Its my PPT presentation about Single Phase Transformer. In it only cover few point about single phase transformer like construction, working principle, emf equation and Transformer on load.

1. SINGLE PHASE TRANSFORMER
3rd semester Electrical Engineering
Sub :- DC Machine & Transformer
Professor name : Ashok D Pateliya
PREPARED BY
1 MAHENDRA RAJPUT 140643109030
2 ANKUR SHAH 140643109031
3 KAUSHAL RANA 140643109021
4 JAY PRAJAPATI 140643109018

2. -: Topic covered :-
• Construction and working principle of
single phase transformer.
• Types of Transformer.
• Emf Equation of Transformer.
• Transformer on NO LOAD.

3. -: Transformer :-

4. -: Construction and working principle
of single phase transformer :-
• A Transformer is a static piece of apparatus by
means of which electric power in one circuit is
transformed into electric power of same
frequency in another circuit.
• The physical basis of a transformer is mutual
induction between two circuit linked by a
common magnetic flux.

5. -:Transformer Construction:-

6. -:Core type transformer:-

7. -: Core-type Transformers :-
The coils used are form-wound and are of the
cylindrical type. The general form of these
coils may be circular or oval or rectangular. In
small size core-type transformers, a simple
rectangular core is used with cylindrical coils
which are either circular or rectangular in
form.

8. But for large-size core-type transformers,
round or circular cylindrical coils are used
which are so wound as to fit over a cruciform
core section as shown in Fig. The circular
cylindrical coils are used in most of the core-
type transformers because of their
mechanical strength. Such cylindrical coils are
wound in helical layers with the different
layers insulated from each other by paper,
cloth, micarta board or cooling ducts. Fig.
shows the general arrangement of these coils
with respect to the core.

9. -:Shell type transformer:-

10. -:Shell-type Transformers:-
In these case also, the coils are form-would but
are multi-layer disc type usually wound in the
form of pancakes. The different layers of such
multi-layer discs are insulated from each other
by paper. The complete winding consists of
stacked discs with insulation space between
the coils–the spaces forming horizontal cooling
and insulating ducts. A shell-type transformer
may have a simple rectangular form as shown
in Fig.

11. A very commonly-used shell-type transformer is
the one known as Berry Transformer–so called
after the name of its designer and is cylindrical
in form. The transformer core consists of
laminations arranged in groups which radiate
out from the centre. It may be pointed out that
cores and coils of transformers must be
provided with rigid mechanical bracing in order
to prevent movement and possible insulation
damage. Good bracing reduces vibration and
the objectionable noise–a humming sound–
during operation.

12. -:EMF Equation of Transformer:-
Let
N1 = No. of turns in primary
N2 = No. of turns in secondary
Φm = Maximum flux in core in webers
= Bm × A
f = Frequency of a.c. input in Hz
flux increases from its zero value to
maximum value Φm in one quarter of the
cycle i.e. in 1/4 f second.

13. ∴ Average rate of change of flux =1/ 4fΦm
= 4 f Φm Wb/s
or volt
Now, rate of change of flux per turn means
induced e.m.f. in volts.
∴ Average e.m.f./turn = 4 f Φm volt
If flux Φ varies sinusoidally, then r.m.s. value of
induced e.m.f. is obtained by multiplying the
averagevalue with form factor.
Form factor =r.m.s. value/ avg. value = 1.11

14. ∴ r.m.s. value of e.m.f./turn = 1.11 × 4 f Φm
= 4.44 f Φm volt
• Now, r.m.s. value of the induced e.m.f. in the
whole of primary winding
= (induced e.m.f/turn) × No. of primary turns
•E1 = 4.44 f N1 Φm = 4.44 f N1 BmA.........(1)
Similarly, r.m.s. value of the e.m.f. induced in
secondary is,
•E2 = 4.44 f N2 Φm = 4.44 f N2 BmA..........(2)

15. -:Transformer on NO Load:-

16. When an actual transformer is put on load,
there is iron loss in the core and copper loss in
the windings (both primary and secondary) and
these losses are not entirely negligible.
Even when the transformer is on no-load, the
primary input current is not wholly reactive.
The primary input current under no-load
conditions has to supply (i) iron losses in the
core i.e. hysteresis loss and eddy current loss
and (ii) a very small amount of copper loss in
primary (there being no Cu loss in secondary as
it is open)

17. Hence, the no-load primary input current I0 is
not at 90° behind V1 but lags it by an angle φ0
<90°.
No-load input power W0 = V1I0 cos φ0
where cos φ0 is primary power factor under no-
load conditions. No-load
condition of an actual transformer is shown
vectorially.

18. primary current I0 has two components :(i) One in
phase with V1. This is known as active or working
iron loss component Iw because it mainly supplies
the iron loss pluss mall quantity of primary Cu loss.
Iw = I0 cosφ0
(ii) The other component is in quadrature with V1
and is known as
magnetizing component Iμ because its function is
to sustain the alternating flux in the core. It is
wattless .Iμ = I0 sin φ0
Obviously, I0 is the vector sum of Iw and Iμ, hence
I0 = (Iμ2 + Iω2).
The following points should be noted carefully

19. 1. The no-load primary current I0 is very small as
compared to the full-load primary current. It is
about 1 per cent of the full-load current.
2. Owing to the fact that the permeability of the
core varies with the instantaneous value of
the exciting current, the wave of the exciting
or magnetizing current is not truly sinusoidal.
As such it should not be represented by a
vector because only sinusoidally varying
quantities are represented by
rotating vectors. But, in practice, it makes no
appreciable difference.

20. 3. As I0 is very small, the no-load primary.
Cu loss is negligibly small which means
that no-load primary input is practically
equal to the iron loss in the transformer.
4. As it is principally the core-loss which is
responsible for shift in the current vector,
angle φ0 is known as hysteresis angle of
advance.