EG501H-EG504B_AC circuit analysis EG501H-EG504B_AC circuit analysis

presentacion de curso

1. Electrical systems for RE (EG501H/EG504B)
AC circuit analysis
Dr. Ali Jamshidi Far
School of Engineering
University of Aberdeen

2. oAC Waveform
oPhasor presentation
oVoltage/current relationship for R, L,
and C
oVoltage/current relationship for RL, RC,
and RLC circuits
oApparent, Active and Reactive power
oResonance
o3-phase system
2
Content

3. 3
Sinusoidal AC waveform1
22
m
v(t)Vsin(t)
f
T


=
== m
v(t)Vsin(t)=+

4. oAssume we want to analyse an electrical circuit
oIn steady-state all currents and voltages of an electrical circuit are in the form of
➔ the frequency of all voltages/currents of the circuit is the same as w
➔therefore, all the voltages/currents can be given as magnitude and phase shift,
which is called as phasor
4x m_x x
x m_x x
v(t)Vsin(t)
i(t)Isin(t)


=+
=+ 22
22
m_x m_x
x rms_x x xrms_x rms_x x x
m_x rms_xm_x rms_x
VI
vV ,iI
VV,II
 == ==
==
Phasor presentation

5. 5
Phasor presentation
Phasor-domain
representation
Phasor-domain
representation
2 & 3 waveforms
Polar
form
Exponential
form
Rectangular
(Cartesian)
form

6. 6
Phasor presentation
oExamples:

7. oSingle resistor R
7m
m
RR
R m R m
V
m
R m R
I
v(t)Ri(t)
i(t)Isin(t)v(t)RIsin(t)
V
or:v(t)v(t)Vsin(t)i(t)sin(t)
R


=
= ===
== === 00
00 22
R rms mm
rms rms
R rms
iII VI
,VV,II
vVV
==
== ==
==

In phasor form:
Voltage/current relationship for R, L, and C

8. oSingle Inductor L
8
Voltage/current relationship for R, L, and C()
2
2
mm
L
L
L m L m m
VV
m m Lm L
di(t)
v(t)L
dt
i(t)Isin(t)v(t)LIcos(t)LIsin(t)
VLIXI,XLLf

 
 
=
= === = +
== == 00
22
22
LL
mm
LL
iI vV
VI
or ,V,I
vV iI

= =

==
= =−


In phasor form:
Current lags voltagedsin(t)
cos(t)
dt
dcos(t)
sin(t)
dt




=
=−

9. oSingle Capacitor C
92
11
mm
C
C
C m C m m
II
m m m m Cm C
dv(t)
i(t)C
dt
v(t)Vsin(t)i(t)CVcos(t)CVsin(t)
ICVVIXI,X
CC

 


=
= === = +
=== = 00
22
22
CC
mm
CC
vV iI
VI
or ,V,I
iI vV

= =

==
= =−


In phasor form:
Current leads voltage
Voltage/current relationship for R, L, and C

10. oRL circuit
The phase shift between voltage and current is something
between 0 and 90 deg
100
2
mm
i(t)Isin(t)v(t)Vsin(t),

 = = +
oAssume the current is with zero phase shift: 2
R L m m
di(t)
v(t)v(t)v(t)Ri(t)LRIsin(t)LIsin(t)
dt

=+=+= + +
oEquations in time domain:
Voltage/current relationship for R, L, and C

11. oRL circuit – Phasor form
110iI,vV == 2
0
2
R L m m
R L R L
v(t)v(t)v(t)RIsin(t)LIsin(t)
VVVVjV




=+= + +
=+=+ 22 L
RL
R
V
VVV,arctan
V
=+=
Voltage/current relationship for R, L, and C0
RL
RL
KVL:
v(t)v(t)v(t)
v(t)v(t)v(t)
−++=
=+
oThe magnitude V and phase Ɵ can be calculated as:L
ZRjX=+
oThe impedance Z can be calculated as:
oThe voltage v(t) in phasor form:

12. oRLC circuit
121
22
R L C
m
R L C m m
di(t)
v(t)v(t)v(t)v(t)Ri(t)L i(t)dt,
dtC
I
v(t)v(t)v(t)v(t)RIsin(t)LIsin(t)sin(t)
C

 

=++=++
=++= + ++ −
 00
22
R L C R L C
iI,VVVV VjVjV

==++−=+−
oIn phasor form:LC
ZRjXjX=+−
Voltage/current relationship for R, L, and C
oThe overall impedance Z can be calculated as:
oThe magnitude V and phase Ɵ can be calculated as:( )
2
2 LC
R L C
R
VV
VVVV,arctan
V

−
=+− = 0
m
i(t)Isin(t)
iI
=
=

13. 13
General equations about Vectors (Vector algebra or
complex number)
•Vector vs Scalar ?()( )
()( )
,axjybzjw
abxzjyw
abxzjyw
=+=+
+=+++
−=−+−
•Sum/Subtraction of
two vectors a & b
Example: Calculate a+b & a-b()()
()()
151364
151342
ab j j
ab j j
+=+++=+
−=−+−=−−
Solution:
We need vector algebra in power systems
Scalar like weight, temperature , etc
Vector like speed, voltage/current in power system, etc. 11,53ajbj=+=+

14. 14
•Multiplication and division of two vectors a & b( ) ( )
( )( )
, ,//
,
aAbBabAB abAB
axjybzjwabxzywjxwyz
  ===+=−
=+=+=−++
Example: Calculate a×b and a/b11,53
:1.4145,5.8331
oo
ajbj
ora b
=+=+
== ( )( )
( )
( )
1513131528
1.415.8345318.2276
/1.41/5.8345310.2414
o
o
ab j j
ab
ab
=−++=+
=+=
= −=
Solution:
General equations about Vectors
Check ??

15. 15() ()
22
cos,sin
,arctan
OPrxjy
xryr
y
rxy
x



→
==+
==

=+=


•Polar <-> cartesian
Example: Convert (2+j8) to polar
Solution:2 2 2 2
282,8
28688.24
8
arctanarctan76
2
o
xjyjxy
rxy
y
x

+=+==
=+=+==

= = =


General equations about Vectors
Example: Convert 8.24∠76 to Cartesian
Solution:() ()
() ()
8.24768.24,76
cos8.24*cos762,
sin8.24*sin768
o
rr
xr
yr



===
= = =
= = =

16. 16
•Calculation of voltage in the RL circuit:
Example: Assume V
R=40∠0 V and V
L=30∠90 V in above RL circuit.
Calculate the magnitude and phase of the supply voltage v(t).
Solution:0
2
RL
VVV

=+ 4003090V?=+= () ()( )
() ()( )
400400
40cos040,40sin00
3090030
30cos900,30sin9030
R
L
Vj
xy
Vj
xy
==+
= == =
==+
= == = ( )( )
( )( )
400030
4000304030
V j j
jj
=+++
=+++=+ 22 30
403050,arctan36.9
40
40305036.9
o
o
V
j


=+== =


+=
General equations about Vectors
We cannot sum two vectors in polar form. Therefore,
we need to convert them to Cartesian form
Adding them:
And convert back to polar form:

17. 17
•Calculation of voltage in the RLC circuit:
Example: Assume V
R=40∠0 V, V
L=30∠90 V and V
C=10∠-90 V in above RLC
circuit. Calculate the magnitude and phase of the supply voltage v(t).
Solution:40030901090V?=++−= 400400
3090030
1090010
R
L
C
Vj
Vj
Vj
==+
==+
=−=− 400003010
4020
V( )j( )
Vj


=++++−
=+ 22 20
402044.7,arctan26.6
40
402044.726.6
o
o
V
j


=+== =


+=
General equations about Vectors
We cannot sum two vectors in polar form. Therefore,
we need to convert them to Cartesian form
Now, they can be added:
And convert back to polar form:0
22
R L C
VVVV

=++−
Try to solve it now, please.

18. 18
oActive, reactive and apparent power (P, Q & S=P+jQ)
oS=V×I
*
where V and I are vectors (have both magnitude and phase)2
2
0,0
,
RR
iIvV
V
VRII
R
SPVIRIIRI
VV
SPVIV
RR
==
==
====
====
Power is always positive,
➔ Energy is always delivered to load
Apparent, Active and Reactive powercos()
sin()
PVI
QVI


=
=
oSingle resistor R
Direct equations to calculate P & Q
(Ɵ is the phase difference between
the voltage and current, which is
ZERO here)

19. 19
oSingle inductor L*2
* * 2
2
*
,,
L
LL
LL
LL
VV
VjXII jIII
jXX
SVIjXIIjXIjQ
VV
SVIVjjjQ
XX
= ==−=
====
====
Apparent, Active and Reactive power
Power is both positive and negative.
➔ Energy goes to load and returns to source2
L
XLfL== *
*
aAaA
axjyaxjy
==−
=+=−
General equations:
(Conjugate of vector a)
Reactive Power for L:
Please try to calculate a
2
and |a|
for above vector acos()
sin()
PVI
QVI


=
= 1
j
j
−
=

20. 20
oRL circuit
Apparent, Active and Reactive powerL
ZRjX=+
Apparent power:()
( )( ) ( )
2
**
2 2 2
**
22
L L L
L
P Q
SVIZIIZI
SVIRjXIIRjXIRIjXI
SPjQRIjXI
===
==+=+ =+
=+=+
P depends on I and R
Q depends on I and X
L ,(X
L depends on L & f)cos()
sin()
PVI
QVI


=
=
Or, you can use the below
direct equations:

21. 21
oExample: Calculate active and reactive power for below RL circuit
Apparent, Active and Reactive power22600.16060.3
6060.3
L
L
XLfL
ZRjX j
====
=+=+
Apparent, active and reactive power:*
12001.41145.1169.345.1169.3cos(45.1)169. 3sin(45.1)119.5119.9
119.5119.9119.5&119.9
o o o o
SVI j j
SPjQ j PWQVAr
==== + =+
=+=+= = *
1200
12001200
1.41145.11.41145.1
6060.38545.1
oo
o
V
V
II
Zj
=

== = =−=
+ cos()1201.411cos(45.1)119.5
sin()1201.411sin(45.1)119.9
o
o
PVI W
QVI VAr


= = =
= = =
Alternative solutions:2 2 2
2
2
2
,1.412,
602120,
60.32120.6
L
P Q
L
SPjQRIjXII
PRI W
QXI VAr
=+=+ ==
===
===
Solution:
First calculate impedance Z
Then voltage and current in phasor form:

22. 22
oExample: Calculate active and reactive power for below RLC circuit
Apparent, Active and Reactive power
Apparent, active and reactive power:*
23008.3968.61929.768.61929.7cos(68.6)192 9.7sin(68.6)
704.11796.6 704.1&1796.6
o o o o
SVI j
S j PjQPWQ VAr
==−=−= −+ −
=−=+= =− *
2300
23002300
8.3968.68.3968.6
1025.5227.468.6
oo
o
V
V
II
Zj
=

== = ==−
− − cos()2308.39cos(68.6)704.1
sin()2308.39sin(68.6)1796.6
o
o
PVI W
QVI VAr


= =−=
= =−=−
Alternative solutions:
Solution:
First calculate impedance Z
Then voltage and current in phasor form:( )( )
6
22500.2006.28
1 1 1
31.8
0.0125010010
106.2831.81025.52
L
C
LC
XLfL
X
C
ZRjXjXjj j


−
====
== ==

=+−=+−=− ( )
2
2
2
2
108.39703.9,
25.528.391796.4
LC
PRI W
QXXI VAr
===
=−=−=−

23. 23
oExample: Calculate the unknown current and voltages
Apparent, Active and Reactive power
Solution:
Impedance Z:
Voltage Vs and current I in phasor form:
And the voltages:

24. 24
oExample:
Apparent, Active and Reactive power
Solution:
Impedance Z:
Voltage Vs and current I in phasor form:

25. 25
oExample:
Apparent, Active and Reactive power
Solution (cont.):
And the currents:
Power drawn from the source:
Alternative wan for power dissipated (considering that power
only dissipates in resistor:

26. 26
•In AC circuits with both capacitor(s) and inductor(s), resonance occurs when
the capacitive reactance (X
C) and inductive reactance (X
L) are equal at a
specific resonant frequency.
•This results in the circuit behaving like a purely resistive component.
•For a series RLC:
Resonance
Example: Calculate the resonant
frequency of the RLC circuit. What is the
impedance seen at that resonant
frequency?
Solution:33
11
1000/
1010
1000
159.1
22
rads
LC
f Hz



−−
== =

===

27. 27
•Parallel RLC circuit:
•At a particular frequency ω:
Resonance
Example: Calculate the resonant
frequency of the RLC circuit. What is the
impedance seen at that resonant
frequency? 33
11
408/
210310
rads
LC

−−
== =

Solution:
The impedance at resonant frequency is Z=R=5Ω

28. 28
•Vs=10V
•Let’s monitor the current for
various voltage frequency in
Simulink

Resonance - Simulink
Freq.=10 Hz
➔ I
max=0.53 A
Freq.=159 Hz (w=1000 rad/s)
➔ I
max=1 A
Freq.=1000 Hz ➔ I
max=0.85 A
Q-factor determines the magnitude of the attenuation. 110.001
0.1
100.001
L
XLL
Qfactor
RRRC

−==== =
f=159 Hz

29. 29
•Previous Simulink example with
R=1 Ω
Resonance - Simulink
Freq.=10 Hz
➔ I
max=0.07 A Freq.=159 Hz (w=1000 rad/s)
➔ I
max=1 A
Freq.=1000 Hz
➔ I
max=0.1 A
Q-factor determines the magnitude of the
attenuation. 110.001
1
10.001
L
XLL
Qfactor
RRRC

−==== =

30. 30
oThe major portion of all electric power presently used in generation,
transmission, and distribution is (balanced) three-phase systems.
oThree phase system provides a balanced system with a constant instantaneous
power (important for smooth operation of electric machines).
oIt delivers more power with less conductor material compared with single-phase
or two-phase systems.
oAdding more phases does not improve efficiency enough to justify the cost.
3-phase AC system

31. 31
oAll the previous discussions apply to 3-phase AC systems as well.
3-phase AC system0
120 120
240 120120
o
Am
oo
Bm
o o o
C m m
v(t)Vsin(t)V
v(t)Vsin(t)V
v(t)Vsin(t)Vsin(t)V



= =
= −=−
= −= +=
oIn a balanced 3-ph AC system,
▪Sum of the phases voltages is zero
▪Sum of the phases currents is zero ()
( )
( )
120 120
120 120
Am
oo
Bm
oo
Cm
i(t)Isin(t)I
i(t)Isin(t )I
i(t)Isin(t )I

 
 
= −=−
= −−=−−
= +−=−

32. 32
oStar (wye or Y) vs Delta (Δ) connection
3-phase AC system3
LP
LP
VV
II
=
= 3
LP
LP
VV
II
=
=
Y-Y Connection
Y-Δ Connection
We also have Δ-Δ andΔ-Y connection.

33. 33
oStar (wye or Y) connection
3-phase AC system

34. 34
oDelta (Δ) connection
3-phase AC system

35. 35
oInstantaneous power
3-phase AC system
Phase voltage Phase current
Y-Connection:
Δ-Connection:

36. 36
oApparent (complex) power
3-phase AC system33
3cos()3cos()
3sin()3sin()
LL PP
LL PP
LL PP
SVIVI
PVI VI
QVI VI


==
==
==
Phase voltage Phase current
In summary:
Active power:
Reactive power:

37. 37
oExample: Assume a balanced 3-phase star (Y) load with phase load impedance
Z=10+j7 Ω is connected to a balanced 3-phase Y source with line voltage 400V.
Calculate phase (and line) current, power factor, and three-phase powers.
3-phase AC system
•Solution: Convert line to phase voltage and Z to polar
form: 400
230.9
33
L
P
V
VV=== 2 2 1 7
10712.21,tan 35
10
o
ZZ
−
=+== =


Power factor:() ()coscos350.819
o
==
Apparent, real and reactive powers:
Phase (line) current:230.9
18.91
12.21
P
LP
V
II A
Z
==== 3340018.9113.1
3cos()340018.91cos(35)10.73
3sin()340018.91sin(35)7.51
LL
o
LL
o
LL
SVI kVA
PVI kW
QVI kVAr


===
= = =
= ==
Check: ????
2 2 2
SPQ=+

38. 38
oExample: Repeat previous example with delta (Δ) connection.
3-phase AC system
•Solution: Convert Z to polar form: 2 2 1 7
10712.21,tan 35
10
o
ZZ
−
=+== =


Power factor:() ()coscos350.819
o
==
Apparent, real and reactive powers:
Phase and line currents:400
32.76 356.7
12.21
L
P L P
V
I AIIA
Z
===== 3340056.739.3
3cos()340056.7cos(35)32.2
3sin()340056.7sin(35)22.5
LL
o
LL
o
LL
SVI kVA
PVI kW
QVI kVAr


===
= = =
= ==
So, for the same line voltage and winding impedance, Δ connection
can deliver more power, but it needs higher insulation.

39. 39
Thanks for your attention!