Electrical Standard Codes is : IS: 7098 / IEC: 60502 / BS: 6622/BS: 7835.
The cable size is said to be proper when the cable is able to carry full load current, able to withstand the short circuit current and should have voltage drop within the acceptable limits as per the standards.
The Permissible Voltage variation of various Transmission and Distribution lines are given below
1. For LT Lines
2. For 11KV and 33KV Distribution Lines
3. For Extra high Voltage Lines
Let we have to Select Size of XLPE Cable for the 3 Phase, 415V, 100KW Load. The Load is supplied from Distribution Transformer 11/0.433KV, 250KVA. The Power factor of The load is 0.8.
Step 1. Find the Load Current
we know that for 3 phase Load
`P = \sqrt{3}VI_{L}Cos \phi`
Where P = Power Rating of Load in kW
V = Voltage in kilo Volts
`I_{L}` = Load Current in Amperes
`Cos \phi` = Power Factor of Load
`I_{L} = \frac{P}{\sqrt{3}V cos\phi}`
`I_{L} = \frac{100}{\sqrt{3}\times0.415 \times 0.8}=173.9A`
Step 2: Calculate Short Circuit Current of Transformer
Full load Current of Transformer, `I_{FT}=\frac{P_{T}}{\sqrt{3}\timesV_{T}} = \frac{250}{\sqrt{3}\times 0.433} = 333.34A`
where
`V_{T}` = Voltage on Secondary Side of Transformer
`P_{T}` = Rating of Transformer
Short Circuit Current of Transformer (`I_{sc}`)
`I_{sc} = \frac{I_{T}\times 100}{Z%} = \frac{333.34 \times 100}{4.5}= 7407.56 = 7.407kA`
Where Z% is percentage impedance of Transformer.
Step 3: Calculate Area of Cross section of Cable
Let "A" is Area of cross section of cable in sqmm
`A=\frac{I_{sc}\times \sqrt(t)}{K}`
Where
`I_{sc}` = short circuit of Transformer in Amperes
t = time of short circuit current. it is generally taken 0.14 seconds to 0.2 seconds.
Always remember take it 10 to 20% above the time take by our protective device to trip the circuit from supply in case of fault.
K = constant, taking account of the resistivity, temperature co-efficient, heat capacity of the conductor material, initial and final temperature
The table below gives the value of "K" as per BS 7671 Table 54C
Putting the values of `I_{sc}`, t and K in above equation
Area of cross section of cable
`A = \frac{7407.56 \times \sqrt(0.2)}{94.48} = 35sqmm`
Step 4: Check Current Rating of 35sqmm cable from catalogue of Cable provided by manufacture.
Below the snap of Catalogue of cable from KEI Industries Limited
From the catalogue the current rating of 35sqmm single core cable is 157A.
but our load current is 173.9 A. So 35sqmm cable is not suitable for our load.
Taking the derating factor (Generally lies between 0.6 to 0.8) into consideration. Let us take derating factor equal to 0.6
The cable should have current rating (I) equal
`I =\frac{173.9}{0.6}=289.8A`
So as per the catalogue 95Sqmm cable has current rating 287A which is near to 289.8 A
So 95Sqmm cable can be used
Now we will check the voltage drop analysis. Whether the 95Sqmm cable has voltage drop within permissible limit or not
Step : Voltage Drop Analysis
Voltage drop (`V_{d}`) For 3 Phase
where
`I_{L}` = Load Current
R = Resistance per meter of cable
X = Reactance per meter of Cable
Putin the values of `I_{L}` , R , X and Number of Runs in the above equation
The Values of R and X can be taken from catalogue
From Catalogue Value of R is 1.110 ohm per meter and X is 0.090 ohm per meter.
power factor , `cos\phi = 0.8`
let we need 100 meters to connect the load with power supply
Thus
`V_{d}=\frac{\sqrt{3}\times 1.2 \times 173.9 \times ((1.110 \times 0.8 )+(0.090 \times 0.6))\times 100}{1000 \times 1} = 19.79 V`
Percentage Voltage drop
`V_{d}% = \frac{V_{d}\times 100}{V_{s}}=\frac{19.79\times100}{415}=4.768%`
For LT Lines The permissible voltage variation is +-6% so that voltage criteria is satisfied
so the cable size is proper.
You Can Download the Excel Sheet for Calculation of Cable Size and Voltage Drop Here (
Click Here)
For calculation of Cable size Using Etap watch the Video below.
