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7/24/2019 All Our Solutions(1)

1/41

ECE 6606PD Distribution Systems Engineering

Assignment 1

Distribution System Load

Characteristics

1-a)Average Power=total annual energy

8760107

= =1141kW8760

1-b)Annual Load Factor= F =Average deand

L!Peak deand

= 1141 =0"#$6#%00

$-a) &ran'(orer kW=1800$000$000

=%04#"48 kW1"1%

&o *alculate &ran'(orer +, calculate

(.r't =Ptanco'-1 /ower (actor)

=1800tanco'-1 /0"%)=%1"6# k2A3

+..lar.ly

$ =1$#"4 k2A3# =68"64 k2A3

&ran'(orer k2A3=1 $ # =$4#4"%78 k2A3

1"1%

&ran'(orer k2A= P$ $ = %04#"48$ $4#4"%78$ =%600"#4 k2A

$-b)Load d.ver'.ty=/1800$000$000)-/%04#"48)=7%6"%$ kW

$-c) F.r't tran'(orer' eergency rat.ng =#1$%51"$%=#06"$% k2A /ot +u.table)

+econd tran'(orer' eergency rat.ng =468751"$%=%8%8"7% k2A /+u.table)

*oo'e'econd tran'(orer

1


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ECE6606PD Distribution Systems Engineering

Assignment # 2

Distribution System Load Forecasting

1. 9la.n br.e(ly te a.n d.((erence between te e9traolat.on, te '.ulat.on, and te econoetr.c

etod' (or load (oreca't" :our an'wer 'ould addre'' te (ollow.ng;

a. &e.r u'e (or d.'tr.but.on 'y'te load (oreca't"

b. &e data re


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&e co'en we.gt' are 0"4, 0"#, 0"$, and 0"1, re'ect.vely"

&able $ and F.g" 1 re'ent te generated re'ult' wen u'.ng te 4-order we.gted ov.ng

average" &e'e re'ult' reveal tat te A .' $#14"41 kW and te 3+ .' $848"8 kW"

&able $ Load (oreca't.ng u'.ng 4-order we.gted ov.ng average

Hour (i) Power (kW)

{e(i)}^2Actual Forecasted e(i)

1 28969.26

2 26392.32

3 24254.58

4 22707.33

5 21787.5 24534.696 2747.20 7547085.86

6 21642.96 23017.347 1374.39 1888939.63

7 22187.94 22160.358 27.58 760.77

8 22463.73 21996.297 467.43 218493.61

9 22850.31 22149.216 701.09 491532.80

10 24301.62 22481.127 1820.49 3314194.7611 26440.89 23287.281 3153.61 9945249.72

12 28536.93 24683.277 3853.65 14850641.44

13 29886.12 26492.394 3393.73 11517376.16

14 30487.05 28233.867 2253.18 5076833.63

15 30160.56 29512.131 648.43 420460.17

16 30874.11 30041.256 832.85 693645.79

17 33837.27 30483.834 3353.44 11245533.01

18 37946.61 31877.958 6068.65 36828537.10

19 37448.58 34520.703 2927.88 8572463.73

20 36800.67 36218.28 582.39 339178.11

21 35871.45 36927.891 1056.44 1116067.59

22 34722.09 36673.158 1951.07 3806666.34

23 32497.41 35755.263 3257.85 10613606.17

24 28453.05 34269.948 5816.90 33836302.34

25 28453.05 31662.006

26 28453.05 29888.826

27 28857.486

$


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40

#%

#0

$%

$0

1%10

%

0

1 $ # 4 % 6 7 8 10 11 1$ 1# 14 1% 16 17 18 1 $0 $1 $$ $# $4 $% $6 $7

"ime #nter$as %&ours'

Actual Foreca'ted

ii) A!"MA

(2$$

F.g" 1 Load (oreca't.ng u'.ng 4-order we.gted ov.ng average

&e (oreca't.ng odel (or te A3>A /$,0,0) .' e9re''ed ateat.cally by,

Y@(i) = a + b1 Y (i 1) + b$ Y

(i $) i = #, 4, %,""""""

were a , b1 ,and

b$ rere'ent te odel coe((.c.ent' tat can be e't.ated u'.ng te Lea't

+


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1"7$6 !b$ " !- 0"8#18"

#


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&able # and F.g" $ re'ent te generated re'ult' wen u'.ng te A3>A /$,0,0)" &e'e re'ult'

reveal tat te A .' 76"71 kW and te 3+ .' 11#4"88 kW"

4%

40

#%

#0

$%

$0

1%

10

%

0

1 $ # 4 % 6 7 8 10 11 1$ 1# 14 1% 16 17 18 1 $0 $1 $$ $# $4 $% $6 $7

"ime #nter$as %&ours'

Actual Foreca'ted

F.g" $ Load (oreca't.ng u'.ng A3>A /$,0,0)

&able # Load (oreca't.ng u'.ng A3>A /$,0,0)

Hour (i) Power (kW)

e i ^2Actual Forecasted e(i)

1 28969.26

2 26392.32

3 24254.58 24466 211.42 44698.42

4 22707.33 22920 212.67 45228.53

5 21787.5 22028 240.50 57840.25

6 21642.96 21727 84.04 7062.72

7 22187.94 22243 55.06 3031.60

8 22463.73 23303 839.27 704374.13

9 22850.31 23326 475.69 226280.98

10 24301.62 23764 537.62 289035.26

11 26440.89 25947 493.89 243927.33

12 28536.93 28433 103.93 10801.44

13 29886.12 30271 384.88 148132.61

14 30487.05 30856 368.95 136124.10

15 30160.56 30771 610.44 372636.99

16 30874.11 29708 1166.11 1359812.53

17 33837.27 31211 2626.27 6897294.11

18 37946.61 35732 2214.61 4904497.45

19 37448.58 40360 2911.42 8476366.42

20 36800.67 36082 718.67 516486.5721 35871.45 35378 493.45 243492.90

22 34722.09 34313 409.09 167354.63

23 32497.41 33102 604.59 365529.07

24 28453.05 30218 1764.95 3115048.50

25 28453.05 25088

26 28453.05 28452

27 28452

4

De

m

an

(

%


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ECE Distribution Systems Engineering

Assignment #

Distribution System

P$anning%1& Lecture #' Sections 1' # and (

%)& it is required to calculate the cost function between dierentsections

Since it is a trans*ortation *rob$em' then +e assume a se*arate cab$eis e,tended -rom each substation to each $oad *oint' no ta**ingo. isa$$o+ed /ta**ing o. is consideredon$y in a transhi*ment *rob$em&

A$so since there is no $imit to the *o+er that cou$d be su**$ied by each

substation' each substation is assumed to su**$y the -u$$ $oad o- a$oad *oint i- associated +ith the $eastcost -unction -or this $oad *oint

or straight distance cost -unction is 1 *u2

or straight ri3er crossing cost -unction

is 10 *u24hus the -o$$o+ing tab$e can be

constructed

Load center Shortestdistance

Shortestdistance

1 1 (/(1& /57(1&

) ) #

/(1)& /58)&

# 8 )

/(756#& /56#&

( 0 #/57(&

8 1 )

/(8& /58&

6 ( 1

/(756& /56&7 1 )

/(7& /57&

) 1

/(7& /5&

5 #/(75&

0


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9/41

Distribution Systems EngineeringECE6606PD

Assignment # 4 So%ution

Distribution System Automation &

Demand Side Management

1. !.'cu'' te a.n (actor' .n(luenc.ng e((.c.ent and rel.able load 'uly to cu'toer'"

Lecture 4, Section )

[30 marks]

2. 9la.n br.e(ly te a.n ba'.c area' re


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Assignment # ' So%ution

Sub-ransmission Lines and )on-echnica%

Distribution Substations Design Factors

1. !.'cu'' br.e(ly te d.((erent tye' o( 'ub-tran'.''.on c.rcu.tB' con(.gurat.on'" :our an'wer 'ould

addre'' te (ollow.ng;

a. &e one l.ne d.agra o( te electr.c c.rcu.t"

b. &e rel.ab.l.ty o( te con(.gurat.on"

c. &e relat.ve co't o( eac con(.gurat.on"

d. &e a.n drawback' o( eac con(.gurat.on"Lecture +, Section 2)

2. !.'cu'' br.e(ly ow te o't ot.al 'ub'tat.on locat.on' /'.te') are deter.ned and te d.((erent

(actor' a((ect.ng te 'elect.on roce''"

Lecture +, Section 4


11/41

Assignment # * So%ution

Distribution Substation Design

As+ects

1. !.'cu'' br.e(ly te d.((erent tye' o( 'ub'tat.on bu' con(.gurat.on'" :our an'wer 'ould addre'' te

(ollow.ng;

a. &e one l.ne d.agra o( te electr.c c.rcu.t"

b. &e o''.ble oerat.ng voltage' (or eac con(.gurat.on"

c. &e a.n drawback' o( eac con(.gurat.on"

d. &e a.n advantage' o( eac con(.gurat.onLecture 6, Section 3

2. A d.'tr.but.on 'ub'tat.on 'erv.ce' a '


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1


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(iii) +ub'tat.on 'ac.ng, bot way'"S

feeder= A

feederx D

1$$"7 = l $ x 1000

l4 = #"0$ mi

Substation Sa!in", bot# Wa$s = $ l4 = 6"04 mi

(iv) &otal ercent voltage dro (ro te (eed o.nt to te end o( te a.n"

Fro F.g" 1, k (or E$ AW coer at $$" k2 .' 0"000$%

HVD=

$x % x D x l

#

4#

4

=$

x 0"00$%x 1000x /#"0$)#

#= 4"% H

(b) *on'.der voltage dro-l..ted (eeder' w.c ave #H voltage dro and (.nd;(i) +ub'tat.on 'ac.ng, bot way'"

HVD=

$x % x D x l

#

4#

4

# =$

x 0"000$%x 1000x l #

#4

l4

= $"6$1 mi

Substation Sa!in", bot# Wa$s = $ l4

= %"$4$ mi

(ii) a9.u load er (eeder"$

Sfeeder

= Afeede

r

x D

= l4$

x 1000

Sfeeder = /$"6$1) x 1000

= 686"641KVA

(iii) +ub'tat.on '.Ge"S

substation= 4x S

feeder

= 4x 686"641

= $7"478%6&VA

(iv) Aere load.ng o( te a.n .n er un.t o( conductor aac.ty"

S feederI =

=686"41

= 17#"16A

4

# x VLL# x $$"


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$


15/41

I u=

II

a9

=17#"1

6$#0

= 0"7%#

F.g" 1 D con'tant curve' (or coer conductor' w.t 0" lagg.ng ower (actor


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#


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????????????????????????????????????????????????????????????????????????????????????????????????????????


Assignment 7

Primary Distribution SystemsDesign and @*eration

1) /(0 mars& E,*$ain brieBy the di.erent *rimarydistribution system congurations2 our ans+er shou$daddress the -o$$o+ing

a)4he sing$e $ine diagram o- each conguration2

b)4he main ad3antage o- each conguration2

c)4he disad3antages o- each conguration2

d)4he degree o- re$iabi$ity o- each conguration2

Lecture 7, Section 3

2) /60


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? or design F 4he 3o$tage dro* +i$$ be summation o-the 3o$tage dro* o- the sho+n three $ines


19/41

Total Volta"e Dro = 1"6 +1"6 + 0"06$% = #"$6$%H

4hus ' design A is better than Design F


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Distribution Systems EngineeringECE6606PD

Assignment # , So%ution

Secondary Distribution Systems Design Services

and

Metering

1. 9la.n br.e(ly te d.((erent 'econdary d.'tr.but.on 'y'te con(.gurat.on'" :our an'wer 'ould

addre'' te (ollow.ng;

a. &e '.ngle l.ne d.agra o( eac con(.gurat.on"

b. &e a.n advantage o( eac con(.gurat.on"

c. &e d.'advantage' o( eac con(.gurat.on"

d. &e degree o( rel.ab.l.ty o( eac con(.gurat.on"Lecture *, Section )2

2. !.'cu'' te a.n coonent' o( te 'econdary d.'tr.but.on 'y'te" :our an'wer 'ould con'.der te

(ollow.ng .''ue';

a. &e 'econdary 'y'te voltage level"

b. &e de'.gn con'.derat.on o( te 'econdary 'y'te"

c. &e degree o( rel.ab.l.ty o( eac coonent .n te 'econdary 'y'te"

Lecture *, Section ))


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Assignment # So%ution

.o%tage Dro+s and /ower Loss

0a%cu%ations

1. F.gure E1 'ow' a 't a' 0 J0"0%

er un.t .edance ba'ed on te tran'(orer rat.ng'" >t .' taed u to ra.'e te low voltage %"0

ercent relat.ve to te .g voltage, ."e", te e


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F.gure E$, ue't.on E1

So%ution1

a- &e k2A rat.ng o( te '


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VDA1 = K x Ss)uarex

l' , K x Slumedx l'

$

= 0.0000- x +000 x *' , 0.0000- x 2000 x 2' = 0.2 .

&e H2! (ro bu' K to bu' * .'

VD1 =K x Slumedx l =0.000*4 x 2000 x 2 = 0.-5 .

&e H2! (ro bu' to bu' * .'

VD/ =VD/A ,VDA1 ,VD1 = 0.+ , 0.2 , 0.-5 = *.- .

&e er un.t voltage at bu' * /'..larly at te lued load ter.nal) .';

V! = Vo 6 VD/ = *.02 60.0*- = *.007 u

&o (.nd te H2! (ro bu' * to bu' !;

&e lued load current .'

I = = $000 %VA

#x (1"00x #4"% %V

)

= ##"17 A

&e ba'e load current .'

Ibase

= S

lumed

# x V ,LL

=$000 %VA

# x #4"%

%V

= ##"47 A

&e er un.t lued load current .'

Iu =

II

base

=##"17

= 0" u##"47

+.ncef = 0.7, tere(ore = 2.5+ andsin = 0.+37"

and te low voltage '.de a' been taed u %H

x V ,LL#

Slumed


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&en te H2! (ro bu' * to bu' ! .'

VD =I /8 co' +9 '.n

)V

base

0"0%=

0"x 0"0%x 0"4#%

1

0"0%

= 62.5+ .3


25/41

b- &e er un.t value' (or te voltage' at bu'e' A, K, *, and ! on te a.n;

VA = V0: V0A = *.02 : 0.00+ = *.02* u

V1F = VA: VA1 = *.02* : 0.002 = *.0*5 u

V = V1F: V1 = *.0*5 : 0.00-5 = *.007

u

VD = V: VD = *.007 : 60.025+' = *.034+ u

c- &e l.ne-to-neutral voltage' at bu'e' A, K, *, and ! on te a.n;

VA = 1,$0x *.02* = 20(335.32 V

V1F = 1,$0x *.0*5 = 20(23+.43 V

V = 1,$0x *.007 = 20(077.25 VVD = 244 x *.034+ = 254.3- V

2. F.gure # 'ow a '.ngle-l.ne rere'entat.on o( a tree-a'e, 6 k2 network" +ub'tat.on 1 'ul.e'

'ub'tat.on' $ and #" +ub'tat.on' $ and # are connected v.a a t.e l.ne" *alculate;

a. &e voltage d.((erence between 'ub'tat.on' $ and # wen te t.e l.ne .' oen"

b. &e l.ne current' wen te t.e l.ne .' connected"

c. &e total ower lo'' wen te t.e l.ne .' connected"

F.gure E#, ue't.on E$

So%ution1

For te 1$% A, PF = 0" loadM

I L1 = 11$"% ; %4"4 = 1$% $%"84 A

For te 1% A, PF = 0"8% loadM I L$ = 16%"7% ; 10$"7$ = 1% #1"78 A

a. wen te t.e l.ne .' oen a' 'own .n F.gure E 4 we aveM


26/41


27/41

I $ =I L$ = 16%"7% ; 10$"7$ = 1% #1"78A

F.gure E 4, ue't.on E$

&e voltage at 'ub'tat.on $ /V2) can be e9re''ed byMV$ = V1 I1 x


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F.gure E %, ue't.on E$


29/41


I1 +I # =I L1 = 11$"% ; %4"4 = 1$% $%"84 A

I $ I# =I L$ = 16%"7% ; 10$"7$ = 1% #1"78

A

and

&ere(ore,

I1 = (11$"% ; %4"4) I#

I $ = (16%"7% ;

10$"7$) +I#

and

I'.ng k2L, we getM

I1 x


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= #x (I $ x r1$ + I $x

r1# + I#x r$#)

= #x

((1$8"16)$x 0"

+ (1$"4)$x 1

+ ($0"#)$x 1)

= 1%6"64 %W

1


31/41


32/41

????????????????????????????????????????????????????????????????????????????????????????????????????????


> ==tan

= #%0"$ tan/co'1

0"7%)

1 1

>1= #%0"$ 0"881 = #0"48 % var


33/41

4he reacti3e *o+er o- the motor at the corrected *o+er-actor is

> ==tan

= #%0"$ tan/co'1

0"88)

$ $

>$= #%0"$ 0"%#7 =18"41% var

4hus' the reacti3e *o+er *ro3ided by the ca*acitor eMua$s

>!

= >1 >$

= #0"48 18"41 = 1$0"07 % var

b) 4he ca*acitor reactance can be ca$cu$ated as

9! ,

#ase

V$

1

= #ase

=>

!,#ase $ f

4hus' the ca*acitance o- the ca*acitor is ca$cu$ated as

=>

!,#ase$

#ase

Nhere

>

=>

!

=1$0"07

% var

!,#ase# #

or de$ta connected ca*acitor' =*hase is eMua$ to (160 =/4he 3o$tage across each indi3idua$ ca*acitor is thetota$ $ine 3o$tage& as sho+n in igure 1

$


34/41


35/41


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Assignment 11

Distribution System =o$tage

9egu$ation1) /#0 mars& =o$tage regu$ators are used to im*ro3e the

Mua$ity o- the distribution system *er-ormance2 DiscussbrieBy the main com*onents o- the automatic 3o$tageregu$ator and their main -unctions2 A$so e,*$ain themain -unction o- the $ine dro* com*ensator /LDC& andthe e$ectric circuit o- the LDC2

N+ect.on' 6 O 7

2) /70 mars& An industria$ customerQs bus is $ocated atthe end o- a # mi$e *rimary $ine +ith a resistance o-02# ROmi and an inducti3e reactance o- 02 ROmi2 4hecustomerQs trans-ormer is rated 8000 =A +ithtrans-ormer im*edance o- 0 ; 0208 *u R based onthe rated =A2 Assume that the industria$ $oad is atthe annua$ *ea o- (000 =A at 0 K $agging *o+er-actor2 Se$ect a *ro*er three?*hase ca*acitor ban si>eto be connected to the industria$ $oad ( = bus to

achie3e the -o$$o+ing goa$sa) Produce a 3o$tage rise o- at $east 020) *u2b) 9aise the @?*ea *o+er -actor to at $east K

$agging *o+er -actor2

Iint

? Tse mu$ti*$es o- three?*hase' 180 3ar ca*acitor unitsin si>ing the

reMuired ca*acitor ban2

? 4he *rimary 3o$tage o- the trans-ormer is 1)2 =2

So$ution

or the origina$ $oad /(000 =A at 0 K $agging *o+er -actor&'

S = 4000co'1 /0"8)

= 4000#6"87%VAo


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n

=o

= So

fo

= 4000 0"8 = #$00 %W

Io+e3e

r

n

= Sn

fn

= #$00 = Sn

0"88

Sn

= #6#6"#64 %VA

4hen' it is reMuired to nd the o$d and ne+ reacti3e *o+ers

> =

/S)$ / )$

== $400 %VA8

o o o

> =

/S)$ / )$=

= 17$7"17 %VA8

n n n

4here-ore' the additiona$ reacti3e *o+er reMuired byinsta$$ing the ca*acitor ban to raise the @?*ea *o+er-actor to at $east K $agging *o+er -actor is

>!

= >o

>n

= $400 17$7"17 = 67$"8# %VA8

Tsing mu$ti*$es o- three?*hase 180 3ar ca*acitor units'then the reMuired reacti3e *o+er o- the ca*acitor ban is780 =A9 /this 3a$ue +i$$ raise the *o+er -actor to 2 K&

Considering the trans-ormer *rimary 3o$tage as 1)2 ='the trans-ormer im*edance can be ca$cu$ated as -o$$o+sU

9

=9/u)9

/1$"810# )$/base) = 0"0% = 1"6#84

tr tr tr %000 10#

4he resu$tant 3o$tage rise -rom insta$$ing the 780 =A9ca*acitor ban is

V8/u) = > 9L

#6#6"#6$ #$00$4000$ #$00$


39/41

1000 /%V1,LL

)

=7%0 /0"8 # +1"6#84)

= 0"018486u1000 /1$"8)$

4his is $ess than the reMuired )K

4hus' 780 =A9 donQt meet the design s*ecications

$


40/41

ncrease the ca*acitor si>e to 500 =A9

V8/u)=

>

9L

1000 /%V1,LL

)

= 00 /0"8 # +1"6#84) = 0"0$$$u1000 /1$"8)$

Chec -or the ne+ *o+er -actor

f = co'/tan1 /

>n )) = co'/tan

1 />

o >

! ) = co'/tan1 /

$400 00)

n

fn

= 0"0%

=n

n

#$00

$


41/41

Assignment 1)

Distribution Ser3ice

9e$iabi$ity1) /80 mars& m*ro3ement o- e$ectric *o+er de$i3ery

re$iabi$ity is an im*ortant tas that is carried out bythe e$ectric uti$ity due to the high cost o- customeroutages2 E,*$ain brieBy the main re$iabi$ity indicesthat are common$y used in measuring distributionsystem re$iabi$ity2 A$so e,*$ain the +ide$y usedre$iabi$ity *ractices in distribution systems2 our ans+er

shou$d address this subect -or the -o$$o+ing customerQsty*e 9esidentia$' $ight $oad commercia$' commercia$'industria$' and agricu$tura$2

N+ect.on' # O %

2) /80 mars& Each com*onent in the system has itsinterna$ ris o- -ai$ure a$ong +ith the e,terna$ -actors2Discuss brieBy ho+ the di.erent e,isting com*onentsin the distribution system can a.ect the system

re$iabi$ity2N+ect.on 6"1

otes

? @ne ty*ed *age *er essay?ty*e Muestion

? Submission Due Date A*ri$' 8' )01#

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All Our Solutions(1)