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21/30431011 DC BS EN 60287-1-3. Electric cables. Calculation of the current rating - Part 1-3. Current rating equations (100 % load factor) and calculation of losses. Current sharing between parallel single-core cables and calculation of circulating curre >
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21/30431011 DC BS EN 60287-1-3. Electric cables. Calculation of the current rating - Part 1-3. Current rating equations (100 % load factor) and calculation of losses. Current sharing between parallel single-core cables and calculation of circulating curre, 2021
- 30431011
- 20_1953e_CD [Go to Page]
- 20/1953/CD
- COMMITTEE DRAFT (CD)
- Project number:
- IEC 60287-1-3 ED2
- Date of circulation:
- Closing date for comments:
- 2021-02-12
- 2021-05-07
- Supersedes documents:
- 20/1946/RR
- IEC TC 20 : Electric cables
- Secretariat:
- Secretary:
- Germany
- Mr Walter Winkelbauer
- Of interest to the following committees:
- Proposed horizontal standard:
- Other TC/SCs are requested to indicate their interest, if any, in this CD to the secretary.
- Functions concerned:
- EMC
- Environment
- Quality assurance
- Safety
- This document is still under study and subject to change. It should not be used for reference purposes.
- Recipients of this document are invited to submit, with their comments, notification of any relevant patent rights of which they are aware and to provide supporting documentation.
- Title:
- Electric cables - Calculation of the current rating - Part 1-3: Current rating equations (100 % load factor) and calculation of losses - Current sharing between parallel single-core cables and calculation of circulating current losses
- Note from TC/SC officers:
- CONTENTS
- FOREWORD 4
- INTRODUCTION 7
- 1 Scope 9
- 2 Normative references 9
- 3 Symbols 9
- 4 Description of method 11
- 4.1 General 11
- 4.2 Outline of method 13
- 4.3 Matrix solution 19
- Annex A (informative) Example calculations 21
- Annex B (informative) Example of the computation of the coefficient for hollow core conductors 35
- Bibliography 37
- Figure B.1 – Representation of a hollow core conductor 37
- Table 1 – Values of for conductors 17
- Table A.1 – Calculated values of di,k 29
- Table A.2 – Calculated values of zz 29
- Table A.3 – Array [Z] including coefficients for currents 31
- INTERNATIONAL ELECTROTECHNICAL COMMISSION
- ____________
- ELECTRIC CABLES –
- CALCULATION OF THE CURRENT RATING –
- Part 1-3: Current rating equations (100 % load factor) and calculation of losses –
- Current sharing between parallel single-core cables and calculation of circulating current losses
- FOREWORD
- 1) The IEC (International Electrotechnical Commission) is a worldwide organization for standardization comprising all national electrotechnical committees (IEC National Committees). The object of the IEC is to promote international co-operation on all questions concerning standardization in the electrical and electronic fields. To this end and in addition to other activities, the IEC publishes International Standards. Their preparation is entrusted to technical committees; any IEC National Committee interested in the subject dealt with may participate in this preparatory work. International, governmental and non-governmental organizations liaising with the IEC also participate in this preparation. The IEC collaborates closely with the International Organization for Standardization (ISO) in accordance with conditions determined by agreement between the two organizations.
- 2) The formal decisions or agreements of the IEC on technical matters express, as nearly as possible, an international consensus of opinion on the relevant subjects since each technical committee has representation from all interested National Committees.
- 3) The documents produced have the form of recommendations for international use and are published in the form of standards, technical specifications, technical reports or guides and they are accepted by the National Committees in that sense.
- 4) In order to promote international unification, IEC National Committees undertake to apply IEC International Standards transparently to the maximum extent possible in their national and regional standards. Any divergence between the IEC Standard and the corresponding national or regional standard shall be clearly indicated in the latter.
- 5) The IEC provides no marking procedure to indicate its approval and cannot be rendered responsible for any equipment declared to be in conformity with one of its standards.
- 6) Attention is drawn to the possibility that some of the elements of this International Standard may be the subject of patent rights. The IEC shall not be held responsible for identifying any or all such patent rights.
- International Standard IEC 60287-1-3 has been prepared by subcommittee 20A: High-voltage cables of IEC technical committee 20: Electric cables.
- This second edition cancels and replaces the first edition published in 2002. This edition constitutes a technical revision.
- This edition includes the following significant technical changes with respect to the previous edition:
- a) Change and update of list of symbols;
- The text of this standard is based on the following documents:
- FDIS
- Report on voting
- XX/XX/FDIS
- XX/XX/RVD
- Full information on the voting for the approval of this standard can be found in the report on voting indicated in the above table.
- This publication has been drafted in accordance with the ISO/IEC Directives, Part 2.
- Annexes A and B are for information only.
- The committee has decided that this publication remains valid until 20XX. At this date, in accordance with the committee’s decision, the publication will be
- • reconfirmed;
- • withdrawn;
- • replaced by a revised edition, or
- • amended.
- INTRODUCTION
- When single-core cables are installed in parallel the load current may not share equally between the parallel cables. The circulating currents in the sheaths of the parallel cables will also differ. This is because a significant proportion of the impedance of large conductors is due to self reactance and mutual reactance. Hence the spacing and relative location of each cable will have an effect on the current sharing and the circulating currents. The currents are also affected by phase rotation. The method described in this standard can be used to calculate the current sharing between conductors as well as the circulating current losses.
- There is no simple rule by which the circulating current losses of parallel cables can be estimated. Calculation for each cable configuration is necessary. The principles and impedance formulae involved are straightforward but the difficulty arises in solving the large number of simultaneous equations generated. The number of equations to be solved generally precludes the use of manual calculations and solution by computer is recommended. For nc cables per phase having metallic sheaths in a three-phase system there are six*nc equations containing the same number of complex variables.
- For simplicity the equations set out in this standard assume that the parallel conductors all have the same cross-sectional area. If this is not the case, the equations may be adapted to allow for different resistances for each conductor. The effect of neutral and earth conductors can also be calculated by including these conductors in the appropriate loops. The method set out in this standard does not take account of any portion of the sheath circulating currents that may flow through the earth or other extraneous paths. In this respect, the effect of earth return path has been excluded for the purposes of the methodology described in the following, as it is concluded that it can affect the magnitude of the resulting circulating currents only by a small extend on a limited number of cases, where both very low soil electrical resistivity values and low earthing conductor resistance values are simultaneously considered.
- The conductor currents and sheath circulating currents in parallel single-core cables are unlikely to be equal. Because of this, the external thermal resistance for buried parallel cables should be calculated using the method set out in 3.1 of IEC 60287-2-1. Because the external thermal resistance and sheath temperatures are functions of the power dissipation from each cable in the group it is necessary to adopt an iterative procedure to determine the circulating current losses and the external thermal resistance.
- ELECTRIC CABLES –
- CALCULATION OF THE CURRENT RATING –
- Part 1-3: Current rating equations (100 % load factor)
- and calculation of losses –
- Current sharing between parallel single-core cables
- and calculation of circulating current losses
- 1 Scope
- This part of IEC 60287 provides a method for calculating the phase currents and circulating current losses in single-core cables arranged in parallel.
- The method described in this standard can be used for any number of cables per phase in parallel in any physical layout. The phase currents can be calculated for any arrangement of sheath bonding. For the calculation of sheath losses, it is assumed that the sheaths are bonded at both ends. A method for calculating sheath eddy current losses in two circuits in flat formation is given in IEC 60287-1-2.
- 2 Normative references
- The following referenced documents are indispensable for the application of this document. For dated references, only the edition cited applies. For undated references, the latest edition of the referenced document (including any amendments) applies.
- 3 Symbols
- dc
- external diameter of the conductor, mm
- ds
- mean diameter of sheath or screen, mm
- f
- system frequency, Hz
- nc
- number of cables per phase
- Dmn
- axial spacing between conductors, mm
- Inc
- current in the conductor of cable nc, A
- Isnc
- circulating current in the sheath of cable nc, A
- R
- resistance of a conducting element, /m
- Rc
- a.c. resistance of conductor at maximum operating temperature, /m
- Rs
- a.c. resistance of the cable sheath or screen at their maximum operating temperature, /m
- Xi,k
- apparent mutual reactance of a pair of conductors
- V
- conductor voltage drop, V
-
- coefficient depending on the construction of the conductor
- ’nc
- sheath loss factor of cable p due to circulating currents
-
- angular frequency of system (2f), s–1
- Note: subscripts m, n, i and k are used in the following only to denote rows and columns of matrices and therefore to identify specific matrix elements. They should not be confused with the corresponding symbols used in other parts of IEC 60287 series of Standards for identifying physical quantities.
- 4 Description of method [Go to Page]
- 4.1 General
- The method calculates the proportion of the phase current carried by each parallel conductor and the circulating current in the sheath of each cable. The loss factor (’) for each case is then calculated as the ratio of the losses in a sheath caused by circulating currents to the losses in the conductor of that cable.
- The method of calculation set out below only considers voltage drop along the conductors. Any unbalance in the load which would lead to unbalanced phase currents is ignored.
- The equations to be solved for the unknown currents in the parallel conductors and their sheaths are built up from a consideration of the basic formulae for the impedance associated with a loop consisting of two long conductors lying parallel to each other and the formulae for the mutual impedance between a loop and an adjacent conductor. Consideration of these equations leads to a system of simultaneous equations for the impedance voltage for all the conductors and sheaths in a three-phase parallel cable system. The impedance voltages for all conductors in parallel in the same phase are equal. Also for the conductors representing the bonded sheaths the voltages are equal. Hence the impedance voltages can be eliminated from the equations. The sum of the currents in the parallel conductors is equal to either the known phase current or zero for the sheaths. This provides the additional information needed for solution of the simultaneous equations.
- It should be noted that all the currents are complex quantities containing both real and imaginary parts.
- The mutual impedance between conductors is a function of their relative positions. Hence, if the relative positions of the cables vary along the route, or the sheaths are cross-bonded, then the impedance for each section shall be calculated individually and the vector results summed in order to obtain the total impedance of each loop. If the route length is very short, then significant errors may occur in the calculated result due to the change in the relative positions of the cables as they approach the terminations.
- The equations set out in this standard can also be used to calculate the current sharing between cables without a metallic sheath or armour and between cables with the sheaths connected together at one end only, single-point bonded. For such calculations, the circulating current in each sheath is zero. Where cable sheaths are bonded at one end only, the standing voltage at the open circuit end of the sheath can also be determined using this method of calculation.
- For the method set out in this standard, it is recommended that the solution of the equations is achieved by a process of matrix algebra. This has the advantage that the solution achieved is unique and not a function of an iterative process. [Go to Page]
- 4.2 Outline of method
- The loss factor for the sheath in a given cable in a parallel circuit is given by:
- (1)
- The currents Isnc and Inc are obtained by solution of equations of the following form where there are nc conductors in parallel and a total of six*nc conductors in a three-phase system. To simplify matters, both the phase conductors and the sheaths are referred to as conductors. The phase conductor currents are I1, I2 etc. The sheath currents are I3nc+1, I3nc+2, I3nc+3, etc.
- For convenience in the calculations, the following notation is used:
- Cable references
- Circuit
- 1
- …
- i
- …
- nc
- Phase R
- 1
- …
- i
- …
- nc
- Phase S
- nc + 1
- …
- nc + i
- …
- 2nc
- Phase T
- 2nc + 1
- …
- 2nc + i
- …
- 3nc
- The conductors can then be identified as follows:
- Reference of a phase conductor = reference of the cable
- Reference of a sheath conductor = reference of the cable + 3nc
- For each phase the current is given by:
- (2)
- The above equations assume forward phase rotation. If the phase rotation is not known, the calculation shall be carried out for both forward and reverse phase rotations.
- For conductor loops representing the sheaths, the current is given by:
- (3)
- The voltage drop in each conductor is then
- – for the conductors of phase R:
- (4)
- for i = 1 to nc;
- – for the conductors of phase S:
- (5)
- for i = nc + 1 to 2nc;
- – for the conductors of phase T:
- (6)
- for i = 2nc + 1 to 3nc;
- – for the sheath conductors:
- (7)
- for i = 3nc + 1 to 6nc.
- Eliminating the voltage drop from this set of equations leads to (6nc – 4) equations having the following form:
- (8)
- where
- and R is defined as follows:
- R = 0 if R = 0 if
- For the phase conductors (refer to array [Z] in Example 1)
- R = Rc if i = k and R = – Rc if i = k – 1 and
- For the sheath conductors (refer to array [Z] in Example 1)
- R = Rs if i = k and R = – Rs if i = k – 1 and
- Xi,k is regarded as a reactance and is defined as follows:
- (9)
- where, if,, then di,k = Dm,n = axial spacing between cables m and n,
- with m = i if m = i – 3nc if
- and n = k if n = k – 3nc if
- If i = k and then
- if i = k and then
- where
- For appropriate values of coefficient α see Table 1.
- Table 1 – Values of for conductors
- Number of wires
- Value of
- 1 (solid)
- 0,779
- 3
- 0,678
- 7
- 0,726
- 19
- 0,758
- 37
- 0,768
- 61
- 0,772
- 91
- 0,774
- 127
- 0,776
- The values given in Table 1 are applicable to non-compacted conductors. For compacted conductors = 0,779 should be used. The values for hollow conductors are dependant on the inner and outer diameter of the conductor. An example of the calculation of for hollow conductors is given in Annex B. [Go to Page]
- 4.3 Matrix solution
- In general the equations developed will be of the form:
- where the values for Q are given by the left-hand side of equations (2), (3) and (8). The value for Z are the coefficients of I in these equations, and the values for I are the unknown currents in the conductors and sheaths.
- In matrix form the equations become:
- where [Z] is a square matrix of the coefficients of I1 to Ik in equations (2), (3) and (8).
- In order to solve for the unknown currents [I] the equation is written as:
- where [Z]–1 is the inverse matrix of [Z].
- Example calculations using the matrix solution are given in Annex A.
- Annex A (informative)Example calculations [Go to Page]
- A.1 Introduction
- The cable dimensions used in these examples are arbitrary and do not represent any particular type of cable.
- It is assumed that the relative positions of the cables do not change over the length of the run. It is also assumed that the bonding conductors have an impedance which is negligible compared with the impedance of the conductors. The skin and proximity effects on a.c. resistance are ignored. The various impedance values calculated in these examples are for 1 000 m long cables.
- These examples assume a supply frequency of 50 Hz.
- The cable and installation parameters are as follows:
- Copper conductor diameter:
- 32,8 mm
- Conductor resistance at 20 °C:
- 28,3 ( 10–6 /m
- Maximum operating temperature:
- 70 °C
- Conductor resistance at 70 °C:
- 33,86 ( 10–6 /m
- Number of wires in conductor:
- 127
- Conductor coefficient for 127 strands:
- 0,776
- Aluminium sheath resistance at 20 °C:
- 0,18 ( 10–3 /m
- Mean diameter of sheath:
- 48 mm
- Sheath temperature:
- 60 °C
- Sheath resistance at 60 °C:
- 0,209 ( 10–3 /m [Go to Page]
- A.2 Example 1
- The cables are laid in flat formation at 200 mm between centres with two cables per phase and no neutrals. The cable arrangement is as follows:
- Cable 1
- Cable 3
- Cable 5
- Cable 6
- Cable 4
- Cable 2
- R1
- S1
- T1
- T2
- S2
- R2
- For convenience in the calculation, the conductors and sheath of each cable are numbered so that the conductors are numbered 1 to 6 and the sheaths 7 to 12. The first cable will have conductor 1 and sheath 7. The second cable being 2, 8 etc. This gives a total of 12 conductors in this example.
- For a single circuit installed in flat formation at 200 mm centres, with one cable per phase, the sheath loss factors calculated in accordance with IEC 60287-1-1 are:
- Outer
- Middle
- Outer
- 1,99
- 1,50
- 2,62
- These values are similar to the values obtained in examples 1 and 2, but significantly different from those obtained in example 4. [Go to Page]
- A.2.1 Calculations
- The zero co-ordinates (0,0) can be fixed at any point, but it is convenient to take the axis of the lower left-hand cable as (0,0). The cable co-ordinates are entered in to the array S below:
- x
- y
- 0
- 0
- Cable 1, phase R
- 1 000
- 0
- Cable 2, phase R
- S =
- 200
- 0
- Cable 3, phase S
- 800
- 0
- Cable 4, phase S
- 400
- 0
- Cable 5, phase T
- 600
- 0
- Cable 6, phase T
- The axial cable spacings are calculated using the following equation:
- m = 1 to 6 n = 1 to 6
- The spacings are given in the array D below:
- 0
- 1 000
- 200
- 800
- 400
- 600
- 1 000
- 0
- 800
- 200
- 600
- 400
- D =
- 200
- 800
- 0
- 600
- 200
- 400
- 800
- 200
- 600
- 0
- 400
- 200
- 400
- 600
- 200
- 400
- 0
- 200
- 600
- 400
- 400
- 200
- 200
- 0
- Clearly this array is symmetrical about its diagonal and it is not necessary to calculate the spacing between cables m and n as well as between cables n and m.
- This array is then modified to include all the values of di,k required to calculate Xi,k. The modified array is given in Table A1.
- The effective reactances Xi,k are calculated using equation (9):
- The coefficients, zz, for the right-hand side of equation (8) are calculated as follows and are given in the array zz, as shown in Table A.2.
- where
- R = 0 if R = 0 if
- R = Rc if i = k and R = –Rc if i = k – 1 and
- R = Rs if i = k and R = –Rs if i = k – 1 and
- The coefficients for the current, I, for the right-hand side of equations (2) and (3) are shown in array H below;
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- Phase R
- H =
- 0
- 0
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- Phase S
- 0
- 0
- 0
- 0
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- Phase T
- 0
- 0
- 0
- 0
- 0
- 0
- 1
- 1
- 1
- 1
- 1
- 1
- Sheath
- For convenience of calculation these coefficients are included in the same matrix as those obtained from consideration of the conductor loops. The new array [Z] is given in Table A.3.
- The values and coefficients for the left-hand side of equations (2) and (3) are given in array [Q] below:
- 0
- 1
- 0
- –0,5 – 0,866j
- 0
- –0,5 + 0,866j
- [Q] =
- 0
- 0
- 0
- 0
- 0
- 0
- The phase and sheath currents in each conductor can then be calculated by solving the simultaneous equations set out in array [Z], Table A.3, and [Q] above. These currents are given below in terms of the resistive and reactive components. Multiplying the inverse of matrix [Z] by [Q] solves the equations.
- 0,5
- 0,5
- –0,25 –0,433j
- –0,25 –0,433j
- –0,25 +0,433j
- [I] =
- –0,25 +0,433j
- –0,216 –0,1892j
- –0,216 –0,1892j
- –0,1309 +0,2164j
- –0,1309 +0,2164j
- 0,3469 –0,0272j
- 0,3469 –0,0272j
- The magnitude of the phase conductor and sheath currents together with the sheath loss factor are given below, assuming a total phase current of 100 A.
- Phase conductor current =; Sheath current =;
- Loss factor =
- Phase current
- Sheathcurrent
- Sheath loss factor
- Cable 1, phase R
- 50
- 28,7
- 2,036
- Cable 2, phase R
- 50
- 28,7
- 2,036
- Cable 3, phase S
- 50
- 25,3
- 1,58
- Cable 4, phase S
- 50
- 25,3
- 1,58
- Cable 5, phase T
- 50
- 34,8
- 2,99
- Cable 6, phase T
- 50
- 34,8
- 2,99
- Table A.1 – Calculated values of di,k
- 12,73
- 1 000
- 200
- 800
- 400
- 600
- 24
- 1 000
- 200
- 800
- 400
- 600
- 1 000
- 12,73
- 800
- 200
- 600
- 400
- 1 000
- 24
- 800
- 200
- 600
- 400
- 200
- 800
- 12,73
- 600
- 200
- 400
- 200
- 800
- 24
- 600
- 200
- 400
- 800
- 200
- 600
- 12,73
- 400
- 200
- 800
- 200
- 600
- 24
- 400
- 200
- 400
- 600
- 200
- 400
- 12,73
- 200
- 400
- 600
- 200
- 400
- 24
- 200
- 600
- 400
- 400
- 200
- 200
- 12,73
- 600
- 400
- 400
- 200
- 200
- 24
- 24
- 1 000
- 200
- 800
- 400
- 600
- 24
- 1 000
- 200
- 800
- 400
- 600
- 1 000
- 24
- 800
- 200
- 600
- 400
- 1 000
- 24
- 800
- 200
- 600
- 400
- 200
- 800
- 24
- 600
- 200
- 400
- 200
- 800
- 24
- 600
- 200
- 400
- 800
- 200
- 600
- 24
- 400
- 200
- 800
- 200
- 600
- 24
- 400
- 200
- 400
- 600
- 200
- 400
- 24
- 200
- 400
- 600
- 200
- 400
- 24
- 200
- 600
- 400
- 400
- 200
- 200
- 24
- 600
- 400
- 400
- 200
- 200
- 24
- Table A.2 – Calculated values of zz
- 0,0339+0,2742j
- –0,0339–0,2742j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0255j
- 0,2343j
- –0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0255j
- 0,0871j
- –0,0871j
- 0,0339+0,2421j
- –0,0339–0,2421j
- 0,0436j
- –0,0436j
- 0,0871j
- –0,0871j
- 0,2022j
- –0,2022j
- 0,0436j
- –0,0436j
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,0339+0,1731j
- –0,0339 – 0,1731j
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,1332j
- –0,1332j
- 0,2343j
- –0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0225j
- 0,209 + 0,2343j
- –0,209–0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0225j
- –0,1011j
- 0,2203j
- –0,2203j
- 0,069j
- –0,069j
- 0
- –0,1011j
- 0,209+0,2203j
- –0,209 – 0,2203j
- 0,069j
- –0,069j
- 0
- 0,0871j
- –0,0871j
- 0,2022j
- –0,2022j
- 0,0436j
- –0,0436j
- 0,0871j
- –0,0871j
- 0,209 + 0,2022j
- –0,209 – 0,2022j
- 0,0436j
- –0,0436j
- –0,0436j
- 0,069j
- –0,069j
- 0,1768j
- –0,1768j
- 0
- –0,0436j
- 0,069j
- –0,069j
- 0,209 + 0,1768j
- –0,209 – 0,1768j
- 0
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,1332j
- –0,1332j
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,209 + 0,1332j
- –0,209 – 0,1332j
- Table A.3 – Array [Z] including coefficients for currents
- 0,0339+0,2742j
- –0,0339 – 0,2742j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0255j
- 0,2343j
- –0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0255j
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0,0871j
- –0,0871j
- 0,0339 + 0,2421j
- –0,0339 – 0,2421j
- 0,0436j
- –0,0436j
- 0,0871j
- –0,0871j
- 0,2022j
- –0,2022j
- 0,0436j
- –0,0436j
- 0
- 0
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,0339 + 0,1731j
- –0,0339 – 0,1731j
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,1332j
- –0,1332j
- 0
- 0
- 0
- 0
- 1
- 1
- 0
- 0
- 0
- 0
- 0
- 0
- 0,2343j
- –0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0225j
- 0,209 + 0,2343j
- –0,209 – 0,2343j
- 0,0871j
- –0,0871j
- 0,0255j
- –0,0225j
- –0,1011j
- 0,2203j
- –0,2203j
- 0,069j
- –0,069j
- 0
- –0,1011j
- 0,209 + 0,2203j
- –0,209–0,2203j
- 0,069j
- –0,069j
- 0
- 0,0871j
- –0,0871j
- 0,2022j
- –0,2022j
- 0,0436j
- –0,0436j
- 0,0871j
- –0,0871j
- 0,209+0,2022j
- –0,209–0,2022j
- 0,0436j
- –0,0436j
- –0,0436j
- 0,069j
- –0,069j
- 0,1768j
- –0,1768j
- 0
- –0,0436j
- 0,069j
- –0,069j
- 0,209+0,1768j
- –0,209 – 0,1768j
- 0
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,1332j
- –0,1332j
- 0,0255j
- –0,0255j
- 0,0436j
- –0,0436j
- 0,209 + 0,1332J
- –0,209 – 0,1332j
- 0
- 0
- 0
- 0
- 0
- 0
- 1
- 1
- 1
- 1
- 1
- 1 [Go to Page]
- A.3 Example 2
- In this example, the same cable data and spacing has been used as in Example 1, but the phase rotation has been reversed.
- The magnitude of the phase conductor and sheath currents together with the sheath loss factor are given below, assuming a total phase current of 100 A.
- Phase current
- Sheath current
- Sheath loss factor
- Cable 1, phase R
- 50
- 34,4
- 2,916
- Cable 2, phase R
- 50
- 34,4
- 2,916
- Cable 3, phase S
- 50
- 24,5
- 1,477
- Cable 4, phase S
- 50
- 24,5
- 1,477
- Cable 5, phase T
- 50
- 29,9
- 2,213
- Cable 6, phase T
- 50
- 29,9
- 2,213 [Go to Page]
- A.4 Example 3
- In this example the same cable data has been used, but the six cables are now arranged in two trefoil groups with 200 mm between centres of the groups. The arrangement is shown below:
- R1
- R2
- S1 T1
- T2 S2
- The cable co-ordinates are as follows:
- x
- y
- 30
- 52
- Cable 1, phase R
- 230
- 52
- Cable 2, phase R
- D =
- 0
- 0
- Cable 3, phase S
- 260
- 0
- Cable 4, phase S
- 60
- 0
- Cable 5, phase T
- 200
- 0
- Cable 6, phase T
- The magnitude of the phase conductor and sheath currents, together with the sheath loss factor are given below, assuming a total phase current of 100 A.
- Phase current
- Sheath current
- Sheath loss factor
- Cable 1, phase R
- 50
- 13,9
- 0,474
- Cable 2, phase R
- 50
- 13,9
- 0,474
- Cable 3, phase S
- 50
- 13,8
- 0,468
- Cable 4, phase S
- 50
- 13,8
- 0,468
- Cable 5, phase T
- 50
- 14,1
- 0,492
- Cable 6, phase T
- 50
- 14,1
- 0,492 [Go to Page]
- A.5 Example 4
- In this example, the same cable data has been used but the cables are now such that the current sharing between phase conductors is not equal. The arrangement is shown below:
- R1 R2 S1 S2 T1 T2
- The cable co-ordinates are as follows:
- x
- y
- 0
- 0
- Cable 1, phase R
- 400
- 0
- Cable 2, phase R
- D =
- 800
- 0
- Cable 3, phase S
- 1 200
- 0
- Cable 4, phase S
- 1 600
- 0
- Cable 5, phase T
- 2 000
- 0
- Cable 6, phase T
- The magnitude of the phase conductor and sheath currents together with the sheath loss factor are given below, assuming a total phase current of 100 A.
- Phase current
- Sheath current
- Sheath loss factor
- Cable 1, phase R
- 46,31
- 38,4
- 4,236
- Cable 2, phase R
- 53,71
- 36,5
- 2,845
- Cable 3, phase S
- 44,59
- 37,4
- 4,346
- Cable 4, phase S
- 55,66
- 34,8
- 2,42
- Cable 5, phase T
- 50,76
- 43,7
- 4,576
- Cable 6, phase T
- 49,62
- 44,4
- 4,947
- Comparison with Example 1 shows that the sheath losses for this cable arrangement are very high. Because of this, arrangements where all the conductors of one phase are placed together should be avoided.
- Annex B (informative)Example of the computation of the coefficient for hollow core conductors
- Consider a hollow core conductor with internal and external diameters di = 17,5 mm and dc = 33,8 mm, respectively. The following procedure can be used to calculate
- Let a = di / dc =17,5 / 33,8 = 0,518. The hollow conductor can be replaced by an equivalent conductor with the inner radius a and the outer radius equal to 1, as shown in figure B.1.
- If a fraction of the total current enclosed within the radius r is denoted by Ir then:
- Figure B.1 – Representation of a hollow core conductor
- The magnetic flux is proportional to Ir/r and the linkage flux, F, is equal to:
- The coefficient is then given by:
- Bibliography
- IEC 60287-1-1:20XX, Electric cables – Calculation of the current rating – Part 1: Current rating equations (100 % load factor) and calculation of losses – Section 1: General
- _____________
- Word Bookmarks [Go to Page]
- HORIZONTAL_STD
- FUNCTION_EMC
- FUNCTION_ENV
- FUNCTION_QUA
- FUNCTION_SAFETY [Go to Page]
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