Size Selection Of Electrical Power Cables

Why Size Selection of Electrical Power cables is Important ?

In order to make sure that Electrical Power cables can function safely and handle a full load without getting damaged, it is crucial to perform cable sizing calculations. These calculations help us determine the right cable size to provide the appropriate voltage for the load.\\Additionally, proper sizing enables us to withstand the most severe short circuit currents and ensures the safety of devices during operation.\\This article will discuss how to choose the correct cable size, using an example to illustrate the process.

Let we have to select the XLPE copper conductor cable for 3 phase, 415V, 3 HP motor, designated on its nameplate as class “A”. The Motor is feed from the Distribution Transformer 11/0.433KV, 100KVA. The Distribution transformer have percentage impedance of 4.5%

Step 1: Short circuit Analysis \\Calculate full load current of Distribution Transformer\\\;\\\textrm{Full load current of Transformer}=\frac{\textrm{Rating of Transformer}}{\sqrt{3}\times\textrm{Voltage on load side (in kV)}}\\\;\\\textrm{Full load current}=\frac{100}{\sqrt{3}\times0.433}=133.33A

Now Calculate the short circuit current of the source i.e Distribution Transformer\\\;\\I_{sc}=\frac{\textrm{Rated current of Transformer}\times 100}{\textrm{Percentage impedance}}\\\;\\I_{sc}=\frac{133.33\times 100}{4.5}=2963A=2.963kA

Area of cross section of cable, A = \frac{I\times \sqrt{t}}{K}\\\;\\\;\;\;Where I = short circuit to withstand cable in Amperes (A)\\\;\;\;t = time of short circuit in seconds (Generally 0.14 to 0.2 sec are consider for motor disconnection in case of fault. )\\\;\;\;K = constant depending upon the material of the cable i.e conductor material and insulation material

Area of cross section of cable, A = \frac{2963\times \sqrt{0.14}}{143.08}=7.75\cong8sqmm\\But 8sqmm is not the standard size of cable so we take nearest standard size of cable that is 10sqmm

Step 2: Calculation of Full load current / Starting Current \\For load other than motor we calculate full load current but incase of motor we calculate short circuit current. \\Starting current of Motor or inrush current or locked rotor current.\\\;\\I_{scm}=\frac{\textrm{Design code letter value}\timesHP\times1000}{\sqrt{3}\times\textrm{Voltage rating (in Volts)}}\\\;\\I_{scm}=\frac{3.14\times3\times1000}{\sqrt{3}\times415}=13.10A

Now from cable catalogue the current carrying capacity of 10sqmm single core cable is 52A\\Current carrying capacity taking derating factor (generally lies between 0.6 to 0.8 or as mentioned in cable catalogue ) into consideration\\\;\\I = 0.6 \times 52 =31.2A\\\;\\Since the starting current of motor should be less than current carrying capacity of cable (31.2A) so full load current / starting current criteria satisfied.

Step 3 : Voltage drop Analysis \\

Voltage drop formulae \\\;\\V=\frac{\sqrt{3}\times 1.2\times I_{full}\times(Rcos\phi+Xsin\phi)\times length\textrm{(in Meters)}}{1000\times\textrm{Number of Runs}}\\\textrm{For single Phase}\\\;\\V=\frac{1.2\times I_{full}\times(Rcos\phi+Xsin\phi)\times length\textrm{(in Meters)}}{1000\times\textrm{Number of Runs}}

The value of Resistance per km (R) and Reactance per km is taken from catalogue of cable. For single core 10sqmm copper conductor cable\\R = 2.24 ohm/km and X = 0.119 ohm/km. \\Generally starting power factor of Motor is 0.25 to 0.3. Instead full load current consider starting current for motor.

Let the lenght of cable is 100 meters.\\Putting the values of above equation.\\\;\\V=\frac{\sqrt{3}\times1.2\times 13.10\times ((2.24\times0.25)+(0.119\times0.96))\times 100}{1000 \times 3}=0.611V\\\;\\ In terms of percentage \\\;\\V =\frac{0.611\times 100}{415}=0.147\%\\Voltage drop should be less or equal to 3% and for starting voltage drop will be between 10 % to 15%. —–criteria satisfied\\All the three criteria are satisfied so the selection of cable is correct

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