculation in minimal mode

S 30 3 30 PS 2

2 2.04+j1.28

110 kV 4000

2.802+j1.611 10

one

3500 PS 1

0.762+j0.331

S _{ras1} = 0.762 + j0.331 – j0.37 = 0.762 + j0.04

S _{ras2} = 2.04 + j1.28- j1.107 = 2.04+ j0.173

S _{2} ^{i} u003d 2.04 + j0.173+ (6.42 + j6.12) = 2.042 + j0.175

S _{3} ^{i} = 0.762 +j0.04+ (2.14+j1.04) = 0.763+j0.039

S _{1} u003d S _{2} ^{i} + S _{3} ^{i} – jQ _{in} u003d 2.805 – j2.448

S _{1} ^{i} u003d 2.805 – j2.448 + (6.42+j6.12) = 2.81 – j2.44

S _{c} u003d S _{1} ^{i} + (- jQ _{in} ) u003d 2.81 + j3.55

Determination of voltage at network points in the minimum mode

U _{3} u003d 110 – = 109.63 kV

U _{2} u003d 109.63 – = 109.49 kV

U _{1} u003d 109.63 – = 109.61 kV

Network calculation in emergency mode

The analysis shows that the most severe accident is considered to be line damage in section 3 – 2 in the maximum mode.

S 30 3 30 PS 2

2

115 kV // 4000

one

3500 PS 1

R/2 X/2 R/2 X/2

With Q _{in1} 3 2

Q _{in3} Q _{in3}

Q _{v1} PS 2

Q _{v2}

8.08+j4.94

R/2

Q _{v2} X / 2

one

PS 1

3.019+j1.175

S _{ras1} = 3.019 + j1.175 – j0.37 = 3.019 + j0.805

S _{ras2} = 8.08 + j4.94- j0.55 = 8.08+ j4.39

S _{2} ^{i} u003d 8.08 + j4.39 + (12.84 + j12.24) = 8.16 + j4.47

S _{3} ^{i} = 3.019 +j0.805+ (2.14+j1.04) = 3.02+j0.806

S _{1} u003d S _{2} ^{i} + S _{3} ^{i} – jQ _{in} u003d 11.18 + j3.25

S _{1} ^{i} u003d 11.18 + j3.25 + (6.42+j6.12) = 11.25+j3.31

S _{c} u003d S _{1} ^{i} + (- jQ _{in} ) u003d 11.25 + j2.206

U _{3} u003d 115 – = 114.25 kV

U _{2} u003d 114.25 – = 112.87 kV

U _{1} u003d 114.25 – = 114.18 kV

The choice of taps on the transformer PS 2

U _{max} u003d 113.66 R _{t} / 2 X _{t} / 2 S _{max} / 2

S _{pr} S _{2} S _{1}

U _{min} u003d 109.49 4.02 + j2.38

2∆S _{xx}

S _{1} R _{t} /2 X _{t} /2 S _{min} /2

2ΔР _{st} 2ΔQ _{w} 1.006+j0.55

TDN 10000 ± 1.77% in VN neutral

U _{2} ^{i} _{max} u003d 113.66 – = 112.07 kV

U _{2} ^{i} _{min} u003d 109.49 – = 109.1 kV

MAX MIN

U _{f} = 10.5 kV U _{f} = 10.0 kV

K _{tzh} u003d U _{2} ^{i} _{max} / U _{f} u003d 10.67 K _{tzh} u003d 10.91

K _{then} u003d 115/11 u003d 10.45

n _{f} % = 100

n _{f} % = 2.1% n _{f} % = 4.4 %

n _{st} u003d 2 * 1% n _{st} u003d 4 * 1%

k _{td} u003d K _{then} * 1- u003d 10.66 k _{td} u003d K _{then} * 1- = 10.9

U _{2} _{d} u003d U _{2} ^{i} _{max} / k _{td} u003d 10.51 kV U _{2} _{d} u003d U _{2} ^{i} _{max} / k _{td} u003d 10 kV

Conclusion: the obtained voltages are close to the desired ones.

Electrical calculation of a 10 kV distribution network with two-sided power supply

DETERMINATION OF THE PERMISSIBLE VOLTAGE LOSS IN THE NETWORK

Voltage deviations and their influence on the operation of electric power receivers. The electrical load never remains constant. As a result, the voltage loss in the line changes, and consequently, the voltage at the consumer. Gradual voltage changes caused by load changes during the day and year are called voltage deviations, in contrast to short-term voltage drops that occur, for example, when starting short-circuited electric motors.

Voltage deviation is the algebraic difference between the voltage at a given point in a given mode and the rated voltage of the network. Voltage deviations are expressed in volts or as a percentage of the rated mains voltage. Voltage fluctuations affect the operation of power receivers. The most sensitive to them are lighting consumers, and first of all, incandescent lamps widely used in agriculture.

**Table 3.1-technical characteristics of transformers**

Installation item | designation | At the most remote TP | At the nearest TP | |||

Under load | ||||||

10 kV busbars | ∆V _{w 10} |
|||||

10 kV network | ∆U _{10} |
-4 | -one | |||

Tr/tor voltage 10/0.4 kV | allowances | ∆V _{th} |
+7.5 | +7.5 | +5 | +5 |

losses | ∆U _{t} |
-4 | -one | -4 | -one | |

Network voltage 0.38 kV | ∆U _{0.38} |
-4.5 | -6 | |||

consumer | ∆V _{p} |
-5 | +5.5 | -5 | +4 |

0% 10 kV +7.5% 0.38 kV

ΔUt=4% ΔUt=4.5%

ab

+5%

0.38 kV

c

ΔUt=6%

d

For the most remote transformer substation, they are set by the transformer allowance equal to + 7.5%. Then the total allowable voltage loss in the network with a voltage of 10 and 0.38 kV will be:

ΔUt = +7.5-4-(-5) = 8.5%

Voltage deviation at the consumer closest to the transformer point:

ΔVp = -1+7.5-1 = 5.5~ +5%

At the nearest transformer point, we accept a surcharge of + 5%, then

ΔUtd = +5-4-(-5) = +6%

ΔUtc = +5-1< = +4

## Be First to Comment