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General CalculationsExcerptfrom
PRYSMIANSWIRE AND CAB LE
ENG IN EER ING GU IDE
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Voltage Rating
The selection of the cable insulation level tobe used in a particular installation shall bemade on the basis of the applicable phase-
to-phase voltage and the general systemcategory as outlined below:
100 Percent Level - Cables in thiscategory may be applied where thesystem is provided with relay protectionsuch that ground faults will be cleared asrapidly as possible, but in any casewithin one minute. While these cablesare applicable to the great majority ofcable installations that are on groundedsystems, they may also be used on othersystems for which the application ofcables is acceptable, provided the above
clearing requirements are met incompletely de-energizing the faultedsection.
133 Percent Level - This insulation levelcorresponds to that formerly designatedfor ungrounded systems. Cables in thiscategory may be applied in situationswhere the clearing time requirements ofthe 100 percent level category cannot bemet, and yet there is adequateassurance that the faulted section will bede-energized in a time not exceeding
one hour. Also, they may be used whenadditional insulation strength over the100 percent level category is desirable.
173 Percent Level - Cables in thiscategory should be applied on systemswhere the time required to de-energize agrounded section is indefinite. Their useis recommended also for resonantgrounded systems. Consult the cablemanufacturer for insulation thicknesses.
In common with other electrical equipment,
the use of cable is not recommended onsystems where the ratio of the zero topositive sequence reactance of the systemat the point of cable application lies between-1 and -40 since excessively high voltagesmay be encountered in the case of groundfaults.
Voltage Regulation
Voltage regulation is often the limiting factorin the choice of either conductor or type ofinsulation. While the heat loss in the cable
determines the maximum current it cansafely carry without excessive deterioration,many circuits will be limited to currents lowerthan this in order to keep the voltage dropwithin permissible values. In this connectionit should be remembered that the highvoltage circuit should be carried as far aspossible so that the secondary runs, wheremost of the voltage drop occurs, will besmall.
The voltage drop of a feeder circuit may becalculated by the following formulae:
( )V
V V
V
S L
L
=10 0
Where:
V = Voltage regulation in percentVL = Voltage across loadVS = Voltage at source
And VS is defined by the formula:
( ) ( )22 sincos IXVIRVV LLS +++=
Where:
= Angle by which the load currentlags the voltage across the load
cos = Power factor of loadR = Total AC resistance of feederX = Total reactance of feederI = Load current
When the power factor angle is nearly thesame as the impedance angle, that is:
tan X
R
The voltage regulation equation above willreduce to the following formula:
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( ) cos sinV V RI XI S L = +
The above formulas apply directly for single-phase lines when resistance and reactanceare loop values and voltage is voltagebetween lines.
For three-phase circuits, use voltage toneutral and resistance and reactance ofeach conductor to neutral. This method willgive the voltage drop to neutral. To obtainthe voltage drop line-to-line, multiply the
voltage drop by 3 .
Note: The percent voltage drop is the samebetween conductors as from conductor to
ground and should not be multiplied by 3 .
Example:
3 single copper conductors, 5 kV cables innon-metallic conduit.
Size conductor = 350 kcmil copper, 90Insulation = 90 mils
J acket = 70 milsOverall diameter = 1.13Voltage (VS) = 5,000 volts (3 phase)Current (I) = 460 amperes
Power Factor = 0.8 (cos )Length = 2550 ft
Resistance = 0.0308 ohms/1000 ftat 25C 0.096 ohmsor 2500 ft at 90C
Reactance (X) = 0.024 ohms/1000 ft(See section onconductor reactance)0.072 ohms for 2500ft including 20%random lay
( ) ( )22 sincos IXVIRVV LLS +++=
22 )]460072(.6[.)]460096(.8[.3
5000+++=
LLVV
Solving for VL: VL = 2,831.6 volts
Line-to-line voltage =2831.6 3 =4904 V
Voltage Drop =5000 - 4904 =96 volts
V =
=5000 4904
4904100 196%( ) .
Approximate Formula:
Voltage Drop = line to neutral
= R I cos +X I sin= 0.096 460 0.8 +0.072
460 0.6= 35.3 +19.8 =55.1
Line-to-line drop =55.1 3 =95.4 V
Conductor Reactance
The reactance of any stranded or solid
conductor can be calculated for a specificfrequency, conductor size, and spacing. Thefollowing equation can be utilized to find thereactance of a given configuration by usingthe concept of geometric mean radius.
3
10 100153.0log1404.02
+=r
gfX
Where:
X = Reactance in ohms/1000 ftf = Frequency in Hertz
r = Radius of conductorg = geometric mean radius between
conductors and is given by thefollowing formula:
g a b znn= ...
Where:n = number of conductorsa = distance from conductor a to bb = distance from conductor b to czn = distance from conductor n-1 to n
Using the equations above, a nomogram (agraphic representation that consists ofseveral lines marked off to scale andarranged in such a way that by using astraightedge to connect known values ontwo lines an unknown value can be read at
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the point of intersection with another line)was constructed (the Nomogram is locatedat later in this document). This diagram canbe used to determine the reactance of anysolid or concentric stranded conductor. Itcovers spacing encountered for conduit
wiring as well as for open wire circuits.Various modifications are necessary for useunder special conditions is covered in noteson the nomogram. The reactances shownare for 60-Hertz operation.
Where regulation is an importantconsideration several factors should be keptin mind in order to obtain the best operatingconditions.Open wire lines have a high reactance. Thismay be improved by using parallel circuitsbut is much further reduced by usinginsulated cable. Three conductors in the
same conduit have a lower reactance thanconductors in separate conduits.
Single conductors should not be installed inindividual magnetic conduit because of theexcessive reactance.
Three conductors in magnetic conduit willhave a somewhat higher reactance than
cables in non-magnetic conduit.
The following table lists equationscommonly used for determining variousparameters of an electrical system where:
E = Phase-to-phase voltageI = Current, in amperes% Eff = Percent efficiency in decimalspf = Power factor in decimalskVA = Kilovolt-amperehp = HorsepowerkW = Kilowatts
Electrical Formulas for Determining Amperes,Horsepower, Kilowatts, and Kilovolt-Amperes
ALTERNATING CURRENTDesired
DataSingle-Phase
Two-Phase*Four-Wire
Three-Phase(per phase)
Amperes whenkVA is known E
kVA 1000 E
kVA
2
1000 E
kVA
3
1000
Amperes whenkilowatts is known pfE
kW1000
pfEkW
21000
pfEkW
31000
Amperes whenhorsepower is known
hp
E Eff pf
746
% hp
E Eff pf
746
2 % hp
E Eff pf
746
3 %
Kilovolt-Amperes
I E1000
I E 21000
I E 31000
KilowattsI E pf
1000 I E pf 2
1000 I E pf 3
1000
HorsepowerI E Eff pf %
746 I E Eff pf 2
746
% I E Eff pf 3746
%
In three-wire, two-phase balanced circuits, the current in the common conductor is 2 timesthat in either of the other conductors.
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Shielding of Insulated Cables
Shielding should be considered for non-metallic covered cables operating at a circuitvoltage above 2000 volts for single
conductor cables and 5000 volts forassembled conductors with a commonoverall jacket.
Definition of shielding
Shielding of an electric power cable is thepractice of confining the dielectric field of thecable to the insulation of the conductor orconductors. It is accomplished by means ofstrand and insulation shields.
Functions of Shielding
A strand shield is employed to precludeexcessive voltage stress on voids betweenconductor and insulation. To be effective, itmust adhere to or remain in intimate contactwith the insulation under all conditions.
An insulation shield has a number offunctions:
1. To confine the dielectric field within thecable.
2. To obtain symmetrical radial distributionof voltage stress within the dielectric,thereby minimizing the possibility ofsurface discharges by precludingexcessive tangential and longitudinalstresses.
3. To protect cable connected to overheadlines or otherwise subject to inducedpotentials.
4. To limit radio interference.5. To reduce the hazard of shock. This
advantage is obtained only if the shieldis grounded. If not grounded, the hazardof shock may be increased.
Use of Insulation Shielding
The use of shielding involves considerationof installation and operating conditions.Definite rules cannot be established on apractical basis for all cases. However,shielding should be considered fornonmetallic covered cables operating at acircuit voltage over 2000 where any of thefollowing conditions exist:
1. Connection to aerial lines.2. Transition from conducting to non-
conducting environment.3. Transition from moist to dry earth.4. Dry soil, such as in the desert.5. Damp conduits.6. Where the surface of cable collects
conducting materials.7. Where electrostatic discharges are of
low enough intensity not to damagecable but are sufficient in magnitude tointerfere with radio or televisionreception.
8. Where safety to personnel is involved.
More specific information on requirementsfor shielding by cable type is provided in thefollowing table.
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Operating Voltage Limits (kV) Above Which Insulation Shieldingis Required
Power Cable - 100 and 133 Percent Insulation Level
1. Single conductor (including assemblies of single conductors)a. With metallic sheath or armor ................................................ 5kVb. All others ................................................................................ 2kV
2. Multiple conductors with common coveringa. With discharge-resisting jacket.............................................. 5kVb. With non-discharge-resisting jacket....................................... 2kVc. With metallic sheath or armor ................................................ 5kV
Multiplying Factors for Equivalent Three-Phase Voltages for
Single- or Two-Phase AC Systems or for DC SystemsSingle- and Two-Phase AC Systems*
Over 5000 VoltsSingle- and Two-Phase ACSystems*
and DC Systems 5000 Volts or Less One Side GroundedUngrounded and Midpoint
Grounded
1 1.73 0.866
* Where it is not definitely specified that a line operates as an isolated single- or two-phase system, it will
be considered as a branch of a 100% insulation level three-phase circuit, and the rating will be the line-
to-line voltage of this 100% insulation level three-phase circuit.
Importance of Shielding
Where there is no metallic covering or shieldover the insulation, the electric field will bepartly in the insulation and partly in whateverlies between the insulation and ground. Theexternal field, if sufficiently intense in air, willgenerate surface discharge and convertatmospheric oxygen into ozone, which maybe destructive to rubber insulations and toprotective jackets. If the surface of the cableis separated from ground by a thin layer of
air and the air gap is subjected to a voltagestress, which exceeds the dielectric strengthof air, a discharge will occur, causing ozoneformation.
The ground may be a metallic conduit, adamp non-metallic conduit or a metallicbinding tape or rings on an aerial cable, a
loose metallic sheath, etc. Likewise,damage to non-shielded cable may resultwhen the surface of the cable is moist, orcovered with soot, soapy grease or otherconducting film and the external field ispartly confined by such conducting film sothat the charging current is carried by thefilm to some spot where it can discharge toground. The resultant intensity of dischargemay be sufficient to cause burning of theinsulation or jacket.
Where non-shielded non-metallic jacketedcables are used in underground ductscontaining several circuits, which must beworked on independently, the external fieldif sufficiently intense can cause shock tothose who handle or contact energizedcable. In cases of this kind, it may be
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advisable to use shielded cable. Shieldingused to reduce hazards of shock shouldhave a resistance low enough to operateprotective equipment in case of fault. Insome cases, the efficiency of protectiveequipment may require proper size ground
wires as a supplement to shielding. Thesame considerations apply to exposedinstallations where personnel who may notbe acquainted with the hazards involvedhandle cables.
Grounding of the Insulation Shield
The insulation shield must be grounded atleast at one end and preferably at two ormore locations. This decreases thereactance to fault currents and increaseshuman safety factor. It is recommended that
the shield be grounded at cable terminationsand at splices and taps. Stress conesshould be made at all shield terminations.
The shield should operate at or near groundpotential at all times. Frequent grounding ofshields reduces the possibility of opensections on nonmetallic covered cable.Multiple grounding of shields is desirable inorder to improve the reliability and safety ofthe circuit. All grounding connections shouldbe made to the shield in such a way as toprovide a permanent low resistance bond.
Using a mechanical clamp (such as aconstant tension spring or a hose clamp) isusually adequate to secure the connection.In some instances, it may be preferable to
solder the connection. The area of contactshould be ample to prevent the current fromheating the connection and melting thesolder. For additional security, a mechanicaldevice, such as a clamp, may be used tofasten the ends of the connection together.
This combination will ensure a permanentlow resistance, which will maintain contacteven if the solder melts.
The wire or strap used to connect the cableshield ground connection to the permanentground must be of adequate size to carrythe fault current. Shielding which does nothave adequate ground connection due todiscontinuity of the shield or to impropertermination may be more dangerous thannon-shielded non-metallic cable andhazardous to life.
Shield Materials
Two distinct types of materials are employedin constructing cable shields:
Nonmetallic shields - consist of either aconducting tape or a layer of conductingcompound. The tape may be conductingcompound, fibrous tape faced or filled withconducting compound, or conducting fibroustape.
Metallic shield - should be nonmagneticand may consist of tape, braid, concentricserving of wires, or a sheath.
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Cable GeometryConfiguration
Induced SheathVoltage Formulae
(V to Neutral per ft)
Metallic Shield LossFormulae for SolidlyGrounded Shields
Where:
Single Phase
A A
S
Phase A:IXM
Phase A:
I2 RS
XM( )
2
RS( )
2X
M( )2
+
Total Loss:
2I2
RS
XM( )
2
RS( )
2X
M( )2
+
Equilateral
A B
S
C
S
Phase A, B & C:IXM
Phase A, B & C:
I2
RS
XM( )
2
RS( )
2X
M( )2
+
Total Loss:
3I2
RS
XM( )
2
RS( )
2X
M( )2
+
P =RS / Y
Q =RS / Z
Rectangular
A B
S
C
S
Phase A & C:
I 3 Y2
XM
A
2
2
+
2
Phase B:IXM
Phase A & C:
I2
RS
P
23Q
2+( )2 3 P Q( ) 4+4 P
21+( ) Q2 1+( )
Phase B:
I2
RS
I
Q2
1+
Total Loss:
3I2
RS
P
2Q
2+ 2+
2 P2
1+( ) Q2 1+( )
Y =xM +A/2
Z =xM - A/6
Formulae continued on Next Page.
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Cable GeometryConfiguration
Induced SheathVoltage Formulae
(V to Neutral per ft)
Loss Formulae forSolidly GroundedMetallic Sheaths
Where:
Flat
A B
S
C
S
Phase A & C:
I 3 Y2
XM
A( )2
+
2
Phase B:IXM
Y =xM +A
Z =xM - A/3
2-Circu it, No PhaseRotation
A B
S
C
S
A B C
Phase A & C:
I 3 Y2
XM
B
2
2
+
2
Phase B:I X
M
A
2+
Y =xM +A +B/2
Z =xM +A/3 - B/6
2-Circuit, PhaseRotation
A B
S
C
S
ABC
Phase A & C:
I 3 Y2
XM
B
2
2
+
2
Phase B:I X
M
A
2+
Phase A & C:
I2
RS
P
23Q
2+( )2 3 P Q( ) 4+4 P
21+( ) Q2 1+( )
Phase B:
I2
RS
I
Q2
1+
Total Loss:
3I2
RS
P2 Q2+ 2+
2 P2
1+( ) Q2 1+( )
Y =xM +A - B/2
Z =xM +A/3 - B/6
xM = 2f (0.1404 log10(S/rm))/ft
A = 2f (0.1404 log10(2))/ft
B = 2f (0.1404 log10(5))/ft
RS = /(8rmt) /ft
RS = Metallic Shield Resistance (/ft)
t = Metallic Tape Thickness (inches)
f = Frequency (Hz)
S = Center to Center Spacing of Cables (inches)
rm = Metallic Shield Mean Radius (inches)
I = Conductor Current (amperes)
= Metallic Shield Resistivity at 50C (cmil/ft)
Values (cmil/ft)Overlapped Copper Tape 30
Overlapped Brass Tape 70
Overlapped Monet Tape 2500
Overlapped Ambrac Tape 350
Lead Sheath 150
Aluminum Sheath 20
Aluminum Interlocked Armor 28
Galvanized Steel Interlocked Armor 70
Galvanized Steel Armor Wire 102
Aluminum (5052 Alloy) 30
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Splices and Terminations
The insulation shields must be removed
completely at splices and terminations,otherwise excessive leakage current andflashover will result. Auxiliary nonmetallicconducting shields may adhere to theinsulation, so that the use of a 100 gritaluminum oxide cloth (or similar material)may be required to assure removal of theconducting material from the insulationsurface (See pages 28 - 30 for generalizeddiscussion of cable preparation for splicingand terminating).
Short-Circuit Currents for
Insulated Cable Conductors
Steady increases in kVA capacity of powerdistribution systems have resulted inpossible short circuit currents of suchmagnitude that the resulting high conductortemperature may seriously damage theconductor insulation. As a guide inpreventing such serious damage, maximumallowable short circuit temperatures thatdamage the insulation to a slight extent only,
have been established for the variousinsulations as follows (based on ICEA P-32-382):
Thermoplastic Compound 150C
Rubber and Varnished Cloth 200C
Impregnated Paper 200C
Thermoset Compound 250C
The charts on the following pages show thecurrents, which, after flowing for the timesindicated, will produce these maximumtemperatures for the given conductor sizes.
These charts are for copper or aluminumconductors operating at 90C or 105C withthermoset insulations. For other, lesscommon charts, reference ICEA P-32-382.
The system short circuit capacity, theconductor cross-sectional area and thecircuit breaker opening time should be suchthat these maximum allowable short circuitcurrents are not exceeded.
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I
0.1
0.2
0.5
1
2
5
10
20
50
100
1cycle
-0.0167
seconds
2cycle
s-0.0333seconds
4cycle
s-0.0667seco
nds
8cycle
s-0.1333seconds
16cycle
s-0.2667seconds
30cycles
-0.5000
seconds
60cycle
s-1.0000seconds
100cycle
s-1.6667seconds
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
ShortCircuitCurrent(thousandsofa
mperes)
Conductor-Aluminum
ThermosetInsulations
Ratedfor90C
Ma
ximumContinuousOperation
CurvesBasedonFormula:
2A2
t=0.0125log
(T2+228.1)
+228.1)
(T1
Where:
I=ShortCircuit,Amperes
A=ConductorArea,CircularMils
t=TimeofShortCircuit,Seconds
T T1 2=MaximumOperatingTemperature(90C)
=MaximumShortCircuitTemperature(250C)
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
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I
1cycle
-0.0167
seconds
2cycle
s-0.0333seconds
4cycles-
0.0667seconds
8cycle
s-0.1333
seconds
16cycle
s-0.2667seconds
30cycles
-0.5000
seconds
60cycles-
1.0000seconds
100cycle
s-1.6667seconds
ShortCircuitCurrent(thousandsofam
peres)
Conductor-Copper
ThermosetInsulations
Ratedfor90C
MaximumContinuousOperation
CurvesBasedonFormula:
2A2
t=0.0297log
(T2+234.5)
+234.5)
(T1
Where:
I=ShortCircuit,Amperes
A=ConductorArea,CircularMils
t=TimeofShortCircuit,Seconds
T T1 2=MaximumOperatingTemperature(90C)
=MaximumShortCircuitTemperature(250C)
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
0.1
0.2
0.5
1
2
5
10
20
50
100
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I
1cycle
-0.0167
seconds
2cycle
s-0.0333seconds
4cycle
s-0.0667seco
nds
8cycle
s-0.1333seconds
16cycles
-0.2667
seconds
30cycle
s-0.5000seconds
60cycle
s-1.0000seconds
100c
ycles
-1.6667
seconds
ShortCircuitCurrent(thousandsofamperes)
Conductor-Aluminum
ThermosetInsulations
Ratedfor105C
MaximumContinuousOperation
CurvesBasedonFormula:
2A2
t=0.0125log
(T2+228.1)
+228.1)
(T1
Where:
I=ShortCircuit,Amperes
A=ConductorArea,CircularMils
t=TimeofShortCircuit,Seconds
T T1 2=MaximumOperatingTemperature(105C)
=MaximumShortCircuitTemperature(250C)
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
0.1
0.2
0.5
1
2
5
10
20
50
100
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I
1cycle
-0.0167
seconds
2cycle
s-0.0333seconds
4cycles
-0.0667seconds
8cycle
s-0.1333se
conds
16cycles-
0.2667seconds
30cycles-
0.5000seconds
60cycle
s-1.0000seconds
100cycle
s-1.6667seconds
Conductor-Copper
ThermosetInsulations
Ratedfor105C
MaximumContinuousOperation
CurvesBasedonFormula:
2A2
t=0.0297log
(T2+234.5)
+234.5)
(T1
Where:
I=ShortCircuit,Amperes
A=ConductorArea,CircularMils
t=TimeofShortCircuit,Seconds
T T1 2=MaximumOperatingTemperature(105C)
=MaximumShortCircuitTemperature(250C)
ShortCircuitCurrent(thousandsofamperes)
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
10AWG
8AWG
6AWG
4AWG
2AWG
1AWG
1/0AWG
2/0AWG
3/0AWG
4/0AWG
250kcm
350kcm
500kcm
750kcm
1000kcm
0.1
0.2
0.5
1
2
5
10
20
50
100
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General CalculationsRevision 4
06/21/2006Page 15 of 20
Shield Short Circuit CurrentFormula
The following simplified formula may beused to calculate allowable sheath currents(based on ICEA P-45-482):
1. The maximum time that a given short-circuit current can flow in a given shieldor sheath, or
2. The maximum short-circuit current thatcan flow in a given shield or sheath for agiven time, or
3. The effective cross-sectional area of ashield or sheath needed to withstand a
given short-circuit current for a giventime.
IMA
N=
60
Where:
I = Short circuit current, amperesA = Shield Area in cmilsN = Number of cyclesM = See Tables on next page
The final temperature the shield or sheathcan reach without damaging the adjacentmaterials limits allowable shield or sheathcurrents. This limiting temperature is definedin ICEA P-45-482 as the variable T2.Various values of T2 are listed below. For
greater detail in regards to the calculationplease refer to ICEA P-45-482.
Values of T2, Maximum Allowable Shield or Sheath TransientTemperature
Cable Material inContact with Shield or
Sheath
T2
(C)
Crosslinked (thermoset)ThermoplasticImpregnated PaperVarnished Cloth
350*200200200
* For lead (Pb) sheaths this temperature is limited to 200C.
NOTE: The material in contact with the shield or sheath shall limit the temperature of theshield or sheath. For example, a cable having a cross linked semi-conducting shield underthe metallic shield and a crosslinked jacket over the metallic shield would have a
maximum allowable temperature of 350C. With a thermoplastic jacket, it wou ld be 200C.
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General CalculationsRevision 4
06/21/2006Page 16 of 20
Values of M Based on Shield Temperature (T2) Above
Values of M(90C)
Values of M(105C)
T2Max
(C)Rated
Voltage(kV) Cu Al Pb Cu Al Pb
200350
200350
200350
200350
200350
55
1515
2525
3535
4646
0.0630.089
0.0630.089
0.0630.089
0.0650.090
0.0650.090
0.0420.059
0.0420.059
0.0420.059
0.0430.060
0.0430.060
0.012----
0.012----
0.012----
0.012----
0.012----
0.0580.085
0.0580.085
0.0600.086
0.0600.086
0.0600.086
0.0390.056
0.0390.056
0.0400.057
0.0400.057
0.0400.057
0.011
----
0.011
----
0.01
1----
0.011
----
0.011
----
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General CalculationsRevision 4
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To calculate the circular mil area of a shield design, use the followingformulas:
Type of Shield or SheathFormulas For
Calculating A(See notes 1 & 2)
1. Wires applied either helically, as a braid or serving; orlongitudinally with corrugations. ndS
2
2. Helically applied tape not overlapped. 127. nwb
3. Helically applied flat tape, overlapped (See note 3).)
(
4100
2 100bd
Lm
4. Corrugated tape, longitudinally applied. ( )[ ]127 50. + + d B bis
5. Tubular Sheath. 4bdm
NOTE 1:Where:
A = Effect ive cross-sect ional area, shield or sheathB = Tape overlap, mils (usually 375)b = Thickness of tape, mils
d is = Diameter over semiconducting insulation shield, milsdm = Mean diameter of shield or sheath, milsds = Diameter of wires, milsw = Width of tape, milsn = Number of serving or braid wires, or tapesL = Overlap of tape, percent
NOTE 2:The effective area of composite shields is the sum of the effective areas of thecomponents. For example, the effective area of a composite shield consisting of ahelically applied tape and a wire serving would be the sum of the areas calculated fromFormula 2 (or 3) and Formula 1.
NOTE 3:The effective area of thin, helically applied overlapped tapes depends, also, upon thedegree of electrical contact resistance of the overlaps. Formula 3 may be used to calculatethe effective cross-sectional area of the shield for new cable. An increase in contactresistance may occur after cable installation, during service exposed to moisture andheat. Under these conditions the contact resistance may approach infi nity, where Formula2 could apply.
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General CalculationsRevision 4
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OTHER CALCULATIONS
Charging Current
The charging current I of a singleconductor insulated power cable can beobtained from the following formula:
ECfI = 2
Where:
I = microamperes per 1000 ft.f = Frequency, HzC = Capacitance, picofarads per ftE = Voltage, phase-to-ground, kV
Capacitance of Cables
The capacitance of a one conductorshielded cable is given by the followingformula:
CSIC
D
d
=7 354. ( )
log
Where:
C = Capacitance, picofarads per ftSIC = Dielectric constant of the
insulation materialD = Diameter over insulationd = Diameter under insulation
Typical Values of SIC
Polyvinyl Chloride......................... 3.5 - 8.0EPR Insulation.............................. 2.5 - 4.0
Polyethylene................................. 3.0 - 6.0Crosslinked Polyethylene............. 2.1 - 2.3Impregnated Paper ....................... 3.3 - 3.7
Dielectric Loss of CableInsulation
The dielectric loss of a single conductorcable can be calculated by the followingformula:
WE SIC
Dd
d =
0 00276 2. ( ) tan
log
Where:
Wd = Dielectric loss, watts per ft.E = Phase-to-neutral voltage, kVSIC = Dielectric constant of the
insulation material
tan = dissipation factor
Maximum Voltage Stress AcrossInsulation
=
cs
ics
g
D
DD
EV
ln
2
Where:
V = Maximum voltage stress, kVEg = Phase-to-neutral voltage, kVDCS = Diameter over conductor shieldDi = Diameter over insulation
Dissipation Factors for InsulationMaterials (tan )
Impregnated Paper..0.002 - 0.0025EPR Insulation. 0.002 - 0.0800Crosslinked Polyethylene..0.0001 - 0.0003
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General CalculationsRevision 4
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Support Grip Length or
Maximum Riser Length for GivenSupport Grip
The following formula may be used todetermine either:
1) Minimum support grip lengths fora given riser section length, or
2) Maximum riser section length fora given support grip workinglength.
( )CW
GLDSL = 8.1
Where:
SL = Riser Section Length, in feetD = Diameter over cable jacket, in.GL = Working length of grip, in.WC= Cable weight, pounds per ft
Prysmian recommends that a recognizedmanufacturer supply the correct grip(s) onan individual application basis, utilizing theformula shown above. Prysmian doessuggest using grips, which will provide
balanced support for the cable (i.e. aDouble Eye grip vs. a Single Eye type).
NOTE: Grips are not suitable for allvertical riser installation and the aboveis offered as a general guide only. If
you have any questions of thesuitability of a grip and/or cableconstruction for a vertical application,please consult the cable manufacturer.
Inductive Reactance to Neutral
The following is a nomogram used todetermine the inductive reactance of solidor stranded (concentric) conductors at 60Hz. It accommodates various spacing ofthe conductors and other unique
parameters as indicated within thenomogram itself.
Special consideration should be given tocables when installed in magnetic ducts.Installing a single cable in a magnetic ductresults in HIGH reactance and a de-ratedampacity. If magnetic ducts must be used,it is recommended that all three cables beplaced in a single duct. While resulting inslightly higher reactances than threecables in a non-magnetic duct, it is optimalrelative to the alternative of a single cable
in a magnetic duct.
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General CalculationsRevision 2
01/18/2005Page 20 of 20
Nomogram for Determining Conductor Reactance at 60 Hz(Series Inductive Reactance to Neutral)
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