--- title: "220 kV double circuit efficiency and regulation" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:double-circuit-220kv-efficiency-regulation" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/double-circuit-220kv-efficiency-regulation site: BCT Engineering --- # 220 kV double circuit efficiency and regulation Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2079 Chaitra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/double-circuit-220kv-efficiency-regulation - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question A double circuit, 220 kV, 50 Hz, three phase transmission line is designed to carry 144 MW for 130 km. The line constants are given as: R = 0.0544 Ω/km (at max allowable temp.), L = 0.494 mH/km, C = 23.1 nF/km. Calculate the following: a) Transmission line efficiency b) Percentage of voltage regulation c) Value of receiving end voltage rise at no load if sending end voltage is held constant. Suggest the options to get efficiency > 94% and voltage regulation < 12%. ## Short answer Using nominal-π model, two parallel circuits, 0.9 pf lag: efficiency is98.93%, voltage regulation2.57%, no-load receiving-end voltage225.65 kV (5.65 kV rise above rated). The line already exceeds the94% efficiency and12% regulation limits; if needed, larger conductors, shunt capacitors, or series compensation could further improve performance. ## Answer The line is 130 km long, so it is a medium line and the **nominal-π** model is used. The two circuits are in parallel. **Assumptions:** power factor 0.9 lagging (not given in the question); receiving-end voltage held at 220 kV. #### Line parameters (per phase) Per circuit: $$ \begin{aligned} R &= 0.0544 \times 130 = 7.072\ \Omega \\ X &= 2\pi(50)(0.494\times10^{-3})(130) = 20.175\ \Omega \\ Y &= 2\pi(50)(23.1\times10^{-9})(130) = 9.434\times10^{-4}\ \text{S} \end{aligned} $$ Two circuits in parallel: $$ Z = \frac{7.072 + j20.175}{2} = 3.536 + j10.088\ \Omega, \qquad Y = j1.887\times10^{-3}\ \text{S} $$ #### Constants and currents $$ \begin{aligned} A &= 1 + \frac{ZY}{2} = 0.9905\angle 0.19° \\ V_R &= \frac{220}{\sqrt{3}} = 127.02\angle 0°\ \text{kV} \\ I_R &= \frac{144\times10^6}{\sqrt{3}\times220\times10^3\times0.9} = 419.9\angle -25.84°\ \text{A} \end{aligned} $$ $$ \begin{aligned} V_S &= AV_R + ZI_R = 129.04\angle 1.59°\ \text{kV/phase} \\ &= 223.51\ \text{kV (line)} \\ I_S &= CV_R + AI_R = 379.1\angle 8.88°\ \text{A} \end{aligned} $$ #### a) Efficiency $$ \begin{aligned} P_S &= \text{Re}(3V_SI_S^*) = 145.56\ \text{MW} \\ \text{Loss} &= 1.56\ \text{MW} \\ \eta &= \frac{144}{145.56}\times100 = 98.93\% \end{aligned} $$ #### b) Voltage regulation At no load, $V_{R0} = V_S/|A|$: $$ \begin{aligned} V_{R0} &= \frac{223.51}{0.9905} = 225.65\ \text{kV (line)} \\ \%VR &= \frac{225.65 - 220}{220}\times100 = 2.57\% \end{aligned} $$ #### c) Receiving-end voltage rise at no load (Ferranti effect) With $V_S$ held at 223.51 kV and the load removed: $$ \Delta V = V_{R0} - V_S = 225.65 - 223.51 = 2.15\ \text{kV} $$ The no-load voltage is 5.65 kV above the rated 220 kV. **Answer: η = 98.93%, VR = 2.57%, no-load rise = 2.15 kV above $V_S$** (receiving end 225.65 kV). The line already meets η > 94% and VR < 12% by a wide margin. #### Options if the limits were not met 1. **Larger conductor** (lower $R$) for higher efficiency, or bundled conductors (lower $X$) for better regulation. 2. **Raise the power factor** with shunt capacitors or SVCs at the load end. At unity pf the regulation falls to about 1.1%. 3. **Series capacitor compensation** to cut $X$ and the voltage drop. 4. **More circuits or a higher voltage** (for example 400 kV), which reduces the current and therefore $I^2R$ and $IX$. 5. **Shunt reactors** at the receiving end to limit the no-load (Ferranti) rise. 6. On-load tap-changing transformers to correct the receiving-end voltage. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which model is used for this 130 km line? A. Short line model B. Nominal π (medium line) C. Long line with hyperbolic functions only D. Lossless model Answer: B. A 130 km line is medium length; nominal π is used. (Key point: Model) 2. How are the parameters of the two circuits combined? A. Z and Y both halved B. Z halved, Y doubled C. Z doubled, Y halved D. Unchanged Answer: B. Series impedance of parallel circuits halves; shunt admittance doubles. (Key point: Parallel circuits) 3. Efficiency is computed as: A. Receiving power divided by sending power, 98.93% B. Sending power divided by receiving power C. 1 minus regulation D. Loss divided by receiving power Answer: A. 144/145.56 = 98.93%. (Key point: Efficiency) 4. Voltage regulation uses the no-load receiving voltage VR0 = VS/|A|, giving: A. 2.57% B. 10% C. 12% D. 0.5% Answer: A. (225.65 - 220)/220 = 2.57%. 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