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Topic 18 | 18-1 | 18-2 | 18-3 | 18-4
Cable length run maximum
Cable length run maximum
Every circuit installed on point can be calculated and it is known as cable length run.
Standard voltage drop according to IEE Wiring Regulation are:
4% (IEE WR 16 Ed)
5% (IEE WR 17 Ed)
Voltage drop percentage calculation
Voltage drop percentage calculation
4% of 240 volt = 240v x 0.04 = 9.6 volt
4% of 415 volt = 415v x 0.04 = 16.6 volt
For reference only
Refer to figure above showing the basic understanding of calculating the cable length run maximum.
This length is to be measured on site from DB to the designated point A. Due to some last minute relocations of the point from A to B, you can calculate the cable size used for the circuit to check what is the maximum length cable you can installed at new point B.
Cable length run maximum formula
Cable length run maximum formula
L max = ( Vc* x 1000 ) ÷ (mV/A.pm x Ib )
where
Lmax length run maximum in meterVc* is maximum voltage drop allowed Note: Vc max ( for 240v = 9.6v ) ( for 415v = 16.6v )mV/A.pm mili-volt ampere per meter Ib current design of load
Table 4D1B
Table 4D1B
Source: IEE Wiring Regulation (16th Edition) Table 4D1B
Example calculation
Example calculation
A machine rated 6.7kW, 415volt AC installed over a length of 7 meter, the cable size used 4.0 mm² find the maximum length run for this circuit ?
a. find the current design, Ib ?
Solution: Ib = P ÷ V = 6700 W ÷ 415v = 16.14 A
b. select the suitable cables used for this circuit and identify the voltage drop properties of this cable.
Cable size select: 4.0mm² with value of 9.5 mV/A.pm
c. calculate the maximum length run for this circuit if extended.
Solution: L max = ( 16.6v x 1000 ) ÷ ( mV/A.pm x Ib )
= (16.6v x 1000 ) ÷ ( 9.5 x 16.14 A ) = 108.26 meter
Conclusion: This circuit cable can be extended from 7 to 108 meter.
Three phase motor rated 9kW/415v with efficiency of 85% and power factor of 0.8 are connected over a distance of 7 meter from 3p isolator point.
a. find the current design, Ib ?
Solution: Ib = P ÷ (415V x Eff x pf x 1.732 )
= 9000 W ÷ ( 415v x 0.85 x 0.8 x 1.732 ) = 18.41 A
b. select the suitable cables used for this circuit and identify the voltage drop properties of this cable.
Cable size select: 4.0mm² with value of 9.5 mV/A.pm
c. calculate the maximum length run for this circuit if extended.
Solution: L max = ( 16.6v x 1000 ) ÷ ( mV/A.pm x Ib )
= (16.6v x 1000 ) ÷ ( 9.5 x 18.41 A ) = 94.91 meter
Conclusion: This circuit cable can be extended from 7 to 94 meter.
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