Power System - Section 5
SECTION 3: INDUCTANCE OF THREE-PHASE TRANSMISSION LINES
Three-phase transmission lines are widely used to transmit electrical power. The inductance of each phase depends on the conductor's physical size, the spacing between the phase conductors, and their arrangement. When the conductors are spaced unequally, the inductances of the three phases may differ. Transposition helps balance the average inductance by exchanging the physical positions of the conductors along the line. This section explains symmetrical and unsymmetrical conductor spacing, geometric mean distance (GMD), geometric mean radius (GMR), and the inductance of transposed three-phase transmission lines.

1. Inductance of a Three-Phase Line
A three-phase transmission line consists of three conductors carrying currents Ia, Ib, and Ic. For a balanced system, the currents have equal magnitudes and are displaced by 120°electrically:
Ia + Ib + Ic = 0
The magnetic field produced by each phase links with the other phase conductors. Therefore, the inductance of each phase depends not only on its own conductor, but also on the currents and positions of the other conductors.
For a solid round conductor, the effective radius used in inductance calculations is:
Ds = r = 0.7788R
where is the physical conductor radius Ds is the self-GMD or GMR of the conductor.
2. Geometric Mean Distance (GMD)
Geometric mean distance represents the equivalent spacing between the phase conductors. For three conductors with mutual distances Dab, Dbc, and , Dca the equivalent GMD is:
Dm = ∛DabDbcDca
Where:
-
Dab= distance between phases and
-
Dbc= distance between phases and
-
Dca= distance between phases and
-
Dm= equivalent GMD (m)
GMD accounts for the mutual spacing of the conductors in a three-phase line.
3. Symmetrical Conductor Spacing
In a symmetrically spaced three-phase line, the three conductors form an equilateral triangle. All phase-to-phase distances are equal:
Dab = Dbc = Dca = D
Therefore,
Dm = ∛D³ = D
For a transposed line, the inductance per phase per unit length is:
L = 2× 10⁻⁷ In ln (Dm/Ds) H/m
For symmetrical spacing, Dm = D , so:
L = 2× 10⁻⁷ In ln (D/Ds) H/m
This formula gives the inductance of one phase conductor, not the sum of all three phases.
4. Unsymmetrical Conductor Spacing
In an unsymmetrically spaced three-phase line, the conductors do not form an equilateral triangle. Thus,
Dab ≠ Dbc ≠ Dca
The inductances of the individual phases may be unequal because the distances between the conductors differ. For a properly transposed line, each phase occupies each physical position for one-third of the line length. The average inductance of each phase can then be represented using the equivalent GMD:
Dm = ∛DabDbcDca
Hence, the average inductance per phase is:
Lavg = 2× 10⁻⁷ In ln (Dm/Ds) H/m
For a solid round conductor, Ds = 0.7788R.
Therefore,
Lavg = 2× 10⁻⁷ In ln (∛DabDbcDca / 0.7788R)
The distances and radius must be expressed in the same units.
5. Transposition of Three-Phase Lines
Transposition is the periodic interchange of the physical positions of the three phase conductors along a transmission line. In a complete transposition cycle, each phase occupies each of the three positions for an equal length.
Each phase occupies all three positions over one complete cycle.
The average inductance is the arithmetic mean of the inductances associated with the three conductor positions:
Lavg = (La + Lb + Lc) / 3
For a transposed three-phase line, this average can be expressed using GMD and GMR:
Lavg = 2× 10⁻⁷ In ln (Dm/Ds) H/m
Transposition equalizes the average inductance of the phases in the idealized line model. It can also reduce the imbalance in induced voltages and the interference caused by unequal conductor positions.
6. Important Equations — Quick Revision
Quantity Equation
GMR of a solid conductor Ds = 0.
GMD for three phases Dm = ∛D
Inductance per phase L = 2× 10⁻⁷ In ln (Dm/Ds) H/m
H/m
GMD for symmetrical spacing Dm = D
Average inductance after transposition L = (La + Lb + Lc) / 3
Inductance for symmetrical spacing L = 2× 10⁻⁷ In ln (D/Ds) H/m
H/m
Five Calculation Problems with Solutions
Problem 1: Calculate the GMD
The distances between the three phase conductors are 3 m, 4 m and 5 m. Calculate the equivalent GMD.
Given:
Dab = 3m, Dbc = 4m, Dca = 5m
Formula:
Dm = ∛DabDbcDca
Solution:
Dm = ∛3×4×5 = ∛60
Dm ≈ 3.9 m
Answer: 3.915 m
Problem 2: Inductance of a Symmetrically Spaced Line
A three-phase line has symmetrical conductor spacing of 4 m. Each conductor has a radius of 0.01 m. Calculate the inductance per phase per kilometre.
Step 1: Calculate GMR
Ds = 0.7788 = 0.7788(0.01)
Ds = 0.007788m
Step 2: Calculate inductance
L = 2×10⁻⁷ In ln (4/0.007788)
L = 2×10⁻⁷ ln(513.61)
L ≈1.248×10⁻⁶ H/m
Convert to H/km:
L = 1.248×10⁻⁶ ×1000
L ≈1.248 mH/km
Answer: 1.248 mH/km per phase
Problem 3: Inductance of an Unsymmetrically Spaced Line
A transposed three-phase line has conductor spacings of 2 m, 3 m and 4 m. The conductor radius is 0.005 m. Calculate the average inductance per phase per kilometre.
Step 1: Calculate GMD
Dm = ∛(2×3×4) = ∛24
Dm ≈2.8845 m
Step 2: Calculate GMR
Ds = 0.7788(0.005) = 0.003894 m
Step 3: Calculate inductance
L = 2×10⁻⁷In ln (208845/0.003894)
L ≈ 2×10⁻⁷ (6.608)
L ≈ 1.322×10⁻⁶ H/m
Therefore,
L ≈ 1.322 mH/km
Answer: 1.322 mH/km per phase
Problem 4: Find the GMD from Inductance
A transposed three-phase line has a conductor GMR of 0.005 m and an inductance of per phase. Calculate the equivalent GMD.
Given:
Ds = 0.005 m, L = 1.0×10⁻⁶ H/m
Formula:
L = 2×10⁻⁷ In ln(Dm/Ds)
Rearranging,
Dm = Dseᴸ/⁽²×¹⁰⁻⁷⁾
Solution:
Dm = 0.005e⁽¹.⁰×¹⁰⁻⁶ ⁾/⁽²×¹⁰⁻⁷⁾
Dm = 0.005e⁵
Dm ≈0.742 m
Answer: Approximately 0.742 m
Problem 5: Total Inductance over a Transmission-Line Length
A transposed three-phase line has an inductance of 1.25 mH/km per phase. Calculate the total inductance of each phase over a 50 km line.
Given:
Lper km = 1025 mH/km
l = 50 km
Formula:
Ltotal = Lper km × l
Solution:
Ltotal = 1.25×50
Ltotal = 62.5 mH
Answer: 62.5 mH per phase
Multiple-Choice Questions (MCQs)
1. What does GMD represent in a three-phase transmission line?
A. The physical radius of a conductor
B. The equivalent mutual spacing between phase conductors
C. The conductor's resistance
D. The distance between towers only
2. Which expression gives the GMD for three phase conductors?
A. (Dab + Dbc + Dca)/3
B. √(DabDbc)
C. ∛(Dab × Dbc × Dca)
D. Dab × Dbc × Dca
3. For a solid round conductor, the GMR used in inductance calculations is approximately:
A. 0.5R
B. 0.7788R
C. 1.25R
D. 2R
4. In a symmetrically spaced three-phase line, the three phase-to-phase distances are:
A. All equal
B. All different
C. Two equal and one zero
D. Always 1 metre
5. For symmetrical spacing with distance D, the GMD is:
A. D/3
B. √3D
C. D
D. 3D
6. What is the formula for inductance per phase of a transposed three-phase line?
A. 2×10⁻⁷ ln(Ds/Dm) H/m
B. 2×10⁻⁷ ln(Dm/Ds) H/m
C. 4×10⁻⁷ ln(Dm/Ds) H/m
D. 2×10⁻⁷(Dm/Ds) H/m
7. What is transposition of a transmission line?
A. Increasing conductor radius
B. Changing the current magnitude
C. Periodically exchanging the physical positions of phase conductors
D. Connecting all phases together
8. In a complete transposition cycle, each phase occupies each physical position for:
A. The entire line length
B. An equal length
C. Zero length
D. A random length
9. Why is transposition used in an unsymmetrically spaced line?
A. To make conductor resistance zero
B. To equalize average phase inductances in the idealized line
C. To eliminate all power losses
D. To increase conductor current
10. If GMD increases while GMR remains constant, the inductance per phase:
A. Increases
B. Decreases
C. Remains constant
D. Becomes zero
Q.
Answer
Explanation
1
B
GMD is the equivalent mutual spacing between the phase conductors.
2
C
The three distances are multiplied and the cube root is taken.
3
B
For a solid round conductor, .
4
A
Symmetrical spacing means all three distances are equal.
5
C
The GMD equals the common spacing .
6
B
The per-phase formula is .
7
C
Transposition periodically exchanges phase positions.
8
B
Each phase occupies each position for an equal length.
9
B
Transposition balances the average inductance of the phases in the ideal model.
10
A
Inductance rises with the logarithm of GMD divided by GMR.