Transfers electric power from one circuit It accomplishes this ...2018/08/09 · transformer has the following resistance and leakage-reactance values: rp = 0.8 ohm, xp = 3.2 ohms,
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Transfers electric power from one circuit
to another without a change of frequency.
It accomplishes this by electromagnetic
induction where the two electric circuits
are in mutual inductive influence of each
other.
The physical basis oftransformer between twocircuits linked by acommon magnetic flux.
It consists of twoinductive coils which areelectrically separatedbut magnetically linkedthrough a path of lowreluctance.
Simple Elements of
a Transformer
Two coils having
mutual inductance
Laminated steel
core
Two types of Transformer
Core-type
Shell-type
Spiral-core or wound-core type
Core-type (L-type)
The windings surround a considerable
part of the core
Does not heat up easily
Bulky
Shell-type (E-I core)
The core surrounds a considerable
portion of the winding
Heat up easily
Compact
Where: Eave = average induced emf in coil
N = number of turns in coil
t = time for flux to reach its maximum value
Φ = flux density in maxwells; weber- drop 10 ‾ ⁸
Volts 10NE 8
AVE
t
Volts 10
41
NE 8
AVE
f
Volts 10N4E 8
AVE
f
Since for a sine wave, the effective voltage E
is equal to 1.11 times the average voltage
Eave,
Volts 10N44.4
E 11.1E
8
ave
f
Primary Induced Voltages
Secondary Induced Voltages
Where: Np = number of primary turns
Ns = number of secondary turns
Volts 10N44.4E 8pp
f
Volts 10N44.4E 8ss
f
The 2300 volt primary winding of a 60 cycle
transformer has 4800 turns. Calculate (a) the
mutual flux, (b) the number of turns in the
230 volt secondary winding.
The maximum flux in the core of a 60 cycle
transformer that has 1230 primary turns and46 secondary turns is 3.76x10⁶ maxwells.
Calculate the primary and secondary induced
voltages.
A single phase transformer has 400 primary and
1000 secondary turns. The net cross-sectional
area of the core is 60 cm². If the primary
winding be connected to a 50 Hz supply at
520 V, calculate (a) the maximum value of
the flux density in the core in Wb/m² (b) the
voltage induced in the secondary winding.
The primary and secondary induced voltages
are related to each other by the ratio of the
number of primary and secondary turns.
S
p
S
p
N
N
E
E
Volts 10N44.4E 8pp
f
Volts 10N44.4E 8ss
f
The volts per turn in a 25 cycle 2400/230 volt
transformer is 8. Calculate (a) the primary
and secondary turns, (b) the maximum flux
in the core.
Since transformer is stationary, it is extremely
efficient because the only losses are those
that occur in the copper windings and those
iron losses.
sssppp PFIEPFIE
sspp IEIE
P
S
S
p
I
I
E
E
P
S
S
p
I
I
N
N
The primary and secondary currents of a
transformer were measured and found to be
3.8 and 152 A, respectively. If the secondary
load voltage is 116 V, what is the primary
voltage?
The turn ratio, induced-voltage ratio and the
current ratio is called the ratio of
transformation, and is represented by the
symbol .
P
S
S
p
S
p
I
I
N
N
E
Ea
a
Step –down transformer
When the primary voltage is reduced to a lower
secondary voltage
> 1
2300/230 volt transformer
Step-up transformer
When the primary voltage is raised.
< 1
13200/66000 volt transformer
a
a
s
s
p
p
Z
E
Z
E 22
ss
s
pp
p
jxr
E
jxr
E 22
2222s ps pp sp s EjxErEjxEr
Real to real Imaginary to imaginary
22s pp s ErEr
s
p
s
p
r
r
E
E2
2
s
p2
s
p
r
r
E
E
s
p2
r
ra
22s pp s EjxEjx
s
p
s
p
x
x
E
E2
2
s
p2
s
p
x
x
E
E
s
p2
x
xa
Referred to the primary side
0V ZIV seqppp az
0V ZIV seqppp az
loaddropapplied VVV
0Vxxrr θIV ss2
ps2
ppp aaja
0VjXR IV seqpeqppp a
Referred to the primary side
Referred to the secondary side
0V ZIV seqssp aa z
0V ZIV seqssp aa z
loaddropapplied VVV
0V
xx
rr θI
Vs
2
ps
2
pss
p
aj
aa
0VjXR IV
seqseqssp
a
Referred to the secondary side
0V ZIV
szeqssp
a
Unity power factor
0V Z0IV szeqppp a
0V90X 0I0R 0IV seqppeqppp a
2eqpp2
eqppsp XIRIVV a
Lagging power factor
0V ZIV szeqppp a
0V90X I0R IV seqppeqppp a
2eqpps2
eqppsp XIsinVRIcosVV aa
Leading power factor
0V ZIV szeqppp a
0V90X I0R IV seqppeqppp a
2eqpps2
eqppsp XIsinVRIcosVV aa
Where: VNL= VP
VF = Vrated
100V
V-VRegulation Voltage
FL
FLNL
A 25 kVA 2300/230 volt distribution
transformer has the following resistance and
leakage-reactance values: rp = 0.8 ohm, xp =
3.2 ohms, rs = 0.009 ohm, xs = 0.03 ohm.
Calculate the equivalent values of
resistance, reactance, and impedance in (a)
in primary side, (b) in secondary side.
Calculate the equivalent resistance and
reactance drops for a secondary load current
of 109 A: (a) in primary side, (b) in secondary
side.
Calculate the percent regulation: (a) for
unity power factor, (b) for a lagging power
factor of 0.8, (c) for a leading factor of
0.866.
A 10 kVA, 2400/240 V single-phase transformer
has the following resistances and reactances.
Find the primary voltage required to produce
240 V at the secondary terminals at full load
when the power factor is 0.8 lagging.
rp = 3 Ω
xp = 15 Ω
rs = 0.03 Ω
xs = 0.15 Ω
Given: 1-10 kVA transformer
2400/240 V
rp=3.66 Ω xp=3.66 Ω
rs=0.0282 Ω xs=0.0413 Ω
Find: (a) Reqp (g) V.R. at 0.8 pf lagging
(b) Xeqp (h) V.R. at unity pf
(c) Zeqp (i) V.R. at 0.8 pf leading
(d) Reqs
(e) Xeqs
(f) Zeqs
Core Loss It is the constant loss.
Where:
kh = proportionality constant that depends upon the volume of the core
ke = proportionality constant that depends upon the thickness of lamination, resistivity of steel
f = frequency
β = flux density available
226.1c
ehc
losscurrent eddy loss hysteresisloss core
P
PPP
PPP
fkfk eh
Hysteresis Loss
Purely magnetic
Results due to the tiny magnetic particles that
produces a kind of molecular friction as they
tend to change alignment with the rapid
reversals of alternating current
Eddy-current loss
Electromagnetic in character
Caused by the flow of currents in the iron in
exactly the same way as in the transformer
windings.
6.1hP fkh
22eP fke
From
Therefore,
10N44.4E 8 f
f
Ek
f
E
NA
f
44.4
10
N44.4
10EA
8
8
Substituting the value of the flux density:
22
6.0
6.1
1
2
2
6.1
226.1cP
Ekf
Ek
f
Ekfk
f
Efkk
fkfk
eh
eh
A 110 V, 30 cps distribution transformer was
tested at its core loss was found to be 480 W.
of the 480 W, 360 W was due to hysteresis
loss. If the applied voltage and frequency are
doubled, what will be its core loss assuming
that there has no considerable changes in
flux.
Voltage ratings must be known.
1. From the circuit shown, the variac should
be set to its minimum output value
2. Adjust the output of the variac until the
reading of the voltmeter is equal to the
rated voltage of the low side.
3. Take the readings of the meters.
_________ I
_________ V
_________ P
oc
oc
oc
Core loss of the transformer
Copper Loss
Variable Loss
eqs2
s
eqp2
p
s2
sp2
pcu
RI
RI
rIrIP
kVA ratings and voltage ratings must be
known.
1. From the circuit shown, the variac should
be set to its minimum output value
2. Adjust the output of the variac until the
reading of the ammeter is equal to the
rated current of the high side.
3. Take the readings of the meters.
_________ I
_________ V
_________ P
sc
sc
sc
Copper loss of the transformer
From the short-circuit test,
22eqpeqpeqp
sc
sceqp
2sc
sceqp
RZX
I
VZ
I
PR
100PPP
P
100PP
P
100P
PE
cucout
output
losses totaloutput
output
input
output
Given: 1-10 kVA transformer
2400/240 V
Reqp = 6.48 ohms
Xeqp = 9.37 ohms
Pc = 120 W
Find:
a. E at H.L. w/ 0.8 pf
b. E at F.L. w/ 0.8 pf
c. E w/ 25% O.L. w/ 0.8 pf
100RIPkI
kIE
cosE k let
100RIPcosIE
cosIEE
100P
PE
eqp2
pcp
p
pp
eqp2
pcppp
ppp
input
output
cuc
eqp2
pc
eqp2
pc
eqp2
pp2
eqp2
pcp2
2eqp
2pcp
eqpppeqp2
pcp
p
PP
RIP
RkIkP
0RkI2IkRkIkPIk
0)RIP(kI
)RI20)(k(kI-)(k)RIP(kI
dI
dE
Maximum efficiency occurs when Pc = Pcu
Since, to have a maximum efficiency, Pc =
Pcu,
100P2P
P
100P
PE
cmaxout
maxout
input
maxoutput
eqs
cs
ceqs2
s
ccu
R
PI
PRI
PP
Multiplying Vs/1000 at both sides,
cu
cRmax
eqsFL
cFLsss
eqs
c
FL
FLsss
P
PkVAkVA
RI
P
1000
IV
1000
IV
R
P
I
I
1000
V
1000
IV
2
Given: 50 kVA, 4600/230 V
O.C.T.: Voc = 230 V, Poc = 285 W
S.C.T.: Vsc = 150 V, Psc = 615 W, Isc = 10.87 A
Find:
a. kVAmax at 0.84 pf
b. Maximum efficiency
Percent Impedance
Percent Reactance
Percent Resistance
100V
ZI100
V
ZI%Z
s
eqss
p
eqpp
100V
XI100
V
XI%X
s
eqss
p
eqpp
100V
RI100
V
RI%R
s
eqss
p
eqpp
Where:
Ip = rated primary current
Vp = rated primary voltage
Is = rated secondary current
Vs = rated secondary voltage
%Z = percent impedance
Zeqp = equivalent impedance referred to the primary side
Zeqs = equivalent impedance referred to the secondary side
%X = percent reactance
Xeqp = equivalent reactance referred to the primary side
Xeqs = equivalent impedance referred to the secondary side
%R = percent resistance
Reqp = equivalent resistance referred to the primary side
Reqs = equivalent resistance referred to the secondary side
A 500 kVA, 13200/2400 V, 60 Hz single-phasetransformer has 4% reactance and 1%resistance. The leakage reactance andresistance of the low voltage winding are0.25 ohm and 0.055 ohm respectively. Thecore loss is 2000 W.
a. Find the leakage reactance and resistanceof the high voltage winding.
b. Calculate the efficiency of this transformerat full load and with 80% power factor.
c. What will be the maximum efficiency withthe same power factor?
It is the ratio of the energy delivered
by the transformer in a 24-hr period
to the energy input in the same
period of time.
)tP(P outhrs 24in output
100
100PPP
PA.D.E.
hrs 24in cu hrs 24in chrs 24in out
hrs 24in output
hrW
hrW
rating)(t)VA )(load of size(P hrs 24in output
)(24)P(P chrs 24in c
)(t)P()load of size(P FLcu2
hrs 24in cu
A 5 kVA, 2300/230, 60 cycle standard
distribution transformer was tested with the
following results: short-circuit test input =
113 W, open-cicuit test input = 40 W. If this
transformer operates with the following
loads: 1 ½ times rated kVA, pf = 0.8, 1 hr; 1
¼ times rated kVA, pf = 0.8, 2 hrs; rated kVA,
pf = 0.9, 3 hrs; ½ times rated kVA, pf = 1, 6
hrs; ¼ times rated kVA, pf = 1, 8 hrs; no
load, 4 hrs. Calculate its all-day efficiency.
It is generally used in low power applications
where a variable voltage is required. The
autotransformer is a special type of power
transformer. It consists of only one winding.
By tapping or connecting at certain points
along the winding, different voltages can be
obtained.
Step-down Autotransformer Step-up Autotransformer
2
1
2
1
N
N
V
Va
a
1
N
N
I
I
1
2
1
2
11input IVP
a
11PP inputdtransforme
dtransformeinputconducted P-PP
Power
transformed
Power
conducted
An autotransformer of given physical
dimensions can handle much more load
power than an equivalent two-winding
transformer.
An autotransformer is to be used to supply a
load of 80 A at 160 V from a 220 V line. If the
transformer is ideal, determine (a) the turns
ratio between the series and the common
windings, (b) current of the series winding,
(c) power conducted to the load, (d) kVA
rating as an ordinary two winding
transformer, (e) kVA rating as an
autotransformer.
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