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Forces Between Bus Bars: FEA Analysis and Theoretical Calculations
1. restart :
with Physics Vectors : Setup mathematicalnotation = true :
FORCE BETWEEN BUS BARS
Camilo Chaves
Electrical Engineer and Physicist
Force between 2 infinite wires (Theory)
dF2
dl2
= i2$dl2 # B1 : Force acting on the wire 2, due to the field B1 generated by the wire1
The field in wire 1 has a simple expression. Hence, the force can be calculated as:
But the wires are infinite and the force everywhere is equal.
So the total Force is: ; note that when the currents has the
same sign, the force is attractive, and when the currents have opposite signs, it repels.
and when the 2 currents are the same, the modulus of the force is:
2. (1.1)(1.1)
(1.3)(1.3)
(1.4)(1.4)
(1.2)(1.2)
Calculating the Constant Value on the Formula
with ScientificConstants :
m0
= GetValue Constant mu 0 , system = SI $GetUnit Constant mu 0
m0
=
p
2500000
kg m
A2
s2
F =
m0
$ i A 2
2$p $d m
$L m
F =
m0
i2
L
2 p d
A 2
Substituting (1.1) in (1.2)
eval (1.2), (1.1)
F =
i2
L
5000000 d
kg m
A2
s2
A 2
Digits d 3 : sets the maximum number of digits to 3
evalf simplify (1.3)
F =
2.00 10K7
i2
L
d
N
3. FEA Analysis of the Force between 2 circular wires with the same current
Force Calculation Between 2 wires with 1 Amp current in the same direction (10cm separation)
The forces calculated on the table are from wire 1 and wire 2 , in this order. They are attractive!
4. In the simulation above, the following parameters were applied:
The current is 1Amp and they are separated by 10cm (0.01m). j0 is the current density.
According to the expression (1.4) we should have:
eval (1.4), i = 1, d = 10$10K2
= F = 2,00 #10 -6
L N
Which differs from the simulated result only because of the size of the mesh chosen!
5. Force Calculation Between 2 wires with 1 Amp current in the opposite direction (10cm separation)
The forces calculated on the table are from wire 1 and wire 2 , in this order. They are repulsive!
6. FEA Analysis of the Force between 2 rectangular conductors with the same current
Let's consider a bus bar of 3/8'' in depth and 1'' in width, it has a current capability of 516Amps.
Wid is the width of the Bus
Hei is the Depth of the Bus
J0_Bus is the current density of the Bus produced by a 1Amp current
Dist is the Distance between the geometric center of the Bus
(the rest of the parameters are for the circular wires simulation)
8. Force Calculation Between 2 Buses with 1 Amp current in the opposite direction
Horizontal Position (10cm separation)
The forces calculated on the table are from Bus 1 and Bus 2 , in this order. They are repulsive!
The result still conform to the equation (1.4), eventhough now the geometry is different.
9. Force Calculation Between 2 Buses with 1 Amp current in the opposite direction
Vertical Position (10cm separation)
The forces calculated on the table are from Bus 1 and Bus 2 , in this order. They are repulsive!
10. Force Calculation Between 2 Buses with 1 Amp current in the same direction
Horizontal Position (10cm separation)
The forces calculated on the table are from Bus 1 and Bus 2 , in this order. They are attractive!
11. Force Calculation Between 2 Buses with 1 Amp current in the same direction
Vertical Position (10cm separation)
The forces calculated on the table are from Bus 1 and Bus 2 , in this order. They are attractive!
12. (1.3.1)(1.3.1)
FEA Analysis of the Force between 3 Bus Bars in a 50KA Short-Current
J0 d 2.9227$108
:
During a 3 Phase short-current , JA = J0$cos 2$p$60$t , JB = J0$cos 2$p$60$tC
2$p
3
, JC = J0$cos 2$p$60$tK
2$p
3
JA = 2.922700000 108
cos 120 p t , JB = K2.922700000 108
sin 120 p tC
1
6
p , JC = K2.922700000 108
cos 120 p tC
1
3
p
assign (1.3.1)
13. plot JA, JB, JC , t = 0 ..
1
60
, color = red, blue, green , gridlines = true, title = "Current density of the Bus-Bars in a Short-Circuit"
t
0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016
K2. #108
K1. #108
0
1. #108
2. #108
Current density of the Bus-Bars in a Short-Circuit
14. FEA Results in comparison with theoretical values
Force Calculation Between 3 Buses with 50KAmp current
Horizontal Position (10cm separation)
The forces calculated on the table are from Bus 1, Bus 2 and Bus 3 , in this order
15. (1.3.1.4)(1.3.1.4)
(1.3.1.5)(1.3.1.5)
(1.3.1.2)(1.3.1.2)
(1.3.1.1.3)(1.3.1.1.3)
(1.3.1.1)(1.3.1.1)
(1.3.1.1.2)(1.3.1.1.2)
(1.3.1.1.4)(1.3.1.1.4)
(1.3.1.3)(1.3.1.3)
(1.3.1.1.1)(1.3.1.1.1)
I1 = 2 $50000$cos 2$p$60$t , I2 = 2 $50000$cos 2$p$60$tC
2$p
3
, I3 = 2 $50000$cos 2$p$60$tK
2$p
3
;
I1 = 50000 2 cos 120 p t , I2 = K50000 2 sin 120 p tC
1
6
p , I3 = K50000 2 cos 120 p tC
1
3
p
evalf subs t = 0, (1.3.1.1)
I1 = 70710.67810, I2 = K35355.33906, I3 = K35355.33906
assign (1.3.1.2)
Theoretical Formulas:
F12 =
2$10K7
$ I1$I2
0.1
, F13 =
2$10K7
$ I1$I3
0.2
, F23 =
2$10K7
$ I2$I3
0.1
F12 = K4999.999998, F13 = K2499.999999, F23 = 2500.000000
F11 = F12 CF13, F22 =KF12 CF23, F33 =KF13KF23
F11 = F12 CF13, F22 = KF12 CF23, F33 = KF13 KF23
eval (1.3.1.4), (1.3.1.3)
F11 = K7499.999997, F22 = 7499.999998, F33 = K0.000001
What is the maximum value of the force over time?
(1.3.1.1)
I1 = 50000 2 cos 120 p t , I2 = K50000 2 sin 120 p tC
1
6
p , I3 = K50000 2 cos 120 p tC
1
3
p
F12 =
2$10K7
$ I1$I2
0.1
, F13 =
2$10K7
$ I1$I3
0.2
F12 = 0.000002000000000 I1 I2, F13 = 0.000001000000000 I1 I3
eval %, (1.3.1.1.1)
F12 = K10000.00000 cos 120 p t sin 120 p tC
1
6
p , F13 = K5000.000000 cos 120 p t cos 120 p tC
1
3
p
eval F11 = F12 CF13, (1.3.1.1.3)
F11 = K10000.00000 cos 120 p t sin 120 p tC
1
6
p K5000.000000 cos 120 p t cos 120 p tC
1
3
p
16. (1.3.1.1.8)(1.3.1.1.8)
(1.3.1.1.5)(1.3.1.1.5)
(1.3.1.1.6)(1.3.1.1.6)
(1.3.1.1.9)(1.3.1.1.9)
(1.3.1.1.7)(1.3.1.1.7)
Diff F11, t = diff rhs (1.3.1.1.4) , t
v
vt
F11 = 1.200000000 10
6
p sin 120 p t sin 120 p tC
1
6
p K1.200000000 10
6
cos 120 p t p cos 120 p tC
1
6
p
C6.000000000 105
p sin 120 p t cos 120 p tC
1
3
p C6.000000000 105
cos 120 p t p sin 120 p tC
1
3
p
t = solve rhs (1.3.1.1.5) = 0, t, allsolutions
t = 0.008333333333 _Z5~C0.0006944444446, 0.008333333333 _Z5~ K0.003472222222
t1 = eval rhs (1.3.1.1.6) , _Z5 = 0 1
t1 = 0.0006944444446
t2 = eval rhs (1.3.1.1.6) , _Z5 = 1 1
t2 = 0.009027777778
First Minimum value: eval (1.3.1.1.4), t = rhs (1.3.1.1.7) = F11 = K8080.127020
Second Minimum value: eval (1.3.1.1.4), t = rhs (1.3.1.1.8) = F11 = K8080.127022
t3 = eval rhs (1.3.1.1.6) , _Z5 = 1 2
t3 = 0.004861111111
First Maximum value: eval (1.3.1.1.4), t = rhs (1.3.1.1.9) = F11 = 580.127018
17. plot rhs (1.3.1.1.4) , t = 0 ..
1
60
, gridlines = true, title = "Force on Bus 1 over time"
t
0.002 0.006 0.010 0.016
K8000
K7000
K6000
K5000
K4000
K3000
K2000
K1000
0
Force on Bus 1 over time
18. Checking Calculated results on the FEA solver:
Parameter time was adjusted to give maximum value on BUS BAR 1
19. Maximum Force Calculation in Bus Bar 1 with 50KAmp current
Horizontal Position (10cm separation) - time = 0.6uS after short circuit
The forces calculated on the table are from Bus 1, Bus 2 and Bus 3 , in this order
evalf convert K8084 N , units, kgf = K824.3385866 kgf
practically this is the equivalt of a mass of 824Kg/m in the Bus Bar.
20. Conclusion
Forces on Bus Bars can be calculated using F12 =
2$10K7
$ i1$i2
d
and F13 =
2$10K7
$ i1$i2
d
Take the distance d from the geometric center of the bus bar (not on the sides of the bus bar!).
The unit d has to be in meters.
With 3 buses the force between Bus 1 and 3 should also be computed.
Calculate the total force on bus 1: F11 = F12 CF13
For Bus 2 , F22 = F21 CF23 ; F21 =KF12 according to Newtows reaction law.
Compute the currents properly, that means, take the phasors in consideration.
In short-circuits the peak transient current is 2 x Irms
The results confirm that the Panel Builders must take seriously the forces involved during a short-circuit.
Next simulation will determine the distances of the isolators , so that the Copper Bus Bars do not bend!