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# 12 transformer

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# 12 transformer

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### 12 transformer

1. 1. TRANSFORMER
2. 2. Φ C vp ip P Np Φ= Npip/S …….(1) where S is the reluctance 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 Φ is produced which is given by an equation
3. 3. dt d Npp ϕ =v )( dt dN p 2 p p iv × ℜ = Since current varied with time , Φ also varied with time. A back electromagnetic force (e.m.f) will be produced which is given by the equation. Substitute Φ = Npip/S into the above equation , then ……(2) ……(3)
4. 4. ( ) dt 2sind Npp ft v m πΦ = Φ = Φm sin 2πft If ip is sinusoidal, the flux produced also sinusoidal, i.e therefore vp = NP2πfΦmcos 2πft = NP2πfΦmsin (2πft + π/2) The peak value = Vpm = NP2πfΦm and vp is leading the flux by π/2. mm pm p f44.4πf2707.0 2 V V Φ=Φ×== PP NNThe rms value ……(4) ……(5) ……(6) ……(7)
5. 5. dt d Nss ϕ =v ……(8) ϕ C vp ip vs is Load P S N s N p 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
6. 6. From (2) and (8) we get p s p s N N V V = …….(9) p s p s N N = v v 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 rearrange p s s p N N I I = …….(10) …….(12) …….(11)
7. 7. 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
8. 8. 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 AIs 625 400 10250 V P -3 s = × == A V P I p p 7.22 11000 10250 3 = × == Full-load secondary current
9. 9. 220011000 400 80 V V N N P s s P =×=×= p s p s N N V V = mf Φ= N44.4E mWb f m 5.22 508044.4 400 N4.44 E s s = ×× ==Φ (b) Number of primary turns recall (c) Maximum flux recall
10. 10. EP VP VSNS 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.
11. 11. 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 2 1 2 I oomo II += o o o I I 1 cos =φPower factor Phasor diagram for no load transfomer Φ IOm IOI EP VP IO VS φο
12. 12. Transformer 1 Transformer 2 Low voltage generator Low voltage load High voltage line 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.
13. 13. 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
14. 14. AB×=Φ recall mf Φ= N44.4E Wb f m 0206.0 5048044.4 2200 4.44N E p p = ×× ==Φ 2 0187.0 1.1 026.0 A m B m == Φ = recall (a) (b) p s p s N N V V =recal l V4132200 480 90 V V N N P s s P =×=×= Practically 10% more allow for insulator
15. 15. AC 7208.1400HC =×= A B a o aa 5.870001.0 104 1.1 H 7 =× × =×= − πµ  A5.8075.87720 =+= A N H Iom 682.1 480 5.807 ===  INH = AIomom 19.1682.1707.0707.0I =×== ∴ (c) magnetomotive force (mmf) for the core is mmf for the airgap is Total mmf is Maximum magnetizing current recal l Rms value
16. 16. 3 0337.00187.08.1 mA =×=×=  kgdensityVol 26378000337.0. =×=×= W4477.1263 =×= A V P I p o 203.0 2200 447 1 === lagging I I o 168.0 21.1 203.0 cos 01 ===φ AIII oomo 21.1203.019.1 222 1 2 =+=+= Volume of core mass of core Core loss= loss rate x mass Core-loss component of current No load current Power factor
17. 17. E1 V1 V2 I1 I2 L(θ2) Loaded transformer L(θ2)= load with power factor of cos θ2 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 I o Φ V 1 , E 1 - I 2 I 1 θ2 θ1 θO V 2 , E 2 I 2 ‘
18. 18. 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. A56280 1000 200 I N N I S P S P =×=×= Recall Equation 12 p s s p N N I I = therefore
19. 19. ooIII φφφ sinsinsin '2'211 += ooIII φφφ coscoscos '2'211 += 6.0sin '2 =∴ φ 98.0sin =∴ oφ2.0c =oosφ 8.0c '2 =φos ( ) ( ) A4.452.038.056 =×+×= '50381 o =φ ( ) ( ) AI 3.5854.364.45 22 1 =+= laggingos o 78.050'38cosc 1 ==φ 805.0 4.45 54.36 tan 1 ==φ ( ) ( ) A54.3698.036.056 =×+×= I o Φ V 1 , E 1 - I 2 I 1 θ2 θ1 θO V 2 , E 2 I 2 ‘ Solve for horizontal and vertical components Power factor
20. 20. Path of leakage V2’ R1 L1 L2 R2 RC Lm V2E1 I1’I1 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.
21. 21. There are four main losses •Dissipated power by wire resistance of the windings (I2 R) •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 E2 R1 L1 L2 R2 RC Lm V2V1 I1’I1 Equivalent circuit of transformer E1
22. 22. Phasor diagram for a transformer on load V1 E1 I2 -I2’ I1X1 V2 E2 I1 I0 I1R1 I1Z1 I2X2 I2Z2 I2R2 φ1 φ2
23. 23. 2 2 1 2 2 2 1 2'2       ≈      = V V X N N XX 2 2 1 2 2 1 2 2'2       ≈      = V V R I I RR 2 2 2'2 2 1 RIRI = We can replace R2 by inserting R2’ in the primary thus the equivalent resistance is Giving us Similarly
24. 24. 2 2 1 21'21       +=+= V V RRRRRe 22 eee XRZ += e e R X tan =eφ eee ZX φsin= eee ZR φcos= then and where 2 2 1 21'21       +=+= V V XXXXXe (b) Transformer simplified circuit Ze V1 E1=V2’ I1 I2 To load E2=V2
25. 25. Phasor diagram of simplified equivalent circuit of transformer φ 2 φ 2 I 1 R e I 1 Z e I 1 X e φ e Magnified Ze portion Φ I 1 X e I 1 E 1 =V 2' V 1 I 1 Z e I 1 R e E 2 , V 2 I 2 φ 2 φ 1 φ e φ 2 Complete φo-φ2
26. 26. voltageload-no voltageload-fullvoltageload-no regulationVoltage − =       = 1 2 12 N N VV p s p s N N V V =recall Secondary voltage on no-load V2 is a secondary terminal voltage on full load       −      = 1 2 1 2 1 2 1 regulationVoltage N N V V N N V Substitute we have
27. 27. ( ) 1 221 sincos V XRI regulationvoltageunitPer ee φφ + =− centper V N N VV 100 1 2 1 21 ×       − = unitper V N N VV 1 2 1 21 regulationVoltage       − =Or From phasor diagram can be proved that ( ) 1 21 cos V ZI regulationvoltageunitPer ee φφ − =− Or
28. 28. A 100kVA transformer has 400 turns on the primary and 800 turns on the secondary. The primary and secondary resistances are 0.3Ω and 0.01Ω respectively, and the corresponding leakage reactances are 1.1Ω and 0.035Ω 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
29. 29. A V P currentprimaryloadFull 45.45 2200 10100 3 = × ==− ( ) unitper0336.0 2200 6.0975.18.055.045.45 = ×+× = Ω=      +=      += 975.1 80 400 035.01.1 22 2 1 21 V V XXXe ( ) ( ) Ω=+=+= 05.2975.155.0 2222 eee XRZ Ω=      +=      += 55.0 80 400 01.03.0 22 2 1 21 V V RRRe (a) (b) (i) ( ) 1 221 sincos V XRI regulationvoltageunitPer ee φφ + =− centper36.3=
30. 30. V78.44678.6440 =+= V N N V P S P 440 400 80 2200 =×=×= centper54.1−= (b) (ii) power factor 0.8 leading ( ) unitperregulationVoltage 0154.0 2200 6.0975.18.055.045.45 −= ×−× = V8.140336.0440 =×= V78.60154.0440 =×= Sec. terminal voltage on no-load The decreasing of full-load voltage is V2.4258.14440 =−=Therefore the secondary full-load voltage The increasing of full-load voltage is Therefore the secondary full-load voltage
31. 31. 1 e1 V RI = 2200 55.045.45 × = 0.0114= 0.0114%= Per unit ltageprimary vo primarytoreferred resistanceequivalent currentprimary load-full       ×      =unitperdropResistance load-noonvoltagesecondary secondarytoreferred resistanceequivalent currentsecondary load-full       ×      =unitperdropResistanceOr dropResistance∴
32. 32. Alternative Full load secondary current Recall Equation 12 p s s p N N I I = Equivalent resistance referred to secondary AS 2.22745.45 80 400 I N N I P S P =×== Ω=      +=      += 022.0 400 80 3.001.0 22 1 2 12 N N RRRe V N N V P S P 440 400 80 2200 =×=×=Secondary voltage on no-load 440 022.02.227 load-noonvoltagesecondary secondarytoreferred resistanceequivalent currentsecondary load-full × =       ×      =unitperdropResistance 1.14%unitper0.0114 ==
33. 33. 2200 975.145.45 ltageprimary vo primarytoreferred resistanceleakageequivalent currentprimary load-full × =       ×      =unitperdropractanceLeakage Ω=      +=      += 975.1 80 400 035.01.1 22 2 1 21 V V XXXe %08.4unitper0408.0 == primarytoreferredresistanceleakageequivalent A V P currentprimaryloadFull 45.45 2200 10100 3 = × ==−