CHITTAGONG UNIVERSITY OF ENGINEERING &
TECHNOLOGY
DEPARTMENT OF ELECTRICAL & ELECTRONIC ENGINEERING
Report On
697 KVA Distribution Transformer design
Course No
Course Title
Group No
Section
Level-Term
EEE-240
Electrical Machine Design
G26
A
2-2
Submitted By:
Remarks
Pritam Bol
Id:1902044
Aritra Sarkar
Id:1902049
Objective:
To design a 697.00 kVA, 3 phase, 50 Hz, 6.6 KV/415 V, delta/star distribution
transformer.Here, we will consider:






Tapping 2.5%, 5% on high voltage side.
Cooling O N (self oil cooled)
Temperature rise over oil 600C
Load loss not more than 8 KW
Percentage impedance %Z= 4.50%
We also Calculated: efficiency at 750C on full load,75% load, and 50% load at a
unity power factor
 Regulation on full load at 750C at unity power factor and 0.8 power factor
lagging.
Solution:
 Voltage per turn(𝑬𝑻) :
An empirical expression that gives voltage per turn fairly
accurately for transformers is :
Et = 𝟏 √
𝟒0
𝑲𝑽𝑨×𝟏𝟎𝟎𝟎
𝑵𝒐 𝒐𝒇 𝒍𝒆𝒈𝒔
[for three-phase core type transformer; no of legs=3]
Therefore,
Et = 12.05 volts/turn
 Specific Magnetic Loading:
Bmax = 1.7 Wb / m2;
Here, material for core is chosen as cold rolled grain oriented (CRGO) steel laminations of
0.35 mm thickness; mitred core construction is used; mitered at 450 Cross Section of the
core:
Et= (4.44 Bm f Ai)
Volts
Where,
Bm = flux density in wb/m2 (taken as 1.7 wb/m2)
f= 50 Hz
Ai= net cross sectional area of the core in the m2
2|Page
10.3×106
4.4×1.7×50
=
=31928.988 mm2
 The diameter of the circumscribing circle for the core, d :
Here, we have chosen 7 step cores.
So, the area should be nearly circular. In the case of a 7 step core,
The core space factor, Ki= 0.88
Stacking factor for laminations, Ks= 0.92
If, d = diameter of the core section,
Ai =0.88 × 0.92 ×
𝜋𝑑2
4
2
d = 50213.88
So,
d= 224.08 mm
We choose it, d = 225 mm ;
∴Ai = 32190.404 mm2
And Bm = 1.686 wb/m2
 Window area Aw :
We know, S = 3.33 Ai Aw Kw δ Bm f ×10-3
Here,
Window space factor (kw) = 0.26[taken approximately]
Aw = Window area;
𝛿 = current density taken as 2.5 A / mm2;
S = output in kVA=697.00 KVA ;
Therefore,
Aw =
697×106×103
3.33×32190.404×0.26×2.5×1.686×50
= 118664.60 mm2
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Now, we choose, window width = 0.88×225=198 mm=200 mm(approx.)
Then, the height of the window =
118664.60
200
= 593.323 mm(assume 595 mm)
Choosing the height of the window = 2 ×width of window approx
= 2×200
= 400 mm
Now, window area = 200× 400 mm2
= 80000 mm2
The main dimensions of the core are therefore:
Diameter, d = 225 mm;
distance between the centers of the adjacent limbs , D = (200+ 213.75 approximate) mm
= 413.75 mm (taken D=415)[with a 7 step
core, the largest width of the core
with d = 200mm is,
0.95×225=213.75 mm=215mm]
Here, Height of window = 595+(30×2) mm
=655mm[adjusted for yoke clearance]
Total width = (2 × 415) + 215 =1045 mm ;
Total height =( 655 + 215+215)=1085 mm
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Figure 1: Dimensions of Core & yoke Assembly(all dimensions are in mm)
 Number of Turns in L.V. Winding :
Voltage per phase = 415= 239.6 V (as the winding is star connected)
√3
& Turns per phase on L.v winding =239.6 = 19.88 turns=20 turns
12.05
 Number of Turns in H.V. Winding :
Turns per phase on h.v. winding =
6600=
12.05
547.71 turns =548 turns;
As the winding is delta connected,
∴Tappings are applied on H.V. side:
5% more 548×1.05 =575 ; 2.5% more = 548×1.025=562
5% less 548×0.95 =521 ; 2.5% less = 548×0.975=534
Thus the turns for H.V. windings are:
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562
575
548
534
521
2.5%
5%
Normal
-2.5%
-5%tappings
 L.V. Winding :
3
Current per phase Ip =697×10 = 969.6702111 A
√3×415
Here, we choose a helical cylindrical coil.
As, Current density, δ = 2.5 A / mm2 ; (assumed)
∴ 𝐴rea of l.v. conductor, a2 = 969.6702111 ÷ 2.5
= 381.868 mm2
=380 mm2
Choosing, rectangular copper conductor from IS : 1.4 : 1 specs.
[For rectangular copper conductors for electrical machines, giving area near about the
required one.]
∴1.4x2 = 194
x = 11.77
∴width = 1.4x = 16.48 mm
∴ thick = 11.8 mm
So, choosing 5 mm thickness × 38 mm width; As thickness can’t be larger than 5mm
for transformer’s copper strip.(**Reference at last page**)
2 conductor strips farming conductor of l.v. area: a2 = 5× 38× 2 =380 mm2
 H.V. Winding: choosing disc coils :
Now, current in h.v. winding per phase Ip
=697×1000 =35.20 A; (being delta connected)
3×6600
Cross section of conductor for h.v. winding , a1 = 35.20 ÷ 2.5 = 14.08 mm2
Choosing round conductor where, d = diameter of conductor
We know,
d=4.234 mm = 4 mm
a1 = πd2 ÷ 4
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Copper area in window = 2 (a1T1 + a2T2)
= 2 (12.57 × 548 + 380 ×20)
= 28972.77 mm2
Now for this dimensions,
we get window space factor,
kw =(𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟)/(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑊𝑖𝑛𝑑𝑜𝑤)
=28972.77 ÷ 118664.60 = 0.244 [which is near about 0.24 chosen]
 Design of the layout of L.V. winding:
The number of turns, T2 =20.
Size of conductor:2 strips of 38 × 5 mm, copper rectangular conductors
With paper insulation for conductors, the size of each conductor
will be:
= (38 + 0.25) mm × (5+ 0.25) mm
= 200.81 mm2
choosing 2 layers for l.v. winding,
Turns per layer = 20/2 =10
Width of conductor 38.25 mm is taken along the winding, with
2 conductor sides
=5.25 + 5.25
= 10.5 mm forming conductor per layer.
For one layer, the dimension of conductors width is 10.5 mm
and height of window wise 38.25 mm for each conductor.
Now, Height of l.v. winding in window = 10 ×38.25 = 382.5 mm =383 mm;
Thickness of l.v. coil= 10.5×2=21 mm
Distance between core and l.v. coil = 3.5 mm
Inside diameter of l.v. coil = 225+ ( 2 ×3.5 ) = 232 mm
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Outside diameter of l.v. winding = 232 + ( 2 × 21)
= 274 mm
Mean diameter of l.v. coil = 232+ 21= 253 mm
Mean length of turn of l.v. coil = π ×253= 794.8248 mm
Figure. 2: Layout of L.V. winding(all dimensions are in mm)
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 Design of the layout of H.V. winding:
The distance between l.v. and h.v. = 12 mm
Inside diameter of h.v. = 274+ ( 12 × 2 ) = 298 mm
Now, Split h.v. winding in 4 coils each with turns =548/4=137
The size of conductor = 4 mm diameter.
With paper insulation on conductor, the diameter = (4 +0.25) mm=4.25 mm
Figure. 3: Layout of H.V. winding(all dimensions are in mm)
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Here we choose 7 layers ;
∴Turns per layer = 𝟏37 = 19.57=20
𝟕
& Height of winding in each h.v. coil = 20 ×4.25=85 mm
Thickness of each coil = 7 ×4.25 = 30 mm
Outside diameter of h.v. coil = 298 + ( 2 ×30 ) = 358 mm
Approximate mean diameter of h.v. coil = 298 + 30 = 328 mm
Mean length of turn = 328 × π = 1030.44 mm
Height of h.v. coils in window = ( 85 × 4 ) +8 +8+8 = 365 mm
The space required between coils and core on either side is taken as 26 mm
The height of window required:= 365 + 26 ×2 = 417 mm ; [Which is acceptable ]
 Percentage Reactance:
l.v. mean length of turn = 794.8248 mm
h.v. mean length of turn = 1030.44 mm
average Lmt =
794.8248+ 1030.44=
𝟐
912.63 mm
Ampere Turn =969.6702111 ×20 = 19393.40 =19393 ;
Approximate mean height of coils=𝟑65+𝟑83
𝟐
=
374 mm
Here, a = 12 mm
b1 = width of h.v. 30 mm;
b2 = width of l.v. =21mm
Now,
a1+ 𝒃𝟏+𝒃𝟐
𝟑
=29 mm
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% reactance, X =
−7
𝟐×𝜋×50x4𝜋×10 ×𝟎.91263×19393×0.029
𝟎.𝟑74×12.05
= 0.044 Pu or 4.4 % [its acceptable as its between 3.5% to 4.5%]
 Percentage Resistance:
Here,
ρ20 = 0.01724ohm/mm2/m
α20 =0.00393
At 75°C, ρ75 = ρ20{1+ α20(75-20)}
=0.021 /mm2/m
Resistance of low voltage (l.v.) winding: (per phase)
=
𝟎.𝟎𝟐𝟏×794.8248×20
380×1000
= 8.78× 10−4 Ω (per phase)
Resistance of high voltage (h.v.) winding: (per phase)
=
𝟎.𝟎𝟐𝟏×1030.44×548
14.08 ×1000
= 0.842 Ω (per phase)
So, Ratio of transformation = (6600) ÷ (239.6) = 27.54 =28
Now,
Equivalent resistance referred to h.v. winding (per phase)
R = 0.842 +8.78× 10−4 × (28)2
= 1.53 Ω
percentage resistance, %R =
35.20x1.53
= 0.0081 pu =0.81%
6600
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Here,
Therefore,
Percentage impedance, %Z = √%R2 + %𝑥2
=√4.42 + 0.812
%
=4.47 %
 Weight of Iron in Core and Yoke Assembly:
From Fig 4
The volume of the core and yoke is given by:
= Ai × { (1045× 2) + (655×3) } mm3
= 32190.404 ×4055mm3 =130532088.2
Weight of iron = 7.85 × 1000 kg /m3. For Cu
Weight of core and yoke = (130532088.2× 7.85 ) ÷ ( 1000 × 1000 )
= 1024.68 kg
Core loss at Bmax = 1.686 wb/m2 is 1.5 watts/kg
∴
Core loss in transformer= 1024.68 ×1.5
= 1537.02 watts ;
 Magnetizing Volt Amperes:
For Bmax = 1.686 wb/m2, VA / kg from the curve is 10.5 VA/kg
Magnetizing volt amperes = 1024.68 × 10.5
= 10759.14 VA
 Weight of low voltage winding:
We know, density of copper =8.89 g/cm3
Number of turns T2 = 20
&
a2 = 380mm2
Mean length of turn = 794.8248 mm
Weight of l.v. winding (per limb) = ( 8.89 × 380× 794.8248 ×20 )÷ 106
=53.7 kg
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 Weight of H.V. winding(per limb):
Number of turns T1 = 562; & normal= 548 ;
a1 = 14.08 mm2 ;
Mean length of turn = 1033.706 mm
= ( 8.89× 12.57× 1030.44× 562) ÷ (106) kg
Weight of 4 coils (one limb)
= 64.7kg ; for all turns
For normal turns,
weight of the coils (one limb) =( 8.89× 12.57× 1030.44× 548) ÷ (106) kg
= 63.1 kg
 Total Weight of Copper in Transformer:
∴ 3× ( l.v. + h.v. )
= 3 (53.7+63.1)
=350.4 kg
 Copper Loss and Load Loss at 750 C:
h.v. current per phase Ip= 35.2 A
Copper loss for 3 phases = 3 × I2 × r
= 3× 35.22 × 1.52
=5650.02 watts
Let, stray load loss about 7%,
Then, load loss (at 750C) = 5650.02×1.07
= 6045.52 watts
Iron loss = 1537.02watts .
Therefore ,
Total loss=(6045.52 +1537.02)
= 7582.54 watts = 7.5 KW
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
Calculation of Performance:
 Efficiency on full load at unity power factor :
Output = 697×1000 watts.
Efficiency,
697×1000
𝜂 = (697×1000)+7582.54 × 100%
= 98.92 %
 Efficiency on 3/4th full load at unity power factor:
Core loss = 1537.02 watts;
Load loss on 3/4 load = 6045.52 ×(3/4)2
=3400.605 watts
Total loss = 1537.02+3400.605
= 4937.625 watts
0.75 ×697×1000
Efficiency on 3/4th of full load ,𝜂 = (0.75 ×697×1000)+4937.625
× 100%
= 99.06%
 Efficiency on ½ of full load at unity power factor:
Total loss = ( 1537.02 + 6045.52 ×0.52 )
= 3048.4 watts
0.5 ×697×1000
Efficiency on 1/2 of full load ,𝜂 = (0.5 ×697×1000)+3048.4
× 100
= 99.13%
 Regulation At Unity Power Factor:
% R = 0.81%,
% X = 4.47%
Now,
( V + IR )2 + ( IX )2 = E2
or, ( 1.0 + 0.0081 )2 + (0.0447)2 = E2
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∴ E = 1.00909 V
%Regulation =(1.00909-1)×100%
=0.909%
 𝑹𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝑨𝒕 𝟎. 𝟖 𝑷𝒐𝒘𝒆𝒓 𝑭𝒂𝒄𝒕𝒐𝒓 ∶
%Regulation = [IR cos φ + IX sin φ] %
= [0.81 ×0.8+4.47 ×0.6] %
=3.33%
 Core loss current, magnetizing current, no-load current :
Core loss = 1537.02 watts.
core loss current, Ic = (1537.02) ÷ ( 3 × 6.6 × 103 )
= 0.0776 A
Magnetizing VA = 10759.14;
magnetizing current, Im = (10759.14) ÷ (3 ×6.6×103)
= 0.543 A
No load current per phase , Io = √(0.07762 + 0.5432)
= 0.549 A
Current per phase = 35.2 A
No load current = (0.549÷35.20)×100%
= 1.56% ; of full load current
 Design of Tank :
Fig: 4 shows the spacing of outside diameters of h.v. coils on the cores.
Outside diameter of h.v. = 358 mm
The distance between coils on adjacent limbs =415-358=55mm(approximate);
Clearance at each end is 38 mm.
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Thus, the length of the tank = 358×3+ 2 ×55 + 2 × 38 = 1260mm
Breadth of the tank = 358+ 55 × 2 mm
= 468 mm;
Choosing 470 mm
Height = 1085 + 50 mm for base + 250 oil level above core +250mm for leads;
= 1385 mm up to oil level + 250 mm for leads ;
=1635 mm
Inside dimensions of the tank of the transformer = (length × breadth ×height)
= (1260× 470 × 1635) mm3
Figure. 4: Tank dimensions (all dimensions are in mm).
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 Temperature Rise :
Now, for dissipation of heat, only 4 surfaces of a tank are taken into consideration.
The top and bottom are not considered.
1635 1260
×
× 2 = 4.121 𝑚2
1000 1000
The surface of the tank =
1635 470
×
× 2 = 1.5369 𝑚2
1000 1000
1635 1260
×
× 2 = 4.121 𝑚2
1000 1000
Total = 5.6579 m2
Full load loss to be dissipated = 7582.54 watts
Now, If 12.5 per m2 per 0C temperature rise is taken as dissipation due to convention and
radiation,
The temperature rise = ( 7582.54) ÷( 12.5 × 5.6579 ) = 107.210C
Now, to maintain the temperature of transformer walls limited to 600C; Then
temperature rise of the oil will be 500 Cand of coils 550C. In that case, the surface
of the tank for cooling has to be increased either by “radiators” or “tubes”
attached to the tank.
If the total surface area is considered, ‘x’ times the tank surface area, we get:
( 5.6579 ) × (x) × (8.8 + 3.7/ x ) 60=7582.54
∴ x = 2.118
Thus,
Additional area to be provided = 5.6579 ×(2.118-1)= 6.326 m2
As, 1385 mm is height up to oil level;
Height of tube is taken as 1000 mm
Surface of 1 tube of 50 mm diameter = π ×50×1000 × l0-6
=0.157 m2
Number of tubes required
=
6.326
0.157
= 41 ; approximately
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 Volume and weight of coil :
Volume of tank up to oil level of 1385 mm
=( 1260 /1000 ) × ( 470 /1000 ) × ( 1635/1000 )
=0.9684 m3
 Volume of transformer core and copper
:
={350.4÷( 8.89 × 103 )} + {(1024.68) ÷ (7.85 × 1000 ) }
=0.17
Volume of oil =(0.9684-0.17)
= 0.7984 m3
Oil required in transformer = 0.7984 ×1000L
=798.4 liters .
Therefore, weight of oil required
= (798.4 × 0.89) kg
= 710.576
kg
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 Weight of the tank :
If the thickness of the tank walls is taken as 5 mm
Weight of tank = 0.005 × [ {(1260/1000 ) × ( 470/1000 ) × 2 }+{( 1635/1000 ) ×( 1260/1000)
×2} + {( 1635/1000 ) × ( 470/1000 ) × 2} ] ×1000 × 7.85
= 268.57 kg
 Volume and Weight of tank :
41 tubes each of 50 mm diameter and 1 m length.
Therefore, Volume
=(π/4) ×(50/1000)2×1×41
= 0.0805 m3,
Volume of oil in tubes = 0.0805 × 1000 = 80.5 liters.
Weight of oil in tubes = 80.5 × 0.89 = 71.64 kg
Weight of tube
= π × (50/1000) × 1 × 0.005 × 41 × 7.85 ×1000 kg
= 252.78 kg
 Total weight of core and Yoke:
Weight of core and yoke assembly = 1024.68 kg
Weight of copper in windings
= 350.4 kg
Weight of tank
= 268.57 kg
Weight of tubes
=252.78 kg
Weight of oil in tank and tubes
= 710.576kg
= 71.64 kg
Total weight= 2678.646 kg
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 Summary:
Specifications:
kVA 697; Volts H.V. 6600 volts; Volts L.V. (no load) 415 volts; Amperes H.V. 35.2 A ;
L.V. 969.67 A (line values); 3 phase, delta/star 50 c/s; temperature rise of oil 107.210C ;
type of cooling ON ; Vector group; percentage impedance 4.47%
±2.5%and ±5% tapings on h.v. side.
 Core and Yoke :
Material: CRGO (cold rolled grain oriented) steel laminations 0.35 mm thick; mitered core
construction 45° cut.
c
Flux density Bmax=1.686 Wb/m2; Net area of cross-section of core, 32190 mm2;
circumscribing circle diameter 225 mm.
Size of core, yoke and frame :width of the window, 200 mm; height of window=595 mm;
Weight of core and yoke assembly 1024.68 kg; core loss at Bmax = 1.686 wb / m2, 1.5 watts
per kg magnetizing VA = 10.5 VA / kg
Windings
L.V
H.V
Type of winding
Helical
Disc
Current density
2.5
2.5 a/mm2
Cross sectional area
ofconductor
380mm2
Conductor : Copper
2 strips of
14.08 mm2
4 mm
5×38 mm
Number Of Layers
perlimb
2
4 discs
Number of turns
20
548 normal
Number of turns per layer
10
137
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Height of winding
inwindow
383 mm
365 mm
Thickness of coil
21 mm
30 mm
Inside diameter of coil
232 mm
298 mm
274 mm
358 mm
Mean length of turn
794.82 mm
1030.44 mm
Resistance at 750c
0.000878 ohms
0.842 ohms
53.7 Kg
63.1 Kg
Outside diameter of coil
Weight of copper for winding
per limb
Total Weight Of Copper
350.4
Kg
 Insulation :
Insulation between core and l.v. winding: pressboard paper
Insulation for conductors: paper
Insulation between layers: Crape paper
Insulation between l.v. and h.v. windings: bakelite paper cylinder; Laminated pressedwood
sticks for spacers for cooling.
Class A insulation for O N-type transformers.Tank:
Temperature rise of oil 107.210C
Inside dimensions of tank : length 1260 mm; breadth :470 mm; height 1635 mm
Tubes 41, each of 50 mm diam; 1000 mm long.
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Oil in the transformer tank
798.4 liters
Oil in tubes
80.5 liters
Weight of oil in the tank
710 kg
Weight of oil in tubes
71.64 kg
Weight of tank
268.57 kg
Weight of tubes
weight of complete transformer
252.78 kg
2678.646 kg
Performance
Percentage resistance
0.81%
Percentage reactance
4.4%
Percentage impedance
4.47%
Iron loss
1537.02
watts
Copper and stray load loss, i.e. load loss at 750C
6045.52 watts
Total loss on full load
7582.54watts
Efficiency on full load at a unity power factor
98.92%
Efficiency on 3/4 the full load at a unity power factor
99.06%
Efficiency on 1/2 full load at a unity power factor
99.13%
Regulation on full load at a unity power factor
0.909%
Regulation on full load at 0.8 power factor lagging
3.33%
Core loss current per phase
0.0776 A
Magnetizing current per phase
0.543 A
No load current per phase
0.549 A
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 Tapings :
562
575
548
534
521
2.5%
5%
Normal
-2.5%
-5%tappings
 Reference : (From internet source and specific books.)
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