1 / 38

Basic Electrical Characteristics

Basic Electrical Characteristics. Carl Landinger Hendrix Wire & Cable. When Electric Current Flows in a Path. There is a voltage (electrical pressure) driving the current An electric field eminates from the current path A magnetic field surrounds the current

montana-wilkerson
Download Presentation

Basic Electrical Characteristics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Basic Electrical Characteristics Carl Landinger Hendrix Wire & Cable

  2. When Electric Current Flows in a Path • There is a voltage (electrical pressure) driving the current • An electric field eminates from the current path • A magnetic field surrounds the current • Except for superconductors, there is some resistance/impedance to the current flow • There is a loop path to-from the source

  3. A Cable Carrying Current has a Magnetic Field Associated with the Current Flow CONDUCTOR INSULATION MAGNETIC FIELD FLUX LINES EXTEND OUT TO INFINITY NOTE THAT ANY COVERING OR INSULATION DOES NOT ALTER THE MAGNETIC FIELD LINES

  4. Two Cables Carrying Current Will Have Magnetic Fields Interacting With Each Other Cable #2 Cable #1 MAGNETIC FIELD (FLUX) FROM EACH CABLE LINKS THE ADJACENT CABLE THIS CAUSES A FORCE TO EXIST BETWEEN THE CABLES. IF THE CURRENTS ARE TIME VARYING, A VOLTAGE IS INDUCED INTO THE ADJACENT CABLE.

  5. Force on Adjacent Current Carrying Conductors I1 dI2 DC: F = lbs./ft. For RMS Symmetrical current Single Phase Symmetrical lbs./ft. AC: F =

  6. Force on Adjacent Current Carrying Conductors I d d A B C I I RMS Symmetrical Current 3F Asymmetrical Fault A or CF Maximum lbs./ft. F = lbs./ft. BF Maximum F =

  7. Force on Adjacent Current Carrying Conductors I d d A C I I RMS Symmetrical Current 3F Asymmetrical Fault Assume: I = 10,000 Amps/Phase, d = 6in. (0.5 ft.) Maximum Force on A or C Phase is: = 689 lbs./ft. This is no small amount of force!

  8. Resistivity Vs Conductivity • Resistivity is a property of every material • Resistivity is a measure of a material to resist the flow of DC current • Resistivity is stated as per unit volume or weight at a specific temperature • Conductivity is a measure of a material to conduct DC current and is the reciprocal of resistivity • Materials having a low resistivity make good conductors. Materials with high resistivities are insulators.

  9. Percent Conductivity • The conductivity of conductor grade annealed copper was established as the standard and given as 100% (IACS) • Other materials are stated as a percentage of being as conductive of this standard • Aluminum is approximately 61% as conductive as annealed copper on a volume basis. However, it is over twice as conductive on a weight basis. • It is possible to exceed 100% i.e. silver is 104.6% • Metal purity and temper effect conductivity

  10. Relationship Between Resistance and Volume Resistivity height = h l = length current flow w = width Area = w X h Volume Resistivity x Length Resistance = Area

  11. Temperature Coefficient of Resistance RT2 = RT1[1 + a(T2-T1) + b(T2-T1)2] where: RT2 = DC resistance of conductor at desired or assumed temperature RT1 = DC resistance of conductor at “base” temperature T2 = Assumed temperature to which dc resistance is to be adjusted T1 = “Base” temperature at which resistance is known a and b = Temperature coefficients of resistance at the base temperature for the conductor

  12. Temperature Coefficient of Resistance (Continued) For the range of temperatures in which most conductors operate the formula reduces to RT2 = RT1[1 + a(T2-T1)] values for a

  13. Effective AC Resistance • “Effective” ac resistance is required for voltage drop calculations • “Effective” ac resistance includes • Skin effect • Proximity effect • Hysteresis and Eddy current effects • Radiation loss • Shield/sheath loss • Conduit/pipe loss

  14. Alternating Current Resistance • For the general case when calculating impedance for voltage drop or system coordination; • Rac = Rdc(1 + YCS + YCP) + DR Where: • YCS is the multiple increase due to skin effect • YCP is the multiple increase due to proximity effect • DR is the apparent increase due to shield loss, sheath loss, armor loss, ……….. Note: The presence of enclosing metallic, magnetic and non-magnetic conduit or raceway will increase these factors as well

  15. Alternating Current ResistanceWhen Calculating for Ampacity Determination • Rac = Rdc(1 + YCS + YCP) Where; • YCS is the multiple increase due to skin effect • YCP is the multiple increase due to proximity effect • Shield loss, sheath loss, armor loss, …are handled as separate heat sources introduced at their location in the thermal circuit. Note; The presence of enclosing metallic, magnetic and non-magnetic conduit or raceway will increase all of these factors.

  16. Insulation Thickness • Cables are voltage rated phase to phase based on a grounded WYE three phase system unless stated • Thus, unless otherwise noted, the insulation thickness is designed for a voltage equal to the cable voltage rating divided by 1.732 • For a 15kV cable the insulation thickness is designed for; 15 kV/1.732 = 8.66 kV • Cables used on other systems must be selected accordingly

  17. Insulation Thickness • For an ungrounded 15 kV delta system the voltage to the neutral point varies from 15 kV/1.732 depending on load balance. For this case, it is common to select insulation thickness based on 1.33 x 15 kV or 20 kV as long as a fault to GRD. is cleared within 1 hour. • This is the origin of the 133% insulation level • The insulation thickness for a 20 kV cable is 215 mils/ICEA, 220 mils/AEIC

  18. Insulation Thickness • When a phase to ground fault occurs on an ungrounded delta system, full phase to phase voltage appears across the insulation • For 15 kV this is equivalent to a 15 X 1.732 = 26 kV cable. • If such a fault is to be allowed to exist for more than 1 hour, it is common to select insulation thickness based on this voltage. • This is the origin of the 173% level • the 173% level is not common and the values are not widely published

  19. Insulation Resistance No insulation is perfect. If the conductor is made into one electrode, and the shield over the insulation, or made shield such as water is used as the other electrode, and a Direct Current Voltage E, applied across the electrodes, a current I, will flow. Using Ohms Law, E = I/R, an insulation resistance can be calculated. . I . E R = insulation resistance (ohms) = E/I

  20. Typical DC Leakage Current With Constant Voltage Applied IG = charging current IA = absorbtion current IL = leakage current IT = total current IL

  21. Insulation Resistance Constant If one uses a 100 to 500 volt DC source to measure the resistance from conductor to shield, or a made shield such as water, of a 1,000 foot length of insulated cable at a temperature of 60°F, the following formula describes the relationship between the insulation thickness, the resistance reading obtained, and a constant which is peculiar to the insulation; R = (IRK) Log10(D/d) Where; R is the resistance in megohms-1,000 feet D is the diameter over the insulation d is the diameter under the insulation IRK is the insulation resistance constant

  22. Insulation Resistance Constants Non Rubber Like Materials Impregnated Paper 2,640 Varnished Cambric 2,460 Crosslinked Polyethylene 0-2 kV 10,000 Crosslinked Polyethylene > 2 kV 20,000 Thermoplastic Polyethylene 50,000 Composite Polyethylene 30,000 60°C Thermoplastic PVC 500 75°C Thermoplastic PVC 2,000

  23. Insulation Resistance Constants Rubber Like Materials Ethylene Propylene Rubber Type I 20,000 Ethylene Propylene Rubber Type II, 0-2kV 10,000 Ethylene Propylene Rubber Type II, >2kV 20,000 Code Grade Synthetic Rubber 950 Performance Natural Rubber 10,560 Performance Synthetic Rubber 2,000 Heat Resistant Natural Rubber 10,560 Heat Resistant Synthetic Rubber 2,000 Ozone Resistant Synthetic Rubber 2,000 Ozone Resistant Butyl Rubber 10,000 Kerite 4,000

  24. Insulation Resistance Constant Important Notes • If the measurement is not made at 60° F but at a temperature not less than 50 or more than 85°F, correction factors must be used to correct to 60° • If the measurement is made on a length other than 1,000 feet, correction to an equivalent 1,000 foot length is necessary • Insulation Resistance Constants (IRK) are published for different classes of insulations. These are minimums and actual values obtained from test measurements should exceed these values or there is an indication of a problem in the material or test • Using IRK to determine the condition of cables in the field is difficult and subject to error

  25. Cable Average Electrical Stress G ave = Voltage to Ground Insulation thickness (mils) G ave = volts/mil T

  26. Cable Radial Electrical Stress at Any Point in the Insulation G x = Vgrd Volts/Mil X Ln(R2/R1) X R1 R2 . Maximum Stress X = R1 Minimum Stress X = R2

  27. STRESS GRADIENT IN #2-7 STRAND175 MIL CABLE AT 7.2 kV ac Maximum Stress = 60.7 V/mil Minimum Stress = 29.2 V/mil

  28. STRESS GRADIENT IN 1/0-19 STRAND345 MIL CABLE AT 20.2 kV ac Maximum Stress = 105 V/mil Minimum Stress = 36.0 V/mil

  29. The Formula for Calculating Per Foot Capacitance For Fully Shielded Cable Is: x 10-12 • where, e is the dielectric constant of the covering • Doc is the diameter over the conductor (or semi conducting shield, if used) • Doi is the diameter over the covering (or insulation in the case of shielded cables)

  30. Shunt Capacitive Reactance • For single conductor shielded primary cables the shunt capacitance may be calculated by where: e = dielectric constant of the insulation Doi =diameter over insulation Dui = diameter under insulation The capacitive reactance may then be calculated as: µµfarad/1000 ft where: f = frequency in Hz j = a vector operator

  31. The Formula for Calculating Charging Current, Per Foot, For A Fully Shielded Cable Is: i = 2pfce i = Charging current f = 60Hz e = Voltage Phase to grd c = Capacitance

  32. Example of Charging Current, per Foot, For Fully Shielded Cable x 10-12 x (14.4 x 103) = 0.539 milliamps/ft e = 2.3 Doc = 1.056 inch Doi = 1.566 inch e = 14.4 kV to ground

  33. Power Factor Vs Dissipation Factor A Cable is Generally a Capacitor Ic Ic should be >>>Ir a b Ir Power Factor = = Cos (b) always < 1.0 Dissipation Factor = Ir/Ic = Tan (a) ranging from 0 to For the normal case where Ic>>>Ir; So, Power Factor and Dissipation Factor are often thought to be the same, but they are very different.

  34. Dielectric Power Dissipation (Dielectric Loss) Ic It Power Dissipation P = E (Ir) = E (It) cos q = E(Ic) tan d BUT; Ic = 2pfCE P = 2pfCE2(Tan d ) d q Ir E

  35. Inductive Reactance henries to neut. per 1000 ft. Where: GMD = Geometric mean distance (equivalent conductor spacing) between the current carrying cables. GMR = Geometric mean radius of one conductor - inches At 60 Hz: 2p(frequency) = 377 or XL = j0.05292 Log10 GMD/GMR ohm to neut. per 1000 ft. j is a vector operator

  36. Geometric Mean Distance C A C B A B B A C Unequal triangle GMD = Equilateral Triangle GMD =A=B=C Right Triangle GMD = 1.123 A

  37. Geometric Mean Distance C C A A B A B Symmetrical Flat GMD = 1.26 A Unsymmetrical Flat GMD = Flat GMD = A

  38. Effective Cross Sectional Area of Sheath/shield (A) B-Tape Lap (mils) n-Number of wires/tapes b-Tape Thickness (mils) L-Tape overlap, % dis-Dia over Ins. Shield (mils) dm-Mean sheath/shield Dia. (mils) ds-Dia. of wires (mils) w-Tape width (mils)

More Related