EEEG215 Lec3 Transformer Principles

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working principle of transformer

Transcript of EEEG215 Lec3 Transformer Principles

Principle of Transformer Action

PHY2049: Chapter 31TransformersPurpose: to change alternating (AC) voltage to a bigger (or smaller) value

input AC voltagein the primaryproduces a fluxchanging fluxin secondaryinduces emf

Principle of Transformer Action

Principle of Transformer ActionPrincipal of Transformer ActionPrinciple of electromagnetic induction.Ideal to inding transformer inding resistances are negligible Fluxes confined to magnetic core Core lose negligible Core has constant permeabilityV1 I1 MMF = N1IeCore flux follos, Ie very closely.Ie & sinusoidal =max sint

Principle of Transformer ActionTransformersNothing comes for free, however!Increase in voltage comes at the cost of current.Output power cannot exceed input power!power in = power out

Transformers: Sample Problem A transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120 V and the output current is 15.0 A. What is the output voltage and input current?

step-up transformer

Ideal Transformer on an inductive load

The exciting current leads the flux by hysteretic angle,

Transformer on LOAD

Equivalent circuit referred to the LT side of a 250/2500 single phase transformer is shown in fig. The load impedance connected to HT is 380+j230. For a primary voltage of 250V, computethe secondary terminal voltageprimary current and power factorPower output and efficiency

Equivalent circuit referred to the LT side of a 250/2500 single phase transformer is shown in fig. The load impedance connected to HT is 380+j230. For a primary voltage of 250V, compute

Z'L = (380+j230) (N1 / N2)2 = (380+j230) (250/2500)2 = 3.8+j2.3Total impedance in the primary

Secondary terminal voltage = I2ZL

the secondary terminal voltageprimary current and power factorPower output and efficiency

Im= V1/jXm = 2500/25090 =1-90 =0-j1I'e = Ic + Im = 0.5+ (0-j1) = 0.5-j1I'1= I'1 +Ie = 40- j30+0.5- j1= 51-37.4 b) Primary current I1 = 51A Primary p.f = cos1 = cos37.4 = 0.794 lagging(c) Load p.fcos2 = 380/ (3802+2302 )= 0.855Power Output = V2I2cos2 = 2220*5*0.855 = 9500 WattsPower Output = I'12RL = 502*3.8 = 9500 WattCore Loss ,PC= v12 / RC = Ic2 RC = 0.52*0.2 =500 WattsPower Input = V1I1cos1 = 250*51*0.794 = 10123.5