Power System

1. 𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄𝒂𝒍 𝑷𝒆𝒓𝒇𝒐𝒓𝒎𝒂𝒏𝒄𝒆 𝒂𝒏𝒂𝒍𝒚𝒔𝒊𝒔 𝒐𝒇
𝑺𝒚𝒏𝒄𝒉𝒓𝒐𝒏𝒐𝒖𝒔 𝒎𝒂𝒄𝒉𝒊𝒏𝒆 𝒊𝒏 𝒂𝒄𝒄𝒐𝒓𝒅𝒂𝒏𝒄𝒆 𝒘𝒊𝒕𝒉
𝒅𝒒𝟎 𝑻𝒕𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 & 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒂𝒕𝒊𝒐𝒏
Prepared by : Pralay Roy
Part time PhD Scholar
Enrollment No. DT21EE002
Department of Electrical &
Electronics Engg.
N.I.T, Mizoram

2. 1)Basic equations of mutual inductance between stator and rotor winding (vary
periodically with θ) as given by following:
𝑙𝑎𝑓𝑑 = 𝐿𝑎𝑓𝑑 cos 𝜃 (1)
𝑙𝑎𝑘𝑑 = 𝐿𝑎𝑘𝑑 cos 𝜃 (2)
𝑙𝑎𝑘𝑞 = 𝐿𝑎𝑘𝑞 cos 𝜃 +
𝜋
2
(3)
The 𝒅𝒒𝟎 Transformation of rotor circuit flux linkages may be explore as
follows:
Ѱ𝑓𝑑 = 𝐿𝑓𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑓𝑘𝑑𝑖𝑘𝑑 − 𝐿𝑎𝑓𝑑[𝑖𝑎 cos 𝜃 + 𝑖𝑏 cos 𝜃 −
2𝜋
3
+ 𝑖𝑐 cos 𝜃 +
2𝜋
3
(1a)
Ѱ𝑘𝑑 = 𝐿𝑓𝑘𝑑𝑖𝑓𝑑 + 𝐿𝑘𝑘𝑑𝑖𝑘𝑑 − 𝐿𝑎𝑘𝑑[𝑖𝑎 cos 𝜃 + 𝑖𝑏 cos 𝜃 −
2𝜋
3
+ 𝑖𝑐 cos 𝜃 +
2𝜋
3
(2a)
Ѱ𝑓𝑑 = 𝐿𝑘𝑘𝑞𝑖𝑘𝑞 + 𝐿𝑎𝑘𝑞[𝑖𝑎 sin 𝜃 + 𝑖𝑏 sin 𝜃 −
2𝜋
3
+ 𝑖𝑐 sin 𝜃 +
2𝜋
3
(3a)

3. Basic equations of stator circuit three phase voltage equations and flux linkages:
𝑒𝑎 =
𝑑Ѱ𝑎
𝑑𝑙
− 𝑖𝑎𝑅𝑎 = 𝑝Ѱ𝑎 − 𝑖𝑎𝑅𝑎 (4)
𝑒𝑏 = 𝑝Ѱ𝑏 − 𝑖𝑏𝑅𝑏 (5)
𝑒𝑐 = 𝑝Ѱ𝑐 − 𝑖𝑐𝑅𝑐 (6)
And
Ѱ𝑎 =
−𝑖𝑎 𝐿𝑎𝑎0 + 𝐿𝑎𝑎2 cos 2𝜃 + 𝑖𝑏 𝐿𝑎𝑏0 + 𝐿𝑎𝑎2 cos 2𝜃 +
2𝜋
3
+ 𝑖𝑐 𝐿𝑎𝑏0 + 𝐿𝑎𝑎2 cos 2𝜃 −
2𝜋
3
+
𝑖𝑓𝑑𝐿𝑎𝑓𝑑 cos 𝜃 + 𝑖𝑘𝑑𝐿𝑎𝑘𝑑 cos 𝜃 − 𝑖𝑘𝑞𝐿𝑎𝑘𝑞 sin 𝜃 (7)
Ѱ𝑏 = 𝑖𝑎 𝐿𝑎𝑏0 + 𝐿𝑎𝑎2 cos 2𝜃 +
𝜋
3
− 𝑖𝑏 𝐿𝑎𝑎0 + 𝐿𝑎𝑎2 cos 2𝜃 −
2𝜋
3
+ 𝑖𝑐[𝐿𝑎𝑏0 +

4. The previous six equations are associated with stator circuits together with the
following equations of rotor circuit which can completely describe the electrical
operation of Synchronous Machine.
Basic equations of rotor circuit voltage equations
𝑒𝑓𝑑 = 𝑝Ѱ𝑓𝑑 + 𝑅𝑓𝑑𝑖𝑓𝑑 (10)
0 = 𝑝Ѱ𝑘𝑑 + 𝑅𝑘𝑑𝑖𝑘𝑑 (11)
0 = 𝑝Ѱ𝑘𝑞 + 𝑅𝑘𝑞𝑖𝑘𝑞 (12)
The Transformation of stator phase currents into new variables as follows:
𝑖𝑑 = 𝑘𝑑 𝑖𝑎 cos 𝜃 + 𝑖𝑏 cos 𝜃 −
2𝜋
3
+ 𝑖𝑐 cos 𝜃 +
2𝜋
3
(13)
𝑖𝑞 = −𝑘𝑞 𝑖𝑎 sin 𝜃 + 𝑖𝑏 sin 𝜃 −
2𝜋
3
+ 𝑖𝑐 sin 𝜃 +
2𝜋
3
(14)
For balanced condition the peak values of 𝒊𝒅 &
𝒊𝒒 𝐚𝐫𝐞 𝐞𝐪𝐮𝐚𝐥 𝐭𝐨 𝐭𝐡𝐞 𝐩𝐢𝐜𝐤 𝐚𝐥𝐮𝐞 𝐨𝐟 𝐬𝐭𝐚𝐭𝐨𝐫 𝐜𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐬 𝐬𝐡𝐨𝐰𝐧 𝐢𝐧 𝐛𝐞𝐥𝐥𝐨𝐰.
For balanced condition,
𝑖𝑎 = 𝐼𝑚 sin 𝜔𝑠𝑡 (15)
𝑖𝑏 = 𝐼𝑚 sin 𝜔𝑠𝑡 −
2𝜋
3
(16)
𝑖𝑐 = 𝐼𝑚 sin 𝜔𝑠𝑡 +
2𝜋
3
(17)

5. Substituting in equation (13), we get,
𝑖𝑑
= 𝑘𝑑 𝐼𝑚 sin 𝜔𝑠𝑡 cos 𝜃 + 𝐼𝑚 sin 𝜔𝑠𝑡

6. Transformed Matrix of 𝒂𝒃𝒄 phase variables to the 𝒅𝒒𝟎 variable can be written in
matrix form as following:
𝑖𝑑
𝑖𝑞
𝑖0
=
cos 𝜃 cos 𝜃 −
2𝜋
3
cos 𝜃 +
2𝜋
3
−sin 𝜃 −sin 𝜃 −
2𝜋
3
− sin 𝜃 +
2𝜋
3
1
2
1
2
1
2
𝑖𝑎
𝑖𝑏
𝑖𝑐
(19)
The inverse transformation is given by,
𝑖𝑎
𝑖𝑏
𝑖𝑐
=
cos 𝜃 −sin 𝜃 1
cos 𝜃 −
2𝜋
3
−sin 𝜃 −
2𝜋
3
1
cos 𝜃 +
2𝜋
3
− sin 𝜃 +
2𝜋
3
1
𝑖𝑑
𝑖𝑞
𝑖0
(20)
The above transformation can also be applied to Stator flux linkages and voltages.

7. Stator flux linkages in 𝒅𝒒𝟎 components :Using the expressions of Ѱ𝒂 , Ѱ𝒃 & Ѱ𝒄
from equation 7,8,9 , transforming the flux linkages into 𝒅𝒒𝟎 components (as in
equation 19)and with suitable reduction in trigonometric term we obtain the
following expressions:
Ѱ𝑑
Ѱ𝑞
Ѱ0
=
2
3
cos 𝜃 cos 𝜃 −
2𝜋
3
cos 𝜃 +
2𝜋
3
−sin 𝜃 −sin 𝜃 −
2𝜋
3
− sin 𝜃 +
2𝜋
3
1
2
1
2
1
2
Ѱ𝑎
Ѱ𝑏
Ѱ𝑐
(21)
Stator Flux linkages in 𝒅𝒒𝟎 components
Ѱ𝑑 = − 𝐿𝑎𝑎0 + 𝐿𝑎𝑏0 +
3
2
𝐿𝑎𝑎2 𝑖𝑑 + 𝐿𝑎𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑎𝑘𝑑𝑖𝑘𝑑 (22)
Ѱ𝑞 = − 𝐿𝑎𝑎0 + 𝐿𝑎𝑏0 −
3
2
𝐿𝑎𝑎2 𝑖𝑞 + 𝐿𝑎𝑘𝑞𝑖𝑘𝑞 (23)
Ѱ0 = − 𝐿𝑎𝑎0 − 2𝐿𝑎𝑏0 𝑖0 (24)
Defining the following new inductances:
𝐿𝑑 = 𝐿𝑎𝑎0 + 𝐿𝑎𝑏𝑜 +
3
2
𝐿𝑎𝑎2 (25)
𝐿𝑞 = 𝐿𝑎𝑎0 + 𝐿𝑎𝑏𝑜 −
3
2
𝐿𝑎𝑎2 (26)
𝐿0 = 𝐿𝑎𝑎0 − 2𝐿𝑎𝑏𝑜 +
3
2
𝐿𝑎𝑎2 (27)

8. Modified expressions of flux linkages with the help of new defined inductances:
Ѱ𝑑 = −𝐿𝑑𝑖𝑑 + 𝐿𝑎𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑎𝑘𝑑𝑖𝑘𝑑 (28)
Ѱ𝑞 = −𝐿𝑞𝑖𝑞 + 𝐿𝑎𝑘𝑞𝑖𝑘𝑞 (29)
Ѱ0 = −𝐿0𝑖0 (30)
Rotor flux linkages in 𝒅𝒒𝟎 components:
Ѱ𝑓𝑑 = 𝐿𝑓𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑓𝑘𝑑𝑖𝑘𝑑 −
3
2
𝐿𝑎𝑓𝑑𝑖𝑑 (31)
Ѱ𝑘𝑑 = 𝐿𝑓𝑘𝑑𝑖𝑓𝑑 + 𝐿𝑘𝑘𝑑𝑖𝑘𝑑 −
3
2
𝐿𝑎𝑘𝑑𝑖𝑑 (32)
Ѱ𝑘𝑞 = 𝐿𝑘𝑘𝑞𝑖𝑘𝑞 −
3
2
𝐿𝑎𝑘𝑞𝑖𝑞 (33)
All the inductances are seen to be constant, i.e. they are independent of the rotor
position.

9. Stator voltage equation in 𝒅𝒒𝟎 components:
Equation 4 to 6 are basic voltage equation in terms of phase flux linkages and
currents.
By applying 𝒅𝒒𝟎 transformation of equation 19,the following equations in terms of
transformed components of voltages, flux linkages and currents results:
𝑒𝑑 = 𝑝Ѱ𝑑 − Ѱ𝑞𝑝𝜃 − 𝑖𝑑𝑅𝑎 (34)
𝑒𝑞 = 𝑝Ѱ𝑞 + Ѱ𝑑𝑝𝜃 − 𝑖𝑞𝑅𝑎 (35)
𝑒0 = 𝑝Ѱ0 − 𝑖0𝑅𝑎 (36)
The terms Ѱ𝑞𝑝𝜃 & Ѱ𝑑𝑝𝜃 are referred to as the Speed Voltages (Due to flux
change in space) and the terms 𝑝Ѱ𝑑 & 𝑝Ѱ𝑞 are referred to as the Transformer
Voltages (due to flux change in time).

10. Electrical Power & Torque:
The instantaneous three phase power output of Stator is
𝑃𝑡 = 𝑒𝑎𝑖𝑎 + 𝑒𝑏𝑖𝑏 + 𝑒𝑐𝑖𝑐
Eliminating phase voltages & currents in terms of 𝑑𝑞0components, we have
𝑃𝑡 =
3
2
𝑒𝑑𝑖𝑑 + 𝑒𝑞𝑖𝑞 + 𝑒0𝑖0 (37)
Under balanced condition , 𝑒0 = 𝑖0 = 0& the expression for power is given by
𝑃𝑡 =
3
2
𝑒𝑑𝑖𝑑 + 𝑒𝑞𝑖𝑞
Using equation 34 to 36 ,to express the voltage components in terms of flux linkages
and currents, by recognizing 𝜔𝑟 as the rotor speed
𝑑𝜃
𝑑𝑡
and rearranging, we have
𝑃𝑡 =
3
2
𝑖𝑑𝑝Ѱ𝑑 + 𝑖𝑞𝑝Ѱ𝑞 + 2𝑖0𝑝Ѱ0 + Ѱ𝑑𝑖𝑞 − Ѱ𝑞𝑖𝑑 𝜔𝑟 − 𝑖𝑑
2
+ 𝑖𝑞
2
+ 2𝑖0
2
𝑅𝑎 (38)
= 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔 +
𝑃𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑖𝑟 𝑔𝑎𝑝 − 𝐴𝑚𝑎𝑡𝑢𝑟𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑙𝑜𝑠𝑠
So, expression of air-gap torque
𝑇𝑒 =
3
2
Ѱ𝑑𝑖𝑞 − Ѱ𝑞𝑖𝑑 ×
𝜔𝑟
𝜔𝑚𝑒𝑐ℎ
𝑇𝑒 =
3
2
Ѱ𝑑𝑖𝑞 − Ѱ𝑞𝑖𝑑 ×
𝑃𝑓
2
(39)

11. The flux linkage equations 28 to 33, associated with stator or rotor circuit,
together with the voltage equations 34 to 36 for the Stator, the voltage equations
10 to 12 for the rotor and the torque equation 39 describe the dynamic
performance of the machine in terms of the 𝒅𝒒𝟎 components
Physical interpretation of 𝒅𝒒𝟎 transformation

12. 𝑖𝑎 = 𝐼𝑚 sin 𝜔𝑠𝑡 + 𝜑 (40)
𝑖𝑏 = 𝐼𝑚 sin 𝜔𝑠𝑡 + 𝜑 −
2𝜋
3
(41)
𝑖𝑐 = 𝐼𝑚 sin 𝜔𝑠𝑡 + 𝜑 +
2𝜋
3
(42)
Using 𝑑𝑞0 transformation
𝑖𝑑 = 𝐼𝑚 sin 𝜔𝑠𝑡 + 𝜑 − 𝜃 (43)
𝑖𝑞 = −𝐼𝑚 cos 𝜔𝑠𝑡 + 𝜑 − 𝜃 (44)
𝑖0 = 0 (45)
For synchronous operation,
𝜃 = 𝜔𝑟𝑡 = 𝜔𝑠𝑡
Therefore,
𝑖𝑑 = 𝐼𝑚 sin 𝜑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑖𝑞 = −𝐼𝑚 cos 𝜑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

13. The analysis of the synchronous machine equations in terms of 𝑑𝑞0 variables is
considerably simpler than in terms of phase quantities, for the following reason:
a) The dynamic performance equations have constant inductances
b) For balanced condition zero sequence quantities disappear
c) For balanced steady state operation, the stator quantities have constant values. For
other mode of operation they vary with time. Stability studies involve also
variations having frequencies below 2-3 Hz.
d) The parameters associated with d and q axes may be directly measured from
terminal test.
Per unit representation:
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑖𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
𝐵𝑎𝑠𝑒 𝑣𝑎𝑙𝑢𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦

14. Per unit system for the stator quantities:
𝑒𝑠 𝑏𝑎𝑠𝑒 = 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑟𝑎𝑡𝑒𝑑 𝑙𝑖𝑛𝑒 − 𝑡𝑜 − 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒, 𝑉𝑜𝑙𝑡.
𝑖𝑠 𝑏𝑎𝑠𝑒 = 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑟𝑎𝑡𝑒𝑑 𝑙𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐴𝑚𝑝.
𝑓𝑠 𝑏𝑎𝑠𝑒 = 𝑅𝑎𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝐻𝑧.
The base values of the remaining quantities are automatically set and depend on the
above as follows:
𝜔𝑏𝑎𝑠𝑒 = 2𝜋𝑓𝑏𝑎𝑠𝑒 electrical radian/sec.
𝜔𝑚 𝑏𝑎𝑠𝑒 = 𝜔𝑏𝑎𝑠𝑒
2
𝑝𝑓
mechanical radian/sec.
𝑍𝑠 𝑏𝑎𝑠𝑒 =
𝑒𝑠 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
, Ohms
𝐿𝑠 𝑏𝑎𝑠𝑒 =
𝑍𝑠 𝑏𝑎𝑠𝑒
𝜔𝑏𝑎𝑠𝑒
, 𝐻𝑒𝑛𝑟𝑦
Ѱ𝑠 𝑏𝑎𝑠𝑒 = 𝐿𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 =
𝑒𝑠 𝑏𝑎𝑠𝑒
𝜔𝑠 𝑏𝑎𝑠𝑒
3- Phase, 𝑉𝐴𝑏𝑎𝑠𝑒 = 3𝐸𝑅𝑀𝑆 𝑏𝑎𝑠𝑒𝐼𝑅𝑀𝑆 𝑏𝑎𝑠𝑒
= 3
𝑒𝑠 𝑏𝑎𝑠𝑒
2
.
𝑖𝑠 𝑏𝑎𝑠𝑒
2
=
3
2
𝑒𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 𝑉𝑜𝑙𝑡 − 𝐴𝑚𝑝𝑒𝑟𝑒

15. 𝑇𝑜𝑟𝑞𝑢𝑒 𝑏𝑎𝑠𝑒 =
3 − 𝑃ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝜔𝑚 𝑏𝑎𝑠𝑒
=
3
2
𝑝𝑓
2
Ѱ𝑠 𝑏𝑎𝑠𝑒𝑖𝑠 𝑏𝑎𝑠𝑒 , 𝑁𝑒𝑤𝑡𝑜𝑛 − 𝑚𝑒𝑡𝑒𝑟𝑠
Per unit Stator voltage equations:
From equation (20),
𝑒𝑑 = 𝑝Ѱ𝑑 − Ѱ𝑞𝜔𝑟 − 𝑖𝑑𝑅𝑎
Dividing throughout by 𝑒𝑏𝑎𝑠𝑒 ,and noting that 𝑒𝑠 𝑏𝑎𝑠𝑒 = 𝑖𝑠 𝑏𝑎𝑠𝑒𝑍𝑠 𝑏𝑎𝑠𝑒 = 𝜔𝑠Ѱ𝑠 𝑏𝑎𝑠𝑒 ,
We get,
𝑒𝑑
𝑒𝑠 𝑏𝑎𝑠𝑒
= 𝑝
1
𝜔𝑠 𝑏𝑎𝑠𝑒
.
Ѱ𝑑
Ѱ𝑠 𝑏𝑎𝑠𝑒
−
Ѱ𝑞
Ѱ𝑠 𝑏𝑎𝑠𝑒
.
𝜔𝑟
𝜔𝑏𝑎𝑠𝑒
−
𝑅𝑎
𝑍𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑑
𝑖𝑠 𝑏𝑎𝑠𝑒
(46)
Expressed in per unit notation,
𝑒𝑑 =
1
𝜔 𝑏𝑎𝑠𝑒
𝑝Ѱ𝑑 − Ѱ𝑞 𝜔𝑟 − 𝑅𝑎 𝑖𝑑 (47)

16. Time can also be expressed in per unit (or radians) with the base value equal to the time
required for the rotor to move one electrical radian at synchronous speed.
𝑡𝑏𝑎𝑠𝑒 =
1
𝜔𝑏𝑎𝑠𝑒
=
1
2𝜋𝑓𝑏𝑎𝑠𝑒
(48)
With time in per unit equation 47 may be written as
𝑒𝑑 = 𝑝Ѱ𝑑 − Ѱ𝑞 𝜔𝑟 − 𝑅𝑎 𝑖𝑑 (49)
Comparing equation 20 with equation 49 , we see that the form of the original equation
is unchanged, when all quantities involved are expressed in per unit
Similarly per unit form of equation 35 and 36 are
𝑒𝑞 = 𝑝 Ѱ𝑑 + Ѱ𝑑 𝜔𝑟 − 𝑅𝑎 𝑖𝑞 (50)
𝑒0 = 𝑝 Ѱ𝑑 − 𝑅𝑎 𝑖0 (51)
Where,
𝑝 =
𝑑
𝑑𝑡
=
1
𝜔𝑏𝑎𝑠𝑒
𝑑
𝑑𝑡
=
𝑝
𝜔𝑏𝑎𝑠𝑒
(52)

17. Per Unit Rotor voltage equation:
From equation 10 dividing throughout by,
𝑒𝑓𝑑 𝑏𝑎𝑠𝑒 = 𝜔𝑏𝑎𝑠𝑒Ѱ𝑓𝑑 𝑏𝑎𝑠𝑒 = 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑍𝑓𝑑 𝑏𝑎𝑠𝑒
Per unit field voltage equation may written as:
𝑒𝑓𝑑 = 𝑝Ѱ𝑓𝑑 + 𝑅𝑓𝑑𝑖𝑓𝑑
(53)
Similarly the per unit form of equation 11 and 12
0 = 𝑝Ѱ𝑘𝑑 + 𝑅𝑘𝑑𝑖𝑘𝑑 (54)
0 = 𝑝Ѱ𝑘𝑞 + 𝑅𝑘𝑞𝑖𝑘𝑞 (55)
Per unit stator flux linkage equations:
Using the basic relationship Ѱ𝑠 𝑏𝑎𝑠𝑒 = 𝐿𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 the per unit form of equations
28,29 & 30 may be written as
Ѱ𝑑 = −𝐿𝑑𝑖𝑑 + 𝐿𝑎𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑎𝑘𝑑𝑖𝑘𝑑 (56)
Ѱ𝑞 = −𝐿𝑞𝑖𝑞 + 𝐿𝑎𝑘𝑞𝑖𝑞 (57)
Ѱ0 = −𝐿0𝑖0 (58)

18. Where by definition:
𝐿𝑎𝑓𝑑 =
𝐿𝑎𝑓𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
(59)
𝐿𝑎𝑘𝑑 =
𝐿𝑎𝑘𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
(60)
𝐿𝑎𝑘𝑞 =
𝐿𝑎𝑘𝑞
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑘𝑞 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
(61)
Per unit rotor flux linkage equations:
Similarly in per unit form equations 31,32 & 33, become
Ѱ𝑓𝑑 = 𝐿𝑓𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑓𝑘𝑑𝑖𝑘𝑑 − 𝐿𝑎𝑓𝑑𝑖𝑑 (62)
Ѱ𝑘𝑑 = 𝐿𝑓𝑘𝑑𝑖𝑓𝑑 + 𝐿𝑘𝑘𝑑𝑖𝑘𝑑 − 𝐿𝑎𝑘𝑑𝑖𝑑 (63)
Ѱ𝑘𝑞 = 𝐿𝑘𝑘𝑞𝑖𝑘𝑞 − 𝐿𝑎𝑘𝑞𝑖𝑞 (64)

19. Where by definition:
𝐿𝑓𝑑𝑎 =
3
2
.
𝐿𝑎𝑓𝑑
𝐿𝑓𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑠 𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
(65)
𝐿𝑓𝑘𝑑 =
𝐿𝑓𝑘𝑑
𝐿𝑓𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
(66)
𝐿𝑘𝑑𝑎 =
3
2
.
𝐿𝑎𝑘𝑑
𝐿𝑘𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑠 𝑏𝑎𝑠𝑒
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
(67)
𝐿𝑘𝑑𝑓 =
𝐿𝑓𝑘𝑑
𝐿𝑘𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
(68)
𝐿𝑘𝑞𝑎 =
3
2
.
𝐿𝑎𝑘𝑞
𝐿𝑘𝑞 𝑏𝑎𝑠𝑒
.
𝑖𝑠 𝑏𝑎𝑠𝑒
𝑖𝑘𝑞 𝑏𝑎𝑠𝑒
(69)
Per unit system of Rotor:
Some Important information:
1) 𝐿𝑎𝑓𝑑 = 𝐿𝑓𝑑𝑎
2) 𝐿𝑎𝑓𝑑 = 𝐿𝑎𝑘𝑑

20. In order to have 𝐿𝑓𝑘𝑑 = 𝐿𝑘𝑑𝑓 so that the reciprocity is achieved, from equation 66 and
68 ,it is necessary to have
𝐿𝑓𝑘𝑑
𝐿𝑓𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
=
𝐿𝑓𝑘𝑑
𝐿𝑘𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
𝐿𝑘𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
2
= 𝐿𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
2
(70)
Multiply by 𝜔𝑏𝑎𝑠𝑒 gives,
𝜔𝑏𝑎𝑠𝑒𝐿𝑘𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
2
= 𝜔𝑏𝑎𝑠𝑒𝐿𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
2
Since,
𝜔𝑏𝑎𝑠𝑒𝐿 𝑏𝑎𝑠𝑒. 𝑖𝑏𝑎𝑠𝑒 = 𝑒𝑏𝑎𝑠𝑒
𝑒𝑘𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑘𝑑 𝑏𝑎𝑠𝑒 = 𝑒𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒 (71)
For mutual inductances 𝐿𝑎𝑓𝑑 & 𝐿𝑓𝑑𝑎 to be equal from equation 59 & 65
𝐿𝑎𝑓𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
=
3
2
.
𝐿𝑎𝑓𝑑
𝐿𝑓𝑑 𝑏𝑎𝑠𝑒
.
𝑖𝑠 𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
Or, 𝐿𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
2
=
3
2
. 𝐿𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒
2

21. Multiply by 𝜔𝑏𝑎𝑠𝑒 and noting that 𝜔𝐿𝑖 = 𝑒, we get
𝑒𝑓𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑓𝑑 𝑏𝑎𝑠𝑒 =
3
2
. 𝑒𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 (72)
=
3
2
𝑒𝑠 𝑏𝑎𝑠𝑒
2
𝑖𝑠 𝑏𝑎𝑠𝑒
2
= 3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴 𝑏𝑎𝑠𝑒 𝑓𝑜𝑟 𝑆𝑡𝑎𝑡𝑜𝑟
Similarly in order 𝐿𝑎𝑘𝑑=𝐿𝑘𝑑𝑎 and 𝐿𝑎𝑘𝑞=𝐿𝑘𝑞𝑎
𝑒𝑘𝑑 𝑏𝑎𝑠𝑒. 𝑖𝑘𝑑 𝑏𝑎𝑠𝑒 =
3
2
. 𝑒𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 (73)
And
𝑒𝑘𝑞 𝑏𝑎𝑠𝑒. 𝑖𝑘𝑞 𝑏𝑎𝑠𝑒 =
3
2
. 𝑒𝑠 𝑏𝑎𝑠𝑒. 𝑖𝑠 𝑏𝑎𝑠𝑒 (74)
This equations imply that the VA base in all rotor circuit must be the same and equal
to the stator 3 phase VA base.

22. The stator leakage inductances in the two axes are nearly equal denoting the leakage
inductance 𝐿𝑙 and mutual inductance by 𝐿𝑎𝑑 and 𝐿𝑎𝑞 :
𝐿𝑑 = 𝐿𝑙 + 𝐿𝑎𝑑 (75)
And
𝐿𝑞 = 𝐿𝑙 + 𝐿𝑎𝑞 (76)
In order to make all the per unit mutual inductances between the stator and rotor
circuits in the d axis equal, from equations 59 and 60, it follows that
𝐿𝑎𝑑 =
𝐿𝑎𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
= 𝐿𝑎𝑓𝑑 =
𝐿𝑎𝑓𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
= 𝐿𝑎𝑘𝑑 =
𝐿𝑎𝑘𝑑
𝐿𝑠 𝑏𝑎𝑠𝑒
.
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
𝑖𝑠 𝑏𝑎𝑠𝑒
Therefore,
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑑
𝐿𝑎𝑓𝑑
𝑖𝑠 𝑏𝑎𝑠𝑒 (77)
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑑
𝐿𝑎𝑘𝑑
𝑖𝑠 𝑏𝑎𝑠𝑒 (78)
Similarly , for the q axis mutual inductances 𝐿𝑎𝑞 and 𝐿𝑎𝑘𝑞 to be equal,
𝑖𝑘𝑞 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑞
𝐿𝑎𝑘𝑞
𝑖𝑠 𝑏𝑎𝑠𝑒 (79)
This completes the choice of rotor base quantities.

23. Per unit Power and Torque :
The instantaneous power at the machine terminal as per equation 37,
𝑃𝑡 =
3
2
𝑒𝑑𝑖𝑑 + 𝑒𝑞𝑖𝑞 + 𝑒0𝑖0
Dividing by base three phase VA=
3
2
𝑒𝑠 𝑏𝑎𝑠𝑒𝑖𝑠 𝑏𝑎𝑠𝑒
The per unit expression may be written as,
𝑃𝑡 = 𝑒𝑑𝑖𝑑 + 𝑒𝑞𝑖𝑞 + 2𝑒0𝑖0 (80)
Similarly, with base torque=
3
2
𝑝𝑓
2
Ѱ𝑠 𝑏𝑎𝑠𝑒𝑖𝑠 𝑏𝑎𝑠𝑒 ,the per unit form of equation 39 is
𝑇𝑒 = Ѱ𝑑𝑖𝑞 − Ѱ𝑞𝑖𝑑 (81)
Per unit reactance
𝑋𝑑 = 2𝜋𝑓𝐿𝑑 Ω
Dividing by 𝑍𝑏𝑎𝑠𝑒 = 2𝜋𝑓𝑏𝑎𝑠𝑒𝐿𝑏𝑎𝑠𝑒
𝑋𝑑
𝑍𝑏𝑎𝑠𝑒
=
2𝜋𝑓
2𝜋𝑓𝑏𝑎𝑠𝑒
.
𝐿𝑑
𝐿𝑏𝑎𝑠𝑒
If 𝑓 = 𝑓𝑏𝑎𝑠𝑒 ,per unit values of 𝑋𝑑 & 𝐿𝑑 are equal. So in case of Synchronous machine
symbols associated with reactance are often used to denote per unit inductance.

24. Summary of per unit equations:
Stator base quantities:
3 phase 𝑽𝑨𝒃𝒂𝒔𝒆= Volt ampere rating of machine, VA
𝑒𝑠 𝑏𝑎𝑠𝑒= Peak phase to neutral related voltage, V
𝑓𝑏𝑎𝑠𝑒 =Rated frequency, Hz
𝑖𝑠 𝑏𝑎𝑠𝑒 = 𝑝𝑒𝑎𝑘 𝑙𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 , 𝐴𝑚𝑝
=
3 𝑃ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
3
2 𝑒𝑠 𝑏𝑎𝑠𝑒
𝑍𝑠 𝑏𝑎𝑠𝑒 =
𝑒𝑠 𝑏𝑎𝑠𝑒
𝑖𝑠𝑏𝑎𝑠𝑒
, Ω
Where,𝜔𝑏𝑎𝑠𝑒 = 2𝜋𝑓𝑏𝑎𝑠𝑒 elect. rad/sec.
𝜔𝑚 𝑏𝑎𝑠𝑒 = 𝜔𝑏𝑎𝑠𝑒
2
𝑝𝑓
mech. Rad /sec.
𝐿𝑠 𝑏𝑎𝑠𝑒 =
𝑍𝑠 𝑏𝑎𝑠𝑒
𝜔𝑏𝑎𝑠𝑒
, Henry
Ѱ𝑠 𝑏𝑎𝑠𝑒 = 𝐿𝑠 𝑏𝑎𝑠𝑒 𝑖𝑠 𝑏𝑎𝑠𝑒, Wb-turns

25. Rotor base quantities:
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑑
𝐿𝑎𝑓𝑑
𝑖𝑠 𝑏𝑎𝑠𝑒 , Amp
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑑
𝐿𝑎𝑘𝑑
𝑖𝑠 𝑏𝑎𝑠𝑒 , Amp
𝑖𝑘𝑞 𝑏𝑎𝑠𝑒 =
𝐿𝑎𝑞
𝐿𝑎𝑘𝑞
𝑖𝑠 𝑏𝑎𝑠𝑒 , Amp
𝑒𝑓𝑑 𝑏𝑎𝑠𝑒 =
3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
, volt
𝑍𝑓𝑑 𝑏𝑎𝑠𝑒 =
𝑒𝑓𝑑 𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
, Ω
=
3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝑖𝑓𝑑 𝑏𝑎𝑠𝑒
2
𝑍𝑘𝑑 𝑏𝑎𝑠𝑒 =
3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝑖𝑘𝑑 𝑏𝑎𝑠𝑒
2 Ω
𝑍𝑘𝑞 𝑏𝑎𝑠𝑒 =
3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝑖𝑘𝑞 𝑏𝑎𝑠𝑒
2 Ω

26. Rotor base quantities:
𝐿𝑓𝑑 𝑏𝑎𝑠𝑒 =
𝑍𝑓𝑑 𝑏𝑎𝑠𝑒
𝜔𝑏𝑎𝑠𝑒
, Henry
𝐿𝑘𝑑 𝑏𝑎𝑠𝑒 =
𝑍𝑘𝑑 𝑏𝑎𝑠𝑒
𝜔𝑏𝑎𝑠𝑒
, Henry
𝐿𝑘𝑞 𝑏𝑎𝑠𝑒 =
𝑍𝑘𝑞 𝑏𝑎𝑠𝑒
𝜔𝑏𝑎𝑠𝑒
, Henry
𝑡𝑏𝑎𝑠𝑒 =
1
𝜔𝑏𝑎𝑠𝑒
, Sec.
𝑇𝑏𝑎𝑠𝑒 =
3 𝑝ℎ𝑎𝑠𝑒 𝑉𝐴𝑏𝑎𝑠𝑒
𝜔𝑚 𝑏𝑎𝑠𝑒
, Nm

27. Complete Set of per unit equations:
In view of the 𝐿𝑎𝑑 𝑏𝑎𝑠𝑒per unit system chosen ,in per unit
𝐿𝑎𝑓𝑑 = 𝐿𝑓𝑑𝑎 = 𝐿𝑎𝑘𝑑 = 𝐿𝑘𝑑𝑎 = 𝐿𝑎𝑑
𝐿𝑎𝑘𝑞 = 𝐿𝑘𝑞𝑎 = 𝐿𝑎𝑞
𝐿𝑓𝑘𝑑=𝐿𝑘𝑑𝑓
Per unit Stator voltage equation
𝑒𝑑 = 𝑝Ѱ𝑑 − Ѱ𝑞𝜔𝑟 − 𝑖𝑑𝑅𝑎 (82)
𝑒𝑞 = 𝑝Ѱ𝑞 + Ѱ𝑑𝜔𝑟 − 𝑖𝑞𝑅𝑎 (83)
𝑒0 = 𝑝Ѱ0 − 𝑖0𝑅𝑎 (84)
Per unit Rotor voltage equation
𝑒𝑓𝑑 = 𝑝Ѱ𝑓𝑑 + 𝑅𝑓𝑑𝑖𝑓𝑑 (85)
0 = 𝑝Ѱ1𝑑 + 𝑅1𝑑𝑖1𝑑 (86)
0 = 𝑝Ѱ1𝑞 + 𝑅1𝑞𝑖1𝑞 (87)
0 = 𝑝Ѱ2𝑑 + 𝑅2𝑑𝑖2𝑑 (88)

28. Per unit stator flux linkage equations:
Ѱ𝑑 = − 𝐿𝑎𝑑 + 𝐿𝑙 𝑖𝑑 + 𝐿𝑎𝑑𝑖𝑓𝑑 + 𝐿𝑎𝑑𝑖1𝑑 (89)
Ѱ𝑞 = − 𝐿𝑎𝑞 + 𝐿𝑙 𝑖𝑞 + 𝐿𝑎𝑞𝑖1𝑞 + 𝐿𝑎𝑞𝑖2𝑞 (90)
Ѱ0 = −𝐿0𝑖0 (91)
Per unit Rotor Flux linkage equations:
Ѱ𝑓𝑑 = 𝐿𝑓𝑓𝑑𝑖𝑓𝑑 + 𝐿𝑓1𝑑𝑖1𝑑 − 𝐿𝑎𝑑𝑖𝑑 (92)
Ѱ1𝑑 = 𝐿𝑓1𝑑𝑖𝑓𝑑 + 𝐿11𝑑𝑖1𝑑 − 𝐿𝑎𝑑𝑖𝑑 (93)
Ѱ1𝑞 = 𝐿11𝑞𝑖1𝑞 + 𝐿𝑎𝑞𝑖2𝑞 − 𝐿𝑎𝑞𝑖𝑞 (94)
Ѱ2𝑞 = 𝐿𝑎𝑞𝑖1𝑞 + 𝐿22𝑞𝑖2𝑞 − 𝐿𝑎𝑞𝑖𝑞 (95)
Per unit Air- gap Torque:
𝑇𝑒 = Ѱ𝑑𝑖𝑞 − Ѱ𝑞𝑖𝑑 (96)

29. Steady state analysis:
Voltage , current & flux relationships
At steady State zero sequenced components are absent and 𝜔𝑟 = 𝜔𝑠 = 1
With 𝑝Ѱ terms are set to zero in equation 86,87 and 88
𝑅1𝑑𝑖𝑖𝑑 = 𝑅1𝑞𝑖1𝑞 = 𝑅2𝑞𝑖2𝑞 = 0 (97)
The per unit machine equations (82 to 96) under balanced steady state conditions,
become
𝑒𝑑 = −Ѱ𝑞𝜔𝑟 − 𝑖𝑑𝑅𝑎 (98)
𝑒𝑞 = −Ѱ𝑑𝜔𝑟 − 𝑖𝑞𝑅𝑎 (99)
𝑒𝑓𝑑 = 𝑖𝑓𝑑𝑅𝑓𝑑 (100)
Ѱ𝑑 = −𝐿𝑑𝑖𝑑 + 𝐿𝑎𝑑𝑖𝑓𝑑 (101)
Ѱ𝑞 = −𝐿𝑞𝑖𝑞 (102)
Ѱ𝑓𝑑 = 𝐿𝑓𝑓𝑑𝑖𝑓𝑑 − 𝐿𝑎𝑑𝑖𝑑 (103)
Ѱ1𝑑 = 𝐿𝑓1𝑑𝑖𝑓𝑑 − 𝐿𝑎𝑑𝑖𝑑 (104)
Ѱ1𝑞 = Ѱ2𝑞 = −𝐿𝑎𝑞𝑖𝑞 (105)

30. Field current:
From equation 101,
𝑖𝑓𝑑 =
Ѱ𝑑+𝐿𝑑𝑖𝑑
𝐿𝑎𝑑
(106)
Substituting for Ѱ𝑑in terms of 𝑒𝑑,𝑖𝑞from equation 99
𝑖𝑓𝑑 =
𝑒𝑞+𝑅𝑎𝑖𝑞+𝜔𝑟𝐿𝑑𝑖𝑑
𝜔𝑟𝐿𝑎𝑑
(107)
Replacing the product of synchronous speed and inductance L by corresponding
reactance X
𝑖𝑓𝑑 =
𝑒𝑞+𝑅𝑎𝑖𝑞+𝑋𝑑𝑖𝑑
𝑋𝑎𝑑
(108)
Phasor representation:
Stator phase voltages in balanced steady state condition may be written as
𝑒𝑎 = 𝐸𝑚 cos 𝜔𝑠𝑡 + 𝛼 (109)
𝑒𝑏 = 𝐸𝑚 cos 𝜔𝑠𝑡 −
2𝜋
3
+ 𝛼 (110)
𝑒𝑐 = 𝐸𝑚 cos 𝜔𝑠𝑡 +
2𝜋
3
+ 𝛼 (111)
Stator phase voltages in balanced steady state condition in 𝒅𝒒𝟎 components we get
𝑒𝑑 = 𝐸𝑚 cos 𝜔𝑠𝑡 + 𝛼 − 𝜃 (112)
𝑒𝑞 = 𝐸𝑚 sin 𝜔𝑠𝑡 + 𝛼 − 𝜃 (113)

31. The angle θ by which the d axis leads the axis of phase 𝑎 is given by
𝜃 = 𝜔𝑟𝑡 + 𝜃0 (114)
Where 𝜃0 is the value of 𝜃 at 𝑡 = 0.
With 𝜔𝑟 equal to 𝜔𝑠 at synchronous speed, substitution for 𝜃 in equation 112 & 113
yields
𝑒𝑑 = 𝐸𝑚 cos 𝛼 − 𝜃0 (115)
𝑒𝑞 = 𝐸𝑚 sin 𝛼 − 𝜃0 (116)
Using 𝐸𝑡 to denote per unit rms value of armature terminal voltage and the per unit rms
and peak values are equal
𝑒𝑑 = 𝐸𝑚 cos 𝛼 − 𝜃0 (117)
𝑒𝑞 = 𝐸𝑚 sin 𝛼 − 𝜃0 (118)