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EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
1
CHAPTER 1
REVIEW OF POWER NETWORK FUNDAMENTALS
1.1 Introduction
The purpose of this course is to introduce a number of engineering concepts involved in
the planning, operation and control of large power systems that are affected by
economical, social and environmental constraints.
Edison established the first electric power company in the US in 1880. The power
generation was d.c. The first a.c. transmission of power was around 1896 in Buffalo, New
York. Currently the installed generation capacity in the US is about 2.5 KW/person.
1.2 Voltage structure of the electric energy system
An electric power system, even the smallest one, constitutes an electric network of vast
complexity. The one factor that determines the system structure more than any others is
the system size.
By pointing out the great diversity in power system magnitude, we wish to make it clear
that there are no general rules regarding system structure that apply to all systems. It is
possible, however, to discern certain similarities characterizing the majority of power
systems.
As we shall see later, power transmittability increases and transmission losses decreases
with increasing voltage level. The larger the blocks of power to be transmitted and the
greater the distance over which they must be wheeled, the higher must be the operating
voltage chosen. The US standard operating voltages are given in Table 1.1.
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Table 1.1
Voltage Class Nominal line voltage
Low 120/240 V (single phase)
208 V
240 V
480 V
600 V
Medium 2.4 KV
4.16 KV
4.8 KV
6.9 KV
12.47 KV
13.2 KV
13.8 KV
23.0 KV
24.94 KV
34.5 KV
46.0 KV
69.0 KV
High 115 KV
138 KV
161 KV
230 KV
Extra High 345 KV
500 KV
765 KV
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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1.3 Generation
Transmission (138 KV and higher)
Subtransmission (23 KV-138 KV)
Distribution (upto 23 KV)
Very large
customers
TRANSFORMERS
Large customers
Small customer
Medium customers
Transmission Level
Transmission Level
Subtransmission
Level
Distribution Level
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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1.4 Generation Mix
Fossil (Coal, gas, oil) 75 %
Hydro 11 %
Nuclear 13 %
Renewable: 1 %
Fossil Units (coal, oil, gas): The heat is obtained from burning fuel from steam of high
temperature and pressure. The thermal energy transformed into mechanical form in
turbine that will drive on electric generator. These units have low efficiency (energy
waste), low reliability (very complex), pollution problem and are difficult to control.
Hydro Units: Hydropower - converts the energy in flowing water into electricity. The
quantity of electricity generated is determined by the volume of water flow and the
amount of head, the height from turbines in the power plant to the water surface created
by the dam. The greater the flow and head, the more electricity is produced. With a
capacity of more than 92,000 MW - enough electricity to meet the energy needs of 28
million households - the U.S. is the world's leading hydropower (including pumped-
storage) producer. Hydropower supplies 49% of all renewable energy used in the U.S.
The dam construction cost is high which is usually combined with irrigation projects.
Nuclear Units: The heat source is the nuclear reactor that generates high grade heat to
produce steam. No air pollution, minute amount of fuel, high initial cost, high
construction time lag, problems with radiation. It is the probable winner in future of
electric power.
Renewable (Green Power Sources): Geothermal, Solar, wind, tidal, geothermal,
nuclear fusion.
Resources such as geothermal, biomass, wind, small hydro and solar generating
technologies could sell their power production into the green energy market, or directly to
consumers with a higher price. A summary of major renewable sources of electricity is
given as follows:
Geothermal energy: Geothermal energy is the heat transferred from the inner part of
the earth to underground rocks or water located relatively close to the earth’s surface.
Molten rock (magma), which is located several miles below the earth’s surface,
produces heat or steam, which heats a section of the earth's crust and warms
underground pools of water (geothermal reservoirs). By making an opening through
the rock to the surface, the hot underground water may flow out to form hot springs,
or it may boil to form geysers. Geothermal reservoirs may provide a steady stream of
hot water that is pumped to the earth's surface by drilling wells deep below the
surface of the earth. Geothermal energy may be used to produce electricity, where
steam is either conveyed directly from the geothermal reservoir or from water heated
to make steam and piped to the power plant. The Geysers Geothermal Field located in
the northern California is one source of geothermal power where this power plant is
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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considered the largest source of geothermal energy in the world and produces as
much power as two large coal or nuclear power plants.
Biomass energy - the energy contained in plants and organic matter - is one of the
most promising renewable energy technologies. Instead of conventional fuels, the
technology uses biomass fuels - agricultural residues, or crops grown specifically for
energy production - to power electric generators. Today, biomass energy account for
nearly 45% of renewable energy used in the United States. Biomass is used to meet a
variety of energy needs, including generating electricity, heating homes, fueling
vehicles and providing process heat for industrial facilities. In the last few decades,
biomass power has become the second largest renewable source of electricity after
hydropower. Hydropower and biomass plants provide baseload power to utilities.
Biomass power plants are fully dispatchable (i.e. they operate on demand whenever
electricity is required). About 350 biomass power plants with a combined rated
capacity of 7000 MW feed electricity into the nation's power lines, while another 650
enterprises generate electricity with biomass for their own use as cogenerators.
National renewable energy laboratory (NREL) researches have helped lower the cost
of ethanol fuel from these sources to $1.22 per gallon. The target of current
researches is 70¢ per gallon.
Photovoltaic (PV) systems - most commonly known as solar cells - convert light
energy into electricity. PV systems are already an important part of our lives. They
power many of small calculators and wrist watches. More complicated PV systems
provide electricity for pumping water, powering communications equipment, and
even lighting our homes and running our appliances. In a surprising and increasing
number of cases, PV power is the cheapest form of electricity for performing many
tasks. Costs have dropped from 90¢ per KWh in 1980 to 22¢ today. Photovoltaics are
cost competitive in rural and remote areas around the world. The National
Photovoltaics Center at NREL is leading federal efforts to improve performance and
lower costs.
Wind energy projects provide cost-effective and reliable energy in the U.S. and
abroad. The U.S. wind industry currently generates about 3.5 billion KWh of
electricity each year, which is enough to meet the annual electricity needs of 1 million
people. Wind energy installations are going up across the country as generating
companies realize the benefits of adding clean, low-cost, reliable wind energy to their
resource portfolios.
Solar thermal electric (STE) technologies - parabolic troughs, power towers, and
dish/engine systems - convert sunlight into electricity efficiently and with minimum
effect on the environment. These technologies generate high temperatures by using
mirrors to concentrate the sun’s energy up to 5000 times its normal intensity. This
heat is then used to generate electricity for a variety of market applications, ranging
from remote power needs as small as a few kilowatts up to grid-connected
applications of 200 MW or more. Solar-thermal electricity is the least cost electricity
for grid-connected applications available today, and it has the potential for further,
significant cost reductions. While not currently competitive for utility applications in
the United States, the cost of electricity from STE can be competitive in international
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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and domestic niche applications, where the price of energy is higher. The goal for
advanced STE technologies is to be below 5¢/kWh. The U.S. annually uses more than
71 trillion BTUs of solar energy (1 million BTU equals 90 pounds of coal or 8 gallons
of gasoline). The residential and commercial sectors use 60 trillion BTUs, the
industrial sector 11 trillion BTUs and utilities 500 billion BTUs.
Ocean thermal energy conversion (OTEC), is an energy technology that converts
solar radiation to electric power. OTEC systems use the ocean's natural thermal
gradient - the fact that the ocean's layers of water have different temperatures - to
drive a power producing cycle. OTEC system can produce a significant amount of
power as long as the temperature between the warm surface water and the cold deep
water differs by about 20°C (36°F). The oceans are thus a vast renewable resource,
with the potential to help us produce 1013
of watts of electric power. The economics
of energy production have delayed the financing of OTEC plants. However, OTEC is
very promising as an alternative energy resource for tropical island communities that
rely primarily on imported fuel.
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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1.5 Fundamentals of Power i(t)
v(t) = Vrms Sin ( t + ) v(t)
i(t) = Irms Sin ( t + )
p(t) = v(t) . i(t) = Vrms . Irms . Sin ( t + ) . Sin ( t + )
V = Vrms , I = Irms (Phasors)
S = V *I Complex Power
= Vrms . Irms - = Vrms . Irms Cos ( - ) + j Vrms . Irms Sin ( - )
= P + jQ
S = Complex Power (Volt Amp)
P = Real Power (Watts) S Q
Q = Reactive Power (Vars)
( - ) = Power factor angle
Cos ( - ) = Power factor
P
Power Triangle
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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IRV
V I
( - ) = 0 P = Vrms . Irms Q = 0
V ILjV )(
I
( - ) = 90 P = 0 Q = Vrms . Irms
I )/( CjIV
V
( - ) = -90 P = 0 Q = -Vrms . Irms
I
V
-90 < ( - ) < 0 P > 0 Q < 0
V
I
0 < ( - ) < 90 P > 0 Q > 0
V I
Lagging
Leading
I V
R
L
C
C
R
L
R
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Example 1.
Assume
)10cos(2
150)( ttv
)50cos(2
5)( tti
a) Determine the complex power drawn by load.
b) Determine load’s power factor.
c) Find the amount of additional reactive shunt capacitance across the load that would change
power factor to 0.9 (lagging).
Load
I
V
1.
)10cos(2
150)( ttv , )50cos(
2
5)( tti
In phase quantities
102
150V V, 50
2
5I A
1007.106V V, 5054.3I A
a)
jQP
j
IVS
97.32462.187
6024.375
)5054.3()1007.106(
*
187.62
375.24
60o
324.97
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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b)
5.0)60cos()]50(10cos[Pf
Since current is lagging the voltage 5.0Pf lagging
c) 84.259.0coscos9.0 1
newPf
Notice that P (real power has not changed upon connecting the capacitor)
86.9084.25tan62.187tan PQnew var
Reactive power supplied by the capacitor is
11.23486.9097.324CQ var
__________________________________________
Example 2.
431 jZ Ω
102Z Ω
If Ps = 1100 W, find Vs and Is, P1 and P2.
Z1 Z2
IS
VS
.
2Z
3 Ω
j4 Ω
10 Ω SV
1Z
SI
Ps = 1100W
16.298.203.3668.31.176.13
13.5350
1043
)10)(43(j
j
jZ T
Ω
03.3621.1921.1998.2
1100S
T
SS I
R
PI
069.70)03.3668.3)(03.3621.19(* TSS ZIV V
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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80060013.53100013.535
69.70
43
)69.70( 22
*1
2
1 jjZ
VS
S
VA
P1 = 600 W
Similarly,
50010
69.70
10
)69.70(2
22
*2
2
Z
VS
S VA
P2 = 500 W
___________________________________________
1.6 3 Phase Network (Balanced network)
ca c ab
3030 VVVV aba
903120 VVVV bcb
a
1503120 VVVV cac
b
bc The sum of three phasors is zero
Per Phase Analysis of balanced system
a
b
c
a a’
y-connected
n ---(neutral line carries zero current)---- n’
c b b’ c’
Source
Load
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Per phase model
a a’
n n’
1.7 - Y Transformation
a ( abbcca ZZZ ,, are given)
Za )/().( cabcabcaaba ZZZZZZ
)/().( cabcabbcabb ZZZZZZ
Zca Zab )/().( cabcabcabcc ZZZZZZ
Zc Zb
c Zbc b
a ( cba ZZZ ,, are given)
)/()...( caccbbaab ZZZZZZZZ
Za )/()...( aaccbbabc ZZZZZZZZ
)/()...( baccbbaca ZZZZZZZZ
Zca Zab
Zc Zb
c Zbc b
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Example: (Balanced system)
a a’
j1
1 0
ZC ZL ZC
n n’
ZL ZL
c b j1 b’ ZC c’
j1
Assume 10jZ L 10jZC Find Ia, ICap, SLoad.
Solution: Transform cap to Ycap,
where 3/103/ jZZ CC LoadZ = (j10) || (-j 10/3) = -j5
aI = 1 0 / (j1 – j5) = .25 90
V a’n = LoadZ aI = -j5 (j.25) = 1.25 0
V a’b’ = 3 (1.25) 30 = 2.165 30
I Cap = (2.165 30) / -j 0.10 = .2165 120
LoadS = 3 V a’n *aI = 3 x (.3125 -90)
V bn = 1 -120 V cn = 1 120
V a’b’ = 2.165 30 V b’c’ = 2.165 -90 V c’a’ = 2.165 150
aI = .25 90 bI = .25 (90-120) cI = 1 (90+120)
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Example: Power Factor Correction:
The aim is to reduce power factor angle to zero in order to minimize the reactive power
consumption.
We can adjust the power factor by adding – to the circuit, inorder to reduce ( - ) = 0
Example:
A 3- load draws 25 KVA (3 phase) at 0.6 lagging power factor from a 110 V (line
voltage) source. Determine the KVA rating of the capacitor in order to increase power
factor to 0.85 lagging. Also, calculate the line current after adding the capacitor.
Solution:
( - ) = Cos –1
(0.6) = 53
S = 25 KVA P = (25)(.6) = 15 KW Q =(25) Sin 53 = 19.96 Kvar
V S Q S Q
ST QT
I P P
QC
S = 3 V *I
I = 25000/ 3 (110) = 131.21 A (without caps)
By adding capacitor to the load, the real power consumption of the system will not
change.
So, ( - ) = Cos –1
(0.85) = 31.78
tan ( - ) = QT / P
So, QT = (15). (tan 31.78) = 9.29 Kvar
And QC = Q - QT = 10.67 Kvar.
After adding the capacitor, TS = P + j QT =15,000 + j 92,900
TS = 3 V *I
*I = TS / 3 (110) = 92.6 A (with caps)
53
53
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Example: Assume Ia = 10 A is in phase with Vbc = 208 volts. Determine the load if
a) Y connected
b) Δ connected
Ia
a
b
c
VLL=208
Balanced
Load
Solution:
VLL = 208 V, 1203
208V V,
Assume 0120anV , then 30208abV then,
9010aI A (since it is in phase with Vbc)
a) Y connected load
129010
0120jZ
Z
VI
ana
Ω
b) Δ connected load
Ia
Ib
Ic
Ica Iab
Ibc
Using KCL, caaba III
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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So if we assume the following relation among the balanced three phase currents,
Ica
Iab
Ibc
-30o
Ia
Ica
Iab
Ibc
Then, 303 abcaaba IIII
Accordingly,
6077.5
303
9010
303
aab
II
Since 30208abV then,
36jI
VZ
ab
ab
______________________________________________________
2. PER UNIT SYSTEM
Definition: Per Unit = actual / base
We summarize the advantages of the pu system as follows:
1) The difference between the single phase and the 3 phase values will be eliminated.
2) The turns ratios of transformers will be eliminated.
3) Actual values of quantities differ significantly depending on the rating or capacity of
the system. However, the pu values are fairly constant. This makes it possible to
estimate unknown pu quantities.
2.1 Single Phase:
If we specify the base values for power (Sb) and voltage (Vb), we will be able to
determine the base values for all quantities. In other words, I = S/ V and Z = V(2)
/ S
All laws of circuits apply to PU quantities as well.
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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For Example:
If V = I (Ohm’s law)
then for base values, Vb = b Ib
if, we divide the first equation by the second one,
V / Vb = I / b Ib then,
V PU = PU IPU (this is the per unit equation which follows the Ohm’s law)
Also, S = VI (amplitude) and Sb = Vb Ib
Hence, S / Sb = VPU IPU
or, SPU = VPU IPU
SPU = PPU + j QPU , where PPU = P / Sb and QPU = Q / Sb
_________________________________________
2.2 Per Unit For a 3-phase system:
base values: Sb(3)
= 3Sb Vb(3)
= 3Vb(1)
per unit values: SPU(3)
= S(3)
/ Sb(3)
= 3S
(1) / 3Sb
(1) = SPU
(1) (indicating that the single
phase value is the same as the three phase value once specified in PU)
VPU(3)
= V(3)
/ Vb(3)
= 3V
(1) / 3Vb
(1) = VPU
(1) (indicating that the single phase value is
the same as the three phase value once specified in PU)
Zb(3)
= Vb(3)
. Vb(3)
/ Sb(3)
= ( 3Vb(1)
) 2
/ 3Sb(1)
= (Vb
(1))2 / Sb
(1) = Zb
(1)
(indicating that the single phase value is the same as the three phase value once specified
in PU)
SO, AS WE SPECIFY THE VALUES FOR PU, WE DO NOT HAVE TO
DISTINGUISH BETWEEN SINGLE PHASE AND 3 – PHASE VALUES.
____________________________________
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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2.3 For a transformer: I1 I2
V1 V2
1:n
Assume losses are negligible (ideal transformer).
If we define Vb and Sb for side 1, then
V2b = nV1b S1b = S2b
Since V1pu = V1 / V1b
Then, V2pu = V2 / V2b = nV1 / nV1b = V1 / V1b = V1pu
Since Z2b = n2Z1b
Then, Z1pu = Z1/ Z1b = Z2/ n2/ Z2b/ n
2 = Z2/ Z2b = Z2pu
So we can determine the pu values on either side of the transformer.
I1pu I2pu
r x
V1pu V2pu
2.4 Change of per unit base:
Z = Z /Zb = Z /(Vb) 2/Sb
If we are planning to determine the pu values in a new base.
Z0 = Z /(Vb0) 2/Sb0 old base
Zn = Z /(Vbn) 2/Sbn new base
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Since Z (actual value) is the same in both cases,
Example: Show the pu diagram
Assume the new base is 138 KV, 10,000 KVA at B. Find the value of impedance on 3
different bases.
1 : 10 2 : 1
R=300
A B C
Solution:
In order to represent the pu diagram of this network, we must consider a common base
for the entire network and determine the pu values of all components using the common
base.
For an ideal transformer, the MVA base will be the same throughout the network.
However, the KV base will change according to the turns ratios of transformers.
The old base values (input dada) on A and C sides are given as follows:
Power Voltage ratio
A - B 10,000 KVA 13.8/138 KV
B - C 10,000 KVA 138/69 KV
Zn = Z0 (Vb0/Vbn) 2
(Sbn/Sb0)
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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The base values for voltage, power and impedance at A, B and C are:
A B C
(138)(13.8/138) = 13.8 KV 138 KV 138 (0.5) = 69 KV
10,000 KVA 10,000 KVA 10,000 KVA
(13.8*1000)2/10,000,000 = 19 (138,000)
2/10,000,000 = 1900 (69,000)
2/10,000,000 = 476
Then the actual value of R:
C: R = 300 Rpu = 300/476 = 0.63
B: R = 300(2)2 = 1200 (actual value) Rpu = 1200/1900 = 0.63
A: R = 1200(.1)2
= 12 (actual value) Rpu = 12/19 = 0.63
So, we can eliminate the transformers and their turns ratios in our computations as the pu
value of R is the same based on 3 different bases.
Again: We list the advantages of the pu system as follows:
4) The difference between the single phase and the 3 phase values will be eliminated.
5) The turns ratios of transformers can be eliminated.
6) Actual values of quantities differ significantly depending on the rating or capacity of
the system. However, the pu values are fairly constant. This makes it possible to
estimate unknown pu quantities.
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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Example: Determine the pu diagram for the following system.
j40
j20 j20
G1 : 20MVA, 18KV X = 0.2
G2 : 20MVA, 18KV X = 0.2
M3 : 30MVA, 13.8KV X = 0.2
: 20MVA, 138 /20 X = 0.1
: 15MVA, 138 /13.8 X = 0.1
Assume the base is 138 KV, 50 MVA in the 40 line.
Solution:
XG1 = XG2 = 0.2 (18/20)2
(50/20) = 0.405 pu
XM3 = 0.2 (50/30) = 0.333 pu
XT1 = XT2 = XT3 = XT4 = 0.1 (50/20) = 0.25 pu
XT5 = XT6 = 0.1 (50/15) = 0.33 pu
Base line impedance = (138)2/50 = 381
X40 = 40/381 = 0.105 pu
X20 = 20/381 = 0.053 pu
1 2
3
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Review of Power Network Fundamentals
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j0.105
j0.053 j0.053
j0.25 j0.25
j0.25 j0.33 j0.33 j0.25
j0.405 j0.33 j0.405
__________________
4. COMPLEX POWER TRANSMISSION
Assume,
0DD VV II Z = jX
Since, *IVS DD
then,
= VD I (cos + sin ) + I +
= VD I cos (1 + tan ) 1V DV
= PD (1 + j )
- -
So, (1)
DS
where
PD = VD I cos
= tan
Also, if 1V = V1
Since *IVS DD
= DV [( 1V – DV ) / jX] *
= VD [ (V1 - VD) / jX] *
2
3
3
3
1
3
DS = PD (1 + j )
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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DS = ((V1 VD - + 90) / X) - (VD2
90 / X)
DS = PD + jQD
PD = (V1 VD sin ) / X
QD = ((V1 VD cos ) / X) - (VD2
/ X) (2)
From (1), QD = PD
So, using (1) and (2),
PD = (V1 VD sin ) / X
PD = ((V1 VD cos ) / X) - (VD2
/ X)
Now, if we eliminate by using sin2
+ cos2
= 1 then,
In this equation, if we assume V1 and X are fixed and plot VD as a function of PD for
different power factors,
VD lagging cos =1 leading
PD
So VD is stable for a leading power factor and as PF changes to lagging, the voltage
collapse occurs for a smaller PD. So, by providing enough capacitance at the load
side, we can control the voltage magnitude.
( PD + (VD2
/ X))2
= (V1 VD / X)2
- PD2
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
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HW # 1
1) If the voltage across the source is 1 volt,
find the Complex Power delivered to the load.
cZ = -j0.2 , R = 10 , LZ = j0.1
cZ LZ R
2) Find naV , nbV , ncV and baV .
a j0.1 a’ j0.1 a
’’
+ +
1 0 -j1 1 0
-
- c’ b
’ c
’’ b
’’
n
c
b
j0.1 j0.1
j0.1 j0.1
3) If Z = r + jx, determine the relation between r and x so that Va’ b
’ > Vab
(balanced system)
a j1 1 a’
+
1 0 Z
-
j1 1
c b c’ b’
j1 1
+
-
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Review of Power Network Fundamentals
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4) Determine bI if caV =10 0 and Z =2 -30. (System is balanced)
aI
Z
bI
5) In the following figure, 912 jZ Assume 0208abV and determine the 3-phase
power and abI
c b
a
abI
aE
aI
Z
6) In the following figure 6.14.11 jZ L , 18.02 jZ L
a) If 4601GV (line voltage), 000,151GP watts , Pf = 0.8 lagging, then determine LV (load)
b) If load is 30 KW at 0.8 Pf lagging, then determine 2GV and 2GP
SourceSource
1LZ 2LZ
2GV1GV
Z
Z
a
c
b
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Review of Power Network Fundamentals
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7) Consider the following case:
50 + j100
T1 T2
Load =
4 MVA at
0.8 pf lagging
G: 15 MVA 13.8 KV X = 0.15 pu
M: 10 MVA 13.2 KV X = 0.15 pu
T1 : 25 MVA 13.2 KV/161 KV X = 0.1 pu
T2 : 15 MVA 13.8 KV/161 KV X = 0.1 pu
Assume that the base MVA is 20 MVA and base voltage is 161 KV across the line.
Determine the pu diagram of the network.
8) Draw the pu diagram (assume base is 100 MVA, 154 KV on 20 + j80 line)
20 + j80
G1 T1 T2 G2
10 + j40 10 + j40
G1 : 50 MVA, 13.8 KV X = 0.15
G2 : 20 MVA, 14.4 KV X = 0.15
T1 : 60 MVA, 13.2/161 KV X = 0.1
T2 : 25 MVA, 13.2/161 KV X = 0.1
Load : 15 MVA, 0.8 pf lag
G
M
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
27
9) In the following figure, assume the base is 138KV, 20MVA on 25 line, and determine the PU
diagram. Also, if Vload = 18KV, determine the load current in p.u.
G: 15MVA, 13.8KV, X = .1 p.u.
M: 12MVA, 13.3KV, X = .12 p.u.
T: 20MVA, 13.5/160KV, X = .11 p.u.
Load = 5MVA at .6Pf
10) Find the real power loss
11) If 26165001 .V and 05002V Find the complete power for each machine and
determine whether they are delivering or receiving real and reactive power. Also find
power losses in the line.
j30
j25
Y/
Y/
/Y
/Y
Load
G
M
V
1
j1
-j1
I2
I1
V1
I
V2
+-
+-
0.7 j2.4
EE 420 – Analytical Methods in Power System
Review of Power Network Fundamentals
28
12) Find the power consumption by 1R and find 1I and 2I
13) Find the per unit diagram.
%XKV,MVA:G 182290
%XKV/,MVA:T 1022022501
%XKV/,MVA:T 611220402
%.XKV/,MVA:T 4611022403
%XKV/,MVA:T 811110404
%.XKV.,MVA.:M 5184510566
4481 .X:Line
43652 .X:Line
KV,MVA:Base 22100 (on the generator side)
14) Find the value of in order to maximize the real power DP received by load. Find the
value of DQ accordingly.
j1 Z= -j1
I2
I1
11R
12R
927 jS
G M
T1
T2
T3
T4
Line 1
Line 2
G
j1
11V 012V
DS