Energy Conversion VI

7/23/2019 Energy Conversion VI

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The University of Oklahoma SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING

1

Lecture VI: Per Unit System (P.U.)

Instructor: Dr. John N. Jiang

MWF 09:30am - 10:20am

Carso E!"##r"! C#$#r: Room 001%

S&r"! 2010




2Co'rs# No$#s o( ECE 3113) E#r!* Co+#rs"o

Summary: A more complicated example

• Example: A simple power system shown in i!ure "elow:

• #ontains a $%& V !enerator connected to an ideal ':'& step uptransormer a transmission line an ideal &:' step*down transormerand a load





E,#$r"a, Po.#r S*s$#m

500 kV TransmissionPower Pant

+eneration

,ransmission

System

-istri"ution

System('.V)

!n"ergroun"

Distri#ution Trans$omer

%esi"entia

&ustomer

&ommercia'

In"ustria

&ustomer

%esi"entia

&ustomer

()erhea"

Distri#ution

Trans$ormer

!r#an

&ustomers

*+ kV ,u#-transmission

230 kV

Transmission

Distri#ution ,u#station

*+'12 kV/

igh-Votage ,u#station

230'*+ kV/

tra-igh-Votage ,u#station

500'230 kV/

Distri#ution ine

!n"ergroun" &a#e

To (ther

igh-Votage

,u#stations





Per Unit System

• Simpliy the analysis o power/ener!y system

with dierent 0olta!e le0els "y a0oidin! the

impudence con0ersion

• Easy to 0eriy the answers (since they all all

into a small ran!e)





T/# P#r-'"$ S*s$#m

• I$ "s 's$omar* $o s#,#$ $.o as# 'a$"$"#s $o #("#a !"+# &#r-'"$ s*s$#m – o,$a!# a Po.#r 4A&&ar#$ Po.#r5

• O# $/#s# as# 'a$"$"#s /a+# ## s#,#$#) a,,

$/# o$/#r as# +a,'#s a # o$a"# * $/# 's'a,#,#$r"a, ,a.s

• T/# as# a&&ar#$ &o.#r r#ma"s os$a$ ./# "$

!o#s $/ro'!/ a $ras(orm#r) o $/# o$/#r /a)+o,$a!# /a!#s a$ #+#r* $ras(orm#r " $/# s*s$#maor"! $o $/# $'rs ra$"o

basebasebasebasebase I V S or Q P =

( )

base

basebase

base

basebase

base

basebase

S

V Z and

V

I Y

I

V Z

2

===




*Co'rs# No$#s o( ECE 3113) E#r!* Co+#rs"o

T/# P#r-'"$ S*s$#m

• I( a &#r-'"$ s*s$#m as# o a #+"#6s 4$ras(orm#r) "'$"omo$or) or s*/roo's !##ra$or5 o. ra$"!s "s 's#) "$s

/ara$#r"s$" .",, o$ +ar* m'/ o+#r a ."# ra!# o( +o,$a!#

a &o.#r ra$"!s7

– S#r"#s r#s"s$a# o( a Tr7 : 0701 &7'7

– S#r"#s r#a$a# o( a Tr7 : 070280710 &7'7

• #a's# &#r-'"$ +a,'#s &ro+"# a o+#"#$ a m#a"!(',

.a* $o om&ar# $ras(orm#r /ara$#r"s$"s ./# $/#* ar# o(

"((#r#$ s"#s) Tr7 Im&#a#s ar# orma,,* !"+# " &#r-'"$ or

as a &#r#$a!# o $/# Tr76s am#&,a$#7





• T/# #$"r# s*s$#m m's$ /a+# $/# sam# as# &o.#r) a $/# as#+o,$a!#s a$ +ar"o's &o"$s " $/# s*s$#m m's$ # r#,a$# * $/#+o,$a!# ra$"os o( $/# $ras(orm#rs7

• O# ommo r',# "s $o /oos# $/# s*s$#m as# 'a$"$"#s $o ##'a, $o $/# as# o( $/# ,ar!#s$ om&o#$ " $/# s*s$#m7

•

P#r-'"$ +a,'#s !"+# $o ao$/#r as# a # o+#r$# $o $/##. as# * o+#r$"! $/#m $o $/#"r a$'a, +a,'#s 4+o,$s)am&#r#s) o/ms) #$75 A,$#ra$"+#,*) $/#* a # o+#r$#"r#$,* * $/# #'a$"os :

( ) ( )2

1/1/2

base

basebase pubase pu

S

S S Q P S Q P =

2

1/1/2

base

basebase pubase pu

V

V V V =

( ) ( ) ( )

( )

( ) ( ) ( ) ( ) pri

basebase

basebasebase pubase pu

Z X Ra

Z X R Z X Ra Z X R

S V

S V Z X R Z X R

1

2sec

2

12

2

22

1/1/2

=″

=′

=

,he Per*unit System





Summary PU System and 1ther 2orms

Quantity Base

Quantity Actual Value PU

[ ]

[ ]

[ ] [ ] [ ]

[ ]

[ ]

[ ] [ ]

[ ]

[ ]

[ ]

base

ohm

PU

base

basebase

base

basebasebase

base

basebasebasebase

basebasebasebasebasebase

base

basebase

Z

Z Z

V

I Y

VA

V

S

V

I

V Z X R

I V VAS Q P

V

VA I

=

=

=====

====

=

22

Per unit s8stem a s8stem o$ "imensioness 9arameters is use" $or

com9utationa con)enience an" $or rea"i8 com9aring the 9er$ormance

o$ a set o$ trans$ormers or a set o$ eectrica machines.

here ;actua <uantit8= an" >#ase <uantit8? are )aues in )ots am9eres ohms etc.@VAAbase an" @V Abase are chosen $irst.

[ ] [ ]

[ ][ ]

[ ][ ] aratioturnsV

V

S S

base

pribase

base pribase

==

=

sec

sec

Bs ong as two si"es are consistent #ase )aue can #e an8thing.

&on)ention is to use rate" )aue




+Co'rs# No$#s o( ECE 3113) E#r!* Co+#rs"o

•Cin" #ase I #ase V #ase #ase , #aseE/

•a)e at east two known )aues

•&acuate P.!. )aue #ase" on the "e$inition o$ P.!. s8stem

•&acuate P.!. acti)e 9ower o$ oa" I2%/

•Cin" the actua 9ower o$ oa"

•Cin" the actua ine oss oa" I2%/

,F100GVB

2indin! Per*unit Value





Solution

• Example: A simple power system shown in 2i!ure "elow:

• #ontains a $%& V !enerator connected to an ideal ':'& step uptransormer a transmission line an ideal &:' step*down transormerand a load





Example (con3t)

• Impedance o line &456&7 and impedance o load

• #hose 8ase Values (I V 9 P) H +enerator: $%& V and '& VA

• I"aseS"ase/V"ase' '&&&&/$%&&.%; A

• 9"ase'V"ase'/I"ase'$%&/&.%;;.&$ 7

H ,ransmission line: '&VA and a+, ':'&. VT <• V"ase V"ase'/a$%&/&.' $%&& V

• S"ase'& VA• I"ase'&&&&/$%&&.&%; A

• 9"ase$%&& V/ .&%; A ;&$ 7

H Load: '&VA and a,L &:' V L <

• V"ase;V"ase/a $%&&/& $& V

• S"ase;'& VA

• I"ase;'&&&&/$&$'.6= A

• 9"ase;$&/$'.6= >.=6 7

• #on0ert this system to its p.u. e?ui0alent cct.

• Sol0e the pro"lems H 2ind power supplied to load in this system

H 2ind power lost in transmission line

• 2ind the true 0alue ("ased on the "ase 0alues)

Ω∠

3010





Example (con3t)

(") to "uild the pu e?ui0alent cct. 1 power system each cct parameter

di0ided "y its "ase 0alue

• V+pu

• 9linepu(&456&)/;&$&.&&%=45&.&6& pu

• 9loadpu• Per unit e?ui0alent circuit can "e o"tained:

pu

00.1470'0470 ∠=∠

pu

063*.16*.5'3010 ∠=∠





Example (con3t)

Acti0e Power• IpuVpu/9totpu

• Per unit acti0e power o load :

Ploadpu Ipu@pu(&.>6B)@('.>&;)&.$%= p.u.• actual power supplied to load:

PloadPloadpuS"ase&.$%= x '&&&&$%=& C

Line Loss• Pline losspu Ipu@ linepu(&.>6B)@(&.&&%=)&.&&%

• Pline Pline losspu S"ase (&.&&%)('&&&&)%. C

..*.305*+.07+4.0512.1

01

/3063*.1/02*.00076.0

01u p

j j

−∠=+

∠=

∠++

∠





Domewor 6

A &*VA %&&&:$%& V distri"ution transormer has the ollowin! resistances andreactance:

,he excitation "ranch impedances are reerred to the hi!h*0olta!e side.

a) 2ind the e?ui0alent circuit o the transormer reerred to the hi!h*0olta!e side aswell as the parameters (all con0erted to the primary side)

")

") 2ind the per unit e?ui0alent circuit o this transormer (you can use the simpliiede?ui0alent circuit).

c) #alculate ,ransormer3s eiciency at ull load (or ull power)a) 8ased on actual resistance

") 8ased on Per Unit Value

c) 2ind actual 0alue o the secondary side load resistanceF.

d) #alculate ,ransormer3s eiciency at G o ull load (or G ull power)a) 8ased on actual resistance

") 8ased on Per Unit Value

c) 2ind actual 0alue o the secondary side load resistanceF.

R P F 32 ohm X PF 45 ohm

RC F 250000 ohm

9rimar8/

RS F 0.05 ohm secon"ar8/ X S F 0.0* ohm

secon"ar8/

X M F 30000 ohm 9rimar8/

Iase"on,im9i$i e

"1<

&ircuit

int: the oa"' im9e"ances o$ the secon"ar8 is not gi)en see eam9es in Notes 22 $or how to "ea with 1'2oa"

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Energy Conversion VI