Toolbox: Electrical Systems Dynamics
Dr. John C. Wright
MIT - PSFC
05 OCT 2010
Introduction
Outline
Outline
Image removed due to copyright restrictions. Please see
"How Electricity Gets from the Power Station to your Home."
PowerWise Teacher's Center, 2007.
AC and DC power transmission
Basic electric circuits
Electricity and the grid
2
SE T-6 Electrical Systems
AC
“The war of the currents”
Pros and Cons
One kills elephants
One has simpler infrastructure
Tesla
Why do we have AC and not
DC?
Look at a simple transmission
circuit to decide.
Use Voltage=120 VDC and
Power=1.2 GW
Edison
3
SE T-6 Electrical Systems
AC
Efficient transmission requires AC
Goals of the analysis
Find
Find
Find
Find
the
the
the
the
generator voltage
power delivered by the generator
power dissipated by the transmission line
ratio PTrans /PLoad
4
SE T-6 Electrical Systems
AC
Simple circuit models
A simple electric circuit
The current I =
VL
RL
The power to the load
VL2
2
PL = I RL = RL
Equating currents (from
Kirchhoff’s Laws), the
transmission line power
RT VL2
2
PT = I RT = RL
RT
I
+
VG
−
+
VT
− +
VL
RL
The power ratio is then the
ratio of resistances: PPTL = RRTL
Generator power �
PG = PL + PT = 1 +
−
5
2
RT VL
RL RL
Generator voltage
� �
VG = PIG = 1 + RRTL VL
�
SE T-6 Electrical Systems
AC
Simple circuit models
Efficiency requires most power is dissipated in
the load
Example for Al and household V
If RT /RL � 1, then
PT � PL
Most of the voltage
appears across the load.
So, we have very small
transmission losses.
For PL = 1.2 GW and VL = 120V
Then RL = PL /VL2 = 1.2 × 10−5 Ω
For transmission assume L = 50 km (a
short distance)
An Aluminum cable with A = 5 cm2
(to minimize sag, Cu not used.)
Resistivity of Al, η = 2.8 × 10−8 Ω-m
∴ RT = ηL/A = 2.8Ω � RL
Conclusion: not so good!
6
SE T-6 Electrical Systems
AC
Simple circuit models
AC can be used to increase the voltage
With AC we can use transformers
Step up the voltage at the generator
Transmit power at high voltage, low current
Step down the voltage at the load
Transmitting at low current should reduce transmission losses
7
SE T-6 Electrical Systems
AC
Simple circuit models
An Ideal Transformer
I1
I2
+
V1
−
n1
+
V2
−
n2
N =n2 /n1 = turns ratio
V2 =NV1
I2 =I1 /N
Physical process is conservation of
magnetic flux/energy
8
SE T-6 Electrical Systems
AC
Simple circuit models
An Ideal Transformer
Common examples of transformers:
I1
I2
+
V1
−
+
V2
−
n1
n2
N =n2 /n1 = turns ratio
V2 =NV1
I2 =I1 /N
Physical process is conservation of
Bottom right: Photo by mdverde on Flickr. Bottom left: Image by Tau Zero on Flickr.
magnetic flux/energy
Top right: Photo by brewbooks on Flickr.
8
SE T-6 Electrical Systems
AC
Analysis
AC transmission reduces losses
RT
+ V
T
I0
+
V1
−
VG
N1
I1
−
IL
+
V2
−
RL
N2
As before, PL = 1.2 GW, VL = 120 V , RL = 1.2 × 10−5 Ω
What are transformer and transmission requirements,
Such that PT � PL ?
9
SE T-6 Electrical Systems
AC
Analysis
From the circuit:
PL =VL IL = RL IL2
PT =VT I1 = RT I12
From the transformer
relation, IL = N2 I1 , it
follows
PT
1 RT
=
2
PL
N2 RL
10
SE T-6 Electrical Systems
AC
From the circuit:
PL =VL IL = RL IL2
PT =VT I1 = RT I12
Analysis
For N2 � 1 a huge reduction in
transmission losses
Practical numbers:
V0 = 12kV , V1 = 240kV (rms)
This implies that N1 = V1 /V0 = 20
From the transformer
relation, IL = N2 I1 , it
follows
PT
1 RT
= 2
PL
N2 RL
Assume small voltage drop across the
transmission line. Then V2 ≈ V1
Second turn ratio becomes
N2 = V2 /VL = 2000
Our transmission loss formula gives
PT /PL ≈ 6%
10
SE T-6 Electrical Systems
AC
Reactive power
The downside to AC: Reactive power
A down side to AC: Reactive power
Why? Load is not pure resistive
Load usually has an inductive component
Resistance absorbs power
Inductor circulates power back and forth
This oscillating power is the reactive power
11
SE T-6 Electrical Systems
AC
Reactive power
Resistors, inductors and capacitors, oh my!
There are three basic circuit elements having different Ohm’s laws.
Element
Resistor
Inductor
Capacitor
Symbol
R
L
C
Ohm’s Law
V = RI
dI
V = L dt
dV
dt
I = sin ωt
V = R sin ωt
V = L cos ωt
V = − C1 cos ωt
Phase shift
0
π/2
−π/2
Impedance
Z [Ω] = V /I
R
jωL
−j
ωC
12
= I /C
SE T-6 Electrical Systems
AC
Reactive power
Phase lags increase reactive power
2πt
Current, I = Im sin( Tperiod
)
2πt
Voltage, V = Vm sin( Tperiod
)
1
I /Im , V /Vm
0.5
0
−0.5
−1
0
1
2
3
4
t/Tperiod
13
SE T-6 Electrical Systems
AC
Reactive power
Phase lags increase reactive power
2πt
Current, I = Im sin( Tperiod
)
2πt
Voltage, V = Vm sin( Tperiod
)
Power,
1
2πt
P =I · V = Im Vm sin2 (
)
Tperiod
„
«
1
2πt
= Im Vm 1 − cos(2
)
2
Tperiod
I /Im , V /Vm
0.5
0
−0.5
−1
0
1
2
3
4
t/Tperiod
13
SE T-6 Electrical Systems
AC
Reactive power
Phase lags increase reactive power
2πt
Current, I = Im sin( Tperiod
+ φ)
2πt
Voltage, V = Vm sin( Tperiod
)
Power,
1
φ
2πt
2πt
P = I · V = Im Vm sin(
) sin(
+ φ)
Tperiod
Tperiod
„
«
1
2πt
= Im Vm cos(φ) − cos(2
+ φ)
2
Tperiod
I /Im , V /Vm
0.5
0
cos(φ) is known as the power factor
−0.5
−1
0
1
2
3
4
t/Tperiod
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SE T-6 Electrical Systems
AC
Reactive power
Reactive power must be supplied.
For parts of the AC cycle the instantaneous power is greater than the
average power
Generator must be able to deliver this higher power even though it is
returned later
Bottom line: generator must have a higher volt-amp rating than
average power delivered: VARs and Watts.
Higher rating → bigger size → higher cost
14
SE T-6 Electrical Systems
AC
Reactive power
Phase shifts are introduced by inductance
A motor will have an inductance, L.
L
I
V cos(ωt)
R
15
SE T-6 Electrical Systems
AC
Reactive power
Phase shifts are introduced by inductance
A motor will have an inductance, L.
It will introduce a phase shift given by tan φ =
ωL
R
L
I
V cos(ωt)
R
15
SE T-6 Electrical Systems
AC
Reactive power
Phase shifts are introduced by inductance
A motor will have an inductance, L.
It will introduce a phase shift given by tan φ =
Amplitude of current will also be reduced.
V
I = 2 2
(ω L + R 2 )1/2
ωL
R
L
I
V cos(ωt)
R
15
SE T-6 Electrical Systems
AC
Reactive power
Phase shifts are introduced by inductance
A motor will have an inductance, L.
It will introduce a phase shift given by tan φ = ωRL
Amplitude of current will also be reduced.
V
I = 2 2
(ω L + R 2 )1/2
This all follows from adding up the voltages for a simple circuit:
dI
L + RI = V cos(ωt)
dt
L
I
V cos(ωt)
R
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SE T-6 Electrical Systems
AC
Reactive power
How to minimize the VA requirement?
To minimize the VA requirements on the generator we want φ → 0
Assume the average power absorbed by the load is �PL �
Calculate the peak generator power [PG (t)]max as a function of �PL �
Note: The peak is 2× (rms volt-amp rating)
Less generator power is cheaper.
16
SE T-6 Electrical Systems
AC
Reactive power
Peak power from inductance
The power dissipated in the load:
�PL � = RI 2
RV 2
=
2 2
ω L + R2
The peak power delivered by the generator
Ppeak = VI (1 + cos φ) =
V2
(ω 2 L2 +
R 2 )1/2
�
1+
R
(ω 2 L2 + R 2 )1/2
�
Using the expression for �PL �, we get:
Ppeak
=
2 �PL �
�
ω 2 L2
+
�1/2
2
R
2R
17
+R
≥1
SE T-6 Electrical Systems
AC
Reactive power
Equivalent cir
Capacitance can balance out reactive power
Recall from our table that the phase
lags are opposite for inductance and
capacitance
tan φL =
ωL
R ,
tan φC =
−1
ωC
Short answer: there is a capacitance
that will keep the current and voltage
in phase (but not eliminate the power
factor)
L
C= 2
R + ω 2 L2
Long answer follows.
† Goal: Find C so there is no re
18
SE T-6 Electrical Systems
Analysis
† I – V relation for a capacitor
dV
Iˆˆ1(t ) = C G
dt
† I – V relation for the load
dIˆˆ2
ˆˆ
VG = RI 2 + L
dt
† Conservation of current
Iˆˆ(t ) = Iˆˆ1(t ) + Iˆˆ2 (t )
31
Solution
† Assume VG = V cos(Xt) (all voltages now rms)
† Current in the capacitor branch
Iˆˆ1(t ) = XCV sin(Xt)
† Current in the load branch (from before)
Iˆˆ2 (t ) =
V
(R
2
2 2 1/ 2
+X L
)
cos(Xt G)
32
The total current
† The total current flowing from the generator
Iˆˆ(t ) = Iˆˆ1(t ) + Iˆˆ2(t )
¯
cos(Xt G)
¡
=V ¡
XC sin(Xt)°°
1/
2
¡ (R 2 + X 2L2 )
°
¢
±
£¦
²¦
¯
¦¦ cos(Xt)cos G
sin
G
¡
° sin(Xt)¦¦
=V ¤
+
C
X
»
°
¦¦(R 2 + X 2L2 )1/ 2 ¡¡ (R 2 + X 2L2 )1/ 2
¦¦
°
¢
±
¦¥
¦
¼
33
The value of C
† Choose C for zero reactive power
† Set sin(t) coefficient to zero
XC =
sin G
(R
2
2 2 1/ 2
+X L
)
† Simplify by eliminating the power factor
L
C = 2
(R + X 2L2 )
34
Calculate the peak power
† Calculate the peak power to learn what has
happened to the VA rating
Ppeak = [PG (t )]max = 2 [VI ]max
= 2V 2
cos G
(R 2 + X 2L2 )1/ 2
RV 2
=2 2
(R + X 2L2 )
= 2 PL
35
The Result
† It worked!!
† The VA requirement has been reduced
Ppeak
VA =
= PL
2
36
AC
Reactive power
Discussion
AC is good for transmission
Have to manage reactive power
Other aspects:
HVDC transmission lines
AC losses from corona discharge
Voltage and frequency tolerances
Stability of the grid to perturbations, eg a power plant going offline
or a transmission line going down.
19
SE T-6 Electrical Systems
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