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
Sir, what is the figure 1.732?
Thanks for sharing.
Sir, what will be the cable derated ampacity incase if i have 3 runs of 1C cable with base ampacity 170amps & cable derating 0.8. do i have to times with 1.732, say 170×0.8×1.732=235.5A. please guide.
Sir, please let us know from which standard (IEC / BS / IS) you have taken the cable ratings.
(1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100) / (415 · 1 · 1000) = 2.2%
the above calculation does not give 2.2% it gives 4.3% please check again. If you use two runs then it will be 2.2%
Dear sir, how did you take system short circuit capacity as 6kA?
GREAT INFORMATIONS
Good day,
Please what is your reference for selecting resistance and reactance values
How we got 139 amperes after getting 170*0.93=158.10, calculation for derated current?
Some step is missing.
Sir. can you please give an idea on how did you get the value of .8 and .6. can you show to me the computation
i need clarification how sin Value is 0.6? any calculation behind it?
yes there is calculation ,with formula that is sinǾ = SQRT(1-( cosǾ * cosǾ))
cos(fi)= o.8
cos inv. of 0.8 is 36.86
then sin of 36.86 is 0.599986 i.e approx 0.6
What the distance between load and supply ? What is the voltage value ? Pf ? And more. Are you serious about your question?
He has considered everything 200m length, 415 V voltage, Pf for converting Kw to KVA.
So what are you asking????
how do i calculate the size of the cable, given the power and voltage, taking installation methods and correction factors into account. e.g. an electrical load is 80KW located at a distance of 200m from supply, voltage is 415V three phase, power is 0.8, permissible voltage drop is 5% and demand factor is 1.
Sir pl supply a cable size calculator EXEL Sheet.
generate 1250 kva and transformer 1200 kva the power haws 1600 A bracer what cable size