Projekt 1 Władysław Kopczyk.doc.docx

Politechnika Rzeszowska

Wydział Elektrotechniki i informatyki

Katedra Energoelektroniki i Elektroenergetyki

Calculation of parameters of long transmission line

Project 1 from subject Transmission of electricity

Name and surname: ŁUKASZ ZYCH L5 EN-DI

Academic year: 2013/2014

Deadline: 21.11.2013

Task 1:

Consider the 400 kV overhead power lines with a length of 200 km (tower is shown on Figure below). At the end of the line, assume a constant voltage U2 = 400 kV.

Parameters of phases conductors:

-  type of conductor: 6x3xAlFe 445/74 (triple bundle conductor)

-  diameter of conductor: dp = 29,63 mm,

-  specific resistance of conductor: 0,0368 W.mm2.m-1

-  conductance of conductor per 1 km of the length: G1 =2,5.10-8 S/km

-  maximal sag of conductor: fp = 9 m.

Parameters of ground conductors:

-  type of conductor: AlFe 185/31

-  diameter of conductor: dg = 19,08 mm,

-  maximal sag of conductor fg = 8 m.

U1

load

U2

l = 200 km

1)   calculate the electrical parameters of the line: resistance (R), inductance L, capacitance (C), conductance (G)

2)                 draw the magnitude of:

- voltage,

- current,

- active power,

- reactive power.

along the line, i.e. U=f(x), I=f(x), P=f(x), Q=f(x) for cases, where the line is loaded with:

a) natural power (surge impedance loading)

b) 2 x natural power

c) 0,5 x natural power

d) 0 MW, 0 MVAr (no load).

3)                 calculate active power losses for cases a), b), c), d) in point 2)

Rozwiązanie:

a1-2= 4,5              [m]             

a1-3= 4,5              [m]             

a2-3= 3,7              [m]             

a1-5= 12,4              [m]             

a1-6= 8,9              [m]             

a2-6= 9,8              [m]             

a1-4= 9,8              [m]             

a2-5= 13,5              [m]             

a3-6= 6,1              [m]             

1.Obliczenie rezystancji (R):

R=ρls Ω

R=0,0368Ωmm2m*200 000 mπ29,632mm2=10,64 Ω

2.Obliczenie induktancji (L):

L1km=0,46*log(aśrre)+0,05m mHkm

re=nrd12

re=329,632mm*400mm*400mm=128 [mm] 0,128m

aśr=3a12*a13*a23*3a15*a16*a263a14*a25*a36

aśr=34,1m*4,5m*3,7m*311,4m*8,9m*9,8m 39,8m*13,5m*6,1m = 4,36 [m]

L1km=0,46*log(4,36m0,128m)+0,053 =0,78mHkm

L=L1km*l

L=0,78mHkm*200 km=156 mH

3.Obliczenie pojemności (C):

C1=1N-N'=1(δ+δN)-(δ'-δN')

δ=10,0242*log(2hśrre) kmµF

  hśr=3h1*h2*h3 [m]

  hśr=317,6m*13,5m*13,5m= 14,8m

h1=h1 '-23 fp=22,9m-23 *8m=17,6 [m]

h2 =h3=h2'-23fp=18,8 m- 23 *8m=13,5[m] δ=10,0242*log2*14,8[m]0,128[m]=82,7kmµF δ'=10,0242*...