# Cable resistance calculation

Update:Jan 20,2022
Summary:The DC standard resistance of the cable can be calculated according to the following formula:R20=ρ20(1+K1)(1+K2)/∏/4×dn×10In the formula: R20——the bra...
The DC standard resistance of the cable can be calculated according to the following formula:
R20=ρ20(1+K1)(1+K2)/∏/4×dn×10
In the formula: R20——the branch current standard resistance of the cable at 20℃ (Ω/km)
ρ20——The resistivity of the wire (at 20℃) (Ω*mm/km)
d——diameter of each core wire (mm)
n——the number of core wires;
K1——The twist rate of the core wire, about 0.02-0.03;
K2——The twist rate of the multi-core cable is about 0.01-0.02.
The actual AC resistance per kilometer of cable at any temperature is:
R1=R20(1+a1)(1+K3)
In the formula: a1——the temperature coefficient of the resistance at t°C;
K3——Coefficient of skin effect and proximity effect, it is 0.01 when the cross-sectional area is below 250mm; 0.23-0.26 when it is 1000mm.
Capacitance calculation
C=0.056Nεs/G
In the formula: C——the capacitance of the cable (uF/km)
εs - relative permittivity (standard is 3.5-3.7)
N - the number of cores of the multi-core cable;
G - shape factor.
Inductance calculation
For underground cables used for power distribution, when the conductor cross-section is circular and the armor and lead clad losses are ignored, the calculation method of the inductance of each cable is the same as that of the conductor.
L=0.4605㏒Dj/r+0.05u
LN=0.4605㏒DN/rN
In the formula: L——the inductance of each phase line (mH/km)
LN - the inductance of the neutral wire (mH/km);
DN - the geometric distance between the phase line and the neutral line (cm);
rN - the radius of the neutral line (cm);
DAN, DBN, DCN - the center distance (cm) between each phase line and the neutral line.
illustration
The measured load current of the 2# living variable load in the work area is 330A, and the existing cable is a 120mm four-core copper-core cable. Check the table to know that its safe current carrying capacity is 260A, the cable is overloaded, and there is a hidden danger. In order to ensure the normal power supply, our work area Plan to split with another cable to ensure normal power supply. (The cables mentioned below all refer to 1KV, VV-type armored polyethylene four-core copper-core cables).
If we look at 330A-260A=70A according to the safe current carrying capacity, we only need to connect a cable with a current carrying capacity of 70A to ensure safe operation in theory (ideally).
According to the calculation method of cable impedance, the impedance values ​​of 16mm, 25mm, 35mm, 50mm, 70mm, 95mm and 120mm are shown in Table 2:
 Nominal cross-sectional area（mm） resistance（Ω/km）（20℃H） Reactance（Ω/km） 16 1.15 0.068 25 0.75 0.066 35 0.53 0.064 50 0.37 0.063 70 0.26 0.081 95 0.19 0.06 120 0.15 0.06

Table 2 Impedance of four-core copper-core cable
According to the above table, we can calculate the impedance mode of the cable. In the case of ignoring the contact resistance of the parallel cable, the parallel cable is understood as two impedances in parallel, and the current distribution value is calculated.
Of course, when two cables are laid in parallel, the safe current carrying capacity of the cable will change. When two cables are used together, the safe current carrying capacity should be 0.92 times the original current carrying capacity.
At this time, the safe current carrying capacity of the 120mm copper core cable is 239A. 25mm is 86A, 35mm is 109A, according to the principle of Iux≥Izmax and a 35mm cable can run safely.
For 2# life and a 35mm four-core copper cable:
︱Z35︱=0.534Ω
The load current is 330A, then I120=︱Z16︱/(︱Z16︱+︱Z120︱)*330
It is obtained that I120=253.19A; I16=76.81A
It is not difficult to see that the 120mm cable is still running over current.

View More

View More

View More

View More

View More

View More

View More

View More

View More

View More