Concept of ac circuit for electrical engineering and this help to study about ac circuit

Ac circuit

1. Power in Single Phase AC Circuits
Let us consider the following circuit.
+v(t)i(t)LoadVIVI
Let
v(t) =
p
2Vsin(!t+V)
i(t) =
p
2Isin(!t+I)

2. Theinstantaneous powerdelivered to the load is
p(t) =v(t)i(t)
p(t) =Vmsin(!t+V)Imsin(!t+I)
p(t) =
VmIm
2
(cos(VI)cos(2!t+V+I))
p(t) =
VmIm
2
cos(VI)
VmIm
2
cos(2!t+V+I)

3. !tp;v;iviVmImIVp
Figure:

4. LetVIbe.
p(t) =VIcosVIcos(2!t+V+I)
p(t) =VIcosVIcos(2!t+VI+ 2I)
p(t) =VIcosVIcos(2!t+ 2I)
p(t) =VIcos(1cos(2!t+ 2I))
| {z }
pI
VIsinsin(2!t+ 2I)
| {z }
pII
pIhas an average value ofVIcoswhich is called the
power.
pIIdoes not have an average. But it's maximum value isVIsin
which is called.

5. !tpppIpIIPQ
Figure:

6. Power
The average powerPis
P=
VmIm
2
cos(VI) =VIcos()
where=VI. Its unit is watts (W).
The reactive powerQis
Q=VIsin
The apparent powerSis
jSj=VI
Its unit isvolt-ampere (VA).
The ratio of real power (P) to apparent power is called as the
power factor (pf).
pf =
VIcos
VI
= cos
Since coscan never be greater than unity,P jSj.

7. Complex Power
Let us dene voltage phasor and current phasor.
V =V\V;I =I\I
The complex power S is
S = VI

S =V\VI\I
=VI\(VI)
S =VIcos+|VIsin
The real part of S is called the average power (P). The imaginary
part of S is called the reactive power (Q).
S =P+|Q

8. ReImVIVIReImPQS
Figure:
IfVleadsI( >0), power factor is lagging.

9. ReImVIV=IReImP=S
Figure:
IfVandIare in phase (= 0), power factor is unity.

10. ReImIVIVReImPQS
Figure:
IfIleadsV( <0), power factor is leading.

11. For two loads (inductive and capacitive) in parallel,
P1Q1S1P2Q2S2STQTPT
PT=P1+P2;QT=Q1+Q2
But
jSTj 6=jS1j+jS2j

12. Power Factor Control
IIfpfdecreases, the current will increase to supply the same
real power.
IThis will increase the line loss. (It is an additional cost to a
utility.)
ICapacitors which supply reactive power are connected in
parallel to improve the power factor.
LoadCPQSQcSnewQnewnew

13. Example 1 : A single-phase inductive load draws to 1 kW at 0.6
power-factor lagging from a 230 V AC supply.
1.
2.
the load to raise the power factor to 0.9 lagging. Determine
the current under this condition.
1.
I=
1000
2300:6
= 7:24 A
Q= 2307:240:8 = 1:332 kVAr
ILoad1 kW1.332 kVAR1.6652 kVA53:13

14. 2.
pfnew= 0:9;new= 25:84

Qnew=QLQc
Qnew=Ptan 25:84

= 484:32 VAr
Qc= 847:68 VAr
Qc=V
2
!C
C= 51F
I=
1000
2300:9
= 4:83A
ILoadC1 kWQLSQcSnewQnew25:84

15. Power in Balanced Three Phase Circuits
Letva,vbandvcbe the instantaneous voltages of a balanced three
phase source.
va=
p
2Vsin(!t+V)
vb=
p
2Vsin(!t+V120

)
vc=
p
2Vsin(!t+V240

)
When it supplies a balanced load,
ia=
p
2Isin(!t+I)
ib=
p
2Isin(!t+I120

)
ic=
p
2Isin(!t+I240

)

16. The instantaneous power is
p=vaia+vbib+vcic
p=
p
2Vpsin(!t+V)
p
2Ipsin(!t+I)
p
2Vpsin(!t+V120

)
p
2Ipsin(!t+I120

)
p
2Vpsin(!t+V240

)
p
2Ipsin(!t+I240

)
p=VpIpcos(VI) +VpIpcos(2!t+V+I)
VpIpcos(VI) +VpIpcos(2!t+V+I120

)
VpIpcos(VI) +VpIpcos(2!t+V+I240

)
p= 3VpIpcos
where=VI.
The instantaneous power in a 3 phase balanced system is constant.

17. !tp;v;ip
Figure:

18. The average/real power in a 3-phase system is
P= 3VpIpcosWatts
In a Y connected load,VL=
p
3VpandIL=Ip,
P=
p
3VLILcos
In a connected load,VL=VpandIL
p
3Ip,
P=
p
3VLILcos
Therefore, the three phase real power is
P= 3VpIpcos=
p
3VLILcos

19. Since the instantaneous power in a 3-phase balanced system is
constant, it does not mean that there is no reactive power. Still
the instantaneous power of individual phases is pulsating.
The 3-phase reactive power is
Q= 3VpIPsin=
p
3VLILsinVAr
The apparent power is
jSj=
p
P
2
+Q
2
= 3VpIp=
p
3VLILVA

20. Per Phase Analysis
If a three phase system is balanced and there is no mutual
inductance between phases, it is enough to analyze it on per phase
basis.
1. Y
connections.
2. avariables using the phaseacircuit with
neutrals connected.
3. avariables
using the symmetry.
4.

21. Synchronous Machine Model
The per phase equivalent circuit of a synchronous machine is
+EXsIaRa+Vt
0

22. Transformer Model
The per phase equivalent circuit of a transformer is
ReqXeqRcXm
ISince the impedance of the shunt path is larger,RcandXm
are neglected.
ISinceReqis much smaller thanXeq,Reqcan be eliminated.
Xeq

23. Example 2: Consider a system where a three phase 440 V, 50 Hz
source is supplying power to two loads. Load 1 is a connected
load with a phase impedance of 10\30

and load 2 is a Y
connected load with a phase impedance of 5\36:87

.
1.
system.
2. F of a three phase
bank of delta connected capacitors to be added in parallel to
the load to improve the overall power factor unity. Find the
line current under this condition.

24. 1. Y
transformation,
+
440
p
3
IL
10
30

3
536:87

IL=
440=
p
3
10=3
30

+
440=
p
3
536:87

IL= 106:64|68:6A
IL= 126:8
32:75

A
pf= cos(32:75

) = 0:84 lag

25. 2.
+
440
p
3
IL
10
30

3
536:87

|
Xc
3
To make overall power factor unity,ILmust be in phase with
the voltage.
)Ic=|68:6A
Xc=
3440
p
368:6
= 11:11
C= 286:52F

26. Example 3: Consider a system where a three phase 400 V, 50 Hz
source is supplying power to two loads. Load 1 draws 5 kW at 0.8
pf lagging and load 2 draws 5 kW at unity power factor. The
voltage across the loads is 400 V.
1.
2.
connected across the loads to improve the overall power factor
to unity. Determine the line current under this condition.
1.
IL1=
5000
p
34000:8
= 9 A
IL2=
5000
p
34001
= 7:2; A
IL= 9
36:87

+ 7:20

= 15:3820:56

A
pf= 0:9363 lag
QT=Q1+Q2= 3:75 + 0 = 3:75 kVAR

27. 2.
QC+QT= 0
QC=3:75 kVAR
(-ve indicates that the capacitor supplies reactive power.)
)QC= 3:75 kVAR
Whenpfis unity,S=P.
IL=
PT
p
3VL
=
1010
3
p
3400
= 14:4 A
IL= 14:4
0

A

28. per unit
The per unit is dened as
per unit =
actual value in any unit
base value in the same unit
There are normally four quantities associated with a power system.
S;V;I;Z
How to nd base quantities?
IChoose any two. NormallySbaseandVbaseare chosen.
IFind the remaining two using their relations.

29. Let us start with single phase.
Sb=S1MVA;Vb=V1kV
Ib=
Sb(MVA)
Vb(kV)
kA
Zb=
Vb(kV)
Ib(kA)

SubstitutingIbinZb,
Zb=
V
2
b
( in kV)
Sb( in MVA)

Zp.u.=
Zactual()
Zb()
)Zp.u.=
Zactual()Sb( 1MVA)
V
2
b
( L- N in kV)

30. For three phase.
Sb=S3MVA;Vb=VLLkV
Ib=
Sb(MVA)
p
3Vb(kV)
kA
Zb=
Vb(kV)
p
3Ib(kA)

SubstitutingIbinZb,
Zb=
V
2
b
( in kV)
Sb( in MVA)

Zp.u.=
Zactual()
Zb()
)Zp.u.=
Zactual()Sb( 3MVA)
V
2
b
(L-L in kV)

31. S
3
b
= 3S
1
b
;VbLL
=
p
3VbLN
S
3
p.u.=
S
3
S
3
b
=
3S
1
3S
1
b
=Sp.u.
Vp.u.=
VLL
VbLL
=
p
3VLN
p
3VbLN
=Vp.u.
IIf the voltage magnitude is 1 p.u., the line-line voltage is 1
p.u. and the line-neutral voltage is also 1 p.u.
ISimilarly, three phase power in p.u. and the single phase
power in p.u. are the same.
Zb=
(VbLL
)
2
S
3
b
=
(
p
3VbLN
)
2
3S
1
b
=
(VbLN
)
2
S
1
b

32. Impedance values of a component when given in per unit without
specied bases are generally understood to be based on the MVA
and kV ratings of the component.
To change p.u. from one base to new base:
Zp.u./
Sb
V
2
b
Z
p.u. (new)=Z
p.u. (given)
Sb(new in MVA)
Sb(given in MVA)

V
2
b
(given in kV)
V
2
b
(new in kV)
Advantages:
IThe per unit values of impedance, voltage and current of a
transformer are the same regardless of whether they are
referred to the HV side or LV side. This is possible by
choosing base voltages on either side of the transformer using
the voltage ratio of the transformer.
IThe factors
p
3 and 3 get eliminated in the per unit power
and voltage the equations.

33. Example 3 : Let us do the same example in p.u.
1.
Sb= 5 kVA;Vb= 400 V
P1p.u. =
5
5
= 1;P2p.u. =
5
5
= 1
SinceIp.u.=
Pp.u.
Vp.u.pf
IL1p.u. =
P1
Vpf
=
1
10:8
= 1:25
IL2p.u. =
P2
Vpf
=
1
11
= 1
IL=IL1+IL2= 1:25
36:87

+ 10

= 2:1420:56

p.u.
Ib=
Sb
p
3Vb
=
510
3
p
3400
= 7:22 A.
IL= 2:14
20:56

7:22 = 15:4420:56

A

34. 2. QC=QT.
QTp.u. =PTp.u.tan= 2tan(20:56

) = 0:75 p.u.
QC=QCp.u.Sb= 0:755 = 3:75 kVAr
Capacitors supply reactive power.
When the problems to be solved are more complex, and
particularly when transformers are involved, the advantages of
calculations in per unit are more apparent .