GE Energy Power Conversion
XXXXXXX XXXXXXX HT CABLE SIZING CALULATION
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ChangeNo Selvan Sakthivelu Initial Issue
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GE Energy Power Conversion
CONTENTS No table of contents entries found.
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GE Energy Power Conversion 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 temperature variation, 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 motor feeders .
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 √Eme k k = 0.094 (constant for Aluminium) Minimum cross section required = 26.2 x √0.16 0.094
Customer Project Document File
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GE Energy Power Conversion = 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 100 Number 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 100 2 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% of normal voltage (V1).
Customer Project Document File
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GE Energy Power Conversion
MODIFICATION RECORD
Revision Date A0 05 June 12
Customer Project Document File
Author Selvan Sakthivelu
: xxxxxxx : xxxxxxx : HT CABLE SIZING CALULATION : xxxxxxx / Document1
Details
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