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CHAPTER 5.0 - TRANSFORMERSCHAPTER 5.0 - TRANSFORMERS
SUBSTATION TRANSFORMER
DISCUSSION TOPICSDISCUSSION TOPICS
SUBSTATION TRANSFORMER
Explain the principles of transformer 1. Explain the operating principles of
transformer by applying the electromagnetic induction concept
2. Describe related formula by the relationship of: a) Primary / secondary windings b) Primary / secondary currents c) Primary / secondary voltages
3. Name various types of transformers and their applications
4. Explain stepped-down and stepped-up transformers
5. Solve problems related to transformers
SUBSTATION TRANSFORMER
Printed circuit board mounted audio
impedance matching transformer, top
right
Audio frequency impedance matching transformer
Photograph of multiple-winding transformer with six windings, a primary and five secondaries
Gas-discharge lighting transformer
SUBSTATION TRANSFORMER
The second generation of power distribution systems (what we are still using) was proposed by Tesla few years later. His idea was to generate AC power of any convenient voltage, step up the voltage for transmission (higher voltage implies lower current and, thus, lower losses), transmit AC power with small losses, and finally step down its voltage for consumption. Since power loss is proportional to the square of the current transmitted, raising the voltage, say, by the factor of 10 would decrease the current by the same factor (to deliver the same amount of energy) and, therefore, reduce losses by factor of 100.
Historically, the first electrical power distribution system developed by Edison in 1880s was transmitting DC. It was designed for low voltages (safety and difficulties in voltage conversion); therefore, high currents were needed to be generated and transmitted to deliver necessary power. This system suffered significant energy losses!
The step up and step down voltage conversion was based on the use of transformers.
Gaulard and Gibbs first exhibited a device with an open iron core called a 'secondary generator' in London in 1882 and then sold the idea to a company Westinghouse. They also exhibited their invention in Turin in 1884, where it was adopted for an electric lighting system.
In 1885, William Stanley, an engineer for Westinghouse, built the first commercial transformer after George Westinghouse had bought Gaulard and Gibbs' patents. The core was made from interlocking E-shaped iron plates. This design was first used commercially in 1886.
Hungarian engineers Zipernowsky, Bláthy and Déri created the efficient "ZBD" closed-core model in 1885 based on the design by Gaulard and Gibbs. Their patent application made the first use of the word "transformer".
Another Russian engineer Dolivo-Dobrovolsky developed the first three-phase transformer in 1889.
Finally, in 1891 Nikola Tesla invented the Tesla coil, an air-cored, dual-tuned resonant transformer for generating very high voltages at high frequency.
a)A transformer is a device that converts one AC voltage to another AC voltage at the same frequency.
b)It consists of one or more coil(s) of wire wrapped around a common ferromagnetic core. These coils are usually not connected electrically together. However, they are connected through the common magnetic flux confined to the core.
c) It has at least two windings, one of them (primary) is connected to a source of AC power; the other (secondary) is connected to the loads.
transformer construction
The invention of a transformer can be attributed to Faraday, who in 1831 used its principle to demonstrate electromagnetic induction foreseen no practical applications of his demonstration.
Russian engineer Yablochkov in 1876 invented alighting system based on a set of induction coils, which acted as a transformer.
Schematic symbol for transformer , separated by lines indicating a ferromagnetic core
Basic transformer operation can be described by two formulae relating the transformation ratio to the turns ratio of the transformer windings.
When the secondary is an open-circuit and an alternating voltage VP is applied to a primary winding, a small current flows which set up a magnetic flux in the core.
This alternating flux links with both primary and secondary coils and induces in them e.m.f.’s of EP and ES. The induced emf, E in a coil of N turns is given by ;
In the ideal transformer, the rate of change of flux is the same for both primary and secondary thus ;
The induced emf per turns is constant. Assuming no losses, EP = EP and ES = VS
fluxofchangeofratethedt
dwherevolts
dt
dNV
;
S
S
P
P
N
V
N
V
S
P
S
P
S
S
P
P
N
N
V
Vor
N
V
N
V
windingondary
theinturnsofnumbertheN
windingprimary
theinturnsofnumbertheN
currentondarytheI
voltageondarytheV
currentprimarytheI
voltageprimarytheV
S
P
S
S
p
P
sec
sec
sec
When the load is connected across the secondary winding, a current I2 flows. In an ideal transformer losses are neglected and a transformer is considered to be 100% efficient.
Hence input powers = output power or VP IP = VS IS ,
the primary and secondary volt-amperes are equal. Thus
With expected transformet ratio as, K ;
Notes ;If K < 1 : Ns < Np : this transformer is called step-down transformerIf K > 1 : Ns > Np : this transformer is called step-up transformerIf K = 1 : Ns = Np : this transformer is called coupling transformer
P
S
S
PSSPP I
I
V
VIVIV
P
S
S
P
S
P
I
I
N
N
V
VK
Example 1;
A transformer is to be used to provide a 60 V output from a 240 V A.C supply. Calculate (a) the turns of ratio required and (b) the number of primary turns, if the secondary is wound with 500 turns. Solution 1; VS= 60 V, VP = 240 V, NS = 500 a) The turns of ratio required ; b) The number of primary turns ;
1:41
4
60
240or
V
VK
S
P
200045005001
4 P
P
S
P
S
P NN
N
N
V
V
Example 2;
The number of windings for the three transformers are ;
a) Np = 100, Ns = 2000b) Np = 3000, Ns = 2000c) Np = 100, Ns =100
Calculate the value of K for each transformer then determine the type of transformer and draw the symbol of transformer to differentiate the number of windings.
Solution 2;
a)
b)
c)
rtransformeupstepKN
NK
P
S 1;20100
2000
rtransformedownstepKN
NK
P
S 1;67.03000
2000
Np
(100)
Ns(2000)
Ns(100)
Np
(100)
Np
(3000)
Ns(2000)
rtransformecouplingKN
NK
P
S 1;00.1100
100
Example 3;
A 2000/200V, 20kVA transformer has 66 turns in the secondary. Calculate i)primary turnsii)primary and secondary full-load currentsNeglect the losses.
Solution 3 ;
i)Primary turns ;
ii)Primary current ;
Sekunder current ;
turnsN
N
N
N
V
VP
P
S
P
S
P 660200
200066
66200
2000
AIIVIV PSSPP 102000
10201020
33
AIIVIV SSSPP 100200
10201020
33
An ideal transformer, i.e. a transformer with no power losses, in which, Primary volt amperes = Secondary volt amperes.
While practical transformers can be extremely efficient, some losses will occur because not all of the magnetic flux produced by the primary winding will link with the secondary winding. The power losses that occur in a transformer are of three types :
1. Copper Losses
These losses can also be called winding losses or I2R losses, because they can occur in windings made from metals other than copper. The losses become evident as heat, generated in the (copper) wire windings as they dissipate power due to the resistance of the wire.
The power loss in a transformer winding can be calculated by using the current in the winding and its resistance, in formula for power, P = I2R. This formula is the reason copper losses are sometimes called I2R losses. To minimise the losses the resistance of the winding must be kept low, using wire of suitable cross sectional area and low resistivity.
2. Eddy Current Losses
The iron or steel core is an electrical conductor as well as a magnetic circuit, the changing current in the primary will tend to set up an EMF within the core as well as in the secondary winding. The currents induced into the core will oppose the changes of magnetic field taking place in the core.
For this reason these eddy currents must be kept as small as possible. This is achieved by dividing the metal core into thin sheets or "laminations", each one insulated from the others by an insulating coat of lacquer or oxide. Laminated cores greatly reduce the formation of eddy currents without affecting the magnetic properties of the core.
In high frequency transformers eddy current losses are reduced by using a core made of a ceramic material containing a large proportion of tiny metal particles, iron dust or manganese zinc. The ceramic insulates the metal particles from each other, giving a similar effect to laminations, and performing better at high frequencies. Due to the ways of reducing losses described above, practical transformers closely approach the ideal in performance.
In large power transformers, efficiencies of about 98% can be achieved. Therefore for most practical calculations, it can be assumed that a transformer is "Ideal" unless its losses are specified.
3. Hysteresis Losses
Each time the alternating current reverses (once each cycle), tiny "magnetic domains" within the core material are reversed. These are physical changes within the core material and take up some energy. The amount of energy used depends on the "reluctance" of the core material; in large cores of power transformers where hysteresis loss maybe a problem it is largely overcome by using special low reluctance "grain oriented" steel as the core material.
Off Load CurrentThe action of a transformer is nearly perfect, the power in both
primary and secondary windings is the same, so when no load is put on the secondary, no secondary current flows and the power in the secondary is zero (VxI=0). Therefore, although a voltage is applied to the primary no current will flow, as the power in the primary must also be zero (very low).
Volts per TurnA transformer with a primary winding of 1000 turns and a secondary
winding of 100 turns has a turns ratio of 1000:100 or 10:1. Therefore 100 volts applied to the primary will produce a secondary voltage of 10 volts.
Since the maximum value of the flux in a normal transformer does not vary by more than about 2 per cent between no load and full load, it is usual to assume the core loss constant at all loads.
Hence, if Pc = total core loss, total losses in transformer arePc + I2
1R1 I22R2
And
Greater accuracy is possible by expressing the efficiency thus :
SSPPcPP
SS
RIRIPfpVI
fpVIEfficiency
lossespowerinput
poweroutput
rinput powe
eroutput powEfficiency
22..
..
powerinput
lossesηEfficiency
powerinput
lossespoweroutput
rinput powe
eroutput powEfficiency
1,
Example ;
The primary and secondary windings of a 500 kVA transformer have resistances of 0.42 and 0.0019 respectively. The primary and secondary voltages are 11 000 V and 400 V respectively and the core loss is 2.9 kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on :
a) full loadb) half load
Solution ;
Therefore secondary I2R loss on full load is (1250)2 0.0019 = 2969 W
and primary I2R loss on full load is(45.5)2 0.42 = 870 W
Total I2R loss on full load = 3839 W = 3.84 kWand Total loss on full load = 3.84 + 2.9 = 6.74 kWOutput power on full load = 500 0.8 = 400 kW Input power on full load = 400 + 6.74 = 406.74 kW
IRPloadfullonlossesTotal
AI
A.I
V
PIIVP
s
p
,
1250400
10500current,secondaryloadFull
54511000
310500current,primaryloadFulla)
3
Solution ;
From equation, = 1 -
= = 0.983 per unit
= 98.3 per cent b) Since the I2R loss varies as the square of the current,
Total I2R loss on half load = 3.84 (0.5)2 = 0.96 kWand Total loss on half load = 0.96 + 2.9 = 3.86 kW
Efficiency on half load = = 0.981 per unit
= 98.1 per cent.
powerinput
losses
74.406
74.61
86.203
86.31
Example ;
In a 50 kVA transformer, the iron loss is 500 W and full-load copper loss is 800W. Find the efficiency at full-load and half-load at 0.8 p.f. lagging. Solution ; Full-load, 0.8 p.f
Full-load output = 50 x 0.8 = 40 kWTotal Full-load losses = 500 + 800 = 1300 W = 1.3 kWFull-load input = 40 + 1.3 = 41.3 kWFull-load η = 100 = 96.85 %
Half-load, 0.8 p.fOutput at half-load = (50 x1/2)x 0.8 = 20 kWTotal loss at half-load = 500 + (1 /2)2 x800 = 700 W = 0.7 kW
Input at half-load = 20 + 0.7 = 20.7 kWFull-load η = 100 = 96.6 %
3.41
40
7.20
20
These two tests enable the efficiency and the voltage regulation to be a calculated without actually loading the transformer and with an accuracy far higher than is possible by direct measurement of input and output powers and voltages.Also, the power required to carry out these tests is very small compared with the full-load output of transformer
1. OPEN-CIRCUIT TEST
The transformer is connected as Figure to a supply at the rated voltage and frequency, namely the voltage and the frequency given on the nameplate.The ratio of the voltmeter readings, V1 / V2, gives the ratio of the number of turns. Ammeter A gives the no-load current, and its reading is a check on the magnetic quality of the ferromagnetic core and joints. The primary current on no load is usually less than 5 per cent of the full-load current, so that the I2R loss on no load is less than 1/400 of the primary I2R loss on full load and is therefore negligible compared with the core loss. Hence the wattmeter reading can be taken as the core loss of the transformer.
2. SHORT-CIRCUIT TEST
The secondary is short-circuited through a suitable ammeter A2, as shown in Figure and the low voltage is applied to the primary circuit. This voltage should, if possible, be adjusted to circulate full-load current in the primary and secondary circuits. Assuming this to be the case, the I2R loss in the windings is the same as that on full load. On the other hand, the core loss is negligibly small, since the applied voltage and therefore the flux are only about one-twentieth to one-thirtieth of the rated voltage and flux, and the core loss is approximately proportional to the square of the flux. Hence the power registered on wattmeter W can be taken as the I2R loss in the windings.
3. AUTO-TRANSFORMER
An auto-transformer is a transformer having a part of its winding common to the primary and secondary circuits. In Figure winding AB has a tapping at C, the load being connected across CB and the supply voltage applied across AB.
I1 and I2 = primary and secondary currents respectivelyN1 = no. of turns between A and BN2 = no. of turns between B and Cn = ratio of the smaller voltage to the larger voltageNeglecting the losses, the leakage reactance and the magnetizing
current.
1
2
2
1
1
2
N
N
I
I
V
Vn
=
Series winding
Common winding
3. AUTO-TRANSFORMER
The nearer the ratio of transformation is to unity, the greater is the economy of conductor material. Also, for the same current density in the windings and the same peak values of the flux and of the flux density, the I2R loss in the auto transformer is lower and the efficiency higher than in the two winding transformer.
Auto transformer are mainly used for interconnecting systems that are operating at roughly the same voltage and starting cage-type induction motors. Should an auto transformer be used to supply a low voltage system from a high voltage system, it is essential to earth the common connection, for example, B in Fig 8.9 otherwise there is a risk of serious shock. In general, however, an auto transformer should not be used for interconnecting high voltage and low voltage systems.
=
MERITS
a. a saving in a cost since less copper is needed.b. less volume, hence less weight.c. a higher efficiency, resulting from lower I2R lossesd. a continuously variable output voltage is achievable if a sliding
contact is used.e. a smaller percentage voltage regulation
DEMERITS
The primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary
=
1. FLUX DENSITY
The strength of the magnetic field (or amount of flux) is directly proportional to the number of parallel conductors, or the number of TURNS around the coil, and to the amount of CURRENT flowing in the coil. The amount of flux, or FLUX DENSITY, Φ (The Greek letter Phi) is proportional to the product of I (amperes) x N (number of turns) or the AMPERE TURNS of the coil. Increasing either the number of turns or the current in them produces an increase in flux.
2. RELUCTANCE
That is to improve the magnetic properties of the core by using a material that has a low Reluctance (Rm ), this is the property of a material that is the magnetic equivalent of the electrical property of Resistance. The lower the reluctance, the easier it is for magnetic flux to flow through the core material. Materials that are easily magnetised have a low reluctance and a high permeability, and none magnetic materials have a high reluctance and a low permeability. The opposite of Reluctance is Permeability, the magnetic equivalent of electrical Conductance.
turnsampereNI ;
3. ELECTRICAL AND MAGNETIC CIRCUITS COMPARED
In the electrical circuit an emf produced by a cell or battery drives a current around the circuit, which consists of a length of wire having some resistance, R.
The magnetic circuit also has a source of power in the form of a coil,supplied by an AC current.
Just as the external electrical source is called an electro motive force(emf), the external magnetic source is called a magneto motive force (mmf), And is measured in ampere turns. An emf produces a current (I), which Has a strength measured in amperes in the electrical circuit; in the magnetic circuit, the mmf produces a magnetic flux, Φ and is measured in units of webers (Wb). The resistance to the flow of magnetic flux in the core is called Reluctance (Rm).
4. MAGNETIC FLUX LINKING PRIMARY AND SECONDARY WINDINGS
A coil (the primary) supplied with an AC current is wound around one side of the core to provide a source of mmf. On the other side of the core, a separate coil (the secondary) is wound which supplies a measuring instrument to measure the amount of current in the coil.
The current in this coil will be proportional to the amount of flux flowing in the core. This arrangement therefore Provides a means of measuring magnetic flux.
5. FLUX IS ALSO AFFECTED BY THE DIMENSIONS OF THE CORE
The fugure showns if that the mmf is kept constant, but the dimensions of the core are altered by changing either the length of the flux path or its cross sectional area, the amount of flux flowing around the core will also change.
Therefore the measured flux (Φ) in the core (and therefore the secondary current) is proportional to the cross sectional area of the core, and inversely proportional to the length of the flux path:
Where:A is the cross sectional area of the core andL is the mean length of the flux path around the core
5. FLUX IS ALSO AFFECTED BY THE DIMENSIONS OF THE CORE (CONT)
The magnetic circuit also has some Reluctance Rm (a type of resistance to flux) ;
Reluctance is measured in Amperes per Weber (A/Wb).
PERMEABILITY (μ)
Electrical resistance also depends not only on the dimensions of the conductor but also on the material of the conductor and its resistivity. Likewise, in magnetic circuits reluctance depends not only on the length and cross sectional area, but also on the Permeability (μ) of the material.
The higher the value for μ the more flux will flow and the more flux that flows, the lower must be the value of reluctance Rm ;Therefore:
So Reluctance increases with the length of the magnetic path (l) and decreases as either the cross sectional area (A) of the core or the Permeability (μ) of the material is increased.
5. FLUX IS ALSO AFFECTED BY THE DIMENSIONS OF THE CORE (CONT)
RELATIVE (μR) AND ABSOLUTE PERMEABILITY (μO)
Permeability is often expressed as:
μ = μR μo
It is normal to find a core material described by its relative permeability (μR) i.e. by how many times the absolute permeability (μ) of the material is greater than the absolute permeability of free space (μO). The absolute permeability of free space;
μo = 4 π x 10-7 H/m = 1.256 x 10-6 H/m
where H is in henrys and m is in metres.
Therefore μ is a simple ratio that does not have any units, The permeability of iron can be many hundreds, so having a magnetic circuit path of iron rather than air greatly increases the flux, which is why iron is a common choice of material for inductor and transformer cores.
6. COMMON TYPES OF TRANSFORMER CORES
The Shell Core is an improvement of the Core type; its magnetic circuit encloses the windings more fully. Notice the centre limb has twice the cross sectional area of the outer limbs, allowing for double the flux within the primary and secondary windings.
The Toroidal core gives an even more efficient coupling, and radiates less electromagnetic energy outside the transformer.
The magnetic circuit of the two part Pot Core, used for smaller high frequency transformers and inductors, totally encloses the windings.
Suppose the maximum value of flux to be m webers and the frequency to be f hertz. From Figure, it is seen that the flux has to change from +m to -m in the half cycle, namely in seconds.
Average rate of change of flux = 2 m =4 f m webers per second
and average e.m.f induced per turn is = 4 f m volts
But for a sinusoidal wave the r.m.s or effective value is 1.11 times the average value, RMS value of e.m.f induced per turn = 1.11 4 f m
Hence r.m.s value of e.m.f induced in primary is
E1 = 4.44 N1 f m volts
And r.m.s value of e.m.f induced in secondary isE2 = 4.44 N2 f m volts
f2
1
f2
1
Example ;
A 250 kVA, 1100 V / 400 V, 50 Hz single-phase transformer has 80 turns on a secondary. Calculate :
a) the approximate values of the primary and secondary currents.b) the approximate number of primary turns.c) the maximum values of flux.
Solution ;
a) Full – load primary current = 250 1000 = 22.7 A 1100
and full – loaded secondary current = 250 1000 = 625 A 400
b) Number of primary turns = 80 11000 = 2200 400
c) From expression ,E2 = 4.44 N2 f m volts
400 = 4.44 80 50 m
m = 22.5 mWb
Example ;
An ideal 25 kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Calculate (i) primary and secondary currents on full-load(ii) secondary e.m.f. and(iii) the maximum core flux
Solution ;
(i)
(ii)
(iii)
50
4
500
40
N
NK
1
2
A 33.83000
1025amperes-VoltI
3
11
VA
K
II 2.10445033.81
2
KE
E
1
2 VKEE 2403000)504(12
mfNE 11 44.4
mwbwbm 2710275005044.4
3000 3
Example ;
1. A 100 kVA, 4000 V/200 V, 50 Hz single phase transformer has 100 secondary turns. Determine :
a) the primary and secondary currentb) the number of primary turnsc) the maximum value of the flux
2. A 4500 V/225 V, 50 Hz single phase transformer is to have an approximate e.m.f per turns of 15 V and operate with a maximum flux of 1.4 T. Calculate :
a) the number of primary and secondary turnsb) the cross sectional area of the core
3. A single phase 2200/250V, 50 Hz transformer has a net core area of 36cm2 and a maximum flux density of 6 wb/m2. Calculate the number of primary and secondary turns.
Solution ; 1. a) I1 = 25 A, I 2 = 500 A b) N1 = 2000 turns
c) m = 9.01 mWb2. a) N1=300, N2= 15 b) A = 0.0483 m2
3. N1 = 459, N2 = 52
Example ;
1.State the function of a transformer.2.List THREE( (3) types of transformer commonly used. Sketch symbol of each transformer and give value to differentiate them. 3.An ideal 25kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000V, 50Hz supply. Determine: Primary and secondary currents
a) The maximum value of fluxb) The type of transformer connections
Solution ; 1.It is a static device that connected two electrical circuit which functions to step-up or step-down voltage and AC current.2.i) Step-up transformer; eg. 1:10 ii) Step-down transformer; eg. 10:1
iii) Coupling transformer; eg. 1:1
Solution (cont) ;
i)Primary and secondary currents
ii) The maximum value of flux
iii) The type of transformer connectionsStep down transformer
Example ;
(a)List THREE (3) advantages of an ideal-transformer and THREE (3) advantages of an auto-transformer.(b)Define auto-transformer and draw step-down auto-transformer and step up auto-transformer.(c)A single phase transformer has 400 primary turns and 1000 secondary turns. The net-cross sectional area of the core is 60 cm2. If the primary winding is connected to 50 Hz at 520V supply, calculate :
i. The transformation ratio and identify the type of transformerii. The voltage induced in the secondary windingiii. The current flow in the secondary winding, if given resistance to the
secondary winding is 100Ωiv. The maximum value of flux density.
Solution ;
(a)List THREE (3) advantages of an ideal-transformer and THREE (3) advantages of an auto-transformer.
Ideal-transformeri. No power in the windings ii. No flux leakage in the core that links the primary and secondary
windings iii. No power losses due eddy current and hysteresis
Auto-transformeri. Simple structure ii. Low cost (less copper)iii. Maximum power transfer
Solution ;
(c)A single phase transformer has 400 primary turns and 1000 secondary turns. The net-cross sectional area of the core is 60 cm2. If the primary winding is connected to 50 Hz at 520V supply, calculate :
i. The transformation ratio and identify the type of transformerii. The voltage induced in the secondary windingiii. The current flow in the secondary winding, if given resistance to the
secondary winding is 100Ωiv. The maximum value of flux density.
Np = 400Ns = 1000f = 50HzVp = 520
i) Transformation ratio K = Ns/Np = 1000/400
= 2.5 Type : Step-up transformer
ii) V2 = V1K = 520 x 2.5 = 1300V
Or using other method : Vp/Vs = Np / Ns = Vs = Vp Ns / Np = (520 x 1000) / 400 = 1300V
Solution ;
iii) Is = Vs / Rs
= 1300 / 100 = 13A
The highest value of voltage is Vs = 1300V, E2 = V2
Therefore, E2 = 4.44fN2Φ Φ = E2 / 4.44 f N2
= 1300 / (4.44 x 50 x 1000) = 5.856 x 10-3 Weber
