Forces between bus bars

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:

(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

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!

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!

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!

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)

Table of current capacity in Bus Bars

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.

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!

Force Calculation Between 2 Buses with 1 Amp current in the same direction
The forces calculated on the table are from Bus 1 and Bus 2 , in this order. They are attractive!

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!

(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)

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

FEA Results in comparison with theoretical values
Force Calculation Between 3 Buses with 50KAmp current
The forces calculated on the table are from Bus 1, Bus 2 and Bus 3 , in this order

(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

(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

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

Checking Calculated results on the FEA solver:
Parameter time was adjusted to give maximum value on BUS BAR 1

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.

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!

Forces between bus bars

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