F5_Introduction to Underground cables

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Multiple conductors: 3 Phase cables: Belted cables- up to 11kV. Screened cables- from 22-66kV. Pressure Cables- beyond 66 kV. Page 2 of 4
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 volt-amperes, 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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online