Post on 09-Oct-2015
description
Transformer Design
PAGE 4
Transformer DesignOutput Equation: - It gives the relationship between electrical rating and physical dimensions of the machines.Let
V1 = Primary voltage say LVV2 = Secondary voltage say HV
I1 = Primary current
I2 = Secondary current
N1= Primary no of turns
N2= Secondary no of turns
a1 = Sectional area of LV conductors (m2)
=
a1 = Sectional area of HV conductors (m2)
=
= Permissible current density (A/m2)
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 =
KVA =
KVA
= KVA-----------(1)
Where
f = frequency
= Maximum flux in the core
= Sectional area of core
= Maximum flux density in the core
Window Space Factor
So
Put equation value of N1I1 form equation (2) to equation (1)
(2)For 1- phase shell type transformer Window Space Factor
Kw
So
Put equation value of N1I1 form equation (4) to equation (1)
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 =
KVA =
KVA
= KVA-----------(6)
Window Space Factor
So
Put equation value of N1I1 form equation (7) to equation (6)
(3)For 3- phase shell type transformer Window Space Factor
Kw
So
Put equation value of N1I1 form equation (9) to equation (6)
Choice of magnetic loading (Bm)(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/mm2
ANAir Natural cooling
ONOil Natural cooling
OFNOil Forced circulated with Natural air cooling (2) Forced Cooling :
2.2---4.0 A/mm2
ABAir Blast cooling
OBOil Blast cooling
OFBOil Forced circulated with air Blast cooling
(3) Water Cooling:
5.0 ---6.0 A/mm2
OWOil immersed with circulated Water cooling
OFWOil Forced with circulated Water cooling
Core Construction:
EMF per turn:We know
and
Q =
KVA
(Note: Take Q as per phase rating in KVA) =
KVA
In the design, the ration of total magnetic loading and electric loading may be kept constant
Magnetic loading =
Electric loading =
So
Or
using equation (2)
Or
Where 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 0.85for 1-phase core type transformer
Kt = 1.0 to 1.2
for 1-phase shell type transformer
Estimation of Core X-sectional area AiWe know
Or
So
Now the core may be following types
d = Diameter of circumscribe circle For Square core
Gross Area
Let stacking factor
Actual Iron Area
(0.45 for square core and take K as a general case)
So
Or
Graphical method to calculate dimensions of the core
Consider 2 step core
Percentage fill
= 0.885 or 88.5%
No of steps1234567911
% Fill63.7%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
So, Window area
where
Kw =Window space factor
For higher rating Kw = 0.15 to 0.20Assume some suitable range for
D = (1.7 to 2) d
Width of the window Ww = D-d
Height of the window
Generally
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) AiDepth of yoke
Dy = aHeight of the yoke hy = Ay/Dy
Width of the core
W = 2*D + d
Height of the core
H = L + 2*hyFlux density in yoke
Estimation of core loss and core loss componet of No load current Ic:Volume of iron in core = 3*L*Ai
m3
Weight of iron in core = density * volume
= * 3*L*Ai
Kg
= density of iron (kg/m3)
=7600 Kg/m3 for normal Iron/steel
= 6500 Kg/m3 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** 3*L*AiWatt
Similarly
Iron loss in yoke
= py** 2*W*AyWatt
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
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
So
MMF required for the core
= 3*L*atcoreMMF required for the yoke
= 2*W*atyokeWe account 5% at for joints etc
So total MMF required
= 1.05[MMF for core + MMF for yoke]Peak value of the magnetizing current
RMS value of the magnetizing current
Estitmation of No load current and phasor diagram:
No load current Io
No load power factor
The no load current should not exceed 5% of the full the load current.Estimation of no of turns on LV and hv winding
Primary no of turns
Secondary no of turns
Estimation of sectional area of primary and secondary windings
Primary current
Secondary current
Sectional area of primary winding
Sectional area of secondary winging
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 R1 & R2 and Cu losses:
Let
Lmt = Length of mean turn
Resistance of primary winding
Resistance of secondary winding
Copper loss in primary winding
Copper loss in secondary winding
Total copper loss
Where
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
%Estimation of leakage REACTANCE:Assumptions
1. Consider permeability of iron as infinity that is MMF is needed only for leakage flux path in the window.
2. 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
Permeance of this elementary cylinder
(Lc =Length of winding)
Leakage flux lines associated with elementary cylinder
Flux linkage due to this leakage flux
Flux linkages (or associated) with primary winding
Flux linkages (or associated) with the space a between primary and secondary windings
We consider half of this flux linkage with primary and rest half with the secondary winding. So total flux linkages with primary winding
Similarly total flux linkages with secondary winding
Primary & Secondary leakage inductance
Primary & Secondary leakage reactance
Total Leakage reactance referred to primary side
Total Leakage reactance referred to secondary side
It must be 5% to 8% or maximum 10%
Note:- How to control XP?
If increasing the window height (L), Lc will increase and following will decrease b1, b2 & Lmt and so we can reduce the value of XP.
Calculation of VolTage Regulation of transformer:
Transformer Tank Design:Width of the transformer (Tank)
Wt=2D + De + 2b
Where
De= External diameter of HV winding
b = Clearance width wise between HV and tankDepth of transformer (Tank)
lt= De + 2a
Where
a= Clearance depth wise between HV and tank
Height of transformer (Tank)
Ht= H + h
Where
h=h1 + h2= Clearance height wise of top and bottom
Tank of a 3-Phase transformerCalculation 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 m2
(Excluding area of top and bottom of tank)Let
= Temp rise of oil(35o C to 50o C)
12.5St=Total full load losses ( Iron loss + Cu loss)
So temp rise in o C
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:Let
xSt= Surface area of all cooling tubes
Then
Losses to be dissipated by the transformer walls and cooling tube
= Total losses
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
Hence
No of cooling tubes
Weight of TRANFORMER:LetWi = 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/m3)
Weight of Oil
= Volume of oil * 880
KgAdd 20% of (Wi+Wc) for fittings, tank etc.
Total weight is equal to weight of above all parts.
L
V
L
V
L
V
L
V
H
V
H
V
H
V
H
V
H
V
1-phase core type transformer with concentric windings
Window
H
V
L
V
L
V
L
V
H
V
L
V
1-phase shell type transformer with sandwich windings
Window
LV
LV
HV
HV
LV
LV
HV
LV
HV
LV
H
V
3-phase core type transformer with concentric windings
H
V
H
V
L
V
L
V
Window
3-phase shell type transformer with sandwich windings
Window
(a) U-I type
(b) E-I type
(c) U-T type
(d) L-L type
(e) Mitred Core Construction (Latest)
45o
d
d/2
1-Step
Or Square- Core
2-Step
Or Cruciform- Core
3-Step Core
4-Step Core
K= 0.45 0.56 0.600.625
3-phase core type transformer
a
2-Step
Or Cruciform- Core
hy
H
d
W
a
b
a
b
Ww=
(D-d)
D
L
b
a
b
Ic
2-Step
Or Cruciform- Core
Im
Io
V1=-E1
E2
0
No load phasor diagram
b1
b2
a
x
dx
N1I1=N2I2
x
Lc
MMF Distribution
EMBED Equation.3
d= 5 Cm
hy
H
Ww
(D-d)
Wt
De
D
D
H
Ht
W
L
lt
a
a
b
b
Tank and Arrangement of Cooling tubes
7.5 Cm
h1
W
h2
EMBED Equation.3
Specific Heat dissipation
6 Watt/m2-0C by Radiation
6.5 Watt/m2-0C by Convection
6 W-Raditon+6.5 W=12.5 Convection
6.5*1.35 W EMBED Equation.3 ( EMBED Equation.3 35% more) Convection only
PAGE http://eed.dit.googlepages.com,
Prepared by: Nafees Ahmed
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