Determine Cable Size

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How to determine the suitable size of cable

for Electrical Wiring Installation with

Solved Examples (in both British and SiSystem

Posted by: Electrical Technology 10/18/2013 in Basic/Important Electrical Formlas!

Electrical "iring #Basic $or %ome&! %o' To! (estions/)ns'ers #Electrical& *+ ,omments

%o' to determine the sitable si-e o$ cable $or Electrical "iring Installation

Voltage drop in Cables

We know that all conductors and cables (except Super conductor) havesome amount of resistance. This resistance is directly proportional to the length and inverselyproportional to the diameter of conductor R !"a #!aws of resistance R $% (!"a)&Whenever current 'ows through a conductor a voltage drop occurs in thatconductor. enerally voltage drop may neglect for low length conductorsbut in a lower diameter and long length conductors we cannot neglectthat voltage drops.*ccording to +,,, rule -/0 at any point between power supply terminal

and installation 1oltage drop should not increase above /.23 of provided(supply) voltage.,xample4 if the Supply voltage is //51 then the value of allowablevoltage drop should be6*llowable 1oltage 7rop $ //5 x (/.2"855) $ 2.21+n electrical wiring circuits voltage drops also occur from the distributionboard to the di9erent sub circuit and :nal sub circuits but for sub circuitsand :nal sub circuits the value of voltage drop should be half of thatallowable voltage drops (i.e. /.;21 of 2.21 in the above case)<ormally 1oltage drop in tables is described in *mpere per meter (*"m)e.g. what would be the voltage drop in a one meter cable which carrying

one *mpere current= There are two methods to de:ne the voltage drop in a cable which we willfollow.+n S+ (System international and metric system) voltage drop is describedby ampere per meter.+n >?S (foot pound system) voltage drop is described in 855feet.

• Update4 <ow you can also use the following @alculators to :nd1oltage drop A the wire siBe in *merican wire gauge.

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Tables

-elow are the important Tables which you should follow for determiningthe proper siBe of cable for ,lectrical Wiring +nstallation.

Click image to enlarge





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http://electricaltechnology.org/wp-content/uploads/2013/10/Table-25282-2529-current-rating-of-flexible-cords-Copper-cables-at-86-25C2-25B0F-or-30-25C2-25B0C.jpg

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To fnd voltage drop in a cable, ollow these simple steps• >irst of all :nd the maximum allowable voltage drop• <ow >ind load current

•

<ow according to load current select a proper cable (which currentrating should be nearest to the calculated load current) from table 8

• >rom Table 8 :nd the voltage drop in meter or 855feet (whatsystem you prefer) according its rated current

(Stay cool we will follow both methods and system for :nding voltagedrops (in meter and 855feet) in our solved example for whole electricalinstallation wiring)

• <ow calculate the voltage drop for the actual length of wiring circuit

according to its rated current with the help of following formulas

(*ctual length of circuit x volt drop for 8m) "855 CD to :nd 1olt dropin per meter(*ctual length of circuit x volt drop for 855ft) "855CD to :nd volt drop in855feet

• <ow multiply this calculated value of volt drop by load factor where

!oad factor $ !oad @urrent to be taken by @able" Rated @urrent of @ablegiven in the table

• This is the value of 1olt drop in the cables when load current 'owthrough it.

• +f the calculated value of voltage drop is less than the valuecalculated in step (8) (Eaximum allowable voltage drop) than thesiBe of selected cable is proper

• +f the calculated value of voltage drop is greater than the valuecalculated in step (8) (Eaximum allowable voltage drop) thancalculate voltage drop for the next (greater in siBe) cable and so onuntil the calculated value of voltage drop became less than themaximum allowable voltage drop calculated in step (8)

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How to determine the proper Cable Size or Given Load with!"amples#>or a given load cable siBe may be found with the help of di9erent tablesbut we should keep in mind and follow the rules about voltage drop.7etermining the siBe of cable for a given load take into account the

following rules>or a given load except the known value of current there should be /53extra scope of current for additional future or emergency needs>rom ,nergy meter to 7istribution board 1oltage drop should be 8./23and for :nal sub circuit voltage drop should not exceed /.23 of Supplyvoltage@onsider the change in temperature when needed use temperaturefactor (Table 0)*lso consider the load factor when :nding the siBe of cableWhen determining the cable siBe consider the wiring system i.e. in openwiring system temperature would be low but in conduit wiringtemperature increases due to the absence of air.

• 4ote: 5eep in mind i6ersity Factor in Electrical "ring Installation 'hile selecting

the proper si-e o$ cable $or electrical 'iring installation

!"amples>ollowing are the examples of determining the proper SiBe of cables forelectrical wiring installation which will make it easy to understand themethod of Fhow to determine the proper siBe of cable for a given loadG.

!"ample $ %%%&& (British / English System)

>or ,lectrical wiring installation in a building Total load is H.2kW and totallength of cable from energy meter to sub circuit distribution board is 02feet. Supply voltages are //51 and temperature is H5I@ (85HI>). >ind themost suitable siBe of cable from energy meter to sub circuit if wiring isinstalled in conduits.

Sol'tion() Total !oad $ H.2kW $ H.2 x8555W $ H255W/53 additional load $ H255 x (/5"855) $ J55W Total !oad $ H255W K J55W $ 2H55W Total @urrent $ + $ ?"1 $ 2H55W "//51 $/H.2*

<ow select the siBe of cable for load current of /H.2* (from Table 8) whichis ;"5.50L (/M *mperes) it means we can use ;"5.50L cable accordingtable 8.<ow check the selected (;"5.50L) cable with temperature factor in Table 0so the temperature factor is 5.JH (in table 0) at H5I@ (85HI>) and currentcarrying capacity of (;"5.50L) is /M* therefore current carrying capacityof this cable at H5I@ (85HI>) would be@urrent rating for H5I@ (85HI>) $ /M x 5.JH $ /L.0/ *mp.Since the calculated value (/L.0/ *mp) at H5I@ (85HI>) is less than that of current carrying capacity of (;"5.50L) cable which is /M* therefore this

siBe of cable (;"5.50L) is also suitable with respect to temperature.





<ow :nd the voltage drop for 855feet for this (;"5.50L) cable from Table Hwhich is ;1 -ut in our case the length of cable is 02 feet. Therefore thevoltage drop for 02feet cable would be*ctual 1oltage drop for 02feet $ (; x 02"855) x (/H.2"/M) $ /.81*nd *llowable voltage drop $ (/.2 x //5)"855 $ 2.21

Nere The *ctual 1oltage 7rop (/.81) is less than that of maximumallowable voltage drop of 2.21. Therefore the appropriate and mostsuitable cable siBe is (;"5.50L) for that given load for ,lectrical Wiring+nstallation.

!"ample * %%& S+ -etric .ecimal S/stem #What type and siBe of cable suits for given situation!oad $ 2.MkW1olts $ /051!ength of @ircuit $ 02meter Temperature $ 02I@ (J2I>)

Sol'tion()!oad $ 2.MkW $ 2M55W1oltage $ /051@urrent $ + $ ?"1 $ 2M55 " /05 $ /2./*/53 additional load current $ (/5"855) x 2./* $ 2* Total !oad @urrent $ /2./* K 2* $ 05./*<ow select the siBe of cable for load current of 05./* (from Table 8) whichis ;"8.5H (08 *mperes) it means we can use ;"5.50L cable according table8

<ow check the selected (;"8.5H) cable with temperature factor in Table 0so the temperature factor is 5.J; (in table 0) at 02I@ (J2I>) and currentcarrying capacity of (;"8.5H) is 08* therefore current carrying capacity of this cable at H5I@ (85HI>) would be@urrent rating for 02I@ (J2I>) $ 08 x 5.J; $ 05 *mp.Since the calculated value (05 *mp) at 02I@ (J2I>) is less than that ofcurrent carrying capacity of (;"8.5H) cable which is 08* therefore this siBeof cable (;"8.5H) is also suitable with respect to temperature.<ow :nd the voltage drop for per ampere meter for this (;"8.5H) cablefrom (Table 2) which is ;m1 -ut in our case the length of cable is 02meter. Therefore the voltage drop for 02 meter cable would be4

*ctual 1oltage drop for 02meter $$ m1 x + x !(;"8555) x 05O02 $ ;.L1*nd *llowable voltage drop $ (/.2 x /05)"855 $ 2.;21Nere the actual 1oltage drop (;.021) is greater than that of maximumallowable voltage drop of 2.;21. Therefore this is not suitable siBe of cablefor that given load. So we will select the next siBe of selected cable(;"8.5H) which is ;"8.02 and :nd the voltage drop again. *ccording to Table (2) the current rating of ;"8.02 is H5*mperes and the volte drop inper ampere meter is H.8 m1 (See table (2)). Therefore the actual voltage

drop for 02 meter cable would be*ctual 1oltage drop for 02meter $



$ m1 x + x !(H.8"8555) x H5O02 $ ;.021 $ 2.;H1 This drop is less than that of maximum allowable voltage drop. So this isthe most appropriate and suitable cable siBe.

!"ample 0>ollowing !oads are connected in a building4Sub-Circuit 1/ lamps each o 8555W andH fans each of M5W/ T1 each of 8/5WSub-Circuit 2L !amps each of M5W and2 sockets each of 855WH lamps each of M55W+f supply voltages are /051 then calculate circuit current and @able siBefor each Sub@ircuit=

Sol'tion() Total load of Sub@ircuit 8$ (/ x 8555) K (H x M5) K (/O8/5)$ /555W K 0/5W K /H5W $ /2L5W@urrent for Sub@ircuit 8 $ + $ ?"1 $ /2L5"/05 $ 88.8* Total load of Sub@ircuit /$ (L x M5) K (2 x 855) K (H x M55)$ HM5W K 255W K 0/55W$ H8M5W

@urrent for Sub@ircuit / $ + $ ?"1 $ H8M5"/05 $ 8M.8* Therefore @able suggested for sub circuit 8 $ 0".5/JG (80*mp) or8"8.0Mmm (80*mp)@able suggested for Sub@ircuit / $ ;".5/JG (/8*mp) or ;"5.M2mm(/H*mp) Total @urrent drawn by both Sub@ircuits $ 88.8* K 8M.8* $ /J./;So cable suggested for Eain@ircuit $ ;".5HHG (0H*mp) 5r ;"8.5Hmm(08*mp)

!"ample 1* 85N.? (;.HLkW) three phase sPuirrel cage induction motor of continuous

rating using Star7elta starting is connected through H551 supply by threesingle core ?1@ cables run in conduit from /25feet (;L./m) away frommultiway distribution fuse board. +ts full load current is 8J*. *veragesummer temperature in ,lectrical installation wiring is 02I@ (J2I>).@alculate the siBe of the cable for motor=

Sol'tion()Eotor load $ 85N.? $ 85 x ;HL $ ;HL5W Q(8N.? $ ;HLW)Supply 1oltage $ H551 (0?hase)!ength of cable $ /25feet (;L./m)

Eotor full load @urrent $ 8J* Temperature factor for 02I@ (J2I>) $ 5.J; (>rom Table 0)



<ow select the siBe of cable for full load motor current of 8J* (from TableH) which is ;"5.0LG (/0 *mperes) Q(Remember that this is a 0phasesystem i.e. 0core cable) and the voltage drop is 2.01 for 855>eet. +tmeans we can use ;"5.50L cable according Table (H).<ow check the selected (;"5.50L) cable with temperature factor in table

(0) so the temperature factor is 5.J; (in table 0) at 02I@ (J2I>) andcurrent carrying capacity of (;"5.50LG) is /0 *mperes therefore currentcarrying capacity of this cable at H5I@ (85HI>) would be4@urrent rating for H5I@ (85HI>) $ /0 x 5.J; $ //.08 *mp.Since the calculated value (//.08 *mp) at 02I@ (J2I>) is less than that ofcurrent carrying capacity of (;"5.50L) cable which is /0* therefore thissiBe of cable (;"5.50L) is also suitable with respect to temperature.!oad factor $ 8J"/0 $ 5.M/L<ow :nd the voltage drop for 855feet for this (;"5.50L) cable from table(H) which is 2.01 -ut in our case the length of cable is /25 feet. Therefore the voltage drop for /25 feet cable would be*ctual 1oltage drop for /25feet $ (2.0 x /25"855) x 5.M/L $ 85.JH1*nd maximum *llowable voltage drop $ (/.2"855) x H551$ 851Nere the actual 1oltage drop (85.JH1) is greater than that of maximumallowable voltage drop of 851. Therefore this is not suitable siBe of cablefor that given load. So we will select the next siBe of selected cable(;"5.50L) which is ;"5.5HH and :nd the voltage drop again. *ccording to Table (H) the current rating of ;"5.5HH is /M*mperes and the volt drop inper 855feet is H.81 (see Table H). Therefore the actual voltage drop for/25feet cable would be*ctual 1oltage drop for /25feet $

$ 1olt drop per 855feet x length of cable x load factor(H.8"855) x /25 x 5.M/L $ M.HL1*nd Eaximum *llowable voltage drop $ (/.2"855) x H551$ 851 The actual voltage drop is less than that of maximum allowable voltagedrop. So this is the most appropriate and suitable cable siBe for ,lectricalwiring +nstallation of given situation.

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Determine Cable Size