This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: P d = VI R = V 2 C tan watts Where is the power supply frequency in radian /sec and is in radians. Since is usually very small tan = P d = V 2 C 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 () 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 = F/km r R 18ln r = 0.329 F/km Total capacitance = 0.329 * 2.5 = 0 . 8225 F (b) Charging current = I C = CV = 2.84A (c) Generated reactive voltamperes, V I C sin90 = 31.24 kVAr (d) Dielectric power loss = V 2 C = 969.24 watts. (e) P d = R V 2 Page 3 of 4 R= d P V 2 = 0.125M CREATED BY: Mahbubur Rahman. Lecturer, EEE, AIUB Page 4 of 4...
View Full
Document
 Spring '11
 dr.sohrabuddin

Click to edit the document details