model calculation

Amtech ProDesign

17th Edition

Model Calculation

Contents Introduction ............................................................................................................................................................ 2

1. Load Summation .................................................................................................................................................. 3

Example 1: Load Summation ......................................................................................................................................... 3

2. Circuit protective device current rating ................................................................................................................ 6

Example 2.1: CPD rating ................................................................................................................................................ 6

Example 2.2: CPD rating: motor circuit ......................................................................................................................... 6

3. Cable sizing .......................................................................................................................................................... 7

Example 3.1: Cable sizing: Grouped cable – ‘subject to simultaneous overload’ ......................................................... 7

Example 3.2: Cable sizing: Grouped cable – ‘not subject to simultaneous overload’ ................................................... 8

Example 3.3: Cable sizing – motor circuit ...................................................................................................................... 9

Example 3.4: Cable sizing: ring final circuit to sockets ................................................................................................ 10

Example 3.5: Cable sizing: cable run through thermal insulation ............................................................................... 10

Example 3.6: Cable sizing: direct in ground ................................................................................................................. 11

4. Voltage Drop ..................................................................................................................................................... 13

Example 4.1: Voltage drop calculation ........................................................................................................................ 13

5. Phase Fault ........................................................................................................................................................ 14

Example 5.1: Maximum phase fault current ............................................................................................................... 15

Example 5.2: Minimum phase fault current ................................................................................................................ 16

Example 5.3: Line conductor adiabatic check ............................................................................................................. 17

6. Earth Fault ......................................................................................................................................................... 19

Example 6.1: earth fault disconnection by CPD .......................................................................................................... 19

Example 6.2: earth fault adiabatic ............................................................................................................................... 20

ProDesign: Model Calculation

ProDesign – Model Calculation v4 09/11 of 21

INTRODUCTION This document provides detailed examples of the principle calculation procedures in ProDesign. Each example is intended to illustrate the method used and the rules applied when performing a particular check or calculation. References made to relevant parts of ‘BS 7671: Requirements for Electrical Installations’ are highlighted in bold and italic text, thus: 433.1.1. The edition referred to is: BS 7671:2008 (2011).

The examples described refer to circuits from the ProDesign project ‘Model Calculation’ which is located in the folder ProDesign ‘projects’ folder.

The project single-line diagram is shown below. The Notes symbols denote which examples apply to which circuits.



1. LOAD SUMMATION The circuit conductors in a distribution network must be protected against the detrimental thermal effects of overload currents (433.1). Therefore, conductors of adequate current-carrying capacity and correctly co-ordinated circuit protective devices need to be selected for each circuit. To ensure this is done correctly, the load current flow in all parts of the network are determined.

Example 1: Load Summation Sub-main SM-3

Calculation for summation of the design load current (Ib) flowing through Cable SM-3.

Method:

o Summarise load currents: Table 1 summarises the load currents for all the circuits connected to DB-1, which in turn is connected to Cable SM-3.

o Loads resolved into resistive and reactive components and diversity applied: using the load currents and power factors, the load currents are resolved into their resistive and reactive components, r and x. The currents are adjusted by the application of diversity settings made in the Load and Distribution Board dialogs.

o Summing loads to give phase loads and power factors: in each phase, the resistive and reactive load components are added (Table 2). These phase load components are then used to give the load current magnitude and power factor for each phase.

o Compare to ProDesign Results: the calculated load currents and power factors are compared to those given in ProDesign for Cable SM-3.

o Summing Phase Loads to Give Neutral Load Current: the vectors of the individual phase load currents are summed to give the current flowing in the neutral conductor. To do this, the phase vectors are stated as complex numbers (Z = r + jX), which are then converted to polar form (Z∠∅). The vectors are then rotated to reflect their relative displacements (0°, 120°, 240°), converted back to complex numbers, and then added to give the r and x components of the neutral current. The r and x components are then used to give the magnitude of the neutral current.

o Compare to ProDesign Results: the calculated neutral load current is compared to that given in ProDesign for Cable SM-3.

Summarise load currents:

Table 1: Summary of phase load currents connected to SM-3 Load Phase L1 Phase L2 Phase L3 Diversity

(A) PF (A) PF (A) PF At Load At DB Load 1 70 0.9 70 0.9 70 0.9 0.8 0.9 Load 2 0 0 0 0 15 0.8 1.0 0.9 Load 3 20 1.0 0 0 0 0 1.0 0.9

Loads resolved into resistive and reactive components and diversity applied:

Using: r = z cosφ = PF; and x = r tan(cosφ-1)

Load 1: r = 70 x 0.9 = 63 A; x = 63 x 0.484 = 30.5122926 A

Diversity for load = 0.8 x 0.9, therefore: r = 63 x 0.72 = 45.36 A; x = 30.5122926 x 0.72 = 21.96885067 A

Load 2: r = 15 x 0.8 = 12 A; x = 12 x 0.75 = 9 A

Diversity for load = 1.0 x 0.9, therefore: r = 12 x 0.9 = 10.8 A; x = 9 x 0.9 = 8.1 A

Load 3: r = 20 x 1.0 = 20 A; x = 20 x 0 = 0 A

Diversity for load = 1.0 x 0.9, therefore: r = 20 x 0.9 = 18 A; x = 0 x 0.9 = 0 A



Summing loads to give phase loads and power factors:

Total load current/power factor:

using: I = √(r2+x2); and PF = r/z

Phase L1: √(63.362 + 21.968850672) = 67.06056963 A; and PF = 63.36/67.06056963 = 0.944817503

Phase L2: √(45.362 + 21.968850672) = 50.4 A; and PF = 45.36/50.4 = 0.90

Phase L3: √(56.162 + 30.06885672) = 63.70307483 A; and PF = 56.16/63.70307483 = 0.881590098

Compare to ProDesign Results:

Rounded to 1 decimal place, gives the same values as the report:

Phase L1: 67.1 A, 0.94 PF;

Phase L2: 50.4 A; 0.90 PF;

Phase L3: 63.7 A; 0.88 PF

Summing Phase Loads to Give Neutral Load Current:

To find the neutral current, the three phase load current vectors are summed.

From Table 2 above, the phase load current components are:

L1: r = 63.36, x = 21.96885067; L2: r = 45.36, x = 21.96885067; L3: r = 56.16, x = 30.0688567.

Expressed as complex numbers, the phase load current vectors are:

L1= 63.36 + j21.96885067; L2= 45.36 + j21.96885067; L3 = 56.16 +j30.0688567.

Before summing the phase currents, the phase load vectors have to be rotated by their phase angle displacements:

L1 + 0 rads (or 0°); L2 + 2π/3 rads (or 120°); L3 + 4π/3 rads (or 240°).

To do this the complex numbers representing the vectors are converted to polar form, as follows:

Complex form: Z = r +jx; Polar form: Z∠∅, where: ∅ = tan-1(x/r)

In polar form: L2 = Z∠∅ = 50.4∠tan-1(21.96885067/45.36)= 50.4∠0.451026811

L3 =Z∠∅ = 63.70307483∠tan-1(30.0688567/56.16) = 63.70307483∠0.491575912

Then the phase angle displacements are added:

L2 = 50.4∠0.451026811 + 0∠2.094395102 = 50.4∠2.545421913

L3 = 63.70307483∠0.491575912 + 0∠4.188790205 = 63.70307483∠4.680366117

Table 2: Summation of resistive and reactive load currents Load Phase L1 Phase L2 Phase L3

r x r x r x

Load 1 45.36 21.96885067 45.36 21.96885067 45.36 21.96885067 Load 2 0 0 0 0 10.8 8.1 Load 3 18 0 0 0 0 0

Total 63.36 21.96885067 45.36 21.96885067 56.16 30.0688567



Convert back to complex numbers and add the vectors:

Using r = Z cos∅; x= Z sin∅

L1 = 63.36 + j21.96885067

L2 = –41.70558274 + j28.29848703

L3 = –2.039606231 – j63.67041502

L1 + L2 + L3 = N = 19.61481103 – j13.40307732

Find the neutral current magnitude from the resistive and reactive components:

Neutral current = √(r2 + x2) = √(19.614811032 – 13.403077322) = 23.75675258 A

Compare to ProDesign Results:

Rounded to 1 decimal place, gives the same value as the report:

Neutral current = 23.8 A



2. CIRCUIT PROTECTIVE DEVICE CURRENT RATING It is necessary to ensure that for each circuit the circuit protective device (CPD) is correctly co-ordinated with the current-carrying capacity of the conductors, so that the circuit is correctly protected against the detrimental thermal effects of overload currents.

The nominal rating (In) of the CPD has to be equal to or greater than the circuit design current (Ib), i.e., (In≥Ib) (433.1.1). Also, the tripping characteristics of the CPD have to comply with 433.1.1 or 433.1.2.

For CPDs with an adjustable overload setting (Ir), it is permissible for the CPD to be adjusted so as to equal or exceed the design current, i.e., Ir≥Ib.

For a final circuit with a motor load, there is a further constraint: the CPD and its rating should be selected such that inadvertent operation of the protection should not occur during motor starting. To this end, ProDesign checks the suitability of any selected CPD for the motor starting duty defined in the circuit, assuming normal starting duty conditions; Example 2.2 below illustrates this.

Note that the motor overload in the motor starter provides overload protection for the circuit conductors and the CPD provides fault protection only. Example 3.3 describes the cable sizing calculations which take into account both the setting of the overload and the CPD energy let-through.

NOTE: for more details on the determination of the design current, see ‘Section 1: Load summation’.

Example 2.1: CPD rating Sub-main SM-4

The selected CPD is: Merlin Gerin, Compact MCCB, NS100H, TM-D, Rating 80 A. The overload is adjusted to 0.8, which gives an overload current setting (Ir) = 0.8 x 80 = 64 A.

The diversified load currents in each phase have been calculated to be: L1 = 39.67 A; L2 = 59.19 A; L3 = 19.78 A.

The design current for the circuit is set to the highest of the phase currents. Therefore the design current Ib = 59.19 A.

Testing the CPD rating using Ir≥Ib; 64≥59.19, therefore the CPD rating is OK.

Failure to comply with this rule would result in an error message.

Example 2.2: CPD rating: motor circuit Final circuit FC-6

The selected CPD type is: Generic, BS 88 Fuse, HRC, gG(General), Rating set to ‘Auto’.

The motor load current = 19.78 A

The recommended ratings for fuses of this type for a Star Delta started motor are:

Fuse rating = 25 A; Maximum Star Delta Motor Rating = 18 A;

Fuse rating = 32 A; Maximum Star Delta Motor Rating = 22 A;

Therefore the fuse rating selected by ProDesign is 32 A (22 ≥ 19.78).



3. CABLE SIZING For each circuit, once the design current Ib has been determined, and then the CPD rating established, cable sizing can proceed.

In general, the suitable cable size is determined by finding, from the tables (Tables 4D1 to Table 4J4), the cable size of the type selected, that has a tabulated rating (It) for the specified installation method, that equals or exceeds the minimum cable capacity (Iz).

The minimum cable capacity Iz is found by dividing the CPD rating ‘In’ by any applicable derating factors: Ca, Cg, etc. [Appendix 4, 5.1.1, equation (1)].

Therefore the general rules are:

Iz = In / (Ca x Cg x etc.) or, for adjustable CPDs: Iz = Ir / (Ca x Cg x etc.) and, It≥Iz

Example 3.1: Cable sizing: Grouped cable – ‘subject to simultaneous overload’ [Appendix 4, 5.1.2, equation (2)]

Sub-main SM-4

The minimum required rating for grouped cables ‘subject to simultaneous overload’ needs to satisfy: It ≥ In/Cg

Therefore, the minimum cable capacity Iz, when grouped, is determined from: Iz = In/Cg (or Iz = Ir/Cg where the CPD is adjustable).

Cable details and installed conditions:

Cable type: Multicore, 90°C thermosetting insulated/sheathed, non-arm Cu Table 4E2; Size = set to ‘Auto’;

Installation Method: Method 8 – In trunking (Reference Method B); Length = 30 m; Ambient Temperature = 35°.

Grouping: Circuit in Group = 6; Grouping – Subject to Simultaneous Overload.

CPD details:

The selected CPD is: Merlin Gerin, Compact MCCB, NS100H, TM-D, Rating 80 A. The overload is adjusted to 0.8, which gives an overload current setting (Ir) = 0.8 x 80 = 64 A.

Determination of minimum cable capacity Iz:

Derating Factors:

Ambient Temperature = 35°, from Table 4B1, for Thermosetting 90° cable: Ca = 0.96

Grouping: from Table 4C1, for Enclosed (Reference Method B), 6 in group: Cg = 0.57

Calculation:

Iz = Ir/(Ca x Cg) = 64/(0.96 x 0.57) = 116.96 A

Determination of cable size:

From Table 4E2A, column 5 (Reference Method B, 1 three or four-core cable, three-phase a.c):

25 mm2 cable, It = 105 A; 35 mm2 cable, It = 128 A

Therefore, ProDesign sizes the cable at 35 mm2, i.e., 128 ≥ 116.96 (It≥Iz)



Example 3.2: Cable sizing: Grouped cable – ‘not subject to simultaneous overload’ [Appendix 4, 5.1.2, equations (3), (4)]

Sub-main SM-3

The minimum required rating for grouped cables ‘not subject to simultaneous overload’ is the larger of the two values resulting from: It ≥ Ib / Cg, or: It ≥ √In2 + 0.48Ib2 [(1 – Cg2) / Cg2]

Therefore, the minimum cable capacity Iz is determined from the larger result of the following two calculations:

Iz = Ib/(Cg x Ca x Ci x Cc), or: Iz = 1/(Ca x Ci) x √In2 + 0.48Ib2 [(1 – Cg2) /Cg2]


Cable type: Multicore, 70°C thermoplastic insulated/sheathed, non-arm Cu Table 4D2; Size = set to ‘Auto’;

Installation Method: Method 34 – On ladder supports (Reference Method E); Length = 10 m. Ambient Temperature = 30°.

Grouping: Circuit in Group = 4; Grouping – Not Subject to Simultaneous Overload.

CPD details:

The selected CPD is: Generic, BS 88 Fuse, HRC, gG(General), Rating set to 100A.


Derating Factors:

Ambient Temperature = 30°: Ca = 1 Protection is not BS3036, cable is not under ground; Ci = 1 Grouping (not subject to simultaneous overload): from Table 4C1, for ladder (Reference Method E), 4 in group: Cg = 0.80

Calculation:

Iz = Ib/Cg = 67.06/0.8 = 83.83 A; or Iz = 1/(1 x 1) √1002 + 0.48 x 67.062 [(1 – 0.82) / 0.82] = 105.9 A Therefore, the minimum cable capacity Iz = 105.9 A


From Table 4D2A, column 9 (Reference Method E, 1 three- or four-core cable, three-phase ac): 25 mm2 cable, It = 101 A; 35 mm2 cable, It = 126 A




Example 3.3: Cable sizing – motor circuit Final circuit FC-6

In ProDesign a motor circuit will normally include a motor starter containing a motor overload. The motor overload can be set to any value equal to or greater than the motor load current (Ib) to reflect its actual setting on site. The cable sizing is then based on the motor overload setting (Irm), with the minimum cable capacity (Iz) being equal to the motor overload setting divided by any rating factors:

Irm ≥ Ib and Iz = Irm/rating factors

Load type:

Motor, Star Delta started, three phase, motor starter at board.


Cable type Multicore, 90°C thermosetting insulated, armoured Cu Table 4E4;

Two three-core cables (2 x 1 x 3c) run together (the motor is star/delta started and the starter is located at the board, therefore six conductors are required to connect the ends of the windings to the starter);

Size = set to ‘Auto’;

Installation Method: Method 31 – On perforated tray (Reference Method E); Length = 10 m; Ambient Temperature = 40°. Grouping = 2; Arrangement = Horizontal Touching.

CPD details:

The selected CPD is: Generic, BS 88 Fuse, HRC, gG(General), Rating set to ‘Auto’. ProDesign has set the CPD rating In = 32 A (see Example 2.2 above).

The motor overload setting = 20 A.


Rating Factors:

Ambient Temperature = 40°, from Table 4B1, for Thermosetting 90° cable: Ca = 0.91

Grouping = 2 in group; from Table 4C4, Cg = 0.88

Calculation:

Minimum cable capacity Iz = Ib/√3(Ca x Cg)

[each conductor is carrying the delta current rather than the phase current – the required rating is divided by √3].

= 19.78/√3(0.91 x 0.88) = 14.3 A


From Table 4E4A, column 5 (Reference Method E, 1 three- or four-core cable, three-phase ac): 1.5 mm2 cable, It = 25 A

Therefore, ProDesign sizes the cable at 1.5 mm2, i.e., 25 ≥ 14.3 (It≥Iz)

Line conductor adiabatic check

A check is required to ensure that, under fault conditions, the line conductors are not subjected to excessive levels of thermal energy that could lead to damage or premature deterioration. The check analyses the fault energy resulting from both the minimum phase fault current and the earth fault current.

From 434.5.2, the general formula used for the adiabatic calculation is:

t = (k2S2)/I2

where: t = the maximum permissible disconnection time (s); k = a factor for the type of conductor. The value of k relates to the conductor’s thermal characteristics; S = the section of the conductor (mm2);



I = the minimum fault current (A).

For disconnection times of less than 0.1 s, the formula is:

I2t < k2S2

where: I2t is the let-through energy (A2s) for the circuit protective device.

In the case of the circuit under consideration, the level of both the phase and earth fault currents are sufficiently high to cause the fuse to rupture in less than 0.1 s. Therefore, the level of energy let-through would be the same for either fault current condition. In such cases, the results are given for the phase fault condition.

The protective device in the circuit FC-6 is a BS 88 Fuse, rated 32 A. From its characteristic curve it can be seen that, at the calculated minimum phase fault current (855 A), the disconnection time would be less than 0.1 s.

From Table 43.1, for a copper conductor with 90 °C thermosetting insulation, k = 143;

The cable section S = 1.5 mm2.

k2S2 = 1432 x 1.52 = 46.01 x 103

The let-through energy for the fuse is 3.4 x 103 A2s.

Therefore the cable complies with the adiabatic requirement: 3.4 x 103 < 46.01 x 106

Example 3.4: Cable sizing: ring final circuit to sockets Final circuit FC-3

For the standard ring final circuit as defined in 433.1.5, protected by a 30 A or 32 A CPD, the minimum cable capacity Iz can be as little as 20 A (assuming the load on any part of the ring is unlikely to exceed Iz for long periods), with a minimum size limit of 2.5 mm2.


Cable type: Multi Core Flat twin and earth PVC Cu Table 4D5; Size = set to ‘Auto’;

Installation Method: Method 20 – Clipped direct, Reference Method C; Length = 10 m; Ambient Temperature = 30°.

Grouping: None.

CPD details:

The selected CPD is: Generic, BS 88 Fuse, HRC, gG(General), Rating set to ‘Auto’. The design current Ib = 20 A.

ProDesign has set the CPD rating In = 32 A as a consequence of the load description: ‘Sockets BS 1363, 32 A ring final circuit to sockets up to 100 m2.


Derating Factors: None

Calculation: Minimum cable capacity Iz = 20 A (433.1.5)


From Table 4D5A, column 6 (Reference Method C): 1.5 mm2 cable, It = 20 A; 2.5 mm2 cable, It = 27 A 1.5 mm2 cable complies with It≥Iz (20≥20) but 433.1.5 requires that the minimum cable section for this circuit type should be 2.5 mm2.

Therefore, ProDesign sizes the cable at 2.5 mm2.

Example 3.5: Cable sizing: cable run through thermal insulation Final circuit FC-2

Where a cable is run through thermal insulation (as defined by 523.7), derating is applied using the derating factor Ci, according to Table 52.2.




Cable type: Single-core, 70°C thermoplastic, non-armoured Cu Table 4D1; Size = set to ‘Auto’;

Installation Method: Run through thermal insulation; Length = 10 m; Distance run through insulation = 150 mm. Ambient Temperature = 30°.

Grouping: None.

CPD details:

The selected CPD is: Generic, BS 88 Fuse, HRC, gG(General), rating set to ‘Auto’. The design current Ib = 15 A, therefore ProDesign has set the CPD Rating In = 16 A.


Derating Factors:

Ambient Temperature = 30°: Ca = 1.

Grouping: not grouped, Cg = 1.

Run through thermal insulation. Distance = 150 mm. Derating factor Ci from Table 52.2: 0.78 for Distance = 100; 0.63 for Distance = 200

By linear interpolation: Ci = [(150 – 100)/(200-100)] x (0.63 – 0.78) + 0.78 = 0.705

Calculation:

Minimum cable capacity Iz = In/(Ca x Cg x Ci) = 16/(1 x 1 x 0.705) = 22.70A


From Table 4D1, column 6 (Reference Method C):

1.5 mm2 cable, It = 20 A; 2.5 mm2 cable, It = 27 A

Therefore, ProDesign sizes the cable at 2.5 mm2. i.e., 27 ≥ 22.70 (It≥Iz)

Example 3.6: Cable sizing: direct in ground Final circuit FC-7

Current ratings for cables installed direct in ground (Table 4A2: Installation methods 72 and 73) are subject to correction factors given in Tables 4B2 (ambient ground temperature), 4B3 (soil thermal resistivity) and 4C2 (spacing).


Cable type: Multicore, 90°C thermosetting insulated, armoured Cu Table 4E4; Size = set to ‘Auto’;

Installation Method: Method 73 – Direct in ground; Length = 10 m; Ambient ground temperature = 15°.

Soil thermal resistivity = 1.0 K.m/W, Grouping: 2 in Group; subject to simultaneous overload, Spacing = 0.125m

CPD details:

The selected CPD is: Generic, BS 88 Fuse, HRC, gG(General), rating set to ‘Auto’. The design current Ib = 310 A, therefore ProDesign has set the CPD Rating In = 315 A.


Derating Factors:

Ground ambient temperature = 15°, Ctg = 1.04 (taken from Table 4B2) Soil thermal resistivity = 1.0 K.m/W, Crg = 1.5 (taken from Table 4B3) Grouping: 2 in group. Spacing = 0.125m, Cgg = 0.85 (taken from Table 4C2) Direct in ground method. Cc = 0.9 (taken from Appendix 4 section 5.1)




The cable ratings are found in Table 4E4, column 7 (Reference Method D).

Calculation:

Iz = Ir/(Ctg x Crg x Cgg x Cc) = 315/(1.04 x 1.5 x 0.85 x 0.9) = 263.95A


From Table 4E4A, column 7 (Reference Method D, 1 three or four-core cable, three-phase a.c):

150 mm2 cable, It = 251 A; 185 mm2 cable, It = 281 A




4. VOLTAGE DROP For each circuit, a per unit voltage drop (mV/A/m or mΩ/v) is found for the cable (from Tables 4D1B, etc.) or busbar (from manufacturer’s data). This data is used to calculate the total circuit voltage drop for the conductor length and load current, corrected for load power factor and operating temperature (Appendix 4, Section 6). See Example 4.1 below.

BS7671:2008 Appendix 12 suggests voltage drop limits for various combinations of source and load type. ProDesign will apply the voltage drop limits found in BS7671:2008 Appendix 12 or user entered limits when defined. Each final load and motor dialog contains a tab showing its voltage drop limit back to source.

Each cable in ProDesign is assigned a voltage drop limit. The voltage drop limit for fixed size cables is taken as being the calculated voltage drop in that cable. Each cable which is set to ‘Auto’ is allocated a voltage drop limit which represents a portion of the overall limit for the leg(s) of the distribution which it forms part.

If the calculated voltage drop in any circuit exceeds the allocated limit, the cable size will be increased to meet the voltage drop requirement. If it is not possible to meet the allocated voltage drop limit by increasing the cable sizes (e.g., where the cable size is fixed), an error message will be given for each circuit where the overall voltage drop limit is exceeded.

Example 4.1: Voltage drop calculation Final circuit FC-1


Cable type: Single-core, 70°C thermoplastic, non-armoured Cu Table 4D1; Size = set to 35 mm2;

Installation Method: Method 20 – Clipped direct, trefoil; Length = 90 m.

Load Conditions:

Design current Ib = 70 A; power factor PF= 0.9; three phase and neutral.

Voltage drop data:

From Table 4D1B, column 7: Size = 35 mm2; r = 1.100; x = 0.170.

Calculating voltage drop:

The load current Ib = 70 A, which when resolved into resistive and reactive components is:

Ir = 63 A; Ix = 30.5123 A [Ir = Ib x PF; Ix = Ib x sin(cos-1(PF))]

The circuit impedance and voltage drop calculation:

r x Z Data from table [mΩ/m] 1.100 0.170 Per phase conductor [÷√3] 0.6351 0.0982 Correction for power factor [cosφ r + sinφ x] 0.5716 0.0428 r corrected for temperature, using Ct = 0.9059*. [Z = r + x] 0.5178 0.0428 0.5606 Total Voltage drop (V) [r = (Ir x Z x length)/1000; x = (Ix x Z x length)/1000; Z = √(r2 + x2)] 3.1785 1.5394 3.5317

* Ct calculated using formula (10)* from Appendix 4, section 6.1: Ct = 230 + tp – [Ca2Cg2 – (Ib2/It2)](tp – 30) / (230 + tp) tp = max. permitted normal operating temperature.

The total voltage drop per phase Vd calculated by ProDesign = 3.5317 V



5. PHASE FAULT Fault currents are calculated at every point in the network. At each point the calculations are performed using the voltage and the impedance between that point and the voltage source, i.e., the active source or sources.

The impedances are calculated and brought together using a Zbus impedance matrix. A separate matrix is constructed for each of the fault types:

• Phase fault: 3-phase circuits, maximum and minimum; 3-phase and neutral, maximum and minimum.

• Earth fault.

A portion of the matrix for 3-phase maximum fault impedances for the Model Calculation project is shown below. The impedances at each point in the network are located on the principal diagonal (the yellow shaded cells). For example, the fault impedance at the load end of Cbl_SM-1 is: resistance = 4.51316663; reactance = 14.44968318.

Per unit values to a 100 MVA base are used in the Zbus matrix calculations, the impedances shown in the matrix are the per unit values used.

The following worked examples for maximum and minimum fault currents, and earth fault current in the next section, use conventional calculation methods to demonstrate the accuracy of the ProDesign calculations, while avoiding the complexities of the construction of the Zbus matrices. Also, each worked example is repeated showing the calculations that are performed by ProDesign using the impedances obtained from the Zbus matrix.

In some versions of ProDesign a Zbus matrix viewer is provided. This is displayed by clicking View Impedance Matrices in the Calculation menu.

Maximum prospective phase fault current

Calculated using the total fault impedance at the source end of the circuit under consideration. The fault impedance includes the source impedance and the conductor impedance of the circuit between the source and the origin of the circuit under consideration. Conductor resistances are calculated at 20 °C. For three-phase and three-phase and neutral circuits, the maximum fault current condition is the three-phase symmetrical fault current (for single-phase circuits, the phase to neutral fault current).

The maximum prospective fault current in a circuit is used to check that the breaking capacity of the circuit protective device (CPD) is adequate (434.5.1).

Example 5.1 below describes the calculation for a three-phase and neutral circuit.

Minimum phase fault current

Calculated using the total fault impedance at the load end of the circuit under consideration. The fault impedance includes the source impedance, the phase conductor impedance of the circuit between the source and the origin of the circuit under consideration, and the impedance of the conductor in the circuit under consideration. Conductor resistances are calculated using a temperature multiplier [Guidance Note 1: Table E2].

The formula used to calculate the minimum fault current (Ifmin) depends on the circuit configuration:

Three-phase only: Ifmin = Iph-ph = UL / 2(Ze + Z1) [Ze = external impedance; Z1 = phase conductor impedance]

Three-phase and neutral: Ifmin = Iph-n = Uph / (Ze + Z1 + Zn) [Zn = neutral conductor impedance]

Single phase: Ifmin = Iph-n = Uph / (Ze + Z1 + Zn)

The minimum prospective fault current in a circuit is used in the adiabatic test to ensure that the maximum phase conductor temperature is not exceeded during a phase fault (434.5.2).



Example 5.2 below describes the minimum phase fault current calculation for a three-phase and neutral circuit.

Example 5.3 below describes the phase fault adiabatic calculation.

Example 5.1: Maximum phase fault current

Final circuit FC-7: conventional calculation

Fault current is calculated by dividing the system voltage by the total fault path impedance. The maximum fault current is calculated at the source end of the circuit, therefore for FC-7 the total fault impedance includes the external source impedance (Ze) and the impedance of the conductors in SM-1.

The maximum phase fault current for three-phase and three-phase and neutral circuits is the three-phase symmetrical fault current Isym = UL / Z

Impedances in circuit for fault at cable FC-7, supply end:

r x r x 250 MVA Transformer primary supply network 0.000064 0.000637 Transformer winding 0.004000 0.019596 Cable SM-1: thermosetting non-armoured, multicore, 240 mm2 Impedance from Table 4E2B, col. 4 [mΩ/m @ 90°] 0.1750 0.1250 Corrected for temperature [r @ 20° = r90 / (1 + (90 – 20) x 0.004)] 0.1367 0.1250 Value for one line conductor (÷√3) 0.0789 0.0722 Calculated for 40 m circuit length [Ω] 0.003157 0.002887 Total 0.00717 0.02312 |Z| = √(R2 + X2)] = √(0.007172 + 0.023122)] = 0.02421 Ω

Calculating fault current:

Ifmax = UL / Z = (400÷√3) / 0.02421 = 9539 A = 9.539 kA

Max. phase fault current calculated by ProDesign = 9.535 kA

Final circuit FC-7: using Zbus impedances

Impedances in circuit for fault at cable FC-7, supply end (per unit values at 100 MVA base):

r x r x 250 MVA Transformer primary supply network 0.0398 0.3980 Transformer winding 2.5000 12.247 Transformer primary supply network + winding 2.5398 12.645 Cable SM-1 1.9734 1.8042 Total 4.5132 14.449 |Z| = √(R2 + X2)] = √(4.51322 + 14.4492)] = 15.1375 pu


Ifmax = MVA Base / (Zpu x √3 x VLL) = 100 / (15.1375 x √3 x 0.4) = 9.535 kA



Example 5.2: Minimum phase fault current

Final circuit FC-7: conventional calculation

Fault current is calculated by dividing the system voltage by the total fault path impedance. The minimum fault current is calculated at the load end of the circuit, therefore for FC-7 the total fault impedance includes the external source impedance (Ze) and the impedance of the conductors in SM-1 and FC-7.

The minimum phase fault current for three-phase and neutral circuits is the single phase and neutral fault current Iph-n: Ifmin = UL/(Ze + Z1 + Zn)

Adding Ze and Z1 and Zn:

r x r x Ze: External impedance from the Transformer dialog 0.0041 0.0202

Z1a: SM-1, 3ph cable impedance from Table 4E2B, col. 4 [r: mΩ/m @ 90°] 0.1750 0.1250 Single phase conductor [√3] 0.1010 0.0722 Corrected for temperature [r @ 20° = r90 / (1 + (90 – 20) x 0.004)] 0.0789 0.0722 Corrected for fault temperature [Using Multiplier 54C = 1.28 in Table E2] 0.1010 0.0722 Calculated for 40 m circuit length [Ω] 0.0040 0.0029

Zna: Neutral conductor – same as phase conductor 0.0040 0.0029

Z1b: FC-7, 3ph cable impedance from Table 4E4B, col. 4 [r: mΩ/m @ 90°] 0.2200 0.1250 Single phase conductor [√3] 0.1270 0.0722 Corrected for temperature [r @ 20° = r90 / (1 + (90 – 20) x 0.004)] 0.0992 0.0722 Corrected for fault temperature [Using Multiplier 54C = 1.28 in Table E2] 0.1270 0.0722 Calculated for 10 m circuit length [Ω] 0.0013 0.0007

Znb: Neutral conductor – same as phase conductor 0.0013 0.0007

Total 0.0274 0.0311 |Z| = √(R2 + X2)] = √(0.02742 + 0.03112)] = 0.0311 Ω


Ifmin = UL / (Ze + Z1 + Zn) = (400÷√3)/0.0311 = 7.390 A = 7.425 kA

Min. phase fault current calculated by ProDesign = 7.418 kA

Final circuit FC-7: using Zbus impedances

Impedances in circuit for fault at cable FC-7, load end (per unit values at 100 MVA base):

r x r x 250 MVA Transformer primary supply network 0.0398 0.3980 Transformer winding 2.5000 12.247 Transformer primary supply network + winding 2.5398 12.645 Cable SM-1 5.0518 3.6084 Cable FC-7 1.5877 0.9021

Total 9.1793 17.156 |Z| = √(R2 + X2)] = √(9.17932 + 17.1562)] = 19.457 pu


Ifmax = MVA Base / (Zpu x √3 x VLL) = 100 / (19.457 x √3 x 0.4) = 7.418 kA



Example 5.3: Line conductor adiabatic check

Final circuit FC-7


From 434.5.2, the general formula used for the phase fault adiabatic calculation is:

t = (k2S2)/I2

where: t = the maximum permissible disconnection time (s); k = a factor for the type of conductor. The value of k relates to the conductor’s thermal characteristics; S = the section of the conductor (mm2); I = the minimum phase fault current (A).

Phase Fault Conditions The protective device in the circuit FC-7 is a BS 88 Fuse, rated 315 A. From its characteristic curve it can be seen that,


I2t < k2S2



The cable section S = 185 mm2.

k2S2 = 1432 x 1852 = 699.87 x 106

The let-through energy for the fuse is 800 x 103 A2s.

Therefore the cable complies with the adiabatic requirement: 800 x 103 < 699.87 x 106

Earth Fault Conditions At the calculated earth fault current (4.4 kA), the disconnection time would be 0.23 s.

For disconnection times greater than 0.1 s, the formula is:

t = (k2S2)/I2

where: t is the maximum permissible disconnection time.

t = (1432 x 1852) / 44002 = 36.15 s

The actual disconnection time = 0.23 s

Therefore the cable complies with the adiabatic requirement: 0.23 < 36.15 s

Line conductor adiabatic check


From 434.5.2, the general formula used for the adiabatic calculation is:

t = (k2S2)/I2

where: t = the maximum permissible disconnection time (s); k = a factor for the type of conductor. The value of k relates to the conductor’s thermal characteristics; S = the section of the conductor (mm2); I = the minimum fault current (A).


I2t < k2S2




In the case of the circuit under consideration, the level of both the phase and earth fault currents are sufficiently high to cause the fuse to rupture in less than 0.1 s. Therefore, the level of energy let-through would be the same for either fault current condition. In such cases, the results are given for the phase fault condition.

The protective device in the circuit FC-6 is a BS 88 Fuse, rated 32 A. From its characteristic curve it can be seen that, at the calculated minimum phase fault current (855 A), the disconnection time would be less than 0.1 s.


The cable section S = 1.5 mm2.

k2S2 = 1432 x 1.52 = 46.01 x 103

The let-through energy for the fuse is 3.4 x 103 A2s.

Therefore the cable complies with the adiabatic requirement: 3.4 x 103 < 46.01 x 106



6. EARTH FAULT Earth fault calculations are performed for each circuit in a network, to check compliance with 411.3: ‘Requirements for fault protection’, as follows:

the earth fault loop impedance (Zs) is calculated;

Using the earth fault loop impedance, the earth fault current (Ief) is calculated;

Using the earth fault current, the earth fault disconnection time is determined from the circuit protective device (CPD) characteristic (or the residual current device (RCD) characteristic where an RCD is used);

A check is made to determine whether the disconnection time is within the maximum for the circuit (411.3.2, Table 41.1);

Where an RCD is used on a TT system, a check is made to ensure the Touch Voltage does not exceed the limit for the circuit (411.5.3);

An adiabatic calculation is performed to ensure that the maximum permitted CPC temperature is not exceeded during an earth fault (543.1.3).

Example 6.1 below describes the calculation for earth fault disconnection by means of the CPD.

Example 6.2 below describes the adiabatic calculation.

Example 6.3 below describes the calculation for earth fault disconnection by means of an RCD.

Example 6.1: earth fault disconnection by CPD

Sub-main SM-1: conventional calculation

Phase conductor type: Multicore, 90°C thermosetting insulated/sheathed, non-arm Cu Table 4E2; Size = 240 mm2; Length = 40 m.

CPC type: XLPE Non-Armoured Cu; Size = 50 mm2.

Circuit Protective Device (CPD): BS 88 fuse; Rating: 500 A.

Calculate the earth fault loop impedance Zs:

Zs = Ze + Z1 + Z2

where: Ze = external earth fault impedance; Z1 = phase conductor impedance; Z2 = CPC impedance

r x r x

Ze: External earth fault impedance from the Transformer dialog 0.0041 0.0202

Z1: 3ph cable impedance from Table 4E2B, col. 4 [r: mΩ/m @ 90°] 0.175 0.125 Single phase conductor [√3] 0.1010 0.0722 Corrected for temperature [r @ 20° = r90 / (1 + (90 – 20) x 0.004)] 0.0789 0.0722 Corrected for fault temperature [Multiplier 54C = 1.28 in Table E2] 0.1010 0.0722 Calculated for 40 m circuit length [Ω] 0.0040 0.0029

Z2: cable impedance from Table 4E1B, col. 5 [r: mΩ/m @ 90°] 0.99 0.27 Single conductor [÷2] 0.495 0.135 Corrected for temperature [r @ 20° = r90 /(1+(90–20) x 0.004)] 0.3867 0.135 Corrected for fault temperature [Multiplier 54C = 1.28 in Table E2] 0.495 0.135 Calculated for 40 m circuit length [Ω] 0.0198 0.0054

Total 0.0279 0.0285 |Z| = √(R2 + X2)] = √(0.02792 + 0.02852)] = 0.03988 Ω



Calculate earth fault current Ief:

Ief = U0/Zs = 230/0.03988 = 5,767 A = 5.767 kA

Earth fault current calculated by ProDesign = 5.78 kA

Determine Disconnection Time:

From the protective device characteristic curve, the maximum disconnection time for a fault current of 5.76 kA is 0.79 s.

The maximum disconnection time for the circuit from 411.3.2.3 is 5 s.

Therefore, the circuit complies with the earth fault disconnection requirement: 0.79 < 5 s

Sub-main SM-1: using Zbus impedances

Impedances in circuit for earth fault at cable SM-1 (per unit values at 100 MVA base):

r x r x Line Impedances (ZL):

250 MVA Transformer primary supply network 0.0398 0.3980 Transformer winding 2.5000 12.247 Cable SM-1: Phase Conductor 2.5259 1.8042

Total ZL 5.0657 14.449 3 ZL 15.1971 43.3491

Earth Impedances (ZG): 250 MVA Transformer primary supply network 0 0 Transformer winding 0 0 Cable SM-1: CPC 37.125 10.125

Total ZG 37.125 10.125

3ZL + ZG 52.3221 53.471

|3ZL + ZG| = √(R2 + X2)] = √(52.32212 + 53.4712)] = 74.814 pu

Calculating earth fault current:

Ifmax = [MVA Base / (3ZL + ZG)] x [1 / (VLL /√3)] = (100 / 74.814) x [1 / (0.4 /√3) ] = 5.78 kA

Example 6.2: earth fault adiabatic Sub-main SM-1

Phase conductor type: Multicore, 90°C thermosetting insulated/sheathed, non-arm Cu Table 4E2; Size = 240 mm2; Length = 40 m.

CPC type: XLPE Non-Armoured Cu; Size = 50 mm2.

Circuit Protective Device (CPD): BS 88 fuse; Rating: 500 A.

From 543.1.3, the general formula used for the phase fault adiabatic calculation is:

S = √(I2t)/k

where: S = the minimum section of the CPC (mm2); k = a factor for the type of conductor (from Tables 54.2 to 54.6).

The value of k relates to the conductor’s thermal characteristics.



For disconnection times of less than 0.1 s: I2t = the let-through energy (A2s) for the circuit protective device.

For disconnection times greater than 0.1 s: I = earth fault current; t = the calculated earth fault disconnection time (s).

As shown in Example 6.1 above: the calculated earth fault current = 5,767 A, the calculated disconnection time = 0.79 s, which is greater than 0.1 s, therefore:

S = √(I2t)/k = √(earth fault current squared x disconnection time)/k = √(57672 x 0.79)/143 = 35.85 mm2

Therefore, the CPC complies with adiabatic requirement: 50 > 35.85 mm2

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