s
BUSBAR SIZING CALCULATION
FOR SWITCHYARD CONDUCTOR
Index of Revisions
1
0
07.04.11
30.12.10
Neeraj Kumar
Neeraj Kumar
S.Jain
S.Jain
Revised as per comments
First Issue
Rev.
Date
Coord.
Checker
Details of Revision
JINDAL INDIA THERMAL POWER LIMITED
2x600 MW/600 MW+10% THERMAL POWER PROJECT AT DERANG, ANGUL,
ORISSA
Date
Prepared by
Coord. by
Checked by
Department
29.12.10
29.12.10
29.12.10
Name
Manish Gupta
Neeraj Kumar
Saurabh Jain sgd.
SPEL-EN 18
Document No.
Type of Document
SIEM-JITPL.5404A-EL-3-2026
Rev. No.
1
Ver.
1
SIEM-JITPL.5404A-EL-3-2026
s
Contents
1
Introduction
3
2
System Data
3
3
Conductor
4
4
Other Referred Inputs
4
5
Attachments
5
6
Conclusion
5
7
References
5
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
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1
Introduction:
The thermal calculation for conductors may be broadly divided into two categories:
a. Continuous Current Carrying Capacity
The current carrying capacity of a conductor is the maximum steady-state current inducing a
given temperature rise in the conductor, for given ambient conditions. The continuous current
carrying capacity depends on the type of conductor, its electrical resistance, the maximum
allowable temperature rise and ambient conditions. The conductor ampacity is limited by the
conductor’s maximum operating temperature. The temperature of a conductor depends upon
the balance of heat input and output. The conductor in an outdoor switchyard gains heat
mainly from solar radiation incident on it and the heat generated by Joule effect. Conductor
looses heat by radiation and convection from its surface. This relationship is expressed in the
Heat balance Equation which in turn gives the value of allowable magnitude of current
through the conductor (for details refer detail calculation).
b. Short Circuit Current Carrying Capacity
A conductor’s temperature will rise rapidly under fault conditions. This is due to the inability of
the conductor to dissipate the heat as rapidly as it is generated. The magnitude and duration
of short circuit fault current that can be sustained by the conductor depends upon the
electrical conductivity and area of cross-section of the conductor withstanding the fault
current. This fault current withstanding capacity is limited to the maximum allowable
temperature rise of the conductor based on its material properties beyond which the
conductors become vulnerable to annealing. In the present calculation, the conductor area
necessary for withstanding the fault current for the given time duration is determined
0
considering final conductor temperature limited to 200 C.
2
System Data:
Switchyard
1
2
3
4
5
6
Nominal System Voltage
System Frequency
Short Circuit Fault Current at Bus
Duration of Fault Current
Design Ambient Temperature
Maximum Operating Temperature
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
400
50
50
1
50
85
kV
Hz
kA
sec
°C
°C
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3
Other Referred Inputs:
A. Calculation for Flexible Conductor
1
Solar Radiation Absorption Coefficient
2
Intensity of Solar Radiation at Site
3
Emissive Coefficient of Conductor with
respect to Black Body
4
Minimum Wind Speed at Site
5
Temperature Coefficient of Electrical
Resistance of Conductor
0.5 (as per Cl. No. 3.10 of IEC 1597 :
1995)
1050 W / m 2
0.5 (as per Cl. No. 3.10 of IEC 1597 :
1995)
1 m/s
0.00403 K  1
B. Calculation for Rigid Conductor
1
2
3
4
4
Solar Radiation Absorption Coefficient
Maximum Operating Temperature of
Conductor
Emissive Coefficient of Conductor with
respect to Black Body
Conductivity of Al Alloy Tube
0.5 ( Table-C.5 IEEE605)
85°C
0.5 (Table-C.5 IEEE605)
55 % IACS
Conductor Data:
A. Flexible Conductor
AAC"TARANTULA"
1
Conductor Diameter
2
Total Cross-sectional Area
3
DC Resistance at 20 °C
36.6
794.8
0.0363
mm
mm 2
 / km
B. Rigid Conductor
4 1/2" Heavy
1
Outer Diameter
2
Wall Thickness
3
Conductivity of Al Tube
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
120
12.00
55
mm
mm
% IACS
Page 4 of 5
SIEM-JITPL.5404A-EL-3-2026
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5
Attachments:
Attachment1 : Calculation of continuous current carrying capacity for flexible conductor
(AAC"TARANTULA")
Attachment2 : Check for circuit current withstand capicity of flexible conductor (Twin
AAC"TARANTULA")
Attachment3 : Calculation of continuous current carrying capacity of Tubular Conductor (4
1/2" Heavy)
Attachment4 : Check for circuit current withstand capicity of Tubular Conductor (4 1/2"
Heavy)
6
Conclusion:
The following table summarize the result:
a. Flexible Conductor
Conductor Type
Twin
AAC"TARANTULA"
Fault
Current Carrying Operating
Temperature Current (kA)
Capacity (A)
2460
85
Duration of Fault
Current (sec)
1
50
b. Rigid Conductor
Conductor Type
4 1/2" Heavy
7
Current Carrying Operating
Fault
Capacity (A)
Temperature Current (kA)
2823.70
85
Duration of Fault
Current (sec)
50
1
References:
1. IEC 1597:1995: Overhead Electrical Conductors - Calculation Methods for Stranded Bare
Conductors.
2. IEC 865:1993: Short-Circuit Currents - Calculation of Effects
3. IEEE Std 605: 1998 IEEE Guide for design of Substation Rigid-Bus Structures
4. Technical Specification for 400 kV Substation
5. ABB T&D Book.
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
Page 5 of 5
SIEM-JITPL.5404A-EL-3-2026
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Attachment 1
Calculation of continuous current carrying capacity for flexible conductor
Reference Document : IEC 1597.
Designation of OHL Conductor
:
AAC"TARANTULA"
Design Inputs :
a.
b.
c.
d.
e.
Overall Diameter of conductor
Cross-sectional area
Design Ambient Temperature
Maximum Operating Temperature
Temperature coefficient of resistance
D
A
Ta
Tf

1. The heat balance equation is :
=
=
=
=
=
36.6
794.8
50
85
0.00403
mm
mm2
0
C
0
C

Pj + Psol = Prad + Pconv
……(1)
where,
Pj is the heat generated by Joule effect
W/m
Psol is the solar heat gain by the conductor surface
W/m
Prad is the heat loss by radiation of the conductor
W/m
Pconv is the convection heat loss
W/m
2. The power loss, Pj, due to Joule effect is given by :
Pj = RT . I 2
……(2)
where,
RT is the electrical resistance of conductor at a temperature T (/m)
I is the conductor current (A)
The value of DC resistance for the conductor at 20°C
=> The value of DC resistance for the conductor at 85°C
=
= R20[1 + (Tf-20)]
=
=
Now,
where,
0.0363 /km
3.6300E-05 /m
4.5809E-05 /m
7.3722E-02 /mile
from Electrical T & D Ref. Book - ABB
X=
0 .063598 f RDC
=
 =
f =
=
=
1.0 (non-magnetic materials)
50 Hz
=
RT =
1.0382 (Table 5, Ch.3 of Ref. 5)
4.7558E-05 /m
permeability
frequency
='> Skin Effect ratio, K, corresponding to the value of X
Therefore AC resistance at 85°C is
1.6563 (Rdc in /mile)
The value of RT will be substituted in equation 2 and value of I will be found. Refer Equation 8 below.
3. The solar heat gain, Psol, is given by :
Psol =  . D . S i
where,
 is the solar radiation absorption coefficient
=
D is the conductor diameter
S i is the intensity of solar radiation
=
=
Hence,
Psol =
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
……(3)
0.5
0.03660 m
2
1050.00 W/m
19.22 W/m
Page 1 of 2
SIEM-JITPL.5404A-EL-3-2026
s
Prad = s .  . D . K e . (T24 - T14)
4. Heat loss by radiation, Prad, is given by :
where,
s is the Stefan-Boltzmann constant
D is the conductor diameter
K e is the emissivity coefficient in respect to black body
T 1 is the ambient temperature
=
50°C
T 2 is the final equilibrium temperature
=
85°C
=
=
=
=
=
Prad =
Hence,
5.67E-08
0.03660
0.5
323
358
……(4)
W.m-2.K-4
m
K
K
18.06 W/m
Pconv =  . Nu . (T2 - T1) . 
……(5)
Nu is the Nusselt number and is given by the equation :
Nu = 0.65Re0.2 + 0.23Re0.61
……(6)
Re is the Reynolds number and is given by equation :
Re = 1.644x109.v .D [T1+0.5(T2-T1)]-1.78
……(7)
5. Convection heat loss is given by :
where,
 is the thermal conductivity of the air film
in contact with the conductor
T 1 is the ambient temperature
=
=
T 2 is the final equilibrium temperature
v is the wind speed (minimum)
D is the conductor diameter
=
-1
-1
0.02585 W.m .K
=
=
=
=
323
358
1.00
0.03660
=>
Re =
1871.606
=>
Nu =
25.725
50°C
85°C
Hence,
Pconv =
K
K
m/s
m
73.12 W/m
I = [(Prad + Pconv - Psol)/RT]1/2
6. The steady-state current carrying capacity is :
……(8)
Where,
Hence, a Single conductor
can carry a continuous current
Prad
Pconv
Psol
RT
=
=
=
=
I

18.06
73.12
19.22
4.7558E-05
W/m
W/m
W/m
/m
1230 A
Therefore,
TWIN AAC"TARANTULA" conductor is capable of carrying a continuous current
=
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
2460 A
Page 2 of 2
SIEM-JITPL.5404A-EL-3-2026
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Attachment - 2
Check for circuit current withstand capicity of flexible conductor
Reference Document : IEC 865
Designation of OHL Conductor
:
Twin AAC"TARANTULA"
I th  I k'' m  n
Thermal equivalent short-time current is given by:
I k''  I k
with
(IEC865-1, equation-63)
See Section 4.6.3, IEC 60909-0
where:
I th is thermal equivalent short-time current (r.m.s, in A)
I k'' is initial symmetrical short-circuit current (r.m.s, in A)
I k is steady state short-circuit current (r.m.s, in A)

m is factor for the heat effect of the d.c. component
n is factor for the heat effect of the a.c. component
Factor m is given by:
m

(IEC865-1,A.8)

1
e 4 fTk ln(  1)  1
2 fk ln(  1)
where:
 is factor for calculating peak short-circuit current
f is frequency in Hz
Tk is duration of short-circuit current (sec)
Factor n is given by:
2
2
''
'

'  I
'  I

T'
I' 
T'
1  d 1  e  20 Tk / Td  k  k   d 1  e  2Tk / Td  k  1 
2Tk
 20Tk

 Ik
 Ik Ik 
 '
''
'
'
'
'
'
 Td 1  e 10 Tk / Td  I k  I k   2Td 1  e Tk / Td  I k  1 

I

I
 5Tk

 k
 k I k  Tk

'
''
'
'
 Td
'  I

I  I
1  e 10.1Tk / Td  k  k  k  1


 I k I k  I k
 5.051Tk

n
I
1
''
k
/ Ik

2



















with
I k'
I k'' / I k

Ik
0 . 88  0 .17 I k'' / I k
T d' 
3 .1
I / Ik
'
k
Thermal equivalent short-circuit current density is given by:
S th 
I th
Ac
………………(6)
where
Ac is the cross-sectional area of aluminum conductor
Bare Conductor have sufficient thermal short-circuit strength as long as the following relation holds:
S th  S thr
T kr
Tk
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
(IEC865-1, equation-67)
Page 1 of 2
SIEM-JITPL.5404A-EL-3-2026
s
with
S thr 
K 
where
K
(IEC865-1, equation-69)
T kr
 20 c 1   20  e  20 C 
ln
 20
1   20  b  20 C 
(IEC865-1, A.9)
S thr is the rated short-time withstand current density (r.m.s) for T kr sec.
Tkr is rated short-time
 e is conductor temperature at the end of a short circuit
 b is conductor temperature at the beginning of a short circuit
For aluminum alloy and ACSR conductor with base temperature of 20 °C
 20
c

 20
=
34800000
=
910
=
2700
=
0.00403
Now for the given system:
Conductor
No. of subconductors
f
e
b
Tkr
Tk

Ik
I k per sub-
conductor
I k'' per subconductor
=
=
Twin AAC"TARANTULA"
2
=
50 Hz
=
200 °C
=
85 °C
=
=
1 sec
1 sec
=
1.8
=
50 kA
=
25 kA
=
25 kA
m
=
I k' / I k
Td'
n
=
I th
Ac
=
25.554038 kA
=
=
794.8 mm 2
0.0007948 m 2
S th
=
32151.533 kA / m 2
K
=
81465472
S thr
=
81465472 A / m
=
81465.472 kA / m 2
=
=
0.045
1
3.1 sec
1
(IEC865-1,1993, Figure 12b)
2
Since S th < S thr , Twin AAC"TARANTULA"Twin conductor can safely withstand fault current of 50kA for 1 sec
with final conductor temperature of 200 °C.
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
Page 2 of 2
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Attachment - 3
Calculation of continuous current carrying capacity of Tubular Conductor
Reference Document : IEEE 605-1998
Designation of tube
:
4 1/2" Heavy
Design Inputs :
a.
b.
c.
d.
e.
g.
Outer Diameter of tube
Wall Thickness
Cross-sectional area
Design Ambient Temperature
Maximum Operating Temperature
Conductivity of Al Tube
=
=
=
=
=
=
Ta
Tf
C'
120
12
4071.50
50
85
55
mm
mm
mm2
°C
°C
% IACS
I2RF+qs=qc+qr+qcond
1. The heat balance equation is :
……(1)
where,
I is the current for allowable temperature rise (A)
R is direct current resistance at the operating temperature (/ft)
F is the skin effect coefficient
qs is the solar heat gain (W/ft)
qc is the convective heat loss (W/ft)
qr is the radiation heat loss (W/ft)
qcond is the conductive heat loss (W/ft)
2. The DC resistance and skin effect coefficient at operating tempreature is calculated as :
Outer Diameter of tube
Thickness of Tube
d =
t =
4.728 inch
0.47 inch
Now, as per C.3.2.9 IEEE 605, the resistance of aluminium conductor at a temperature is expressed as :
-4
RT2
= (8.145*10 /C'A2)*[1+0.00403*C'(T2-20)/61]
where
C'
A2
=
=
Conductivity of Al conductor in % IACS
Cross-sectional area of conductor in in2
2
6.31 in
85 °C
=
T2
=
Hence for 85°C temperature, the value of RT2 is calculated as :
RT2
Skin effect ratio
=
2.85E-06 /ft
F  1  3 .409  10 13  2.078  3.92  3 .32  10 21  4.55  
……(2)
valid for 0.05  t / d  0.4
with
  t/d ,  
f / R 
where,
f is system frequency
=
50
X
Y
Z
=
=
=
0.1
2313.568
6.499406
and hence skin effect ratio
T2
and
  6 .50  1 .27  3.33
2000 
 f / R   3000
T2
Hz
(RT2 in /m)
F
=
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
1.4323
Page 1 of 2
SIEM-JITPL.5404A-EL-3-2026
s
qs = 0.00695*6*Qs*A9*K*sin()
3. The solar heat gain, qs, is given by :
where,
6 is the solar radiation absorption coefficient
= effective angle of incidenceof Sun =
cos-1*[cosHc*cos(Zc-Z1)]
considering
Hc
=
0.5
=
90 deg
=
88 deg
Zc
=
180 deg
Z1
=
90 deg
2
97.55 W/ft
2
56.70 in /ft
Q s is the intensity of solar radiation
A' is projected area of conductor,Square inches per foot
=
=
= 12*sin* conductor size
K = Heat multiplying factors for high altitudes
=
Hence,
qr = 36.9*10-12**A (Tc4 - Ta4)
where,
A is the conductor surface area
 is the emissivity coefficient in respect to black body
T a is the temperature of
50 °C
surrounding bodies in kelvin
T c is the maximum operating
85 °C
temperature of conductor
Hence,
……(4)
=
=
=
2
178.24 in /ft
0.5
323 K
=
358 K
qr =
18.22 W/ft
qc =  d-0.4*A*T
5. Forced convection heat loss is given by :
……(5)
heat transfer is considered at wind speed 2fps and 1 atm pressure
where,
T difference in temperature in deg C
=
between conductor surface and ambient
qc =
Hence,
35.0 °C
33.51 W/ft
I = [(qr + qc - qs)/RF]1/2
6. The steady-state current carrying capacity is :
qcond is neglected for practical purposes
Where,
qr
qc
qs
R85F
Hence, a Tubular Bus Conductor
can carry a continuous current
withstanding a 35 °C temperature rise above ambient
I
=
=
=
=

2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
(As per TableC.1, Pg-57
: IEEE 605 1998)
1.00 (Table-C.1, Pg-58
IEEE 605-1998)
19.22 W/ft
qs =
4. Heat loss by radiation, qr, is given by :
……(3)
18.22
33.51
19.22
4.08E-06
……(6)
W/ft
W/ft
W/ft
/ft
2824 A
Page 2 of 2
SIEM-JITPL.5404A-EL-3-2026
s
Attachment - 4
Check for circuit current withstand capicity of Tubular Conductor
Reference Document : IEE 605-1998
Designation of tube
:
4 1/2" Heavy
Maximum Fault Current that can be allowed to flow through the tubular conductor with it's final temperature within
the limits determined by it's mechanical properties is given by :I
where,
C x 106 A [1/t log10{Tf - 20 + (15150/G) / Ti - 20 + (15150/G)}]1/2 A
………………(1)
I = Maximum allowable root mean square value of fault current (A)
C
=
0.144
A = Conductor cross-sectional area (in2)
G = Conductor conductivity (% International Annealed Copper Standard, IACS)
t = Duration of fault (s)
Tf = Allowable final conductor temperature (°C)
=
Ti = Conductor temperature at fault initation (°C)
Therefore, for a system with given fault level, the cross-sectional area of the conductor suited to withstand the fault
current for the defined duration and within the allowable final temperature is determined from the equatioin above a
From eqn ..(1), the expression for area of conductor is
A
=
(1/C ) x 10-6 I [t / log10{Tf - 20 + (15150/G) / Ti - 20 + (15150/G)}]1/2 in2
Now, for the given system the values are :I
=
G = C'
=
t
=
Tf
=
=>
50
55.00
1
200
kA
% IACS
sec
°C
Ti
=
50
°C
A
=
0.83
in2
Therefore, required Cross-sectional Area of the conductor suited to withstand the fault current for the
defined duration and within the allowable final temperature of the conductor
A
=
537.81
mm2
4071.50
mm2
Now, Cross-sectional Area of selected Al Tube
At
=
At
>
A
Hence, actual cross-sectional area of the tube, At (4071.5 mm2) is more than the required cross-sectional area
A ( 537.81 mm2) Thus selected Al. Tube is capable of withstanding the Short Circuit Fault Current safely.
2x600 MW Thermal Power Plant Project at Derang, Angul, Orissa
Page 1 of 1