ABHISHEK (EN 4th Year)
HARSH VARDHAN (EN 4th Year) HEMANT (EN 4th Year) VIDYA COLLEGE OF ENGINEERING
CONTENT
REFRENCES
Transformer Basics A transformer is an electrical device that transfers energy from one circuit to another purely by magnetic coupling. coupling. Relative motion of the parts of the transformer is not required for transfer of energy. Transformers are often used to convert between high and low voltages voltages,, to change impedance impedance,, and to provide electrical isolation between circuits
TWO WINDING TRANSFORMER CONNECTION SINGLE
PHASE
BASIC TRANSFORMER ACTION
50 8
INVENTION
Michael Faraday, Faraday, who invented an 'induction ring' on August 29, 1831. This was the first transformer, although Faraday used it only to demonstrate the principle of electromagnetic electr omagnetic induction and did not foresee the use to which it would eventually be put.
NEED FOR REDUCTION IN LOSS Very high speed growth of industries in our
country. High transmiss transmission ion loss from generation point
to distribution point as well as more demand compared to generation of electrical energy. Progress of new projects for generation of
power is very slow as compared to the growth of the requirement
Transformer losses categorized as follows.
No load Loss (Excitation Loss) Hysteresis Loss Eddy current loss
Load
Loss (Impedance Loss)
I²R loss. Stray loss
HYSTERESIS LOSSES Hysteresis loss is that energy lost by reversing the magnetic field in the core as the magnetizing AC rises and falls and reverses direction.
MECHANICAL LOSS
This loss consist of bearing friction,brush friction and windage loss.The windage loss includes the power required to circulate air through the machine. machine.
Increment in core loss caused by distortion of air-gap flux plus the increment in I 2R loss due to non uniform distribution of conductor current is called stray load loss. loss.
br ought in close to When the probe is brought a conductive material, the probes changing magnetic field generates current flow in the material.
The eddy currents produce their own magnetic fields that interact with the primary magnetic field of the coil.
? ? ? ?? ? WHY WE USE LAMINATED CORE
?
Working Methodology
The losses which occur in transformer are: (a):- Copper loss or I2 R. (b):- Iron loss or Core loss. Copper loss Pc :- Here we will calculate copper loss for one embedded bar similarly losses can be calculated for all other bars Pc=I2R Where R is resistance of coil coil and I is current in coil coil R=ƍL/A Where ƍ is specific resistance of material, L is length of coil and A is area of coil, so copper loss can be calculated .
Iron loss or core loss Pi:Pi :- Iron loss occurs in the magnetic core of
the transformer. This loss is the sum of hysteresis loss(Ph) and Eddy current loss(Pe). Pi=Ph + Pc Pi=Kh fBmn + Kef2B2m Kh= Proportionality constant which depends upon the volume and quality of the core material and the units used. Ke= Proportionality constant whose value depends upon the volume and resistivity of the core material,thickness of laminations and units u nits used . Bm=Maximum flux density in the core. f=Frequency of the alternating flux.
Separation of Hysteres Hysteresis is and Eddy-Curre Eddy-Current nt losses:- The
transformer core
loss Pi has two components namely hysteresis loss P and Eddy-current loss Pe. Pi=Ph + Pc Pi=Kh fBmn + Kef2B2m The exponents n varies in the rage 1.5 to 2.5 depending upon the ferromagnetic material for a given Bm the hysteresis loss varies directly as the frequency and the Eddy current loss varies as the square of the frequency. That is, Ph α f or Ph=af and Pe α f2 or Pe=bf2 Where a and b are constants . Pi=af+bf2
For separation of these two losses the no load test is performed on the transformer. However, the primary of the transformer is connected to a variable frequencyand variable sinusoidal supply and the secondary is open circuited. Now V=4.44fǾmT Or V/f=4.44BmAiT For any transformer T and Ai are constants. Therefore Bm will remain constant If the test is conducted so that the ratio (V/f) is kept k ept constant Pi/f=a+bf During this test, the applied applied voltage V and frequency f are varied together so that (V/f) is kept constant. The core loss is obtained at different frequencies
This graph is a straight line AB of the t he form y=mx+c,as shown in figure. The intercept of the straight line on the vertical axis gives a and slope of line AB gives b. Thus, knowing the constants a and b, hysteresis and Eddy current losses can be separated.
DC Shunt Motor Generator Generato r Wattmeter A R h R
h
Y
R
M
V0
G
V B
N
f
v
NEC 1
Frequency meter
d l e i f
Rheostat
Circuit Diagram of Setup
EC
Calculation
F o r N o n Em Em b e R=Resistance
For Embedded Core R=Resistance of Winding=80.9 ohms Pc=Copper loss=I^2* R= 1.06^2*80.9 ( )=89.99 W Pi= Pt–Pc Pi=Input Power (Variable) Pt=Total Power (Pi)1 =11-2 8 9.99=.01 22 W (Pi/f )1 =.489 =.489 W/Hz (Pi) 5=14 89.99=50 -0 .01 W (Pi/f )5=W/Hz W 1 /Hz From Graph 2
B=0062*50 0 . 062*50 2
2
He=bf = .0 06 *50 2=
25.5W
25.4W
S.NO.
FREQUENCY F (hz)
CURRENT A(amper)
WATTMETER (watts)
GENRATED VOLTAGE (volts)
POWER (Pi)
Pi/F
01
45
1.05
125
340
35.01
.778
02
46
1.05
128
352
03
47
1.05
133
358
04
48
1.05
135
369
05
50
1.05
147
386
57.01
1.14
Table for Non Embedded Core
S.NO.
FREQUENCY F (hz)
CURRENT A(amper)
WATTMETER (watts)
GENRATED VOLTAGE (volts)
POWER (Pi)
Pi/F
01
45
1.07
112
336
22.01
.489
02
46
1.07
117
345
03
47
1.07
121
354
04
48
1.07
128
363
05
50
1.07
140
388
5o.01
1.00
Table for Embedded Core
OUTPUT OF PROJECT 1.4 c y c/
Non Embedded Core Embedded Core
el
1.2
t r
ni e w o P
u p
Slope b1=0.007
1 0.8 0.6
a1
Slope b2 =0.0102 a2
0.4 45
46
47 48 Frequency
49
50
CONCLUSION
CHALLENGES FACED
Easy
to fabricate
Increased
Transformer Efficiency
Reduced Economy
Humming Loss
How eddy current loss minimize by using laminated core in transformer ? Member since: June 26, 2007
About me: WILLIUM TERROSE I am located
in the USA central time zone (UTC -6)
Assume that a changing magnetic flux is passing through a certain square cross sectional area of the transformer core. Look at a loop of current enclosing that flux. The power dissipated in that particular loop is proportional to the square of the area enclosed by that loop (A) divided by the length of the path (L). If you divide that square into two rectangles by laminating the core, the area enclosed in the loop will be cut in half while the length will be reduced to 3/4 of the original length. The result will be two loops of current with a total power dissipation of 2X(.5A)^2/.75L. That makes the sum of the power dissipated in the two smaller loops two thirds of the power dissipated in the original loop. More laminations reduce the dissipation
References • Dr.P.S. Dr.P.S. BIMBHRA Electrical Machines, Machines, Macmillan. ISBN 0-333-19627 0-333-19627-9 -9.. •Heathcote, Heathcote, MJ (1998). J&P (1998). J&P Transformer Book, 12th ed., ed. , Newnes. ISBN 0-7506-1158 0-7506-1158-8 -8.. •Hindmarsh, J. (1984). Electrical Machines and their Applications, 4th ed., ed., Pergamon. ISBN 0-08-030572-5. 0-08-030572-5. •Shepherd,J; Shepherd,J; Moreton,A.H; Moreton,A.H; Spence,L.F. (1970). Higher Electrical Engineering, Engineering , Pitman Publishing. ISBN 0-273-40025 0-273-40025-8 -8..
Vote of Thanks
Mr. A.K. Singha Mr. Ramveer Singh
THANK YOU
THANK YOU
