elements of electrial design

1. Sub :ELEMENTS OF ELECTRICAL DESIGN
TOPIC : DESIGN OF CHOKE COIL
1
GROUP NAME :
1) Ankit patil(150990109009)
2) Ravijeet vashi(150990109010)
3) Raviraj solanki (150990109011)
Guided by : Mr.Sourav choubey

2. Introduction of Choke
A coil of copper wire
wound on laminated iron
core has negligible
resistance is known as
choke coil.
When an ac voltage is
applied to the purely
inductive coil, an emf
known as self-induced emf
is induced in the coil due to
self-inductance of the coil
which opposes the applied
voltage.

3. In the case of tube light, the
sudden stop of current through
the choke produces voltage of
several hundreds volts (approx.
900 to 1000V) across it, because
of self inductance of coil.
The high voltage starts flow of
electrons from one filament to
the other through the gas filled
inside the tube.

4. Design Procedure of choke(No air- gap)
The Design Procedure of choke coil is similar to the single phase small
transformer design.
Similar terms are use in this case also.
The only different is that , it has only one winding support on the iron
core.
Therefore, the Turns per volt is found on the basis of 2 output VA this
means if voltage V is applied across choke and I current through the
choke, then is taken to find Tefrom the reference table.
VI

5. Steps in designing the choke
EMF equation E = 4.44Φmf
N
m mB is given. If not take B = 1Wb/m2
Stacking factor Ks=0.9 is assumed
Gross area Ai /0.9
2
VI
2
VA Q
= = 2 ,find Tefor this value from the table.
Bm
mAi 
.

6. Square section is used. Width of central limb A =
For winding, Turns T=V x Te
δ is given = 2.5A/mm2Area of the conductor wire =I/δ
From this area (a) = π/4 d2
From above equation find d.
Refer standard conductor size table and find the matching size d of conductor and
with insertion diameter d1
Space in window for winding = and
Increase the space area by 20% to accommodate former insulation, packing etc.
The total area required = 1.2
Ag
Sf
T a
d1 
 d 
2
Sf 0.8
Sf
T a

7. 2

8. Design of variable air-gap single phase and 3
phase choke coil
In this section we have
study design procedure
of choke operating on (I)
single phase (II)Three
phase with adjustable air-
gap between limbs of
iron core.

9. Construction
Iron core
variable by
of one
rotating
part
wheel
and one part is fixed.
Copper coils with N/2
turns.
Rotating wheel to adjust air-
gap.

10. In circuit
R= Resistance of coil
L= Inductance of
coil
N=Number of turns of
coil
V=voltage with frequency f
According to basic equation, dt
V  iR  L
di
dt
L  N
dand
dt
v  iR  dtN
 iR  N
d

11. Step in Designing a Variable choke
A. For Magnetic part
B. For Electric part
C. Mechanical dimension
D. To find R,L,Z of the coil

12. For Magnetic part
1. To find constant k
K  Bg.Ai 
0VphIph
2fLg
where µ0=Permeability
f= frequency
Vph=phase voltage in r.m.s
Iph=phase voltage current in r.m.s

13. 2. Knowing the relation
Bg
2
Ai 
K
Bg
2
.Ai  K
Bg
K
Ai 

14. 3. Assume staking factor Ks=0.9
Agi
Ks 
Ai
 0.9
Agi 
Ai

Ai
; Agi  Ai
Ks 0.9

15. 4. AT is Amp.Turns (m.m.f) for the air gap lg.
(lg is total length of each air gap
if lg’ is the length of each air
gap
for single phase chokethen lg= 2 lg’
and lg= 1.5 lg’ for 3 phase values
AT required for iron part = ATi
ATi << ATg
ATi = 10 to 20 % of ATg
so that ATtotal= ATi + ATg
0
ATg 
Bg.lg

16. For Electric part
1. Number of Turn per
coil = AT/I where I= current
There are 2 coil of N/2 turns for single phase variable choke
There are 3 coil each of N turns for a 3-phase variable choke.
thus number of turns for the coil are decided.

17. 2. Current density (δ)
Generally enameled copper conductors are used for the choke
winding.
δ=2.3 to 2.5 Amp/mm2
for single phase choke current is I
for 3-phase choke current is taken i.e. Iph

18. 3. To find diameter (d) of bare conductor
C.S. area of bare conductor
Hence
4
a = 
d 2
4.a

d 

19. 4. Use of standard size conductor tables
From these table exact or nearest size of conductor is selected.
(d and a )
d’ = diameter of insulated
conductor So, area of insulated conductor
a’ =

d1
2
4
mm2

20. Mechanical dimension
these is for single phase choke
Window area for single phase choke
Window area for 3-phase choke
 
 d1
 d 
2
Aw'
1. Window size
average value of space factor sf=0.8
sf= active area/gross area
2 
N
a'
2
sf
Aw'
N a'
Aw'
2 N a'
sf

21. coil to be2. Depth df and height hf of
accommodated in the window space.
hf=actual height of coil
2
Aw
Ww 
hf’ =available height
df =actual depth of coil
df’ =depth of coil
Window area AAww
=H2wW. Ww
2 w

22. Some clearance on both side 10+10=20mm be kept
for formar etc. as shown in the figure.
Available height for winding hf’=Hw-20mm
Height wise turns per larger Nh=
Depth wise turns per layer Nd=
Actual depth of coil, df=df’+5mm
Actual height of coil, hf= hf’+
hf'
Nh
d1
N

23. 3. To find and distance between the two coils in the
window (dc) and overall dimension.
dc=Ww-2df
D =
D = Ww+ A
Total width = D+A
= Ww+2A
Distance (D) betweenoftwo lines.
Ww  AA
2 2

24. To find R,L,Z of the coil
•
where
•
• Reactance XLof coil = Ω
AND XL= 2πfL Ω
L= Henry
R 
l

Iph
a=C.S area of conductor
Impedance of coil with maximum air-gap Z 
Vph

Z2
 R2
Xl
2f
a
=Resistivity of copper = 1.73 x10-8

25. Thank You….