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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 medium OHTL. 2- Medium OHTL : ( from 80 km to 240 km) The medium OHTL can be represented either by:

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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 Linemedium OHTL Dr. Sahar Abd El Moneim Moussa

  3. 2- Medium OHTL: ( from 80 km to 240 km) The medium OHTL can be represented either by: • - equivalent circuit • T- equivalent circuit Dr. Sahar Abd El Moneim Moussa

  4. j X IS Ir R ICR Ics I VS VR C/2 C/2 i- - Circuit: • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa

  5. The General equation is: • In Matrix Form : VS = 1+ Z VR IS= 1+IR Dr. Sahar Abd El Moneim Moussa

  6. The ABCD Parameters of a short T.L: A=D= 1+, B=Z, C= • Efficiency: x100% • Voltage Regulation: = x100 Dr. Sahar Abd El Moneim Moussa

  7. i- T- Circuit: • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa

  8. The General equation is: • In Matrix Form : VS= 1+Z VR IS= 1+IR Dr. Sahar Abd El Moneim Moussa

  9. The ABCD Parameters of a short T.L: A=D= 1+, B= Z, C= • Efficiency: x100% • Voltage Regulation: = x100 Dr. Sahar Abd El Moneim Moussa

  10. Example 9 A 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/km And C= 0.005 F/km. the line is supplying a load of 30MW at 0.85 lagging pf. Find using approximated -model: • A,B,C and D constants of the line • V and I at the sending end of the line •  and VR of the line Dr. Sahar Abd El Moneim Moussa

  11. 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 10-6mho= 314x10-690 mho YZ= -0.016 + j 0.00628 ( 1) • A= D= 1 B= Z= 20 + j 52  = 55 69  C= Y = j 314 x 10-6mho= 314x10-6 90 mho Dr. Sahar Abd El Moneim Moussa

  12. 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 Dr. Sahar Abd El Moneim Moussa

  13. 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% = Dr. Sahar Abd El Moneim Moussa

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