Power System - Section 4
SECTION 2: INDUCTANCE OF SINGLE-PHASE TRANSMISSION LINES
The inductance of a single-phase transmission line is determined by the magnetic flux linkages produced by current flowing through its conductors. These flux linkages consist of internal flux, which exists within the conductor, and external flux, which surrounds it. By calculating both components, the inductance of a two-wire transmission line can be determined. This section explains internal and external flux linkages, inductance per unit length, and the equivalent radius of a solid conductor used in transmission-line calculations.
1. Magnetic Field of a Two-Wire Line
A single-phase two-wire transmission line consists of two conductors separated by a distance . The conductors carry equal currents in opposite directions: one carries current , while the other carries the return current
For a conductor of radius , assuming uniform current distribution and a non-magnetic conductor, the magnetic field intensity at a distance from its centre is:
Inside the conductor(r<R):
Hᵣ = Ir/2πR²
Outside the conductor (r≥R):
Hᵣ = Ir/2πR
The corresponding magnetic flux density is:
B = μ₀H
where μ₀ = 4π×10⁻⁷ H/mfor air and
approximately non-magnetic materials.
The magnetic field inside the conductor increases linearly with distance from the centre. Outside the conductor, it decreases inversely with distance.
2. Internal Flux Linkages
Internal flux linkages are produced by the magnetic field within the conductor itself. Since the current is assumed to be uniformly distributed across the conductor's cross-section, only a fraction of the total current is enclosed by a circular path of radius r, where r<R.
The current enclosed by this path is:
Iᵣ = I(r/R)²
Using Ampere’s law,
Hᵣ = Iᵣ/2πr = Iᵣ/2πR²
For a conductor of unit length, the small flux through a thin cylindrical shell of radius and thickness dr is:
DΦ = B dr
Substituting B = μ₀ Hᵣ,
DΦ = μ₀Ir/2πR² dr
Only the fraction Iᵣ/I = (r/R)² of the conductor's current links this internal flux. Therefore, the differential flux linkages are:
dλ = (r/R)² dΦ
Integrating from the centre to the conductor surface gives the total internal flux linkages:
λint = μ₀Ir³/2πR² dr
λint = μ₀I/8π
The internal flux linkages per unit current are independent of the conductor radius under the stated assumptions.
3. External Flux Linkages
External flux linkages are produced by the magnetic field outside the conductor. For a two-wire line, consider the region between the surface of one conductor and the other conductor.
At a distance from the centre of the first conductor, the magnetic field due to that conductor is:
Hᵣ = I/2πr
Hence,
B = μ₀I/2πr
For a unit length of conductor, the flux through a thin strip of thickness is:
dΦ = B dr = μ₀I/ 2πr dr
Integrating from the conductor surface to the other conductor at distance , the external flux linkages are:
λext = μ₀Ir/2πR dr
λext = μ₀I/2π In ln (D/R)
The logarithm is the natural logarithm. This expression assumes D is much larger than , so the integration limit can be taken as approximately D.
4. Total Flux Linkages and Inductance
The total flux linkages of one conductor are the sum of internal and external flux linkages:
λ = λint + λext
λ = μ₀/8π + μ₀/2π In ln (D/r)
For a linear magnetic system, inductance is flux linkages per unit current. Thus, the inductance per conductor is:
Lconductor = μ₀/8π + μ₀/2π In ln (D/r)
Using μ₀ = 4π×10⁻⁷ H/m
Lconductor = 2 ×10⁻⁷ [1/4 + In ln (D/r)] H/m
For a complete two-wire circuit, the inductances of the two conductors add. Therefore, the loop inductance per unit length is:
Lloop = 4 ×10⁻⁷ [1/4 + In ln (D/r)] H/m
This distinction is important: use the per-conductor formula when the question asks for one conductor, and the loop formula when it asks for the complete single-phase two-wire line.
5. Equivalent Radius of a Conductor
The expression for inductance can be simplified by defining an equivalent radius that accounts for the internal flux linkages of a solid round conductor.
Starting from:
Lconductor = 2×10⁻⁷[1/4 + In ln(D/R)]
Since , 1/4 = (e¹/⁴)
Lconductor = 2×10⁻⁷ In ln (D/Re¹/⁴)
Define the equivalent radius as:
r = Re⁻¹/⁴ = 0.7788R
Therefore,
Lconductor = 2 ×10⁻⁷ In ln (D/r) H/m
And for the complete two-wire loop:
Lloop = 4 ×10⁻⁷ In ln (D/r) H/m
The equivalent radius is also called the effective radius or geometric mean radius (GMR) of a solid round conductor in this inductance calculation. It is smaller than the physical radius because it accounts for the contribution of internal flux linkages.
Multiple-Choice Questions (MCQs)
Questions
1. The total flux linkages of a conductor consist of:
A. Resistance and capacitance
B. Internal and external flux linkages
C. Voltage and current
D. Conductance and resistance
2. For a solid conductor carrying uniformly distributed current, the magnetic field intensity inside the conductor is proportional to:
A. 1/r
B. 1/r²
C. r
D. r²
3. The magnetic field intensity outside a long straight conductor is:
A. I/2πr
B. Ir/2πr²
C. I/2πR
D. 2πr/I
4. The internal flux linkages per unit length of a solid conductor with uniform current distribution are:
A. μ₀I/2π
B. μ₀I/8π
C. μ₀I/4π
D. μ₀I/π
5. The external flux linkages per unit length depend on:
A. Only the conductor current
B. Only the conductor radius
C. The current and the ratio of conductor spacing to radius
D. Only the conductor material
6. The equivalent radius of a solid round conductor is approximately:
A. 1.25R
B. 0.5R
C. 0.7788R
D. R
7. Which expression gives the inductance per unit length of one conductor in a single-phase two-wire line?
A. 4 × 10⁻⁷ ln(D/r)
B. 2 × 10⁻⁷ ln(D/r)
C. 2 × 10⁻⁷ ln(r/D)
D. 4 × 10⁻⁷ ln(D/r)
8. The loop inductance of a two-wire line is:
A. Half the inductance of one conductor
B. Equal to the inductance of one conductor
C. Twice the inductance of one conductor
D. Four times the inductance of one conductor
9. The equivalent radius is used to account for:
A. The effect of internal flux linkages
B. The conductor’s insulation thickness
C. The line’s capacitance
D. The resistance of the return conductor
10. If conductor spacing increases while conductor radius remains constant, the inductance:
A. Decreases
B. Increases
C. Remains unchanged
D. Becomes zero
Q.
Answer
Explanation
1
B
Total flux linkages include internal and external components.
2
C
Inside the conductor, .
3
A
Outside the conductor, .
4
B
Internal flux linkages are per metre.
5
C
External flux linkages contain and .
6
C
The equivalent radius is .
7
B
This is the per-conductor inductance formula.
8
C
The two conductor inductances add to form the loop inductance.
9
A
The equivalent radius incorporates the internal-flux contribution.
10
B
Inductance increases with .
Five Calculation Problems with Solutions
Problem 1: Calculate Internal Flux Linkages
A solid conductor carries a current of 100 A. Calculate its internal flux linkages per metre, assuming uniform current distribution.
Given:
I = 100 A
Formula:
λᵢₙₜ = μ₀I / 8π
Solution:
λᵢₙₜ = (4π × 10⁻⁷)(100)/8π
Answer: 5 μWb - turn/m
Problem 2: Calculate External Flux Linkages
A conductor of radius 0.01 m carries a current of 50 A. The distance to the return conductor is 2 m. Calculate the external flux linkages per metre.
Given:
I = 50A, R = 0.01 m, D = 2m
Formula:
λₑₓₜ = μ₀I / 2π In ln (D/R)
Solution:
λₑₓₜ = 2 × 10⁻⁷(50) In ln (2/0.01)
= 10⁻⁵ln(200)
λₑₓₜ ≈5.298 ×10⁻⁵ Wb - turn/m
Answer: 52.98 μWb - turn/m
Problem 3: Find the Equivalent Radius
A solid round conductor has a physical radius of 0.02 m. Calculate its equivalent radius.
Given:
R = 0.02m
Formula:
r = 0.7788R
Solution:
r = 0.7788(0.02)
r = 0.015576 m
Answer: 0.015576 m or 15.576 mm
Problem 4: Calculate Inductance per Conductor
A single-phase two-wire line has a conductor radius of 0.01 m and conductor spacing of 2 m. Calculate the inductance per metre of one conductor.
Step 1: Calculate equivalent radius
r = 0.7788(0.01) = 0.007788 m
Step 2: Apply the inductance formula
Lconductor ≈2 × 10⁻⁷ In ln (D/r)
= 2 × 10⁻⁷ In ln (2/0.007788)
= 2 × 10⁻⁷ In ln (256.806)
Lconductor ≈1.11 ×10⁻⁶ H/m
Answer: per conductor
Problem 5: Calculate Loop Inductance
A single-phase two-wire transmission line has conductor spacing of 1.5 m and conductor radius of 0.005 m. Calculate the total loop inductance per metre.
Step 1: Calculate equivalent radius
r = 0.7788(0.005) = 0.003894 m
Step 2: Apply the loop inductance formula
Lₗₒₒₚ = 4 × 10⁻⁷ In ln (D/r)
= 4 × 10⁻⁷ In ln (1.5/0.003894)
4 × 10⁻⁷ ln(385.208)
Lₗₒₒₚ ≈2.38×10⁻⁶ H/m
Answer: for the complete loop