Handbook of Electrical Engineering

498 HANDBOOK OF ELECTRICAL ENGINEERING

Similarly (20.25) becomes:









Vd1

Vq1

0

0







=

           

R 1 +j

ω

ωn

XL 1 0 j

ω

ωn

Xm 0

0 R 1 +j

ω

ωn

XL 1 0 j

ω

ωn

Xm

j

ω

ωn

Xm

ωr

ωn

Xm R 2 +j

ω

ωn

XL 2

ωr

ωn

Xm

−

ωr

ωn

Xm j

ω

ωn

Xm −

ωr

ωn

Xm R 2 +j

ω

ωn

XL 2

           









Id1

Iq1

Id2

Iq2







 (^20.^31 )

WhereVd1,Vq1,Id1,Iq1,Id2andIq2are the phasor equivalents of their instantaneous values,
XL 1 is the total reactance of the primary andXL 2 that of the secondary. For the balanced three-phase
operation of the induction motor the following discussion applies.









Vd1

−jVd1

0

0







=

           

R 1 +j

ω

ωn

XL 1 0 j

ω

ωn

Xm 0

0 R 1 +j

ω

ωn

XL 1 0 j

ω

ωn

Xm

j

ω

ωn

Xm

ωr

ωn

Xm R 2 +j

ω

ωn

XL 2

ωr

ωn

Xm

−

ωr

ωn

Xm j

ω

ωn

Xm −

ωr

ωn

Xm R 2 +j

ω

ωn

XL 2

           









Id1

−jId1

Id2

−jId2









( 20. 32 )

Consider the first and third rows in (20.32), and define the rotor slip speed assω=ω−ωr.The
comments following (20.25) regarding pairs of equations also apply here.

These two rows become:-

Vd1=

[

R 1 +j

ω

ωn

XL 1

]

Id1+j

ω

ωn

XmId2 (20.33)

0 =

sω

ωn

XmId1+

[

R 2 +j

sω

ωn

XL 2

]

Id2 (20.34)

Divide the secondary equation by the slips:-

0 =j

ω

ωn

XmId1+

[

R 2

s

+j

ω

ωn

XL 2

]

Id2 ( 20. 35 )

The equations (20.33) and (20.35) represent the familiar stationary coupled circuit shown in
Figure 20.3.

The magnitude of the axes variables is equal due to the symmetry of the winding construction,
as shown in (20.32). Hence|Vq1|=|Vd1|,|Iq1|=|Id1|and|Iq2|=|Id2|. The relationship between