HT Cable Sizing Calculation

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HT CABLE SIZING CALULATION

Change Number: ChangeNoAuthor: Selvan SakthiveluDesign Office:Reason for new revision:(see last page for modification record)

Initial Issue

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Customer : xxxxxxxProject : xxxxxxxDocument : HT CABLE SIZING CALULATIONFile : xxxxxxx / document.docx©COPYRIGHT 2012 GE ENERGY (USA), LLC AND/OR ITS AFFILIATES. All rights reserved. GE Energy Power Conversion Proprietary Technical Information

CLASS II (GE INTERNAL NON-CRITICAL)Ref : ReferenceRevision : A0Date : 05 Jun 12Page : 1/5

CONTENTS

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1.0 OBJECTIVE:

The objective of this document is for calculation of HT power cables.

2.0 REFERENCES:

The following references are used for this calculation:

2.1. Gloster cable catalogue.

2.2. Fault current considered as 26.2 kA for 0.16 second(Operating time for VCB and Numerical Relay)

Operating time for VCB = 65 ms = 0.065 s ; Numerical Relay = 40 ms = 0.040 s,

Total time = 0.065+ 0.040 =0.105 s

Fault duration considered = 0.16 s

2.3. Derating factor = Temperature factor x grouping factor

Temperature factor = 1.0 for 40 deg C ambient

Grouping factor = 0.68 for 6 layers of cable trays with 6 cables in each tray touching each other

Therefore, Derating factor = 1.0 x 0.68 = 0.68

3.0 DESIGN CRITERIA:

The cable selected shall fulfill the following criteria:

3.1. To be able to withstand the short circuit current.

3.2. To be able to carry full load current after considering appropriate derating factor for ambient temperaturevariation, method of laying (in air and in trays in our case).

3.3. Voltage drop limited to 1% for cables to switchgear and transformers feeders and 3% for cables to motorfeeders .

4.0 INCOMING CABLE FOR 33 kV SWITCHGEAR

4.1 SHORT CIRCUIT WITHSTAND CAPABILITY OF CONDUCTOR

Fault current is taken as 26.2 kA for 0.16 second

Minimum cross section required = Fault current x √Emek

k = 0.094 (constant for Aluminium)

Minimum cross section required = 26.2 x √0.16 0.094




= 111.49 sq.mm

Therefore, minimum cable cross section selected is 120 sq.mm

4.2 CURRENT CARRYING CAPACITY :

Full load current of switchboard = 347.39 A (Assumption)

Cable selected = 2 runs of 3C x 120 sq.mm (Earthed cable)

Current rating of 3C X 120 sq.mm cable = 270 A

Derating factor = 0.68

Therefore, thermal ampacity of the conductor = Current Rating x Deration Factor x Number of runs

= 270 x 0.68 x 2

= 367.2 A

Hence, thermal ampacity is more than the full load current.

4.3 VOLTAGE DROP (OPTIONAL)

Let PF Cos θ = 0.8 , Sin θ = 0.6

Voltage drop in % = √3 x Full load Current x (Rc cos θ + Xc sin θ ) x L x 100Number of Runs x 1000 x V

Here, L = 800 m (assumed)

Rc = 0.325 W/km

Xc = 0.124 W/km

Substituting the values in the above equation, we get,

Voltage drop in % = √3 x 347.39 x ((0.325 x 0.8)+ (0.124 x 0.6)) x 800 x 1002 x 1000 x 33000

= 0.244 %

Permitted percentage voltage drop = 1%

From above results, we see that maximum steady state voltage drop calculated is less than permitted 3% ofnormal voltage (V1).




MODIFICATION RECORD

Revision Date Author DetailsA0 05 June 12 Selvan Sakthivelu




HT Cable Sizing Calculation

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