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N N N N: - PF Is Not Given, So Assume 0.85 and - Efficiency 88 %

A 125 KW heater requires a cable conductor size to be calculated. The nominal current is 232.48A. Selecting a 25% larger size of 290.61A or 15% larger of 267.36A. The cable type is enclosed trunking with ambient temp of 40C. The required current for the cable is 417.52A or 384.14A. A 150mm^2 copper twin cable is selected which has a current rating of 400A, satisfying the requirements. The calculated voltage drop of 8.55V is below the permissible limit of 16.6V. The minimum required cable size based on I^2*t/k formula is 52.63mm^2, which is smaller

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778 views6 pages

N N N N: - PF Is Not Given, So Assume 0.85 and - Efficiency 88 %

A 125 KW heater requires a cable conductor size to be calculated. The nominal current is 232.48A. Selecting a 25% larger size of 290.61A or 15% larger of 267.36A. The cable type is enclosed trunking with ambient temp of 40C. The required current for the cable is 417.52A or 384.14A. A 150mm^2 copper twin cable is selected which has a current rating of 400A, satisfying the requirements. The calculated voltage drop of 8.55V is below the permissible limit of 16.6V. The minimum required cable size based on I^2*t/k formula is 52.63mm^2, which is smaller

Uploaded by

Rane Siddesh
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PPT, PDF, TXT or read online on Scribd
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A 125 KW heater,415V, 3Φ, 50 Hz, connected to PCC length=100m two

other circuit is running in enclosed cable trunk ambient temperature is


40 C. Fault level=40KA. Grouping factor= 0.8
Calculate the size of conductor

• PF is not given, so assume = 0.85 and


• Efficiency = 88 %

Nominal current rating is obtained as,


KW 125
In    232.48A
3×V×η×PF 3×415  0.88  0.85
Selecting 25% extra current rating
 I nnw  1.25  I n  290.61A or
15% extra current rating
I n  1.15  I n  267.36A
• Assumption for cable conductor size calculations
– Type B ( Enclosed cable trunking) type selected
– Selecting copper conductor

Given,
Grouping Factor G = 0.8
Ambient temp. factor for 400C is t = 0.87 ( table 20)

Assuming
Thermal insulation factor T=1(no thermal insulation)
Semi enclosed fuse factor S=1(no rewirable fuse)
Tabulated current required for conductor
In new 290.61
It = = = 417.52A
GATS 0.8×0.87×1×1
OR
In new 267.36
It = = = 384.14A
GATS 0.8×0.87×1×1

• But installation method of type b does not have Ic =417.5


A conductor size for copper conductor. ( Table 14)
• So selecting 150 mm sq. PVC insulated, non armoured
Aluminium twin cable (table 20)
In table 20 find current carrying capacity
of 417 or 384A
But, for cond. cs area of 150 mm 2 , I = 200
So selecting twins cable
( i.e. 2 cables running in parallel)
So, for twin cable I=200  2 = 400A
and hence selected.
Permissible voltage drop for this cable
is given as, = 0.64 mV/A/m
For runnung condition permisible voltage drop
should be 4 to 5% of rated voltage.
i.e. 4% of 415 = 16.6 V
And with the cable selected total voltage drop is
calculated as,
V= mV/A/ m at 40 0C × length × current I n
 0.64 10 3   232.48 
=  100   
 0.87   2 
= 8.55 V (2 as twin cable used)
As, 8.55 < 16.6V so, voltage drop is permissible
• Now checking for conductor size selection, with the
2 2
condition I sc t = k A
I 2sc ×t
A=
k

K=76 as, Cable size is below 300 Sq. mm and


cable is PVC insulated Aluminium
let t= 10 m sec assumed

 40K 
2
× 10 m sec
A =
762
A = 52.63 m 2

A cable size selected is 150m 2 which is more than 52.63.


Hence design is permisible.

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