Selection of Location for Distribution Substation

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Parameters

Number of consumers

Load 1 Load 2 Load 3 Load 4 Load 5

P, kW 25 400 630 63 1000

X, km 0.3 0.4 1.2 1.6 2.8

Y, km 0.3 1.2 1.6 0.3 1.0

Cos φ 0.6 0.7 0.8 0.75 0.9

Finding load center coordinates from the data given below

Find: a) Determine the coordinates of active electrical load centre

b) Determine the coordinates of reactive electrical load centre c) Plot the given data on the load topology graph. I take 3x2km topology with power loads consumers (1 div. = 0.1 km).

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Find the highest load (load 5) radius using ma;

1000 0.633.14 800

km

Ra5 =√(P5/π*ma5)

Solution: a) Plotting the electrical load centers of each consumer on the load

topology, with scale of mg=0.2km/cm (Fig.1)

b) Calculating the radius of a circles of active and reactive load of each consumer.

c) Calculating the scale for active load (ma) using the the following

formulaTaking the lowest load(load-1) radius Ra1= 0.1km,

then = 25/3.14*0.12=796≈800 kW/km2min. 2

min

;active pPmR

min. 2

min

;active pPmR

Rai =√(Pi/π*ma) and Rri =√(Pi/π*ma),

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if the max. load Ra5 is possible to plotted on the topology, then the scale is accepted for the rest of active load to plot their radius, therefore the scale is accepted the rest of the active loads to plot their radius.Therefore,

Calculating the radius of circles for active loads with the scale I get that is ma=800kw/km2

Ra1= 0.0199*√P1 =0.0199*√25=0.10kmRa2= 0.0199*√P2 =0.0199*√400=0.40kmRa3= 0.0199*√P3 =0.0199*√630=0.50kmRa4= 0.0199*√P4 =0.0199*√63=0.16kmRa5= 0.0199*√P5 =0.0199*√1000=0.63km

d) Calculating the reactive load of each consumer using the following relation: Q1=Pi*tan(φi)where, tanφ – the angle which can be found using cosφ. φi= cos -

1(cos( φi) ), cos(φi) is the power factor for Pi load

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Load1 φ1=cos -1(0.6 )= 53.13o

Load2 φ2=cos -1(0.7 )= 45.57o

Load3 φ3 =cos -1(0.8 )= 36.87o

Load4 φ4=cos -1(0.75 )= 41.41o

Load5 φ5 =cos -1(0.9 )=25.84o

Therefore, the reactive power isQ1 = 25*tan 53.13 = 33.33KvarQ2 =400*tan 45.57= 408.04KVarQ3 =630*tan 36.87= 472.50KVarQ4 =63*tan 41.41= 55.56KVarQ5 =1000*tan 25.84= 484.28KVar

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e) Calculating the radius of circles for reactive loads with the same scale that is ma=800kw/km2

Rr1= 0.0199*√Q1 =0.0199*√33.33= 0.11kmRr2= 0.0199*√Q2 =0.0199*√408.04=0.402kmRr3= 0.0199*√Q3 =0.0199*√472.50=0.43kmRr4= 0.0199*√Q4 =0.0199*√55.56=0.15kmRr5= 0.0199*√Q5 =0.0199*√484.28=0.44km

=

We need to use the following formula to determine coordinates of all consumer loads

10

1

;

n

i ii

a n

ii

P XX

P

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10

1

;

n

i ii

a n

ii

PYY

P

Centre A (Xa0 ; Ya0) – location of the main Distr.SS

10

1

;

n

i ii

r n

ii

Q XX

Q

1

0

1

;

n

i ii

r n

ii

QYY

Q

Centre B (Xr0 ; Yr0) – location of the Comp. equipment

Where, Xa0 ; Ya0 – coordinate of the active electrical load centreXr0 ; Yr0 – coordinate of the reactive electrical load centre

f) Finding the approximate centre of active and reactive electrical loads:

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=

Therefore , the distribution sub station will be allocated near by point A (1.80 ; 1.20)

10

1

;

n

i ii

a n

ii

P XX

P

xao =

xao = 1.80km

10

1

;

n

i ii

r n

ii

QYY

Q

Yro =

Yro = 1.20km

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10

1

;

n

i ii

r n

ii

QYY

Q

Yro = Yro = 1.20km

Therefore , the synchronous compensator will be allocated near by point B (1.50 ; 1.20).

xro = 1.50km

10

1

;

n

i ii

r n

ii

Q XX

Q

xro =

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Parameters Number of consumers

Load 1 Load 2 Load 3 Load 4 Load 5

P, kW 25 400 630 63 1000

Ra , km 0.10 0.40 0.50 0.16 0.63

Cos φ 0.6 0.7 0.8 0.75 0.9

tanφ 1.33 1.02 0.75 0.88 0.48

Q , kVar 33.33 408.04 472.50 55.56 484.28

Rr , km 0.11 0.402 0.43 0.15 0.44

Answer: Point A(1.80 ; 1.20) will be the place where the DSS is to be located and centre of active load. Point B (1.50 ; 1.20) will be the place where the Comp. is to be located and centre of reactive load.

Load center coordinates output data

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Fig.1. Load Topology of all group of consumers

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ConclusionLocation of transformer substation and compensating equipment is near to the larger loads.

When the power factor increases, the active load radius increases and the reactive load radius decreases.

THANK YOU