8/11/2019 Solved Problems Ch2

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Solved Problems: Chapter 2

2.5A source which can be represented by a voltage source of 8 V rms in series with an internal resistance of 2k is connected to a 50- load resistance through an ideal transformer !alculate the value of turns ratio for

which ma"imum power is supplied to the load and the corresponding load power# $sing %A&'A() plot the thepower in milliwatts supplied to the load as a function of the transformer ratio) covering ratios from *0 to *00

Solution:

+or ma"imum power transfer) the load resistance ,referred to the primary must be e.ual to the source resistance

2

2*

2

20002000 /2

50L L L

NR R n R n

N

= = = = =

&he primary current1

2 2

2

* * 2ower supplied to the load1 8 m3

2 4 4

s s sload L s

s s s

V V VI P R I R

R R R= = = = =

+or a general turns ratio n1

2

2 2

* *2 2

s s sload L L

s L s L s L

V V VI P R I n R

R R R n R R n R

= = = =

+ + +

2.10A 4/0-V12400-V transformer has a series leakage reactance of 2 as referred to the high-voltage side Aload connected to the low-voltage side is observed to be absorbing 25 k3) unity power factor) and the voltage is

measured to be 450 V !alculate the corresponding voltage and power factor as measured at the high-voltageterminals

Solution:

6econdary current1 2 *25000 4/0

5555 A rimary current 5555 *0/5 A

450 2400

load

load

PI I

V

= = = = =

rimary voltage1 ( )* * 2 22400

2 450 248 V4/0

V j I V V = + = =

( ) ( )* * 22 2 *0/5 248 248 7/*8 28*0 758 VV j I V j j = + = + = + = o

ower factor at primary terminals1 cos,758 078/* lagging

2.11&he resistances and leakage reactances of a 0-kVA) /0-9:) 2400-V1240-V distribution transformer are

;* 0/8 ;2 000/8 raw the e.uivalent circuit referred to ,i the high- and ,ii the low-voltage sides 'abel theimpedances numerically

b !onsider the transformer to deliver its rated kVA to a load on the low-voltage side with 20 Vacross the load ,i +ind the high-side terminal voltage for a load power factor of 085 lagging ,ii+ind the high-side terminal voltage for a load power factor of 085 leading

c !onsider a rated-kVA load connected at the low-voltage terminals operating at 240V $se

%A&'A( to plot the high-side terminal voltage as a function of the power-factor angle as the load

power factor varies from 0/ leading through unity power factor to 0/ pf lagging

8/11/2019 Solved Problems Ch2

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j 8 0/8 0/8 j 8 240 12400

?

V 9

@

?

V '

@

A 9

?

V 9

@

?

V '

@

240 1 2400j 008 j 008 000/8 000/8

A '

?

V send

@

feeder

Z f = 75? j /0

5 1 24

?

V 9

@

?

V load

@

Z eq= 02 + j *2

?

V load

@

I load

?

V '9

@

Solution:

,a,i referred to the 9V side

,ii referred to the 'V side

,b $sing the e.uivalent circuit referred to the 9V side) 20 0 VLV = o

'oad current10000

78 A where is the pf angle , 0 for leading pf20

loadI = = >

;eferred to the 9V side1 78 A 200 0 ,*/ *5/78H H L H HI V V Z I j = = + = + + o

200 *25/8cos *4/28sin ,*4/28cos *25/8sin HV j = + + +

pf 085 leading *7 22/ *** 22/ / VHV j= = + = o o

pf 085 lagging *7 287 **// 2708 282 VHV j= = + = o o

2.13A single-phase load is supplied through a 5-kV feeder whose impedance is 75 ? j/0 and a 5-kV12400-V transformer whose e.uivalent impedance is ,02 ?j*2 referred to its low-voltage side &he load is */0

k3 at 087 leading power factor and 240 Va !ompute the voltage at the high-voltage terminals of the transformer

b. !ompute the voltage at the sending end of the feederc !ompute the power and reactive power input at the sending end of the feeder

Solution:

,a =.uivalent circuit for the transformer and load1

0

*

*/0 *0 cos /8 A

240 087

cos 087 2* leading /8 2* A

L load load load

load

P V I I

I

= = =

= = = o o

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&he 9V side voltage referred to the 'V side1

,02 *2,/8 2* 240 2**2 747 V

5 * *84 kV 4 kV

24

H eq load L

H H H

V Z I V j j

V V j V

= + = + + = +

= = + =

o

,b 'oad current referred to the 9V side124

/8 2* 52/8 2* A5

load feed I I = = = o o

,75 /0 52/8 2* * *84 28/ kV

45 kV

send f feed H

send

V Z I V j j j

V

= + = + + + = +

=

o

2.17&he following data were obtained for a 20-kVA) /0-9:) 24001240-V distribution transformer tested at /09:1 Voltage, Current, Poer, V A 3

3ith high-voltage winding open-circuited 240 *08 *223ith low-voltage terminals short-circuited /* 8 25

a !ompute the efficiency at full-load current and the rated terminal voltage at 08 power factorb Assume that the load power factor is varied while the load current and secondary terminal voltage are held

constant $se a phasor diagram to determine the load power factor for which the regulation is greatest 3hat is this

regulation#Solution:

,a ;ated current on the 9V side 20 kVA B 2400 8 A &herefore) total power loss at full load current1' *22 ? 25 7 3 'oad power at full load) 08 pf 08 C 20 k3 */ k3 &herefore) input power

*/ ? 07 */7 k3 efficiency ,*/ B */7 C *00D 7 D

,b &he e.uivalent impedance of the transformer1 ) ) )eq H eq H eq H Z R jX= +

) )2 2

2 2

) ) )

25 /* /

8 8

, , //

sc sceq H eq H

sc sc

eq H eq H eq H

P VR Z

I I

X Z R

= = = = = =

= =

'et load current and voltage referred to the 9V side1 0lH lH V V I I = = o

2 2 2 2

, and

, cos sin

, cos , sin 2 cos

s lH eq lH eq lH Z d d eq lH Z

d d

s d d d d

V V Z I V Z I V V V Z I

V V jV

V V V V V VV V

= + = + + = + +

= + +

= + + = + +

@ @

2 2 2

2

2 cos 2 cos;egulation * *s lH d d d d

lH

V V V VV V V V V

V V V V

+ + = = = + +

&herefore) regulation is ma"imum when cosis ma"imum

*cos * 0 tan 578*;

eqH

Z Z

eqH

X

= = + = = = =

o

%a"imum regulation1

/ 8 /** VdV = =

3hen /**0 ;egulation 002/ 2/D2400

ds d VV V V

V= = + = = = =o

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?

?

V 9

@

IsE E

?

V '

@

Z G= j *//8 Z eq= 0005 + j 008

2.37A three-phase generator step-up transformer is rated 2/-kV145-kV) 850 %VA and has a series impedance of0005 ? F008 per unit on this base Gt is connected to a 2/-kV) 800-%VA generator) which can be representedas a voltage source in series with a reactance of j*5 per unit on the generator base

,a !onvert the per unit generator reactance to the step-up transformer base

,b &he unit is supplying 00 %3 at 45 kV and 075 power factor lagging to the system at the transformerhigh-voltage terminals,i !alculate the transformer low-side voltage and the generator internal voltage behind its reactance in

kV

,ii +ind the generator output power in %3 and the power factor

Solution:

,a Hn the transformer base850

*5 *//8 pu800

genX = =

,b er-unit e.uivalent circuit1

(i) &ransformer low-side voltage and generator internal voltage1

) )

)

)

45 kV) 2/ kV) 850 %VA

*0 0 pu

00 850kA *2 kA kA *4225 kA

45 075 45

*2 08//8 pu 08//8 *82 pu*4225

H;

base H base L base

H

bases base H

base H

s s

V V VA

V

VAI I

V

I I

= = =

=

= = = = =

= = =

o

o

00 08250825 pu 08//8 pu

850 cos * 075s p

H p

PP I

V = = = = =

,0005 008 *02/4 00* pu *0287 74

*0287 pu 2/5 kV

, *//8 *48 *4442 pu 20//4 444 pu 2/ 20//4 5 kV

L H s

L

G L s G

V V j I j

V

E V j I j E

= + + = + =

= =

= + = + = = =

o

o

,ii Eenerator output power ,at its terminals

*

*0287 74 08//8 *82 082/* 0/* pu082/* 850 02*7 %3

0/*power factor cos,tan 072/ lagging

082/*

G L s

G

! V I jP

= = = += =

= =

o o