Power Systems Test 4

For economic sizer the ratio of the outer diameter to the conductor diameter should be ‘e’.

\(V = r\;{g_{max}}\ln \left( {\frac{{{r_1}}}{r}} \right) = r\;{g_{max}}\ln e = r{g_{max}}\)

Where r is the radius of the conductor in cm,

⇒ 100 = 80 r

⇒ r = 1.25 cm

Diameter of the conductor = 2 × 1.25 = 2.5 cm

Given that,

Capacitance between two conductors (C_ab) = 1 μF

Capacitance per phase = C/ph = 2 C_ab

The capacitance between a pair of cores with third core earthed for a length of 20 km is

C₃ = 0.18 × 20 = 3.6 μF

Core to neutral capacitance (C_N) = 2C₃ = 2 × 3.6 = 7.2 μF

Charging current (I_C) = 2πfV_phC_N

\( = 2\pi \times 50 \times \frac{{3300}}{{\sqrt 3 }} \times 7.2 \times {10^{ - 6}} = 4.31\;A\)

kVA taken by the cable = 3V_phI_C

\( = 3 \times \frac{{3300}}{{\sqrt 3 }} \times 4.31 = 24.63\;kVA\)

Concept:

The insulation resistance a single core cable is given by

\(R = \frac{\rho }{{2\pi l}}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)\)

Where ρ is the resistivity of insulation

l is the length of cable

r₁ is the radius of the conductor

r₂ is the internal sheath radius

Calculation:

Given that,

Insulation resistance (R) = 495 MΩ

Length of cable (l) = 1 km = 10⁵ cm

The diameter of the conductor = 2.5 cm

Radius of the conductor (r₁) = 1.25 cm

Resistivity of insulation (ρ) = 4.5 × 10¹⁴ Ω-cm

\(495 \times {10^6} = \frac{{4.5 \times {{10}^{14}}}}{{2\pi \times {{10}^5}}}\ln \left( {\frac{{{r_2}}}{{1.25}}} \right)\)

⇒ r₂ = 2.495 cm

Capacitance of cable, \(C = \frac{{{\varepsilon _r}l}}{{41.4{{\log }_{10}}\left( {\frac{D}{d}} \right)}} \times {10^{ - 6}}F\)

ε_r = 4, l = 2000 m

d = 14 cm

D = 14 + 2 (10) = 34 cm

\(C = \frac{{4 \times 2 \times {{10}^3}}}{{41.4{{\log }_{10}}\left( {\frac{{34}}{{14}}} \right)}} \times {10^{ - 9}} = 0.5\;\mu F\)

Voltage between core and sheath is

\({V_{ph}} = \frac{{66 \times {{10}^3}}}{{\sqrt 3 }} = 38.1\;kV\)

Charging current = 2πfC V_ph

= 2π × 50 × 0.5 × 10^-6 × 38.1 × 10³

Concept:

The capacitance a single core cable is given by

\(C = \frac{{2\pi {\varepsilon _0}{\varepsilon _r}}}{{\ln \left( {\frac{D}{d}} \right)}}\)

Where ε_r is the relative permittivity of the dielectric

D is the diameter of the cable with insulation

d is the diameter of the core

Dielectric loss in a cable = V²ωC tan δ

Calculation:

Given that,

Diameter of the core (d) = 2.5 cm

Radial thickness of insulation = 0.5 cm

The diameter of the thickness of insulation = 1 cm

Diameter of the cable = 2.5 + 1 = 3.5 cm

The relative permittivity of the dielectric (ε_r) = 3.0

The capacitance of a cable \( = \frac{{2\pi \times 8.854 \times {{10}^{ - 12}} \times 3}}{{\ln \left( {\frac{{1.75}}{{1.25}}} \right)}} = 4.96 \times {10^{ - 10}}F/m\)

For 4 km long, the capacitance is

= 4.96 × 10^-10 × 4000 = 1.984 μF

Power factor at no load = cos ϕ = 0.02

⇒ ϕ = 88.854°

Loss angle, δ = 90 – ϕ = 90 – 88.854 = 1.146°

Dielectric loss per phase is,

\( = 2\pi \times 50 \times {\left( {\frac{{33 \times {{10}^3}}}{{\sqrt 3 }}} \right)^2} \times 1.984 \times {10^{ - 6}} \times \tan 1.146 = 4.52\;kW\)