dc mmotor formulas basics

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CHAPTER1

D.C.Machine Problems

with Solutions

Problems Sub-topics

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

1.10

1.11

1.12

1.13

1.14

takes you to the start page after you have read this Chapter.

Start page has liks to other Chapters.

!ro".1.1 DC generator-bac em!

#he a d$ geerator is "eig drive at 1200 %!&' the geerated e(f is 125 ). #hat*ill "e the geerated e(f if +a, the field flu- is de$reased "y 10 *ith the speed

re(aiig u$haged 'ad +",if the speed is redu$ed to 1100 %!& 'the field flu-

re(aiig u$haged /

Solutio +a,

1 125

1= 1

2= .91

112.5 )olts s*er

+", 1 1200 %!&

2 1100 %!&2121

114.5833 )olts s*er

Prob.1." #. c. generator- generator $ motor action

shut (a$hie has ar(ature ad field resista$es of 0.04 oh( ad 100 oh(s

respe$tively. #he $oe$ted to a460 ) d$ supply ad drive as a geerator at 600

%!&' it delivers 50 k#. Cal$ulate it speed *he ruig as a (otor ad

d.$.geerator "a$k e(f

d.$ (a$hiegeerator (otor a$tio

d.$. (otorouput po*er torue

d. $. shut (otorffe$ts of flu- torue o $urret ad speed

d.$.geeratorlog shut $o(poud

d.$.geeratorsparallel operatio

d.$.geerators i paralleleffe$t of $hagig the e-$itatio of oe

d.$. (otor hysterisis eddy $urret losses

d.$. (otor effi$ie$y

uiversal (otor

d.$. (otor field *eakeig

d.$.geerator :rasiet "ehaviour

;ra$tioal k# series (oor

Speed $otroller of a d.$ (otor

#


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takig 50 k# fro( the sa(e supply. Sho* that the dire$tio of rotatio of the (a$hie

as a geerator ad as a (otor uder these $oditios is u$haged.

Solutio !1'>eerator po*er 50000 #

)' ter(ial )oltage 460 )

=' >eerator lie $urret!1)

108.6957 =' &otor lie $urret 108.6957

=f2&otor field $urret )%f geerator field $urret

4.6

=a1'>eerator ar(ature $urret=?=f2

113.2957

%a 0.04 oh(

%f 100 oh(s

1>eerator "a$k e(f)?=a1%a

464.5318 )

=a2&otor ar(ature $urret==f2

104.0957

2'&otor "a$k e(f)=a2%a455.8362

Speed is proportioal to "a$k e(f.

1'>eerator speed 600 %!&

2'&otor speed 1+21,

588.7685 %!&

:he dire$tio of field flu- re(ais the sa(e *he operatig as a (otor or a geerator.

@o*ever' the dire$tio of the ar(ature $urret $hages i the t*o (odes of operatio.

:herefore' a$$ordig to ;le(igAs Beft had %ule for (otor operatio ad the %ight

@ad %ule for the geerator operatio' the dire$tio of rotatio is u$haged.

Prob.1.% #. c. motor- output power $ tor&ue

240 ) d$ (otor has a ar(ature resista$e of 0.68 oh( ad dra*s aa fullload $urret of 24 at a speed of 100 %!& .Cal$ulate+a, the "a$k e(f +", the

output po*er developed' ad+$, the torue developed.

Solutio

) 240

=a 24

%a 0.68

' "a$ke(f )=a%a 223.68 )olts s*er

!' po*er developed =a

5368.32 #atts s*er

' speed 1000 %!&

:':orue60!+23.1416,

51.26356 ( s*er

Prob.1.' #. c. shunt motor-E!!ects o! !lu( $ tor&ue on current an# spee#

shut (otor $oe$ted a$ross a 440 ) supply takes a ar(ature $urret of 20 ad

rus at 500 %!&. :he ar(ature resista$e is 0.6 oh(s. =f the (ageti$ flu- is redu$ed

"y 30 ad the torue developed "y the ar(ature i$reases "y 40 '*hat are the

values of the ar(ature $urret ad of the speed/

Solutio


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) 440

=a1 20

%a 0.6

1)=a1%a 428 )olts

1= 1

2= 0.71

2e* "a$k e(f299.6 )olts s*er

:1'rigial torue :1

:2'e* torue 1.4:1

=a2e* $urret

1origial speed 500 rp(

1'origial "a$k e(f 1

:' :orue is proportioal to =a. :herefore

:2:1 +21,+=a2=a1,+12,'or

=a22+:2:1,+=a11,+12,

0.08 .1

2)=a2%a.:herefore'=a2+)2,%a 234 (ps

500++440=a2.6,428,+10.7,

or'

+440=a20.6,2 0.5992 .2

;ro( euatios 1 ad 2'

+440=a2.6,=a2 7.49

or'

=a2 54.38813 (ps

;ro( .1'

2 679.8517 %!& s*er

Prob.1.) #.c.generator-long shunt compoun#

4 pole 500) 25 k# logshut $o(poud geerator delivers fullload at the rated

voltage.Cal$ulate the e(f geerated' if the ar(ature resista$e is 0.03 oh('series field

resista$e is .04 oh(' ad the shut field resista$e 200 oh(.Cota$t drop per "rush

is .9 volts. egle$t ar(ature rea$tio.

Solution*

Shut field =sh

Series field

>

=B?=sh

=B

Boad

212/1=10.7

is proportioal to ad . :herefore

21+21,+1/2,1++)=a2%a,1,+1/2,

=B?=sh

=B

Boad


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! >eerator po*er

25000 #

) >eerator voltage

500 )olts

=B Boad $urret!)50

%sh Shut field resista$e

200 oh(s

=sh shut field $urret)%sh

2.5

=B?=sh 52.5

%s Series field resista$e

0.04 oh(

)olt drop i series field+=B?=sh,%s

2.1 )

%a r(ature resista$e

0.03 oh()olt drop i ar(ature+=B?=sh,%a

1.575 )

)olt drop i "rushes2.9

1.8 )

:otal voltage drop2.1?1.575?1.8

5.475 )

:er(ial voltage 500 )

(fter(ial voltage ?total voltage drop

505.475 ) s*er

Prob.1.+ #.c.generators-parallel operation

:*o shut geerators ru i parallel o a "us"ar ad supply 1000 to $osu(ers.a$h

(a$hie has a ar(ature resista$e of .02 oh( ad a field resista$e of 30 oh(s' the

e(fAs geerated "eig 465 ) ad 460 ) respe$tively. Cal$ulate

+a, the output of ea$h (a$hie' ad +", the "us voltage

Solution* Bet

)


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=1 +1),%a .2

=2 +2),%a .3

Su"stitutig s.2 3 i .1'

++1),%a ,?++2),%a, =B ?+2)%f,

fro( *hi$h'

) +1%f?2%f%f%a=B,+2%?2%f,

452.1985 )olts

=1 +1),%a

640.0733

=2 +2),%a

390.0733

:otal $urret640?3901030 1000 of load 30 for the fields

Prob.1., # . c. generators in parallel-e!!ect o! changing the e(citation o! one

;our geerators are paralleled o a 240 ) "us"ar. all the (a$hies share eually a load

of 800 ' the idu$ed e(f of ea$h "eig 250).

operator iadvertetly alters the positio of a field regulator ar( there"y i$reasig

the e(f of oe (a$hie "y 4 . Cal$ulate the $urret of ea$h (a$hie ad state

*hether ay har( has "ee doe.

Solution* 1 origial e(f of (a$hie1 250 )

1A e* e(f of (a$hie11?.041 260 )

=a1 $urret of (a$hie1 8004

=a1 200 ) "us"ar voltage 240 )

%a ar(ature resista$e

=a1%a1). :herefore'

%a +1),=a1

0.05 oh(

)1 e* "us"ar voltage

=a1A e* $urret of (a$hie 1

=a2A $urrets of ea$h of the other three (a$hies

%B load resista$e240800 0.3 oh(

1A%a=a1A)1

2%a=a2)1

%B+=a1A?3=a2,)1

Solve the three euatios ad o"tai

=a1A 352 s*er

=a2 152 s*er

)1 242.4 )olts


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ssu(ig that 200 is 100 full load $urret of ea$h (a$hie '(a$hie1 is loaded

(ore tha 175 of its full load $apa$ity.


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= =put $urret 4

=a ar(ature $urret 3

%a ar(ature resista$e 0.32 oh(

g )=a%a

239.04 )olts

# stray loss g=a

# 717.12 # s*er 0b

= =put $urret 25

=a ar(ature $urret ==f

24

g )=a%a

232.32 )olts

!out g=a#

4858.56 #

lteratively'

!out!i!loss)==a=a%a=f=f%f#4858.56#

!i )=6000 #

ffi$ie$y !out100!i

80.976 s*er

Prob.1.12 uni3ersal motor

@. #he $oe$ted to a 230 ) E.C. supply ad loaded to take 1 it rus at 2000 rp(.

Eeter(ie the speed ad po*er fa$tor *he $oe$ted to a 230 ) '50 @F supply ad

loaded to take the sa(e $urret.

D.C.

) supply voltage 230 volts

% 20 oh(s

= 1

"a$k e(f )%= 210 volts

speed 2000 rp(

4n a.c o! same 3alue 0r.m.s

B 0.318309 @

f 50 @F

G 23.14fB 99.94907 oh(s

=G 99.94907 )olts

=% 20 )olts!hasor diagra(

=G =% ?r

)


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=

;ro( phasor diagra(' rea$ta$e drop leads $urret "y 90 deg' =% drop ad rotatioal

e(f r are i phase *ith the $urret+egle$tig efffe$t of iro loss,

@e$e

r srt+))=G=G,207.1477 )olts

;or the sa(e peak flu-' the e(f of rotatio+r(s, is 1srt+2, of E.C. valueH "ut peak flu-

i a.$ is srt+2, ti(es flu- o E.C. for sa(e r(s $urret' therefore r is sa(e for sa(e

speed ad is proportioal to speed.

he$e e* speed r

1972.836 rp( s*er

!f +=%?r,)

0.987599 lag s*er

Prob.1.11 5iel# weaening

d.$ series (otor havig a overall resista$e of 0.15 oh( has the follo*ig

$hara$teristi$ at 600 ).

Curret' 0 40 80 120 160 200

Speed' rp( 3600 3600 2500 2100 1900 1780

=f the field is *eakeed "y tappigs *hi$h redu$e the u("er of turspole

to 70 of or(al'at *hat speed the *ould the (otor ru *he takig 160

fro( the 600) supply. egle$t ar(ature rea$tio.

Solution*

:he (agetisatio $hara$teristi$+flu- gaist $urret at $ostat speed, is

derived fro( the give torue$urret $urve as follo*s

:he e(f $a "e $al$ulated fro( the resista$e ad $urrets.

Si$e the flu- is proportioal to ' the pu flu- +takig 200 as produ$ig

1 pu flu-, $a "e foud fro(

Ipu 1.0+200,+200,

$urret'= (ps 0 40 80 120 160 200

=% drop' ) 0 6 12 18 24 30' ) 0 594 588 582 576 570

Ipu 0.00 0.52 0.73 0.87 0.95 1.00

$os


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: 160 ' the flu- is 0.95.:he tappigs redu$e the effe$tiv e$urret to 160.7 112 .

:he flu- fro( the a"ove graph at 112 is a"out 0.84 pu.

:he e(f is ot altered.

:he e* speed is 1900.95.84 2149 rp(

:he tappig $hages the overall (otor resista$e '"ut this is egle$ted.

Co(pare the use of diverter resista$es ad field tappigs for series (otor speed $otrol.

Prob.1.1" D.C generator -Transient beha3iour

:he field *idig of a separately e-$ited d.$.geerator has a idu$ta$e of 75 @

ad a resista$e of 150 oh(s. :he rotatioal (utual idu$ta$e $oeffi$iet

"et*ee the field ad ar(ature is 1 @rads.

#ith the (a$hie ue-$ited ad drive at 1000 rp( ' a ra(p fu$tio voltage

v+t, 100t volts

is applied at t 0 s to the ter(ials of the field *idig. Eeter(ie the ar(ature

e(f at t 0.5 s.

Solution*

:he voltage euatios for the field ad ar(ature $ir$uits are

vf rf . =f ? Bf difdt+ rf?Bfp,uf

va *r. Baf. if

vf 100t

Explain why series motors are used in preference to shunt motors for traction applications.

#hat *ould " ethe ffe$t of ar(ature rea$tio/


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:herefore' va *r. Baf . 100t+rf?Bf.p,

:rasfor(ig to sdo(ai for Bapla$e trasfor(s'

va+s, J100Ks.s+s?rfBf,L. M. +*rBafBf,

*r1000- 23.141660 104.72 rads

rf 150

Bf 75

alpharfBf 2

Baf 1

"eta100 *r.BafBf 139.6267

vas "etaKs.s+s?alpha,L "etaKJ+1s.s,,1+s+s?alph,,M

vas +1+alph.alph,J+alphs.s,alphs+s?alph,,

Covertig to ti(edo(ai'

va+t, K"eta+alphaalpha,L+alpha.t 1?e-p+alpha.t,

*ith t 0.5 se$

va+t, 12.84145

!ro".1.13 5ractional 8 series motor

fra$tioal k# series (otor has a resista$e of 30 oh(s ad a idu$ta$e of

0.5 @. *he $oe$ted to a 250 ) d.$ supply ad loaded to take 0.8 it rus

at 2000 rp(.Eeter(ie the speed ad po*er fa$tor *he $oe$ted to a 250 ) '

50 @F supply ad loaded to take the sa(e $urret.

Solution*

E.C.supply

= 0.8

r oh(s 30

=r 24) volts 250

)=r volts


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=r24 ) r

) 250 )

=G126 )

I =

!hasor diagra(

%ea$ta$e drop leads $urrt "y 90 degH =r drop ad rotatioal e(f r are

i phase *ith $urret + effe$t of iro loss egle$ted,

rsrt+))+=G,+=G,,=r 192.122 volts

;or the sa(e peak flu-' the e(f of rottaio+r.(.s, is 1srt+2, of d.$ valueH

"ut peakflu- o a.$ is srt+2, ti(es flu- o d.$ foer sa(e r.(.s $urret.:herefore'r is sa(e for sa(e speed ad is proportioal to speed.

@e$e e* speed 2000r 1700.191 rp(

p.f ++r?=r, ), 0.86 lag

Prob. 1.1' Spee# controller o! a #.c motor

d. $ (otor is drivig a loadd havig a iertia of 200 kg(N2 ad has its

speed $otrolled "y a $losedloop syste(.;ri$tioal torue 10 (rads adthe (otor torue produ$ed "y the error sigal is 100( per rads of error.

=f the (otor is at rest ad the $otroller is suddely set to de(ad a speed of 20

rads' develop a e-pressio for the speed i ter(s of ti(e ad deter(ie

the steadystate speed *hi$h *ill evetually rea$h.

Solution*

uatio of (otio is

OdN2PodtN2 ? ; dPodt D +dPidt dPodt,

; ;ri$tioal torue

Po outputPi de(aded agular positio

@e$e'

dN2PodtN2 ? +;?D dPodt ,Odt D dPidt

Chagig to s do(ai'


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Po +s, .sN2 ? +++;?k,O,Po+s,.s +D+s.O,,dPidt

+1s, dPidt' is a sudde $hage of iput i the for( of a step fu$tio 20 rads

+;?D,O 0.55

DO 0.5

+D+s.O,,dPidt 10

:herefore'

Po+s,.SN2 ?0.55 Po+s, .s 10s

Po+s, 10+sN2+s?0.55, +100.55,K +1sN2, +1.55,+.55+s?.55,L

:akig the iverse trasfor(' *e get

Po+t, +10t.55 , +10.55N2 , +1e-p+.55t,,

@e$e speed dPodt +10.55, +10e-p+.55t,.55N2,

Steady state speed +at t, 10.55 18.2 rad.s


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dc mmotor formulas basics

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