REPRESENTATION OF TRANSMISSION LINES PERFORMANCE OF TRANSMISSION LINES EE-PC 4111: POWER SYSTEM ANALYSIS Constants of Transmission Lines • A transmission line has three constants: R, L and C. These constants are distributed uniformly along the whole length of the line. • R and L form the SERIES IMPEDANCE. • While the capacitance C () existing between the conductors forms the SHUNT PATH. Classification of Overhead Transmission Lines • Depending upon the matter in which capacitance is taken into account, overhead transmission lines are classified as: 1. Short Transmission Lines 2. Medium Transmission Lines 3. Long Transmission Lines Short Transmission Lines • When the length of an overhead transmission line is upto about 50 mi and the line voltage is comparatively low (< 20 kV), it is usually considered as a short transmission line. • Due to smaller length and lower voltage, the capacitance effects are small and hence can be neglected. • Therefore, while studying the performance of a short transmission line, only resistance and inductance of the line are taken into account. Medium Transmission Lines • When the length of an overhead transmission line is about 50- 150 mi and the line voltage is moderately high (>20 kV < 100 kV), it is considered as a medium transmission line. • Due to sufficient length and voltage of the line, the capacitance effects are taken into account. • For purposes of calculations, the distributed capacitance of the line is divided and lumped in the form of condensers shunted across the line at one or more points. Long Transmission Lines • When the length of an overhead transmission line is more than 150 km and line voltage is very high (> 100 kV), it is considered as a long transmission line. • For the treatment of such a line, the line constants are considered uniformly distributed over the whole length of the line and rigorous methods are employed for solution. Important Terms • VOLTAGE REGULATION. When a transmission line is carrying current, there is a voltage drop in the line due to resistance and inductance of the line. The result is that receiving end voltage (ππ ) of the line is generally less than the sending end voltage (ππ ). This voltage drop (ππ − ππ ) in the line is expressed as a percentage of receiving end voltage ππ and is called voltage regulation. Percent voltage regulation: π½ πΊ − π½πΉ %π½πΉ = × πππ% π½πΉ Important Terms • TRANSMISSION EFFICIENCY. The power obtained at the receiving end of a transmission line is generally less than the sending end power due to losses in the line resistance. The ratio of receiving end power to the sending end power of a transmission line is known as the transmission efficiency of the line. Percent transmission efficiency: π·πΉ %πΌπ» = × πππ% π·πΊ π½πΉ π°πΉ cos π½πΉ %πΌπ» = × πππ% π½πΊ π°πΊ cos π½πΊ Single-Phase Short Transmission Lines • For short transmission lines, only resistance and inductance of the line are taken into account. • Here, the total line resistance and inductance are shown as concentrated or lumped instead of being distributed. Analysis of Single-Phase Short Transmission Lines By KVL, ππ − πΌ π + πππΏ − ππ = 0 Rearranging, ππ = ππ + πΌ π + πππΏ For short transmission lines, πΌ = πΌπ = πΌπ Since π + πππΏ is represented as the series impedance of the line, then π½πΊ = π½πΉ + π°πΉ π In proper phasor notation, π½πΊ ∠πΆ = π½πΉ ∠π + π°πΉ π Three-Phase Short Transmission Lines • This system may be regarded as consisting of three single phase units, each wire transmitting one-third of the total power. As a matter of convenience, we generally analyze 3-phase system by considering ONE PHASE ONLY. Analysis of Three-Phase Short Transmission Lines • Since we consider one phase only, then we represent it by one of its phases to the neutral. • The equations we use in the analysis of single phase lines can be used in the per-phase calculations for three-phase transmission lines. • However, we represent the quantities as their per phase values. Hence, π½πΊ(πππ πππππ) = π½πΉ(πππ πππππ) + π°πΉ(πππ πππππ) π(πππ πππππ) • Interpreting these equations in their simple phasor forms, if ππ is the reference phasor then π½πΊ ∠πΆ = π½πΉ ∠π° π + π°πΉ πππ πππππ π πππ πππππ [π/πβππ π] Example Problem No. 1 1. A single phase overhead transmission line delivers 1100 kW at 33 kV at 0·8 p.f. lagging. The total resistance and inductive reactance of the line are 10 β¦ and 15 β¦ respectively. Determine: (i) sending end voltage (ii) sending end power factor and (iii) transmission efficiency. Solution: First we solve for πΌ, ππ 1100 × 103 πΌ = πΌπ = = ππ cos ππ 33 × 103 0.8 πΌ = πΌπ = 41.67 π΄ πππππππ In phasor form, πΌ = 41.67∠ − 36.87° π΄ Example Problem No. 1 (i) sending end voltage The sending end voltage is given by ππ = ππ ∠0° + πΌπ Computing its value, ππ = 33000∠0° + 41.67∠ − 36.87° 10 + π15 π½πΊ = πππππ. ππππ∠π. ππ° π½ (ii) sending end power factor Since πΌ = 0.42° and ππ = 36.87°, the sending pf angle is ππ = πΌ + ππ = 0.42° + 36.87° ππ = 37.29° So cos ππ = cos 37.29° = π. ππππ πππππππ Example Problem No. 1 (iii) transmission efficiency We know that ππ %π π = × 100% ππ where ππ = ππ πΌπ cos ππ . Plugging in the values, we get ππ %π π = × 100% ππ πΌπ cos ππ 11000 × 103 %π π = × 100% 33709.3176 41.67 0.7956 %πΌπ» = ππ. ππ% Example Problem No. 2 2. An overhead 3-phase transmission line delivers 5000 kW at 22 kV at 0·8 p.f. lagging. The resistance and reactance of each conductor is 4 β¦ and 6 β¦ respectively. Determine: (i) percentage regulation and (ii) transmission efficiency. Solution: Solving for ππ , we get ππ = ππ = ππ ∠0° 3 22000∠0° 3 + + πππππ ∠ − cos −1 πππ 3ππ cos ππ 5000 × 103 3 × 22000 × 0.8 π + ππΏ ∠ − cos −1 0.8 π½πΊ = πππππ. ππ∠π. π° π½/πππππ 4 + π6 Example Problem No. 2 (i) percentage regulation Solving the percent voltage regulation can either be in phase or line quantities, in this solution, let us use per phase quantities ππ − ππ %ππ = × 100% ππ 22000 13822.65 − 3 %ππ = × 100% 22000 3 %π½πΉ = π. πππ% (ii) transmission efficiency Solving for πΌ, πΌ = πΌππππ = 5000 × 103 3 × 22000 × 0.8 = 164.02 π΄/πβππ π Example Problem No. 2 (ii) transmission efficiency πππππ 5000 × 103 % ππ = × 100% = 2 πππππ + 3πΌ π 5000 × 103 + 3 164.02 % πΌπ» = ππ. ππ% 2 4 × 100% That’s all. Thank you!