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TRANSFORMER. . C. i p. N p. v p. P.

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C

ip

Np

vp

P

A transformer consists of a core made of laminated iron separated by insulators and a coil of Np turns wound around the core. This coil is supplied with an a.c voltage supply vp which then produces a current ip. Due to this current , a flux F is produced which is given by an equation

- = Npip/S…….(1)
where S is the reluctance

Since current varied with time , F also varied with time. A back electromagnetic force (e.m.f) will be produced which is given by the equation.

……(2)

Substitute = Npip/S into the above equation , then

……(3)

If ip is sinusoidal, the flux produced also sinusoidal, i.e

= msin 2ft

……(4)

therefore

vp = NP2fmcos 2ft= NP2fmsin (2ft + /2)

……(5)

The peak value = Vpm = NP2fm

and vp is leading the flux by p/2.

……(6)

The rms value

……(7)

primary

C

is

ip

Np

Ns

vs

Load

vp

P

S

When another coil is wound on the other side of the core with no of turns Ns , then the fux will induce the e.m.f vS as given by

……(8)

…….(9)

…….(10)

Or in rms value

With load , is will flow in the load, mmf at load will equal to the mmf at input, then (in rms value)

NpIp = NsIs

…….(11)

rearrange

…….(12)

Secondary

NP : NS

VP

VS

Symbol for ideal transformer

For ideal transformer, the energy transferred will be the same as input. Thus power at primary is same power at secondary.

Pp = Ps

or IpVp = IsVs

A 250 kVA,11000V/400V, 50Hz single –phase transformer has 80 turns on the secondary. Calculate

(a) The appropriate values of the primary and secondary currents;

(b) The approximate number of primary turns;

(c) the maximum value of the flux.

(a) Full-load primary current

Full-load secondary current

IO

NP

EP

VP

NS

VS

Ideal transformer with no load

Io is the no load current when the secondary is open circuit. This current consists of Iomthat isrequired to produce the flux in the core (it is in phase) and Io1 is to compensate the hysteresis and eddy current losses.

fo

EPVP

IO

IOI

IOm

VS

Phasor diagram for

no load transfomer

VP= emf of supply to the primary coil and 90o leads the flux.

EP=emf induced in the primary coil and same phase as VP.

VS=emf induced in the secondary coil and 90o lags the flux.

Iom=magnetizing current to produce flux and it is in phase with flux.

Io1=current to compensate the losses due to hysteresis and eddy current.

Io=the no load current and given by

Power factor

Highvoltageline

Transformer 2

Low voltage generator

Low voltage load

Application of transformer

Transformer converts the energy to high electrical voltage and transmits in the high voltage line. At the load, the high voltage energy is converted to low voltage. In this way, it will compensate the losses during transferring of the voltage energy.

A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in core is 1.8m and the joints are equivalent to the airgap of 0.1mm. The value of the magnetic field strength for 1.1 T in the core is 400A/m, the corresponding core loss is 1.7W/kg at 50Hz and the density of the core is 7800kg/m3.

If the maximum value of the flux is to be 1.1T when a p.d of 2200V at 50Hz is applied to the primary, calculate:

(a) the cross-sectional area of the core;

(b) the secondary voltage on no load;

(c) the primary current and power factor on no load

magnetomotive force (mmf) for the core is

mmf for the airgap is

Total mmf is

recall

Maximum magnetizing current

Rms value

mass of core

Core loss= loss rate x mass

Core-loss component of current

No load current

Power factor

‘

I1

I2

E1

V1

V2

L(2)

Loaded transformer

Loaded transformer

L(q2)= load with power factor of cos q2

V1 = emf at supply

E1=induced voltage at primary

V2=emf at load

E2=induced voltage at secondary

I1= primary current

I2=secondary current

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a power factor 0.2 lagging when secondary current is 280A at a power factor of 0.8 lagging. Calculate the primary current and the power factor. Assume the voltage drop in the windings to be negligible.

Recall Equation 12

therefore

I1

R1

L1

I1’

L2

R2

RC

Lm

V2

E1

V2’

Equivalent circuit of transformer

Leakage fluxes

Path of leakage

Flux leakage is due to secondary current that produce flux which encounter the primary flux. Some of the flux will link to its own windings and produce induction. This is represented by inductance L1. Similarly with the flux in secondary and represented by L2.

E1

I1

R1

L1

I1’

L2

R2

RC

Lm

V2

V1

E2

Equivalent circuit of transformer

- There are four main losses
- Dissipated power by wire resistance of the windings (I2R)
- Power due to hysteresis
- Power due to eddy current
- Power via flux leakages.

R1= wire resistance of primary windings

L1=inductance due to leakage flux in primary windings

RC=resistance represent power loss due to in hysteresis and eddy current

Lm= inductance due to magnetizing current Iom

L2=inductance due to leakage flux in secondary windings

R2=wire resistance of secondary windings

V1

-I2’

I1

E1

f1

I1Z1

I1R1

I1X1

I0

I2R2

I2X2

f2

I2Z2

I2

V2

E2

Phasor diagram for a transformer on load

Simplified/approximate equivalent circuit

We can replace R2 by inserting R2’ in the primary thus the equivalent resistance is

Giving us

Similarly

Voltage regulation of a transformer

recall

Secondary voltage on no-load

V2 is a secondary terminal voltage on full load

Substitute we have

- A 100kVA transformer has 400 turns on the primary and 800 turns on the secondary. The primary and secondary resistances are 0.3W and 0.01W respectively, and the corresponding leakage reactances are 1.1W and 0.035W respectively. The supply voltage is 2200V. Calculate:
- The equivalent impedance referred to the primary circuit;
- The voltage regulation and the secondary terminal voltage for full load having a power factor of (i) 0.8 lagging and (ii) 0.8 leading.
- The percentage resistance and leakage reactance drops of the transformer

(b) (i)

Sec. terminal voltage on no-load

The decreasing of full-load voltage is

Therefore the secondary full-load voltage

(b) (ii) power factor 0.8 leading

The increasing of full-load voltage is

Therefore the secondary full-load voltage

Per unit

Recall Equation 12

Full load secondary current

Equivalent resistance referred to secondary

Secondary voltage on no-load

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