Power System

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.



2

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



3

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



4

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



5

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



6

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.



7

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



8

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



9

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



10

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



11

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



12

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



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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)



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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



15

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



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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.



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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.

____________________________________



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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



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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)



20

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.



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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



22

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 )



23

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



24

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

+

-



25

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



26

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



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



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

Documents

Power System