## 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**.

### 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 Temperature | Insulation | |

PVC | XLPE/EPR | |

10 | 1.22 | 1.15 |

15 | 1.17 | 1.12 |

20 | 1.12 | 1.08 |

25 | 1.06 | 1.04 |

35 | 0.94 | 0.96 |

40 | 0.87 | 0.91 |

45 | 0.79 | 0.87 |

50 | 0.71 | 0.82 |

55 | 0.61 | 0.76 |

60 | 0.5 | 0.71 |

65 | 0 | 0.65 |

70 | 0 | 0.58 |

75 | 0 | 0.5 |

80 | 0 | 0.41 |

#### Ground Temperature Correction Factor (K2)

Ground Temperature Correction Factor: K2 | ||

Ground Temperature | Insulation | |

PVC | XLPE/EPR | |

10 | 1.1 | 1.07 |

15 | 1.05 | 1.04 |

20 | 0.95 | 0.96 |

25 | 0.89 | 0.93 |

35 | 0.77 | 0.89 |

40 | 0.71 | 0.85 |

45 | 0.63 | 0.8 |

50 | 0.55 | 0.76 |

55 | 0.45 | 0.71 |

60 | 0 | 0.65 |

65 | 0 | 0.6 |

70 | 0 | 0.53 |

75 | 0 | 0.46 |

80 | 0 | 0.38 |

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

Soil Thermal Resistivity: 2.5 KM/W | |

Resistivity | K3 |

1 | 1.18 |

1.5 | 1.1 |

2 | 1.05 |

2.5 | 1 |

3 | 0.96 |

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

Nature of Soil | K3 |

Very Wet Soil | 1.21 |

Wet Soil | 1.13 |

Damp Soil | 1.05 |

Dry Soil | 1 |

Very Dry Soil | 0.86 |

#### Cable Depth Correction Factor (K5)

Laying Depth (Meter) | Rating Factor |

0.5 | 1.1 |

0.7 | 1.05 |

0.9 | 1.01 |

1 | 1 |

1.2 | 0.98 |

1.5 | 0.96 |

#### Cable Distance correction Factor (K6)

No of Circuit | Nil | Cable diameter | 0.125m | 0.25m | 0.5m |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 0.75 | 0.8 | 0.85 | 0.9 | 0.9 |

3 | 0.65 | 0.7 | 0.75 | 0.8 | 0.85 |

4 | 0.6 | 0.6 | 0.7 | 0.75 | 0.8 |

5 | 0.55 | 0.55 | 0.65 | 0.7 | 0.8 |

6 | 0.5 | 0.55 | 0.6 | 0.7 | 0.8 |

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

No of Cable/Tray | 1 | 2 | 3 | 4 | 6 | 8 |

1 | 1 | 1 | 1 | 1 | 1 | 1 |

2 | 0.84 | 0.8 | 0.78 | 0.77 | 0.76 | 0.75 |

3 | 0.8 | 0.76 | 0.74 | 0.73 | 0.72 | 0.71 |

4 | 0.78 | 0.74 | 0.72 | 0.71 | 0.7 | 0.69 |

5 | 0.77 | 0.73 | 0.7 | 0.69 | 0.68 | 0.67 |

6 | 0.75 | 0.71 | 0.7 | 0.68 | 0.68 | 0.66 |

7 | 0.74 | 0.69 | 0.675 | 0.66 | 0.66 | 0.64 |

8 | 0.73 | 0.69 | 0.68 | 0.67 | 0.66 | 0.64 |

**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:**

- Cable derating amp should be
**higher than full load current of load**. - Cable voltage drop should be
**less than defined voltage drop**. - No. of cable runs
**≥**(Full load current / Cable derating current). - 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%**

**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.**

**Cable derating Amp (187 Amp)**is higher than full load current of load (139 Amp) =**O.K****Cable voltage Drop (2.2%)**is less than defined voltage drop (5%) =**O.K****Number of cable runs (1)**≥ (139A / 187A = 0.78) =**O.K****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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A 2 core cable supplies current to a 240v single phase load of 18kw at 0.78 power factor. The cable is 40m long and each conductor has a cross-sectional area of 35mm.calculate the voltage drops in the cable at the load,ignoring the reactance of the cable?

Voltage drop of cable =

(1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%

answer not coming, explain pl.

Yeah, voltage drop Calculation for 3.5C x 95 mm² is incorrect. it should be 4.3% instead of 2.2 %

This information is fantastic.

Thanks

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

Cable size?

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

KVA/1.732*VL =amp , if we include PF then it wont be kVA which is Apparent power, but Real Power kW.

so we will have, kW/1.732*VL*PF = amp

hope this clears the confusion.

Amp = KVA/(1.732*415) or

Amp = KW/(1.732*415*0.8)

Hope you understand the difference.

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

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 .

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

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

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

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)

It’s tabulated value. BS 7671:2018

U CAN FIND IT IN CURRENT RTING CHART

Why I couldn’t find current ratings chart

(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

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.

I need DC motor current rating chart

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