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Pharos University EE-385. Electrical Power & Machines “Electrical Engineering Dept” Prepared By: Dr. Sahar Abd El Moneim Moussa. OHTL OVERHEAD TRANSMISSION LINE (Part 2). CLASSIFICATION OF T.L ACCORDING TO LENGTH. According to length , T.L. can be classified as follows:
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Pharos UniversityEE-385 Electrical Power & Machines “Electrical Engineering Dept” Prepared By: Dr. SaharAbd El MoneimMoussa Dr. Sahar Abd El Moneim Moussa
OHTLOVERHEAD TRANSMISSION LINE(Part 2) Dr. Sahar Abd El Moneim Moussa
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
Representation of a transmission line by a two port Network Dr. Sahar Abd El Moneim Moussa
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 Dr. Sahar Abd El Moneim Moussa
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. Dr. Sahar Abd El Moneim Moussa
1- Short OHTL: (up to 80 km) • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa
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
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
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
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
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: • Line to neutral voltage at the sending end • Efficiency of the line • Voltage regulation of the line Dr. Sahar Abd El Moneim Moussa
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 Dr. Sahar Abd El Moneim Moussa
b) %= s=Vs - Is = 4.4 –(-31.8)= 36.2 %= c) = x100 VR%= Dr. Sahar Abd El Moneim Moussa
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
j X IS Ir R ICR Ics I VS VR C/2 C/2 i- - Circuit: • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa
The General equation is: • In Matrix Form : VS = 1+ Z VR IS= 1+IR Dr. Sahar Abd El Moneim Moussa
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
i- T- Circuit: • Equivalent Circuit: Dr. Sahar Abd El Moneim Moussa
The General equation is: • In Matrix Form : VS = 1+Z VR IS= 1+IR Dr. Sahar Abd El Moneim Moussa
The General equation is: • In Matrix Form : VS = 1+VR IS= 1+IR Dr. Sahar Abd El Moneim Moussa
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
Example 10 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
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= 314x10-690 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
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 10-690 x 76210 0 + 154 -31.8= 143 -23.5 A Dr. Sahar Abd El Moneim Moussa
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