ump electrical machine all quiz (1)

8/12/2019 UMP Electrical Machine All Quiz (1)

1/11

BEE/BEP/I213 I

U niversitihllalmysiarAh{Ar$G

Facult_v of Electrical & Electronics EngineeringBEE2L23 ELECTRICAL MACHINES

Name: ft1&7rotlSection:

Question 1

Figure L shows a coil of wire wrapped around an iron core. The flux in the core is given by the equationO=0.25 sin 299t Wb. There were 100 turns on t}re core and the flux density obtained from the experiment is1,08 T. Sketch the magnetic flux pattern with ciear indication of flux direction, and then determine

il The magnetic flux linkage with respect to tiiJ The inducedvoltageattheterminal forf = 1s and t= S siii) The induced voltage within the conductor if a conductor with a length of 3.5 cm is inserted

:. through iron core at 45 mm/s

If the coil of 0.07 H were connected to 25 V rrppiy, rrd *" wire in the circuit has a resistance of 2 fl andiumped resistance of 10 O, find the new magnetic flux linkage.

[10 Marks]

i

:'t;

'l --+.1

Figure 1

ID:

___

Mappins CO,PO: CO1,PO1CO 01:Acquire fundamental principles of electromagnetism, transformerand electrical machinesPOI-: Ability to acguire and apply knowledge of sciences and electrical & electronics engineering fundamentals.

NLR (A FKEE UMP


2/11

BEE/BEP/12I3I

(a) The nagSretic flu: pattarl

i) Thrrragredcflux linkage, r1=x*

= 100(0. ?5sirr2Pgt)

= ?5sin299t l/y'eber - tum

ii) The ir:rduced valtqge at ita teflninal at t = 1 s and t= 5 s

tr'=Al& =l/ =iI#=too+(0.25 sin z,off)ntd'= 100(0.15i crrs l99f

= i+tS cos.lg$.{

dt t = I :+ tr' : ?475 cos 199 (l) = 3623.96V"_af f = 5 *f =T4T5cos ?99(5) = l?gT.W

(iii) E = ay

= 1.0(0.035X0.U45)

= l.70irr trr'

The new m agret.ic flirr lirrkage.

t: Lif?

but i=l 4L,,11

- " : ?.08,4IUTJ:

Tnere.fire

A = 0.07(2.fi8) = 0.15trfleb er -fr;r:n

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3/11

BEE/BEP/BECI1213 I

Univensit;l#i*laysi*.PAHANG

Faculty of Electrical & Electronics EngineeringBEE? L23 ELECTRI CAL MACHINES

Name:

Section:

Question 1

The followings are the reading of the test on the transformer.i.1 Determine tlie transformer parameters. Assume the calculationii) Sketch the corect equivalent circuit witlr detail parameters.

ID:

referred.to low voltage side.

Ma CO,PO,Domain : CO1,PO1,C3C0 01: Acquire fundamental principies of electromagnetism, transformer and electrical machinesP01: Ability to acquire and apply kBowledge of sciences and electrical & electronics engineering fundamentalsC3

Transformer ratirtg

Open Circait Test:Short Circuit Test:

Voc= 240 V;Vsc = 30 V:

Ioc=5A;Isc = 24.1 A;

Poc = 150 WPsc:250 W

10 kVA 240 / 415 V 50 Hz l{t transformer0. 866 lagg ing power factor

[10 Marks]

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4/11

Open Circuit Test- Performed at l,rrv voltage side i. e . L't;

Po" = l'o"Io, cos {,

^ -rl' 150'1

-,=--=cos l=82.82'l.(240)(s i.i

i = icosd

.' of oc

= 5 cos82.83o

= q414

I*= fo,sin4"= 5 sin82.82"

= 4.96A:Henee-

., ;". ?40ti.=:=-=384C #" i" 0.63 :N*=r*--24o =4s.4e #- n I* 4.96

Short Circuit Test- Performed at high voltage side i.e. 2nd;- Check:

, -VA - 10x103 _.A .t A _ lrFr" -::- -ifila-,sec'/ v2 415-

Hence, reading are taken fromhigh voltage side *2d winding- -,( zso ) --^

[ (30x24.1),J:

BEE/BEP/BECII2I3 I

2",=(\)"," =( 'o \rur.rr"\ 24.1,t= 143j_y'.17 * .'. fi* = 0.43f), ,Y* = 1.17 fJ u

Refer tg low voltage side . i.e. primary

R.-,= R--El' =0.43e \'= 0.14n"aq ""[sr)-":(+rs] -=--* #

x.^,= x ^^[q]' =r.17(4 J' = o.rn*-.q ,. * I. E, ) -. ., [qrs ) - :::::: #Equivalent circuit,

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5/11

BEE/BEP/BECII2I3 I

Univ*rsitiPrliaiay*iatrAi-{AruG

Faculty of Electrical & Electronics EngineeringBEEZ L23 ELECTRICAL MACHINES

CO,PO,Domain : CO1,PO1,C3C0 01: Acquire fundamental principles of electromagnetism, transformer ana eiecricai .n-ocninesPO1": Ability to acquire and apply knowledge of sciences and electrical & electronics engineering fundamentalst-')

The following are the test result from a 300 kVA 33001415 V, 50 Hz Y-A 30 transformer,

Ps6 = 3000 WV66:415 Vl6s =20A

P5g: 5000 WV55: 150 V156 :90.91 A

From short circuit test, the transformer parameters of R61 and Xsl are 0.2O and 0.930 respectively.(i) Determine the other transforrner parameters refered to high voltage side.(ii) Sketch the refen'ed approxirnate equivalent circuit of the transformer and properly iabel all the

appropriate parameters.

[10 Marks]SOLUTION:

(Open Cirarit Test- Performedai lourroltage side. Hence, knnrnrthai, at.lr:urwltage sifu, &Connected

I:. nobthttt,I,:.l3Ii Ia, =ft',-I* = I..tl J

Fo = ,i3l'* *cosq(n -, [ :uttr ]

-tftf,-t-t -- tJi (415X20)J4? nas

= I t.9J

II.p = *cosd*

- .tn.rLl

= ? crrs77.95"..J3I lt i

= t.+ lA

J-* = *r*4P +g ---i?tl: -I urr?7 95'

= 1 1.?9rt

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6/11


7/11

BEE/BEP/BEC/1112 TI

# ffi L,lniver"sitiW $ trxataysiaref* trAHANGtt.

Faculty of Electrical & Electronics EngineeringBEEZ L23 ELECTRICAL MACHINES

Name: 5c:i_uilii it

An induction maohines draws 29.52-57.73 A at full load from 415 V 3-phase y-connected 50 Hzdistribution system.

i) Select a suitable fixed shunt capacitor for power factor corection for all loading of the motor ifthe power factor is improved to 0.85.ii) Draw ttre phasor diagram.

SoU,*f,on

i) Grr-en,>:

1.-. f,q,i L_5T.T3 flV..r {ts ,

Pf= cos o

= aE (-51.1a) =|^\r={sVrT.ts0

=Js(4rs-)Ceq.s)= tl3at w

&n., = P&n G-s ltga I ton 3t'Tt "

F"p,lno* = ll o q5 :

[10 Marks]

,1''t\J

aa/e. Qoid - Q nu,Ieogl - Tort

= ]orb vAx,//'r",/ i: ([ ( t,

a= tro1s VA/

{/i

36s{ Vae P/1:' "-l ::a

- fi) 3 Lstv.--l------------J+ts/E (rrl * 5 o) {-/ '|- a= t{S'Q n^f '/

tt/':--. , , -; .r,.1, . ,OLl.I

0-53'

t=

S.otdScoe &

P&Cor Q

--, i

ii3 alo.53

a t 3trc

I'c- = Q.

-

tr l-Va r^)/

i)or6= SsnG r],.= f,r36c S'n (_q 1y': lEost vAR id rmgro\rL {s o,EE ffP.C,*- = tos e

Q= sr,rr;"rll

c 104*{f attthf

Qur+ fMappine CO n: CO2,P0

Co02:Analyzethetransformerandmachinesequivalentcircuits'under steady state conditions

PO1: Ability to acquire and apply knowledge of sciences and electrical & electronics engineering fundamentaisC3

NLTT @ FKEE UMP


8/11

BEE/BEP/I2I3I

Unlversitih/IalaysiaPAHAI{G

Faculty of Electrical & Electronics EngineeringBEE2I23 ELECTRICAL MACHI NES

Name:

Section: Date:

A motor in Figure 1 rotates at its full rated speed of 1500 rpm when supplied by a 120 Vps source. The full load iineculrent is 5l A. The compensating winding has an armature resistance of 0. I Q and field resistance of 120 O. Find

i) The current in the armatureii) The motor back emfiii) The net power and net torqueiv) The efficiency of the machine

[0 Marks]

Figure I

ID:

ins CO,PO,Dsmain : COZ,PO1CO 0Z: Analyze the transformer and machines equivalent circuits and the operating conditions for electricai ;acilpes

under steady state corditionsP01: Ability to acquire and apply knowledge of sciences and electrical & electronics engineering fundamentals

NLR@ FKEE UMP


9/11

Solution:

(r) L:110V/110Q:1A11- i1 - 11

_ql r-Jl_ I:50 A

(ii)Eo : Vr

-IoF-o

: 12tr - (50) (ll.1):115 V(iii) Total power suppl]',

F11: V1 x 11-130x5i:6120 W

_ 604-,Lo =:-JfrN

60(5750)2JI(1500)

= 1[ .14&

11:(Pqr/Ph) x 108 %=(5750/61?E)x 100%:93.9$ o/o

P ower^di s apated in armatrre,Pn: L'&

- 50'x 0.1:250W

Porryer- dissipaied in fiel{Pr: Ii'&

- 120x 12-120W

Therefore, flefi povuer: Pin - Pa - Pf

:f+;fwzs' - tzo

OR

Po,r - P.*o -P,1 (F": 0)--t-xIo:115:15[:5?50W

(\4

BEE/BEP/I213.I

NLR @ FKEE UMP


10/11

BEE/BEP/BECIIII2II

UniversitiMalaysiaPAHANG

Faculty of Electrical & Electroniqs EngineeringBEE2L23 ELECTRI CAL MACHINES

Sectionr Date:

Qtlle 6

A 440 V 50 hp Y-connected synchronous motor has per phase resistance and reactance of 0.5 O and 1 Qrespectively. The efficiency of the motor is approximate to 95 %o.

i) Draw the power flow diagram of the synchronous motor.If the motor has a 0.866 lagging power factor, determine the;

ii) Input power.iii) Per phase armature current.iv) Per phase induced emf.v) Copper losses.

ID:

Lr,a66r.,,.

rwrr , P" = 5C hp\h= 4\oRo

= o,S.cLXq = (rr-.pf'= c,trr, loOfTrL * ?sT

fi [I0 Marks]j: V, l, g/o"P,n /

=___- =3i. e6r t

6;GVTOSQ

=:P

fii) p,^Tr

Pf=c*-t& = O,BLt(L = ZOo ,i

3.g= 5.1.r+q L-30' fr;/"

blo =

=

F,n =

=

?rtil17

W

mhPsoxT+L = 3JaF

,v) V1 , Lo* [f*.n_]LB"t1XrI /E% =,v1,-- [ rere] tR*+lxrl i.J'\^>-- (/ag'w = {lc, - (tT,ilq r*3or( o,s *Ji) ,0P-L x roo%

373oo x\oo7oo.c"< 17r< 1G *

I51

"63kw

I

'::J,/

= 3B*,s -\Br,rs or 3s6.aq 1-, U*oJj lr{rjri-'ji6t: t:i 3-r:i'5'f a-tc.*,.-r,v) P* . 3r.. R;

= 3(sq"\q)'(os),.O= E . 3oB kW ,r(

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,

C0 02: Analyze the transformer and machines equivalent circuits and the operating conditions for electrical machinesunder steady state conditions

PO1: Ability to acquire and appiy knowledge of sciences and electrical & electronics engineering fundamentalsC3


11/11

BEE/BEP/BECIIIIz IIUniversitiMalaysiaPAHANG

f Ent{ixir4.i.c*t.ir *.::r.,firtt

Faculty of Electrical & Electronics EngineeringBEEZ I23 ELECTRICAL MACHINES

Name:(-i-\9,:Lr I lf

Section:

Qurt ]QAru.6

Question 1

Figure 1 show the 3-phase synchronous generator and motor which are connected in A-A connection. Theinformation about the machine is given below;

Svnchronous senerator Synchronous motorVr:415

Pu:550 W&=0.5 Cl

pFO.85 laeeine

P"=20hp

If the power loss in copper wire from the generator to motor isabout 3550 W, find the;

a) Output power of the synchronous generator.b) Input power of the synchronous motor.c) Overall efficiency of the generator-motor set.

15% losses

P,, = Po,, + total losses

ID:

l5Yo of the generated power which is

Fn':mln =

""'*rooyoll' D I

| '* I

SoUr4;'on;

e) Po.v lSTo- roo7"

Figure I

e)r 1,9-'

Po, =Pin, =

[10 Marks]x 100 /o

;

3550 W -,'",iY)roc (?e")

t\ //tf,3bu+ w

I

D p.,n _, ,1oofo - \s7" . g57.

,;i' Bs /, x 3ss0 =;/ou1 vlle7" /or ,f luu7 -3syo = )orrTw

r rrg ,"1.,,fl

a0 x 7+6

=t+dro

wP.** + loscesPo+P*1*W

fro' %J = G.V.. l. d( u y*T. = &-- - asb6_F = 3u*i++ QG Vr cos G. l-< u=> G - "rcv'r.*T. g = 38 T+ : ,=,,4t q

.h r-1-,j'R, = 3(:a:t)'(o,s)=+.iq qK"@FKEE u**,e/

" Pn, = 1366J t ssc -i--+qq,qs . ?T rtbT ..d.tl+ qeo o --;i J : 5 l

in : CO2,PO1C0 02: Analyze the transformer and machines equivalent circuits and the operating conditions for electrical machines

under steady state conditionsPOL: Ability to acquire and apply knowledge of sciences and eiectrical & electronics engineering fundamentalsC3

ump electrical machine all quiz (1)

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