Introduction

A three phase synchronous motor is an electrical device which is used to convert electrical power

into mechanical movement. The motor consist of three windings on the stator which is called the

stator windings and one winding on the rotor which is called the field winding. Movement is

produced by the motor by interaction between magnetic fields between the windings. In order to

model the motor, a set of differential equations which has time independent variable is used.

These differential equations have current and also angular velocity as the dependent variable.

Objective

1. To plot the speed versus time of the rotor given the parameters of the motor with the

motor not loaded and frequency of 50Hz by using Euler method.

2. To evaluate the value of the steady state speed w as the frequency of the voltage supply

to the stator windings is varied.

3. To determine how the current in the windings vary when the load changes and determine

the load value that cause the rotor to stall.

Discussion

1. To solve the first task, we let wr = rotor speed and the initial conditions to be t=0, θ=0, i=0

and wr = 0

Euler method is choose to be used to determine the new value of λa, λb , λc λf , θ and wr with time

step of 0.001s

Equation:

i. λ i+1=λ i+d λdth

d λdt

=v−ri

ii. θi+1=θi+dθdth

dθdt

=wr

iii. w r+1=w r+dwr

dth

dwr

dt=T e−Bmwr

I

Linear equations below is used to find the new value of ia ib ic and if by using matrix to solve it.

Laa ia+Lab ib+Lac ic+Laf if=λa

Lba ia+Lbb ib+Lbc ic+Lbf if=λb

Lca ia+Lcb ib+Lcc ic+ Lcf if=λc

Lfa ia+L fb ib+Lfc ic+L ff if=λ f

Laa=Ls+Lmcos (2θ)

Lbb=Ls+Lmcos (2(θ−2π3 ))

Lcc=L s+Lmcos (2(θ+2 π3 ))

Lab=−M s−Lmcos (2(θ+ π6 ))

Lbc=−M s−Lmcos (2(θ−π2 ))

Lca=−M s−Lmcos (2(θ+ 5 π6 ))

2. The result for task 1 did not reach steady state speed because the time step chosen is not

small enough to show it.

3. The actual steady state speed through calculation

The V will constant at time step, ∆ t=0.001

va=240 cos (ωt ) ;ω=2πf

¿240cos (2π∗50∗0)

¿240V

vb=240 cos (ωt−2π3 )

¿240 cos (2πx 50 x0−2 π3

)

¿−120V

vc=240 cos(ωt+ 2 π3 )

¿240 cos (2πx 50 x0+ 2π3

)

¿−120V

The Laa , Lbb , Lcc also is constant when ∆ t=0.001

Laa=Ls+Lmcos (2θ)

¿0.000309+0.00321 cos (2∗2π∗50∗0 )

¿0.003519

Lbb=Ls+Lmcos (2(θ−2π3 ))

¿0.000309+0.00321 cos (2(2 π∗50∗0−2 π3

))¿−0.001296

Lcc=L s+Lmcos (2(θ+2 π3 ))

¿0.000309+0.00321 cos (2(2 π∗50∗0+ 2π3

))¿−0.001296

The Lab , Lbc , Lca also is constant when ∆ t=0.001

Lab=−M s−Lmcos(2(θ+ π6 ))=−0.004715

Lbc=−M s−Lmcos (2(θ−π2 ))=0.0001

Lca=−M s−Lmcos (2(θ+ 5π6 ))=−0.004715

The equations for the magnetic flux are show below:

Laa ia+Lab ib+Lac ic+Laf if=λa

Lba ia+Lbb ib+Lbc ic+Lbf if=λb

Lca ia+Lcb ib+Lcc ic+ Lcf if=λc

Lfa ia+L fb ib+Lfc ic+L ff if=λ f

In order to find all of the current, i, we need to solving the λ

v=ri+ dλdt

λ=∫ (v−ri )dt

λa= ∫0

0.001

(va−r ia )dt ; va=240 cos (ωt ) ;ω=2πf

¿ ∫0

0.001

(240 cos (ωt)¿−r ia)dt ¿

¿ 240ω

[sin (2πft )]00.001−¿

¿0−0.00000204 ia

¿−0.00000204 ia

λb= ∫0

0.001

(vb−r ib )dt ; vb=240 cos (ωt−2π3 )

¿ ∫0

0.001

(240 cos (ωt−2 π3 ))−r ib¿ dt

¿−0.00000204 ib

λc= ∫0

0.001

(vc−r ic )dt ; vc=240 cos(ωt+2 π3 )

¿ ∫0

0.001

(240 cos (ωt+ 2π3 ))−r ic ¿dt

¿−0.00000204 ic

λ f= ∫0

0.001

(v f−r if )dt ;v f=100

¿10−0.00000204 if

Substitute all of the finding into the magnetic flux equations

0.003519 ia−0.004715 ib−0.004715ic+0.00311 if=−0.00000204 ia

0.00352104 ia−0.004715 ib−0.004715 ic+0.00311 if=0−−−−−−−−(1)

−0.004715 ia−0.001296 ib+0.0001 ic−0.001555 if=−0.00000204 ib

−0.004715 ia−0.00129396 ib+0.0001 ic−0.001555 if=0−−−−−−−−(2)

−0.004715 ia+0.0001 ib−0.001296 ic−0.001555 if=−0.00000204 ic

−0.004715 ia+0.0001 ib−0.00129396 ic−0.001555 if=0−−−−−−−−(3)

0.00311 ia−0.001555ib−0.001555 ic+0.0003 if=10−0.00000204 if

0.00311 ia−0.001555ib−0.001555 ic+0.00030204 if=10−−−−−−−−(4)

Solve the equations simultaneously

(0.00352104 −0.004715 −0.004715−0.004715 −0.00129396 0.0001−0.004715 0.0001 −0.00129396

0.003111 −0.001555 −0.001555

0.003111−0.001555−0.0015550.00030204

| 000

10)

ia=2586.068831 A

ib=−1293.613883 A

ic=−1293.613883 A

if=−6848.097304 A

Therefore, the magnetic flux are given by

λa=−0.00000204 ia

λa=−0.00000204∗2586.068831 A=−0.00527558

λb=−0.00000204∗−1293.613883 A=0.00263897

λc=−0.00000204∗−1293.613883 A=0.00263897

λ f=10−0.00000204 i f

λ f=10−0.00000204∗−6848.097304 A=10.01397012

Substitute all of the λ to find the torque produced by the rotor T e , T e will constant when

∆ t=0.001

T e=λa λ f cosθ+ λb λ f cos(θ−2π3 )+λc λ f cos(θ+2 π

3 )

¿−0.00527558 (10.01397012 ) cos (2πft )+0.00263897 (10.01397012 ) cos (2πft−2 π3 )

+0.00263897 (10.01397012 ) cos (2πft+ 2π3 )

¿−0.07925Nm

The mechanical equation governing the movement of the rotor is given as follows:

Id ωr

dt=Te−T l−Bmωr

The question mention that, the motor is not loaded T l=0

d ωr

dt=T e−Bmωr

I

w r=(Te−Bmw r

I ) t=0

4. For second task, the results show the higher the frequency, the higher the steady state speed

ω. The change in frequency only effect the change of steady state speed, it does not affect the

voltage of voltage supply.

Conclusion

In this project, the Excel implementation of Euler methods was described and the speed versus

time of the rotor was plotted. It was given the parameters of the motor with frequency of 50Hz

and the motor was not loaded by using Euler method. Time step of 0.001 was chosen in the

stimulation so that the solution of ordinary differential equations was converging. The results did

not reach steady state speed shown that the time step that used in the stimulation was too large.

The value of the steady state speed as the frequency of the voltage supply to the stator windings

is varied was evaluated. The current in the windings will change as the load changes. The load

value that will cause the rotor to stall was determined.