EE6401 Notes.pdf

UNIT I

INTRODUCTION

Magnetic Circuits

Ampere’s law

� ò

� � H. dl =

C

� ò

�

B .d a� =

�

0

B:

ò

� J.da�

�

H:

magnetic field intensity vector,

�

J : current density.

S

m a g n e t ic f lu x d e n s i t y v e c t o r .

magnetic flux density is conserved

Þ

S

�

�

m = m r m0 : magnetic permeability of medium.

B = mH

m0 : permeability of free space m0 =4p ´ 10

-7

m r : r e l a ti v e p e r m e a b il i ty

Simple magnetic circuit. � ò C

� � H. dl =

ò

�

J .da �= Ni �

: magnetomotive force (mmf, ampere-turns).

F

S �

Magnetic flux crossing surface S:

=ò .

�

(Weber, Wb)

a

S

fc @ Bc Ac

� ò

� � H . dl @

fc : flux in core,

Bc : flux density in the core Ac : cross-se ctional area of the core.

Hc lc

Þ

C

Þ f

F Â

Â=

Bc m lc m Ac

lc = Ni = F

Þ

: reluctance

f m Ac

lc = F

Fig. 1.2 Magnetic circuit with air gap.

Flux is the same in the magnetic core and the air-gap. Bc = Þ Bg =

mmf

flux density in the magnetic core.

Ac

flux density in the air-gap.

Ag

� ò

� � H. dl = Hc lc

+ Hg g = Ni = F

C

Þ

F=

Þ

m 0 Ag

F = f(Â c + Â g )

Þ f

=

÷

Bc m

g

lc +

m0

g =f ç è

+ç

ö ÷

m Ac ø

F

 c : reluctance of core,  g : reluctance of air-gap.

Analogy between electric and magnetic circuits. (a) Electric circuit, (b) magnetic c ir cuit. The magnetic circuit shown in Fig.1.2 has dimensions Ac = Ag = 9 cm2, g = 0.05 0 cm, lc = 30 cm, and N = 500 tums. Assume the value m r = 70,000 for core material. (a) Find the reluctances Âc and Âg . For the condition that the magnetic circuit is operating with Bc = 1.0 T, find (b) the flux f and (c) the current i. Example 1.1

Solution: (a)

g

0.3 = 7000 0 ´ 4p ´ 10-7 ´ 9 ´10 -4

lc

Â=c

=

mr m0 Ac

=

3.79 ´ 103 A.tu rns/Wb

g 5 ´ 10 -4 = = 4.42 ´ 10 m0 Ac 4p ´ 10 -7 ´ 9 ´ 10 -4

(b )

f = Bc A c = 1.0(9´1 0-4 ) = 9 ´10-4 W b

(c )

i=

F = N

c

g

N

=

A.turns/Wb

9 ´10-4 (4.46 ´105 ) = 0.8 A 500

Air-gap fringing fields. Example 1.2 The magnetic structure fo a syn ch ronous machine is shown schematically in Fig. 1.5. Assuming that rotor and stator iron have infinite permeability ( m ®¥), find the air-gap flux f and flux density Bg. For this example I = 10 A, N = 1000 turns, g = 1 cm, and Ag = 2000 cm2.

Solution: total air-gap length = 2g. Reluctance of iron core is negligible ((m ®¥). Rg = B =g

2g

Þ

m0 Ag f

= Ag

f =

F Rg

=

Ni m0 Ag 2g

1000 ´ 10 ´ 4p

2 ´ 0.01

´ 10 -7´ 0.2

= 0.13 Wb

0.13 = 0. 65 T 0. 2

Simple synchronous machine. FLUX LINKAGE, INDUCTANCE AND ENERGY � E . ds � = -

Faraday’s law.

� ò C

dl dt

d

d

dt ò

B.�da � S

e : induced voltage,

dt

l=

N f : flux linkage (Wb-turns).

Linear magnetic circuit: flux linkage is proportional to current. Inductance: L = l i

l =

Nf

=

NF Rtot

2 N 2i Þ L= Rtot Rtot

If the reluctance of the core is negligible compared to that of the air-gap 2

2

L=

æ g ç m0 Ag è

ö ÷÷ ø

=

m0 Ag

g

(Henry, Weber-turn /A)

Example 1.3

The magnetic circuit of Fig. 1.6a consists of an N-turn winding on a magnetic core of infinite permeability with two parallel air gapos f lentg hs g1 and g2 and areas A1 and A2, respectively. Find (a) The inductance of the winding and (b)The flux density Bl in gap 1 when the winding is carryingacu rtren i. Neglect fringing effects at thea igr ap. Solution: (a)

Req = L=

(b)

1

=

R1R2 R1 + R2 N 2 ( R1 + R2 ) R1R2

Ni m0 A1 Ni = R1 g1

R1 =

g1

= m0 N

Þ

R2 =

m0 A1 æ

1

è g1

B1 =

f1

A1

+

=

g2 m0 A2

A2 ö ÷ g2 ø m0 Ni

g1

(a) Magnetic circuit and (b) equivalent circuit for Example 1.3.

Magnetic circuit with two windings. The total mmf:

F = N1i1 + N 2i2

Neglect the reluctance of the core, and assume Ac = Ag, the resultant core flux produced by the total mmf is =

F N 1i1 + N 2i2 m A = =g( N 1i1 + N 2i2 ) 0 c Rg m0 Ac

g

Flux linkage of coil 1, æ m0 Ac ö

l1 = N1f = N12

ç

è

Þ

g

÷ i1

ø

æm A ö + N1 N0 cç è g ø

÷ i2

2

l1 = L11i1 + L12i2 æ m0 Ac ö

L11i1 : flux linkage of coil 1 du e to its owncu rer nt.

L11 = N1 2 ç

L12i2 : flux linkage of coil 1 duet ocu rren t i2 . L12 = N1 N an d 2

2

ç

æ m0 Ac ö

÷

g

: self-inductance of coil 1

÷

g ø

è

: mutual-inductance between coils 1

ø

è

Similarly, flux linkage of coil 2,

l2 = N 2f = N 1 N 2

æ m0 Ac ö

ç

è

Þ

g

ø

m A ö + N02 cç è g ø

2æ

÷ i1

÷ i2

l2 = L21i1 + L22i2

L21 = L12 = N1 N

2

ç

æ m0 Ac ö è

g ø

÷

: mutual-inductance,

æ m Ac ö

L22 2= N 20 è

ç

g ø

÷

: self-inductance of coil 1

Induced voltage (magnetic circuit with single winding):

e=

dl d di dL = ( Li ) = L + i dt dt dt dt

In electromechanical energy conversion devices, inductances are often time-varying. Power and Energy

Power delivered to the winding: = ei = i

dl dt

(Watts, W= Joules/ second)

Change in magnetic stored energy DW in the magnetic circuit in the time interval tl to t2 t2

DW =

l2

òt p dt = lò i dl 1

1

Single- winding system of constant inductance, thce h ang e in magneit c stored energy as the flux level is changed from l 1 to l2 l2

=

=

ò l1

l2

=

ò

l 1 2 2 1 (l - l1 ) Þ W (l ) = or W (i ) = Li 2 2L 2L 2

l

l1

: total stored magnetic energy.

Example 1.6

For the magnetic circuit of Example 1. 1 (Fig1., . 2) find (a) the inducat nce L, (b) the magnetic stored energy W for Bc = 1.0 T, and (c) the induced vlo tage e for a 60-Hz time-varyingco er flux of the form Bc = 1.0 sin(ωt) T where ω = 2πf = 2π(60)= 377r ad/sec. l

N

L= i = i

(a) (b) From Example 1.1,

F = Âc + Âg

Bc = 1.0 T

F = Ni

Þ

L=

Þ i = 0 .8 A

2 N2 500 = 10 ´ Âc + Âg 4.46 5 = 0.56 H

1 2

1 2

Þ W = Li 2 = (0.56)(0.82 ) = 0.18 J

(c) dl df = dB = NAc c dt = 170 cos(377t ) V

e=

= 500 ´ 9 ´ 10-4 ´ 377 ( ´ 1.0 cos(377t) )

PROPERTIES OF MAGNETIC MATERIALS ·

Magnetic materials are used to constrain and direct magnetic fields in well-defined paths.

·

Inatr ansmfor erh t ey ar e used to maximize the coupling between wthe i ndgn i s, adn to lower the excitation current required for transformer operation.

·

In electric machinery, theya re usde tos h ape the magnetfiic seld to obtain desired electrical and mechanical characteristics.

are composed of iron and alloys of iron with cobalt, tungsten, nickel, aluminum, and other metals, are the most common magnetic materials. Ferromagnetic materials

Relationship between B and H for a ferromagnetic material is nonlinear and multivaluBed: curve or hysteresis loop.

H

B-H loops for M-5 grain-oriented electrical steel 0.012 in thick. Only the top halves of the loops are shown here.

Dc

magnetization curve for M-5 grain-oriented electrical steel 0.012 in thick.

AC EXCITATION

Sinusoidal coref lu:x

j (t ) = fmax sinwt = AcBmax sinwt

Voltage induced in the N-turn winding, e(t ) = wNfmax coswt = Emax coswt

Emax = wNfmax = 2p fNAcBmax

Rms value of a periodic function of time, Frms =

Erms =

1 T

T

ò f (t )dt 2

0

Emax = wN fmax = 2p fNAcBmax 2

Exciting current:

current required to produce magnetic flux in the core; nonsinusoidal because of the nonlinear magnetic properties of tcher o e. Sources of power loss in the core: 1-) E ddy currents: currents inducedby eth time- va rying flux in the core due to nonzero conductivity of the material. These currents cause ohmic I2R loss. To reduce the effectsof d e dy cu rrents, magnetic structures are built of thin sheets of laminations of the magnetic material. These laminations are aligned in the direction of the field lines, and are insulated from each other by an oxide layer. 2-) Hysteresis loop: The time-varying excitation causes the magnetic material to undergo a cyclic variation described by a hysteresis loop. Energy input W to the magncetric o e of Fig. 1.1 as the material undergoes a single cycle æ i W=� ò d l �ò è

H c lc ö( A N ÷ cø

NdBc ) = Aclc

Aclc : volume of the core. � ò

H

dBc c

: area of the ac hyst eresis loop.

Hysteresis power loss = W´ f

ò

c

�

H

dBc

Excitation phenomena. (a) Voltage, flux, and exciting currentcorre ; (b) hysteresis loop.

spogndin

Exciting rms voltamperes per kilogram at 60 Hz for M-5 grani -oriented electrical steel 0.012 in thick.

Hysteresis loop; hysteresis loss is proportional to the loop area (shaded).

Core loss at 60 Hz in watts per kilogram for M-5 grain-oriented electrical steel 0.012 in thick.

Laminated steel core with winding for Example 1.8. The magnetic core in Fig. 1.15 is made from laminations of M-5 grain -oriented electrical steel. The winding is excited with a 60-Hz voltage to produce a flux densityni the steel of B = 1.5 sin ωt T, where ω = 2π60 = 377 rad/sec. The steoc el cups ie 0.94 of the core crosssectional area. The mass-density fo the stee l is 7.65 g/cm3. Find (a) the applied voltag(b e, ) the peak current, (c) the rms exciting current, and (d) the core loss. Example 1.8

a.

Voltage induced dj dB = NAc c = 200 ´ 4 in dt dt = 274 cos(377t ) V

e=N

b.

2

æ 1m ´ 0.94 ´ ç è 39.4in

ö

2

÷´ 1.5 ´ 377 cos( 37 7t )

ø

Bmax = 1.5 T

Þ

lc = 0.71m.

Þ

Bmax = 1.5 T Vc = 105.5 in3.

Þ Þ

Total rms volt-amperes:

1 .1

H max = 36A-t urnsm/ H l 36 ´ 0.71 = 0.13 A. peak current I = max c = N 200 Pa = 1.5 VA/kg Wc = 13.2 kg. Pa = 1.5 VA/kg ´13.2 kg =20 AV

Ij , rms =

Pa Erms

=

20 = 0.10 A 275/ 2

A magnetic circuit witah agsingle ir ap is shown in Fig. 1.24. The core dimensions are: -3

2

Cross-sectional area Ac = 1.8 × 10 m Mean core length lc = 0.6 m Gap length g = 2.3 x 10-3 m N = 8 3 t u rn s

Assume that the core is of infinite permeability ( m ®¥ ) and neglect the effects of fringing fields at the air gap and leakage flux. (a) Calculate the reluctance of the core Rc and that of the gap Rg . For a current of i = 1.5 A, calculate (b) the totalf l ux f, (c) the flux linkages λ of the coil, and (d) the coil inductance L.

Solution:

(a)

Rc = 0 since m ® ¥

(b)

=

(c)

l =

(d)

=

Rg=

g m0 Ac

2.3 ´ 10-3 = 4p ´ 10-7 ´ 1.8 ´ 10 -3

1.017 ´ 10 A/Wb

Ni 83 ´ 1.5 = = 1.224 ´ 10- Wb Rc + Rg 1.017 ´ 106

Nf = 1.016 ´ 10-2 Wb l 1.016 ´ 10-2 = = =. m i 1.5

e 1.1. Assuming 1. 3 Consider the magnetic circui tof F.ig 1.24 with the dimensions foPr oblm infinite core permeabilcity, alceulat (a) the number of turns required to achieve an inductance of 12 mH and (b) the inductor current which will result in a core flux density of 1.0 T.

Solution:

(a) L =

2

Rg

= 12 ´ 10-3 mH

N = 12 ´ 10-3 ´1.017 ´ 106 = 110.47 Þ N = 110 turns

Þ

(b) Bc = Bg = 1.0 T i=

1 .1 3

l

L

Þ

f=

Bg Ac = 1.8 ´ 10 -3 Wb -3

=

N f 110 ´ 1.8 ´ 10 = = 16.5 A L 12 ´ 10 -3

The inductor of Fig. 1.27 has the following dimensions:

Ac = 1.0 cm 2 lc = 1 5 c m g = 0.8 mm N = 4 8 0 t u rn s

Neglecting leakage and fringing and assuming

m r = 1000 ,

calculate the inductance.

Solution:

L=

l

i

l =

N f = NBc Ac

mmf equation: H clc + H g lg = Ni Þ

Bc = Bg

Þ

Bc =

m0 Ni

g + ( lc / m r )

Þ

Bc lc + m g = Ni 0

m r m0

L=

m0 N 2 Ac

g + ( lc / m r )

=

4p ´ 10 - ´ 480 ´ 10 = 3 0 .4 7 7 m H [ 0.08 + (15 / 1000)] ´ 10 -2

The inductor fo Problem 1.13 is to be op erated from a 60-Hz vol tage source. (a) Assuming negligible coil resistance, calculate eh t rms inductor voltage corresponding tao p kea core flux density of 1.5 T. (b) Un der this operating condition, calculate the rms current and the peak stored energy. 1 .1 4

Solution:

(a) dl dBc v(t ) = dt = NAc dt Bc = Bmax sinwt Þ v(t ) = wNAcBmax coswt 1 1 (2p ´ 60) ´ 480 ´ 10-4 ´1.5 = 19.2V w c max = rms = 2 2

(b) I rm s =

1 .1 6

Vrms = 1.67 A wL

Wpeak =

1 2 1 LI peak = ´ 30.477 ´10-3 ´ ( 2 ´1.67)2 = 85.0 mJ 2 2

A square voltage wave having a fundamental frequency fo 60 Hz and equal positive and negative half cycles of amplitude E is applied to a 1000-turn winding surrounding a closed iron core of 1.25 x 10-3m2 cross section. Neglect both the winding resistance andany effetc s of leakagef lux .

(a) Sketch the voltage, the winding flux linkage, and the core flux as a function of time. (b) Find the maximum permissible value of E if the maximum flux density si not to exceed 1.15 T.

(a)

e E λm a x

voltage

T

λ

Φ

t λmax

E

(b )

dl lmax - ( -lmax ) Þ l ò e( t ). t Þ E = = 4 f lmax = 4 fN fmax = 4 fNAcBmax dt T/2 -3 E = 4 ´ 60 ´ 1000 ´ 1.25 ´ 10 ´1.15 = 345 V

e(t ) = Þ

1.24 The reciprocating generator fo Fig. 1.34 has a movable plunger (position x) which is supported so that it can slide in and out of ht e magnetic yokemwhilea intaignin a constant air gap of length g on each side adjacent to theyo k.e Both the yoke and the plunger can be considered to be of infinite permeability. The mo tio n of the plungeis rco nstradine such thatits position is limited to 0 £ x £ w . There are two windings on this magnetic circuit. The first has N1 turns and carries a constant dc current I0. The second, which has N2 turns, is open-circuited and can bec on nected to a load. (a) Neglecting any fringing effects, find the mutual inductance between windings 1 and 2 as a function of the plunger position x. (b ) The plunger is driven by an external source so that its motion is given by x

where generated as a result of

=

w(1+ e sin wt ) 2

< 1. Find an expression for the sinusoidal voltage which is

e

this motion.

(a)

L21 = N1 N ÷

(b)

v2 =

d l2 dt

æ m0 Ac( x ) ö

2

ç

Ac ( x ) = D ( w - x )

2g l2 = L21i1

dx 1 = we w cosw t dt 2

Þ

Þ

v2 = I 0

v2 = - I 0

dL21 dL dx æ m N N D ö dx = I 0 21. = -I 0 ç 0 1 2 ÷ . dt dx dt 2g è ø dt

æ m0 N1N 2 Dwe w ö t ç ÷ c osw 4g è ø

UNIT II TRANSFORMERS

A transformer is a static machine whitrch anssfer ac electrical power from one circuit to another wht i out any el ectrical link between. It essentially consisst of two windings- the primary and secondary, wound on a common laminated magnetic core. The winding connected to the ac source is called primary winding and the one connected to load is called secondary windin. g The alternating voltage V1 is applied to the primary. Dependinon ge th no. of turns of primary(N1) and secondary (N2) , tal he ternagtin emf, E2 is induced in the secondary.

Working Principle

When a sinusoidally v rying voltage V1 is applied to tpr he imaryn, a alternating flux Φ is set up in the core. Thisflxu links both the windings and induces emfs E1 and E2 in them according to Faraday’s laws of electromagnetic induction. According to Lenz’s law, induced emf acts in opposite direstion to the appl ied voltage V1. Ie. E1

=

N1 df

E2

=

- N2

Therefore

dt

df dt

E2 N2 = = K where K is known as voltage transformation ratio. E1 N 1

If N2>N1 then E2>E1, it will be a step up transformer. If N1>N2 then E1>E2, it will be a step down transformer.

Ideal Transformer

An ideal transformer is one that has

1. no winding resistance 2. no leakage flux 3 . n o i ro n l o s s . EMF Equation of a Transformer

Consider an alternating voltage V1 of frequencyf appdlie to primary of the transformer. This develops a primary current which sets up an alternating flux Φ The instantaneous emf induced in thes primary i e1= -N1

d df = - N 1 m s in w t dt

= -ω N1 Φm Cosωt

=-2 П f N1 Φm Cosωt = 2 П f N1 Φm sin(ωt-90) It is clear from the above equation that maximum value of induced emf in the primary is Em1=2 П f N1 Φm the rms value E1=Em1/ 2 = 4.44 f N1 Φm E2=4.44 f N2 Φm Transformer Construction

The main elements of a transformeare r o tw windings and a core. The two coils are insulated from each other as well as from the core. The core is constructed from laminations of sheet steel or silicon steel assembled to provide a continuous magnetic pat.h Silicon steel offers low hysteresis loss and the laminations minimizes eddy cu rrent loss. The laminations are insulated from each other by a lighco t atgn i of varnish. According to the core construction and

the manner in which the primary sande conad rya er placed around it, transformerare s classdifie as 1. core type

2. Shell type

LOSSES IN TRANSFORMER

The losses in transformer are 1. iron losses or core losses 2. Copper losses Iron losses

Since iron core is subjected to alternating fluxth , re e occurs eddy cur raentndh ystesresi loss in it. These two losses together known as iron losses and core losses. Both hysteresis and eddy current losses depends on maximum flux density Bm in the core and supply frequency f. Iron loss

= Hysteresis loss + Eddy r cur ent loss

Copper Losses

`

The primary andseoc ndary of the transf ormer have winding resistances of R1 and R2 respectively.

Total copper loss

=

I12 R1 + I22 R2

Where I1 and I2 are primary and secondary currents.

Total losses in a transform = ePi cr +

P

EFFICIENCY OF A TRANSFORMER

Transformer Efficiency

=output power Input power

Iron loss of a transformer

= Pi

Full load copper loss

= Pc

Total full load loss

= Pi + Pc

Full load efficiency

= Full dloa VA *P.f Full load VA * P.f + Pi+Pc

For any fraction x of the full load efficiency

=

x Full load VA * P.f

x*Full load VA * P.f + Pi+x2 Pc Power Transformer

Power Transformers are used in generating stations or sub stations for transforming voltage at each end of transmissioln i n.e They are put in operation during load hours and thrown off during light load hours. These transformers are designed to have maximum efficiencyt a or near full load. Normally the power transformers are rated in MVA.

Distribution Transformer

Distribution Transformers are used for stepping down the voltage toast and ar voltage and kept near or at the consumer’s premises.They are continuously n i circuit whether they are carryingany dloa or not. He core losses would occur for all tha time where copper losses occur only when they are loaded. So they are designed ot reduce the core losses compared to copper losses. They must be designed for good all day efficiency and not for efficiency at full load.

Instrument Transformer

Instrument transformers are used to extend the range of instruments for the measuring purposes. They are of two types 1. Current transformers for measuring large ac currents 2. Potential transformers for measuring high ac voltages

Fig.2.1 Schematic views of (a) core-type and (b) shell-type transformers.

Cutaway iv ew of self-protected distribution transformt er yplica of sizes 2 to 25 kVA, 7200:240/120 V. Only o ne high-voltage insulator and lightning arresteisr nedede because one side of the 7200-V line and one side of the prrima y are grounded. (General Electric Company.) Figure 2.2

No-Load Conditions

Exciting current

ij

establishes an alternating flux in the core. Voltage induced in the primary

w in d in g

e1 =

d l1 dj = N1 dt dt

KVLe qn . for the primary winding : v1 = R1ij + e1

R1: primary resistance

(primary leakage flux neglected)

Resistance drop is very small Þ induced voltage e1 is very nearly equal to the applied voltage. Hence, it is almost sinusoidal. Therefore, the flux is also sinusoidal. j = jmax sinwt

Þ

dj = wN1j max coswt dt V1 E1 @ V1 Þ j max = 2p

e1 = N1

Rms value: E1 =

1 2p fN1j max = 2p fN1j max 2

The core flux is fixed by the applied voltage. The required exciting current is determinby ede magnetic properties ofco thee r .

th

Figure 2.4 Transformer with open secondary.

Iˆc : core-loss component of exciting current; supplies the power

due to hysteresis and eddy current losses. Pc = E1 Ij cosqc

Iˆm : magnetizing component;

Figure 2.5 No-load phasor diagram.

The Ideal Transformer

Figure 2.6

v1 = e1 = N1

Ideal transforamer nd .load

dj dt

Core flux links the secondary and induces the voltage e2 : Þ

v2 = e2 = N 2

dj dt

v1 N1 = v2 N 2

When a current in the secondary windgn i flows, the total mmf should be zero since the reluctance of the core is very large. Þ

N1i1 - N 2i2 = 0

Instantaneous power:

Þ

i1 N 2 = i2 N1

vi = v i 11

2 2

Figure 2.7 Three circuits which are identical at terminals ab when the transformer is ideal. Example 2.2

The equivalent circuit of Fig. 2.8a shows an ideal transformer with ain mpe danec R2 + j X2 = 1 + j4 Ω connectedin se ri s with the secondary. The tu ms ratio N1/N2 = 5:1. (a) Draw an equivalent circuit with the series impedanrce efeerr d to the primary sid.e (b) For a primary voltage of 120 V rms and a short connected across the terminAals , B calculate the primary cturren and the current flowing in the short.

F repe2d.8anEcqeurievfaelrernetdctiorctuhietspfroim r Earxy.ample 2.2. (a) Impedance in series with the secondary. (bi)guIm

Figure 2.9 Schematic view of mutual and leakage fluxes in a transformer.

Iˆ2¢ is the component of the primary current which exactly counteracst the mmf of the secondary current Iˆ2 . The net mmf is produec d by theex citign current I (in the primary winding).

Therefore the net mmf is N 1Iˆ = N1Iˆ1 - N 2 Iˆ2 = N1( Iˆ + Iˆ2¢ ) - N 2 Iˆ2 Þ

Iˆ2¢ =

N1Iˆ2¢ = N 2 Iˆ2

Þ

N2 ˆ I2 N1

The equivalent sinusoidal current Iˆ that represents ethex citgn i current can be resolved into a core-loss component Ic in phase with the emf Ê1 ,maand by 90°.

aggnetizin component Im lagging Ê1

Rc : core-loss resistance Lm : magnetizing inductance

Þ

magnet2izingr eac tance: X m = 2p f Lm Ê1

core loss due torethe t sul ant mutual flux =

Rc

excitation branch

Exciting impedance Zj =Rc / / jX m Ê1 N 1 = Ê2 N 2

Figure 2.10 Steps in the development of thee travnsformer qui alent circuit. X l¢2 =

æ è

ç

ö1 N

N2 ø

2

Xl 2

2

N1 ö ÷ R2 N è 2ø N V2¢ = 1 V2 N2 æ

R2¢ = ç

Example 2.3

A 50-kVA 2400:240-V 06 -Hz distribution transformer hasale akga e impedancef o 0.72 + j0.92 Ω in the high-voltage winding and 0.0070 + j0.0090 Ω in the low-voltage winding. At rated voltage and frequency, the impedance Zj of the shunt branch (equal to the impedance of Rc and jXm in parallel) accounting for the exciting current is 6.32 + j43.7 Ω when viewed from the lowvoltage side. Draw tehe quivatlen circuit referred to (a) the hgi h-voltage side and (b) the lowvoltage side, and label the impedances numerically.

Approximate equivalent circuits

Equivalent series impedance = R eq + jX eq Example 2.4

Consider the equivalent-T circuit of Fig. 2.11a of the 50-k VA24 00:04 2 V distribution transformer oEx f amel p 2.3 in which the impedances are referred to the high-voltage side.(a ) Draw the cantileeq ver uivt alen circuit with the shunt branch at the high-voltage terminal. Calculate and label Req and Xeq. (b) With the low-voltage terminaopenl circiu t and 2400 V applied to the high-voltage terminal, calculate the voltage at the low-voltage terminal as predicted by each equivalent circuit.

Req = 0.72 + 0.7 0 = 1.42 Ω Xeq = 0.92 + 0.90 = 1.82 Ω

Vcd = Vab = 2400 V

Thee quivl a ent T-circuit æ Zj ö Vc¢d ¢ = 2400 ç ç Zj + Z l 1 ÷ è ø

632 + j 4 3 7 0 ö 2 400 ç ÷ è 632.72 + j 4 3 7 0 .9 2 ø æ

÷=

= 2399.4 + j 0.315 V

Example 2.5

The 50-kVA 2400:240-V transfmr o er w o h se pa rameters are given in Example 2.3 is used to step Vs datow l odafeeednedr oisf a24fe0e0dV.er wFdn hoi seim thne tsheenvdoinltgage endta tohe f the the pveodaltangcee aist 0th.3e0s+ecjo1n.d6a0ryte. rThe minvasloltaofgethe transformwh er n e the load connected to its secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops ni the transformer and feeder caused by the exciting current.

Zeq = 1 .42 + j 1.82 Ω Z t o t = 1 .7 2 + j 3 . 4 2 Ω = R + j X Iˆ =

50 000

2400

q = cos-1(0.8) = 36.87

= 20.83A

Þ

V ¢ = 2328.3V

Þ

lagging (i.e. current lags voltage) �

Î = 20.83Ð - 36.87 A = 16.66 - j12.5 A

Þ

Vˆ2¢ + (R + jX ) Iˆ = Vˆs = 2400Ðd

�

(V ¢+ 71.41)2 + 35.482 = 24002

Þ

æ

2

è

N1 ø

V=ç

ö ÷V ¢ =

232.83

Short-Circuit Test

Zj ( R2 + jX l 2 ) Zj � R2 + jX l 2 Zj + R2 + jX l 2 Z sc � R1 + jX l1 + R2 + jX l 2 = R e q + jX eq

Z sc = R1 + Þ

jX l1 +

V ¢= V2ˆ ¢

Vsc , I sc and Psc measured Z eq = Z sc =

Vsc

I sc

Þ

, Req =

Psc

,

I s2c

Xe =

Ze

- Re 2

Open-Circuit Test

2.16 Equivalent circuit with open-circuited secondary. (a) Complete equivalent circuit. (b) Cantilever equivalent circuit with the exciting branch at the transformer primary. Figure

Vo c, I oc and Poc measured

Zoc = Zj =

Rc ( jX m )

Rc

Þ

Rc =

Voc2

Poc

+ jX m

,

Zj =

Voc

I oc

,

1

Xm =

2

(1/ Z ) - (1/ R ) j

c

2

Example 2.6

With the instruments located on the high-voltage side and the lw o -voltage side short-circuited, the short-circuit test readings for the 50-kVA 24 00:240-V transf ormer of Example 23. are 48 V, 20.8 A, and 617W. nA open-circuit test with the low-voltage sideen re gized gives instrument readings on that side of 240 V, 5.41 A, and 186 W. Determine the efficiency adn the voltage regulation at full load, 0.80 power factor lagging. From the short-circuit test, Z eq ,H =

48 617 = 2.31 W , R e q , H = = 1.42 W , X eq,H = 2.312 - 1.42 2 = 1.82 W 20. 8 20.82

At full load (transformer supplying 50 kVA to the load at 240 V), Power factor = 0.8 Þ

IH =

Pload = Pout = 50 000 × 0.8 = 40 000 W

Resistive power loss on winding resistances: PR = R eq, H From open-circuit test, Pcore= 186 W

I H2 = 1.42 ´ 20.8 = 617W

5000 0 = 20. 8 A 24 00

T o t al lo s s e s ,

Ploss = PR + Pcore = 803 W

Total power supplied from high-voltage winding, Pin = Pout + Ploss=4 08 03 W Pout 40000 = ´ 100 % = 98 % Pin 40803

Efficiency

=

VoltaRe ge

gulat:ion

A t full load, Vˆ2¢ = 2400Ð0 V, �

IˆH = 20.8Ð - cos-1 0.8 = 20.8Ð - 36.87 A = 16.64 - j12.48 A �

Vˆ1 = Vˆ2¢+ (Req + jX eq ) IˆH = 240(12 +0 .4 + j1.82)(16.64 - j12.48) = 2446 + j13 V Regulation =

2446 - 2400 ´ 100% = 1.9 2% 24 00

TRANSFORMERS IN THREE-PHASE CIRCUITS

2.19 Common three-phastr e ansfmr o er connectio;ns the trna sformer windings are indicated by the heavy lines. ( N1 / N 2 = a ) Figure

Example 2.8

Three single-phase, 50-kVA 2400:240-V transof rmers, eaid ch entlica with thatof Exapm le 2.,6 are connected Y-D in a three-phase 150-kVA ba nk to step down the voltage at the load end of a feeder whose impedance is 0.15+j 10.0 Ω/phase. The voltage at the sending edn of the feedeisr 4160 V line-to-line. On their secondary sid,es the transformers supplya balancde trh ee-phase load through a feeder whose impedance is0. 0050 + j0.0020 D/phase. Find the line- to -line voltage at the load when the load draws rated current from the transformers at a power factor of 0 .80 l ag gi n g .

Threephase load

LV feeder HV feeder

Single-phase equivalent circuit: 0.15+ j 1.0 Ω

1.42 + j 1.82 Ω

0.15+ j 0.6 Ω

+

VH

Load

_

The voltage at the sending end of the feeder is Vs =

4160 = 2400 V line-to-neu tr al 3

The low-voltage feeder impedance referred to the high voltage side, 2

æ 4160 ö Zlv, H = ç ÷ ´ (0.0005 + j 0.0020) = 0.15 + j 0.60 W è 240 ø

Combined series impedance of ht e high- andlow- voltaeg feeders referred to the high-voltage side, Z eeder, H = 0.30 + j1.6 W/phase-Y

The equivalent single-phase series impedance of the transformer is equal to the single-phase series impedance of each single-phase transformer as referred to its high-voltage side Z eq, H = 1.42 + j1.82 W/phase-Y

Therefore, the single-phase equivalent circuit for this system is identical to that in Example 2.5.

Vload = 2329V

elin -to-neutral referred to the HV side. 240 ö Referred to the LV side: Vload = 2329 ´ æç V = 134 V line-to-neutral. ÷ è 4160 ø

\

Line-to-il ne Vload = 3 ´ 134 = 233 V

The per-unit system

Quantity in per-un it =

Actual quan tit y Base value of quan tti y

Pbase, Qbase , VAbase = Vbase I base

Rbase , X base , Z base = é VA

Vbase I base

ù

( P, Q ,VA) pu on base 2 = ( P, Q ,VA) pu on base 1 ´ ê base 1 ú ë VA base 2 û æ Vbase 1 ö æVA base 2 ö ÷ ç ÷ è Vbase 2 ø è VA base 1 ø

( R, X , Z ) pu onbaes 2 = ( R , X , Z ) pu onbaes 1 ´ ç

Example 2.12

The equivalent circuit for a 100-MVA, 7.97-kV:79.7-kV transformer issh own in Fig. 2.2 2a. The equivalent-circuit parameters are: X L = 0.040 W, X H = 3.75 W, X m = 114 W, RH = 0.085 W

RL = 0.76 mW,

Note that the magnetizing inductance has been referred to the lw o -voltage side of the equivalent circuit. Convert the equivalent circuit parameters to per unit using the transformre ra ting as base. Base quantities: LV side:

V A base= 100 MV A, V base= 7.97 kV

Rbase = X ba=se

Vb2ase =0 3 .6 5 W VAbase

HV side:

Þ

XL =

0.040 7.6 ´10 -4 114 = 0.063 p.u., RL = = 0.0012 p.u., X m = = 180 p.u. 0.635 0.635 0.635

V A b as e = 1 0 0 M V A , V b a se = 7 9 . 7 k V V2 3.75 0.085 Rbase = X base = base = 63.5 W Þ X H = = 0.0591 p.u., RH = = 0.0013 p.u. VAbase 6 3.5 63.5

Example 2.13

The exciting current measured on the low-voltage side ofa50- kV,A 2400:240-V transfomr er s i 5.41 A. Its equivaimn lent peda ce r eferred to the hgi h-voltage sideis 124 . +j 12.8 Ω. Using the transformer rating as the base, express in per unit on the low- and high-voltage sides (a) the exciting current and (b) the equivalent impedance. Base values: Vbase,H = 2400 V, Vbase,L = 240 V, I base,H = 20.8 A, I base,L = 208 A 240 0 240 Þ Z base,H = = 115.2 W, Z base,L = = 1.152 W 20.8 20 8 (a )

(b )

Per-unit value of exciting current referred to the LV side:

Ij , L =

Per-unit value of exciting current referred to theHV si:de

Ij , H =

Z eq , H

5.41

= 0.026 p.u.

208 0. 54 1 2 0 .8

= 0.026 p.u.

1.42 + j1.82 = = 0.0123 + j 0.0158 p.u. 115.2

The equivalent impedance referred to the LV side, æ ö Zeq, L = ç 10 ÷ ´ (1.42 + j1.82) = 0.0142 + j 0.0182 W è ø

Per-unit value

Z eq , L =

0.0142+ j0.0182 = 0.0123 + j0.0158 p.u. 1.152

3.7 Approximate Equivalent Circuits

The voltage drops

I1 R1 and I1 X 1 (Fig.3.11e) are normallys mall

then the shunt branch (composed o

c1 and

and

E1 @ V1. If this is true

X m ) can be moved to the supply terminal, as

shown in Fig.3.12a. This approximate equivalent circuit simplico fies

mtputa ion of currents,

because both the exciting branch impedance and the load branch impedanceare i d rectly connected across the supply voltage . Besides, the winding resistances and leakage reca tances can be lumped together. This equivalent circuit (Fig.3.12a) is frequently used to determine the performance characteristics of a practical transformer.

In a transformer, ethex citgn i current

Io

is a small percentage of the rated current of the

transformer (less than 5%). A further approximation of the equivalent circuit can be maed by removing the excitation branch, as shown in Fig.3.12b. The equivalent circuit referred to side 2 is also shown in Fig.3.12c.

Fig.3.12 Approximate equivalent circuits.

3.8 Transformer Rating

The kA V r atign andv oltaeg ratings of a tran sforem rarema rde k on sit nameplate. For exatmpa le,l

ypica transformmay er r ca ry the following information on the nameplate: 10 kVA,

1100/ 110 volts. What are hme e aninsg of these ratings? The voltage ratingin s diceat that the transformer has twwi o

ndin, gs one rated for 1100 volts and the other for 110 volts. These

voltages are proportional to their respective numbers of turns, and therefore the voltage ratio also represents the turns ratio (a = 1100/ 110 = 10). The 10 kVA rating means that each winding is designed for 10 kVA. Therefore the current rating for the high-voltage winding is 10,000/ 1100 = 9.09 A and for the lower-voltage winding is 10,000/110 = 90.9 A. It may be noted that when the rated current of 90.9 A flows through the lowvoltage winding, the rated current of 9.09 A will flow through the highvoltage winding. In an actual case, howeverthe , wi

ndgn i that is connected

to the supply (callde the primary winding) will carry an additional component of current (excitation current), which is very small compared to the rated current of the winding.

3.9 Determination Of Equivalent Circuit Parameters

The equivalent circuit model (Fig.3.12(a)) for the actual transformer can be used to predict the behavior of the transformer. The parameters

R1, X l1, Rc1, X m1, R2 , X l 2

and a = N1 / N 2

must be known so that the equivalent circuit model can be used. If the complete design data of a transformer are available, these parameters can be calculated from the dimensions and properties of the materials used. For example, the winding resistances

(R1, R2 )

can be calculatedfrom e th resistivity f o co ppre wires, the total length, and the

cross-sectional area of the winding. The magnetizing inductances

Lm can be calculated from the

number of turnsof eth winding and the reluctance of the magnetic path. The calculation oft h e leakage inductance

(Ll )

will involve accounting for partial flux linkages and is therefore

complicated. However, formulas are available frmo

which a reliable determination of these

quantities can be made. These parameters can be directly adn more easily determpiby ned

gerformin tests that involve

little power consumption. Tw o tests,ano- ao l d test (or open-ic rcuit test) and a short-circuit test, will provide information for determining the parameters of the equivalent circuit of a transformer. 3.9.1 No-Load Test (Or Open-Circuit Test)

This test is performed by applying a voltage to either the high-voltage side or low-voltage side, whichever is convenient. Thus, if a 1100/ 110 volt transformer wereto eb tested, the voltage

would be applied to the low-voltage winding, because a power supply fo 110 volts is more readily available than a supply of 1100 volts. A wiring diagram foropne circuit test of a transformer is shown in Fig.3.13a. Note that the secondary windgn i is kept open. Therefore, from the transformer equivalten circuit of Fig.3.12a the equivalent circuit under open-circuit conditions is as shown in Fig.3.12b. The primary current is the exciting current and the losses measureby d e th wattmeter are essentially teh co er losses. The equivalent circuiof tF

ig1.3. 3b shows that pthea ramt e ers Rc1 and Xm1 bcande

eteermin d

from the voltmeter, ammeter, and wattmeter readings. Note that the core losses will be the same whether 110 volts are applied to the low-voltage winding having the smaller number of turns or 1100 volts are applied to the high-voltage winding having the larger number fo turns. The core loss depends on the maximum value of flux in the core.

(a )

(b )

Fig.3.13 No-load (or open-circuit) test. (a) Wiring diagram for open-circuit test. (b) Equivalent circuit under open circuit 3.9.2 Short-Circuit Test.

This test is performed by short-circuiting one winding and applying rated current to the other winding, as shown in Fig.3.14a. In the equivalent circuit of Fig.3.12a for the transformer, the impedance of the excitation branch (shunt branch composed of Rc1 and X m1) is much larger than

that of the seriesbr anhc (composed of

e

1

and Re 1) . If the secondatry ermisnal arshe orted,

the high impedance of the shunt branch can be nelg ected. The equivalent circuit with the secondary shtor -circuited can thusbe reepr sented by the ci rcuit showin nF 2

1ig.3. 4b. Note that

is small, only a sm all supply voltae g is required to pass rated current

since Zeq1 = Req1 + X eq1 2

through the windings. It is convenient to perform this test by ap lying a voltage to the high-voltage winding. As can be seen from Fig.3.14b, the parameters Req1 and Xeq1 can be determined from the readingofs

voltrmete , ammeter, and wattmeter. In a well designed transformer, R1 = a 2 R2 = R2¢

and X l1 = a 2 X l 2 = X l ¢2 . Note that because the voltage applied under the short-circuit condition is small, the core losses are neglected and the wattmeter reading can be taken entireoly t represent the copper losses in the 2

windings, represented by I1 Req1 .

Fig.3.14 Short-circuit test. (a) Wiring diagramfor r sho t-circuit test. (b). Equivalent circuit at short-circuit condition.

The following example illustrates the computation of the parameters of the equivalent circuit of a transformer

Example 3.4

Tests are performed on a 1 f , 10 kVA, 2200/220 V, 60 Hz transformer and the

following results are obtained.

(a) Derive the parameters for the approximate equivalent circuits referred to the low-voltage side and the high-voltage side. (b) Express the excitation current as a percentage of the rated current. (c) Determine the power factor for the no-load and short-circuit tests. Solution:

Note that for the no-load test the supply voltaeg (full-rated voltageof22 0)V is applied to th e low-voltage winding, and for the shor t-circuit test the supply vlo tage is applied to the high-voltage winding with the low-voltage Equivale winding shorted. The ratings of the windings are a s follo:ws

V1(rated ) = 2200 V

V2(rated ) = 220 V 1 00 0 0 =4 5 .5 A 2200 10000 I 2( rated ) = = 4 5. 5 A 2 20 The equivalent circuit and the phasor diagram for the open-circuit test are shown in Fig.3.15a.

I1( rated=)

Power , Poc =

V22 c2

Then

Rc 2 =

I c2 = m2

2 20 2 = 4 84 W 100

22 0 = 0. 45 A 48 4

(I 22 - Ic22 ) = (2.52 - 0.452 ) = 2.46A

=

X m2 =

V2

=

m2

22 = 89.4 W 2.46

The corresponding parameters for the high-voltage side are obtained as follows: Turns ratio

a=

2200 = 10 22 0

Rc1 = a 2 Rc2 = 102 * 484 =48 400 W X m1 = a 2 X m2 = 102 *89.4 = 8940 W The equivalent circuit with the low-voltage winding shorted is shown in Fig.3.15b. Power Psc

Then,

= I12 Req1 2 15

Req1 =

= 1 0 .4 W

4.55 2 Z eq1 =

Vsc1 I sc 1

=

Then, X eq1 =

150 = 32.97 W 4.55 Ze2 1 - Re2 1 = 31.3 W

Fig.3.15

The corresponding parameters for the low-voltage saide are s f: ollows

Req1 = a2

Req=2

X=eq1 a2

X=eq2

10.4

= 0 0 .1 4 W 10 2 31.3

= 10 2

0.313 W

The approximate equivalent circuits referred to the low-voltage side and the high-voltage side are shown in Fig.3.15c. Note that the impedance of the shuntbrna ch is muchla reg r than that of the series branch. (b) From the no-load test the excitation current, with rated voltage applied to the low-voltage w i n d i n g , i s:

I o = 2 .5 A T hi s i s

2.5 *100% = 5.5% of the ratedcu rr ent of the winding 45.5

c power factor at no load =

P o w er volt ampere

=

100 = 0. 18 2 202 * 2.5

Power factor at short circuit condition =

21 5

150 * 4. 55

= 0 .3 1 5

Obtain the equivalent circuit of a 200/400-V, 50 Hz, 1 phase transformer

Example 3.5

from the following test a :-O.C. test : 200 V, 0.7 A, 70W-on LV side S.C. test : 15 V, 10 A, 85 W-on HV side

Calculate the secondary voltage when delivering 5 kW at 0.8 power factor lagging, the primary voltage be ing 20V0 . Solution:

From O.C. Test

Po = Vo I o * cos j o P \ cos j o = o = Vo I o

Then j o

70 = 0.5 200 * 0.7

= cos -1 0.5 = 60 o

Then I c1 = I o

cosj o = 0.7* 0.5 = 0.35 A

Im1 = I o sin j o = 0.7 *0.866 = 0.606A Then

Rc1 =

And

X m1 =

Vo1 I c1

Vo1 I m1

=

2 00 = 571.4 W 0.35

=

20 0 = 330 W 0.606

As shown in Fig.3.16, these values refer to primary i.e. low-voltage side From Short Circuit test:

It may be noted that inthsi test instruments have been placed inthese con dary i.e. hig hvoltage winding and the low-voltage winding i. e. primary has been short-circuited. Now,

Ze

2

=

V2 sc 15 = = 1.5W I 2 sc 10

2

æ1ö Z eq1 = a 2 * Z eq 2 = ç ÷ *1.5 = 0.375W è2ø Also, Psc

= I 22sc Req2 85 = 0.85 W 100

Then,

Req2 =

Then,

Req1 = a 2 * Req2 = æç 1 ö÷ * 0.85 = 0.21 W

2

Then,

è2ø

Xeq1 = Zeq2 1 - Req2 1 = 0.3752 - 0.212 = 0.31 W

Fig.3.16 Output KVA =

r e a l p o wer 5 = = . Power factor 0.8

Output current

I2 =

Now,

f ro m

5000 = 15.6 A 0 .8 * 4 0 0 aproximate

th e

o

o

equivalent

circuit

refeared

o

V Ð0 = V Ðd I Ð j * Z eq 2 o o o Then, V2 Ð 0 = 400Ðd - 15.6Ð- 36.87 * (0.85 + j1.2 ) 2

1¢

-

2

V2Ð 0o = 400Ðd o - 15.6Ð - 36.87o *1.5Ð 54.7o

to

secondery

:

V2 Ð0o = 400Ðd o - 23.4Ð18.17 o From the above equation we have two unknown variables

V2 and do

it need two

equations to get both of them. The above equation is a complexoneo s we can get two equations out of it. If we equate the real parts together and the equate the imaginary parts: So from the Imaginary parts:

V2 sin (0 ) = 400 sin (d o )- 23.4* sin (18.17 o ) 0 = 400* sin (d o )- 7.41o T h en , d

o

= 7.4o

So from the Real parts:

V2 cos (0 ) = 400 * cos( 7.41o )- 23.4* cos (18.17 o ) Then,

V2 = 374.5 V

Example 3.6

A 50 Hz, 1 - f transformer has a turns ratio of 6. The resistances are 0.9 W,

0.03 W and reactances are 5W

adn 0.13 W for high-voltage and low- vol tage, windings

respectively. Find (a) the voltage to be applied to the HV side to obtain full-load current of 200 A in the LV winding on short-circuit (b) the power factor on short-circuit.

Solution:

The turns ratio is

a=

N1

=6

2

Req1 = R1 + a 2 R2 = 0.9 + 62 * 0.03 = 1.98 W X e q 1 = X 1 + a 2 X 2 = 5 + 6 2 * 0. 13 = 9 . 6 8 W Zeq1 = Req2 1 + X eq21 = 1.982 + 9.682 = 9.88 W

1

=

2

=

a

2 00 = 33. 3 3 A 6

(a)

Vsc = I1 * Zeq1 = 9.88*33.33 = 329.3V

(b)

cos j =

Req1 1. 9 8 = = 0 .2 Zeq1 9.88

Example 3.7 A 1 phase, 10 kVA, ,500/250-V, 50 Hz transformer has the following constants:

Resistance: Primary 0.2 W ;e.S condary 0.5W Reactance: Primary 0.4W ; Secondary 0.1 W Resistance of equivalent exciting circuit referred to primary,

c1

=

1 50 0 W

Reactance of equivalent exciting circuit referred to primary,

X m1 = 750

W. What would be the readings of the instruments when the transformer is connected for the open-circuit and-short-circuit tests?

Solution: O.C. Test:

m1

V1

=

m c1

=

V1

=

c1

æ1 ö

Io

=

500 2 = 750 3

50 0 1 = A 150 0 3

2

2

æ 2ö + = èç 3÷ø èç 3 ÷ø = 0.745 A

No load primary input V1 * I c1 = 500*

1 = 16 7 W 3

Instruments usedin r p imary circuit are: voltmeter, ammeter and wattmeter, theire r adinsg being 500 V, 0745 A and 167 W respectively.

S.C. Test

Suppose S.C. test is performed by short-circuiting the LV, winding i.e. the secondaryos that all instruments are in primary.

Req1 = R1 + R2¢ = R1 + a 2 R2 = 0.2 +4 * 0.5= 2.2 W X eq1 = X 1 + X 2¢ = X 1 + a 2 X 2 = 0.4 +4* 0.1= 0.8 W Then, Ze

1

= Req2 1 + Xe2q1 = 2.22 + 0.82 = 2.341 W

Full-load primary current

=

1

R a t ed kV A 100 00 = = 20 Rated Pr imaryvoltage 500

Then Vsc

A

= I1 * Zeq1 = 20* 2.431 = 46.8V

Power absorbed

= I12 * Req1 = 202 * 2.2 = 880 W

Primary instruments will read: 468 V, 20 A, 880 W.

3.10 Efficiency

Equipment is desired to operate at a high efficiency. Fortunate l y, losses in transformers are small. Because the transformer is a static device, there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer the efficiency can be as high as 99%. The efficiency is defined as follows:

h=

output power ( Pout) *100 = Pout *100 Input Power (Pin) Pout + Losses

The losses in the transformer are the core loss

h=

Pout Pout = Pout + Losses Pout + Pc + Pcu

(3 . 1 4 )

(Pc) and copper loss (Pcu ) . Therefore, (3 . 1 5 )

The copper loss can be determined if the winding currents and their resistances are known:

Pcu = I 12R1 + I 22R2

(3 . 1 6 )

= I 12 Req1 = I 22Req2 The copper loss is a function of the load current.

The core loss dependson eh t peak flux density ni the core, which in turn depends on the voltage applied to the transformer. Since a transformer remains connected to an essentially constant voltage, the core loss is almost constant and can be obtained from theno- loda test of a transformer. Therefore , if the parameters of the equivalent circuitofa a tr nsformeare r knnow , the efficiency of the transformer under any operating condition may be determined. Now,

Pout = V2 I 2 cosj 2 Therefore,

h=

V2 I 2 cosj 2 V2 I 2 cos j 2 + Pc + I 22 Req2

h=

V2¢ * I 2¢ * cos j 2 * 1 00 V2¢ * I 2¢ * cos j 2 + Pc + I 2¢2 Req1

* 10 0

(3.17)

(3 . 1 8 )

3.11 Maximum Efficiency

For constant values of the terminal voltage V2 and load power factor angle j 2 , the maximu m efficiency occurs when:

dh =0 dI 2

(3.19)

If this condition is applied to Eqn. (3.17) the condition for maximum efficiency is:

Pc

=I

2 2 Req 2

(3 . 2 0 )

That is, corce loss = ospper los . For full load condition,

Pcu, FL = I 2,2FL Req 2

(3 . 2 1 )

Let

x=

I2

= per unit loadi ng

(3 . 2 2 )

2, FL

From Eqn s. (3.20), (3.21) and (3.22).

Pc = x 2 Pcu, FL Then,

(3.23)

æ P x = çç ö c ÷ è

(3.24)

ø

For constant values of the terminal voltage

V2 and load current I 2 , the maximum efficiency

occurs when:

dh =0 dj 2

(3.25)

Fig.3.17 Efficiency of a transformer.

If this condition is applied to Eq.(3.17), the condition for maximum efficiency is

j 2 = 0 Then, cosj 2 = 1 that is, load power factor = 1 Therefore, maximum efficiencyni a transformer occurs when the loapo d wr e factor is unity (i.e., resistive load) and load current is su ch that copper los s equalscoer loss. The variatioof n efficiency with load current and load power factor is shown in Fig.3.17. Example 3.8 For the transformer in Example 3.4, determine

(a)

Efficiency at 75% rated output and 0.6 PF.

(b) Power output at maximum efficiency and the value of maximum efficiency. At what percent of full load does this maxinmum efficie cy occur?

Solution:

(a )

Pout = V2 I 2 cos j 2 . = 0.75 *10 000 * 0.6 = 4500W

Pc = 100W , 2

Pcu

h=

= I Req1 2 = (0.75* 4.55) *10.4 =121 W 1

4500 *100 = 95.32% 4500 + 100 + 121

(b) At maximum efficiency

Pcore = Pcu and PF = cos j 2 = 1 Now, Pcore = 100W

= I 22Req 2 = Pcu 1/ 2

Then=, I 2

Po u t

h max

æ ö ÷= 31 è 0. 104 ø

A

= V2 I 2 cos j 2 = 220 *31*1 = 682 0 W

hmax =

Pout h max Pout h max + Pc + Pcu

=

6820 6820 + 100 + 100

= 9 7. 1 5 % output kVA=6.82 and Rated kVA=10 Then, h max occurs at 68.2% full load. Anther Method

From Example 3.4

Then

Pcu, FL = 215W

æ P X = çç c Pcu, è

Example 3.9

FL

÷ ø

÷

æ 100 ö ÷ = 0.68 è 215 ø

= ç

Obtain the equivalent circuit of a

transformer from the following testa:-

V8k A 200/400 V, 50 Hz, 1 phase

O.C. test : 200 V, 0.8 A, 80W,

S.C . test : 20 V,

20 A, 100 W Calculate the secondary voltage when delivering 6 kW at 0.7 power factor laggi ng, the primary voltabe ge

ing 20V0 .

From O.C. Test

Po = Vo I o * cos j o Po 80 \ cos j o = = 0.5 V o I o 2 0 0 * 0 .8 Then j o

= cos -1 0.5 = 60 o

Then I c1

= I o cos j o = 0.8* 0 .5 = 0 .4 A

I m1 = I o sin j o = 0.8* 0.866 = 0.69282A Vo1 200 Then, Rc1 = = 50 0 W = Ic1 0.4

And

X m1 =

Vo1 200 = 288.675 W = I c1 0.69282

From Short Circuit test:

It may be noted that in this test instruments have been placed in the secondary i..e high voltage winding and the low voltage winding i.e. primary has been short-circuited. Now,

Z eq 2 = VI 2 sc = 20 20 = 2 sc Also,

Psc = I 22sc Req 2

Then,

Req 2 =

Then,

1 00 = 0.25 W 20 2

X eq 2 = Z eq2 2 - Req2 2 = 12 - 0.25 2 = 0.968246 W

Output current

6000 I 2 = 0.7*400 = 21.4286 A

Now, from the aproximate equivalent circuit refeared to second:ery

V2 Ð0 o = V1¢Ðd o - I 2 Ð j o * Z eq 2 T h en ,

V2 Ð0o = 400Ðd

o

-

21.4286 Ð - 45.573o * (0.25 + j 0.968246)

V2 Ð0o = 400Ðd o - 21.43 Ð 29.9495o From the above equation we have two unknown variables

V2 and do

it need two

equations to get both of them. The above equation is a complexoneo s we can get two equations out of it. If we equate the real parts together and the equate the imaginary parts:

So from the Imaginary parts:

V2 sin (0 ) = 400 sin (d o ) - 21.43* sin (29.9495 o )

( )

0 = 400* sin d o - 10.6986 T h en , d

o

= 1.533o

So from the Real parts:

)

V2 cos (0 ) = 400 *cos(1.533 o - 21.43* cos(29.9495 o ) Then,

V2 = 381.288 V

Example:3.10 A 6kVA, 250/500 V, transformer gave the following test results

short-circuite De I.

20 V ; 12 A, 100 W and Open-circuit test : 250 V, 1 A, 80 W

termine the trm ansfor er equivalent circuit.

II. calculate applied voltage, voltage regulation and efficiency when the output is 10 A at 500

volt and 0.8 power factor lagging. III. Maximum efficiency, at what percent of full load does this maximum efficiency occur? (At

0.8 power factor lagging). IV. At what percent of full load does the effeciency is 95% at 0.8 power factor lagging.

Solution:

(I) From O.C. Test

Po = Vo I o * cos j o Po 80 \ cos j o = = =0. 2 3 Vo I o 205 *1.0 -1

o

Then j o = cos 0.32 = 71.3371 Then I c1 = I o cosj o = 1.0 * 0.32

= 0 .3 2 A

Im1 = I o sin j o = 1.0*0.7953 = 0.7953 A Then

Rc1 =

Vo1 250 = = 781. 25 W Ic1 0.32

And

X m1 =

Vo1 250 = 314.35 W = Im1 0.7953

As shown in Fig.3.16, these values refer to primary i.e. low-voltage side From Short Circuit test:

The rated current of the secondary side is:

I2 =

6000

= 12 A

5 00 It is clear that in this test instruments have been placed in the secondary i.e. highvoltage winding and the low-voltage winding i. e. primary has been short-circuited. Now,

Zeq 2 =

V2sc 20 = = 1.667W I 2 sc 12

Zeq1 = a Also, Psc

Then,

Then,

Then,

2

2

* Z eq2 = æ ö÷ *1.667 = 0.4167W è2ø

= I 22sc Req2

Req 2 =

10 0 = 0.694 W 12 2 2

Req1 = a * Req2 =

æ ö

÷

è2ø

2

* 0. 694 = 0 .174 W

X eq1 = Ze2q1 - Req21 = 0.41672 - 0.1742 = 0.3786 W

As shown in the following figure, these values refer to primary i.e. low-voltage side

j0.3786 0. 17 4 I0

V1 314.35

V2¢ 781.25

The parameters of series branch can be obtained directlyby moid fying the short circuit test data to be referred to the primary side as following: SC test 20 V ; 12 A, 100 W (refered to secondery) SC test 20*a=10 V ; 1 2/a=24A, 100 W (refered to Primary) S o,

Zeq1 =

V1sc 10 = = 0.4167W I1sc 24

Also,

Psc = I12sc Req1

Then,

Req1 =

Then,

1 00 = 0. 17 4 W 24 2

X eq1 = Ze2q1 - Req2 1 = 0.41672 - 0.1742 = 0.3786 W

It is clear the second method gives the same results easly.

(II) Output KVA = 10 *500*0.8

= 4 kVA

Now, from the aproximate equivalent circuit refeared to second:ery

V1 Ðd o = V2¢ Ð 0 o + I 2¢ Ð j o * Z eq1 T h en ,

V1 Ðd o = 250 Ð 0o + 20 Ð- 36.87o * (0.174 + j 0.3786) = 257.358 Ð0.89o

VR =

V1 - V2¢ 257.358 - 250 = *100 = 2.943% V2¢ 250

Pout = 10* 500* 0.8 = 4kW ,

Pi = Poc = 80W , and , Pcu = 102 * Req 2 = 100 * 0.694 = 69.4W

Pcu

æ ö = Psc * çç 2 ÷ è I 2 SC ø

2

or

2

÷

= 1ç00 *÷æ 10 ö = 69.4 W è 12 ø

Pout

4000

h = Pout + Pi + Pcu =4000 + 80 +

(III) maximum effeciency ocures when Pc

694. *100 = 964. %

= Pcu = 80W

th e

The percent of the full load at which maximum efficiency occurs is :

=

æ

P

c ÷ ç ç Pcu, FL ÷ è ø

=

80 = 0.8945% 10 0

Then, the maximum efficiency is :

6000 * 0.8945 *80. 6000 * 0.8945 * 0.8 + 80 + 80

=

=

(IV)

Pout = 0 .9 5 Pout + Pi + Pcu

h=

=

6000 *.08* x = 0. 95 6000 *.08* x + 80 + 100* x 2

T h en ,

95 x 2 - 240 x + 76 = 0

.

Then, Or

x = 2.155

(Unacceptable)

x = 0.3712

Then to get 95% efficiency at 0.8 power factor the transformer must work at 37.12% of full load.

3.12 All-Day (Or Energy) Efficiency, h ad

The transformer inapo wr e plant usually opear tes near its full capacity adn is taken out of circuit when it is nore t quire.d Such transformers are called power transformers, and they are usually designedform

axmimu

efficiencyo ccurrgn i ne r a the rated out put. A transformer

connected to the utili tyt hta supplies power toyo ur house andthe locail tysi called a distribution tra nsfo rmer . Such transformers are connected to the power s ystem for 24 hoursaday dn a operate

well below the ratedpo wre output for mo ts of theti m.e It is therefore desirable to design a distribution transformer for maximum efficiency occurring at the average output power. A figure of merit that will be more appropriate to represent the efficiency performance of a distribution transformer is the "all-day" or "energy" efficiencyof eh t transformer. This is defined as follows:

h ad =

h ad =

energy output over 24 hours * 100 energy i nput over 24 hours

(3.26)

energy output over 24 hours energy output over 24 hours + Losses over 24 hours

If the load cycle of the transformer is known, the all day effeciency can be deteremined.

Example 3.11 A

50 kVA, 2400/240 V transformer has a core loss P, = 200 W at rated voltage

and a copper loss Pcu = 500 W at full load. It has the following load cycle. % L oa d

0 .0 %

Power Factor H ou r s

6

50 %

75%

100 %

110%

1

0 .8 L a g

0.9Lag

1

6

6

3

3

Determine the all-day efficiency of the transformer.

Solution

Energy output 24 hours is 0.5*50*6+0.75*50*0.8*6+1*50*0.9*3+1.1*50*1*3=630 kWh Energy losses over 24 hours: Core loss =0.2*24=4.8 kWh Copper losses = 0.52 *0.5*6 + 0.752 *0.5*6 + 12 * 0.5 * 3 + 1.12 * 0.5*3 =5.76 kWh Total energy loss=4.8+5.76=10.56 kWh Then, h AD

=

63 0 * 1 0 0 = 9 8 .3 5 % 630 + 10.56

3.13 Regulation of a Transformer

(1) When a transformiser d loade with a constant primary voltage, then the secondat ry erminal voltage drops because of its internal resistance and leakagereac tanc e. Let. V2 o

= Secondary terminal voltage at no-load = E2 = E1 / a = V1 / a Because at no-lo ad the impedance drop is negligible.

V2 = Secondary terminal

voltage on full-load. The change ni secondaryte rminla voltage from no-load to full-lead is = V2o - V2 . This change dividedby

V20

is known as regulation down. if this change is dividedby

full-load secondary terminal voltage, then it is called regulation up.

% r eg =

Vno- load - Vload *100 Vload

- V2 ) V2 ) % reg = ( no - load ( load *100 (V2 )load

% re g =

V1¢- (V2 )load V1 -(V2¢) load *100 = *100 (V2 )load (V2¢) load

(3 . 2 7 )

(3 . 2 8 )

(3 . 2 9 )

V2

i .e .

As the transformer is loaded, the secondary te rminl a voltage falls (foar la gggn i power factor). Hence, to keepthe outptu voltage constant, the primary voltage must be increased. The risein primary voltage required to maintain rated output voltage fromno- loda to full-load at a given power factor expressed as percentage of rated primary voltage gives the regulatioof ne

th

transformer. Vectodir

agma r

for the vol tage dropin e th tran sformre for different loda power factor is

shown in Fig.3.18. It is clear that the onlyway ot get V1 less than V2¢ is when the power factor is leading which means the load has capacitive reac tanc e (i.e.the dor p on

Zeq1

will be negative,

which means the regulation may be negative).

V1 2¢

I 2¢ Zeq1

V2¢

X eq 1

I 2¢ Req1

I 2¢

(a)

V1

I 2¢ X eq1

I 2¢ Zeq1

V1 I 2¢

I 2¢ Req1

V2¢

I 2¢

2¢

(b)

eq1

I 2¢ Z eq1

I 2¢ Req1 V2¢

(c)

Fig.3.18 Vector diagram for transformer for different power factor (a)la ggign PF (b) Unity PF (c) Leading P F.

Example 3.12 A 250/500 V, transformer gave the following test results

Short-circuit test : with low-voltage winding shorted. short-circuited

20 V ; 12 A, 100W

Open-circuit test : 250 V, 1 A, 80 W on low-voltage side. Determine the circco uit

nsta, nts insert these on the equivalent circuit diagram and calculate

applied voltage, voltage regulation and efficiencyw hne the output is 5 A at 500 volt and 0.8 power fa ctor lagging.

Solution Open circuit test

cos j o =

Poc Voc Ioc

80 = 0. 32 250* 1

I c1 = I o c o s j o = 1 * 0 . 3 2 = 0 . 3 2 A

Im1 = I o2 - I c2 = 12 - 0.322 = 0.95A c1

=

m1

V1oc 250 = = 781. 3 W I c 0.32

=

V1oc Im

=

250 = 263.8 W 0.95

Short circuit test

As the primary is short-circuited, all values refer to secondary windgn i . So we can obtain and then refer them to primary to get Req1 and X eq1 as explained before

Req2 and X eq 2

in Example 3.5 or we can modify the short circuit data to the primary and then we can calculate

Req1 and X eq1 directly. Here will use the two method to compare the results. First method

Psc

Re q2 =

2 2 sc

1 00 = 0.694 W 122

Vsc 20 = = 1.667 W I 2 sc 12

Z eq 2 = Then,

=

X e 2 = Zeq2 2 - Req2 2 = 1.6672 - 0.6942 = 1.518 W

As

c

a nd X m

(RTeqh2e,n X eq2 , and

refer to

primary, hence

we

w ill

transfer th ese

values

Z eq 2 ) to primary with the help of transformation ratio.

Req1 = a 2 * Req2 = 0.52 * 0.694 = 0.174 W X eq1 = a 2 * X eq 2 = 0.52 * 1.518 = 0.38 W Z eq1 = a 2 * Z eq 2 = 0.52 * 1.667 = 0.417 W Second method

Short-circuited results refeard to secondery are20 ,V 12 A, 10W 0T

,hen Short-circuited results

refeard to primary are 10 V, 24 A, 100 W Then Req1 =

Zeq1 = Then,

Psc 100 = = 0.174 W I12sc 242

V1sc 10 = = 0.417 W I1sc 24

Xeq1 = Zeq2 1 - Req2 1 = 04172 - 0.1742 = 0.38 W

pplied voltage

V1 d o = V2¢ 0o + I 2¢ j o * Zeq1 Then, V1

d o = 250 0o + 10 - cos -1 0.8* (0.174 + j 0.38)

V1 d o = 250 0o + 10 - 36.24o * 0.418 65.4o

V1 d o = 250 0o + 4.18 29.16o

V1 d o = 250 0o + 3.65 + j 2.04 = 25 3. 65 + j 2.04 = 253.7 0.47 o

Voltage regulation

% re g =

(V1 ) - (V2¢) load (V2¢)load

100

(V2¢)load = 250 00 % reg =

253.7 - 250 * 1 0 0 = 1 .4 8 % 250

Effeciency

h=

V2¢ * I 2¢ * cos j *1 00 V2¢ * I 2¢ * cos j + Pcu + Piron

h=

250 *01 * 0.8 250 *10* 0.8 + 102 *0.174 + 80

Example 3.13

A 1f, 10 kVA, 2400/24V, 00

=

.

6 Hz distribution transformer has the following

characteristics: Core loss at full voltage =100 W na d Copper lossat hl a f load =60 W (a) Determine the efficiency of the transformer when it delivers full load at 0.8 power factor lagging. (b) Determine the rating at which the transformer efficiency is a maximum. Determine the efficiency if the load power factor is 0.9. (c) The transformer has the following load cycle: No load for 6 hours, 70% full load for 10 hours at 0.8 PF and 90% full load for 8 hours at 0.9 PF

Solution: ( a)

Pout = 10* 0.8 = 8 kW Pcore = 100 W , Pcu, FL == 60 * 2 2 = 240W h=

(b)

80 00 *100 = 95.92% 8000 + 100 + 240

x = 100 = 0.6455`` \ 240

h max =

10 *103 * 0.6455 *90. = 96.67% 104 * 0.6455 *90. ) + 100 + 100

(

Output energy in 24 hours is:

E24hrs = 0 + 10* 07. * 0.8 *10 +10 * 0.9 * 0.9 *8 = 120.8kWh Energy losses in the core in 24 hours is

Ecore = 100*2 4* 10 - 3 = 2.4 kWh Energy losses in the cupper in 24 hours is

Ecu = (204 * 0.72 *10 + 240 * 0.92 *8)*10 - 3 = 2.7312kWh Then, hall

day

=

1 2 0 .8 *100 = 95.93% 120.8 + 2.4 + 2.7312

3.14 Percentage Resistance, Reactance and Impedance

These quantities are usually measurdeby e th voltage drop at full-load current expressed as a percentage of the normal voltage of the winding on which calculations are made. (i) Percentage resistance at full load

I1*Req1 I12 Req1 *1 0 0 = * 10 0 V1 V1I1

%R =

I 22 Req2 = *100 = %Cu Loss at full load V2 I 2

(3 . 3 0 )

Percentage reactance at full load:

%X =

I1 * X eq1

*100 =

V1 %Z =

I 2 X eq2

*100

I 1Z e q 1 I 2 Z eq 2 * 100 = * 100 V1 V2

%Z = % R 2 + % X

(3 . 3 1 )

V2 (3 . 3 2 )

(3 . 3 3 )

3.15 Autotransformer

This is a special connectioofne

th transformer from which a variablAC e

voeltag can be

obtained at the secondary. A common winding as shown in Fig.3.19 is mounted on core and the secondarys i taken from a tpa on the windingIn . t con rast to the two-winding transformer discussed earlier, the primary and secondaryf o an autotransformer are physically connce ted. However, the basic principle of operation is the same as that of the two-winding transformer.

Fig.3.19 Step down autotransformer. Since all the turns link the same flux in the transformer core,

V1

N1

(3.34)

V2 = N 2 = a If the secondaryt appni g is replacedbya range 0

srlide , the output voltage can be va ride ov re t h e

< V2 < V1 .

The ampere-turns provided by the upper half (i.e., by turns between points a and b) are:

æ1

( N1 - N 2 ) * I 1 =

è

-

ö N I1 ÷ aø 1

(3.35)

The ampere-turns provided by the lower half (i.e., by turns between points b and c) are:

N 2 ( I 2 - I1 ) =

1

a

( I 2 - I1 )

(3.36)

æ

from amper tunr balance, from equations (3.35) and (3.36) ç1

è

-

1öN ÷ aø

(3 .37 ) T h en ,

1

I2

=

1 a

(3 . 3 8 )

I =

1

N1 ( I 2 - I1 ) a

Equations (3.34) and (3.37) indicath te , at viewed from the terminaof ls e th autotransformer, the voltages and currents are related by the same turns ratio as in a two-winding transformer. The advantages of an autotransformer connection are lower leakage reactances, lower losses, lowex er

cigtin current, increased kVA rating (see Example 3.11), and variable outpuvt lo tage

when a sliding contacist duse for sthee conad ry. The disadvantage is the direct connection between the primary and secondary sides.

A 1 f , 100 kVA, 2000/200 V two-winding transformer is connected as an autotransformer as shown in Fig.E2.6 such that more than 2000 V iob s tainde at the secondary. Example 3.14

The portion ab is the 200 V windin,g and thepot r ion be is the 20wV 00 kVA rating as an autotransformer.

Fig.3.20

Solution:

The craurrent tings of the windings are

.inding Compute the

Therefore, for full-load operation of the autotransformer, the terminal currents are:

A single-phase, 100 kVA, two-winding transformer when connected as an autotransformer can deliver 1100 kVA. Note that this higher rating of an autotrans former results from the conductive connection. Not all of the 1100 kVA is transformed by electromagnetic induction. Also note that the 200 V winding must have sufficient insulation to withstand a voltage of 2200 V to ground. Example 3.15

A single phase, 50 kVA, 2400/460 ,V 50 Hz transformer has an efficiency of

0.95% when it delivers 45kW at 0.9 power factor. This transformer is connected as an aut otransformer to supply load to a 2400 V circuit from 2860 V source. (a) Show the transformer connection. (b) Determine the maxi mum kVA the autotransformer can supply to 2400 V circuit.(c) Determine the efficiency of the autotransformer for full load at 0.9 power factor. Solution:

(a)

460

2860

2400

(b) I s ,2 w Then,

3

=

50 *10

= 108.7

A

2460

kVA)Auto = 108.782860 = 310.87 kW 3

(c) h 2 w

=

50*10 * 0.9 = 0. 95 50 *103 * 0.9 + Pi + Pcu, FL

Then, Pi

+ Pcu, FL = 2368.42 W

h Auto =

310870 * 0.9 = 99 . 61 % 310870* 0.9 + 2368.42

3.16 Three-Phase Transformers 3.16.1 Introduction

Power is distributed throughout The worlby dm sean of3-ph asetr anms ission lines. In order to transmit this power efficiently and economically, the voltages must be at appropriate lev.els These levels (13.8 kV to 100 kV) depend upon the amount of power that has to be transmitted and the distance it has to be earned. Another aspect is the appropriate voltage levels used in factories and homes. These are fairly uniform, ranging from 120/240 V single-phase s ystemsto 480 V3, phaes s ystems. Clearl y, this requires the use o3f phaes transformers to transform the

voltages from one level to another. The transformers may be inherently 3- pha,se having three primary windings and three secondary windinsg mounted ona3- lge ged core. However, the same result can be achieveby d gusin three single-phase transformers connected together to form a 3-phase transformer bank. 3.16.2 Basic Properties Of 3-Phase Transformer Banks

When three single-phase transformers areusde to transforma3-ph e as voltage, the windings can be connected in severawa l .ys Thus, the primaries may be connected in delta and the secondaries in wye, or vice versa. As a result, the ratio of the 3-phase input voltage to the 3-phase output voltage dependns t o only upo n the turns ratio of thtre ansfmr o ers, but also upon how they arec on ected. A3- phaes transformer bank can also produceap

heas shift between3the- phes a input voltage

and the3- pah se output voltage. The amount of phase shift depends again upon the turns ratio of the transformers, and on how the primaries and secondaries are interconnected. Furthermore, the phaseshift featuren e lab es us to change the number of phases. Thus3a, converted intoa2- aph se,a6-

pehas s ystem can be

pha, se or a 12-phase system. Indeed, if there were a practical

application for it, we could even conveart 3-

pheas s ystem intoa5-

pehas system by an

appropriate choice of single-phase transformers and interconnections. In making the varco ious

nnec,tions it is importato nt

oebserv transformer polarities. An error

in polarity may produce a short-circuit or unbalance the line voltages and currents. The basic behavior of ba lance3-d aph se transformer banks can be understoodbym following simplifying assumptions:

agkin the

1.The exciting currents arne e gliglib e. 2.The transformer impedances, due to the resistance and leakage reactanec of the windings, arene gligilb e. 3.The total apparent input power to the transformer bank is equal to the to tal apparent output power. Furthermore, when single-phase transformers are connected intoa3- aph se s ystem, they retai n all their basic single-phase properties, such ascu rretn ratio, voltage ratio, and flux in the core. Given the polarity marks

X 1, X 2 a n d H 1 , H 2 ,

the phase shift between primary adn

secondary is zero. 3.16.3 Delta-Delta Connection

The three single-phase transformers P, Q, and R of Fig.3.21 transform the voltage of the incoming transmission line A, B, C to a level appropriatefor e th outgoing transmission line 1, 2, 3. The incoming line is connected to the source, and the outgoing line is connected to the load.

The transformers are connected in delta-delta. Terminal terminal

2

H 1 of each transformer is connectetod

of the next transformer. Similarly, terminals

transformers are connected together. The actuph al

1 an d

2

of successive

yasic la l yotu of thetr anf s ormers is shown in

Fig.3 .21 . The corresponding schematic diagram is given in Fig.3.22. The schemdim atic

agra

is

drawnin h suc away ot show not only th e connections, btu also theph aos r relationship between the primary and secondary voalt ges. Thus, each secondary windgn i is drawn parallel to the corresponding primary winding to which it is coupled. Furthermore, if source G produces voltages

E AB , EBC , ECA according to the indicaph ted

raso diagram, the primary windings are

oriented the samwa e ,y phaseby aph se. For example, the primary of transformer P between lines A and B is oriented horizontally, in the same direction as phasor E AB .

Fig.3.21 Deltac-deltao

nnecntio oft h re single-phase transformers. The incomili ng

(source) are A, B, C and the outgoing lines (load)are 1, 23, .

Fig.3.22 Schematic diagram of a delta-delta connection and associated phasor diagram.

nes

Insu ch a delta-delta connection, the voltages between the respective incoming and outgoing transmission lines are in phase. If a balanced load is connected to lines1-2- 3, the resulting line currents are equal in magnitude. This produces balanced line currents in the incoming lines A-B-C.

As in any delt a connection, the linecu rrenst are 43 times greater than the respective

currents

S flowing in the primary and secondary windings (Fig.3.22). The power

P and

rating of the transformer bank is three times the rating of a single transformer. Note that although the transformer bank constit3uates e phas arrangement, each transformer, considered alone, actsas f i it were placed inasi nlg ephase circuit. Thus, a current I P flo win g

H 1 H 2 in the primary winding is associated with a current I S flowing from X 2 to X1

from

in the secondary. Example 3.16

Three single-phase transformers are connected in delta-delta to step down a

line voltage of 138kV ot 4160 V tso u- ppyl power to a manufacturing plant. The plant draws 21 MW at a lagging power factor of 86 percent. Calculate a. The apparent power drawn by the plant b. The apparenpo t wre furnished by the H V li n e

c.The current in theHVl isne d. The current in theLVli nse e. The currents in the primary and secondary windings of each transformerf. The load carried by each transformer Solution:

a. The appearent power drawn by the plant is:

S = P / cosj

= 21/0.86 = 24.4 MVA

b. The transformebank r f itsel abnsoarbs

because the

e egligibl amount of active rande actev i power

I 2 R losses and the reactive power associated with the mutual flux and thele

aak ge

fluxes are small. It follows that the apparent power furnished by the HV line is also 24.4 MVA. c.The current in each HV line is:1

=

S 2 4. 4 * 10 6 = = 1 02 A 3 * 138 00 3 * V1

d.The current in the LV lines is:-

2

=

S 24. 4 * 10 6 = = 3386 A 3 * 4 1 60 3 V2

e. Referring to Fig.3.19, the current in each primary winding is:

I =

102 3

= 58.9 A

The current in each secondary winding is:

IS =

3 386 3

f. Becausethe

= 1955 A ptlan load is balanced, each transformer carries one-ht ird of the total load,or

24.4 /3 = 8.13 MV A.

The individual transformer load can also be obtained by multiplying the primary volteag times the primary current:

S = E p I = 138000*58.9 = 8.13 MVA Note that we can calculate the line currents and thcu e rrenstin e thtr anfs ormer windings even though we do not know how3the- phes a load is connected. In effect, the plant load (shown as a box in Fig.3.22) is composed of hundrein dofs

dilvidua loads, some of whichare conn ected in

delta, others in wye. Furthermore, someare sinlg e-phase loads operatingat mhuc lower voltages than 4160 V, powered by smaller transformers located inside the plant. The sum total of these loads usually results in a reasonably well-balanced 3-phase load, represented by the box. 3.16.4 Delta-wye connection

When the transformers are connected in delta-wye, the three primary wi ndings are connected the sameway sa in Fig.3.21. However, the secondary windinsg are connected so that all the X 2 terminals are joined together, creating a common neutral N (Fig.3.23). In such a delta-wye connection, the voltage across each primary winding is equal to the incoming line voltage. However, the outgoing line voltage is 3 times the secondary voltage across each transformer. The relative values of the currents in the transformer windings and transmission lines are given in Fig.3.24. Thus, the line currents in phases A, B, and C are

3 times the currents in the

primary windings. The line currents in phases 1, 2, 3 are the same as eth currents in the secondary windings. A delta-wye connection produces a 30 phase shift between the line voltages of the incoming and outgoing transmissionli ne.s Thus, outgoing line voltage E12 is 30 degrees aheadof incomgin line voltage EAB, as canbes ee n from thph e asrodi arg am. If the outgoinlig e n feeds an isolated group of loads, the phase shift creates no problem. But, if the outgoing line has to be connected in parallel with a line coming from another source, the 03 degrees shiftma y make such a parallel connection impossible, even if the line voltages are otherwise identical.

One of the important advantages of the wye connection is that it reducesthe

amot un of

insulation needed inside the transformer. The HV winding has to be insulated for only 1/

3 , or

58 percent of the line voltage.

Fig.3.23 Delta-wye connection of three single-phase transformers.

Fig.3.24 Schematic diagram ofad ealt -wey connection and associatedph asro diagram(The . phasor diagrams on the primary and secondary sides are not drawn to the same scale.) Example3.17

Three single-phase step-up transformers rated at 90 MVA, 13.2 kV/80 kV are

connected in delta-wye on a 13.2kV transmissno i line (Fig.3.25). If they fee d a 90 MVA load, calculate the following:

a.The secondavry line oltag b.The currents in the transformer windings c.The incoming and outgoing transmission line currents

F i g .3 . 2 5 .

Solution

The easiest way to solve this problem is to consider the windings of onl y one transformer, say, transformer P. a. The voltage across the primary winding is obviously 13.2 kV The voltage across the secondary is, therefore, 80 kV. The voltage between the outgoing lines 1, 2, and 3 is:

V2 = 80* 3 = 13 9 kV b. The load carried by each transformer is

S = 90 / 3 = 30 MVA

3.16.5 Wye-delta connection

The currents and voltages in a wye-d elat connection are identical to those in the delta-wye connection. The primary and secondary conneict ons are simply interchange.d In other words, the

H2

terminals are connected togethr e to create a neutr,al and the

X 1, X 2

terminals are

connected in delta. Again, there results a 30 degrees phase shift between the voltages of the incoming and outgoing lines. 3.16.6 Wye-wye connection

When transformers are connected inwye-w ye,sp ecla i precautions have to be taken topr evne t severe distortion of the line-ot -neutral voltages. One way to per vent the distortion is to connect the neutral of the prima ry to the neutraof l ethso urec , usually by way of teh ground (Fig.3.26). Anotherways i to provide each transformer with a third winding, called tertiary winding. The tertiary windinsg of the tthrere ansm for ers aco re

nnecdte in delta (Fig.3.27). They ofte n provide

the substation service voltage where the transformlers are instal ed. Note that there i s no phase shift between thein comign and outgoing transmissionl i ne voltages of a wye-wye connected transformer.

Fig.3.26 Wye-w ye connection with neutral of the primary connectedto eth neutral oft h e source.

Fig.3.27 Wye-wye connection using a tertiary winding. Example 3.18

Three single phase, 30 kVA, 2400/240 V, 50 Hz transformers are connected to

form 3 j, 2400/41trs V6

an former bank. The equivalent impedance of each transformer referred

to the high voltage side is 1.5+j2 S2. The transformer delivers 60 kW at 0.75 power factor (leading). (a) Draw schematic diagram showing the transformer connection. (b) Determine the transformer wiWing current (c) Determine the primary voltage. (d) Determine the voltage regulation. Solution: (a )

(b)

kVA =

60 = 80kVA 0.75 3

80 *10

= 111.029

A

s

= 3 *416 24 0 0 a= = 24 0 111.029 I1 ph = = 1 1 .1 0 3 A 10

I1L = 11.103*=3

191.23 A

V2¢ = 2400Ð0o V , I 2¢ = 11.103Ð 41.41o A V1 = V2¢ + I 2¢ * (Z eq1 ) o

o

=12-4V002¢ Ð 0+11 .10 3 Ð 41.41 * (1.5 + j 2 ) = 2397.96 Ð0.66 V V * 10 0 V2¢ 2397.96 - 2400 = =- . 240 0

VR =

Problems:

1

A 1 0, 100 kVA, 1000/ 100 V transformer gave t he following test results: onpe -circuit

test 100 V, 6.0 A, 400 W short-circuit test 50 V, 100 A, 1800 W (a) Determine the rated voltage and rated current for the HV and LV sides. (b) Derive an approximate equivalent circuit referred to theHV sed i . (c) Determine the voltage regulation at full load, 0.6 PF leading. (d) Draw the phasor diagram cfor ondit(ion c). 2

A 1 ¢,25 kVA, 220/440 V, 60 Hz transformer gave the following test results.

Open circuit test

: 220 V, 9.5 A, 650 W Short-circuit test

: 37.5 V, 55 A, 950 W

(a) Derive the approximate equivalent circuit in per-unit values. (b) Determine the voltage regulation at full load, 0.8 PF lagging. (c) Draw the phasor dicagram for bondition ( ). 3

A1

parameters:

f

10 kVA, 2400/ 120 V, 60 Hz transformer has the following equivalent circuit Zeq1 = 5 + 5 j2 W, Rc1 = 64 kW and Xm1 = 9.6 kW Standard no-load and

short-circuit tests are performed on this transformer. Determine the following: No-load test results: Voc , I oc , 4-

Poc Short-circuit test results:

Vsc , Isc , Psc

A single-phase, 250 kVA, 11 kV/2.2 kV, 60 Hz transformer has the following

parameters. RHV= 1 .3 W XHV=4.5W, RLV = 0.05 W, XLV = 0.16, Rc2= 2.4 kW X m2 = 0.8 kW (a) Draw the approximate equivalecint trcui (i.e., magnetizing branch, with Rc1 and Xm connected to the supply terminals) referred to the HV side and show the parameter values. (b) Determine the no load current in amperes (HV side) as well as in per unit. (c) If the low-voltage winding terminals are shorted, determine

(d)

(i )

The supply voltage required to pass rated current through the shorted winding.

(i i )

The losses in the transformer.

The HV winding of the transformer is connected to the 11 kV supply and a lo da ,

Z L = 15Ð - 90 o W is connected to the low voltage wg indin

. Determine:

(i)

Load voltage. (ii) Voltre age gulnatio .

5

A 1-f , 10 kVA, 2400/240 V, 60 Hz distribution transformer has the following

characteristics: Core loss at full voltage = 100 W Copper loss at half load = 60 W (a)

Determinthe e

lagging. (b)

eeffici ncy fo the transf ormer when it delivers full load at 0.8 powe r factor

Determine the per unit rating at which the transformef er feici ncy si a ma xi mum.

Determine this efficiencyf i the load power factor is 0.9. The transformer has the following load cycle: No load for 6 hours 70% full loadfor 01 hours at 0.8 PF 90% full load for 8 hours at 0.9 PF Determine the all-day efficiency of the transformer. 6

The transformer of Problem 5 is to be used as an autotransformer (a)

connection that will result in maximum kVA rating. (b) high-voltage and low-voltage sides. (c)

S ho w

th e

Determin e the voltage ratings of the

Determine the kVA rating of the autotransformer.

Calculate for both high-voltage and low-voltage sides. 7

A 1 f , 10 kVA, 460/ 120 V, 60 Hz transformer has anef ficne i cy of 96% whne it

delivers 9 kW at 0.9 powerf actor. This transformer is connected as an autotransformer to supply load to a 460 V circuit from a 580 V source. (a)

Show the autot ransformer connection.

(b )

Determine the maximum kVA the autotransformer can supply to the 460 V circuit.

(c)

Determine the efficien cy of the autotransformer for full load at 0.9 power factor.

8

Reconnect the windings of a 1 f , 3 kVA, 240/120 V, 60 Hz transformer so thatit nca

supply a load at 330 V from a 110 V supply. (a) Show the connection. (b) Determine the maximum kVA the reconnected transformer can deliver. 9

Three 1¢, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a 3 f 460/208

V transformer bank. The equivalent impedance of eahc transformerre ef rred to the hhig -voltage sid e is 1.0 + j2.0 tra W. The

nsformer delivers 20 kW at 0.8 power factor (leading).

(a) Draw a schematic diagram showing the transformer connection. (b) Determine the transformer windicu ng

rr.ent (c) Determine the primary voltage . (d) Determine the voltage

regulation. 10

A 1 f 200 kVA, 2100/210 V, 60 Hz transformerh sa the following characteristics. The

impedancofe e

th high-voltage winding is 0.25 + j 51. W with the lowvoltage winding

short-circuited. The admittance (i.e., inverse fo impedance) of the lw o -voltage winding is 0.025 - j O.075 mhos with the high-voltage winding open-circuited. (a) Takin g thtre ansfmr o er ratgn i as base, determine teh base values ofpo we,r voltage, current, and impedance for both the high-voltage and low-voltage sides of the transformer. (b )

Determine the per-unit value of the equivalent resistance and leakage reactance of the

transformer. (c) (d )

Determine the pe r-unit value of the excitation current at rated voltage.

Determine the per-unit value of the total power loss in the transformer at full-load output

condition.

UNIT III

ELECTROMECHANICAL ENERGY CONVERSION

Introduction

Electromechanical energy conversions – use a magnetic field as the medium of energy conversion.Electromechanical energy conversion device: Converts electrical energy into mechanical energy

or

Converts mechanical energy into electrical energy Three categories of electromechanical energy conversion devices:  Transducers (for measurement and control)- s mall motion Transform the signals of different forms. Examples: microphones, sensors and speakers. 

Force producing devices (translational force)- limited mechanical motion. Produce forces mostly for linear motion drives, Example Actuators - relays, solenoids and electromagnets.



Continuous energy conversion equipment. Operate in rotating mode. Examples: motors and generators.

Energy Conversion Process The principle of conservation of energy:

Energy can neither be created nor destroyed. It can only be changed from one form to another. Therefore total energy in a system is constant . An electromechanical converter system has three essential parts:

① An electrical system (electric circuits such as windings) ② A magnetic system (magnetic field in the magnetic cores and air gaps) ③ A mechanical system (mechanically movable parts such as a rotor in an electrical machine).

The energy transfer equation is as follows: n rease in æ Ele c tric a l ö æM ec han ica lö æ Ic ö ç ÷ ç ÷ ç ÷ æ Energy ö energy in pu t stored energy i n energy ç ÷ =ç ÷ + ç ÷ + ç los ses ÷ è ø ç from sources÷ ç o u tp ut ÷ ç mag neti c f iel d ÷ ø è ø è è ø

The energy balance can therefore be written as:

æ Electricalenergy ö æ M echanicalenergy ö æ Increasein ö ç ÷ ç ÷ ç ÷ input f romsources ÷ = output + f rict ion + sto r e d field ç ç ÷ ç ÷ ç - resis tan celoss ÷ ç and windag eloss ÷ ç energy +coreloss ÷ è ø è ø è ø For the lossless magnetic energy storage system in differential form,

dWe = dWm + dW f dWe = i d l = differential change in electric energy input dWm = fm dx = differential change in mechoaunicaltenergy

tpu

dWf = differential change in magnetic stored energy

Energy in Magnetic System

Consider the electromechanical system below:

Axial length (perpendicular to page) = l

The mechanical force fm is defined as acting from the relay upon the external mechanical system and the differential mechanical energy output of the relay is

dWm = fm dx

Then, substitution dWe = id l, gives d W f = id l – f m d x

Value of Wf is uniquely specified by the values of l and x, since the magnetic energy storage system is lossless.

W = ò idl

dWf = differential change in magnetic stored energy E n erg y a n d C o en erg y

The l- i characteristics of an electromagnetic system depends on the air-gap length and B-H characteristics of the magnetic material. For a larger air-gap length the characteristic is essentially linear. The characteristic becomes non linear as the air-gap length decreases. l

l

l-i

Wf Increased air- gap l e n g th

Wf’

i

i

For a particular valrue of i -gap length, the field energy is represented by the red area between l axis and l-i characteristic. The blue area between i axis and l- i characteristic is known as the coenergy The coenergy is defined as

Wf = ò l

From the figure of l - i characteristic, Wf’ + Wf = l i

Note that Wf’ > Wf if the l - i characteristic is non linear and Wf’ = Wf if it is linear. The quantity of coenergy has no physical significance. However, it can be used to derive expressions for force (torque) developed in an electromagnetic system Determination of Force from Energy

The magnetic stored energy Wf is a state function, determined ubyniquely variables λ and x. This is shown expliby citly

thee independent stat

dWf (λ, x) = id l – f m dx

For any function of two independent variables F(x1,x2), the total diffre ential equation of F with respect to the two stva atei r ables x1 and x2 can bewr ittne

dF(x1, x2 ) =

¶F ( x1, x2 ) ¶F( x1, x2 ) dx1 + dx2 ¶x 2 x ¶x1 x 2

1

Therefore, for the total differential of Wf

dWf (l , x)

¶Wf (l , x) ¶l

dl +

¶Wf (l , )

dx l

And we know that

dWf (l , x ) = idl - f m dx By matching both equations, the current:

i=

¶Wf (l , x ) ¶l

x

where the partial derivative is taken while holding x constant and the me chanical force: m

=-

¶Wf (l , x ) ¶x l

Where the partial derivative is taken while holding l constant. Determination of Force from Energy: Linear System

For a linear magnetic system for which l=L(x)i: l

l

0

0

Wf (l , x) = ò i(l , x) dl =ò

l

L( x )

dl

1 2 L( x )

and the force, fm can be found directly:

m

=-

¶Wf (l , x ) ¶ æ 1 l2 ö ÷ l2 dL ( x ) = - çç ÷ = 2 L( x ) 2 dx x L ( x ) ¶x ¶ 2 l è øl

Determination of Torque from Energy

For a system with a rotating mechanical terminal, the mechanical terminal variables become the angular displacement θ and the torque T. Therefore, equation for the torque:

¶W (l , q ) T=- f ¶q

l

where the partial derivative is ta ken while holding l constant. Determination of Force from Coenergy The coenergy Wf’ is defined as

Wf' (i, x) = il -Wf (l , x) and the differential coenergy dWf’:

dWf' (i, x) = d (il ) - dWf (l, x) We know previously that

dWf (l , x ) = idl - f m dx By expanding d(iλ):

d (il ) = idl + ldi So, the differential coenergy dWf’:

dWf' (i, x) = d (il ) - dWf (l, x) = idl + ldi - (idl - f m dx) = ldi + f m dx By expanding dWf’(i,x): dWf' (i, x) =

¶Wf' (i, x) ¶W' (i, x) di + f dx ¶i x ¶x i

and, from the previous result:

dWf' (i, x) = ldi + fmdx

By matching both equations, l: l=

¶Wf' (i, x) ¶i

x

where the partial derivative is taken while holding x constant and the me chanical force: m

=

¶Wf' (i, x) ¶x i

where the partial derivative is tak en while holding i constant. For a linear magnetic system for which l=L(x)i: i i i Wf (i, x) = ò l (i, x)di =ò L( x)idi =L( x) 0

2

0

and the force, fm can be found directly: m

=

i ¶Wf (i, x) ¶ æ ç L( x) = ¶x i x çè

2

÷ö

÷ = øi

i 2 dL( x) x

For a system with a rotating mechanical terminal, the mechanical terminal variables become the angular displacement θ and the torque T. Therefore, equation for the torque: '

T = ¶Wf (i,q ) ¶q

i

where the partial derivative is tak en while holding l constant.

Determination of Force Using Energy or Coenergy?

The selection of energy or coenergy as the function to find the force is purely a matter of convenience. They both give the same result, but one or the other may be simpler analytically, depending on the desired result and characteristics of the system being analyzed. Direction of Force Developed 1. By using energy function:

m

¶W (l , = -x) f

l

The negative sign shows that the force acts in a direction to decrease the magnetic field stored energy at constant flux.

2. By using coenergy function:

m

=+

¶Wf' (i, x ) ¶x

i

The positive sign emphasizes that the force acts in a direction to increase the coenergy at constant current. 3. By using inductance function: m

=+

i 2 dL(x) 2 dx i

The positive sign emphasizes that the force acts in a direction to increase the inductance at constant current.

B-H Curve and Energy Density

In a magnetic circuit having a substantial air gap g, and high permeability of the iron core, nearly all the stored energy resides in the gap. Therefore, in most of the cases we just need to consider the energy stored in the gap. The magnetic stored energy, l

W f = ò idl 0 in w h ich

=

Hg an N B

Therefore, W = ò

0

dl = d ( Nf ) = d ( NAB) = NAdB B Hg 0 N NAdB = Ag ò H dB

However, Ag is volume of thae igr ap. Dividing both sides of the above equation by the volume Ag results in

wf =

W B = ò HdB 0 Ag B

Where i wf = ò0 HdB s energy per unit volume wf is known as energy density. The area between the B-H curve and B axis represents the energy density in the air gpa .

In the same manner,

H

w f = ò BdH is coenergy per unit volume. 0 The area between the B-H curve and H axis represents the coenergy density in the air gap.

For a linear magnetic circuit, B = mH or H = B/m, energy density:

w f = ò0 H dB = ò0

B m

dB =

B 2m

and coenergy density: H H mH 2 w 'f = ò B dH = ò m HdH = 2 0 0

In this case, it is obvious that w f = wf’. Doubly-excited Systems Rotating Machines 

Most of the energy converters, particularly the higher-power ones, produce rotational m o ti o n .



The essential part of a rotating electromagnetic system is shown in the figure.



The fixed part is called the stator,



The rotor is mounted on a shaft and is free to rotate between the poles of the stator



Let consider general case where both stator & rotor have windings carrying current ( is an d ir )

the moving part is called the rotor.



Assume general case, both stator and rotor have winding carrying currents (non-uniform air gap – silent pole rotor)



The system stored field energy, Wf can be evaluated by establishing the stator current is and rotor current ir and let system static, i.e. no mechanical output



Stator and rotor flux linkage l is expressed in terms of inductances L (which depends on position rotor angle q, L(q)

Stored field energy

T o rq u e

In linear system, coenergy = energy ’ W f = Wf 

First two terms represents reluctance torque; variation of self inductance (exist in both salient stator and rotor, or in either stator or rotor is salient)



The third term represents alignment torque; variation of mutual inductance.



Reluctance Torque – It is caused by the tendency of the induced pole to align with excited pole such that the minimum reluctance is produced. At least one or both of the winding must be excited.



Alignment Torque – It is causeby dta n ende cyof eth excited rotor to align with excited stator so as to maximize the mutual inductance. Both winding must be excited.



Forces and Torques

Lorentz force law F = q( E + v ´ B )

F: N,

q: Coulom bs,

E:

V/m, B : T or Wb/m2

large numbers of charged particles are in motion, Fv

Current density,

= r (E + v ´ B) J = rv

Fv:fo

rce denis ty (f orce per nu it volume), N/m 3

r : c ha rge de nsit(C y m/

3

)

A/m2

Right-hand rule for determining the directma io n

gi net c-fico e ld

mpotnenoft e h Lorentz force F = q(v × B).

Ex a m pl e 3. 1

A nonmagnetic rotor containsi inag l ng e-turn coil is placed in a uniformma gen tic field of magn itude B0, as showninFi. g 3.2. The coilsi des are at rad ius R and thewi er carries current I as indicated. Find the θdirected torque aas f ucn tion of rotor position α when I = 10 A, B0 = 0.02 T and R = 0.05m. Assuem that the rotor is of length l = 0.3 m.

Force per unit length of the wire, F = I × B For wire 1 carrying current I into the paper,

F1q = - I B0 l sina

For wire 2 carrying current I out of the paper,

F2q = - I B0 l sin a

Torque, T = F1q R + F2q R = -2 I B0 l R sina = -2(10)(0.02)(0.05)(0.3)sin a = -0.006sina N.m

magnetic-field electromechanical-energy-conversion device

simple force-producing device

The energy method, Pin(electrical) = Pstored (magnetci f iel d) + Pout (mechanical) ei = ´dt

dW fl d

dt

+

d

dt

Þ

losslesssy stm e

dl dt dWfld = i dl - f ld dx

( f ld x )

e=

E n erg y B a l a n ce

æ energy input ö æ mechanical ö æ increase in energy ö æ energy ç from electric ÷ ç ÷ ç converted ÷ = + energy ÷÷ +çç stored in ç ÷ ç ÷ ç ç sources ÷ ç ouu ÷ ç tp t ø è mag netic fiedl ø÷ èç into heat ø è è

ö ÷ ÷ ø

Conversion of energy into heat: ohmic heating due to current flow in windings + mechanical friction Lossless magnetic energy storage system: dWelec = dWmech + dWfld

dWelec : differe ntial electric energy in put dWmech : diffe rential mec ha nical energy output

dW fld : differential change in storedene rgy

Singly Excited Systems

An electromagnetic relay e i dt

dWele c =

e=

dl dt

dWmech =

f ld dx

Þ

dW fl d = i d l - f

Þ

ld

dWelec =

i dl

dx

Lossless magnetic energy storage system: conservative system: value of Wfld is uniquely specified by the values of λ and x (λ, x are called the state variables)

Þ

Wfld is uniquely specifiedby λ and x

Two different integration paths for Wfld

W fld =

ò

path 2a

dW fld +

ò

dW fld

path 2b

On 2a: d l = 0, and f ld = 0 since l = 0 On 2b: dx = 0

For a linear system (λ proportional to i):

l = L( x)i

Þ

dW fld = 0 ü ý þ

l0

Þ W fld (l0, x0 ) =

ò i ( l , x0 ) d l 0

l

l

0

0

Wfld ( l , x ) = ò i ( l ¢ , x ) d l =

¢l ¢

L( x )ò

dl

1 l 2 L( x )

¢

Example 3.2

The relay shown in Fig.3.6a is made from infinitely-permeable magnetic material with a movab le plunger, also of infinitely-permeable material. The height of the plunger is much greater than the air-gap length (h g) >> . Calcuelat the magnetic stored energy Wfld as a function of plunger position (0 < x < d) for N = 1000 turns, g = 2.0 mm, d = 0.15 m, l = 0.1 m, and i = 10 A. 1 Wfld = L( x )i 2 2

L( x ) =

Agap = l ( d - x ) = ld æç 1 - x ÷ö dø è

fld

=

m0 N 2Agap

Agap : gap cr oss-sectional are a

2g

2 L( x ) = m0 N ld (1 - x / d ) 2g

Þ

1 (10002)(4p ´10-7 )(0.1)(0.15) 1 2 2(0.00 2)

xö x = 236 çæ1 - ö÷ J d ÷ø dø è

ç è

Relay with movable plunger Magnetic Force and Torque from Energy

Wfld is a state function determined uniquely by the values of the independent state variables λ and x. dW fld = i d l - f ld dx

For any function of two independent variables F(x1, x2), dF ( x1, x2 ) = Þ

¶F ¶F dx1 + dx2 ¶x1 ¶x2

i = ¶W fld¶(ll , x)

f

ld

Þ

dW fld =

¶ W fld ¶l

dl +

¶ W fl d dx ¶x

= - ¶W fld (l , x )

¶x

For linear magnetic systems for which λ = L(x)i, f

Example 3.3

ld

=-

¶ æ1 l ö l dL ( x ) = ç ¶x è 2 L( x ) ÷ø 2 [ L( x ) ] 2 dx

l =

L( x )i

Þ

f

ld

1 dL ( x ) = i2 2 dx

Table 3.1 contains data from an experiment in which the inductance of a solenoid wa s measured as a function of position x, where x = 0 corresponds to the solenoid being fully retr act ed.

x (cm) L(mH)

0 2 .8

0.2 2.26

0 .4

1.78

0.6 1.52

Table 3.1 0 .8 1 .0

1.34

1.26

1. 2

1 2. 0

1.4 1.16

1 .6

1 .8

2 .0

1.13

1.11

1.10

Plot the solenoid force as a function of position for a current of 0.75 A over the range 0.2 < x < 1.8 cm. Using the MATLAB function polyfit, a fourth-order polynomial fitoft eh inductancas fea c un tion of x is obtained: L( x) = a (1)x 4 + a(2) x 3 + a(3)x 2 + a (4) x + a(5) 1 2 dL( x) 1 Þ f ld = i = i ëé4a (1)x 3 + 3a(2) x 2 + 2a (3)x + a (4) û ù dx 2 2 2

MATLAB script: cl le car c % He r e i s th e data : x i n c m , L i n m H xdata = [ 0 0. 2 0 . 4 0. 6 0 . 8 1. 0 1 . 2 1. 4 1. 6 1. 8 2.0] ; Ldata = [2. 8 2.2 6 1.7 8 1.5 2 1.3 4 1.2 6 1.2 0 1.1 6 1.1 3 1.1 1 1.10] ; %Convert t o S I unit s x = xda ta*1. e-2 ; L = Ldata*1. e-3 ; len = length( x ) ; x max = x(len ) ; % Us e polyfi t t o perfor m a 4't h orde r fi t o f L t o x . Store % th e polyno mia l coefficient s i n vecto ra . eTh fi t wil l b e % o f th e form : % % Lfi t : a(1)*x^ 4 + a(2)*x^ 3 + a(3)*x^ 2 + a(4)*a +x ) (5 ; % a = polyfit(x,L,4 ) ; % Let' s chec k th e fi t for n = 1:10 1 xfit(n) = x max*( n-1)/100 ; Lfit(n) = a(1)*xfit(n)^ 4 + a(2)*xfit(n)^ 3 + a(3)*xfit(n)^ 2 ... + a(4)*xfit(n ) + a(5) ; end % Plo t th e dat a an d the n th e fi t t o co mpar e (conver t xfi t t o c m a nd Lfi t t o mH ) figure(1) plot (xdata, Ldata, ' * ' )

hold plot (xfit*100 , Lfit*1000 ) xla bel('x [c m ] ') yla bel('L [ mH ] ')

% No w plo t th e forc e. Th e forc e wil l b e give n b y % % (1/2)i^ 2 dL/d x =(1/2 ) i^ 2 ( 4*a(1)*x^ 3 + 3*a(2)*x^ 2 + 2*a(3)* x + a(4) ) % %Se t curren t t o 0.7 5 A I = 0.75 ; for n = 1:10 1

xfit(n) = 0.00 2 + 0.016*( n-1)/100 ; F(n) =4*a(1)*xfit(n)^3+3*a(2)*xfit(n)^2+2*a(3)*xfit(n)+a(4) ; F(n) = (I^2/2)*F(n) ; end figure(2) plot (xfit*100,F ) xla bel('x [c m ] ') yla bel('Force [ N] ')

System with rotating mechanical terminal: x ® q : angular displacement, dWfld (l ,q ) = i d l - Tfld dq Þ Tfld = -

Linear magnetic systems: l = L(q )i

Þ

T fld = -

W fld =

f

ld

® T fld torque

¶Wfld (l , q ) ¶q

1 l 2 L(q

¶ æ1 l ö 1 l dL(q = ç 2 ¶q è 2 L(q ) ÷ø [ L(q ] )

l = L (q )i

Þ

T fld =

1 dL(q i 2 )

Example 3.4

The magnetic circuit of Fig. 3.9 consists of a single-coil stator and an oval rotor. Because the air-gap is nonuniform, the coil inductance varies with rotor angular position, measured between the magnetic axis of the stator coil and the major axis of the rotor, as

L(q ) = L0 + L2 cos(2q )

L0 = 10.6 mH L2 = 2.7 mH

Find the torque as a function of θ for a coil current of 2 A.

Solut ion :

T fld =

2

i2

d

= i 2 ( -2 L2 sin(2q ) ) = -4 ´ 2.7 ´ 10 -3 sin(2q ) = -1.08 ´ 10 -2 sin(2q ) N.m

Magnetic Force and Torque from Coenergy

The coenergy W ¢ld is defined as: W f¢ld (i , l ) = il - W fld dWf¢ld (i, l ) = d (il ) - dWfld (i , l ) d (il ) = id l + ldi dW fld = i dl - f ld dx Þ dW fl¢d = id l + ldi - (i dl - f ld dx ) = ldi + f fld dx

¶Wfl¢d ¶Wfl¢d di + dx ¶i ¶x ¢ ¶Wfld (i, x) ¶Wfl¢d (i, x ) l= f ld = ¶i ¶x

dWfld¢ (i, x ) =

The coenergy can be found from the integral of λ di i

W f¢ld (i , x ) =

ò

i¢, x ) di

0

For linear magnetic systems, l = L( x )i Wfl¢d (i , x ) =

1

L( x )i 2

f

Þ

ld

=

¶Wfl¢d (i, x ) ¶x

2 For a rotating electromechanical system, i

W f¢ld (i , q ) = ¢

=

1 2 dL( x ) i dx 2

i¢,q ) di

ò 0

¢

Magnetically linear system,

¶Wfld (i, q ) T fld = ¶q

Wf¢ld (i,q ) =

1

L(q )i 2

¶W fl¢d (i,q ) 1 2 dL(q = i ) ¶q 2

Tfld =

Example 3.5

For the relay of Example 3.2, find the force on the plunger as a function of x when the coil is driven by a controller which produces a current as a function of x of the form x

i( x ) = I æç ÷ èd ø

ö

A

0

From Example 3.2, 2

L( x ) =

( -

0

)

2g

1 dL( x) 1 æ m N 2l ö f ld = i 2 =- i ç 0 2 dx 2 è 2g ø

I 02m0 N 2l æ x ö ÷= 4 g çè d ÷ø

2

2

The coenergy for this system is W f¢ld ( i , x ) =

1 1 m N 2ld (1 - x / d ) m0 N 2ld (1 - x / d ) æ x ö L ( x )i 2 = i 2 0 = ç ÷ 2 2 2g 4g èd ø

The force cannot be found by taking the partial derivativotef s respect o

x, because in the expression for

differentiating with respect to x.

ld

hi expression for W fl¢d (i , x ) with

the current must be kept constantw hile

1 2 1 l / L = Li 2 . For a 2 2 opoartion tl, eh two functions are not

For a magnetically-linear system, the energy and coenergy are nume rilca ly equ al: nonlinear system in which x and i or B and H are not linepr a rl y equal.

Energy = ò i dλ Co-energy = ò λ di sum of the energy and coenergy, W f¢ld (i, l ) + W fld (i , l ) = il Example 3.6

The magnetic circuit shown inFi g. 3.12 is made of high-ep rmeability electrical steel. The rotor is free to turn about a vertical axis. The dimensions are shown in the figure. (a) Derive an expressino for thetor que acting on the rotor in et rms of thedi mesn ions and the magnetic fieldinte h two air gaps. As sume the reluctancof e e th steel to be negligible ( i.e. μ ® ¥) and neglect the effects of fringing. (b) The maximum flxu density in the overlapping portions of the air gap s is t o be limited to approximately51.6 T to avoid exces ive saturation of the steel. Compute the maximum torque for r1 = 2.5cm , h = 1.8 cm, and g = 3 mm.

H ag =

( a)

Ni 2g

l=

N f = NBag Aag = NAagm0 H ag = m0 NAag 2

Þ

L=

0

Aag = æç r1 + g ö÷ q . h 1

ag

2

T fl d =

2 ø

è

Ni = 2g

2 0

2g

L(q ) =

Þ

ag

i

1 m0 N 2 æ r1 + g ö÷ h.q 2 ø 2 g çè

1 2 dL(q ) m0 N 2 i 2 æ 1 i r1 + g ÷ö h = 2 dq 4 g çè 2 ø

(b)

Bmax = 1.65 T

Þ

H max =

Bmax m0

=

Ni 2g

-3

Þ

Ni = 2g Bmax = 2 ´ 3 ´ 10 ´-17 .65 = 7878.2 A-t ursn m0 4p ´ 10

Maximum torque, T fld =

1 4p ´10- ´ (7878.2) m0 N 2i 2 æ r1 + g ö÷ h = (0.025 + 0.0015)0.018 = 3.1 N-m 4 g çè 2 ø 4 ´ 0.003

Multiply-Excited Magnetic Field Systems

Differential energy dWfld (l1, l2 ,q ) = i1 d l1 + i2 d l2 - Tfld dq Þ Tfld = -

¶Wfld (l1, l2 , q ) ¶q

Current and torque in terms of energy i1 =

¶Wfld (l1 , l2 ,q ) ¶l1

i2 =

¶ Wfld (l1, l2 ,q )

T fld = -

¶l2

¶Wfld (l1, l2 ,q ) ¶q

To find the energy by integrating the differential energy: W fld ( l10 , l20 ,q0 ) =

l2 0

l1 0

0

0

ò i2 (l1 = 0, l2 ,q = q0 ).d l2 + ò i1(l1, l2 = l20 ,q = q0 ).d l1

Magnetically linear system: l1 = L11i1 + L12i2 l2 = L21i1 + L22i2

L12 = L21

Solving for i1 and i2 in terms of l1, l2

L12 ö -æ1 lö1 L22 ÷ø è ç lø ÷

where

D = L11L22 - L12 L21 l20

W fld ( l10 , l20 ,q0 ) =

.d l1

i1 =

æ i1 ö æ L11 çi ÷ = ç è 2 ø è L21

Þ

2

i2 =

l10

L11(q0 )

ò D(q0 ) l2 .d l2 + ò 0

1 D

( L22l1 - L12l2 )

1 ( - L21l1 + L11l2 ) D

L22 (q0 )l1 - L12 (q0 )l20 D (q0 )

0

= L (q ) l202 + L (q ) l102 - L (q ) l10l20 0 0 0 11 22 12 2 D(q0 ) 2 D (q0 ) D(q0 )

The coenergy is defined as, ¢

W fld (i1, i2 ,q ) = l1i1 + l2 i2 - W fld Þ dW fld (i1, i2 , q ) = d ( l1i1) + d ( l2i2 ) ¢

¢

(

i d l1 + i2 d l2 - T fld d q

)

= l1di1 + l2di2 + T fld d q1

¶Wfld (i1 , i2 , q )

dW fld (i1 , i2 ,q ) = ¶ Wfld (i1, i2 ,q ) ¶i1

¢

¶ Wfld (i1 , i2 , q )

¶Wfld (i1 , i2 , q )

¢

¶i1

1 2 ¶Wdifld+(i1 , i2 , q )¶i ¶ i2

¢

di2 +¶W

¶iq ¶q

f ld (i1 , 2 ,q )

dq

The coenergy can be found by integrating the differential, i 20

W f¢ld ( i10 , i20 , q0 ) =

i10

ò l2 (i1 = 0, i2 ,q = q0 ).di2 + ò l1(i1, i2 = i20 , q = q0 ).di1 0

0

For a linear system, Wf¢ld (i1, i2 ,q ) =

1 1 L11(q )i1 2 + L22(q )i22 + L12(q )i1i2 2 2

Torque, Tfld =

¶W fl¢d (i1, i2 ,q ) 1 2 dL11(q ) 1 2 dL22(q ) dL (q ) = i1 + i2 + i1i2 12 ¶q 2 dq 2 dq dq

Example 3.7

In the system shown in Fig. 3.15, the inductances in henrys are given as L11 = (3 + cos 2θ) × 10-3; L12 =0.3cosθ; L22 = 30 + 10cos2θ. Find and plot the torque Tfld(θ) for current il =0.8 A and i2 = 0.01 A.

dL11(q ) 1 2 dL22 (q ) dL12 (q ) 2 1 2 d q + 2i dq + i i dq 1 1 = i12 (-2 ´ 10 -3 sin 2q ) + i22( -20sin 2q ) + i1i2 (-0.3 sin q ) 2 2 = -(1.64 sin 2q + 2.4 sin q ) ´10 -3 N-m 1

2

T fld = 2 i1

Dynamic Equations

Model of a singly-excited electromecha nical system . KVLeqn . for the electrical system,

di dL( x ) dl l = L ( x )i Þ v 0 = R i + L ( x ) + i dt dt dt dL ( x ) dL ( x ) dx di dL( x ) dx = Þ v0 = Ri + L ( x ) + i dt d x dt dt dx dt

v0 = Ri +

di : self-inductance voltage term dt dL( x ) dx dx i : speed voltage ( : mechanicalp s eed) dx dt dt L( x )

Forces in the mechanical system in terms of mechanical position (and its derivatives): = - K ( x - x0 ) dx D = dt d 2x M = -M 2

Spring:

K

Damper : Acceleration of mass :

K : springco sn tant B: damping constant M : mass of moving part

dt Force equilibrium (f0: external mechanical excitation force): f

ld

Þ

= f0 - ( fK + fD + f M )

d2x dx -B - K ( x - x0) + f ld( x , i ) dt 2 dt

f 0 (t ) = - M

Note: If x > x0 and increasing (with positive second derivative: M acceleratgn i ), then the forces fK , fD , and fM all oppose ffld. The differential equations fsy or the :overall

v0 (t ) = Ri + L( x )

0(

stem

di dL( x ) dx +i dt dx dt

d 2x dx t ) = -M dt 2 - B dt - K ( x - x0) + f ld ( x , i )

Given the i npu ts(exc itat ions) v0 (t ) and f 0 (t ) these e qu ations ca n be solved to find x (t ) and i (t ). Example 3.10

Figure 3.24 shows in cross section a cylindrical solenoid magne tin hw ich the cylindricalpl uneg r of mass M moves vertically in brass guide ringsof thicknses g and meadi n aem ter d. The permeabilityofbrs as is the sa me as tha t of fer e space andis 0 = 4π × 10 -7 H/m in SI units. The plunger is supported by a spring whose spring constant is K. Its unstretched length is l0. A mechanical load force ft is applied to the plunger from the mechanical system connectedto ,it as shownin Fig. 3.24. Asseum that frictional force is linearly proportional to the velocity adn that the coefficienof t r f iction is B. The coil has N turns and resistance R. Its terminal voltage is vt and its current is i. The effects of magnetic leakage and reluctance of the steel are negligible. Derive the dynamic equations of motion of the electromechanical system, i.e., the differential equations expressing the dependent variables i and x in terms of vt, ft, and the given constants and dimensions.

The reluctance of the upper and lower gaps,

g

R1 =

R2 =

m0p xd

g m0p ad

Total reluctance,

R = R1 + R2 =

g

+

m0p

g m0p ad

=

g æ1 1ö ç + ÷ m0p è x a ø

Inductance,

L( x ) =

N2

=

m0 p adN 2 æ x ö æ ç ÷ = L ¢ç

x

÷

èa+ xø èa+xø R g The magnetic force acting upward on the plunger in the positive x direction is

f

ld

=

¶Wf¢ld (i, x ) 1 2 dL( x ) 1 2 aL¢ = i = i 2 ( a + x )2 ¶x 2 dx

The induced emf in the coil is

di dL di L d dx ( Li) = L + i = L + i dt dt dt d x dt ai dx x ö di = ¢çæ ÷ + (a + x )2 dt è a + x ø dt The dynamical equations of the system: e=

d

dt

d2x

1 ai 2 K ( x - l0) + L ¢ dt 2 dt 2 ( a + x )2 ai dx æ x ö di v(t ) = Ri + L ¢ç ÷ +L ( a + x ) 2 dt è a + x ø dt (t ) = -M

-

B

dx

-