Pharos University EE-385

Dr. Sahar Abd El Moneim Moussa 1

Pharos UniversityEE-385

Electrical Power & Machines“Electrical Engineering Dept”

Prepared By:Dr. Sahar Abd El Moneim Moussa


OHTLOVERHEAD TRANSMISSION

LINE(Part 2)


CLASSIFICATION OF T.L ACCORDING TO LENGTH

According to length , T.L. can be classified as follows:

1- Short T.L. : up to 80 km( 50 miles) 2- Medium T.L. : >80 km <240 km 3- Long T.L. : > 240 km


Representation of a transmission line by a two port Network

ABCD phSV phRV

RI SI


The General equation is :

Vs = A VR + B IR

IS = C VR + D IR

In Matrix Form :

VS A B VR

IS = C D IR


Where,

VS = sending- end phase (line –to-neutral) voltage

VR = receiving - end phase (line –to-neutral) voltage

Is = sending-end phase current.

IR = receiving-end phase current.


1- Short OHTL: (up to 80 km)

Equivalent Circuit:


The General equation is:

Vs= VR+ Z IR

Is=IR

In Matrix Form :

VS = 1 Z VR

IS = 0 1 IR


The ABCD Parameters of a short T.L: A=D=1, B=Z, C=0

Where,Z = R+ jXL = zL = rL +jxL Ω

Z= total series impedance per-phase in ohms.z = series impedance of one conductor in ohms per unit lengthXL = total inductive reactance of one conductor in ohms.

X = inductive reactance of one conductor in ohms per unit length L = length of the line.


Efficiency:

x100%

Voltage Regulation:It is the change of voltage at the receiving end of the line when the load varies from no-load to a specified full load at a specified power factor while the sending end voltage is held constant.

= x100


Where,

VRNL = magnitude of receiving end voltage at no-load

VRFL = magnitude of receiving end voltage at full-load with

constant Vs


Example 9:A three-phase, 50 Hz overhead 40 km T.L. has a line voltage of 23 kV at the receiving end, a total impedance of 2.48+j6.57 per phase, and a load of 9 MW with receiving end lagging pf of 0,85. Calculate:a) Line to neutral voltage at the sending endb) Efficiency of the linec) Voltage regulation of the line


Solution:a)

Cos R= 0.85 lagging , R=31.8 lagging= -31.8

Is=IR=265.8 -31.8 A= 225.9 –j140.1 A

Vs= VR+Z IR= 132790 + (2.48 +j 6.57)* (225.9 –j 140.1)

=13279 + 560.2 –j347.5 +j1484.2 +920.5= 14759.7 + j 1136.7 = 148034.4 V


b) %= s=Vs - Is = 4.4 –(-31.8)= 36.2

%=

c) = x100

VR%=


2- Medium OHTL: ( from 80 km to 240 km)

The medium OHTL can be represented either by:

.i - equivalent circuit ii. T- equivalent circuit


i- - Circuit:Equivalent Circuit:

j X

VS VR

IS R

C/2C/2

Ics I ICR

Ir



In Matrix Form : VS = 1+ Z

VR

IS = 1+ IR


The ABCD Parameters of a short T.L:A=D= 1+, B=Z, C=

Efficiency: x100%

Voltage Regulation: = x100


i- T- Circuit:Equivalent Circuit:



In Matrix Form : VS = 1+ Z VR

IS = 1+ IR



In Matrix Form : VS = 1+ VR

IS = 1+ IR


The ABCD Parameters of a short T.L:A=D= 1+, B= Z, C=

Efficiency: x100%

Voltage Regulation: = x100


Example 10A 50 Hz 200 km, transmission line has 132 kV between the lines at the receiving end and has per phase R=0.1 /km , L=0.828 mH/kmAnd C= 0.005 F/km. the line is supplying a load of 30MW at 0.85 lagging pf. Find using approximated -model:

a) A,B,C and D constants of the lineb) V and I at the sending end of the line)c and VR of the line


Solution:Z= R+ j L= 0.1 x 200 + j 2 x x 50 x 0.828 x 10-3 x 200 = 20 + j 52 = 55 69 Y= j C= j 2 x x 50 x 0.005 x 10-6 x 200 = j 314 x 10-6 mho= 314x10-6 90 mhoYZ= -0.016 + j 0.00628 ( 1)

a) A= D= 1

B= Z= 20 + j 52 = 55 69

C= Y = j 314 x 10-6 mho= 314x10-6 90 mho


b) Vs= A VR + B IR = VR + Z IR

VR (ph)=

IR= =

Vs= 76210 0 + 55.7 69 x 154 -31.8= 83200 3.6 V

Is= C VR= + DIR= Y VR + I R

= 314 x 10-6 90 x 76210 0 + 154 -31.8= 143 -23.5 A


c) PR= 30 MW

PS = 3 x VSph x Is x Cos S

S= VS - IS= 3.6 –(23.5) = 27.1

PS= 3 x 83200 x 143 x cos 27.1= 31.77 MW

%=

VR% =

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