Cable sizing

For cable sizing we should know
1-SHORT CIRCUIT CRITERIA
Cable Short circuit capacity should be higher than system Short circuit capacity. It is also known as minimum cross-section requirement that can withstand with fault current.
2-AMPACITY RATING
Cable is having base current carrying capacity based on its cross-section area (square mm). But based on installations there are some de-rating factor that affect its current carrying capacity of the cable. So After applying de-rating factor on the cable the current which we get that is known as ampacity rating of cable. Ampacity rating of cable must be higher than Full load current of the load.
3-VOLTAGE DROP CRITERIA
Cable voltage drop should be less than defined voltage drop

SHORT CIRCUIT CRITERIA

1-short circuit criteria (It is also known as minimum cross-section requirement)
Area required=(I √t)/k sq mm
here
I= fault current in Ampere
t= time for which cable must withstand with fault current
K= Constant which depend on cable conductor material and insulation material
generally value of k are

K	Cu	Al
Xlpe	143.08	94.48
PVC	115	76.03

Note:- PVC can withstand at 70°C whereas XLPE can withstand at 90°C. As XLPE insulated cable can withstand at 90°C therefore it can carry more current.

Now Let suppose we have a LT motor which operate at 415 Volt and its fault current=50KA and we need to find a cable size for this
Now first we need to find fault time
Generally motor operate by instantaneous relay which have fault clearing time=0.05second and breaker operating time=0.06second.
So total Fault time= Fault clearing time+ Breaker operating time=0.05+0.06=0.11
and some additional margin=0.09
Now,
   total time=total Fault time + additional margin time=0.11+0.09=0.2sec
let suppose we using Cupper cable with XLPE Cable where k=143.08

Area required=(I √t)/k sq mm
=50,000*√0.2/143.08
=156.28 Sq mm
Nearest size around this is 185 Sq mm

AMPACITY RATING

Cable is having base current carrying capacity based on its cross-section area (Sqmm). But based on installations there are some de-rating factor that affect its current carrying capacity of the cable. So After applying de-rating factor on the cable the current which we get that is known as ampacity rating of cable.

For understanding ampacity we take an example, let we have a 3-phase motor which operate at 415V and 80KW power, who has power factor (P.F)=0.8,demand factor=1.

Now first we need find full load current for our system.
First Method
As we know P=√3*VICosθ
I=P/VCosθ
I=1.73X80,000X415X0.8
I=139
So for running this motor we need a cable which as easily withstand with 84.5A
therefore 84.5A=Full load current

2nd Method
Consume Load=Total load*Demand factor=80*1=80KW
In KVA=KW/Power factor =80/0.8=100KVA
Full load current=(KVA*1000)/(1.732*Voltage)=(100.1000)/(1.732*415)=139.12A

Once we have full load current, now we need to find de-rating factor(K). Since atmospheric condition always effect the current carrying capacity of a cable.

De-rating Factor (K) Concept

As when we can see manufacture datasheet the current carrying capacity is generally given at 40°C. But generally we have to consider Ambient temperature is around 50°C for New Delhi summer condition and at 50°C the current carrying capacity of cable is decrease.
Similarly there are several factors which affect the current carrying capacity.

1)Temperature
2)Cable laying(Where we are laying the cable)
  In ground (Depth and thermal resistance of soil)
  In open air
  In Duct(It is a type of pipe or closed table try)
3)Distance between the cables.(As distance between the cable decrease the mutual inductance increase).It is also known as Cable distance Correction
4)Number of cable are grouping together(grouping factor)-->As sometimes we use cable tray in which as number of cable are group together.

So, Generally we have 7 de-rating factors such as, Air temperature(K1), Ground Temperature(K2), Soil Thermal resistance(k3), Soil correction(K4), Cable depth(K5), Cable distance Correction(K6), Cable grouping(K7)
Soil Thermal resistance(k3)->if we measure it, if or someone tell us about it. Only then we include its value it. The common value of resistivity are 1, 1.5, 2, 2.5 and 3. The corresponding value for these resistivity are given in Kelvin meters per watt( K m /W) or degree Celsius centimeter per watt (°C.cm/W).
Soil correction(K4)-->The value for soil correction is almost fixed such as, Very Wet Soil(K4=1.21), Wet Soil(K4=1.13), Damp Soil(K4=1.05), Dry Soil(K4=1), Very Dry soil(K4=0.86)

Now let consider our cable is 1 meter inside the ground and ground temperature is 35°C where Damp soil is present.
First we need Calculate the Value of "K"

De-rating factor(K)
K=K2(Rating factors from manufacture datasheet ground temperature=0.96)Xk4(for Damp soil k4=1.05)xK5(from datasheet inside depths of 1050mm=1.05 meter and rating value=0.96)Xk6(K6=1 as there is only one cable we are using and there is no other cable is present there) K=0.96X1.05X0.96X1=0.9776
total de-rating factor(K)=0.9776
As we get value of K, Now we need to divide value of full load current by K and result will be our needed amperes in the cable according to datasheet of manufacture
=full load current(139)/k(0.9776)
=142 Amps

So we need a cable which can carry more than 142 Amps in datasheet of manufacture.
In manufacture datasheet 70mmX3.5core XLPE insulated aluminium armoured can carry 164 ampere current inside the ground so it is sufficient according to our requirement.

Now we check the total current carrying capacity of 70mmX3.5core XLPE insulated aluminium armoured cable after de-rating factor=164X0.97=159.08Amps
So, ampacity of 70mmX3.5core XLPE insulated aluminium armoured cable=159.08Amps

As this cable is sufficient for our current requirement therefore now we will need to check its voltage drop.

VOLTAGE DROP

First Method
Lets we take length=200 meter and voltage=415
Now As we know the size of our cable that is 70mmX3.5core we can easily find resistance=0.568 and reactance=0.0742 of cable from manufacturer datasheet

For 3 phase Voltage drop=[1.732*Current*(Resistance* Cosθ+ Reactance* Sinθ)*Cable Length]/[line voltage * No of run * 1000(for Km) ]
=1.732*current*(R Cosθ + j Sinθ)*Length*100/[line voltage * No of run * 1000(for Km) ]

=[1.732 * I * (0.568*P.F+0.0742* Sinθ)*200]/[415X1X1000]
Sinθ=√(1-cosθ^2)
   =√(1-0.8^2)
   =√(1-0.64)
   =√(0.36)
Sinθ=0.6

V.D ratio=[1.732X 139 X (0.56X0.8+0.074*0.6)X200]/[415X1X1000]=57.129/1000=0.0571
for convert it into percentage=V.D ratioX100
=0.0571X100
=5.71%
Note:-The NEC recommends that the maximum combined voltage drop for both feeder and branch circuit should not exceed 5%.
so we need to increase the size of cable now we take 95mm cable where R=0.410 and X=0.0725
Ampacity of 95mm=197*0.97=191.09A
V.D ratio=[1.732* 139 *(0.41*0.8+0.0725*0.6)*200]/[415*1*1000]=43.102/1000=0.0431
for convert it into percentage=V.D ratio*100
0.0431*100
4.31%

2nd Method
From Datasheet aluminium armoured XLPE Insulated the approx Voltage Drop
70mm=0.99Mv/amp/mtr
95mm=0.72Mv/amp/mtr

voltage drop=[ (milli-volt drop/A/M)*full load current in Ampere* Length in meter ]/1000(to convert mill-volt into volt)
V_d=(0.99*139*200)/1000=27,522/1000=27.522
Voltage drop after 200 meter=27.522

Now we calculate 5% of our applied voltage=Voltage*5%
=415V*5/100
=2075/100=20.75
As for 5% maximum voltage drop=20.75V, But in 70mm for 200 meter the voltage drop=27.522V. It is greater than 5%. Therefore we go for next size that is 95mm

for 95mmX3.5core
V_d=(0.72*139*200)/1000=20.01
As 20.01 is less than 20.75 therefore 20.01 is acceptable for our cable.
Therefore we use 95mmX3.5core XLPE insulated Aluminium Armoured cable.

3rd Method
All cable has a resistance that is given as ohms per metre – “W/m”. Some manufacturers give this figure, however others give the voltage drop expressed as “mV/A/m” ) milli-Volts per Amp per metre .

Lets, We have a 1.0 mm² cable connected to a battery that is at 12 metres distance and that needs to supply two light bulbs that have a total load of 50 watts at 12 volts,Now we need to find the voltage drop?
Let suppose that manufacturer give us cable resistance 0.038 W/m.
As we know that,V=IR.Therefore first we need to find out current
And, we know, P / V = I
50 watts / 12 volts = 4.17 Amps
V_d=IR*length
V_d=4.17A x (12m x 0.038W/m) = 1.901 V