Transformer Design

Transformer Design:

• Power rating [MVA]

•Core

• Rated voltages (HV, LV)

• Insulation coordination

• Loss evaluation

• Temperature rise limits, Temperature limits

•Cooling, cooling method

Transformer Design

Question: Design a 2500 kVA, 6.6/66 kV, 50Hz, 3 – phase Delta/star core type oil immersed natural cooled power transformer. Maximum temperature rises not to exceed 45ºC.

Design of core:

We have,

E t = k Q

= 0 . 6 2500

= 30 V / turns

Then for 3 – phase transformer,

E t = 4 . 44 f Φ m

Φ m = 30 4 . 44 × 50 = 0 . 1351 Wb

Using Hot Rolled Silicon steel (HRS),

B m = 1 . 3 Wb / m 2

for core type power transformer.

So,

B m = Φ m A i

A i = 0 . 1351 1 . 3 = 0 . 10395 m 2

Using 3 – stepped core,

A i = 0 . 6 d 2

d = 0 . 10395 0 . 6 = 0 . 4162 m

∴ a = 0 . 9 d = 0 . 9 × 0 . 4162 = 0 . 3746 m

∴ b = 0 . 7 d = 0 . 29136 m

∴ c = 0 . 42 d = 0 . 1748 m

Design of Window:

For 3 – phase power transformer, we have,

Q = 3 . 33 f Φ m K w A w δ × 10 - 3 kVA

………. (i)

Then for natural cooled power transformer,

δ = 2 . 3 A / mm 2

So, equation (i) becomes,

2500 = 3 . 33 × 50 × 0 . 1351 × 0 . 125 × A w × 2 . 3 × 10 6 × 10 - 3

∴ A w = 0 . 3865 m 2

And let us take,

H w W w = 2

So,

A w = H w × W w = 2 W w 2

W w = 0 . 3865 2

∴ W w = 0 . 4396 m

∴ H w = 0 . 87920 m

Design of Yoke :

Depth of yoke (Dy)

= a = 0 . 3746 m

And,

A y = 1 . 2 A gi

(for HRS)

Also,

A gi = A i K i = 0 . 10395 0 . 9 = 0 . 1155 m 2

A y = 1 . 2 × 0 . 1155

H y × 0 . 3746 = 0 . 1386

∴ H y = 0 . 3699 m

Overall Dimension of Frame:

D = W w + d = 0 . 4396 + 0 . 4162 = 0 . 8558 m

W = 2 D + a = 2 × 0 . 8558 + 0 . 3746 = 2 . 0862 m

H = H w + 2 H y = 0 . 87920 + 2 × 0 . 3699 = 1 . 619 m

Design of Low Voltage (LV) Winding:

We have,

T p = V p E t

(6.6/66kV delta/ star)

T p = 6 . 6 30 = 220 turns

And,

I p = Q 3 × V p = 2500 × 10 3 3 × 6600 = 126 . 26 A

So,

ρ = I p a p

a p = 126 . 26 2 . 3 = 54 . 896 mm 2
a p = π 4 × d p 2
d p = 4 π × 54 . 896 = 8 . 36 mm

∴ d p ' = d p + 2 × 0 . 025 = 8 . 41 mm

Now,

Maximum number of turns in 1 layer

= H w d p '

= 0 . 87920 8 . 41 × 10 - 3 ≈ 105 turns

And, minimum number of layer

= 220 105 = 2 . 09 ≈ 2 layers .

Number of turns per layer
= 220 2 = 110 = 110 × 2 turns / layer

So, taking 2 layers of 110 turns

Now,

Radial depth of LV winding

b p

=No.of layers

× dp '

+(No.of layer-1)

× thickness of layer

Radial depth of LV winding

b p = 2 × 8 . 41 + 0 . 5 × 2 - 1 = 17 . 32 mm

Also,

Axial height of LV winding

L cp = No . of turns × diameter of insulated conduct or

Axial height of LV winding

L cp = 110 × 8 . 41 = 925 . 1 mm

Then,

Insulation thickness

y = 5 + 0 . 9 × kV

y = 5 + 0 . 9 × 6 . 6 - 0 = 10 . 94 mm

Inner diameter of LV winding

= d + 2 y

= 0 . 4162 + 2 × 10 . 94 × 10 - 3 = 0 . 43808 m

Outer diameter of LV winding=

0 . 43808 + 2 × 0 . 01732 = 0 . 47272 m i . e Inner daimeter + 2 bp

∴ L mtp = π × 0 . 4727 + 0 . 43808 2 = 1 . 43 m

Design of High Voltage (HV) Winding:

We have,

T s = V s - ph V p - ph × T p = 66000 3 × 6600 × 220 ≈ 1270 turns

I s = Q 3 × V s = 2500 × 10 3 3 × 66000 3 = 21 . 86 A

So,

a s = I s ρ = 9 . 50 mm 2

d s = 4 π × 9 . 50 = 3 . 479 mm

∴ d s ' = ds + 2 × 0 . 025 = 3 . 529 mm

Then, maximum number of turns per layer

= H w d s ' = 0 . 87920 3 . 529 × 10 - 3 ≈ 249 turns

Minimum number of layers

= 1270 249 ≈ 5 layers

So, number of turns per layer

= 1270 5 = 254 = 254 × 2 + 255 × 1 + 253 × 1

We shall take 2 layers of 254 turns and 1 layer of 255&253 turns.

Now,

Radial depth of HV winding

b s

=No.of layers

× dp '

+(No.of layer-1)

× thic kness of layer

Radial depth of HV winding

b s = 5 × 3 . 529 + 5 - 1 × 0 . 5 = 19 . 645 mm

Also,

Axial height of HV winding

L cs = No . of turns × diameter of insulated conductor

Axial height

L cs = 254 × 3 . 529 = 896 . 3 mm

Insulation thickness

( a ) = 5 + 66 × 0 . 9 = 64 . 4 mm

So, Inner diameter

= 0 . 4162 + 64 . 4 × 2 × 10 - 3 = 0 . 545 m i . e ( d + 2 × a + 2 × bs )

And, outer diameter

= 0 . 545 + 2 × 0 . 019645 = 0 . 58429 m i . e ( Inner daim of HV + 2 × bs )

∴ L mts = π × 0 . 58429 + 0 . 545 2 = 1 . 7738 m

Then,

L mt = L mtp + L mts 2 = 1 . 7738 + 1 . 43 2 = 1 . 60 m

Also,

L c = L cp + L cs 2 = 0 . 9251 + 0 . 8963 2 = 0 . 9107 m

So,

X p = 2 π f μ 0 T p 2 × L mt L c × a + b p + b s 3

= 2 π × 50 × μ 0 × 220 2 × 1 . 60 0 . 9107 × 64 . 4 × 10 - 3 + 0 . 01964 + 0 . 01732 3

∴ X p = 2 . 5 Ω

Per unit reactance

= I p X p V p = 126 . 26 × 2 . 52 6600

∴ X pu = 0 . 0492 pu

Also,

Resistance of Primary winding

r p = ρ × L mtp a p × T p

= 0 . 021 × 10 - 6 × 1 . 43 × 220 54 . 896 × 10 - 6 = 0 . 1203 Ω

Resistance of secondary winding

r s = 0 . 021 × 10 - 6 × 1 . 7738 × 1270 9 . 50 × 10 - 6 = 4 . 979 Ω

Then,

Total resistance as referred to LV

R p = r p + r s × T p T s 2

= 0 . 1203 + 4 . 979 × 220 1270 2

∴ R p = 0 . 2697 Ω

Per unit resistance (

R pu ) = I p × R p V p = 126 . 26 × 0 . 2697 6600

∴ R pu = 0 . 005159 pu

V.R=

R pu cos Φ + X pu sin Φ

=0.005159

× 0 . 8 + 0 . 0492 × 0 . 6

=3.33%

Losses calculation:

Copper loss

Copper losses=3

× I p 2 × R p =

× 126 . 26 2 × 0 . 2697 = 12898 . 33 W

Stray loss=15%of copper loss =1934.74W

Total copper loss=copper loss+ stray loss=12898.33+1934.74=14833.0795W

Iron loss

Density of lamination (

ρ ) = 7 . 6 × 10 3 Kg / m 3

Weight of 3 limbs (core) =3

× ρ × v

:( v=

H w × A i )

× 7 . 6 × 10 3 × 0 . 10395 × 0 . 87920

=2083.756Kg

Weight of 2 yoke=2

× ρ × v

× ρ × A y × w ) = 2 × 7 . 6 × 10 3 × 0 . 1386 × 2 . 0862

=4395.039Kg

Flux density in limb=1.3wb/

m 2

Specific core loss=2.5w/Kg

Iron loss in limb=specific core loss

× weight of 3 limbs

=2.5

× 2083 . 756 = 5209 . 39 W

Since the area of yoke is taken 20% more than that of core, value of flux density is less than that of core.

i.e flux density=1.3-(0.2

1.3)=1.04

Flux density in yoke=1wb/

m 2

Specific core loss=1.2w/Kg

Iron loss in yoke=specific core loss

× weight of 2 yoke

=1.2

× 4395 . 039 = 5274 . 046 W

Total iron loss=

5209 . 39 W

+5274.046=10483.436W

Efficiency=

Output output loss × 100 %

2500000 × 0 . 8 2500000 × 0 . 8 + 10483 . 436 + 14833 . 0795

=98.74%

Condition for maximum efficiency (ŋ)

p i = ŋ 2

× p c

ŋ =

10483 . 436 14833 . 0795 = 0 . 84

Design of Tank:

Here,

Height of frame (H) = 1.619 m

Height of yoke (Hy) = 0.369 m

Distance between two limbs (D) = 0.8558 m

Allowing 50 mm for base and 150 mm for oil,

Height level for oil (Ho) = H + (50 + 150) mm

= 1.619 m + 200 mm

= 1.619 +0.2 = 1.819 m.

Outer diameter of HV winding (De) = 0.5842 m

Width of tank (Wt) = D + De + 2b = 0.8558 + 0.5842 + 2 × 85

× 10 - 3

= 1.61 m

Length of tank (Lt) = De +2l= 0.5842+ 2 × 125

× 10 - 3

= 0.8342 m

Height of tank (Ht) = Ho +h = 1.819+ 550

× 10 - 3

= 2.369 m

Surface area of tank (st) = 2Ht (Lt + Wt) = 2 × 2.369(0.8342+1.61) = 11.58

m 2

Let the area be

xs t .

Then,

Total heat dissipating area= (1+x)

s t .

And, specific loss dissipation (⋋) =

12 . 5 + 8 . 8 x 1 + x

Here, temperature rise (Ө) =

total losses 12 . 5 s t .

25316 . 5155 12 . 5 × 11 . 58

=174.89 ºC

Here the temperature rise in the tank is 174.89 º C exceeding maximum temperature rises. So tubes are needed in tank.

Now, temperature rise, (Ө) =

tot al losses ( 12 . 5 + 8 . 8 x ) s t .

Or, 45=

25316 . 5155 12 . 5 + 8 . 8 x 11 . 58

x=4.10

Then, tube area=

xs t . = 4 . 10 × 11 . 58 = 47 . 48 m 2

Taking length of tube ( lt) =2.1 m

Diameter of tube (dt) =50 mm=0.05m

Area of each tube=

A = π × lt

dt=

π × 0 . 05 × 2 . 1 = 0 . 33 m 2

Then,

Number of tubes required=

total tube area area of each tubes = 47 . 48 0 . 33 = 140 tubes

Design sheet

Ratings of transformer:

100 kVA
Three phase
Frequency-50HZ
6.6/66kV
Core type
Distribution transformer

Design of core, window, yoke &frame:

S.N	Design parameters	Notations	Dimensions
1.	Design of core Output constant	K	0.6
2.	Voltage per turns	E t	30v/turn
3.	Flux density	B m	1.3Wb/m^2
4.	Net iron area	A i	0.10395m^2
5.	Flux	Φ m	0.1351Wb
6.	Core dimensions	a b c	0.3746m 0.29136m 0.1748m
1.	Design of window Window space factor	K w	0.125
2.	Current density	Δ	2.3A/mm^2
3.	Window area	A y	0.3865m^2
4.	Height of window	H w	0.87920m
5.	Width of window	W w	0.4396m
1.	Design of yoke Depth of yoke	D y	0.3746m
2.	Height of yoke	H y	0.3699m
3.	Net yoke area	A y	0.1386m^2
4.	Flux density	B m	1.3Wb/mm^2
1.	Overall dimension of frame Height of frame	H	1.619m
2.	Distance between centre of two limbs	D	0.8558m
3.	Width of limbs	W	2.0862m
1.	Distance between core &LV winding	Y	10.94mm
2.	Distance between L.V&HV winding	A	64.4mm

Design of LV and HV winding

S.N	Windings parameters	LV winding	HV winding
1.	Turns per phase(Ts,Tp)	220turns	1270turns
2.	Current(Is,Ip)	126.26A	21.86A
3.	Current density	2.3A/mm^2	2.3A/mm^2
4.	Area(As,Ap)	54.896mm^2	9.50mm^2
5.	Diameter of conductor(Ds,Dp)	8.36mm	3.479mm
6.	Diameter of insulated conductor(d)’	8.41mm	3.529mm
7.	Maximum no.of turns in layer	105turns	249turns
8.	Minimum no of layers	2layers	5layers
9.	Total no.of turns in 1 layer	110	254
10.	Radial depth(bs,bp)	0.01732m	0.019645m
11.	Axial depth(Lcs,Lcp)	0.9251m	0.8963m
12.	Inner daimeter	0.43808m	0.545m
13.	Outer diameter	0.4727m	0.58429m
14.	Mean diameter	0.4553m	0.5646m
15.	Length of mean turn(Lmts,Lmtp)	1.43m	1.7738m

Calculations of per units values& voltage regulations:

S.N	Parameters	Notations	dimensions
1.	Length of mean turn	Lmt	1.60m
2.	Axial depth	Lc	0.9107m
3.	Resistivity	ρ	0.021*10^-6 Ωm
4.	Resistance at primary side	Rp	0.1203 Ω
5.	Resistance at secondary side	Rs	4.979 Ω
6.	Total resistance referred to primary side	Rp	0.2697 Ω
7.	Per unit resistance	Rpu	5.159*10^-3 Ω
8.	Reactance	Xp	2.5 Ω
9.	Per unit reactance	Xpu	0.0492pu
10.	Voltage regulation	V.R	3.33%

Efficiency calculation:

S.N	Design parameters		Dimensions
1.	Weight of 3 limbs		2083.756kg
2.	Flux density in limbs		1.3wb/ m 2
3.	Specific core loss of limbs		2.5w/Kg
4.	Iron loss in 2 limbs		5209.39W
5.	Weight of 2 yokes		4395.039kg
6.	Flux density in yoke		1wb/ m 2
7.	Specific core loss of yoke		1.2w/Kg
8.	Iron loss in yoke		5274.046W
9.	Total iron loss		10483.436W
10.	Copper loss		12898.33W
11.	Stray loss		1934.74W
12.	Total copper loss		14833.0795W
13.	Total losses		25316.515W
14.	Efficiency	ŋ	98.74%
15.	Condition for maximum efficiency		0.84

Design of tank:

S.N	Design parameters	Notations	Dimensions
1.	Height level for oil	Ho	1.819 m.
2.	Width of tank	Wt	1.61m
3.	Length of tank	Lt	0.8342m
4.	Height of tank	Ht	2.369m
5.	Surface area of tank	St	11.58 m 2
6.	Tube area	Xst	47.48 m 2
7.	Length of tube	Lt	2.1m
8.	Diameter of tube	Dt	0.025m
9.	Area of each tube	At	0.17 m 2
10.	Number of tubes required	N	279

$C:\Users\User\Desktop\IMG_20171218_190255.jpg$

Voltage(kV)	Rating kVA	clearance
> to 33kV	Less than 1000	b	l	h
	Less than 1000	75	100	550
	1000-5000	85	125	550

Figures:

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Fig: overall dimension of frame

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Fig: operating characteristics of design

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Pic sources:Three-phase Current Transformers In Industrial Warehouse Stock …

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