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Pharos University EE-272. Electrical Power Engineering 1 “Electrical Engineering Dep ” Prepared By: Dr. Sahar Abd El Moneim Moussa. OverHead Transmission Line Short OHTL. CLASSIFICATION OF T.L ACCORDING TO LENGTH. According to length , T.L. can be classified as follows:

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pharos university ee 272

Pharos UniversityEE-272

Electrical Power Engineering 1

“Electrical Engineering Dep”

Prepared By:

Dr. SaharAbd El MoneimMoussa

Dr. SaharAbd El MoneimMoussa

overhead transmission line short ohtl
OverHead Transmission LineShort OHTL

Dr. Sahar Abd El Moneim Moussa

classification of t l according to length
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

Dr. Sahar Abd El Moneim Moussa

slide4

1- Short OHTL: (up to 80 km)

  • Equivalent Circuit:

Dr. Sahar Abd El Moneim Moussa

slide5

The General equation is:

Vs= VR+ Z IR

Is=IR

  • In Matrix Form :

VS = 1 Z VR

IS = 0 1 IR

Dr. Sahar Abd El Moneim Moussa

slide7

The ABCD Parameters of a short T.L:

A=D=1, B=Z, C=0

Where:

  • Z = R+ jXL= zL = rL +jxLΩ
  • Z= total series impedance per-phase in ohms.
  • z = series impedance of one conductor in ohms per unit length
  • XL = total inductive reactance of one conductor in ohms.
  • X = inductive reactance of one conductor in ohms per unit length
  • L = length of the line.

Dr. Sahar Abd El Moneim Moussa

slide8

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

Dr. Sahar Abd El Moneim Moussa

slide9

Where:

  • VRNL = magnitude of receiving end voltage at no-load
  • VRFL = magnitude of receiving end voltage at full-load with

constant Vs

Dr. Sahar Abd El Moneim Moussa

slide10

Example 8:

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:

  • Line to neutral voltage at the sending end
  • Efficiency of the line
  • Voltage regulation of the line

Dr. Sahar Abd El Moneim Moussa

slide11

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= 132790 + (2.48 +j 6.57)* (225.9 –j 140.1)

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

= 148034.4 V

Dr. Sahar Abd El Moneim Moussa

slide12

b) %=

s=Vs - Is = 4.4 –(-31.8)= 36.2

%=

c) = x100

VR%=

Dr. Sahar Abd El Moneim Moussa

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