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Basic Electrical Characteristics
Carl Landinger
Hendrix Wire & Cable
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When Electric Current Flows
in a Path
There is a voltage (electrical pressure)
driving the current
An electric field eminates from the currentpath
A magnetic field surrounds the current
Except for superconductors, there is someresistance/impedance to the current flow
There is a loop path to-from the source
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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 NOTALTER THE MAGNETIC FIELD LINES
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Two Cables Carrying Current Will Have
Magnetic Fields Interacting With Each Other
Cable #1 Cable #2
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.
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Force on Adjacent Current
Carrying Conductors
I1
d
I2
DC: F = 54101 2
7
.I xI x
d
lbs./ft.
For RMS Symmetrical current Single Phase Symmetrical
AC: F = 108101 2
7
.I xI x
d
lbs./ft.
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Force on Adjacent Current
Carrying Conductors
A B C
d d
I I
I
RMS Symmetrical Current
3F Asymmetrical Fault
A or CF
Maximum34 9
102 7.
I x
d
F = lbs./ft.
BF
Maximum
F = 37 5102 7
.I x
d
lbs./ft.
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Force on Adjacent Current
Carrying Conductors
A C
d dI I
I
RMS Symmetrical Current
3F Asymmetrical Fault
34 9 10 1005
4 2 7.
.
x x = 689 lbs./ft.
Assume: I = 10,000 Amps/Phase, d = 6in. (0.5 ft.)
Maximum Force on A or C Phase is:
This is no small amount of force!
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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 resistivityMaterials having a low resistivity make good
conductors. Materials with high resistivities are
insulators.
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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 beingas conductive of this standard
Aluminum is approximately 61% as conductive as
annealed copper on a volume basis. However, it isover 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
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Relationship Between
Resistance and Volume Resistivity
l = lengthheight = h
current flow w = width Area = w X h
Resistance = Volume Resistivity x LengthArea
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Temperature Coefficient of Resistance
RT2 = RT1[1 + aTT + bTT]
where:
RT2 = DC resistance of conductor at desired orassumed temperature
RT1= DC resistance of conductor at base temperature
T2 = Assumed temperature to which dc resistance is
to be adjustedT1 = Base temperature at which resistance is known
a and b = Temperature coefficients of resistance
at the base temperature for the conductor
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Temperature Coefficient of Resistance
(Continued)
For the range of temperatures in which most conductors
operate the formula reduces to
RT2 = RT1[1 + aTT]
values for a
Conductor 0C 20C 25C
61.2% Aluminum 0.00440 0.00404 0.00389
100.0% Copper 0.00427 0.00393 0.00378
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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
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Alternating Current Resistance
For the general case when calculating impedance forvoltage 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, sheathloss, armor loss, ..
Note: The presence of enclosing metallic, magnetic and non-magnetic conduit or raceway will increase these factors as
well
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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 asseparate heat sources introduced at their location inthe thermal circuit.
Note; The presence of enclosing metallic, magnetic and non-magnetic conduit or raceway will increase all of these factors.
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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 isdesigned for; 15 kV/1.732 = 8.66 kV
Cables used on other systems must be selected
accordingly
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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 on1.33 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
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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 kVcable.
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
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Insulation Resistance
No insulation is perfect. If the conductor is made intoone 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.
E
I
. .
R = insulation resistance (ohms) = E/I
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Typical DC Leakage Current
With Constant Voltage Applied
IG = charging current
IA = absorbtion current
IL = leakage current
IT = total current
IL
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Insulation Resistance Constant
If one uses a 100 to 500 volt DC source to measure theresistance from conductor to shield, or a made shield suchas water, of a 1,000 foot length of insulated cable at atemperature of 60F, the following formula describes the
relationship between the insulation thickness, theresistance reading obtained, and a constant which ispeculiar to the insulation;
R = (IRK) Log10(D/d)
Where;R is the resistance in megohms-1,000 feetD is the diameter over the insulation
dis the diameter under the insulation
IRK is the insulation resistance constant
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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 20,000Thermoplastic Polyethylene 50,000
Composite Polyethylene 30,000
60C Thermoplastic PVC 500
75C Thermoplastic PVC 2,000
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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 950Performance 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
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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 85F, 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
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Cable Average Electrical Stress
G ave = Voltage to Ground
Insulation thickness (mils)
G ave = volts/mil
T
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Cable Radial Electrical Stress at
Any Point in the Insulation
G x = Vgrd Volts/MilX Ln(R2/R1)
.
R1X
R2
Maximum Stress X = R1
Minimum Stress X = R2
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STRESS GRADIENT IN #2-7 STRAND
175 MIL CABLE AT 7.2 kV ac
Maximum Stress = 60.7 V/mil
Minimum Stress = 29.2 V/mil
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STRESS GRADIENT IN 1/0-19 STRAND
345 MIL CABLE AT 20.2 kV ac
Maximum Stress = 105 V/mil
Minimum Stress = 36.0 V/mil
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The Formula for Calculating Per Foot
Capacitance For Fully Shielded Cable Is:
CD
D
oi
oc
=7 354
10
.
log
x 10-12
where, 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)
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Shunt Capacitive Reactance
For single conductor shielded primary cables the shuntcapacitance may be calculated by
where:
= dielectric constant of the insulation
Doi =diameter over insulation
Dui = diameter under insulation
The capacitive reactance may then be calculated as:
C
Log DD
oi
ui
=7354
10
farad/1000 ft
X j fcc=
1
2
where:
f = frequency in Hz
j = a vector operator
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The Formula for Calculating Charging Current,
Per Foot, For A Fully Shielded Cable Is:
i = 2fce
i = Charging current
f = 60Hz
e = Voltage Phase to grd
c = Capacitance
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Example of Charging Current, per
Foot, For Fully Shielded Cable
ix
Log
=
2 607 354 2 3
1566
105610
. .
.
.
x 10-12 x (14.4 x 103) = 0.539 milliamps/ft
= 2.3
Doc = 1.056 inchDoi = 1.566 inch
e = 14.4 kV to ground
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Power Factor Vs Dissipation Factor
A Cable is Generally a Capacitor
Ic
ab
Ir
Ic should be >>>Ir
Power Factor =Ir
I Ir c( ) ( )2 2+
= Cos (b) always < 1.0
Dissipation Factor = Ir/Ic = Tan (a) ranging from 0 to For the normal case where Ic>>>Ir;
Ic Ir Ic +( ) ( )2 2
So, Power Factor and Dissipation Factor are often thought to be
the same, but they are very different.
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Dielectric Power Dissipation
(Dielectric Loss)
Ic It
Power Dissipation
P = E (Ir)
= E (It) cos q= E(Ic) tan d
BUT;
Ic = 2fCE
P = 2fCE2(Tan d )
Ir E
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Inductive Reactance
L LogGMD
GMRX= 01404 1010
3. henries to neut. per 1000 ft.
Where:GMD = Geometric mean distance (equivalent conductorspacing) between the current carrying cables.
GMR = Geometric mean radius of one conductor - inches
At 60 Hz: 2(frequency) = 377
orXL = j0.05292 Log10 GMD/GMR ohm to neut. per 1000 ft.
j is a vector operator
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Geometric Mean Distance
Equilateral Triangle
GMD =A=B=C
Right Triangle
GMD = 1.123 A
Unequal triangle
GMD = AxBxC3
A C
B A
CB BA
C
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Geometric Mean Distance
A B
C
Symmetrical FlatGMD = 1.26 A
A B
C
Unsymmetrical FlatGMD = AxBxC3
A
FlatGMD = A
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Effective Cross Sectional Area of
Sheath/shield (A)
Type of Shield/Sheath Formula to Calculate (A)
Wires/Braid nds2
Helical Tape, no lap 1.27 nwb
Helical Tape, lapped 4100
2 100bd Lm ( ) Corrugated Tape, LCS 127 50. [ ( ) ] d B bis + + Tubular 4bdm
B-Tape Lap (mils) n-Number of wires/tapesb-Tape Thickness (mils) L-Tape overlap, %
dis-Dia over Ins. Shield (mils)
dm-Mean sheath/shield Dia. (mils)
ds-Dia. of wires (mils)
T idth ( il )
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