10. Core type Shell type
1. Easy in design and construction
2.Has low mechanical strength due to
non – bracing of windings
3. Reduction of leakage reactance is
not easily possible
4. The assembly can be easily
dismantled for repair work.
5. Better heat dissipation from
windings.
6. Has longer mean length of core and
shorter mean length of coil turn.
Hence best suited for EHV ( Extra
High Voltage) requirements.
1. Comparatively complex
2. High mechanical strength
3. Reduction of leakage reactance is
highly possible.
4. It cannot be easily dismantled
for repair work.
5. Heat is not easily dissipated from
windings since it is surrounded by
core.
6. it is not suitable for EHV ( Extra
High Voltage) requirements.
11. Distribution transformers
The load on the distribution transformer varies from time to
time and the transformer will be on no-load most of the time.
Hence distribution transformer are designed with less iron loss
and designed to have the maximum efficiency at a load much
lesser than full load.
Also it should have good regulation to maintain the variation of
supply voltage within limits
Power transformers
Used in power stations and substations
During heavy load periods all the transformers are put in
operation and during light load periods some transformers are
disconnected.
Designed to have maximum efficiency at or near full load
12. Output Equation of Single Phase Transformer
“The equation which relates the rate kVA output of transformer to the area of core and
window is called output equation”
13. Output Equation of Single Phase Transformer
The induced emf in a transformer, E =4.44f𝝓m T Volts
Emf per turn Et=4.44f𝝓m (a)
Window space factor, Kw= Conductor area in window = Ac
Total area of window Aw
Conductor area in window, Ac = Kw Aw (1)
The current density δ is same in both the windings.
Current density, δ = Ip = Is
ap as
Area of cross - section of primary conductor, ap = Ip
δ
Area of cross – section of secondary conductor, as = Is
δ
14. If we neglect magnetizing mmf then primary ampere turns is equal to
secondary ampere turns.
Ampere turns, AT = Ip Tp = Is Ts
Total copper area in window Ac = Copper area of + Copper area of
primary winding secondary winding
= Number of primary Number of secondary
turns x area of + turns x area of
cross – section of cross – section of
primary conductor secondary conductor
= Tp ap + Ts as = Tp Ip + Ts Is (since ap = Ip and as = Is )
δ δ δ δ
= 1 (Tp Ip + Ts Is ) = 1 ( AT + AT) (Since AT = Ip Tp = Is Ts )
δ δ
ie, Ac= 2AT (2)
δ
15. Equating (1) and (2)
Kw Aw = 2 AT / δ
Ampere turns, AT = 1 Kw Aw δ (b)
2
The kVA rating of single phase transformer is given by,
kVA rating, Q = Vp Ip x 10-3
≈ Ep Ip x10-3 (since Ep ≈ Vp)
= Ep Tp Ip x 10-3 (Et = Ep and AT = Tp Ip )
Tp Tp
= Et AT x 10-3 (c)
16. On substituting from Et and AT from (a) and (b) in C
Q = 4.44 f𝝓m Kw Aw δ x 10-3
2
= 2.22 f𝝓m Kw Aw δ x 10-3 (since B m = 𝝓m )
Ai
= 2.22 f Bm Ai Kw Aw δ x 10-3
18. The induced emf in a transformer, E =4.44f𝝓m T Volts
Emf per turn Et=4.44f𝝓m (a)
Window space factor, Kw= Conductor area in window = Ac
Total area of window Aw
Conductor area in window, Ac = Kw Aw (1)
The current density δ is same in both the windings.
Current density, δ = Ip = Is
ap as
Area of cross - section of primary conductor, ap = Ip
δ
Area of cross – section of secondary conductor, as = Is
δ
19. If we neglect magnetizing mmf then primary ampere turns is equal to
secondary ampere turns.
Ampere turns, AT = Ip Tp = Is Ts
Total copper area in window Ac = Copper area of + Copper area of
primary winding secondary winding
= 2 X Number of primary 2 X Number of secondary
turns x area of + turns x area of
cross – section of cross – section of
primary conductor secondary conductor
= 2Tp ap + 2Ts as = 2 Tp Ip + 2 Ts Is (since ap = Ip and as = Is )
δ δ δ δ
= 2 (Tp Ip + Ts Is ) = 2 ( AT + AT) (Since AT = Ip Tp = Is Ts )
δ δ
ie, Ac= 4AT (2)
δ
20. Equating (1) and (2)
Kw Aw = 4 AT / δ
Ampere turns, AT = 1 Kw Aw δ (b)
4
The kVA rating of single phase transformer is given by,
kVA rating, Q =3 Vp Ip x 10-3
≈3 Ep Ip x10-3 (since Ep ≈ Vp)
= Ep Tp Ip x 10-3 (Et = Ep and AT = Tp Ip )
Tp Tp
= 3Et AT x 10-3 (c)
21. On substituting from Et and AT from (a) and (b) in C
Q = 3x 4.44 f𝝓m Kw Aw δ x 10-3
4
= 3.33 f𝝓m Kw Aw δ x 10-3 (since B m = 𝝓m )
Ai
= 3.33 f Bm Ai Kw Aw δ x 10-3
22. Emf per turn
The transformer design starts with selection of
an appropriate value of emf per turn. Hence an equation for
emf per turn can be developed by relating output kVA,
magnetic and electric loading. In transformer the ratio of
specific magnetic and electric loading is specified rather
than actual value of specific loadings.
Let, ratio of specific magnetic
and electric loading = r = фm
AT
23. The volt ampere per phase a transformer is given by the product of voltage
and current per phase. Considering the primary voltage and current per
phase we can write,
kVA per phase, Q = Vp Ip x 10-3
= 4.44 fфm Tp Ip x 10-3 (Vp ≈ Ep = 4.44 fфm Tp )
= 4.44 fфm ATp x 10-3 (Tp Ip = AT )
= 4.44 fфm фm x 10-3 ( AT = фm )
r r
фm
2 = ___ Q r _____
4 .44f x 10-3
фm = ___ Q r _____
4 .44f x 10-3
24. Transformer type K
Single phase shell type
Single phase core type
Three phase shell type
Three phase core type,
distribution transformer
Three phase core type, power
transformer
1. to 1.2
0.75 to 0.85
1.3
0.45
0.6 to 0.7
