Power System Representation

1
 Problem 1.1
If 𝑣 = 141.4 sin(𝜔𝑡 + 30) 𝑉 and 𝑖 = 11.31 cos(𝜔𝑡 − 30) 𝐴, find
for each (a) the maximum value, (b) the rms value and (c)
the phasor expression in polar and rectangular form if the
voltage is reference. Is the circuit inductive or capacitive?
❖ Notations:
▪ Lowercase letter to indicate instantaneous values
▪ Capital letters to indicate phasors
▪ Vertical bars to indicate magnitudes (rms values)
▪ Single subscript notations
• Voltages with respect to reference node
• Currents from a point in the direction of arrow
▪ Double subscript notations
• To avoid polarity markings of voltages and arrows for currents.
• vab=va-vb, voltage of node a with respect to node b.
• Iab is current flowing from node a to node b
▪ Eulers identity e±jθ = 1(±𝜃) = Acosθ ± jAsinθ θ = ω𝑡?
 Problem 1.4
A single-phase AC voltage of 240 V is applied to a series circuit
whose impedance is 10∠60𝑜 Ω. Find R, X, P, Q and power factor of
the circuit.
If a capacitor is connected in parallel with the circuit, supplies 1250
var, find the P and Q supplied by the 240 V source, and find
resultant power factor.
Vm Im V2
2 Zcosθ
P= cos θ = V I cosθ = I = cosθ
2 Z
Vm Im 2 V2
Q= sin θ = V I sinθ = I Zsinθ = sinθ
2 Z
𝑃2 + 𝑄2 = V I
Q P
pf = cosθ = cos tan−1 =
P P 2 + Q2
Complex power: if V=|V|α and I=|I|β
𝑆 = 𝑃 + 𝑗𝑄 = 𝑉𝐼 ∗ = 𝑉 𝑒 𝑗𝛼 𝐼 𝑒 −𝑗𝛽 = 𝑉 𝐼 𝑒 𝑗(𝛼−𝛽) = 𝑉 𝐼 (𝛼 − 𝛽)
= V I cos α − β + j V I sin α − β
Example 1.1
 The Power Triangle

S1: inductive load

S2: Capacitive load
Helpful to evaluate the pf of
combined loads
In general: SR≠S1+S2
 Voltages and Currents in Balanced 3 Circuits
Operator ‘a’
Operator ‘a’ causes the rotation of 120𝑜 in counterclockwise direction.
It is complex number of unit magnitude with an angle of 120𝑜 .
𝑎 = 1∠120𝑜
Problem 1.12:
Evaluate the following expressions in polar form:
▪ 𝑎 −1
▪ 1 − 𝑎2 + 𝑎 Further:
▪ 𝑎2 + 𝑎 + 𝑗 Van=V, Vbn=V240˚, Vcn=V120˚
▪ 𝑗𝑎 + 𝑎2 Express as a function of operator a
and the evaluate following:
Vab, Vbc and Vca
 Per Unit system
❖ kV is the most convenient unit to express the transmission
voltage.
❖ Electrical power system is interconnected system with
various voltage levels and various capacity equipment.
❖ It is quite convenient to work with per unit system of
quantities with reference to common base.
❖ Per Unit (p.u) value of any quantity is defined as:
𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑖𝑛 𝑎𝑛𝑦 𝑢𝑛𝑖𝑡
𝑇ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑢𝑛𝑖𝑡
❖ Convention for per unit system:
▪ The value of 𝑆𝑏𝑎𝑠𝑒 is same for entire system.
▪ The ratio of 𝑉𝑏𝑎𝑠𝑒 on either side of transformer is same as the ratio of
transformer voltage.
 Base Quantities
𝑉𝑎𝑐𝑡𝑢𝑎𝑙
❖ 𝑉𝑝𝑢 =
𝑉𝑏𝑎𝑠𝑒
❖ We define Base kV and Base MVA.
❖ In Single Phase:
𝑆𝑏𝑎𝑠𝑒
▪ 𝐵𝑎𝑠𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑏 =
𝑉𝑏𝑎𝑠𝑒
𝑉𝑏𝑎𝑠𝑒 (𝑏𝑎𝑠𝑒 𝑘𝑉𝐿𝑁 )2
▪ 𝐵𝑎𝑠𝑒 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 Ω = 𝐼𝑏𝑎𝑠𝑒
= 𝐵𝑎𝑠𝑒 𝑀𝑉𝐴∅

❖ In three phase [∴ 𝑆3∅ = 3𝑆∅ , 𝑎𝑛𝑑 𝑉𝐿𝐿 = 3𝑉𝐿𝑁 ]

𝑏𝑎𝑠𝑒 𝑘𝑉𝐴3∅
▪ 𝐵𝑎𝑠𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑏 =
3 ∗𝐵𝑎𝑠𝑒 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑘𝑉𝐿𝐿
(𝑏𝑎𝑠𝑒 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑘𝑉𝐿𝐿 )2
▪ 𝐵𝑎𝑠𝑒 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 =
𝐵𝑎𝑠𝑒 𝑀𝑉𝐴3∅

❖ 𝐵𝑎𝑠𝑒 𝑃𝑜𝑤𝑒𝑟 𝑘𝑊 = 𝐵𝑎𝑠𝑒 𝑘𝑉𝐴

 Example 1.3 & 1.4
❖ The terminal voltage of a Y-connected load consisting of
three equal impedances of 20∠30𝑜 Ω is 4.4kV line to line.
The impedance of each of three lines connecting the load
to a bus at substation is 1.4∠75𝑜 Ω. Find the line to line
voltage at the substation bus.

❖ Repeat above problem by working with per unit system on

a base of 4.4kV, 127A.
 Changing of the Base
❖ Sometimes Impedances are expressed in some given
base.
❖ Computations are more convenient in some other base.
❖ Since all quantities should be expressed on same base,
therefore it is necessary to change the base of
impedances.
2
𝑏𝑎𝑠𝑒 𝑘𝑉𝑔𝑖𝑣𝑒𝑛 𝑏𝑎𝑠𝑒 𝑘𝑉𝐴𝑛𝑒𝑤
𝑍𝑝𝑢,𝑛𝑒𝑤 = 𝑍𝑝𝑢,𝑔𝑖𝑣𝑒𝑛
𝑏𝑎𝑠𝑒 𝑘𝑉𝑛𝑒𝑤 𝑏𝑎𝑠𝑒 𝑘𝑉𝐴𝑔𝑖𝑣𝑒𝑛
 Problem 1.22 & 1.23
❖ A generator is rated at 500MVA, 22kV. Its Y-connected
winding have a reactance of 1.1 per unit. Find the ohmic
value of the reactance.

❖ If above generator is connected in a circuit for which the

bases are specified as 100MVA, 20kV. Find per unit value
of reactance on new specified base.
 Advantages of Per Unit System
❖ Circuits are simplified
❖ Voltages have same range in per unit in all parts of the
system from EHV system to distribution and utilization
❖ Voltages, currents and impedances expressed in per unit
do not change when they are referred from one side of
transformer to the other side. This is a great advantage.
❖ Manufacturers usually specify the impedances of
machines and transformers in per unit or percent of name
plate ratings.
 Practice Problem
❖ 1ϕ
▪ Solve for the current, load voltage and load power in the circuit shown
below using per unit analysis with an SB of 100 MVA, and voltage base is
8 kV in the left part.

j1Ω 1:10 j24Ω 5:1 j1Ω

8kV 10Ω
 Practice Problem
❖ 3ϕ
▪ Solve for the current, load voltage and load power in the previous
circuit, assuming a 3f power base of 300 MVA, and line to line voltage
base of 13.8 kV on left side (square root of 3 larger than the 1ϕ example
voltages). Also assume the generator is Y-connected so its line-to-line
voltage is 13.8 kV.

▪ Convert to per unit as before. Note the system is exactly the same!
 Use of Symbols
❖ Symbols are used to represent the power system.
❖ Purpose is to simplify the power network.
❖ ANSI and IEEE has published standard symbols for
electrical diagrams.
❖ Electrical elements are shown by standardized schematic
symbols.
❖ Power system simulators uses symbols for electrical
components.
Use of Symbols
 One Line Diagram
❖ Conventional Diagram:
▪ Power systems are extremely complicated electrical networks.
▪ In three phase network, each power circuit consists of three conductors.
▪ A complete conventional diagram showing all connections is very
complicated and impractical.
❖ One Line Diagram:
▪ Simplified notation for representing a three-phase power system.
▪ It has its largest application in power flow studies.
▪ Instead of representing each of three phases with a separate line or
terminal, only one conductor is represented.
▪ A concise way of representing the basic arrangements of power system
components.
 One Line Diagram
❖ Draw OLD for following system
▪ Two Generators, one grounded through reactor, one through resister,
and load A are connected at bus 1.
▪ Bus 1 connected to transmission line through step-up transformer.
▪ One generator, grounded through reactor and load B connected to bus 2
▪ Bus 2 connected to opposite end of transmission line through
transformer.
▪ Every component is attached with other through circuit breaker.
 Impedance Diagram
❖ Used to calculate the performance of circuit under load
conditions.
❖ All components are represented by their equivalent circuits
to form per phase impedance diagram.
❖ Procedure:
▪ The single-phase transformer equivalents are shown as ideals with
impedances on appropriate side (LV/HV).
▪ The generators are represented as constant voltage sources with series
impedance.
▪ The transmission lines are approximated by their equivalent Π-Models.
▪ The loads are assumed to be passive and are represented by a series
branch of resistance or reactance
▪ Since the balanced conditions are assumed, the neutral grounding
impedances do not appear in the impedance diagram
 Reactance Diagram
❖ Reactance diagrams are commonly used in fault analysis.
❖ We simplify our calculations by omitting parameters which
have little impact on fault calculations.
❖ Parameters to be omitted in reactance diagram
▪ Resistances
▪ Static Loads
▪ Shut admittances of transformers
▪ Transmission line capacitances
Reactance Diagram

Impedance diagram

Reactance diagram
 Practice Problem

1:5 6:1

25kW+j10kVar
❖ Suppose rated voltage at load is 340V, and actual voltage
is 330 V.
❖ Impedances of both transformers are 0.5 + 𝑗1Ω referred to
HV side, and impedance of line 𝑙𝑚 is 1 + 𝑗2Ω.
❖ Determine sending end voltage by Working with
▪ Actual System
▪ Per unit system taking rated voltage as base, and base kVA of 100.
 Matrix Representation of Power System
❖ Junctions formed by connecting two or more circuit
elements connected to each other are called nodes.
❖ Systematic formulation of equations by applying
Kirchhoff’s current law is the basis of excellent computer
solutions of power system problems.
❖ Node equations are formed by applying Kirchhoff's current
law at each bus of the network.
❖ Resulting expressions lead to the development of network
matrix.
Matrix Representation of Power System
 Admittance and Impedance matrices
❖ General form of Admittance matrix for four bus network:
𝑌11 𝑌12 𝑌13 𝑌14
𝑌21 𝑌22 𝑌23 𝑌24
𝑌𝑏𝑢𝑠 =
𝑌31 𝑌32 𝑌33 𝑌34
𝑌41 𝑌42 𝑌43 𝑌44
▪ Node Admittance:
• 𝑌𝑗𝑗 = σ 𝑌𝑖𝑗 = 𝑆𝑢𝑚 𝑜𝑓 𝑎𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑛𝑜𝑑𝑒 𝑗
▪ Branch Admittance:
• 𝑌𝑖𝑗 ൧𝑖≠𝑗 = −𝑁𝑒𝑡 𝑎𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑏𝑟𝑎𝑛𝑐ℎ 𝑏𝑒𝑡𝑦𝑤𝑒𝑒𝑛 𝑛𝑜𝑑𝑒 𝑖 𝑎𝑛𝑑 𝑗

❖ Inverting 𝑌𝑏𝑢𝑠 Yeilds 𝑍𝑏𝑢𝑠

𝑍11 𝑍12 𝑍13 𝑍14
𝑍21 𝑍22 𝑍23 𝑍24
𝑍𝑏𝑢𝑠 = 𝑌𝑏𝑢𝑠 −1 =
𝑍31 𝑍32 𝑍33 𝑍34
𝑌= 𝑍 −1 =
1 𝑍41 𝑍42 𝑍43 𝑍44
𝑅 + 𝑗𝑋
 Example 2.2
A single-phase transformer has 2000 turns on the primary
winding and 500 turns on the secondary. Winding
resistances are r1=2.0Ω and r2=0.125Ω. Leakage reactances
are x1=8.0Ω and x2=0.5Ω. The resistance load Z2 is 12Ω. If
the applied voltage at the terminals of the primary windings is
1200V, find V2 and the voltage regulation. Neglect
magnetizing current
Example 2.6
Problem 2.2
Problem 2.10
 Homework 1
❖ Examples:
▪ 1.1,1.3, 1.4, 1.5, 2.2, 2.6, 2.7, 2.8, 2.9
❖ Problems:
▪ 1.1, 1.4,1.5,1.8,1.9, 1.12,1.13, 1.22, 1.23, 2.1, 2.2, 2.7, 2.10, 2.11, 2.12,
2.14, 2.15