lec3 power system

8/18/2019 lec3 power system

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POWER SYSTEMS ILecture 3

06-88-590-68

Electrical and Computer Engineering

University of Windsor

Dr. Ali Tahmasebi


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Per Unit Calculations

l A key problem in analyzing power systems is the

large number of transformers.

– It would be very difficult to continually have to refer

impedances to the different sides of the transformers

l This problem is avoided by a normalization of all

variables.

l This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity=


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Per Unit Conversion Procedure, 1

1. Pick a 1f VA base for the entire system, SB

2. Pick a voltage base for each different voltage level,

VB. Voltage bases are related by transformer turns

ratios. Voltages are line to neutral.3. Calculate the impedance base, ZB= (VB)

2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unit

Note, per unit conversion on affects magnitudes, not

the angles. Also, per unit quantities no longer have

units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)


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Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are

already in per unit)

2. Solve

3. Convert back to actual as necessary


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Per Unit Example

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 bases of

8 kV, 80 kV and 16 kV.

Original Circuit


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Per Unit Example, cont’d

2

2

2

80.64

100

8064

100

162.56

100

Left B

Middle

B

Right B

kV Z

MVA

kV Z

MVA

kV Z

VA

= = W

= = W

= = W

Same circuit, with

values expressed

in per unit.


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L

2*

1.0 0 0.22 30.8 p.u. (not amps)3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L L L L

G

I j

V S V I

Z

S

Ð °= = Ð - °+

= Ð °- Ð - °́ 2.327Ð 90°

= 0.859Ð - 30.8°

= = =

= Ð °́ Ð °=0.22Ð °


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To convert back to actual values just multiply the

per unit values by their per unit base

LActual

ActualL

ActualG

MiddleB

ActualMiddle

0.859 30.8 16 kV 13.7 30.8 kV

0.189 0 100 MVA 18.9 0 MVA

0.22 30.8 100 MVA 22.0 30.8 MVA

100 MVAI 1250 Amps

80 kV

I 0.22 30.8 Amps 275 30.8

V

S

S

= Ð - °́ = Ð - °

= Ð °́ = Ð °

= Ð °́ = Ð °

= =

= Ð - °́ 1250 = Ð - °A


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Per Unit Equivalent of Transformers


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Three Phase Per Unit

1. Pick a 3f VA base for the entire system,

2. Pick a voltage base for each different voltage

level, VB. Voltages are line to line.

3. Calculate the impedance base

Procedure is very similar to 1f except we use a 3f

VA base, and use line to line voltage bases3

BS f

2 2 2, , ,3 1 1

( 3 )3

B LL B LN B LN B

B B B

V V V Z S S S

f f f = = =

Exactly the same impedance bases as with single phase!


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Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I

3 3 3

B B B

B LL B LN B LN

S S S

V V V

f f f f f

= = = =

Exactly the same current bases as with single phase!


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Three Phase Per Unit Example

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 bases of 13.8 kV,

138 kV and 27.6 kV (square root of 3 larger than the1f 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 (per phase) is

exactly the

same!


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3 Per Unit Example, cont'd

L

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L L L L

G

I j

V S V I

Z

S

Ð °= = Ð - °

+

= Ð °- Ð - °́ 2.327Ð 90°

= 0.859Ð - 30.8°

= = =

= Ð °́ Ð °=0.22Ð °

Again, analysis is exactly the same in p.u.!


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3 Per Unit Example, cont'd

L

Actual

ActualL

ActualG

MiddleB

ActualMiddle

0.859 30.8 27.6 kV 23.8 30.8 kV

0.189 0 300 MVA 56.7 0 MVA

0.22 30.8 300 MVA 66.0 30.8 MVA

300 MVAI 125 (same cur 0 Amps

3138 kVI 0.22 30.

rent!)

8

V

S

S

= Ð - °́ = Ð - °

= Ð °́ = Ð °

= Ð °́ = Ð °

= =

= Ð - °́ Amps 275 30.81250 = Ð - °A

Differences appear when we convert back to actual values


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3 Per Unit Example 2

• Assume a 3f load of 100+j50 MVA with VLL of 69

kV is connected to a source through the per-phase

network below:

What is the supply current and complex power?

Answer: I=467 amps, S = 103.3 + j76.0 MVA


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Per Unit Change of MVA Base

l Parameters for equipment are often given using

power rating of equipment as the MVA base

l When only one component is considered, the

nameplate values of that component are selected as

base values

l In a larger system, all per-unit data must have the

same power base


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Per Unit Change of MVA Base

• To convert from an “old” base to a “new” base:

= , ×,

, =

, =

, ×,

,

but =

→ , = , ×

,

,

,

,

or

, = , ,

,

,

,


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Per Unit Change of Base Example

• A 30 MVA, 240/480 kV transformer has a leakage

reactance or 2.71%. What is the reactance on a 100

MVA and 230 kV base?

= 0.0271×240

230

100

30 =0.098 ..

= 0.098×

=52.032 W


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Transformer Reactance

l Transformer reactance is often specified as a

percentage, say 10%. This is a per unit value

(divide by 100) on the power base of the

transformer.l Example: A 350 MVA, 230/20 kV transformer has

leakage reactance of 10%. What is p.u. value on

100 MVA base? What is value in ohms (230 kV)?

2

1000.10 0.0286 p.u.350

2300.0286 15.1

100

e X = ´ =

´ = W


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Three Phase Transformers

• There are 4 different ways to connect 3f transformers

Y-Y -

Usually 3f transformers are constructed so all windings

share a common core


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3 Transformer Interconnections

-Y Y-


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Y-Y Connection

Magnetic coupling with An/an, Bn/bn & Cn/cn

1, , An AB A

an ab a

V V I a a

V V I a= = =


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Y-Y Connection: Per Phase Model

Per phase analysis of Y-Y connections is exactly the

same as analysis of a single phase transformer.

NOTE: In this simplified circuit, RC

is ignored.


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Y-Y Connection: Per Phase Model

Y-Y connections are common in transmission

systems.

Key advantages are the ability to ground each sideand there is no phase shift is introduced.

Main disadvantage is that the undesirable 3rd harmonic

exciting current will cause difficulties to the rest of thenetwork.


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- Connection

Magnetic coupling with AB/ab, BC/bb & CA/ca

1 1, , AB AB A

ab ab a

V I I a

V I a I a= = =


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- Connection: Per Phase Model

Per phase analysis similar to Y-Y except impedances

are decreased by a factor of 3.

To use the per phase equivalent we need to use

the delta-wye load transformation

=

Ð -30°


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- Connection

• Advantage of a Dwinding (in any combination) is

that it can trap the 3rd harmonic exciting current

• Key disadvantage is D-Dconnections can not be

grounded; not commonly used.


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-Y Connection

Magnetic coupling with AB/an, BC/bn & CA/cn


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-Y Connection V/I Relationships

, 3 30

30 30Hence 3 and 3

For current we get

1

13 30 303

130

3

AB ABan ab an

an

AB Anab an

ABa AB

ab

A AB AB A

a A

V V a V V V

V a

V V V V

a a

I I a I

I a

I I I I

a I

= = ® = Ð °

Ð ° Ð= =

= ® =

= Ð - °® = Ð °

I = Ð °


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-Y Connection: Per Phase Model

Note: Connection introduces a 30 degree phase shift!

Common for generator step-up since there is a neutralon the high voltage side which can be grounded,

reducing insulation requirement of transformer.

Even if a = 1 there is a sqrt(3) step-up ratio


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Y- Connection: Per Phase Model

Exact opposite of the D-Y connection, now with a

phase shift of -30 degrees.


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Load Tap Changing Transformers

l LTC transformers have tap ratios that can be varied

to regulate bus voltages

l The typical range of variation is ±10% from the

nominal values, usually in 33 discrete steps(0.0625% per step).

l Because tap changing is a mechanical process, LTC

transformers usually have a 30 second deadband to

avoid repeated changes.

l Unbalanced tap positions can cause "circulating

vars"


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Phase Shifting Transformers

l Phase shifting transformers are used to control the

phase angle across the transformer

l Since power flow through the transformer depends

upon phase angle, this allows the transformer toregulate the power flow through the transformer

l Phase shifters can be used to prevent inadvertent

"loop flow" and to prevent line overloads.


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Autotransformers

l Autotransformers are transformers in which the

primary and secondary windings are coupled

magnetically and electrically.

l This results in lower cost, and smaller size and weight. Also smaller leakage impedances results

in smaller series voltage drop (advantage) but also

higher short-circuit current (disadvantage).

l Another disadvantage is loss of electrical isolation between the voltage levels. This can be an

important safety consideration when a is large.

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