Transcript Location of Voltage on Bare Conductor

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
Except for superconductors, there is some
resistance/impedance to the current flow
There is a loop path to-from the source
2
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
3
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.
4
Force on Adjacent Current
Carrying Conductors
I1
d
I2
I1 xI 2 x107
DC: F = 5.4
lbs./ft.
d
For RMS Symmetrical current Single Phase Symmetrical
AC: F =
I1 xI 2 x107
lbs./ft.
10.8
d
5
Force on Adjacent Current
Carrying Conductors
I
d
I
A
d
B
C
I
RMS Symmetrical Current
3F Asymmetrical Fault
A or CF
Maximum
F=
I 2 x107 lbs./ft.
34.9
d
BF
Maximum
F=
I 2 x107 lbs./ft.
37.5
d
6
Force on Adjacent Current
Carrying Conductors
I
I
A
d
d
C
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:
34.9 x10  x107
0.5
4 2
= 689 lbs./ft.
This is no small amount of force!
7
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.
8
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
9
Relationship Between
Resistance and Volume Resistivity
height = h
current flow
l = length
w = width
Area = w X h
Resistance = Volume Resistivity x Length
Area
10
Temperature Coefficient of Resistance
RT2 = RT1[1 + aTT + bTT]
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
11
Temperature Coefficient of Resistance
(Continued)
For the range of temperatures in which most conductors
operate the formula reduces to
RT2 = RT1[1 + aTT]
values for a
Conductor
0°C
20°C
25°C
61.2% Aluminum 0.00440 0.00404 0.00389
100.0% Copper
0.00427 0.00393 0.00378
12
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
13
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 nonmagnetic conduit or raceway will increase these factors as
well
14
Alternating Current Resistance
When 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 nonmagnetic conduit or raceway will increase all of these factors.
15
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
16
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
17
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
18
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.
E
I
.
.
R = insulation resistance (ohms) = E/I
19
Typical DC Leakage Current
With Constant Voltage Applied
IG = charging current
IA = absorbtion current
IL = leakage current
IT = total current
IL
20
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
21
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
22
Insulation Resistance Constants
Rubber Like Materials
Ethylene Propylene Rubber Type I
20,000
Ethylene Propylene Rubber Type II, 0-2kV
Ethylene Propylene Rubber Type II, >2kV
Code Grade Synthetic Rubber
Performance Natural Rubber
Performance Synthetic Rubber
Heat Resistant Natural Rubber
Heat Resistant Synthetic Rubber
10,000
20,000
950
10,560
2,000
10,560
2,000
Ozone Resistant Synthetic Rubber
Ozone Resistant Butyl Rubber
Kerite
2,000
10,000
4,000
23
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
24
is difficult and subject to error
Cable Average Electrical Stress
G ave = Voltage to Ground
Insulation thickness (mils)
G ave = volts/mil
T
25
Cable Radial Electrical Stress at
Any Point in the Insulation
G x = Vgrd
Volts/Mil
X Ln(R2/R1)
X
R2
R1
.
Maximum Stress X = R1
Minimum Stress X = R2
26
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
27
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
28
The Formula for Calculating Per Foot
Capacitance For Fully Shielded Cable Is:
7.354
C=
Doi
log10
Doc
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)
29
Shunt Capacitive Reactance

For single conductor shielded primary cables the shunt
capacitance may be calculated by
7354
C=
Doi
Log10
Dui
µµfarad/1000 ft
where:
 = dielectric constant of the insulation
Doi =diameter over insulation
Dui = diameter under insulation
The capacitive reactance may then be calculated as:
1
Xc =
j 2fc
where:
f = frequency in Hz
j = a vector operator
30
The Formula for Calculating Charging Current,
Per Foot, For A Fully Shielded Cable Is:
i = 2fce
i = Charging current
f = 60Hz
e = Voltage Phase to grd
c = Capacitance
31
Example of Charging Current, per
Foot, For Fully Shielded Cable


 7.354 x 2.3 
 x 10-12 x (14.4 x 103) = 0.539 milliamps/ft
i = 2 60
.

  Log 1566

10
 
 
1056
.
 = 2.3
Doc = 1.056 inch
Doi = 1.566 inch
e = 14.4 kV to ground
32
Power Factor Vs Dissipation Factor
A Cable is Generally a Capacitor
Ic
a
Ic should be >>>Ir
b
Ir
Power Factor =
Ir
( Ir ) + ( Ic )
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 ) 2 + ( Ic) 2
So, Power Factor and Dissipation Factor are often thought to be
the same, but they are very different.
33
Dielectric Power Dissipation
(Dielectric Loss)
Ic
It
Power Dissipation
P = E (Ir)
= E (It) cos q
= E(Ic) tan d
d
q
Ir
E
BUT;
Ic = 2fCE
P = 2fCE2(Tan d )
34
Inductive Reactance
GMD
L = 01404
.
Log10
X 10 3 henries to neut. per 1000 ft.
GMR
Where:GMD = Geometric mean distance (equivalent conductor
spacing) between the current carrying cables.
GMR = Geometric mean radius of one conductor - inches
At 60 Hz: 2(frequency) = 377
or
XL = j0.05292 Log10 GMD/GMR ohm to neut. per 1000 ft.
j is a vector operator
35
Geometric Mean Distance
A
C
B
Equilateral Triangle
GMD =A=B=C
B
C
A
A
Right Triangle
GMD = 1.123 A
B
C
Unequal triangle
GMD = 3 AxBxC
36
Geometric Mean Distance
C
C
A
B
Symmetrical Flat
GMD = 1.26 A
A
B
Unsymmetrical Flat
GMD = 3 AxBxC
A
Flat
GMD = A
37
Effective Cross Sectional Area of
Sheath/shield (A)
Type of Shield/Sheath
Wires/Braid
Helical Tape, no lap
Helical Tape, lapped
Corrugated Tape, LCS
Tubular
B-Tape Lap (mils)
b-Tape Thickness (mils)
dis-Dia over Ins. Shield (mils)
dm-Mean sheath/shield Dia. (mils)
ds-Dia. of wires (mils)
w-Tape width (mils)
Formula to Calculate (A)
nds2
1.27 nwb
4 bd m
100
2 (100  L )
1.27[ ( d is + 50) + B ]b
4bdm
n-Number of wires/tapes
L-Tape overlap, %
38