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Home / Technical Articles / An example how to calculate voltage drop and size of electrical cable

Input information

Electrical details:

Electrical load of 80KW, distance between source and load is 200 meters, system voltage 415V three phase, power factor is 0.8, permissible voltage drop is  5%, demand factor is 1.


Cable laying detail:

Cable is directed buried in ground in trench at the depth of 1 meter. Ground temperature is approximate 35 Deg. Number of cable per trench is 1. Number of run of cable is 1 run.

An example how to calculate voltage drop and size of electrical cable
An example how to calculate voltage drop and size of electrical cable (photo credit: 12voltplanet.co.uk)

Soil details:

Thermal resistivity of soil is not known. Nature of soil is damp soil.


Ok, let’s dive into calculations…

  • Consumed Load = Total Load · Demand Factor:
    Consumed Load in KW = 80 · 1 = 80 KW
  • Consumed Load in KVA = KW/P.F.:
    Consumed Load in KVA = 80/0.8 = 100 KVA
  • Full Load Current = (KVA · 1000) / (1.732 · Voltage):
    Full Load Current = (100 · 1000) / (1.732 · 415) = 139 Amp.

Calculating Correction Factor of Cable from following data:

Temperature Correction Factor (K1) When Cable is in the Air

Temperature Correction Factor in Air: K1
Ambient TemperatureInsulation
PVCXLPE/EPR
101.221.15
151.171.12
201.121.08
251.061.04
350.940.96
400.870.91
450.790.87
500.710.82
550.610.76
600.50.71
6500.65
7000.58
7500.5
8000.41

Ground Temperature Correction Factor (K2)

Ground Temperature Correction Factor: K2
Ground TemperatureInsulation
PVCXLPE/EPR
101.11.07
151.051.04
200.950.96
250.890.93
350.770.89
400.710.85
450.630.8
500.550.76
550.450.71
6000.65
6500.6
7000.53
7500.46
8000.38

Thermal Resistance Correction Factor (K4) for Soil (When Thermal Resistance of Soil is known)

Soil Thermal Resistivity: 2.5 KM/W
ResistivityK3
11.18
1.51.1
21.05
2.51
30.96

Soil Correction Factor (K4) of Soil (When Thermal Resistance of Soil is not known)

Nature of SoilK3
Very Wet Soil1.21
Wet Soil1.13
Damp Soil1.05
Dry Soil1
Very Dry Soil0.86

Cable Depth Correction Factor (K5)

Laying Depth (Meter)Rating Factor
0.51.1
0.71.05
0.91.01
11
1.20.98
1.50.96

Cable Distance correction Factor (K6)

No of CircuitNilCable diameter0.125m0.25m0.5m
111111
20.750.80.850.90.9
30.650.70.750.80.85
40.60.60.70.750.8
50.550.550.650.70.8
60.50.550.60.70.8

Cable Grouping Factor (No of Tray Factor) (K7)

No of Cable/Tray123468
1111111
20.840.80.780.770.760.75
30.80.760.740.730.720.71
40.780.740.720.710.70.69
50.770.730.70.690.680.67
60.750.710.70.680.680.66
70.740.690.6750.660.660.64
80.730.690.680.670.660.64

According to above detail correction factors:

– Ground temperature correction factor (K2) = 0.89
Soil correction factor (K4) = 1.05
– Cable depth correction factor (K5) = 1.0
– Cable distance correction factor (K6) = 1.0

Total derating factor = k1 · k2 · k3 · K4 · K5 · K6 · K7

– Total derating factor = 0.93


Selection of Cable

For selection of proper cable following conditions should be satisfied:

  1. Cable derating amp should be higher than full load current of load.
  2. Cable voltage drop should be less than defined voltage drop.
  3. No. of cable runs  (Full load current / Cable derating current).
  4. Cable short circuit capacity should be higher than system  short circuit capacity at that point.

Selection of cable – Case #1

Let’s select 3.5 core 70 Sq.mm cable for single run.

  • Current capacity of 70 Sq.mm cable is: 170 Amp,
    Resistance = 0.57 Ω/Km and
    Reactance = 0.077 mho/Km
  • Total derating current of 70 Sq.mm cable = 170 · 0.93 = 159 Amp.
  • Voltage Drop of Cable =
    (1.732 · Current · (RcosǾ + jsinǾ) · Cable length · 100) / (Line voltage · No of run · 1000) =
    (1.732 · 139 · (0.57 · 0.8 + 0.077 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 5.8%

Voltage drop of cable = 5.8%

Here voltage drop for 70 Sq.mm Cable (5.8 %) is higher than define voltage drop (5%) so either select higher size of cable or increase no of cable runs.

If we select 2 runs, than voltage drop is 2.8% which is within limit (5%) but to use 2 runs of cable of 70 Sq.mm cable is not economical, so it’s necessary to use next higher size of cable.


Selection of cable – Case #2

Let’s select 3.5 core 95 Sq.mm cable for single run, short circuit capacity = 8.2 KA.

  • Current capacity of 95 Sq.mm cable is 200 Amp,
    Resistance = 0.41 Ω/Km and
    Reactance = 0.074 mho/Km
  • Total derating current of 70 Sq.mm Cable = 200 · 0.93 = 187 Amp.
  • Voltage drop of cable =
    (1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%

To decide 95 Sq.mm cable, cable selection condition should be checked.

  1. Cable derating Amp (187 Amp) is higher than full load current of load (139 Amp) = O.K
  2. Cable voltage Drop (2.2%) is less than defined voltage drop (5%) = O.K
  3. Number of cable runs (1) ≥ (139A / 187A = 0.78) = O.K
  4. Cable short circuit capacity (8.2KA) is higher than system short circuit capacity at that point (6.0KA) = O.K

95 Sq.mm cable satisfied all three condition, so it is advisable to use 3.5 Core 95 Sq.mm cable.

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Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

199 Comments


  1. Hardeep
    Sep 15, 2023

    You explain it very well, really informative. But please consider cable type ( aluminium or copper)
    Thanks.

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