INDEX S.No. 1
Description CALCULATIONS FOR BUSBAR & CONDUCTOR SIZING
1
Rev A
REV A
INPUT LIST FOR CALCULATION OF CONTINUOUS CURRENT CARRYING CAPACITY OF PANTHER CONDUCTOR. S.No. Description 1
Conductor Type
2
The value of DC resistance RT at 200 C is
Symbol
Value
Unit
-
Panther
-
RT
0.136
Ω/km
k-1
(As per IS 398(P-II) 3
Temperature coefficient of electrical resistance (As per IEC 1597 clause 4.2-Page-23)
α
0.00403
4
Initial temperature
T1
20 293
0
5
Final temperature
T2
75 348
0
6
Frequency
f
50
7
Factor determined by conductor construction 1 for circular, stranded, compacted and sectored.
µr
1
8
The solar radiation absorption coefficient
γ
0.5
9
Diameter of conductor
D
0.021
m
10
Intensity of solar radiation
Si
900
W/m2
11
Stefan - Boltzmann constant
S
5.67E-008
W.m-2.K-4
12
Emissivity coefficient in respect to black body
Ke
0.45
T1
50 323
C K C K
Hz
1.0 for Black body 0.5 for average oxidized copper 0.35-0.45 for naturally weathered (i.e. dulled surface) Increase with aging 0.15-0.25 for extruded surface Increase with aging 13
Ambient Temperature
14
Cross Wind speed (minimum-90' to the Line)
v
1
m/s
15
Thermal conductivity of the air film in contact with the conductor
λ
0.02585
W.m-1.K-1
(As per IEC 1597 clause 3.7)
2
C K
0
CALCULATION FOR CONTINUOUS CURRENT CARRYING CAPACITY OF PANTHER CONDUCTOR. 4.1
General The current carrying capacity (CCC) of a conductor is the maximum steady state current inducing a given temperature rise in the conductor , for given ambient conditions.
4.2
The heat balance equation is : The steady state temperature rise of a conductor is reached whenever the heat gained by the conductor from various sources is equal to the heat losses. This is expressed by equation (1) Pj + Psol = Prad + Pconv ----------------- (1) (As per IEC 1597 equation no (1)) Where, Pj is the heat generated by joule effect Psol is the solar heat gain by the conductor surface Prad is the heat loss by radiation of the conductor. Pcov is the convention heat loss.
4.3
Joule effect Power losses Pj (W) , due to Joule effect are given by equation (2) Pj = RT I2 ----------------- (2) (As per IEC 1597 equation no (2))
Where,
RT is the electrical resistance of conductor at a temperature T (Ω/m) I is the conductor (A) 1.36E-04 Ω/m
The value of DC resistance RT at 20 0C is The DC resistance of a conductor at a temperature T2 is given by :
RT2
4.4
----------------- (3) (As per IEC1597 equation (9) )
RT2 =RT1 [ 1+α (T 2−T 1 )]
Hence ,
1.66E-04 Ω/m 2.67E-01 Ω/miles
=
Calculation of AC resistance. If the conductor is composed of one or more concentric circular elements, then the centre portion of the conductor will be enveloped by a greater magnetic flux than those on the outside. Consequently the self induced back-emf will be greater towards the centre of the conductor, thus causing the current density to be less at the centre than the conductor surface. This extra concentration at the surface is known as skin effect, and results in an increase in the effective resistance of the conductor.
Now for the calculation of skin effect coefficient ,a factor X is defined as
X S =0 . 063598×
√
μr × f R dc
----------------- (4)
Hence, Xs
=
8.70E-01
For this value of Xs the skin effect coefficient obtained from Table X
K
X
K
X
K
X
K
0.0 0.1 0.2 0.3 0.4 0.5
1.00000 1.00000 1.00001 1.00004 1.00013 1.00032
1.0 1.1 1.2 1.3 1.4 1.5
1.00519 1.00758 1.01071 1.01470 1.01969 1.02582
2.0 2.1 2.2 2.3 2.4 2.5
1.07816 1.09375 1.11126 1.13069 1.15207 1.17538
3.0 3.1 3.2 3.3 3.4 3.5
1.31809 1.35102 1.38504 1.41999 1.45570 1.49202
3
0.6 0.7 0.8 0.9
1.00067 1.00124 1.00212 1.00034
1.6 1.7 1.8 1.9
1.03323 1.04205 1.05240 1.06440
2.6 2.7 2.8 2.9
1.20056 1.22753 1.25620 1.28644
For X =1.2
K
=
1.01071
For X =1.3
K
=
1.01470
For X = K =
8.70E-001 1.01271
Therefore AC resistance at T2 temperature
Rt2
=
K X RT2 =
4.5
1.68E-04 Ω/m
The Solar heat gain Solar heat gain ,Psol is given by equation Psol = γ D Si (As per IEC1597 equation (3) )
----------------- (5)
Where , γ is the solar radiation absorption coefficient D is the diameter of the conductor Si is the intensity of solar radiation Hence Psol = 4.6
9.45 W/m
Radiated heat Loss Heat loss by radiation , Prad (W) is given by equation Prad = s π D Ke (T24 - T14) (As per IEC1597 equation (4) )
----------------- (6)
Where , s is the Stefan - Boltzmann constant ( 5.67 X 10 -8 W.m-2.K-4) D is the conductor diameter (m) Ke is the emissivity coefficient in respect to black body T is the temperature (K) T1 is the ambient temperature (K) T2 is the final temperature Hence , Prad 4.7
3.6 3.7 3.8 3.9
=
6.3624522 W
Convention heat loss Only forced convention heat loss, Pconv (W) , is taken into account and is given by equation Pcov = λ Nu (T2-T1) π (As per IEC1597 equation (5) )
----------------- (7)
Where, λ is the thermal conductivity of the air film in contact with the conductor, assumed constant . Nu is the Nusselt number , given by equation Nu = 0.65 Re 0.2+0.23.Re 0.61
----------------- (8) 4
1.52879 1.56587 1.60314 1.64051
(As per IEC1597 equation (6) ) Where, Re is the Reynolds number given by equation Re = 1.664 . 109 v D [(T1+0.5(T2-T1)]-1.78 (As per IEC1597 equation (7) )
----------------- (9)
Where, v is the wind speed in m/s D is conductor diameter (m) T is the temperature (K) T1 is the ambient temperature (K) T2 is the final temperature Hence, Re
=
1115.9376
By putting Re value in equation No 8 Nu
=
19.271111
By putting Nu value in equation no 7 Pconv = 4.8
39.105421 W
Steady state current carrying capacity (CCC) The steady -state current capacity can be calculated by : Imax = [ (Prad+Pconv-Psol)/RT2]0.5
----------------- (10)
Where, Prad =
6.36245223 W
Pconv=
39.1054207 W
Psol = Rt2
=
9.45 W/m 1.68E-04 Ω/m
Hence single Panther conductor can carry continuous current I =
4.9
i ii iii
462.67
A
3.2 3.2 33
MVA MVA kV
56
A
Required continuous current rating of 33 KV Bus For Trafo Bay Maximum Power Output of Two Trafo Net Power Output of Two Trafo. Secondary Rated Voltage Total current =
MVA RATING X 1000 RATED VOLTAGE X 1.73
4.10 Conclusion As per Above calculation the required continous current for bus is 56Amps. And Single PANTHER conductor can take 462Amps current so the conductor size is sufficent to take such load.
5
SHORT TIME WITHSTAND CAPABILITY 1) SHORT TIME WITHSTAND CAPABILITY FOR PANTHER CONDUCTOR Reference Document : (IEC 865 PART-I) Short time withstand capability for aluminium conductor (ACSR) can be calculated by equation
S thr =
K I = √T kr A
(As per IEC 865 part-I clause A9 figure 13 page no.113) Where , Sthr is rated short -time withstand current density (r.m.s.) for 1 sec. Tkr is Rated short time. K is factor for calculating Sthr which is given by equation
√
κ 20 ×C× ρ 1+ α 20(θ c−20 0 C ) K= In 0 α 20 1+ α 20 (θ b−20 C ) Where, θb is the conductor temperature at beginning of short circuit.
75
0
C
θc is the conductor temperature at end of short circuit.
200
0
C
Where , As per IEC 865 part-I clause A9 figure 13 page no.113 table for aluminium conductor) 7
Κ20
3.48 x 10
/Ωm
Κ20
=
C
=
910 J/kg0C
ρ
=
2700 kg/m3
Specific Mass
α 20
=
0.004 1/0C
Temp. Coefficient
hence K is
3.48E+07 1/Ωm
Specific Conductivty at 20 deg.C. Specific Thermal Capacity
8.57E+07 A*s/m2
For Single Panther Cross-sectional Area for Single Panther
2.61E-04 m2
Short time withstand capability for single Panther
2.24E+04 A
for 1 sec
Short time withstand capability for Single Panther
22373
for 1 sec
A
Conductor Fail In Short Circuit Required Short time withstand capability 25kA for 1 sec. (As per Tech. Specification)
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