Cable Sizing for Fast Transient Loads

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Cable Sizing for Fast Transient

Loads

Robert

E.

Henry,

P.

E.

Life Member

Bechtel National Inc.

P.O. ox

21 19 Richland WA 99352 USA 509-371-2775 (W) 371-2762 (F)

[email protected]

Abst rac t

-

The most recognized standards for

sizing

US

manufactured cables a r e ampacity

tables published in the National Electrical Code.

These tables ar e only for steady-state loads. This

paper addresses the t ransient tempera ture

response of cables undergoing fast heavy loading.

Exponential functions are analytically derived

that express responses to step changes

io

cable

currents . Formulas for coeffrcients and t ime

coostants a re derived. Cable manufac turers‘ da ta

an d othe r sources for physical proper ties ar e used

to calculate the coefficients and time constants for

thre e coppe r condu ctor sizes. Steady-state end

tempe ratures ar e based on National Electrical

Code ampacity tables. The der ived formulas are

limited to cables in open tray or free air, and in

condu it in air. The methods to develop da ta are

detailed for other conductor sizes and aluminum

conductors. Also presented a re the means to size

power cables based on calculated transient

tempe ratures. These calculated tempe ratures are

comp ared with cable temp eratu re ratings to select

a cable size. An examp le is included that

illustrates sizing feeders

for

five 2000 hp c rnsber -

conveyors.

Index Terms - Power cables, transient heating,

transient tem pera ture functions, cable sMng.

I. INTRODUCTION

The m ost widely accepted standard for cable sizing is

based on “The Calculation of the Temperature Rise

and Load Capability of Cable Systems,” by Neber

and McGrath, AIEE Paper 57-660. It is the basis for

the ampacity tables of the National Electric Code [l ]

Article 310, and IEEE 835-1994 (IPCEA Pub. P-46-

426) Standard Power Cable Ampacity Tables. For

applications where transient loads are encountered,

recommended practice, embodied in IEEE guides,

generally addresses cyclic loads varying on a daily

basis. Amp acity recommen dations do not exist for

fast, heavy loading of cables.

One load of this type is a combustion-turbine starting

system in a combined cycle power plant, wherein the

generator is

used

as a starting motor for the

0-7803-777 1-0/03/$17.00 2003

IEEE.

combustion turbine. The combustion turbine’s

exhaust is ducted to a heat-recovery boiler. A

variable-frequency drive is employed that ramps up

the shaft to a low speed,

an&

holds at this load,

consisting of the engine’s air compressor, for a

purging cycle. Then the variable-frequency drive

further ramps up the

shaft

speed to about seventy

percent, when fuel is introduced and ignited. At this

point the turbine assumes the load and the electric

motor current falls to zero. The duration

of

this

starting cycle can be about half

an

hour for a large

machine, and the motor current may reach 1600

amperes during the short period when fuel is

introduced and ignited.

This half-hour cycle is too short a period for

conductors to reach their steady-state ratings. Even

with consecutive start attempts, cables might not

reach their steady-state ratings. Also considering

that a turbine in power plant service is started rarely,

a machine’s starting circuit conductors may employ

their higher emergency temperature ratings for this

service, and more prolonged cable heating would be

necessary to reach this higher te m p er a m rating.

Other examples

of

fast heating might be found io

large crushers and conveyors. These can require

short bursts of large currents on starting and during

periodic loading, when coarse aggregate is initially

crushed. Also, electric furnaces in steel mills have

short bursts of large currents when scrap steel is first

melted. Usually, fast heating is found with variable

frequency drives that are em ployed to deliver varying

amounts of non-periodic high torque to rotating

machinely, or with rectifiers to deliver occasional

heavy currents to equipment.

This paper addresses the transient temperature

response of cables undergoing fast transient heating.

Because there are too many possible criteria, a

general method and tables for rating cables are not

developed, but a calculation guide is outlined that can

be used to size cables for fast transient loads in some

installations.

Sponsor

of

author’s p resentation is Bechtel National Inc.

of

Richland WA.

18

mailto:[email protected]

mailto:[email protected]



11. TRANSIENT

EATINGF

POWER CABLES

If after a long time, a three-core power cable is

energized, its temperature will climb to a steady state

value. Depending on how it is installed and in what

it is installed, it will take a few or many hours to

reach a steady temperature. Considering a

small

cable in free air without wind, this cable will heat up

in 1 1/3 hours. A large cable will take 6 hours. If in

conduit buried in the earth, a large cable will take 16

hours.

Cables

in

Air

A single small bare cable in free air carrying a

current will heat up exponentially with time:

where

AT

‘D

t

K

R

C

Steady state temperature rise on

Kelvin or Celsius scales (“C).

Time varying rise

“ 2).

Time brs).

Time constant equal to R times

C

(hrs).

Thermal resistance between wire and

free

air (“C I

W).

Thermal capacitance ofw ire (Whr/”C).

This neglects any temperature difference within the

cable’s cross section, and assumes that the cable is

long enough that the temperature is uniform along the

cable’s length.

Here AT = Q R

(2)

where Q constant heat

loss

in the cable

(W)

.

If the current is constant, in (2), constant Q would be

assumed constant by neglecting the rise in cable

electrical resistance with temperature increase. But

over a moderate value

of

temperature rise, this is a

small change, and the average value of resistance

over the value of AT may usually be employed, or

altematively, the resistance at an average

temperature.

If the cable is insulated, and if the small radial

temperature difference through the insulation is

neglected or taken as averaged, equation

I )

is true,

but

C

must also include the insulation’s thermal

capacitance. Also, if the amount of cable is a unit

length, then the values of R, C and Q are per unit

length. The effective value of

Q

is said to be equal

to AT when time, t, reaches a value of 4K, or four

time constants. That is, when the value of the

exponential is 0.982. This is less than

2%

difference

from the valu e of the exp onentia l at infinite time.

If there are three conductors spaced together, with

each carrying the same current as the conductor

above, there is less than three times the exposed cable

surface area in the bundle. Then the value of R for

the bundle is greater than 4 the thermal resistance of

one wire, so the bundle heats up to a slightly higher

tempe rature rise than for the single wire. But the

capacitance for the bundle is three times the thermal

capacitance of one wire,

so

the value of K is slightly

greater than for the single wire. Thus, the bundle

heats up with a slightly longer heating time to the

higher temperature rise. Also, this assumes that the

small

temperature differences through the cross

section of the bundle are averaged.

Given the rated temperature rise found in ampacity

tables (see [ I ] , Table 310-67) the conductor current

rating is found in these tables. The temperature rise

with a lesser constant current value is proportional to

the square of the current ratio, if cable dielectric loss

and shield

loss

are negligible. The applicable values

for C can

be

found from manufacturers’

data and the

properties of the materials making up the cable.

Installing in Conduit

If the bundle is installed in a conduit in free air, (1)

becomes more complicated. Now there is a

resistance between the bundle and the conduit, and a

second resistance between fhe conduit and free air.

Also, the conduit’s thermal capacitance affects the

heating time. First, the resistance has increased,

increasing the heating time. Second, the conduit

doesn’t begin to heat up in the beginn ing. It doesn’t

have heat generated in it like the bundle,

so it

begins

to heat up after the bund le starts to transfer heat to

it,

which happens only as the bundle heats up. And the

conduit’s thermal capacitance adds

to

that of the

system. With both resistance and system capacitance

increased, the time constant is increased. (Here

again, the small radial temperature difference through

the conduit wall is neglected.) The exponential

function for the temperature rise (18b) is derived in

Appendix A. There are now two time constants, K’

and K”, with

K

greater than K”. The formulas for

the time constants

are

(13), (14) and (15) in

Appe ndix A. In Appe ndix B, a simplification is

developed based on the criterion, c h , which states

that, if

19



.

. .

the exponential function converges to

Q

=

A T ( 1 -

e K')

and the effect of K ecomes insignificant.

o(

s the

system time constant and CI, I and R2 are defined

in Appendix A.)

Installing in Underground Ducts

Burying the conduit containing the bundle of three

wires increases the system thermal resistance, adding

now, instead of the resistance between the conduit

and free air, the increased resistance between the

conduit and through the path of heat flow to the

earth's surface. Also, the thermal capacitance of the

earth in this path increases the overall amount of

capacitance. These additions of thermal resistance

and capacitance increase yet again the system time

constant and further complicate the system

exponential response. If the added complication

from installing the cables in a conduit in free air is

projected to buying the conduit, then more

coefficients and more time constants in the

expression of the time-varying temperature rise can

be expected. The treatment

of

this response is not

addressed in this paper. Howev er, if another

simplifying criterion is postulated that is similar to

Ch, and if this criterion has the same simplifying

property, then the approximate simple exponential

response

for

the temperature rise would again result.

Such a derivation

of

this response would yield

formulas for the time constants. One o f these would

be the longest and be embodied in the simple

exponential response.

Wiseman, in his paper, An empirical method for

determining transient temperatures of buried cable

systems

[Z],

addressed this formula. His paper

reviews a large body of work by many engineers on

the subject of buried cables that was done in the

decades during and following

WW

U. Their major

concem

was

with large high voltage cables. Another

technique that may be employed to calculate transient

temperature response for buried cables is to use

digital computation.

Temperaiure Rise Values and Heat Losses

The NEC ampacity tables with their listed values of

temperature rise define values of conductor current.

Also

in the

NEC,

Table

9

lists electrical resistance

for

the standard su es of cables. Using values from this

table, generated heat can be calculated

as

the

ampacity current squared times the listed electrical

resistance for each conductor in the bundle.

(Dielectric and shield losses must be negligible).

Using this temperature rise and generated heat, the

value for thermal resistance can be obtained from

(2).

Another source for conductor electrical resistance

data is the cable manufachuer's published cable

characteristics.

The NEC table ampacities are based on published

electrical resistance values. The code states that

adjustments must be made for shield and dielectric

losses. Also, adjustments must be made for heating

from harmonic currents.

nI.

APPLICATION

F CABLE DATA

Example

of

Table Values

A

500

kcmil

5

kV single unshielded copper

conductor is used. A one-foot length is considered.

The ampacity for this cond uctor in air is 685 amperes

(Table 310-69). The resistance is 0.027 ohms per

thousand feet. Based on these values Q is 12.65

watts (per foot) and R is 3 69OCI W (per foot).

Based

on

manufacturer's data sheets, the weight of

copper is 1.54 Ibs and the weight of the insulation

and jacket is

0.511

Ibs (both per foot). The specific

heat of copper is 0.092 and the specific heat of

polyethylene is

0.50.

Converting units, the value of

C fo r the copper is 0.0749 w-h/ C and the value of C

for the insulation is 0.1347 w-h/ C. Their sum ,

0.2097 Whr/OC.

Multiplying these values for

R

and C for this large

single conductor in free air, the value for K is 0.77

hours. It would take a period of about four time

constants, or

3.1

hours, for this conductor to heat up.

Considering a one-foot length of three bund led

conductors of this sam e cable in free air, the

ampacity listed in Tab le 310-67 is

645

amperes. Q is

27.3 Watts, and R is 1 8O0C/ W. The value of C is

three times the value of one conductor: C is 0.629

whr/oc. With these values for R and C for these

three large single bundled conductors in free air, the

20



value for K is 1.13 hours. It would take a period of

about four time constants,

or

about four and a half

hours, fo r this three-conductor bund le to heat up.

Empirical values for K are listed in Table 43 of

ANSUTEEE Std. 242-86 [3]. Table 43 applies to

three-conductor cab les and is repeated here:

Table 43

K Factors for Equations in

8.5.2.4

Cable Air UG Direct

No In Duct Buried

Cond Cond

<

#2 0.33

0.67

1.00 1.25

#2-4/0

1.00

1.50 2.50 3.0

2 2 5 0 MCM 1.50 2.50 4.00 6.00

The application for these table values in this standard

is similar to transient temperature calculations. The

table values are suitable for the application in the

standard, in section 8.5.2.4, which makes use of the

values for K.

The value of 1.13 hours above compares with the

table value of

1.5

hours for

500

kcmil cables.

Cables in Conduit

In Appendix

A,

formulas for the transient

temperature rise of cables within a conduit mounted

in free air are derived. The properties of these

equations are discussed in Appendix B. Values for

coefficients and time constants have been calculated

for

250

kcmil,

500

kcmil and

750

kcmil, with

5

kV

unshielded and 8 kV shielded insulation. The results

are listed in the calculation sheet included as Table

BI, and discussed in Appendix B. Referring to Table

BI, the results show that the larger the cable size, the

longer are the time constants, and time constants with

the cables in conduit are longer than the time

constants in free

air

or spaced in tray. In Table BI,

values for R and C discussed above are listed under

Rt for R, and C l for C. For

500

kcmil and

750

kcmil cable sizes, not much difference is seen

between the two similar cable insulations listed in the

table. Comparing the table time constant values with

the values in Table

43,

the values in the first two

columns in Table 43 are a fair summary for the

calculated constants in Table BI.

A criterion, introduced above, is developed in

Appendix B that converts the cable temperature rise

equation to the simple exponential form (see

Appendix B equation

20).

Values for this

Ch

criterion are listed in T able BI, and show that as cable

size increases, the simple exponential form becomes

more accurate. With the largest value of ch, 1.276

for the

250

kcmil conductor size, the simple

exponential does represent a Fdirly accurate m eans to

calculate the conductor temper ame. (For a

discussion of the error see Appendix B). The values

suggest that for smaller cables the simple exponential

form would be less accurate. A short form of the

equation, (18d) based on Fh, the fractional part of

Ch, is exact.

Applications to Fast

Transient

Loads

The best way to apply the data in Table BI is to

illustrate through an example:

Examole: Five crusher-conveyors are located two

miles from the processor plant. Each i s driven by a

2000

Hp motor fed fiom a variable frequency drive,

having rated output of

2000

hp at

4160 V

at

60

Hz.

The drive equipment is located in the processor plant.

Rated motor voltage is

4000V;

ameplate current is

250

amperes. The motor current profile for this

crusher application is shown in Table I below.

To overcome the conveyor’s inertia and stiction

during starting, an initial current of 2000 amperes at

15 Hz

s required. Th e current ramps up to this value

in

2

seconds and holds until conveyor movement

results. This takes a maximum of 30 seconds.

Conveyor loading then commences, feeding the

crusher almost immediately. The motor frequency

ramps to 60 Hz during initial loading. Biting into the

first few loadings requires a series of transient motor

current surges that average 1500 amperes, or less,

and last no more than 20 seconds each. Typically, a

couple of surges are usually required, but no more

than four are ever necessary. Each time a transient

occurs, the drive automatically lowers the frequency

to

30Hz

until the crushing is completed, thereby

providing the large torque required for the transient

crushing loads.

These transient motor currents occur with a minimum

of

2

minutes betw een each transient. The motor

current between these transients is at

230

amperes.

Afler the last transient, motor current decreases to a

nominal

80

percent, or

200

amperes at

60 Hz.

This

lasts until the next loading. The next loading requires

21



the same series of

1500

ampere transients, repeating

the first cycle, except for the starting 2000 ampere

surge. These loading cycles continue

through

two

shifts. Each loading cycle takes a minimum of two

hours.

For voltage regulation considerations, a 250 kcmil

conductor is, first selected. Its impedance

characteristics.are adequate for the 10,000 foot feeder

length from the processing plant out to the motors. It

is believed that, because of the cable heating from

these large current surges, two 500 kcmil feeders in

parallel will he required.

The drive manufacturer recommends that the feeders

be installed in conduit, to provide shielding for the

high-frequency current elements, composing the

square current profiles that the drive generates for the

motor.

The square current profiles contain harmonics,

having the six-pulse characteristic profile. The six-

pulse profile is the fundamental, of unit amplitude,

and the odd harmonics except the triplens. These are

the harmonics that are divisible by three. The

amplitude of each harmonic is the reciprocal of the

harmonic number. Thus, the seventh harmonic's

amplitude is 1/7th. By taking the square of each

harmonic current, multiplying

it

by the 'increased

resistance from skin effect at the harmonic frequency,

and adding these products, the total heating effect of

the six-pulse

current

camed by the

conductor

is

calculated. This assumes that the resistance to each

harmonic current changes only

from

skin effect; that

is, there are no significant changes due to frequency

from stray induced couplings that result in increased

resistance. This is usually the case. (This is not true

in transformers where losses are frequency dependent

and harmonic losses are greater.) And here, where

the heavy loading is at 30 Hz, it is the case. The six-

pulse harmonic heating effect amounts

to

15 percent

of the heating from the fundamental current. Thus,

the equivalent current would be the square root of

1.15, or 1.073, times the fundam entalvalue.

The first consideration is the temperature rise

resulting from the initial 2000 ampere starting current

that lasts 30 seconds. The 250 kcmil size will be

considered first. The temperature rise can be

calculated from equation 18d in Appendix B. Listed

in Table B I are the values for the constants and time

constants for one three-conductor bundle

of 250

kcmil conduc tors. The value for AT is 2372OC,

calculated from (2) above. The heat dissipated in the

conductors is 746 watts, with a corrected current of

2146 amperes. The values

for

the exponentials after

30 seconds are 0.972 for the sub-transient, and

0.9953

for

the transient. This calculatio n shows that

the temperature rise from the initial 2000 amperes for

30 se con ds is 14.1'C.

The temperature rise from one 1500 ampere surge

can similarly be calculated. This calculation shows

that the temperature rise from an initial 1500 amperes

for 20 seconds would be 54°C .

The two-hour current profiles ar e listed in Table I. If

the rms equivalent current of these profiles

is

calculated, the temperature rise kom these profiles

can he found. The rms equivalent currents would

result in the same heating

as

with these profiles.

Each transient heating profile would plot as a

temperature rise varying about the same exponential

curve

as

that generated by its

rms

equivalent current.

The rms values in Table I were calculated by

summing the squares of the corrected currents

multiplied by their intervals. Then, dividing this sum

by the two-hour period determined the mean squared

current. Taking the square root of this mean squared

current yielded the rms value.

TABLE

I

MOTOR

LOADING

PROFILES FOR CRUSHER-

CONVEYOR

First Cycle

T i m e p

urrent

0 .5min 2000A

0.5 min 230A

0.33 miu 1500 A

2

min 230

A

0.33 min 1500

A

2 min 230 A

0.33 min 1500 A

2 min 230 A

0.33 min 1500 A

111.67min 200A

Recurring Cycles

0 .33min 1500A

2 m i n 2 3 0 A

0.33min 1500A

2 m i n 2 3 0 A

0.33min 1500A

Z m i n 2 3 0 A

0.33 min

1500

A

112.67 min 200 A

i me

Current

rms value is 286 A

Using this same m ethod, the rise from the four 1500

ampere surges in the second column, based on their

r equivalent surge current of 672.5 amperes, is

2 I . P C .

At the end of the first cycle, the. cables reach a

-temperature rise of 31.6 C. After the second cycle,

rms

value is 255 A

22



the temperature rise is 35.8 C. This is from using the

combined

rms

value of both cycles, of 271 amperes.

The

full

response to steady state takes 7.

I

hours. By

this time a temperature rise of 38.0 C is reached.

Then, additional rises from each set of four 1500

ampere surges adds spikes of 21.8 C.

This makes a

peak rise of

59.8

'C for the se spikes.

With these moderate values of temperature rise with

one 250 kcmil feeder for each motor, it is clear that

only one three-conductor cable, instead of two

500

kcmil cables, is required.

The savings in installing

only five of these feeders, instead o f the assumed ten,

is estimated at 1.1 million dollars

(US).

IV. ROCEDURE

FOR

SIZING

CABLES

The steps given in the examp le above

are

as follows:

Determine the worst-case current profile.

This includes the number of repetitive

cycles.

Adjust the currents for harmonics, shield

losses and dielectric

losses.

Calculate the rms value for the profile.

Determine the maximum twenty-four hour

and seven-day-average design ambient

temperatures. Max imum peak temperature

ever recorded has little meaning for the

cable time constants.

From voltage regulation requirements, select

a trial cable size. Calculate the peak rise for

high-current surges as if they occur at the

start of th e response.

Calculate the steady-state temperature rise

fiom the

rms

current, using repeated profiles

if required.

Add the steady state and peak rises

to

the

design ambient temperature and compare

with the temperature rating o f the trial cable

size. For seldom repeated current profiles,

having some days between energizations,

compare with the cable's emergency

temperature rating.

Select a larger (or smaller) conductor size

and calculate steady-state and peak

temperatures if the first selection is

unsatisfactory.

Determine the fast heating capability of the

drive or power source. The cables' fast

heating capability should no t greatly exceed

the capability of the source to provide the

currents.

Cables that are rated for 105OC continuous operating

temperature have an emergency and an instantaneous

temperature rating of 1 3 0 T and 25OOC. This is

listed in manufacturers' data.

Limitations of the Methods

The equations and

data

do not apply to three-

conductor cable assemblies. No r do they apply to

more than three conductors in a conduit. Their

limitations with underground installations are

discussed above. If groups of three condu ctors

installed in a cable tray are not space d apart, the data

do not apply. The equations do apply to PVC and

aluminum conduit, as well

as to

other sizes of

conductors, and to aluminum conductors. Data for

these materials can be developed as was done for the

data in Table BI.

IEC

sizes can be calculated in the same way as

discussed above. Ampacity values for

IEC

sizes,

once established for local conditions, can be used in

the same way

as

hose in the NEC.

The problem with three conductors in a cable

assembly (see NEC Tables 310.71, .72, .75 and .76)

is the error in attempting to represent the cables with

one schematic node. An additional node in the

schematic, Fig.Al in Appendix A, would be required

to represent the outer covering layers and the fillers

within. With three nodes, solvin g a third-order

polynomial would be required, instead of the

quadratic,

(IO)

in Appendix A. This would yield

three time constants, with one longer than the other

two, and equations for their values similar to (12),

(13) and

(14).

Just as with a three-node

representation of an underground installation

discussed above, a criterion similar to

ch

could be

postulated, and a simplified form, like I ) above,

would be exp ected.

With some error, the solutions of Appendix A might

be extended

to

represent the three-conductor cable

assembly by adjustments in

C I

and

RI.

Similarly,

six conductors might also be represented.

The rub

23



might be in adjusting RI. The NEC lists the

ampacity

of

six conductors in a conduit as 80% of

the ampacity of three conductors. Six conductors,

carrying

71%

of

the

current of three conductors,

generate the same heat as do the three conductors.

Does this mean that R I is 9% less for all conductor

and conduit sizes?

Some value may be found for other installations in

knowing the calculated peak rise for the same

conductors installed in a conduit in air. For example,

the calculated peak rise might be greater than the

peak rise with the conductors installed in an

underground duct.

The method used

to

divide the value of

RI +

RZ

in

Table BI may not be accurate. The effect on the

values of

K’

and

K”

has not been explored.

v.

CONCLUSIONS

This paper has presented the means to size power

cables based

on

calculating transient temperatures

resulting from fast transient heating. Analytical

methods are employed. Cable temperature ratings

are compared with these calculated temperatures to

make a cable size selection.

Data for three conductor sizes are developed and

presented. The methods to develop data for other

conductor sizes are developed. The formulas derived

herein are limited to cables spaced apart in open tray

or free air, and in conduit in air.

A

procedure for sizing cables is presented. An

example is included that sizes the feeders for five

2000 Hp crusher-conveyors.

Methods to extend the usefulness of the data to other

installations of cables are discussed. Further work is

recommended to develop more simplified rating

procedures.

VI. APPENDIX . DERIVATIONF

TRANSIENT HEA T

FLOW

EQUATION

The model for the derivation is shown in Fig. Al. In

this model, the three conductors are considered as

spaced togethcr and located in the axial center

of

the

conduit. The conduit is assumed to be suspended in

free air. The conduit and cables are regarded as long

enough to have no axial heat flow, and a unit length

of the conduit and cables is represented. Thereby, the

heat flow from the conductors is more

or

less only

radial to the inside conduit wall, and to the

surrounding air. This simp lifies the heat flow as one

dimensional. Also, any temperature differences

among the three conductors and their coverings are

ignored and assumed

as

negligible, and the small

radial temperature difference through the conduit

wall

is

neglected. The conductors are represented

in

Fig.

A I

as the left node.

CONDUCTORS CONDUIT

R I

“O+J =Jqc t R 2

T

1

Ambient Heat Sink and

Temperature Reference

Fig. A l . Schematic representation ofhea t transfer h m

three conductors in a conduit in air

The thermal resistances in this model are assumed

constant and do not change with temperature. This is

not strictly true, but the fourth-power radiant heat

transfer can be essentially proportional to

temperature difference over a moderate temperature

range, and conductive and convective coefficients

can be similarly treated. The thermal capacitances

also muqt not change with temperature. These

qualifications allow the differential equations that

follow to be of the linear type with constant

coefficients.

In this analysis, the three conductors generate over

time a constant heat loss, represented as Qo. Prior to

zero time, the conductors are de-energized, and the

conductors and conduit are at ambient temperature.

ARer zero time the currents

in

the

three

conductors

are equal and such that the heating generated in the

conductors, Qo, is constant. Also, the heat generated

in the conductors consists

of

Joule heating, dielectric

losses and shield losses in the cables. Any small

amount of heating in the conduit generated by

induced currents is neglected. For the duration of the

responses derived below, the ambient temperature is

constant in this model.

24



The elements in Fig. 1A represent the following:

Constant heat flow generated in

unit leng th of the three cables (W).

Thermal capacitance of the three

cables,

Wh / “C).

Thermal capacitance of theconduit

(wh

/

“C).

Thermal resistance betwee n cables

and con duit (“C

/

W).

Thermal resistance between conduit

and air (“C / W).

That time-va rying portion of Qo

stored in mass of the cables (W).

That portion of the time-varying

heat flowing t?om the cables to

conduit that

is

stored in the

conduit’s mass (W).

Time-varying heat flowing

&om

he

conduit to air (W).

In the equations that follow:

AT

CD

OC

‘ d

QC

KI

K2

K3

Solutions

Steady-state co nductor temperature

rise abo ve ambient reached after

the transie nt responses (“C).

Timevarying temperature

r ise or

difference (“C).

Temperature degree difference of

the Kelvin or Celsius scales.

@c on dui t (“C).

ables (“C).

Time constant equal to

R1

C1

(hr).

Time constant equal to R2 C2

(hr).

Time constant equal to R2 CI (hr).

dt

= CP

condu i t, a t t im e t

=

T (2)

d t , a t t im e t

=

T (4)

(5)

Taking the Laplace transforms of equations

I ) through 5 ) :

1

-eo

S

The L aplace transform of Qo is denoted by

-

The Laplace transform o f

ql

s denoted by

q l

:

The Laplace transform of

7 )

1 -

= R , q ,

is:

Q d = -q2

=

R ,q ,

sc

The Laplace transform of aCs:

1 -

’

-q

I ’

e sc,

Solving for q2 using

(6 ) .

7), and

8 ) :

1 - -

I - I - R -

-q,

= - q 2

+

R , q ,

+

sc, sc, sR,C,

q2

.

Multiplying by

sc,

:

Substituting in

(6):

25



1 -

q 2 + -

sR2C,

q2

1

Qo

=

S

1 -

) q 2

1

+-

2

sR,C,

+

c R 2 C 2 - sR2C2

C RI c,

+ S +

~ = ~ - + S ~ R , C , + S -, RIG,

c2 R2C2

1 -

- ) 4 2

=

R2C2

Within the quadratic, taking the coefficient of s

changing some terms and adding fractions:

Substituting the expressions for

Rs:

Rewriting ( IO ) :

K ,

i

,

i

,

Qo

= [

K , s 2 + K , S

KIK2

1 -

K , K , q 2

+ K,-

K 2 + K , + K , 1 -

s + -

K , K 2

q

o = K , [ s 2+

KlK2

The quadratic coefficient

I

, + K , +

K ,

1

[ S 2

+

s + -

KIK2 KIK2

1 1

K

K

has two roots expressed as (s +-) (s +

,

which are real and

one

is larger than the other. Let

1

1

e the lesser root and e the greater root.

K' K

Then R is larger than K .

(1 1)

Qo

1

472

=

K l (s++)(si+)'

The solution to (1

1)

can

now

be found as

the anti-transform of q2

:

Qo 1

K 1 1

q 2 =

K'

K

(See [41, Table

ofkplplace Transforms p189,

number

77.)

The expressions

for

the roots are found from the

classical quadratic solution. The second term in the

solution for the roots, which is under the radical, is

( K , + K 2 i K , ) 2-- 4

K , ~ K , ~ KIK2

*I12

Rewriting this term as

- , / ( K , + K 2 + K , ) 2 - 4 K , K ,

1

I

the expression for the roots becomes:

2KIK2

1 - K , i K , + K ,

--

K 2KIK2

26



and

1

P

K ' K , K , '

K ' K = K , K , ; K ' -K = p .

P

1 -

q 3 =

2

sR2c

I

-

q = -q,, and

q3

=

dt

=

sK2 K2 =02

21



VII. APPENDIX

B.

PROPERTIES OF THE

Time Constants

HEAT

FLOW

EQUATIONS

Initial Conditions

Both expressions for

92

and 43 above contain the

exponentials with time constants R and K , given in

Appendix A (IZ),

(13)

and (14). And as stated,

R

s

larger than

K ,

such that K'

- K = p

Equation

(12)

Appendix A:

P

~~

defines p. The expression for the cable temperature

rise,@.,, emb odies both heat flows, and can be

Considering the expressions for

q

and

4

from

divided into two periods of time, the sub-transient

and transient. Taking the value of time equal to

t = 4 K ,

16)

d

Y

At time t

= 0,

t

e

K'

and e

K

are each equal to one,

K

q 2

Qo

- 2 - ( I - I ) = O

P

P

43

=eo-

( K ' ( l ) - K ( l ) ) =

Qo---@ -O.

Thus,

with both

92 and

43 equal

to

zero,

Qo -

P

is equal

to

zero.

Final Conditions

At time t

= -

I

e K' nd e

K

are each equal to zero.

,

In

1

6), e

K

has disappeared before

-..

- e

K' =

0.982

This is less than

2

percent from a value

of

one minus

the exponential, reached a t infinite time. Afler t

=

(17)

4K , the value

of

Letting time from zero

to

about t

=

4 K be defined

as

the sub-transient period, then the period from this

time to about

t =

4

K' is

defined as he transient

region. To be consistent, the time beyond about t

=

4 R is considered steady state.

Sub-Transient Response

The expression for the cable temperature rise, given

in Appendix

A,

can be re-written as

( K ' - - K 2 R 1

e $ ; ] (18c)

A T [ I --

K'-K

RI +R2

e

K ,and

e

K

is zero, making

42

equal to zero.

In (17), e

'

as likewise disappeared before transient term, and the second is the transient term.

e K'

and

e K are zero, making the steady-state expone ntial with

two

coefficients,

K

and

Here, the first expone ntial term in AT

is

the sub-

Within the sub-transient term is a decaying

value

O f 43

equal to

Qo

.. These subtract from each other. The

difference in these terms responds in a way to

K2Rl

RI +R2

28



account for the delay in the temperature rise of the

conduit. Until the cables have become warm and

have started to transfer heat to the conduit, the

conduit does not start to get warm because the

conduit, unlike the cables, doesn't have heat

generated in it. Thus, the conduit lags the

temperature rise of the cables. The difference in

these terms is added to the transient term, and

disapp ean b y the end of the sub-transient period.

Transient R esponse

Atter the initial sub-transient portion o f the response,

QC becomes

_

1

__

K , R , e

'1,

K'-K RI +R 2

with both e ''erms having evaporated

Q c = A T [ l -

RIK, -

RI +R2

It

can be shown that

Then, if K

(RI+R 2 )cl

= K ,

nd

Q c

collapses into the simple

IK2

R, +R2

expression:

Q c = A T (

1

- e T ) .

20)

This is the usual, expected and em pirical result that is

found, and it is a much simplified form of the

transient response. Also, to the extent that

K' =

(RI

+

R 2 )C,,

his transient response is useful

to calculate the cable temperature atter the short-

duration transient has evaporated.

Not only is the expression for the

transient

response

simplified, if the approximation is true, the

sub-

transient function is likew ise simplified. If the

approximation is true, the two sub-transient

coefficients disappear:

I

Qc=A T ( 1 -

e

K ' ) +

. .

With this approximation, the sub-transient response is

negligible, and 20) is the entire sub- and transient

response. Hence, the following criterion is def ined

AISO,

with

(RI

+

R2

)Cl

= K , +

K3

Returning to (lec), at zero time

Qc

is zero, as

discussed above. Therefore the transient and sub-

transient coefficients, with the exponentials equal to

1 at zero time, are equal and subtract. This can be

shown

by substituting

ch

in

(19)

and (18c) at zero

time, and reducing the two coefficients.

CoolDown

The model in Appendix A can be used to derive the

expressions for cool down. These will be of the form

of decaying exponentials with time constants K and

K' and coefficients similar (if not identical) to those

in (18c). The ch criterion will still be valid,

however. Just as Qc reduces to ZO), the cool-down

expression would also reduce to the approximation:

I

Q c =

AT e

K

T o the extent that ch is

nor

true, the decay can be

expressed as the sum of

two

exponentials, with time

constants

of

K' and K ' , and with coefficients that add

to

the initial cable temperature.

29



The two exponential expressions can be illustrated by

constructing a semi-log plot of the cable cool-down

from steady state temperature. Starting from initial

time, the plot is the sum of

two

lines, each an

exponential decay, with slopes of

- K

and -K'. The

latter is similar to (22), except the s u m of the initial

temperatures o f both lines equals AT.

The derivation of Appendix A uses initial values of

zero for cable and conduit temperature rises. If the

derivation of Appendix A were redone, but with non-

zero constants for initial values for cable and conduit

temperatures, it would yield formulas for cable and

conduit temperatures during warm up from any

values of initial temperature rises. If these same

constants were included in a derivation for cool

down, then a complete set of analytical equations for

warm up and cool down would result. With this

full

heat-up and cool-down set of equations, the full

dynamic response to any fast transient electrical

current profile could then be calculated.

Calculation of Constants

The calculation table, Table BI, illustrates the

equations of Appendix A and the simplifications

above..

In

these calculations, Q, the heat flow

generated in the three conductors, is based

on

the

electrical resistance, Re in the table, which is taken

from the NEC, Chapter

9

Table 9.

For triplexed cables in tray or free air, R in the main

text is labeled Rt in the table. C in the main text is

C l in the table, because

C l

represents the thermal

capacitance of the conductors and their coverings

whether in tray or conduit. Thus, the time constant,

RC, is RtC l in the table.

For triplexed cables in conduit, the value for R I is

obtained by calculating R1 +

R2,

by using the

ampacity table cited in the calculation sheet, and

subtracting R2. The value of R2 is found by

multiplying Rt by the ratio of the triplexed

conductors' diameter divided by th e conduit diameter.

The diameter of a circle that circumscribes the three

conductors is used to represent the triplexed

conductors' diameter. Thi s circle's diameter is 2.155

times one conductor's diameter. No adjustments

were made for emissivity, or diameter in the conduit's

convective coefficient, comprising

R2,

because these

were considered as almost the same as for the

circumscribing circle. Th e value for R I determined

in this manner is almost always lesser than Rt. (See

the 750 kcmil size,

8

kV insulation, where they are

about equal.) This implies that the snug fit of the

cables in the conduit results in an air gap that is

thinner than the convective boundaty layer around

the warm cables in free air. Nevertheless, R I+ R2 s

based on the cited ampacity table.

The conduit size is determined as for 40% fill or the

next larger size. Thus, a

3

1R-size Schedule 40

conduit is used for 250 kcmil, a 4 size for

500

kcmil,

and a 5 size is used for 750 kcmil.

No

adjustment in

ampacity from using diameters larger than these sizes

is made in the ampacity tables.

However, a greater

value for C2, and therefore

K',

results with a larger

heavier conduit, or with Schedule

80

wall conduit as

well. This is a means to increase the time constant if

slower heating is desired.

The values for specific heat in the table are converted

from water, having a specific heat o f 0.52 74 Wh per

lb 'C.

The relative specific heat of polyethylene, of

0.50, was used in the tables for insulation and jacket.

This is not representative for PVC, which has a

relative specific heat of 0.25 [9]. Thus, lesser values

for

K'

can be expected with PV C cab le jackets.

Error Using Simple Exponential

Near the bottom of the Cable Heating Tabulation,

Table

BI,

are listed the values for the criterion,

Ch.

These are greater than unity, and the largest is for the

250 kcmil conductors, having

the

value of 1.276.

The error for the 250 kcmil conductors is shown in

the following Table

B

11.

As shown in this table, the error is large in the first

few minutes into the sub-transient period, and

decreases with time.

Therefore, if accurate cable

temperatures are needed for the early minutes, the

full expression for temperature rise must be used.

Otherwise a value smaller than the actual value is

calculated using the simplified expression.

TABLE B

I1

ERROROF

SIMPLE EXPONENTRL

250 kcmil conductor size in conduit

Time

t =K

t=K'/2 t=4K

t - Z R

17.51~1 53m 70m 3.5h

@ c i ~ ~ 0.177 0.4239

0.5090 0.8720

f

1

- e

K

0.151 0.3935 0.4819 0.8647

EKOI -14.7% -7.2% -5.3% -0.84%

30



The values for Ch in Table BI increase with smaller

conductor size. With cable sizes smaller than 250

kcmil, the error will be greater than in the table

above.

Let: 1 < c h <2,hen c h

=

1 + a fraction, or

c h 1 + F h .

This is an expression of the simple exponential plus

an adjustment, calculated fro mF h, the fractional part

of

ch.

REFERENCES

[ I ]

[2]

N ation al Fire Pro tection A m . , W A 0

National Electrical Code, 2002 Ed. NFPA

Wiseman, R. J.

An

empirical method for

determining transient temperatures of buried

cable systems, AIEE Transactions on Power

Apparatus and Systems, pt.III, vo l72 , pp

545

-

562,

June

1953

IEEE Std. 242-86 Recommended Practice

for Protection and Coordination of Industrial

and Commercial Power Systems

-

Buff

Book, IEEE

G. E. Roberts and H. Kaufman, Table of

Laplace Transforms, W. B. Saunders Co.,

1966

Bateman Manuscript Staff,Table o f Integral

Transforms,

Vol.

1, McGraw Hill Book Co.,

1954

[3]

[4]

[ 5 ]

[6]

W. E. Boyce and R. C. DiPrima, Elementary

Differential Equations and Boundary Value

Problems, 2nd Ed., J. Wiley and Sons, 1969

Houston Wire and Cable Catalog,

1987,

Houston Wire and Cable Co., Houston TX.

Hubbel Anderson, Technical Data - A

Reference for the Power Industry, Hubbel

Anderson Power Systems, AEC lll

[9] Salamander Ceramic Infrared Emitters

Technical Manual, Physical Properties of

Materials, p

13,

Mor Electric Heating

Assoc. Inc.

Engineering Data for Copper and Aluminum

Conductor Electrical Cables, Bulletin EHB-

90,

The Okonite Company, Ramsey NJ

[7]

[8]

[ IO]

31



Constant

Re Ohms1IOM)'

13

(105

DegC)

3 13sqrd Re = Q3

Temp 1 Q3

= RI

Wconductor

Ibs

Wcu

Ibs

CC

wcu

EC w-h/Deg

C

Wcond -Wcu

=

Wi

Ibs

ei w-hilb DegC

Wi ci w-hiDeg C

Wceu + wici = c

Rt I

4RtCI

Conductor Dia.

tX

2.155

Dc=

De

2.74 Dc

= Dd

min

Conduit Dia. ODd

DeJODd RI

=

U2

Id(lO5 DegC)

Source

TABLE

I

CABLE

HEATPIG

TABULATIONF CALCULATED CONSTANTS

Code: Ch 9 Tbl

9

Code:Tb1310-

67

Code: Tbl310-

67

Dam Sht.

Data Sht.

Data Sht.

Data Sht.

Data Sht.

Data Sht.

250 kcmil 500kr.mil 500 kcmil

5 kV 5 kV 8 kV Shlded

I n

open tray

or free air

0.05411000' 0.02911000' 0.02911ow

415 645

A

650

A

27.90 W 36.19 W 36.76 W

23 3 DDgnv ISOoegnV 1.77Deg/W

3.555

Ibs

6.165 Ibs 6.336 Ibs

2.316 Ibs 4.632 Ibs 4.632 Ibs

0.0485

0.0485 0.0485

0.1123 0.2247 0.2247

1.239 Ibs 1.533 Ibs 1.704 Ibs

0.264 0.264 0.264

0.3271 0.4047 0.4499

0.436 W-WC 0.629 w-h/C 0.675 w-WC

1.02 hrs 1.13 h n

1.19 hrs

4.1 h n 4.5

hra

4.8 hrs

In

steel conduit in free

air

1.14in. 1.40 in.

1.27

in.

2.451

in

3.017

in.

2.737

in.

3.12in 3.84 3.48 in.

4.000 in. 4.500 in.

4.500

in.

1.43 oeglw 121DDgnv 1.08Orgnv

750 kcmil

5 kV

0.02lIlOW

835

A

43.93 w

1A8

Dc w

9.060 Ibs

6.848 Ibs

0.0485

0.3321

2.212

Ibs

0.264

0.5839

0.9161 w-

WC

13 6 hrs

5.4 h n

1.U

n.

3.534

in.

4.49

in.

5.563

in.

0.940 Oegnu

750 kcmil

8 kV Shlded

0.02111000'

820

A

42.36 W

1.53 Deg/W

9.099 Ibs

6.848 Ibs

0.0485

0.3321

2.251 Ibs

0.264

0.5943

0.9263 w-WC

1.42 h n

5.7

hrs

1.47 n.

3.103 in.

4.03 in.

5.563 in.

0.853

D e e

Code: Tbl310-

355

A 530 A 535

A

665 A 655

A

77

.

3

Id

ylrd Re

=

Qo

20.416 W 24.44 W

24.90W

27.86 W 27.03 W

Temp 1Qo=

R I + R 2

Code: Tbl3lO- 3.18

De@

2.66

OegnV

2.61 D e w 2.33 D e w 2.40

Dew

77

RI

Is RI > R t ?

Conduit Wt. Wd

cd

Wd cd

=

C2

R I C I = K I

R2 C2

=

K 2

R2 CI = K 3

2KIK2

K l

+ K 2 + K 3

X I +KZ+K.z)sqrd

4KIK2

Rho

x

K

K ' / ( K I + K 3 ) = C h

4 K

4 K

._

1.75 DrgnV

no

Data Sht. 8.31 Ibs

Data Sht. 0.0567

O.47lw-WC

0.763

0.674

0.623

1.0285

2.060

4.2436

2.0570

1.4787

1.77

hrs

0.291

hrs

1.276

l l h n

7.1 hm

1.45 O W 1.53 oegn

no no

9.82 Ibs 9.82 Ibs

0.0567 0.0567

0.557~-WC

0.557w-NC

0.912 1.033

0.674 0.602

0.761 0.729

1 3 9 v 1.55DeglW

no Y

13.44Ibr

13.44Ibs

0.0567 0.0567

0.762

w-h/C

0.762 w-WC

1.273 1.436

0.716 0.650

0.861 0.732

~~ ~ ~~~ ~~

1.229

1.244

1.823 1.867

2.347 2.364 2.850 2.818

5.508 5.588 8.123 7.941

2.458 2.488 3.646 3.734

1.747 1.761 2.116 2.051

2.05

h r s

2.06 hrs

0.30

hn

0.30

hrs

1.225 1.169

1.2 hn 1.2hn

8.2

hrs

8.2

hn

2.48 hrs 2.43 hn

0.37

hrs

0.38 h n

1.162

1.121

1.5 hrs

1.5

hrs

9.9

hrs

9.7 hrs

32

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Cable Sizing for Fast Transient Loads

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