Características básicas de cables

7/30/2019 Caractersticas bsicas de cables

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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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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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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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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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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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Características básicas de cables