Lecture #1.
Introduction to Power System Operation and
Control
“Review of Three-Phase Power Fundamentals”
Dr. Salman Harasis
Basic Power Notation
• S = MVA 3- = 3-phase MVA
• S = MVA 1- = 1-phase MVA
• Q = MVAR 3- = 3-phase MVAR
(reactive)
• Q = MVAR 1- = 1-phase MVAR
(reactive)
• P = MW 3- = 3-phase MW (real)
• P = MW 1- = 1-phase MW (real)
• VL-L = Line-to-line voltage
• VL-G = Line-to-ground voltage
• I = Line (phase) current
•  = Phase angle between voltage
and current phasors (related to
power factor)
• R, L, C = Resistance, inductance
and capacitance
• Z, R, X = Impedance, resistance,
reactance
• f = Frequency
•  = Phase angle of voltage
phasors
2
Definitions of Voltage
Instantaneous Peak Line-To-Ground Voltage
(VL-G,Inst) = VL-L,RMS√ 2/ √ 3
RMS Voltage line-to-ground
(VL-G,RMS) = VL-L,RMS/ √ 3
Phase A – Black
Phase B – Red
Phase C – Blue
RMS Voltage - Green
Three-phase voltage each displaced
by 120 degrees
3
Definitions of Voltage (Cont.)
Instantaneous Peak Line-To-Ground Voltage in
per-unit = VL-G,Inst / (VL-L,RMS√ 2/ √ 3)
RMS Voltage line-to-ground in per-unit =
VL-G,RMS / (VL-L,RMS/ √ 3)
Phase A – Black
Phase B – Red
Phase C – Blue
RMS Voltage - Green
Three-phase voltage each displaced
by 120 degrees
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Basic Relationships
Power Angle Curve
VS at 
VR at 0
P
P
X
VS VR
P=
sinδ
X
0
90

180
5
Control of Real and Reactive Power Flow
V1+
V20
Note:
• Reactive power (VAR) flows “downhill” from higher voltage magnitude
to lower voltage magnitude.
• Real power (P) flows “downhill” to a more lagging angle
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Basic Relationships
" Impedance"
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Basic Relationships
MVA 3 = 3VL−LI = P3 + Q 3
2
2
P3 = MW3 = 3VL −LIcos() = MVA cos()
Q 3 = MVAR 3 =
3VL −LI sin() = MVA sin()
2
VL-L
MVA 3 =
Z
2
VL-L
MW3 =
R
MVAR 3
VL-L
=
X
2
power factor(pf ) = cos(tan −1 ( Q ))
P
Z = R2 + X2
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Basic Relationships
MVA 1φ = VL−GI
2
VL-G
MVA1 =
Z
2
VL-G
MW1 =
R
2
VL-G
MVAR1 =
X
VL −G =
VL-L
1
XC =
2fC
3
B C = 2fC
X L = L2f
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What is Short-Circuit Strength?
• A substation/bus can be designated by its “available” short-circuit
strength.
• Short-circuit strength can be expressed in terms of fault (short-circuit)
MVA, for example:
MVASys = 800 MVA
• For a substation at a given voltage, this can be expressed as an
equivalent fault (short-circuit) current:
I3LG =
10
What is Short-Circuit Strength? (Cont.)
• Short-circuit strength can also be represented in terms of a fault (shortcircuit) impedance (reactance) or most simply as it’s equivalent
inductance:
XL =
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What is Short-Circuit Strength? (Cont.)
• The short circuit impedance can be considered as a “thevenin”
impedance looking back into the system as viewed at the substation.
• Resistance “R” is often neglected in hand calculations, but, can be easily
modeled in simulation programs.
– Ratio of L and R is also referred to as X/R ratio.
L
60 Hz
Source
R
Circuit Breaker
Equivalent
Source impedance
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Capacitive Reactance (XC)
• Consider a 345 kV rated substation with a 100 Mvar 3φ capacitor bank:
Mvar3φ = 100 Mvar
• For this 345 kV substation, this can be described in terms of capacitive
reactance (XC), capacitance (C), and current (IC):
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Inductive Reactance (XL)
• Consider a 500 kV rated substation with a 180 Mvar 3φ shunt reactor:
Mvar3φ = 180 Mvar
• For this 500 kV substation, this can be described in terms of inductive
reactance (XL), inductance (L), and current (IL):
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Per Unit Calculations
• For many reasons, it is convenient to put electrical power quantities in terms
of “per unit” values of a selected “base” value
• The per unit value is simply the ratio of that quantity to the “base” value
• Base values are used for:
–
–
–
–
MVA, MW, MVAR (S, P, Q)
Voltage
Current
Impedance, Resistance, Reactance (Z, R, X)
• Two out of these four are specified (usually voltage and MVA) and the other two are
calculated
• Per unit bases and physical impedances do not change for variations in
operating conditions
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Some Basic
Per Unit Relationships
Sbase = MVAbase = Specified
MVA pu
(usually 100 or 1000 MVA 3)
MW and MVAR use the MVA base
Vbase = Specified L - L or L - G
Ibase =
Sbase1
VbaseL−G
=
Sbase3
Vpu =
3VbaseL−L
Ipu
2
Zbase
MVA
=
MVA base
VbaseL−G VbaseL−L VbaseL−L
=
=
=
Ibase
MVAbase
3Ibase
V
Vbase
I
=
Ibase
Z pu =
Z
Z base
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Example Problem 1
• 210 MW 3 is flowing in a line through a 230 kV L-L 60 Hz substation operating at rated voltage at 0.9
power factor. There is a 79 Mvar 3 capacitor bank in the substation. Consider the following:
– Select MVA base = 100 MVA 3
– Select V base = 230 kVL-L
230 kVL-L rated
210 MW3 at 0.90 PF
79 MVar3
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Example Problem 1 (Cont.)
a) What is Ibase?
b) What is Zbase?
c) What is the P, Q, and MVA flowing thru the substation on the line in
physical units and p.u.?
d) What is the p.u. operating voltage?
e) What is the reactance of the capacitor bank in physical units and p.u.?
What is the capacitance in physical units (Farads)?
f) What is the current flowing in the capacitor bank in physical units and
in p.u.?
i.
What if the voltage is operating at 1.045 p.u.?
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Example Problem 1 (Cont.)
a) What is Ibase?
b) What is Zbase?
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Example Problem 1 (Cont.)
c) What is the P, Q, and MVA flowing thru the substation on the line in
physical units and p.u.?
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Example Problem 1 (Cont.)
d) What is the p.u. operating voltage?
e) What is the reactance of the capacitor bank in physical units and p.u.?
What is the capacitance in physical units (Farads)?
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Example Problem 1 (Cont.)
f) What is the current flowing in the capacitor bank in physical units and
in p.u.?
i.
What if the voltage is operating at 1.045 p.u.?
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Summary
• For many reasons, it is convenient to put electrical power quantities in terms
of “per unit” values of a selected “base” value
• The per unit value is simply the ratio of that quantity to the “base” value
• Base values are used for:
–
–
–
–
MVA, MW, MVAR (S, P, Q)
Voltage
Current
Impedance, Resistance, Reactance (Z, R, X)
• Two out of these four are specified (usually voltage and MVA) and the other two are
calculated
• Per unit bases and physical impedances do not change for variations in
operating conditions
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