Saturday, December 26, 2020

Important Formula of Design of Transformer - 1

 

EMF induced in the transformer winding

E = 4.44fΦmN

Where

E = Induced emf

F = Frequency

Φm = Maximum flux density

N = Number of turns

 

Volts per turn in the transformer

ES / EP = 4.44fΦmNS / 4.44fΦmNP

ES / EP = NS / NP

ES / NS = EP / NP

Volts per turn in the primary winding = Volts per turn in the secondary winding

 

The minimum winding diameter Dm of continuously transposed conductor ( CTC )

Dm = ( Pitch of transposition ) × ( Number of conductors ) / π

Where pitch of transposition is distance between adjacent cross overs measured on one side of conductors and it is normally 15 times strip width.

 

Temperature rise

θ = QC / S

Where

Q = Losses

S = Heat dissipating area

C = Cooling co – efficient

 

Current density

δ = I / a

Where δ = Current density

I = Current

a = Area

 

Conductor area in window AC

= 2AT / δ

Where AT = Ampere turns of windings

                  = IPNP + ISNS

               δ = Current density

 

Window space factor

Kw = Conductor area in window / Total conductor area of window

     = AC / AW

 

kVA rating of single phase transformer

Q = 2.22fBmax δ KwAwAi × 10 – 3 kVA

Where

f = Frequency

Bm = Maximum flux density

δ= Current density

Kw = Window space factor

Aw = Window area

Ai = Net core area

 

Rating of three phase transformer

Q = 3.33fBmax δ KwAwAi × 10 – 3 kVA

kVA rating of one phase

Q = 4.44fΦmNPIP × 10 – 3

    = 4.44fΦm ( AT ) × 10 – 3

Where Φm /  ( AT ) is constant for given transformer

Voltage per turns Et = k √ Q

Where k is constant = √ ( 4.44fr × 103 )

Where r = Φm /  ( AT ) Constant

 

Copper loss per meter3

= I2R

= I2 ( ρL / a )

= ρ ( I2 L / a )

= ρ ( a2 δ2 L / a )    ( Current density δ = I / a )

= ρ δ2 ( aL )

= ρδ2

Where δ = Current density and

           ρ = Resistivity of copper

Copper loss per meter3 = ρ δ2

 

Specific iron loss is equal to iron loss per kg

 

Ratio of iron loss to copper loss

Wi / Wc = Pi Gi / Pc Gc

Where

Pi = Iron loss per kg

Pc = Copper loss per kg

Gi = Weight of iron

Gc = Weight of copper

 

Condition for minimum cost of transformer

Weight of iron = Weight of copper

Gi = Gc

 

Condition for minimum losses ( Maximum efficiency ) in the transformer

Iron losses = Copper losses

Wi = x2 Wc

Where x = load at maximum efficiency occurs

                    OR

Wi / Wc = PiGi / PcGc

x2 = PiGi / PcGc

Gi / Gc = x2 ( Pc / Pi )

It is a condition for maximum efficiency in the transformer.

 

You may also like :

Switching surges MCQs – 1

Switching surges MCQs – 2

Switching surges MCQs – 3

Switching surges MCQs – 4

Switching surges MCQs – 5

Switching surges MCQs – 6


No comments:

Post a Comment

Competition Exam Papers

MCQs on Electrical Topics

Current Affairs