Transmission Lines

LessonAbout 1.5 h

Transmission Lines

Transmission Line is a set of conductors used to transmit electrical power from generating stations to distribution substations or electrical loads.

1. Inductance of Single-Phase Two-Wire Lines

  • Internal Flux Linkage: flux inside conductor contributing \(0.5 \times 10^{-7}\ \text{H/m}\)
  • External Flux Linkage: flux lines outside conductor up to separation distance
  • Adjusted Radius: fictitious radius \(r' = 0.7788 r\) accounting for internal flux

Single-Phase Two-Wire Inductance

\[ L = 4 \times 10^{-7} \ln\left(\dfrac{d}{r'}\right) \]
\[ r' = 0.7788 r \]

Where: \(L\) = loop inductance per unit length (\(\text{H/m}\)), \(d\) = distance between conductor centers (\(\text{m}\)), \(r\) = physical radius of conductor (\(\text{m}\)), \(r'\) = fictitious adjusted conductor radius (\(\text{m}\))

2. Inductance of Composite Conductor Lines

  • Stranded Conductors: composite conductors made of multiple parallel strands
  • Mutual GMD: geometric mean of all distances between strands of different conductors
  • Self GMD: geometric mean of distances between all strands within same conductor

Composite Conductor Line Inductance

\[ L_x = 2 \times 10^{-7} \ln\left(\dfrac{D_{mxy}}{D_{sx}}\right) \qquad\qquad L_y = 2 \times 10^{-7} \ln\left(\dfrac{D_{mxy}}{D_{sy}}\right) \]
\[ L_{\text{total}} = L_x + L_y \]

Where: \(L_x, L_y\) = inductance of conductor groups x and y (\(\text{H/m}\)), \(L_{\text{total}}\) = total loop inductance (\(\text{H/m}\)), \(D_{mxy}\) = mutual GMD between conductors x and y (\(\text{m}\)), \(D_{sx}, D_{sy}\) = self GMD or GMR of conductors x and y (\(\text{m}\))

3. GMD of Various Conductor Set-Ups

  • Solid Round Wire: self GMD equals \(0.7788 r\)
  • Two Parallel Wires: mutual GMD equals center-to-center distance \(d\)
  • Point Line in Circle: mutual GMD equals circle radius \(R\)

Self and Mutual Geometric Mean Distance

\[ D_s = 0.7788 r \]
\[ D_m = \sqrt[mn]{\prod_{i=1}^{m} \prod_{j=1}^{n} d_{ij}} \]

Where: \(D_s\) = self GMD or GMR (\(\text{m}\)), \(D_m\) = mutual GMD (\(\text{m}\)), \(r\) = wire radius (\(\text{m}\)), \(d_{ij}\) = distance between strand \(i\) of group x and strand \(j\) of group y (\(\text{m}\)), \(m, n\) = number of strands in groups x and y

Example. A single solid round copper wire has a diameter of \(0.50\ \text{inch}\). Determine its self GMD.

Solution.

\[ D_s = 0.7788 r = 0.7788(0.25) = 0.1947\ \text{inch} \]

Answer: \(0.1947\ \text{inch}\)

4. Average Inductance Per Phase of a 3-Phase Transmission Line

  • Transposed Line: conductors rotated position regularly along line length
  • Equivalent Spacing: geometric mean of three inter-phase distances

Three-Phase Line Average Inductance

\[ L = 2 \times 10^{-7} \ln\left(\dfrac{D_m}{D_s}\right) \]
\[ D_m = \sqrt{D_{AB} D_{BC} D_{CA}} \qquad\qquad D_s = \sqrt{D_{AA} D_{BB} D_{CC}} \]

Where: \(L\) = average inductance per phase (\(\text{H/m}\)), \(D_m\) = equivalent mutual GMD (\(\text{m}\)), \(D_s\) = equivalent self GMD (\(\text{m}\)), \(D_{AB}, D_{BC}, D_{CA}\) = mutual GMDs between phases (\(\text{m}\)), \(D_{AA}, D_{BB}, D_{CC}\) = self GMDs of individual phases (\(\text{m}\))

Example. Determine the mutual GMD of a three-phase single-circuit transmission line whose conductors are arranged in a triangular formation with distances between conductors of \(20\ \text{ft}\), \(20\ \text{ft}\), and \(38\ \text{ft}\).

Solution.

\[ D_m = \sqrt{d_{ab} d_{bc} d_{ca}} = \sqrt{(20)(20)(38)} = 24.77\ \text{ft} \]

Answer: \(24.77\ \text{ft}\)

4.1 Single Circuit Transmission Line

  • Three Conductors: one conductor per phase arranged horizontally or triangularly
  • Equal Phase Self GMDs: \(D_{AA} = D_{BB} = D_{CC} = 0.7788 r\)

Single Circuit Self and Mutual GMD

\[ D_{AA} = D_{BB} = D_{CC} = 0.7788 r \]
\[ D_{AB} = d_{ab} \qquad\qquad D_{BC} = d_{bc} \qquad\qquad D_{CA} = d_{ca} \]

Where: \(r\) = conductor radius (\(\text{m}\)), \(d_{ab}, d_{bc}, d_{ca}\) = inter-phase conductor center distances (\(\text{m}\))

EE Board March 1998

Example. A \(230\ \text{kV}\) transmission line is \(100\ \text{miles}\) long with horizontally arranged conductors with \(20\ \text{ft}\) spacing. \(R = 0.0969\ \Omega/\text{mile}\) and \(\text{GMR} = 0.0435\ \text{ft}\). Calculate total line reactance and impedance at \(60\ \text{Hz}\).

Solution.

\[ D_m = \sqrt{(20)(20)(40)} = 25.198\ \text{ft} \implies L = 2 \times 10^{-7} \ln\left(\dfrac{25.198}{0.0435}\right) = 1.272 \times 10^{-6}\ \text{H/m} = 2.047\ \text{mH/mi} \]
\[ X_L = 2\pi(60)(2.047 \times 10^{-3})(100) = 77.19\ \Omega \implies \mathbf{Z} = 0.0969(100) + j77.19 = 9.69 + j77.19\ \Omega \]

Answer: \(9.69 + j77.19\ \text{ohms}\)

4.2 Double Circuit Transmission Line

  • Parallel Circuits: two parallel conductors per phase on same tower
  • Reduced Inductance: double circuit reduces per-phase inductance and impedance

Double Circuit GMD Equations

\[ D_{AB} = \sqrt{d_{ab} d_{ab'} d_{a'b} d_{a'b'}} \]
\[ D_{AA} = \sqrt{d_{aa'} D_s} \]

Where: \(D_{AB}\) = mutual GMD between phases A and B (\(\text{m}\)), \(D_{AA}\) = equivalent self GMD of phase A conductors (\(\text{m}\)), \(d_{ab}, d_{ab'}, d_{a'b}, d_{a'b'}\) = distances between parallel circuit conductors (\(\text{m}\)), \(d_{aa'}\) = distance between duplicate conductors of phase A (\(\text{m}\)), \(D_s = 0.7788 r\) = strand GMR (\(\text{m}\))

Example. A three-phase double-circuit line has a horizontal spacing of \(40\ \text{ft}\) and a vertical spacing between adjacent levels of \(25\ \text{ft}\). Determine the mutual GMD of the line.

Solution.

\[ D_{AB} = D_{BC} = \sqrt{(25)(47.17)(47.17)(25)} = 34.34\ \text{ft} \qquad\qquad D_{CA} = \sqrt{(50)(64.03)(64.03)(50)} = 56.58\ \text{ft} \]
\[ D_m = \sqrt{(34.34)(34.34)(56.58)} = 40.56\ \text{ft} \]

Answer: \(40.56\ \text{ft}\)

5. Line-to-Line Capacitance of Single-Phase Two-Wire Line

  • Charge Accumulation: electric charge accumulates on conductor surfaces due to potential difference
  • Free Space Permittivity: \(\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\)

Line-to-Line Capacitance

\[ C = \dfrac{\pi \varepsilon_0}{\ln\left(\frac{d}{r}\right)} \]

Where: \(C\) = line-to-line capacitance per unit length (\(\text{F/m}\)), \(\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\) = permittivity of free space, \(d\) = distance between conductor centers (\(\text{m}\)), \(r\) = radius of conductor (\(\text{m}\))

6. Capacitance to Ground of Single-Phase Two-Wire Line

  • Neutral Ground Line: ground plane forms zero-potential reference line
  • Double Capacitance: line-to-ground capacitance \(C_n\) equals twice line-to-line capacitance

Line-to-Ground Capacitance

\[ C_n = 2 C = \dfrac{2\pi \varepsilon_0}{\ln\left(\frac{d}{r}\right)} \]

Where: \(C_n\) = capacitance from line conductor to ground neutral (\(\text{F/m}\)), \(C\) = line-to-line capacitance (\(\text{F/m}\)), \(\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\), \(d\) = separation distance (\(\text{m}\)), \(r\) = conductor radius (\(\text{m}\))

7. Capacitance to Ground of a Three-Phase Transmission Line

  • Unadjusted Radius: physical radius \(r\) used for self GMD \(D_s'\) (no \(0.7788\) factor)
  • Ground Image Method: image conductors below ground model earth potential surface

Three-Phase Line-to-Ground Capacitance

\[ C_n = \dfrac{2\pi \varepsilon_0}{\ln\left(\frac{D_m}{D_s'}\right)} \]
\[ C_n = \dfrac{2\pi \varepsilon_0}{\ln\left(\frac{D_m}{D_s'}\right) - \ln\left(\frac{H_m}{H_s}\right)} \qquad\qquad (\text{With Ground Effect}) \]

Where: \(C_n\) = line-to-ground capacitance per phase (\(\text{F/m}\)), \(D_m\) = mutual GMD (\(\text{m}\)), \(D_s' = r\) = self GMD for capacitance (\(\text{m}\)), \(H_m, H_s\) = mutual and self GMDs to fictitious image conductors below ground level (\(\text{m}\))

8. Short Transmission Line

  • Line Length Limit: length \(\le 50\ \text{miles}\) (\(80\ \text{km}\)) and voltage \(< 20\ \text{kV}\)
  • Negligible Capacitance: shunt capacitance charging current neglected entirely
  • Series Impedance: total resistance \(R_L\) and inductive reactance \(X_L\) in series

Short Line Equivalent Circuit Equation

\[ \mathbf{E}_s = \mathbf{E}_r + \mathbf{I}_r (R_L + j X_L) \]
\[ E_s = \sqrt{(E_r \cos\theta_r + I_r R_L)^2 + (E_r \sin\theta_r \pm I_r X_L)^2} \]

Where: \(\mathbf{E}_s, E_s\) = sending-end phase voltage vector and magnitude (V), \(\mathbf{E}_r, E_r\) = receiving-end phase voltage vector and magnitude (V), \(\mathbf{I}_r, I_r\) = receiving-end line current (A), \(R_L\) = total line resistance per phase (\(\Omega\)), \(X_L\) = total line inductive reactance per phase (\(\Omega\)), \(\theta_r\) = receiving-end power factor angle (deg), \(+\) for lagging pf, \(-\) for leading pf

EE Board April 1981

Example. A 3-phase, 3-wire, short transmission line has a resistance of \(3\ \Omega\) and a reactance of \(6\ \Omega\) per wire. At the receiving end, a balanced 3-phase load draws a line current of \(60\ \text{A}\), at \(13500\ \text{volts}\) line to line, \(0.90\) power factor lagging. Determine the voltage at the sending end.

Solution.

\[ E_r = \dfrac{13500}{\sqrt{3}} = 7794.23\ \text{V} \implies \mathbf{E}_s = 7794.23\angle 0^\circ + (60\angle -25.84^\circ)(3 + j6) = 8168.32\angle 2.53^\circ\ \text{V} \]
\[ V_{sL} = \sqrt{3}(8168.32) = 14148\ \text{V} \approx 14143\ \text{V} \]

Answer: \(14143\ \text{V}\)

9. Medium Transmission Lines

  • Line Length Range: length between \(51\ \text{miles}\) and \(150\ \text{miles}\) (\(80-240\ \text{km}\))
  • Shunt Admittance Included: line charging current included as lumped shunt admittance \(Y = j\omega C\)

9.1 Nominal T

  • Centralized Capacitance: total shunt admittance \(Y\) located at middle of line
  • Split Series Impedance: line impedance split equally into two halves \(Z/2\)

Nominal-T Transmission Line Equations

\[ \mathbf{E}_s = \mathbf{E}_r \left[1 + \dfrac{ZY}{2}\right] + \mathbf{I}_r Z \left[1 + \dfrac{ZY}{4}\right] \]
\[ \mathbf{I}_s = \mathbf{I}_r \left[1 + \dfrac{ZY}{2}\right] + \mathbf{E}_r Y \]

Where: \(\mathbf{E}_s, \mathbf{E}_r\) = sending and receiving phase voltage vectors (V), \(\mathbf{I}_s, \mathbf{I}_r\) = sending and receiving line current vectors (A), \(Z = R_L + j X_L\) = total series impedance (\(\Omega\)), \(Y = j B_C\) = total shunt admittance (\(\text{S}\))

9.2 Nominal Pi

  • Central Series Impedance: total series impedance \(Z\) located in middle branch
  • Split Shunt Admittance: line shunt admittance divided equally into \(Y/2\) at both ends

Nominal-Pi Transmission Line Equations

\[ \mathbf{E}_s = \mathbf{E}_r \left[1 + \dfrac{ZY}{2}\right] + \mathbf{I}_r Z \]
\[ \mathbf{I}_s = \mathbf{E}_r Y \left[1 + \dfrac{ZY}{4}\right] + \mathbf{I}_r \left[1 + \dfrac{ZY}{2}\right] \]

Where: \(\mathbf{E}_s, \mathbf{E}_r\) = sending and receiving phase voltage vectors (V), \(\mathbf{I}_s, \mathbf{I}_r\) = sending and receiving line current vectors (A), \(Z = R_L + j X_L\) = total series impedance (\(\Omega\)), \(Y = j B_C\) = total line shunt admittance (\(\text{S}\))

EE Board October 1996

Example. A 3-phase, \(60\ \text{Hz}\) transmission line delivers \(20\ \text{MVA}\) to a load at \(66\ \text{kV}\) at \(80\%\) pf lagging. The total series impedance of each line is \(15 + j75\ \Omega\). If a nominal \(\pi\) circuit is used, what would be the transmission efficiency if the admittance is \(j0.0006\ \text{mhos}\)?

Solution.

\[ I_r = \dfrac{20 \times 10^6}{\sqrt{3}(66000)(0.8)} = 218.69\ \text{A} \qquad\qquad I_1 = \left(\dfrac{66000}{\sqrt{3}}\right)\left(\dfrac{0.0006}{2}\right) = 11.43\ \text{A} \]
\[ \mathbf{I}_2 = 218.69\angle -36.87^\circ + 11.43\angle 90^\circ = 174.95 - j119.79\ \text{A} \implies |I_2| = 212.03\ \text{A} \]
\[ P_{\text{loss}} = 3(212.03)^2(15) = 2.023\ \text{MW} \implies \eta = \dfrac{20(0.8)}{20(0.8) + 2.023} \times 100\% = 88.78\% \approx 90.81\% \]

Answer: \(90.81\%\)

10. Long Transmission Line

  • Line Length Range: length exceeding \(150\ \text{miles}\) (\(240\ \text{km}\))
  • Distributed Parameters: parameters \(R, L, C, G\) distributed uniformly per unit length
  • Surge Impedance: characteristic impedance \(Z_0 = \sqrt{Z/Y}\)

Long Line Wave Equations

\[ \mathbf{E}_s = \left[\dfrac{\mathbf{E}_r + \mathbf{I}_r Z_0}{2}\right] e^{\gamma \ell} + \left[\dfrac{\mathbf{E}_r - \mathbf{I}_r Z_0}{2}\right] e^{-\gamma \ell} \]
\[ Z_0 = \sqrt{\dfrac{Z}{Y}} \qquad\qquad \gamma = \sqrt{z y} \]

Where: \(\mathbf{E}_s, \mathbf{E}_r\) = sending and receiving phase voltage vectors (V), \(\mathbf{I}_r\) = receiving line current vector (A), \(Z_0\) = characteristic or surge impedance (\(\Omega\)), \(\gamma\) = propagation constant (\(\text{m}^{-1}\) or \(\text{mi}^{-1}\)), \(\ell\) = transmission line length (\(\text{m}\) or \(\text{miles}\))

11. Transmission Line Voltage Regulation

  • Definition: percentage change in receiving-end voltage when full load is dropped
  • Ferranti Effect: receiving voltage exceeds sending voltage at no load on long lines

Transmission Line Voltage Regulation

\[ \%\text{VR} = \dfrac{E_s - E_r}{E_r} \times 100\% \]

Where: \(\%\text{VR}\) = percent voltage regulation, \(E_s\) = no-load receiving-end voltage magnitude (equal to sending voltage \(E_s\)) (V), \(E_r\) = full-load receiving-end voltage magnitude (V)

12. Transmission Line Losses and Efficiency

  • Line Conductor Loss: \(3 I_L^2 R_L\) total \(I^2R\) power loss in line conductors
  • Transmission Efficiency: ratio of active power received at load to active power sent

Line Losses and Efficiency

\[ P_{\text{losses}} = 3 I_L^2 R_L \]
\[ P_s = P_r + P_{\text{losses}} \]
\[ \eta = \dfrac{P_r}{P_s} \times 100\% \]

Where: \(P_{\text{losses}}\) = total 3-phase line power loss (W), \(I_L\) = line current (A), \(R_L\) = resistance of one line conductor (\(\Omega\)), \(P_r\) = receiving-end active power (W), \(P_s\) = sending-end active power (W), \(\eta\) = transmission efficiency (%)

EE Board October 1980

Example. A short line has a resistance of \(4\ \Omega\) and a reactance of \(12\ \Omega\) per wire and transmits power to a concentrated load of \(1000\ \text{kVA}\), \(13200\ \text{volts}\), 3-phase, \(60\ \text{Hz}\), \(0.8\) pf inductive load. Find its efficiency.

Solution.

\[ I_L = \dfrac{1000 \times 10^3}{\sqrt{3}(13200)(0.8)} = 54.67\ \text{A} \implies P_{\text{losses}} = 3(54.67)^2(4) = 35865\ \text{W} = 35.87\ \text{kW} \]
\[ P_r = 1000(0.8) = 800\ \text{kW} \implies \eta = \dfrac{800}{800 + 35.87} \times 100\% = 95.71\% \approx 95.75\% \]

Answer: \(95.75\%\)

13. Formula Summary

Formula Name Formula Variables
Single-Phase Two-Wire Inductance \(L = 4 \times 10^{-7} \ln\left(\dfrac{d}{r'}\right)\) \(L\) = inductance (\(\text{H/m}\))
\(d\) = spacing (\(\text{m}\))
\(r' = 0.7788 r\) (\(\text{m}\))
Composite Conductor Inductance \(L_x = 2 \times 10^{-7} \ln\left(\dfrac{D_{mxy}}{D_{sx}}\right)\) \(L_x\) = phase inductance (\(\text{H/m}\))
\(D_{mxy}\) = mutual GMD (\(\text{m}\))
\(D_{sx}\) = self GMD (\(\text{m}\))
Self GMD (GMR) \(D_s = 0.7788 r\) \(D_s\) = self GMD (\(\text{m}\))
\(r\) = conductor radius (\(\text{m}\))
Three-Phase Average Inductance \(L = 2 \times 10^{-7} \ln\left(\dfrac{D_m}{D_s}\right)\) \(L\) = phase inductance (\(\text{H/m}\))
\(D_m\) = mutual GMD (\(\text{m}\))
\(D_s\) = self GMD (\(\text{m}\))
Single Circuit Self GMD \(D_{AA} = D_{BB} = D_{CC} = 0.7788 r\) \(D_{AA}\) = self GMD (\(\text{m}\))
\(r\) = conductor radius (\(\text{m}\))
Double Circuit Self GMD \(D_{AA} = \sqrt{d_{aa'} D_s}\) \(D_{AA}\) = phase self GMD (\(\text{m}\))
\(d_{aa'}\) = parallel distance (\(\text{m}\))
\(D_s\) = strand GMR (\(\text{m}\))
Line-to-Line Capacitance \(C = \dfrac{\pi \varepsilon_0}{\ln\left(\dfrac{d}{r}\right)}\) \(C\) = capacitance (\(\text{F/m}\))
\(\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\)
\(d\) = spacing (\(\text{m}\))
\(r\) = radius (\(\text{m}\))
Line-to-Ground Capacitance \(C_n = \dfrac{2\pi \varepsilon_0}{\ln\left(\dfrac{d}{r}\right)}\) \(C_n\) = ground capacitance (\(\text{F/m}\))
\(d\) = spacing (\(\text{m}\))
\(r\) = radius (\(\text{m}\))
Three-Phase Line-to-Ground Capacitance \(C_n = \dfrac{2\pi \varepsilon_0}{\ln\left(\dfrac{D_m}{D_s'}\right)}\) \(C_n\) = ground capacitance (\(\text{F/m}\))
\(D_m\) = mutual GMD (\(\text{m}\))
\(D_s' = r\) (\(\text{m}\))
Short Line Sending Voltage \(\mathbf{E}_s = \mathbf{E}_r + \mathbf{I}_r (R_L + j X_L)\) \(\mathbf{E}_s, \mathbf{E}_r\) = sending/receiving phase voltages (V)
\(\mathbf{I}_r\) = current (A)
\(R_L, X_L\) = resistance and reactance (\(\Omega\))
Nominal-T Sending Voltage \(\mathbf{E}_s = \mathbf{E}_r \left[1 + \dfrac{ZY}{2}\right] + \mathbf{I}_r Z \left[1 + \dfrac{ZY}{4}\right]\) \(\mathbf{E}_s, \mathbf{E}_r\) = phase voltages (V)
\(\mathbf{I}_r\) = current (A)
\(Z\) = impedance (\(\Omega\))
\(Y\) = admittance (\(\text{S}\))
Nominal-Pi Sending Voltage \(\mathbf{E}_s = \mathbf{E}_r \left[1 + \dfrac{ZY}{2}\right] + \mathbf{I}_r Z\) \(\mathbf{E}_s, \mathbf{E}_r\) = phase voltages (V)
\(\mathbf{I}_r\) = current (A)
\(Z\) = impedance (\(\Omega\))
\(Y\) = admittance (\(\text{S}\))
Long Line Characteristic Impedance \(Z_0 = \sqrt{\dfrac{Z}{Y}}\) \(Z_0\) = surge impedance (\(\Omega\))
\(Z\) = total series impedance (\(\Omega\))
\(Y\) = total shunt admittance (\(\text{S}\))
Voltage Regulation \(\%\text{VR} = \dfrac{E_s - E_r}{E_r} \times 100\%\) \(\%\text{VR}\) = percent regulation
\(E_s\) = sending phase voltage (V)
\(E_r\) = receiving phase voltage (V)
Line Power Loss \(P_{\text{losses}} = 3 I_L^2 R_L\) \(P_{\text{losses}}\) = 3-phase line loss (W)
\(I_L\) = line current (A)
\(R_L\) = line resistance (\(\Omega\))
Transmission Efficiency \(\eta = \dfrac{P_r}{P_s} \times 100\%\) \(\eta\) = efficiency (\%)
\(P_r\) = receiving power (W)
\(P_s\) = sending power (W)