F5_Introduction to Underground cables

F5_Introduction to Underground cables - P d = VI R = V 2 C...

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American International University Bangladesh EEE 3103: Electrical Power Transmission and Distribution Introduction to underground cables Advantages: 1. Safer. 2. Less liable to damage due to environmental hazards. 3. Low maintenance cost. 4. Less chance of fault. Disadvantages: 1. Fault detection is difficult. 2. High leakage current. 3. Expensive. Usage: Long water crossing (England-France). Densely populated area. Construction Conductor Insulator Sheath (Protecting Layer) Figure: Construction of under ground Cables. Insulators: XLPE (Cross linked Poly Ethylene) PVC Rubber. Page 1 of 4
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Multiple conductors: 3 Phase cables: Belted cables- up to 11kV. Screened cables- from 22-66kV. Pressure Cables- beyond 66 kV. Page 2 of 4
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Dielectric Power loss: Let V= line to neutral supply voltage I R = current through R I C = current through C = charging current. I = Supply Current δ = loss angle. Dielectric Power loss = P d = VI cosφ = VI R From Phasor diagram, I R = I C tanδ = (VωC) tanδ
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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 volt-amperes, 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...
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F5_Introduction to Underground cables - P d = VI R = V 2 C...

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