Current Carrying Capacity

The Current Carrying Capacity (Current Rating)

4-1 The Fundamentals:

In the first place the current carrying capacity is determined by the max.

permissible conductor temperature and the ambient conditions as far as they are

relevant for the dissipation of heat. With mass-impregnated paper-insulated cables

the permissible difference between the temperature reached under full-load

conditions and the no-load temperature has also to be taken into account.

Inadmissibly high conductor temperatures and excessive temperature differences

will speed up the ageing process. The current carrying capacity of H.T. Cables has

to be calculated with particular care.

The cable is heated up by the ohmic losses occurring in the conductors and-

if the cable is operated with a.c. – in the metallic covering. The dielectric losses are

negligible with Protodur cables up to Uo/ U = 5.8 / 10 kV, mass-impregnated

cables up to Uo /U = 5.8/10 kV, mass-impregnated cables up to Uo/U = 34.7/60 kV

and protothen cables up to Uo/U = 64 / 110 kV operating voltage. Under steady-

state conditions the heat dissipated is equal to the sum of all the losses in the cable.

The heat flows by conduction to the surface of the cable and, if this is arranged in

free air, is transferred to the surrounding air by convection and radiation. With

cables buried in the ground the heat generated by losses flows from the surface of

the cable by thermal conduction via the ground into the atmosphere. Whilst the

cable itself is heated up. The difference between conductor temperature and

ambient temperature is nearly proportional to the total losses. The law of heat flow

is analogous to ohm’s law, with the heat flow (ohmic losses) V, corresponding to

electric current I, the difference in temperature corresponding to voltage U and

the total thermal resistance S corresponding to electrical resistance R:

Thus - V S analogous to U = I R. (1)

The heat flow is the sum of all losses generated in the cable. To reach the

surrounding air from its point of origin the heat flow must overcome the thermal

resistance Sk of the cable and the thermal resistance of the air SL or of the ground

SE caused by the transfer of heat from the surface of the cable to the surrounding

media.

In accordance with the analogy existing between the heat flow and the

electric current (equation 1) an equivalent circuit diagram (Fig. 1) may be drawn

for the heat losses flowing from the cable and the temperature rise they produce.

The heat transfer by radiation and convection, if the cable is installed in free air, is

represented by two resistors connected in parallel to each other and in series with

the remaining thermal resistances of the cable. The heat losses due to the current

(ohmic losses) are developed in the conductor, in the metal sheath and in the

armour (see also equation 6). They are represented by electric currents fed in at

these points. As a result of the heat losses, the conductor temperature L is

increased by and the surface temperature of the cable o by a as compared

with the ambient temperature u.

Fig. (1) Equivalent Circuit Diagrams for the Heat Flow in a Cable.

With the equation for the ohmic losses

V = n I2 RW 10-3, W/m (2)

we obtain from equation (1) the current carrying capacity for operation with a.c. or

three-phase current with the cable installed in free air, for example:

I = √ Δϑ 105

n RW (SK' + S L)

A

(3)

where

= given temperature rise of the conductor with respect to the

surroundings, C.

n = number of current carrying conductors,

RW = effective resistance of a conductor at operating

temperature /km.

SK'

= fictitious thermal resistance of the cable, C cm/W

SL = thermal resistance of the air, C cm/W.

The effective resistance (a.c. resistance) RW is arrived at as follows:

RW = R + R /km, (4)

Where R = d.c. resistance at operating temperature, /km.

The extra resistance:

R = R [y + y1 + (1 + y + y1) ( + 1)], /km (5)

indicates the measurable increase in the conductor resistance due to current-

dependent a.c. losses. These losses are produced in the conductor by the skin and

proximity effects (y and y1), inductive currents and eddy currents and reversal of

magnetic polarity in the armour (1). By introducing these factors into equation (1)

we obtain for the temperature rise of the conductor:

= nI2R(1 + y + y1) [Sis + (1 + ) SCi + (1 + + 1) (SCa + SL) C (6)

For the thermal resistance of the cable we have:

Sk = Sis + SCi + SCa C cm/W (7)

The component resistance are those of the insulation Sis and of the inner and outer

protective coverings SCi and SCa. (The thermal resistance of the metallic

components can be assumed to be negligible.)

The fictitious thermal resistance of the cable is obtained from equations (3)

and (6):

Sk' =

Sis + (1 + λ ) SCi

1 + λ + λ1

+ SCa ,∘C cm /W

(8)

By substituting the thermal resistance of the soil SE for the thermal resistance of the

air SL in equation (3) the current carrying capacity IE of the cable for burial in the

ground is obtained.

For the special case of operation with d.c. it is necessary to substitute the d.c.

resistance R at operating temperature for RW and the actual thermal resistance of

the cable SK for SK'

in equation (3).

The thermal resistance of the cable SK and the effective resistance RW

depend mainly on the values of the components of the cable and its design. The

permissible temperature rise and the thermal resistance of the surroundings SE

or SL are determined by the ambient conditions at the place of installation of the

cable. Generally acceptable values may be agreed upon for the conditions which

are most important i.e. ambient temperature, arrangement of cables, mode of

operation etc. With the help of these the current carrying capacity may be

calculated and planning work is made easier. The values in tables are based on the

standard conditions incorporated in VDE 0255, VDE 0265 and VDE 0271. These

correspond to the climatic conditions prevailing in Central Europe as well as the

methods of laying and the operating conditions most common in Germany.

In the following, these conditions are called “normal” and are indicated by

the index “n” added to the symbol.

With installation in free are the current carrying capacity is:

I n = √ Δ ϑ 105

n RW (Sk' + SLn )

A

(9)

For burial in the ground

I n = √ Δ ϑ 105

n RW (Sk' + SEn )

A

(10)

applies

It must be remembered that for normal conditions several current carrying

capacity values, which depend on the methods of laying and operation, are given

for one and the same cable (six values each for a single-core cable 0.6/1 kV with

copper and aluminum conductors). The extent to which the current carrying

capacity depends on the ambient conditions makes it impossible to arrive at a

“rated current carrying capacity” similar to the rated voltage.

The current carrying capacity is determined by:

(1) Conductor resistance and losses due to the current (ohmic losses),

(2) Thermal resistance of the cable,

(3) Conductor temperature and ambient temperature (temperature difference),

(4) Installation conditions (in free air or in the ground).

(1) Conductor resistance and losses due to the current (ohmic losses):

Certain planning problems can only be solved with knowledge of the ohmic

losses. These losses for cables in the ground are shown in tables for permissible

current In. For operating currents I (other than current rating under conditions In)

the losses may be calculated as follows:

V = V n ( II n )

2

, W/m (11)

(2) Thermal Resistance of the Cables:

The fictitious thermal resistance SK'

required for carrying out special

planning work may be easily obtained for a single cable according to equation

(12):

SK' =

Δϑ n 102

V n

− SEnC cm/W (12)

where

n = temperature rise under normal conditions (see table 1).

Vn = losses depending on the current under standard conditions.

SEn = thermal resistance of the soil at E = 100 C cm/W

The thermal resistance of the soil SEn may be taken from Fig. 6 or calculated as

shown on page 109.

(3) a. Conductor Temperature:

The conductor temperature has been fixed with regard to the life of the cable

and is stated in VDE specifications for nearly all of cables. Please refer to table 1

for a summary.

(3) b. Ambient Temperature:

Measuring the ambient temperature is not possible in every case; it has to be

estimated quite often. The value used for planning should be exceeded only on a

few days during the year or a few hours during the day.

The following ambient temperature values may be taken as a basis, unless

higher values have been established by measurements or experience:

Unheated cellars 20C

Normal conditioned rooms

(unheated during the summer) 25C

Factories, workshops etc. 30C

The aforementioned ambient temperatures are valid under the conditions

prevailing in Central Europe.

Temperatures of above 30C are normally reached in rooms, which are not

sufficiently protected against solar radiation, or badly ventilated or contain

machinery with high heat losses etc.

Table 1: Permissible Conductor Temperature of Mass-impregnated Cables (with

Paper Insulation and Metal Sheath) and Protodur Cables.

Type of CableRated

Voltage

Uo/U

kV

Permissible

Conductor

Temperature

C

Temperature Rise at the

Conductor

Max. Permissible

Temperature ifference1)1)

in the Grounded

at 20C

degC

in Air

at 30C

degC

in the

Ground

degC

in Air

degC

Mass-impregnated Cables

Belted Cables

0.6/1

3.5/680 60 50 65 55

5.8./10 65 45 35 45 35

Single-core

Cables,

H.S.L. Cables

and

H-type Cables

0.6/1

3.5/680 60 50 65 55

5.8/10 70 50 40 55 45

8.7/15

11.6/2065 45 35 45 35

17.3/30 60 40 30 40 30

Protodur Cables

All Types

0.6/1

3.5/670 50 40 55 55

5.8/10 65 45 35 50 50

8.7/15

11.6/2060 40 30 45 45

17.3/30 55 35 25 40 40

Conversion factors for ambient temperatures higher or lower than those

shown may be taken from table 2.

Under certain conditions the heat generated by the losses in the cables may

lead to an increase in the ambient temperature. This applies to cables in tunnels in

particular (see page 94).

1)1) These values have not been fixed for plastic-insulated cables (protodur cables), but it is recommended that they should not be exceeded in consideration of the longitudinal of the cables.

(3) c. Temperature Rise:

The permissible temperature rise of a cable is determined by the permissible

conductor temperature and the ambient temperature – other sources of heat

excluded – (see page 80 and table 1). In cases where the ambient temperature of

cable installation differs from the normal 20C underground or 30C in air, for

example with low outside temperatures over an extended period or due to heating

wires running parallel to the cable route, the permissible current I is calculated as

follows:

I = I n √ (Δϑ / Δϑ n) A (13)

For direct burial = n + 20 - u C (14)

For installation in free air = n + 30 - u C (15)

where

In = Current rating under normal conditions

n = Temperature rise under normal conditions

= actual ambient temperature

= permissible temperature rise according to equations (14) and (15)

With mass-impregnated paper-insulated cables the temperature variation

may not exceed the max. permissible values shown in table 1, which means that

the load may not be increased even at lower ambient temperatures.

The temperature rise within the conductor for any current I, with the ambient

temperature remaining constant and the change in resistance with changing

conductor temperature being neglected, is calculated as follows:

Δϑ = Δϑ n ( II n )

2∘C

(16)

(4) a. Installation in free Air (Thermal Resistance of the Air):

Installation “in free air” is taken as the normal condition for determination of

the current carrying capacity of a cable. By “in free air” is meant that the heat

losses are freely dissipated from the surface of the cable to the surroundings by

natural convection and radiation under exclusion of outside sources of heat,

without the surrounding media being media being heated up noticeably (infinite

heat capacity of the surroundings).

The pre-conditions are:

The clearance between the cables and walls, floor or ceiling must be at least

2 cm. With cables arranged in flat formation in one layer the clearance between

them should not be less than twice the diameter of the cables. Several layers of

cables arranged in flat formation should be installed with a vertical clearance of at

least 30 cm.

Protection against direct solar radiation etc. Sufficiently large or ventilated

rooms such that the ambient temperature is not increased by the heat generated by

the losses in the cables.

If the cable is installed touching a wall or directly on the floor, the current

carrying capacity has to be reduced by 0.95. Rating for grouping of cables are

given in tables 4 and 5. In these tables the reduction by 0.95 for direct installation

on a wall has been taken into account where necessary.

The thermal resistance of the air for a cable installed in free air may be

determined according to page 104. As a rule, however, knowledge of the same is

not required for planning.

With cables of very small diameter the thermal resistance of the air is

considerably greater than the thermal resistance of the soil. The current rating of

cables with a small diameter in free air is therefore lower than in the ground. With

increasing diameter of the cable the thermal resistance of the air decreases faster

than that of the soil. The result is that the ratio of current rating for installation in

free air and in the ground increases together with the diameter of the cable and that

with large diameters the current carrying capacity in free air is greater than that for

installation in the ground. Tests have confirmed that this is actually the case.

(4) b. Direct Burial in the Ground (Thermal Resistance of the Soil):

In the ground, cables are normally embedded in sand or riddled soil and

covered with bricks. This method of laying forms the basis for the current rating

values, whereby the depth of laying h is assumed to be 70 cm and the thermal

resistivity of the soil E to be 100C. cm/W. The equation for the thermal resistance

of the soil SE of, for example, a multicore cable with diameter d (see page 111)

shows that SE increases with increasing h and E, whilst the current rating

decreases:

SE =σ E

2 πln ( 4 h

d ) C cm/W (17)

The influence of the depth of laying is small. With increasing depth the

ambient temperature and normally also the thermal resistivity decrease as the

deeper layers of the ground normally contain more moisture and stay more

uniformly moist than the upper layers. For the normal depths of laying for L.T. and

medium tension cables (70 to 120 cm) a conversion of the current rating values is

not necessary if the ambient temperature is taken as being 20C and the thermal

resistivity of the soil as being 100C cm/W.

Of much greater importance is the thermal resistivity of the soil, which

depends on a large number of factors. There is especially the drying-out of the soil

caused by a continuous load according to the rating tables, which leads to an

increase in the thermal resistance of the soil in the course of time. Therefore the

values in tables (e.g. table 6) for direct burial are to be reduced to 75% if the

current load is constant all the time. This is not necessary, if the possibility of the

soil drying out has been taken into account by calculating with a sufficiently high

value for the soil thermal resistivity (table 3).

4-2 Calculating the Load Current:

The load current is the result of operating voltage U in kV and load to be

transmitted P in kW:

For d.c.:I = P

UA

(18)

For single-phase a.c.: I = P / (U cos ) A (19)

For three-phase current: I = P / (√3 U cos ) A (20)

0

Examples:

1) 3 Protodur cables type NYFGBY 3 x 185 sm 3.5/6 kV laid directly in the ground,

clearance 7 cm, covered with bricks; public utilities load; thermal resistivity of the

soil 150C cm/W, ambient temperature 30C.

Tables of Rating Factors

Rating Factors for the Ambient Temperature

Table 2:

Max. Permissible

Conductor

Temperature

of the Cable C

Ambient Temperature

15 20 25 30 35 40 45 50 55 60

Insulation of

Plas

-tics

Pa-

pe

r1)1)

Pas-

tics

Pa-

per1)

Plas

-tics

Pa-

per1) Plastics and Paper

For Cables in Ground

85 1.04 1 0.96 0.92 0.88 0.83 0.78 0.73 .068 0.62

80 1.04 1 0.96 0.91 0.87 .082 0.76 0.71 0.65 0.58

70 1.05 1 0.95 0.89 0.84 0.77 0.71 0.63 0.65 0.45

65 1.05 1 1 0.94 0.88 0.82 0.75 0.67 0.58 0.47 0.33

60 1.06 - 1 0.93 0.87 0.79 0.71 0.61 0.50 0.35 -

55 1.07 - 1 0.93 0.85 0.76 0.65 0.53 0.38 - -

For Cables in Air

85 1.13 - 1.09 - 1.04 1 0.95 0.90 0.85 0.80 0.74 0.67

80 1.14 1.05 1.09 1.05 1.05 1 0.95 0.89 0.84 0.77 0.71 0.63

70 1.17 1.06 1.12 1.06 1.06 1 0.94 0.87 0.79 0.71 0.61 0.50

65 1.19 1 1.13 1 1.07 1 1 0.93 0.85 0.76 0.65 0.53 0.38

60 1.22 1 1.15 1 1.08 1 1 0.91 0.82 0.71 0.58 0.41 -

55 1.26 - 1.18 - 1.10 - 1 0.89 0.77 0.63 0.45 - -

Thermal Resistivity of the Soil

1)1) With regard to the max. permissible temperature difference for mass-impregnated paper-insulated cables (see page 83) it is not possible in every case to increase the current rating at low ambient temperature.

Table 3 Multi-core Cables

Rated

Voltage

Uo/U

kV

Cross- sectional

Area of

Conductors

mm2

Thermal Resistivity of the Soil E

degC-cm/W

701)1) 100 120 150 200 250 300

Rating Factor

Twin-core Plastic (P.V.C.) and Paper-insulated Cables

0.6/1

up to 25 1.09 1 0.95 0.88 0.80 0.73 0.69

35 – 95 1.11 1 0.94 0.87 0.78 0.71 0.66

120 – 240 1.12 1 0.94 0.86 0.78 0.70 0.65

300 – 500 1.13 1 0.93 0.86 0.77 0.69 0.65

3- and 4-core Plastic (P.V.C.) and Paper-insulated Cables

0.6/1

up to 25 1.11 1 0.94 0.87 0.78 0.72 0.67

35 – 95 1.13 1 0.93 0.86 0.76 0.70 0.64

120 – 240 1.14 1 0.93 0.85 0.76 0.69 0.63

300 – 500 1.15 1 0.92 0.85 0.75 0.68 0.63

3-core Plastic (P.V.C.) –insulated Cables

3.5/6

and

5.8/10

up to 25 1.11 1 0.94 0.87 0.78 0.72 0.67

35 – 95 1.13 1 0.93 0.86 0.76 0.70 0.64

120 – 240 1.14 1 0.93 0.85 0.76 0.69 0.63

300 – 500 1.15 1 0.92 0.85 0.75 0.68 0.63

3-core Paper-insulated Cables (Belted, H-type and H.S.L. Cables)

3.5/6

and

5.8/10

up to 25 1.09 1 0.95 0.88 0.80 0.73 0.69

35 – 95 1.11 1 0.94 0.87 0.78 0.71 0.66

120 – 240 1.12 1 0.94 0.86 0.78 0.70 0.65

300 – 500 1.13 1 0.93 0.86 0.77 0.69 0.65

3-core Cables with Paper Insulation (H.S.L. and H-type cables) and Plastic (P.V.C.) – insulated Cables

8.7/15

11.6/20

17.3/30

up to 25 1.08 1 0.96 0.9 0.81 0.75 0.7

35 – 95 1.09 1 0.95 0.89 0.79 0.73 0.67

120 – 240 1.1 1 0.95 0.88 0.79 0.72 0.66

300 – 500 1.11 1 0.94 0.88 0.78 0.71 0.66

Table 3 (Continuation) Single-core Cables

1)1) The factor in this column may be used only if the thermal resistivity of the soil has been actually measured.

Rated

Voltage

Uo/U

kV

Cross- sectional

Area of

Conductors

mm2

Thermal Resistivity of the Soil E

degC-cm/W

701)1) 100 120 150 200 250 300

Rating Factor

3 Unarmoured Cables in Flat Formation or in Trefoil Formation

0.6/1

3.5/6

5.8/10

up to 25 1.12 1 0.94 0.85 0.76 0.70 0.65

35 – 95 1.14 1 0.94 0.84 0.74 0.68 0.62

120 – 240 1.15 1 0.93 0.83 0.74 0.67 0.61

300 – 500 1.16 1 0.92 0.83 0.73 0.66 0.61

8.7/15

11.6/20

17.3/30

up to 25 1.11 1 0.94 0.87 0.78 0.72 0.67

35 – 95 1.13 1 0.93 0.86 0.76 0.70 0.64

120 – 240 1.14 1 0.93 0.85 0.76 0.69 0.63

300 – 500 1.15 1 0.92 0.85 0.75 0.68 0.63

1 Separately Installed D.C. Cable with Plastic Insulation (P.V.C.)

0.6/1

up to 25 1.09 1 0.95 0.88 0.80 0.73 0.69

35 – 95 1.11 1 0.94 0.87 0.78 0.71 0.66

120 – 240 1.12 1 0.94 0.86 0.78 0.70 0.65

300 – 500 1.13 1 0.93 0.86 0.77 0.69 0.65

1 Separately Installed D.C. Cable with Paper Insulation

8.7/15

11.6/20

17.3/30

up to 25 1.08 1 0.95 0.89 0.80 0.75 0.7

35 – 95 1.1 1 0.95 0.88 0.78 0.73 0.67

120 – 240 1.11 1 0.94 0.87 0.78 0.72 0.66

300 – 500 1.12 1 0.93 0.87 0.77 0.71 0.66

Grouping in Air:

1)1) The factor in this column may be used only if the thermal resistivity of the soil has been actually measured.

Table 4: Multi-core Cable in Three-phase Systems and Single-core Cables in D.C.

Systems. These factors are valid only under the condition that the

ambient temperature is not perceptibly increased by the heat generated

by the cable.

Distance from the wall 2cm

Clearance between cables = diameter d

Number of Cables

1 2 3 6 9

Rating Factor

Cable laid on the ground in flat formation

0.95 0.90 0.88 0.85 0.84

Cables laid on troughs (circulation of air is restricted)

Number of

troughs

1 0.95 0.90 0.88 0.85 0.84

2 0.90 0.85 0.83 0.81 0.80

3 0.88 0.83 0.81 0.79 0.78

6 0.86 0.81 0.79 0.77 0.76

Cables laid on racks in flat formation

Number of

racks

1 1 0.98 0.96 0.93 0.92

2 1 0.95 0.93 0.90 0.89

3 1 0.94 0.92 0.89 0.88

6 1 0.93 0.90 0.87 0.86

Cables arranged on structures or on the wall

1 0.93 0.90 0.87 0.86

Arrangements for which a reduction of the current rating is not necessary

Distance form the wall

2cm

Clearance between cables

= 2d

Any number of cables

Grouping in Air:

Table 5: Single-core Cables in Three-phase Systems. These factors are valid only

under the condition that the ambient temperature is not perceptibly

increased by the heat generated by the cable.

Distance from the wall 2 cm

Clearance between cable = diameter d

Number of System

1 2 3

Rating Factors

Cables laid on the ground in flat formation

0.92 0.89 0.88

Cables laid on troughs (circulation of air is restricted)

Number of

troughs

1 0.92 0.89 0.88

2 0.87 0.84 0.83

3 0.84 0.82 0.81

Cables laid on racks in flat formation

Number of

racks

1 1 0.97 0.96

2 0.97 0.94 0.93

3 0.96 0.93 0.92

6 0.94 0.91 0.90

Cables arranged on structures or on the wall

0.94 0.91 0.89

Touching the wall 0.89 0.86 0.84

Current rating under normal conditions according to table 6 for a single cable: In =

370 A.

Permissible conductor temperature according to table 6 : 70 C.

Rating factors:

For higher ambient temperature according to table 2:0.89

For higher thermal resistivity of the soil according to

Table 3 0.85

For grouping according to table 0.75

The permissible load current for 3 cables is therefore:

I = 3.370. 0.89. 0.85. 0.75 = 630 A.

The transmission power (complex PS) of the cable connection is:

PS = I √3 U 10-3

= 630 √3 6 10-3

= 6.55 MVA

2) 5 MVA at an operating voltage of 10 kV have to be transmitted by Protodur

cables.

Ambient conditions:

Cables laid in flat formation in the ground, clearance 7 cm covered with

bricks; continuous load; thermal resistivity of non-dried-out soil 100C cm/W

ambient temperature 25C.

Load Current:

I = PS 103 / (√3 U) = 5 103 / (√3 10) = 290

Estimate: 2 cables installed parallel will be required.

Table 6: Three-and Four-core Cables for 0.6/1 kV, e.g. NYY, NYCY, NYCWY,

NYKY, YTY, Three-core Cables for 3.5/6kV and 5.8/10 kV, e.g.

NYFGbY, NYSEY, NYHFGbY.

NominalCross-sectional

Areamm2

Rated Voltage Uo/U in kV0.6/1 3.5/6 5.8/10

Permissible Conductor Temperature70C 70C 65C

Three-and Four-core Three-core Three-coreAmbient Temperature and Installation

20C 30C 20C 30C 20C 30CGround Air1)1) Ground Air Ground Air

Current Rating in ACopper Conductors1.5 27 18 -- -- -- --2.5 36 25 -- -- -- --4 46 34 -- -- -- --6 58 44 58 48 -- --10 77 60 76 65 72 6316 100 80 98 86 93 8325 130 105 125 110 120 11035 155 130 150 135 145 13050 185 160 175 165 170 15570 230 200 220 205 210 19595 275 245 260 250 245 235120 315 285 295 285 280 270150 355 325 335 325 320 310185 400 370 370 370 355 350240 465 435 425 430 405 400300 520 500 475 490 450 450400 600 600 540 570 520 530

Aluminum Conductors25 100 82 97 87 92 8435 120 100 115 105 110 10050 145 125 135 130 130 12070 175 155 170 160 160 15095 215 190 200 195 190 185120 245 220 230 220 215 210150 275 250 260 250 245 240185 310 285 290 285 280 270240 360 340 330 340 315 310300 410 390 380 390 355 355400 470 460 425 450 410 420

Rating factors:

1)1) For the current rating of cables of Uo/U = 0.5/1 kV in air please refer to the preface also.

For continuous load: see page 0.75

For grouping according to table: 0.85

For higher ambient temperature according to table 2 with the max

permissible conductor temperature of 65C for protodur cables for 5.8/10kV:

0.94

The fictitious current per cable, with 2 cable installed in parallel, is:

I f =2902 × 0 .75 × 0. 85 × 0 . 94

= 242 A

According to table 6 cables with a cross-sectional area of copper of 95 mm2

have to be installed (permissible current rating under normal conditions: 245 A).

4-3 Cables in Tunnels:

Unventilated Tunnels:

In unventilated covered tunnels the heat generated in the cables is dissipated

only via the walls of the tunnel in all directions. The accumulating heat increases

the temperature of the air surrounding the cable within the tunnel, which means

that the current rating is reduced as compared with the values for installation in

free air. The heating of air in the tunnel depends solely on the magnitude of the

power loss of the cables, whilst the number of cables causing this power loss and

their arrangement within the tunnel is of no influence.

Figure 2 shows the increase in the air temperature within the tunnel as a

function of the power per 1 m length of the tunnel with the circumference of the

tunnel as parameter. This diagram, together with the current rating values of cables

in free air as given table 6, enables the cable for a certain load to be determined, or

if the number of cables, the cross-sectional area and load as well as the dimension

of the tunnel are given, the temperature rise of the cables to be worked out.

When calculating the circumference of a tunnel, only those surfaces which

permit dissipation of heat should be taken into account; walls which border on

warm operational rooms, transformer cells etc. may not be included.

The Installation of Cables in Tunnels:

The cables are either fixed to the walls, using clamps etc., or are laid out on

steel structures or racks. The necessary clearance between racks depends on their

depth, but should not be less than 300 mm. The clearance between the cables

should always be at least equal to the cable diameter in order to reduce to a

minimum the direct transmission of heat between them, regardless of whether the

cables are fixed directly to the walls or laid out on structures or racks.

The head room should not be less than 2 m and the width sufficient to leave

a free passage of at least 60 to 80 cm. The depth of racks arranged with a vertical

clearance of 30 cm should not exceed 50 cm in order to facilitate installation of the

cables.

The planning of such an installation may be carried out as follows: First the

cross-sectional area of each cable is estimated, working with values approximately

30% higher than those required for installation in free air. Heavy currents may

make the installation of several cables in parallel necessary. Next a rough drawing

of the tunnel will be made, showing the required height, width, number of racks

and arrangement of the cables in accordance with the rules given above.

Fig. 2 Increase of the Air Temperature inside a Tunnel as a Function of the Power

Loss per 1 m Length of Tunnel

The reduction factor for grouping in free air f’ may now be taken from table

4 or 5 for the arrangement laid down in the sketch. Next the total losses of all

cables in the tunnel and the increase in the tunnel temperature are determined

according to Fig. 2. The temperature of the air in the tunnel with no load on the

cables is to be increased by this value and the reduction factor for higher ambient

temperature f’’ is to be taken from table 2 or calculated according to equation (13).

Multiplication of the current rating under normal conditions In with these factors

must not lead to a result lower than the required current currying capacity:

In f f I

Thus

f f I/In (21)

The number of cables, their cross-sectional areas or the circumference of the

tunnel should be increased if this is not the case (see also the example on page

).

The time constant of a tunnel is very large compared with that of the cables.

The rise in temperature of the air within the tunnel may therefore be determined

from the losses calculated on the basis of the r.m.s. value Iq of the currents over a

period of 24 hours:

I q = √ I 12 t1 + I2

2 t2 + . ..

t1 + t2

A(22)

t1 + t2 + … = 24 hr

where I1, I2 .. = the currents flowing during the periods t1, t2 …

Tunnels with Forced Ventilation:

If the heat dissipation through the walls of the tunnel is not sufficient and the

temperature of the air inside the tunnel become too high, causing the maximum

conductor temperature to be exceeded, then forced ventilation has to be provided,

unless other methods may be resorted to as, for example, increasing the

circumference of the tunnel and thus the heat-dissipating surface.

In most cases the calculation is based on the total heat generated by the

losses in the cables, neglecting the fact that heat is still dissipated through the walls

of the tunnel. The fans selected on this basis will more than fully meet the

requirements and there will be ventilation power to spare for further extensions of

the installation.

The quantity of cooling air Q required depends on the total heat generated by

the losses in the cables V, the length of the tunnel 1 and the temperature rise of

the cooling air ku flowing through the tunnel:

Q = 0.77 10-3 V 1 / ku m3/s (23)

The speed of the flow v is calculated on the basis of the cross section of the

tunnel A in m2:

V = Q / A, m/s (24)

The speed of the air flow should not exceed 5 m/s if annoying noise is to be

avoided.

The temperature rise of the cooling air has to be determined with ranged to

its original temperature and the permissible temperature at the point of exit from

the tunnel. In most cases the original temperature of the cooling air is the same as

the ambient temperature u. For the relatively warmest cable, taking into account

its maximum permissible conductor temperature Ln, we obtain for the temperature

rise of the cooling air:

Ku Ln - u - , C (25)

with

= n (I/In)2 C (26)

The grouping factor f need not be applied as the flowing air considerably

improves the heat dissipation of the cables.

Example:

The cables shown in table 7 for the loads indicated are to be installed in a

tunnel 2.2 m high and 1.5 m wide. In the first stage 8 h of full load daily are to be

taken into account. Operation at full load for 16 h per day, for which forced

ventilation may be provided, must be possible also. The ambient temperature

inside the tunnel with no load on the cables is taken to be 35C. The intended

arrangement of the cables is shown in Fig. 3.

Fig. 3

Table 7:

Serial a b c d

Type of Cable

Uo/U kV

NYFGbY

3 150

sm

3.5/6

NYSEY

3 240

rm

5.8/10

NEKBY

3 70 rm

11.6/20

NEKBY

3 120 rm

11.6/20

Number of Cables 14 8 6 7

Load I A 205 240 120 170

Normal Values

in Air In A 325 400 205 280

in the Ground In A 335 410 215 290

in Air n degC 40 35 35 35

In the Ground Vn W/m 51 47 45 48

Ln C 70 65 65 65

II n ( Air ) 0.63 0.6 0.585 0.607

Full Load 8 h Daily

V W/m 6.35 5.35 4.83 5.5

V W/m 89 42,8 29 38.5

f 0.76 0.715 0.715 0.715

f f 0.66 0.62 0.62 0.62

degC 16 12.6 12 12.9

(ku)max. degC 19 17.4 18 17.1

A reduction factor f 0.87 for the grouping of 5 cables on each of 7 racks has to

be used (see also table).

With 8 h full load daily the r.m.s. value of the currents of the cables (a) is:

I q = √ I12 t1

t1+ t2

= I √ 824

A = 205 √ 824

A = 118 A(22)

The losses (with In for underground installation according to page inserted into

the equation) are:

V = V n ( I q

In )2

= V n8

24 ( II n

)2

(11)

= 518

24 (205335 )

2

=6 . 35 W /m

and

V = 14 6.35 = 89 W/m

According to table 7 the total losses of all cables are:

89 + 42.8 + 29 + 38.5 = 199.3 W/m

and the temperature rise of the air in the tunnel according to Fig. 2 amounts to

12C. Thus the temperature of the air increases to 35 + 12 = 47C.

The factor for the cables (a) is obtained by equations (13) and (15)

f ' ' =√ ΔϑΔϑ n

= √40+30−4740

= 0 .76

The change-over to operation with 16 h full load daily leads to:

I q = I √1624

A(22)

and

V = V n162 ( I

I n)2

, W /m(11)

The total losses in the tunnel are thus doubled

V = 2 199.3 = 398.6 W/m

and the temperature of he air within the tunnel increases by 20C (Fig. 2) to:

35 + 20 = 55C

Forced ventilation has to be provided.

Thus for cables (a)

= 40 (205/325)2 = 16 C (16)

KU 70 – 35 – 16 19 C (25)

The values for the remaining cables may be taken from Table 7. ku = 10C

is chosen. With a tunnel length of 20 m and a cross-section of 1.5 2.2 m = 3.3 m2

the quantity of cooling air required is:

Q = 0 .77 × 10−3 390 × 2010

= 0 .6 m3 /s(23)

and speed of the air flow

= 0.6 / 3.3 = 0.182 m/s (24)