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Unformatted text preview: P d = VI R = V 2 ωC tanδ watts Where ω is the power supply frequency in radian /sec and δ is in radians. Since δ is usually very small tanδ = δ P d = V 2 ωCδ watts Problem: An 11kV 50Hz. single phase cable has a diameter of 20mm, and an inter sheath radius of 15mm. If the dielectric has a relative permittivity of 2.4 and a loss angle (δ) of 0.031 radians, determine for 2.5 km long: (a) capacitance (b) charging current (c) Generated reactive volt amperes (d) dielectric loss (e) equivalent insulation resistance. Solution: (a) C = μF/km r R 18ln ε r = 0.329 μF/km Total capacitance = 0.329 * 2.5 = 0 . 8225 μF (b) Charging current = I C = ωCV = 2.84A (c) Generated reactive voltamperes, V I C sin90 = 31.24 kVAr (d) Dielectric power loss = V 2 ωCδ = 969.24 watts. (e) P d = R V 2 Page 3 of 4 R= d P V 2 = 0.125MΩ CREATED BY: Mahbubur Rahman. Lecturer, EEE, AIUB Page 4 of 4...
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 Spring '11
 dr.sohrabuddin
 Volt, power supply, Electric power transmission, Dielectric

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