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Pharos University EE-272

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

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  1. Pharos UniversityEE-272 Electrical Power Engineering 1 “Electrical Engineering Dep” Prepared By: Dr. SaharAbd El MoneimMoussa Dr. SaharAbd El MoneimMoussa

  2. OverHead Transmission LineShort OHTL Dr. Sahar Abd El Moneim Moussa

  3. 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

  4. 1- Short OHTL: (up to 80 km) • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa

  5. 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

  6. Representation of a transmission line by a two port Network Dr. Sahar Abd El Moneim Moussa

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. b) %= s=Vs - Is = 4.4 –(-31.8)= 36.2 %= c) = x100 VR%= Dr. Sahar Abd El Moneim Moussa

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