## 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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Thank u so much sir, such a wonderful messages.And please sir I need to know about how prepared load calculation and schedule?

You must know about your load in system sir, classification about motor and feeder loads, service unit (continuous, intermittent, spare), and then calculation consumed load, demand load, peak load. After that, you can sizing trafo rating, generator rating, Ishort circuit, cable sizing, and more.

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. I hope your earliest reply. Thanku

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thanks a lot

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Hi All,

How to select the value of cosǾ & sinǾ in the formula.

Voltage Drop of Cable= (1.732x Current x (RcosǾ+jsinǾ) x Cable Lengthx100) / (Line Voltage x No of Runx1000)

Can anyone help me in this please?

While calculating the voltage drop, cosǾ = starting p.f.(0.2), which is constant.

sinǾ = SQRT(1-( cosǾ * cosǾ)).

Example ; cosǾ=0.2 , then sinǾ = SQRT(1-( 0.2 * 0.2)) =0.979

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Voltage Drop of cable at transient = (1.732*current *((resistance*cosΦst)+(reactance *sinǾst)) * cable length * 100) / (line voltage * No.of runs * 1000)

where, cosΦst = 0.2

sinǾst = SQRT (1-(0.2*0.2))

cospi is power factor from there u can find sinpi using inverse cos formula

Ǿ depends on the load.

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