# Transformer Design

 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

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

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,

b s

=No.of​​ layers

× dp '

+(No.of​​ layer-1)

× thic kness of layer

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 =

3

× 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 )

​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ =3

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

​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ =2083.756Kg

Weight​​ of​​ 2​​ yoke=2

×  ρ × v

​​ ​​ ​​ ​​ ​​​​ =2

×  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 abc 0.3746m0.29136m0.1748m 1. Design​​ of​​ windowWindow​​ 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​​ yokeDepth​​ 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 ​​​​ frameHeight​​ 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

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

Figures:

​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ Fig:​​ overall​​ dimension​​ of​​ frame

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