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 INFINITYNOTE THAT ANY COVERING OR INSULATION DOES NOTALTER THE MAGNETIC FIELD LINES

4

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 CABLETHIS CAUSES A FORCE TO EXIST BETWEEN THE CABLES.IF THE CURRENTS ARE TIME VARYING, A VOLTAGE IS INDUCEDINTO THE ADJACENT CABLE.

5

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 = 10 8101 2

7

.I xI x

d

lbs./ft.

6

Force on Adjacent Current Carrying Conductors

A B Cd d

I I

I

RMS Symmetrical Current3 Asymmetrical Fault

A or CMaximum

34 9102 7

.I x

d

F = lbs./ft.

BMaximum

F = 37 5102 7

.I x

d

lbs./ft.

7

Force on Adjacent Current Carrying Conductors

A Cd dI I

I

RMS Symmetrical Current3 Asymmetrical Fault

34 9 10 10

0 5

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!

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

l = lengthheight = h

current flow w = width Area = w X h

Resistance = Volume Resistivity x LengthArea

11

Temperature Coefficient of Resistance

RT2 = RT1[1 +

where:RT2 = DC resistance of conductor at desired or

assumed temperatureRT1 = DC resistance of conductor at “base” temperatureT2 = Assumed temperature to which dc resistance is

to be adjustedT1 = “Base” temperature at which resistance is known

and Temperature coefficients of resistanceat 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 +

values for

Conductor 0°C 20°C 25°C61.2% Aluminum 0.00440 0.00404 0.00389100.0% Copper 0.00427 0.00393 0.00378

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) + RWhere: YCS is the multiple increase due to skin effect YCP is the multiple increase due to proximity effect R 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 ResistanceNo 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.

EI

..

R = insulation resistance (ohms) = E/I

20

Typical DC Leakage Current With Constant Voltage Applied

IG = charging currentIA = absorbtion currentIL = leakage currentIT = total current

IL

21

Insulation Resistance ConstantIf 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 2,640Varnished Cambric 2,460Crosslinked Polyethylene 0-2 kV 10,000Crosslinked Polyethylene > 2 kV 20,000Thermoplastic Polyethylene 50,000Composite Polyethylene 30,00060°C Thermoplastic PVC 50075°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 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

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)

.

R1

X

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/milMinimum Stress = 29.2 V/mil

28

STRESS GRADIENT IN 1/0-19 STRAND345 MIL CABLE AT 20.2 kV ac

Maximum Stress = 105 V/milMinimum Stress = 36.0 V/mil

29

The Formula for Calculating Per Foot Capacitance For Fully Shielded Cable Is:

C DD

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)

30

Shunt Capacitive Reactance For single conductor shielded primary cables the shunt capacitance

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:

CLog

DD

oi

ui

7354

10

µµfarad/1000 ft

Xj fcc

1

2

where:f = frequency in Hzj = a vector operator

31

The Formula for Calculating Charging Current, Per Foot, For A Fully Shielded Cable Is:

i = 2fce

i = Charging currentf = 60Hze = Voltage Phase to grdc = Capacitance

32

Example of Charging Current, per Foot, For Fully Shielded Cable

ix

Log

2 607 354 2 3

1566105610

. .

.

.

x 10-12 x (14.4 x 103) = 0.539 milliamps/ft

= 2.3Doc = 1.056 inchDoi = 1.566 inche = 14.4 kV to ground

33

Power Factor Vs Dissipation Factor A Cable is Generally a CapacitorIc

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.

34

Dielectric Power Dissipation (Dielectric Loss)

Ic It

Power DissipationP = E (Ir) = E (It) cos = E(Ic) tan

BUT;

Ic = 2fCEP = 2fCE2(Tan )

Ir E

35

Inductive Reactance

L LogGMD

GMRX 01404 1010

3. 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: 2(frequency) = 377orXL = j0.05292 Log10 GMD/GMR ohm to neut. per 1000 ft.

j is a vector operator

36

Geometric Mean Distance

Equilateral TriangleGMD =A=B=C

Right TriangleGMD = 1.123 A

Unequal triangleGMD = AxBxC3

A C

B A

CB BA

C

37

Geometric Mean Distance

A B

C

Symmetrical FlatGMD = 1.26 A

A B

C

Unsymmetrical FlatGMD = AxBxC3

A

FlatGMD = A

38

Effective Cross Sectional Area of Sheath/shield (A)

T y p e o f S h i e l d / S h e a t h F o r m u l a t o C a l c u l a t e ( A ) W i r e s / B r a i d n d s 2 H e l i c a l T a p e , n o l a p 1 . 2 7 n w b H e l i c a l T a p e , l a p p e d 4

1 0 0

2 1 0 0b d

Lm ( )

C o r r u g a t e d T a p e , L C S 1 2 7 5 0. [ ( ) ] d B bi s T u b u l a r 4 b d m

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)w-Tape width (mils)