Forces between bus bars

restart :with Physics Vectors : Setup mathematicalnotation = true :

FORCE BETWEEN BUS BARSCamilo 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 Formulawith ScientificConstants :m

0= GetValue Constant mu 0 , system = SI $GetUnit Constant mu 0

m0

=p

2500000kg m

A2 s2

F =m

0$ i A 2

2$p $d m$L m

F =m

0i2 L

2 p dA 2

Substituting (1.1) in (1.2)eval (1.2), (1.1)

F =i2 L

5000000 dkg m

A2 s2 A 2

Digitsd 3 : sets the maximum number of digits to 3evalf simplify (1.3)

F =2.00 10K7 i2 L

dN

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 NWhich 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 BusHei is the Depth of the BusJ0_Bus is the current density of the Bus produced by a 1Amp currentDist 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 directionHorizontal 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 directionVertical 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 directionHorizontal Position (10cm separation)

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 directionVertical 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

J0d 2.9227$108 :

During a 3 Phase short-current , JA = J0$cos 2$p$60$t , JB = J0$cos 2$p$60$tC2$p3

, JC = J0$cos 2$p$60$tK2$p

3

JA = 2.922700000 108 cos 120 p t , JB =K2.922700000 108 sin 120 p tC16p , JC =K2.922700000 108 cos 120 p tC

13p

assign (1.3.1)

plot JA, JB, JC , t = 0 ..160

, color = red, blue, green , gridlines = true, title = "Current density of the Bus-Bars in a Short-Circuit"

t0.002 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 valuesForce 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

(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$60$tC2$p3

, I3 = 2 $50000$cos 2$p$60$tK2$p3

;

I1 = 50000 2 cos 120 p t , I2 =K50000 2 sin 120 p tC16p , I3 =K50000 2 cos 120 p tC

13p

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$I30.2

, F23 =2$10K7$ I2$I3

0.1F12 =K4999.999998, F13 =K2499.999999, F23 = 2500.000000

F11 = F12CF13, F22 =KF12CF23, F33 =KF13KF23

F11 = F12CF13, F22 =KF12CF23, F33 =KF13KF23

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 tC16p , I3 =K50000 2 cos 120 p tC

13p

F12 =2$10K7$ I1$I2

0.1, F13 =

2$10K7$ I1$I30.2F12 = 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 tC16p , F13 =K5000.000000 cos 120 p t cos 120 p tC

13p

eval F11 = F12CF13, (1.3.1.1.3)

F11 =K10000.00000 cos 120 p t sin 120 p tC16p K5000.000000 cos 120 p t cos 120 p tC

13p

(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

vtF11 = 1.200000000 106 p sin 120 p t sin 120 p tC

16p K1.200000000 106 cos 120 p t p cos 120 p tC

16p

C6.000000000 105 p sin 120 p t cos 120 p tC13p C6.000000000 105 cos 120 p t p sin 120 p tC

13p

t = solve rhs (1.3.1.1.5) = 0, t, allsolutionst = 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 ..160

, gridlines = true, title = "Force on Bus 1 over time"

t0.002 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 currentHorizontal 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 orderevalf convert K8084 N , units, kgf =K824.3385866 kgfpractically 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

dand F13 =

2$10K7$ i1$i2d

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 = F12CF13

For Bus 2 , F22 = F21CF23 ; 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!