TRANSFORMER

where S is the reluctance  C vp ip P Np 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

Substitute  = Npip/S into the above equation , then 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  = m sin 2ft ……(4) therefore vp = NP2fmcos 2ft = NP2fmsin (2ft + /2) ……(5) The peak value = Vpm = NP2fm and vp is leading the flux by p/2. ……(6) The rms value ……(7)

 C vp ip vs is Load P S Ns Np primary secondary 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)

From (2) and (8) we get Or in rms value …….(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)

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 NP : NS VP VS Primary Secondary Symbol for ideal transformer

Example 1 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

(b) Number of primary turns recall (c) Maximum flux recall

Ideal transformer with no load EP VP VS NS NP IO Ideal transformer with no load Io is the no load current when the secondary is open circuit. This current consists of Iom that is required to produce the flux in the core (it is in phase) and Io1 is to compensate the hysteresis and eddy current losses.

VP= emf of supply to the primary coil and 90o leads the flux. Phasor diagram for no load transfomer  IOm IOI EPVP IO VS fo 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

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. Transformer 1 Transformer 2 Low voltage generator Low voltage load High voltage line

Example 2 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

(a) recall recall Practically 10% more allow for insulator (b) recall

(c) magnetomotive force (mmf) for the core is mmf for the airgap is Total mmf is recall Maximum magnetizing current Rms value

Volume of core mass of core Core loss= loss rate x mass Core-loss component of current No load current Power factor

Loaded transformer E1 V1 V2 I1 I2 L(2) Loaded transformer Phasor diagram ‘ 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

Example 3 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

Solve for horizontal and vertical components ‘ Solve for horizontal and vertical components Power factor

Equivalent circuit of transformer Leakage fluxes V2’ R1 L1 L2 R2 RC Lm V2 E1 I1’ I1 Equivalent circuit of transformer 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.

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. E2 R1 L1 L2 R2 RC Lm V2 V1 I1’ I1 Equivalent circuit of transformer E1 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

Phasor diagram for a transformer on load V1 E1 I2 -I2’ I1X1 V2 E2 I1 I0 I1R1 I1Z1 I2X2 I2Z2 I2R2 f1 f2 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

Transformer simplified circuit then Ze V1 E1=V2’ I1 I2 To load E2=V2 then where and

Magnified Ze portion Complete Phasor diagram of simplified equivalent circuit of transformer Complete fo-f2 Magnified Ze portion

Voltage regulation of a transformer recall Secondary voltage on no-load V2 is a secondary terminal voltage on full load Substitute we have

Or From phasor diagram can be proved that Or

Example 4 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

(a) (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

Or Per unit

Alternative Recall Equation 12 Full load secondary current Equivalent resistance referred to secondary Secondary voltage on no-load