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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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author-pic

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 13 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

172 Comments


  1. S K MANDAL
    Sep 03, 2021

    Allowable voltage drop : 5%, Cable Length: 1300M, Motor power: 110KW. cable to be laid on tray in air
    Cable size?


  2. Hennah S. Kamara
    Sep 03, 2021

    From the above calculations of Full Load current, I still have a doubt.
    Why did we ignored the PF in getting our full Load current?

    KVA/ 1.732*415 =amp

    Amp =KVA/1.732*415*0.8


  3. Afnitha
    Sep 01, 2021

    How we can calculate number of cables that can accommodate in a cable tray ?


  4. ABHIMANYU BHARDWAJ
    Jul 27, 2021

    how to calculate cable sizes when the total load is 18 kw but it is divided into 6 towers each at a distance of 1 km and each tower has a load of 3 kw ,with 3 towers on one side and 3 towers on the other side , and placing one feeder pillar between them (i.e . 3 towers on one side then feeder pillar and then 3 towers on other side), which will feed them supply ,this feeder will given main supply from LT PANEL placed at a distance of 700 mtrs . please do tell . this is the problem i am facing .


  5. Dirshe
    Jul 16, 2021

    95sq.mm 3.5 core cable which standard is this?


  6. Adegoke Adeyemi Adewara
    Jul 07, 2021

    I love this platform. Looking forward to learning more from you genius.


  7. NASIR SHAH
    Jul 02, 2021

    I WANT TO BA AN ENGINEER AND WANT TO MAKE HUGE MACHINERY AND MAKE FREE ENERGY GENERATOR SYSTEM I LIVE IN GULSHAN RAVI


  8. Durga
    May 04, 2021

    How to identify the value for R and j
    For 95 sq mm R=0.41 · 0.8 + j=0.074 · 0.6
    Where as
    For 70 sq mm R=0.57 · 0.8 + j=0.077 · 0.6)


    • Nuwan
      Jun 03, 2021

      It’s tabulated value. BS 7671:2018


    • SURAJ
      Jul 06, 2021

      U CAN FIND IT IN CURRENT RTING CHART


      • Suraj yadav
        Jul 19, 2021

        Why I couldn’t find current ratings chart


  9. siva
    Apr 26, 2021

    (1.732 · 139 · (0.57 · 0.8 + 0.077 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 5.8%

    sir from above how we got 139


    • Shivaram
      Apr 29, 2021

      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.


      • Suraj yadav
        Jul 19, 2021

        I need DC motor current rating chart


      • zeeshan raza
        Jul 26, 2021

        Instead of this, we can use a simple formula for 3 core cables – total connected load x 1.74 = load current

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