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Transformer Design
OUTPUT EQUATION: - It gives the relationship between electrical rating and physical dimensions
of the machines.
Let
V 1 = Primary voltage say LV
V 2 = Secondary voltage say HV
I 1 = Primary current
I 2 = Secondary current
N 1 = Primary no of turns
N 2 = Secondary no of turns
a 1 = Sectional area of LV conductors (m
2 )
I 1
a 1 = Sectional area of HV conductors (m
2 )
2
I
(^) = Permissible current density (A/m
2 )
Q = Rating in KVA
We place first half of LV on one limb and rest half of LV on other limb to reduce leakage flux.
So arrangement is LV insulation then half LV turns then HV insulation and then half HV turns.
(1) For 1-phase core type transformer
Rating is given by
Q =
3 1 1
V I KVA
= ^ ^
3 1 1
f N I m
KVA
1 1 V 4. 44 f N m
= ^ ^
3 1 1
fA B N I i m KVA^ -----------(1)^
( ) m i m AB
Where
f = frequency
m (^) = Maximum flux in the core
i A (^) = Sectional area of core
m B (^) = Maximum flux density in the core
Window Space Factor
( ) w
w WindowArea A
ActualCuSectionAreaofWindingsin Window K
w
A
a N a N 1 1 2 2
1 1 2 2
1 1 2 2
a I a I
A
I N I N
w
w
A
I N I N
1 1 2 2
So
L
V
L
V
L
V
L
V
H
V
H
V
H
V
H
V
1-phase core type transformer with
concentric windings
Window
1 1 2 2
1 1 ForIdealTransformerIN IN A
I N
w
( 2 ) 2
1 1
w w K A N I
Put equation value of N 1 I 1 form equation (2) to equation (1)
KVA
K A Q fA B
w w i m
3 10 2
- 44
- 22 10 ( 3 )
3
Q fA B K A KVA i m w w
(2) For 1- phase shell type transformer
Window Space Factor
Kw
w
A
a N a N 1 1 2 2
1 1 2 2
1 1 2 2
a I a I A
I N I N
w
w
A
I N I N
1 1 2 2
1 1 2 2
1 1 ForIdealTransformerIN IN
A
I N
w
So
( 4 ) 2
1 1 ^
w w K A N I
Put equation value of N 1 I 1 form equation (4) to equation (1)
KVA
K A Q fA B
w w i m
3 10 2
- 44
- 22 10 ( 5 )
3
Q f A B K A KVA i m w w
Note it is same as for 1-phase core type transformer i.e. equ (3)
(3) For 3-phase core type transformer
Rating is given by
Q =
3 1 1
V I KVA
= ^ ^
3 1 1
f N I m (^) KVA ^ V 1 (^) 4. 44 f mN 1
=
3 1 1
fA B N I i m KVA -----------(6) (^ m AiBm )
Window Space Factor
L
V
L
V
H
V
H
V
L
V
L
V
H
V
H
V
3-phase core type transformer with
concentric windings
L
V
L
V
H
V
H
V
Window
1-phase shell type transformer with
sandwich windings
LV
HV
LV
HV
LV
LV
HV
LV
HV
LV
Window
(1) Normal Si-Steel 0.9 to 1.1 T
(0.35 mm thickness, 1.5%—3.5% Si)
(2) HRGO 1.2 to 1.4 T
(Hot Rolled Grain Oriented Si Steel)
(3) CRGO 1.4 to 1.7 T
(Cold Rolled Grain Oriented Si Steel)
(0.14---0.28 mm thickness)
CHOICE OF ELECTRIC LOADING ^
This depends upon cooling method employed
(1) Natural Cooling: 1.5---2.3 A/mm
2
AN Air Natural cooling
ON Oil Natural cooling
OFN Oil Forced circulated with Natural air cooling
(2) Forced Cooling : 2.2---4.0 A/mm
2
AB Air Blast cooling
OB Oil Blast cooling
OFB Oil Forced circulated with air Blast cooling
(3) Water Cooling: 5.0 ---6.0 A/mm
2
OW Oil immersed with circulated Water cooling
OFW Oil Forced with circulated Water cooling
CORE CONSTRUCTION:
(a) U-I type (b) E-I type
(c) U-T type
(d) L-L type
EMF PER TURN:
We know
V 1 (^) 4. 44 f mN 1 ( 1 )
1
1 t m f N
V
SoEMF Turn E
and
Q =
3 1 1
V I KVA ( Note: Take Q as per phase rating in KVA)
= ^ ^
3 1 1
f N I m
KVA
10 ( 3 )
3 1 1
EN I KVA t
In the design, the ration of total magnetic loading and electric loading may be kept constant
Magnetic loading = m
Electric loading = N 1 I 1
So tan^ ( " ") 11 (^3 )
11
put in eqution r
cons t say r N I N I
m m
KVA
r
Q E
m t
3 10
Or KVA f r
E Q E
t t
3 10
- 44
using equation (2)
E fr Q
t
(^23)
Or E^ t Kt Q Volts / Turn
Where
3
- 44 10
Kt fr is a constant and values are
Kt = 0.6 to 0.7 for 3-phase core type power transformer
Kt = 0.45 for 3-phase core type distribution transformer
Kt = 1.3 for 3-phase shell type transformer
Kt = 0.75 to 0.85 for 1-phase core type transformer
Kt = 1.0 to 1.2 for 1-phase shell type transformer
ESTIMATION OF CORE X-SECTIONAL AREA Ai
We know
E (^) t Kt Q ( 1 )
(e) Mitred Core Construction (Latest)
o
% Fill 63.7% 79.2% 84.9% 88.5% 90.8% 92.3% 93.4% 94.8% 95.8%
ESTIMATION OF MAIN DIMENSIONS:
Consider a 3-phase core type transformer
We know output equation
Q f A B K A KVA i m w w
3
- 33 10
So, Window area
2 3
- 33 10
m
fA B K
Q
A
i m w
w (^)
where Kw = Window space factor
forupto KVA HigherKV
K w 10 30
8
forupto KVA HigherKV
Kw 200 30
10
forupto KVA HigherKV
Kw 1000 30
12
For higher rating Kw = 0.15 to 0.
Assume some suitable range for
D = (1.7 to 2) d
Width of the window Ww = D-d
Height of the window
w
w
widthof window W
A
L ( )
L Ww Aw
Generally 2 to 4 W
L
w
Yoke area Ay is generally taken 10% to 15% higher then core section area (Ai), it is to reduce the iron loss in
the yoke section. But if we increase the core section area (Ai) more copper will be needed in the windings
and so more cost through we are reducing the iron loss in the core. Further length of the winding will
increase resulting higher resistance so more cu loss.
Ay = (1.10 to 1.15) Ai
Depth of yoke Dy = a
http://eed.dit.googlepages.com, Prepared by: Nafees Ahmed
d
L
W
D
(D-d)
W w
=
H
h y
3-phase core type transformer
2-Step
Or Cruciform- Core
b (^) a
b
a
Height of the yoke hy = Ay/Dy
Width of the core
W = 2*D + d
Height of the core
H = L + 2*hy
Flux density in yoke
m y
i y B A
A B
ESTIMATION OF CORE LOSS AND CORE LOSS COMPONET OF NO LOAD CURRENT IC:
Volume of iron in core = 3LAi m
Weight of iron in core = density * volume
= i^ * 3LAi Kg
i (^) = density of iron (kg/m3)
=7600 Kg/m
3 for normal Iron/steel
= 6500 Kg/m
3 for M-4 steel
From the graph we can find out specific iron loss, pi (Watt/Kg ) corresponding to flux density Bm in core.
So
Iron loss in core =pi* i^ * 3LAi Watt
Similarly
Iron loss in yoke = py* ^ i * 2WAy Watt
Where py = specific iron loss corresponding to flux density By in yoke
Total Iron loss Pi =Iron loss in core + Iron loss in yoke
Core loss component of no load current
Ic = Core loss per phase/ Primary Voltage
Ic 1
3 V
P
i
ESTIMATION OF MAGNETIZING CURRENT OF NO LOAD CURRENT Im:
Find out magnetizing force H (atcore, at/m) corresponding to flux density Bm in the core and atyoke
corresponding to flux density in the yoke from B-H curve
B at m B at m m core c yoke / , /
So
MMF required for the core = 3Latcore
Sectional area of secondary winging
2 2
I
a
Where is current the density.
Now we can use round conductors or strip conductors for this see the IS codes and ICC (Indian Cable
Company) table.
DETERMINATION OF R 1 & R 2 AND CU LOSSES:
Let Lmt = Length of mean turn
Resistance of primary winding
2 1
(^61)
1 , , 75 a m
L N m R
mt dc o
o o ac dc
R to R 1 , , 75 1 , , 75
( 1. 15 1. 20 )
Resistance of secondary winding
2 2
(^62)
2 , , 75 a m
L N m R
mt dc o
R (^) 2 , ac , 75 o ( 1. 15 to 1. 20 ) R 2 , dc , 75 o
Copper loss in primary winding I R Watt 1
2 1
Copper loss in secondary winding I R Watt 2
2 2
Total copper loss (^2)
2 1 2
2 1
3 I R 3 I R
' 1 2
2 1
I R R
I R p
2 (^31)
Where Total resis cereferred toprimaryside
R Rp R R
tan
' 01 1 2
Note: On No load, there is magnetic field around connecting leads etc which causes additional stray losses
in the transformer tanks and other metallic parts. These losses may be taken as 7% to 10% of total cu losses.
DETERMINATION OF EFFICIENCY:
Efficiency InputPower
Output Power
OutputPower Losses
Output Power
100
OutputPower IronLoss Culoss
Output Power (^) %
ESTIMATION OF LEAKAGE REACTANCE:
Assumptions
- Consider permeability of iron as infinity that is MMF is needed only for leakage flux path in the
window.
- The leakage flux lines are parallel to the axis of the core.
Consider an elementary cylinder of leakage flux lines of thickness dx at a distance x as shown in following
figure.
MMF at distance x
http://eed.dit.googlepages.com, Prepared by: Nafees Ahmed
x
x
a b 1
b 2
dx
N
1
I
1
=N
2
I
2
L
c
MMF Distribution
x b
N I
M
x 1
11
Permeance of this elementary cylinder
L
A o
c
mt o L
L dx (Lc =Length of winding)
S
Permeance A
L S
o
1 &
1
Leakage flux lines associated with elementary cylinder
d M Permeance x x
Flux linkage due to this leakage flux
x x d Noof trunswithwhichitisassociated d
c
mt o L
L dx x b
N I
b
N I
1
11
1
11
dx b
x I L
L N
c
mt o
2
1
1
2 (^1)
Flux linkages (or associated) with primary winding
(^)
1
0
2
1
1
2 1
' 1
b
c
mt o dx b
x I L
L
N
1 1
2 1
b I L
L
N
c
mt o
Flux linkages (or associated) with the space ‘a’ between primary and secondary windings
I a L
L
N
c
mt o o 1
2 1
We consider half of this flux linkage with primary and rest half with the secondary winding. So total flux
linkages with primary winding
' 1 1
o
1 1
2 1 1
b a I L
L
N
c
mt o
Similarly total flux linkages with secondary winding
' 2 2
o
2 2
2 2 2
b a I L
L
N
c
mt o
Primary & Secondary leakage inductance
3 2
(^21) 1 1
1 1
b a
L
L N I
L
c
mt o
(^22) 2
2
2 2
b a
L
L
N
I
L
c
mt o
Primary & Secondary leakage reactance
3 2
2 2
(^21) 1 1 1
b a
L
L X fL f N
c
mt o
(^22) 2 2 2
b a
L
L
X fL f N
c
mt o
c
mt o L
L dx x
b
N I
1
11
Tank of a 3-Phase transformer
CALCULATION OF TEMPERATURE RISE:
Surface area of 4 vertical side of the tank (Heat is considered to be dissipated from 4 vertical sides of the
tank)
St= 2(Wt + lt) Ht m
2 (Excluding area of top and bottom of tank)
Let
= Temp rise of oil (
o C to 50
o C)
12.5St =Total full load losses ( Iron loss + Cu loss)
So temp rise in
o C t 12.5S
Totalfullload losses
If the temp rise so calculated exceeds the limiting value, the suitable no of cooling tubes or radiators must be
provided
CALCULATION OF NO OF COOLING TUBES:
H
W
h 1
h 2
H
t
Specific Heat dissipation
6 Watt/m
2
0 C by Radiation
6.5 Watt/m
2
0 C by Convection
Let xSt= Surface area of all cooling tubes
Then
Losses to be dissipated by the transformer walls and cooling tube
= Total losses
12. 5 8. 5 Totallosses t t S xS
So from above equation we can find out total surface are of cooling tubes (xSt)
Normally we use 5 cm diameter tubes and keep them 7.5 cm apart
At= Surface area of one cooling tube
d (^) tubeltube , mean
Hence
No of cooling tubes
t
t
A
xS
WEIGHT OF TRANFORMER:
Let
Wi = Weight of Iron in core and yoke (core volume* density + yoke volume* density) Kg
Wc= Weight of copper in winding (volume* density) Kg
(density of cu = 8900 Kg/m
3 )
Weight of Oil
= Volume of oil * 880 Kg
Add 20% of (Wi+Wc) for fittings, tank etc.
Total weight is equal to weight of above all parts.
d= 5 Cm
7.5 Cm
Tank and Arrangement of Cooling tubes
6 W-Raditon+6.5 W=12.5 Convection (^) 6.5*1.35 W 8. 5 ( 35% more) Convection only
