#### Transcript Cct component - Universiti Sains Malaysia

TRANSFORMER 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 is produced which is given by an equation = Npip/S …….(1) where S is the reluctance Since current varied with time , also varied with time. A back electromagnetic force (e.m.f) will be produced which is given by the equation. d vp N p dt ……(2) Substitute = Npip/S into the above equation , then vp Np2 d (ip ) dt ……(3) If ip is sinusoidal, the flux produced also sinusoidal, i.e = m sin 2ft therefore ……(4) d m sin 2ft vp N p dt vp = NP2fmcos 2ft = NP2fmsin (2ft + /2) The peak value = Vpm = NP2fm ……(5) ……(6) and vp is leading the flux by /2. The rms value Vp Vpm 2 0.707 N P 2πf m 4.44 N P f m ……(7) C secondary primary is ip N vp p P N Ss vs Load 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 d vs Ns dt ……(8) From (2) and (8) we get vs Ns vp N p Vs N s Vp N p Or in rms value …….(9) …….(10) With load , is will flow in the load, mmf at load will equal to the mmf at input, then (in rms value) NpIp = NsIs rearrange Ip Ns Is N p …….(11) …….(12) 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 Primary NP VP Secondary : NS VS Symbol for ideal transformer 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 P 250 103 Ip 22.7 A Vp 11000 Full-load secondary current P 250 10-3 Is 625 A Vs 400 (b) Number of primary turns recall NP Vs N s Vp N p Ns 80 VP 11000 2200 Vs 400 (c) Maximum flux recall E 4.44 N f m Es 400 m 22.5mWb 4.44 N s f 4.44 80 50 IO EP VP NP NS VS 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. EP VP= emf of supply to the primary coil VP IO and 90o leads the flux. IOI o EP=emf induced in the primary coil and same phase as VP. IOm VS=emf induced in the secondary coil VS and 90o lags the flux. Phasor diagram for Iom=magnetizing current to produce flux no load transfomer and it is in phase with flux. Io1=current to compensate the losses due to hysteresis and eddy current. 2 2 I I I Io=the no load current and given by o om o1 I o1 Power factor cos o Io 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 Low voltage generator High voltage line Transformer 2 Low voltage load 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 E 4.44 N f m (a) recall Ep 2200 m 0.0206Wb 4.44N p f 4.44 480 50 recall B A m 0.026 A 0.0187m 2 B 1.1 (b) recall Practically 10% more allow for insulator Vs N s Vp N p Ns 90 NP VP 2200 413V Vs 480 (c) magnetomotive force (mmf) for the core is H C C 400 1.8 720 A mmf for the airgap is H a a B a 1.1 7 0.0001 87.5 A o 4 10 Total mmf is recall 720 87.5 807.5 A H N I Maximum magnetizing current Rms value I om H 807.5 1.682 A N 480 Iom 0.707 I om 0.707 1.682 1.19 A Volume of core A 1.8 0.0187 0.0337m3 mass of core Vol. density 0.0337 7800 263kg Core loss= loss rate x mass 2631.7 447W Core-loss component of current No load current Power factor P 447 I o1 0.203 A V p 2200 2 I o I om I o21 1.192 0.2032 1.21A cos I 01 0.203 0.168 lagging Io 1.21 E1 I1 I2 V1 V2 L(2) V1 , E 1 Loaded transformer L(2)= load with power factor of cos 2 V1 = emf at supply E1=induced voltage at primary V2=emf at load E2=induced voltage at secondary I1= primary current I2=secondary current Phasor diagram - I2‘ I 1 1 O 2 I2 V2 , E 2 Io 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 Ip Is IP Ns Np NS 200 IS 280 56 A NP 1000 cos2' 0.8 coso 0.2 sin 2' 0.6 V1 , E 1 - I 2‘ sin o 0.98 O I1 cos 1 I 2' cos 2' I o cos o 56 0.8 3 0.2 45.4 A 2 I2 56 0.6 3 0.98 36.54 A I1 1 1 Solve for horizontal and vertical components I1 sin 1 I 2' sin 2' I o sin o I V2 , E 2 45.42 36.542 58.3A o 36.54 38 50' 1 tan 1 0.805 45.4 Power factor cos1 cos 38o50' 0.78 lagging Io I1 R1 E1 Path of leakage I1’ L1 RC Lm L2 R2 V2’ Equivalent circuit of transformer 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. V2 There are four main losses •Dissipated power by wire resistance of the windings (I2R) •Power due to hysteresis I1 R1 L1 I1’ L2 R2 •Power due to eddy current •Power via flux leakages. V1 RC Lm E1 E2 Equivalent circuit of transformer 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 V2 V1 -I2’ E1 I1 1 I1Z 1 I1R1 I0 I1X 1 I2R2 I 2X2 2 I2 I2Z2 V2 E2 Phasor diagram for a transformer on load We can replace R2 by inserting R2’ in the primary thus the equivalent resistance is I 2R I 2R 1 2 2' 2 2 2 Giving us I2 V1 R2' R2 R2 I1 V2 Similarly N1 V1 X 2 X 2 ' X 2 N2 V2 2 2 V1 Re R1 R2 ' R1 R2 V2 2 V1 X e X 1 X 2 ' X 1 X 2 V2 then where 2 Z e Re2 X e2 Re Z e cos e (b) Ze I1 V1 I2 E1=V2’ E2=V2 X e Z e sin e and Xe tan e Re Transformer simplified circuit To load Phasor diagram of simplified equivalent circuit of transformer V1 o-2 E1=V2' I1Ze I1Ze I1 e I1Xe I1Re I1Xe e I1Re Magnified Ze portion I2 E2,V2 Complete Voltage regulation recall no - load voltage full - load voltage no - load voltage Vs N s Vp N p Secondary voltage on no-load N2 V2 V1 N1 V2 is a secondary terminal voltage on full load N2 V2 V1 N1 Voltage regulation N2 V1 N1 Substitute we have N1 V1 V2 N2 Or Voltage regulation V1 per unit N1 V1 V2 N2 100 V1 per cent From phasor diagram can be proved that I1 Z e cose 2 Per unit voltage regulation V1 Or I1 Re cos 2 X e sin 2 Per unit voltage regulation V1 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: (a)The equivalent impedance referred to the primary circuit; (b)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. (c)The percentage resistance and leakage reactance drops of the transformer (a) 2 2 2 2 V1 400 Re R1 R2 0.3 0.01 0.55W 80 V2 V1 400 X e X 1 X 2 1.1 0.035 1.975W 80 V2 Z e Re2 X e2 (b) (i) 0.552 1.9752 2.05W P 100 103 Full load primary current 45.45 A V 2200 I1 Re cos 2 X e sin 2 Per unit voltage regulation V1 45.450.55 0.8 1.975 0.6 0.0336 per unit 2200 3.36 per cent Sec. terminal voltage on no-load VP The decreasing of full-load voltage is NS 80 2200 440V NP 400 440 0.0336 14.8V Therefore the secondary full-load voltage 440 14.8 425.2V (b) (ii) power factor 0.8 leading Voltage regulation 45.450.55 0.8 1.975 0.6 0.0154 per unit 2200 1.54 per cent The increasing of full-load voltage is Therefore the secondary full-load voltage 440 0.0154 6.78V 440 6.78 446.78V Or full - load equivalent resistance primary current referred to primary Resistance drop per unit primary vo ltage full - load equivalent resistance secondary current referred to secondary Resistance drop per unit secondary voltage on no - load Resistance drop I1R e 45.45 0.55 0.0114 V1 2200 0.0114% Per unit Alternative Recall Equation 12 Full load secondary current Equivalent resistance referred to secondary Ip Is Ns Np NP 400 IS IP 45.45 227.2 A NS 80 N2 Re R2 R1 N1 Secondary voltage on no-load VP 2 80 0.01 0.3 0.022W 400 2 NS 80 2200 440V NP 400 full - load equivalent resistance secondary current referred to secondary Resistance drop per unit secondary voltage on no - load 0.0114 per unit 1.14% 227.2 0.022 440 equivalent leakage resistance referred to primary 2 V 400 X e X 1 X 2 1 1.1 0.035 1.975W 80 V2 2 P 100 103 Full load primary current 45.45 A V 2200 full - load equivalent leakage resistance primary current referred to primary Leakage ractance drop per unit primary vo ltage 0.0408 per unit 4.08% 45.45 1.975 2200