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Insulation resistance numerical
- 1. EE504.54 1 OBJECTIVE • Solve problems on insulation resistance of a cable
- 2. EE504.54 2 Problems 1. A single core cable has a conductor diameter of 1 cm and insulation thickness of 0.4 cm. If the specific resistance of insulation is 5 X 1014 -cm, calculate the insulation resistance for a 2 km length of the cable?
- 3. EE504.54 3 Solution: • Conductor diameter = 1 cm = 1 x 10-2 m • Conductor radius (r1) = (1 x 10-2)/2 = 0.005 m • Length of the cable (l) = 2 km = 2000 m • Resistivity of insulation ( ) = 5 x 1014 -cm =5 X 1012 -m • Internal sheath radius (r2) = (0.005 + 0.004) m = 0.009 m
- 4. EE504.54 4 Insulation resistance of the cable is 1 2 i log 2 R r r l e Fig.1
- 5. EE504.54 5 Insulation resistance of the cable is 005 . 0 009 . 0 log 2000 x 2 10 x 5 R 12 i e 10 x 234 . 0 9 M 234
- 6. EE504.54 6 2. The insulation resistance of a single core cable is 495 M per km. If the core diameter is 2.5 cm and resistivity of insulation is 4.5 X 1014 -cm, calculate the insulation thickness.
- 7. EE504.54 7 • Solution: • Conductor radius (r1) = (2.5 x 10-2)/2 = 0.0125 m • Length of the cable (l) = 1000 m • Resistance of insulation (Ri) = 495 X 106 • Resistivity of insulation ( ) = 4.5 X 1012 -m
- 8. EE504.54 8 Insulation resistance of the cable is 2 i 1 R log 2 e r l r Here r2 is the Internal sheath radius to be calculated. Fig.2
- 9. EE504.54 9 0125 . 0 r log 1000 x 2 10 x 5 . 4 10 x 495 2 12 6 e 12 6 2 10 x 5 . 4 10 x 495 x 1000 x 2 0125 . 0 r log e 69 . 0 0125 . 0 r log 2 e
- 10. EE504.54 10 2 0125 . 0 r2 0125 . 0 x 2 r2 m 025 . 0 r2 2 1 I r -r nsulation thickness 0.0125 - 0.025 0.0125 m (or) 1.25 cm