Find the inductive reactance per mile of a single-phase overhead transmission line operating at 60 Hz, given the conductors to b
1 answer:
Answer:
The inductive reactance is 0.8281 Ω/mile
Explanation:
Given;
frequency, f = 60 Hz
space between the center, GMD = 20 ft
The inductive reactance per mile is calculated as;
Where;
GMR is geometric mean radius (obtained from manufacturer's table)
GMD is geometric mean distance
For Partridge conductor, GMR = 0.0217 ft;
Xa = 2.022 x 10⁻³ x 60 x ln (1 / 0.0217)
Xa = 0.4647 Ω/mile
= 2.022 x 10⁻³ x f x ln GMD
= 2.022 x 10⁻³ x 60 x ln(20)
= 0.3634 Ω/mile
Inductive reactance;
Therefore, the inductive reactance is 0.8281 Ω/mile
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The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)