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%
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.
Related electrical guides & articles
I’m developing a multifamily subdivision. (12 buildings ) consists of 18 units each buildind.
I’m planning on running unground eletric, 8 feet deep.
Each build will have a inside transformation. 1,100kva. Each building unit will have the strength of 60,000 volts.
I need to know what size wire I use to jump style of the transformers to each build from one power source of the main supply. So basically a line that can support 13,000 kva. Can someone help?
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Eric
In which standard can I find these tables for the correction factors?
Different topics on improvement of electric distribution system and related calculations etc.
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