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- 1. CHAPTER 2 TRANSFORMER2.1 INTRODUCTION• Transformer allow voltage level to be changed throughout the electrical distribution system so that the most economical voltage can be used in each part of system .• Generators are limited to about 25kV due to the size of insulation required, but transmission losses at 25kV would be unacceptable.• Thus, voltage are stepped up for transmission.• Transformer allow the voltage level to be changed to the most economical level.• Compared to rotating machines, the transformer is relatively simple.• Transformer comprises two or more electric circuit coupled by magnetic circuit. Figure 2.1: An autotransformer with a sliding brush contact 17
- 2. 2.2 TRANSFORMER CONSTRUCTION Primary Secondary coil coil Figure 2.2 Figure2.3: An ideal step-down transformer showing magnetic flux in the core• Consists of two or more electrical windings that are linked together by a magnetic field.• Except for special-purpose transformers, the coupling is enhanced with a ferromagnetic core.• When AC voltage is applied to the primary winding, magnetic flux is established, which links the secondary winding.• If the flux is sinusoidal, a sinusoidal voltage will be induced in the secondary. 18
- 3. • The primary is the side that is connected to the source (generator or power system), and the secondary is the side that is connected to the load.• Either side may be the high-voltage or low-voltage side. Figure 2.4: Step up transformer (hydro station)Core Material• The core constructed of a high-permeance (low-reluctance) material to minimize the magnetixing current.• To keep eddy current losses down, the core is made of laminations, the thickness of which is inversely proportional to the rated frequency of the transformer.• Eddy current losses are proportional to the lamination thickness squared.• Thus, halving the thickness would reduce the eddy current losses by 75%. Figure 2.5: Three-phase pole-mounted step-down transformer. 19
- 4. Core Configuration• There are two types of transformers cores are used:- (i) Core type (ii) Shell type• In the core type, there are winding on each leg of the core (the winding surround the core).• In the shell type, the winding are on the center leg of the core, and the core surrounds the windings. Figure 2.6. Two types of transformer core.Conductors• Copper provides the best conductivity and, therefore the minimum volume for the coil.• Aluminum is usually used to reduce the cost.• The conductor must carry current without overheating.• The conductors may be round, square, or rectangular, and there may be several conductors in parallel to reduce the I2R losses. 20
- 5. • The operating temperature of the coil is extremely important because the insulation may deteriorate at increased temperature and the resistance of the coil also increases with temperature. Figure 2.7: Three-phase oil-cooled transformer with cover cut away2.3 IDEAL TRANSFORMER OPERATION• A transformer makes use of Faradays law and the ferromagnetic properties of an iron core to efficiently raise or lower AC voltages.• It of course cannot increase power so that if the voltage is raised, the current is proportionally lowered and vice versa. Figure 2.8 21
- 6. Faradays Law• Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be "induced" in the coil.• No matter how the change is produced, the voltage will be generated. dΦFaraday’s law: e = N where N = number of turns dt dΦ = rate of change of flux dt e = instantaneous induced voltage e(t ) = 2 Erms sin( wt ) dΦ e = dt N t 1 Φ (t ) = ∫ 2 Erms sin( wt )dt N0 − 2 Erms Φ (t ) = cos( wt ) wN − 2 Erms Φ (t ) = cos( wt ) 2πfN Φ (t ) = −Φ max cos( wt ) 2πfNΦ maxTherefore Erms = = 4.44 fNΦ max 2Voltage• For an ideal transformer, all the flux is confined to the iron core and thus links both the primary and secondary.• Therefore: Ep = 4.44fNpΦmax Es = 4.44fNsΦmax Ep Np E1 N1 = =a = =a Es Ns E2 N 2• A transformer is called a step-down transformer if the primary side has more turns than the secondary side (a>1).• A transformer is called a step-up transformer if the primary side has fewer turns than the secondary side (a<1).• A step-down transformer has the high voltage facing the power system and low voltage facing the load. 22
- 7. • The opposite is true for the step-up transformer.Current• Because the losses are zero in the ideal transformer, the apparent power in and out of transformer must be the same: │Sin│=│Sout│=│Vp││Ip│=│Vs││Is│ Ip Vs Ns 1 I1 V2 N2 1Therefore = = = = = = Is Vp Np a I2 V1 N1 a• The ratio of the currents is the inverse of the voltage ratio.• If we raise the voltage level to a load with a step-up transformer, then the secondary current drawn by the load would have to be less than the primary current, since the apparent power is constant.Impedance Vs Zs = Is• From Ohm’s Law, Vp Vp ZS = a = aI p a (a ) I p Zp Zs = a2 2 Zp Np =a = N 2 Zs s Example 1The nameplate on a single-phase, step-down transformer indicates that is rated 33.33kVA,7967V/120V. Find the rated current on the primary and secondary sides.Solution 23
- 8. S 33.33kVAIs = = = 277.8 A Vs 120V S 33.33kVAIp = = = 4.184 A Vp 7967VExample 2An ideal, step-down transformer has 1500 turns in the primary coil and 75 turns in thesecondary coil. If 2400 V is applied to the primary, what is the voltage on the secondary?SolutionVp Np =Vs Ns Ns 75 Vs = V p = 2400 = 120V Np 1500 Example 3An ideal, step-down transformer has 1500 turns in the primary coil and 75 turns in thesecondary coil. If 10.4 A flows in the primary of the transformer, calculate the loadcurrent.SolutionIp Ns =Is Np Np 1500 Is = I p = 10.4 = 208 A Ns 75 Example 4A transformer is rated 75kVA, 7200/240 or 120V. Find the rated current on each side ofthe transformer. Find the exciting current if it is 1.7% of rated current.SolutionThe rated current on each side of the transformer :- 24
- 9. S rated 75kVAIp = = = 10.42 A Vp 7200V S rated 75kVAI s120V = = = 625 A Vs120 120V S rated 75kVAI s 240V = = = 312.5 A Vs 240 240VThe exciting current is on the primary side, so:-I ex = 0.017Ip = 0.017(10.42) = 0.177AExample 5A transformer is rated 75kVA, 7200/240V. A load having an impedance of 1.0Ω isconnected to the secondary.(i) What is the current on each side?(ii) What is the power delivered to the load?(iii) What is the input impedance on the primary side?Solution(i) V = IR V 240V Is = s = = 240 A R 1Ω Ip N = s Is N p 240 Ip = 240 = 8 A 7200(ii) P = I 2 R = 240 2 (1) = 57.6kW 2 Zp Np (iii) = Zs Ns 2 7200 Zp = (1Ω) = 900Ω 240 25
- 10. 2.4 PRACTICAL TRANSFORMERThe Equivalent Circuit• All transformer have winding resistance, a core with finite permeability, leakage flux and hysterisis and eddy current losses and are thus nonideal.• The transformer as Figure 2.9 which represented by the core and coils, is still not ideal because it has magnetizing current flowing into it. Figure 2.9• Adding the magnetizing branch as shown in Figure 2.10, the transformer is now become an ideal transformer. Figure 2.10 Rp jXp a:1 jXs Rs jXm Rm 1.0k 1.0m 1.0m 1.0m Figure 2.11:Equivalent T-circuit 26
- 11. 2 2 Rp jXp jXs =a jXs Rs =a Rs a:1 jXm Rm 1.0k 1.0m Figure 2.12: T-circuit referred to the primary Rp jXp Rp= 2 jXp= 2 a a jXs Rs a:1 TR1 jXm Rm 1.0k 1.0k 1.0m 1.0m 1.0m Figure 2.13: T-circuit referred to the secondaryReflection In Transformer• One simplification that introduces only a small error is to move the magnetizing branch to the primary terminals and then combine the primary and secondary resistance and leakage reactance.• The combination of the winding resistance is called the equivalent resistance, and the combination of the leakage reactance is called the equivalent reactance.• The combination can be referred to one side only, either primary or secondary.• The following Figure 2.14 and 2.15, illustrates transformer reflection. Re q p jXe q p a:1 1.0k 1.0m 1.0k 1.0m jXm Rm Figure 2.14: Cantilever circuit referred to the primary Req,p = Rp + a2Rs = Rp+Rs’ Xeq,p = Xp + a2Xs = Xp+Xs’ 27
- 12. Re q s jXe q s a:1 jXm Rm 1.0m 1.0m 1.0k 1.0k Figure 2.15: Cantilever circuit referred to the secondary Req,s = (Rp/a2 ) + Rs = R’p + Rs Xeq,s = (Xp/a2 ) + Xs = X’p+ Xs Figure 2.16Example 6A step-down transformer is rated 100kVA, 7200/277 V and has the following equivalentcircuit parameter: Rp = 2.92Ω Rs = 0.00432Ω Rm = 51840Ω Xp = 14.6Ω Xs = 0.0216Ω Xm = 12960Ω 28
- 13. Find the equivalent winding impedance referred to the high voltage side. Repeat for thelow voltage side.SolutionEquivalent impedance referred to the high/primary side: Vp 7200a= = = 25.99 Vs 277Req , p = R p + a 2 RsReq , p = 2.92 + (25.99) 2 (0.00432) = 5.839ΩX eq , p = X p + a 2 X sX eq , p = 14.6 + (25.99) 2 (0.0216) = 29.19ΩZ eq , p = (5.839 + j 29.2)ΩEquivalent impedance referred to the low/secondary side:a = 25.99 RpReq ,s = 2 + Rs a 2.92Req ,s = 2 + (0.00432) = 8.643x10 −3 Ω 25.99 XpX eq ,s = + Xs a2 14.6X eq ,s = + (0.0216) = 0.0432Ω 25.99 2Z eq , s = (0.86 + j 4.3) x10 −2 ΩExample 7A step-down transformer is rated 140kVA, 400/1000 V and has the following equivalentcircuit parameter: Rp = 10Ω Rs = 50Ω Rm = 5000 Ω Xp = 20Ω Xs = 80Ω Xm = 12000Ω 29
- 14. Draw the equivalent winding impedance referred to the high voltage side and low voltageside.Solution N1 400a= = = 0.4 N 2 1000Refer to the high voltage side/secondary side R1 10Req = 2 + R2 = + 50 = 112.5Ω a 0.4 2 X 20X eq = 21 + X 2 = + 80 = 205Ω a 0.4 2Z eq = 112.5 + j 205Ω 112.5 ohm j205 ohm 1,0m 1,0kRefer to the low voltage side/primary sideReq = R1 + a 2 R2 = 10 + 0.4 2 (50) = 18ΩX eq = X 1 + a 2 X 2 = 20 + 0.4 2 (80) = 32.8ΩZ eq = 18 + j 32.8Ω 18 ohm j32.8 ohm 1,0m 1,0k 30
- 15. 2.5 DETERMINING CIRCUIT PARAMETERThere are 2 simple test that can provide the data required to calculate values for theelements of the transformer equivalent circuit:- (i) Short-Circuit Test (ii) Open-Circuit Test(i) Short-Circuit Test (normally at low voltage side) Rp jXp jXs Rs HV side jXm Rm 1.0k 1.0m 1.0k Figure 2.17: Short-circuit test arrangement• One-side of the transformer is shorted, and voltage is applied on the other side until rated current flows in the windings.• Measured the 1. applied voltage (Vsc), 2. winding current (Isc), 3. input power (Psc)• Generally, the low-voltage side of the transformer is shorted and voltage is applied to the high-voltage side.• The measurements are used to calculate Req and jXeq which referred to the high voltage side.• With the low-voltage winding shorted, the input impedance is:- Zin = (Rp + jXp) + [(R’s + jX’s)║[Rm║jXm]] ,without ignored Rm and jXm• If Rm and jXm>>>R’s and jX’s, therefore Rm and jXm ignored. Zin = (Rp + jXp) + [(R’s + jX’s)• From short circuit test, Vsc, Isc and Psc are measured. Therefore:- Vsc │Z eq │= │Z sc │= Isc R eq = Psc/(I2sc) X eq =√ (│Z eq│2 – R2eq) 31
- 16. (ii) Open-Circuit Test(normally at high voltage side) Rp jXp jXs Rs LV side jXm Rm 1.0k 1.0m 1.0k Figure 2.18: Open-circuit test arrangement• High-voltage side of the transformer is opened and rated voltage is applied to the low-voltage side.• Readings of Voc, Ioc and Poc are taken on the low voltage side. Then Rm and Xm can be calculated.• Measured the 1. applied voltage (Voc), 2. winding current (Ioc), 3. input power (Poc)• Therefore Rm and Xm which referred to the low voltage side can be calculate.• The input impedance during the open-circuit test is the primary winding in series with the exciting branch: Zin = (Rp + jXp) + (Rm║jXm)• The leakage reactance and winding resistance on the primary side can be negligible as they are too low compare to Rm and Xm, Rm and jXm>>>Rp and jXp, Therefore, Zin = (Rm║jXm)• From open-circuit test, Voc, Ioc and Poc are measured. Therefore:- Poc = VocIoccosθoc θoc = cos-1(Poc/IocVoc) IR = Ioccos θoc IX = Iocsin θoc 32
- 17. Rm and Xm can be calculated by: Rm = Voc/IR Xm = Voc/IXExample 8Short-circuit and open-circuit tests were performed on a 100kVA transformer, rated7200V/277V, with the results listed below. Assuming step-down operation, determine theequivalent circuit parameters of the transformer referred to the high voltage side. Vsc = 414 V Voc = 277 V Isc = 13.89 A I oc = 14.88 A Psc = 1126 W Poc = 1000 WSolutionShort circuit(HV) (Req,Xeq) Vsc 414 Zeq = = = 29.8Ω I sc 13.89 Psc 1126W Req = = = 5.836Ω I sc 13.89 2 2 2 2 Xeq = Z eq − Req = 29.23ΩOpen circuit(LV) (Rm,Xm) P 1000W θ oc = cos −1 oc = cos −1 V I 277(14.88) = 75.96° oc oc I R = I oc cos θ oc = 14.88(cos 75.96°) = 3.61AI x = I oc sin θ oc = 14.88(sin 75.96°) = 14.435 A V 277Rm = oc = = 76.73Ω I R 3.61 V 277X m = oc = = 19.19Ω I x 14.435 33
- 18. Referred to the high voltage side/primary sideReq ,hv = 5.836ΩX eq ,hv = 29.23ΩRm ,lv = 76.73Ω 2 7200 Rm ,hv = (76.73Ω) = 51.84kΩ 277 X m ,lv = 19.19Ω 2 7200 X m ,hv = (19.19Ω) = 12.97 kΩ 277 Example 9Short-circuit and open-circuit tests were performed on a 100kVA transformer, 50 Hz,rated at 120V/2400V, and the results are listed as follows: Vsc = 40 V Voc = 120 V Isc = 41.67 A I oc = 6 A Psc = 380 W Poc = 40 W(i) Draw the equivalent circuit with the necessary parameters of the transformer referredto the low voltage side.(ii) Draw the equivalent circuit with the necessary parameters of the transformer referredto the high voltage sideSolutionShort circuit(HV) (Req,Xeq) Vsc 40 Zeq = = = 0.96Ω I sc 41.67 Psc 380W Req = = = 0.22Ω I sc 41.67 2 2 2 2 Xeq = Z eq − Req = 0.96 2 − 0.22 2 = 0.93Ω 34
- 19. Open circuit(LV) (Rm,Xm) P 40W θ oc = cos −1 oc = cos −1 V I 120(6) = 86.82° oc oc I R = I oc cos θ oc = 6(cos 86.82°) = 0.33 AI x = I oc sin θ oc = 6(sin 86.82°) = 5.99 A V 120Rm = oc = = 363.64Ω I R 0.33 V 120X m = oc = = 20.03Ω Ix 5.99(i) Referred to the low voltage side/primary sideRm ,lv = 363.64ΩX m,lv = 20.03ΩReq ,hv = 0.22Ω 2 120 Req ,lv = (0.22Ω) = 0.55mΩ 2400 X eq ,hv = 0.93Ω 2 120 X eq ,lv = (0.93Ω) = 2.325mΩ 2400 The equivalent circuit:- 0.55m ohm j2.325m ohm 1,0m 1,0k 363.64 ohm 1,0k j20.03 ohm 1,0m 35
- 20. (ii) Referred to the high voltage side/secondary side Req ,hv = 0.22ΩX eq ,hv = 0.93ΩRm,lv = 363.64Ω 2 2400 Rm,hv = (363.64Ω) = 145456Ω 120 X m,lv = 20.03Ω 2 2400 X m,hv = (20.03Ω) = 8012Ω 120 0.22 ohm j0.93 ohm 1,0m 1,0k 145 456 ohm 1,0k j8012 ohm 1,0m 36
- 21. 2.6 TRANSFORMER LOSSESGenerally, in any machine there will be two types of losses namely: (i) iron losses P iron = IRm2Rm = VocIoccosθoc (ii) copper losses P copper = I12R1 + I22R2 = I12R eq,p = I22Req,s2.7 TRANSFORMER EFFICIENCY η = OutputPower = InputPower Pout = Pin −Ploss Pin PinOutput Power , P out = VAcosθ2Input Power , P in = Output Power + Total LossesTotal Losses , P loss = P copper losses + P core losses Energy dissipated in Hysteresis and the resistance of eddy current losses winding2.8 VOLTAGE REGULATIONThe purpose of voltage regulation is actually to see what is the voltage being dropped inthe secondary winding between no-load and full-load condition. Vnl − Vfl VR = Vnl 37
- 22. If transformer equivalent circuit referred to:- (i) primary side, R 01 X01 1 2 aV2 V 1 − aV 2 VR = V1 (ii) secondary side, R 02 X02 1 2 V2 V1 − V 2 a VR = V1 aExample 10By referring to the Example 9,calculate the terminal voltage, V1 and voltage regulation,VR if a load at 0.8 power factor lagging is connected to 2400 V side. (neglect themagnetizing impedance).SolutionLoad is connected to secondary side:- 38
- 23. S 100kVAI2 = = = 41.67∠ − cos −1 (0.8) = 41.67∠ − 36.87° = 33.34 − j 25 A V2 2400 120a= = 0.05 2400If referred primary side: I1 0.55m ohm j2.325m ohm I2=I2/a V1 V2=aV 2 1,0m 1,0k I2V 1 = I 2 (0.55m + j 2.325m) + V 2 = (0.55m + j 2.325m) + aV 2 a (33.34 − j 25)= (0.55m + j 2.325m) + 0.05(2400) 0.05= 121.53 + j1.28V = 121.54∠0.6°V V 1 − aV 2 121.54 − 0.05(2400)VR = = x100% = 1.27% V1 121.54If referred to secondary side: I1=aI1 0.22 ohm j0.93 ohm I2 V1=V1/a V2 1,0m 1,0kV 1 = I 2(0.22 + j 0.93) + V 2= (33.34 − j 25)(0.22 + j 0.93) + 2400= 2430.58 + 25.51V 1 = V 1 (a) = 0.05( 2430.58 + 25.51)= 121.53 − j1.28V = 121.54∠0.6°V V 1 − V 2 121.54 − 2400 a 0.05VR = = x100% = 1.27% V1 121.54 a 0.05 39
- 24. 2.9 AUTOTRANFORMERS OR 1.0m 1.0m LOAD LOAD Figure 2.19: Autotransformer schematic• Figure 2.19 show the autotransformer schematic.• Autotransformer is a transformer with only one winding.• The low-voltage coil is essentially placed on top of the high-voltage coil and is called the series coil.• The connection is called an autotransformer and can be used as a step-up or a step- down transformer.• The advantages of autotransformer are:- (i) cheaper (ii) more efficient, because losses stay the same while the rating goes up compared to a conventional transformer (iii) lower exciting current (iv) better voltage regulation• The disadvantages of autotransformer are:- (i) larger short circuit current available (ii) no isolation between the primary and secondary 40
- 25. Example 11A 220/440V, 25kVA and 50 Hz transformer is connected as an autotransformer totransform 660V to 220V.(i) Determine the ratio ‘a’.(ii) Determine the kVA rating of the auto transformer.(iii) With a load of 25 kVA, 0.8 lagging power factor connected to 220 V terminals, determine the currents in the load and the two transformer windings.Solution I1 1,0m 1,0m 1,0m N1 I2 V1=660V N2 I2-I1 V2=220V N 1 660(i) a= = =3 N 2 220 25kVA(ii) I1 = = 37.87 A 660 660(37.87) kVAauto = = 25kVA 1000 25kVA(iii) I2 = = 113.64 A 220 I 2 − I 1 = 113.64 − 37.87 = 75.77 A 41
- 26. 2.10 THREE PHASE TRANSFORMERSIntroduction: • Three phase transformers are used throughout industry to change values of three phase voltage and current. • Since three phase power is the most common way in which power is produced, transmitted, an used, an understanding of how three phase transformer connections are made is essential.Construction: • A three phase transformer is constructed by winding three single phase transformers on a single core. • These transformers are put into an enclosure which is then filled with dielectric oil. • The dielectric oil performs several functions. • Since it is a dielectric, a nonconductor of electricity, it provides electrical insulation between the windings and the case. • It is also used to help provide cooling and to prevent the formation of moisture, which can deteriorate the winding insulation.Connections: • There are only 4 possible transformer combinations: 1. Delta to Delta - use: industrial applications 2. Delta to Wye - use : most common; commercial and industrial 3. Wye to Delta - use : high voltage transmissions 4. Wye to Wye - use : rare, dont use causes harmonics and balancing problems. • Three-phase transformers are connected in delta or wye configurations. • A wye-delta transformer has its primary winding connected in a wye and its secondary winding connected in a delta (see Figure 2.20). • A delta-wye transformer has its primary winding connected in delta and its secondary winding connected in a wye (see Figure 2.21). 42
- 27. Figure 2.20:Wye-Delta connection Figure 2.21:Delta-Wye connectionDelta Conections: • A delta system is a good short-distance distribution system. • It is used for neighborhood and small commercial loads close to the supplying substation. • Only one voltage is available between any two wires in a delta system. • The delta system can be illustrated by a simple triangle. • A wire from each point of the triangle would represent a three-phase, three-wire delta system. • The voltage would be the same between any two wires (see Figure 2.22). Figure 2.22 43
- 28. Wye Connections: • In a wye system the voltage between any two wires will always give the same amount of voltage on a three phase system. • However, the voltage between any one of the phase conductors (X1, X2, X3) and the neutral (X0) will be less than the power conductors. • For example, if the voltage between the power conductors of any two phases of a three wire system is 208v, then the voltage from any phase conductor to ground will be 120v. • This is due to the square root of three phase power. • In a wye system, the voltage between any two power conductors will always be 1.732 (which is the square root of 3) times the voltage between the neutral and any one of the power phase conductors. • The phase-to-ground voltage can be found by dividing the phase-to-phase voltage by 1.732 (see Figure 2.23). Figure 2.23Connecting Single-Phase Transformers into a Three-Phase Bank: • If three phase transformation is need and a three phase transformer of the proper size and turns ratio is not available, three single phase transformers can be connected to form a three phase bank. • When three single phase transformers are used to make a three phase transformer bank, their primary and secondary windings are connected in a wye or delta connection. • The three transformer windings in Figure 2.24 are labeled H1 and the other end is labeled H2. • One end of each secondary lead is labeled X1 and the other end is labeled X2. 44
- 29. Figure 2.24• Figure 2.25 shows three single phase transformers labeled A, B, and C.• The primary leads of each transformer are labeled H1 and H2 and the secondary leads are labeled X1 and X2.• The schematic diagram of Figure 2.24 will be used to connect the three single phase transformers into a three phase wye-delta connection as shown in Figure 2.26. Figure 2.25 45
- 30. Figure 2.26• The primary winding will be tied into a wye connection first.• The schematic in Figure 2.24 shows, that the H2 leads of the three primary windings are connected together, and the H1 lead of each winding is open for connection to the incoming power line.• Notice in Figure 2.26 that the H2 leads of the primary windings are connected together, and the H1 lead of each winding has been connected to the incoming primary power line.• Figure 2.24 shows that the X1 lead of the transformer A is connected to the X2 lead of transformer c.• Notice that this same connection has been made in Figure 2.26.• The X1 lead of transformer B is connected to X1, lead of transformer A, and the X1 lead of transformer B is connected to X2 lead of transformer A, and the X1 lead of transformer C is connected to X2 lead of transformer B.• The load is connected to the points of the delta connection. 46
- 31. Tutorial 21. A step-down transformer is rated 25kVA, 660-240 V and has the following equivalentcircuit parameter: Rp = 15Ω Rs = 10Ω Rm = 50000Ω Xp = 20Ω Xs = 8Ω Xm = 13000ΩFind the equivalent winding impedance referred to the high side and the low side.2. Short-circuit and open-circuit tests were performed on a 10kVA transformer,50 Hz,rated 400V-240V, with the results listed below. Vsc = 10 V Voc = 240 V Isc = 10kVA/400V=25A I oc = 4 A Psc = 120 W Poc = 80 W(i) Draw the equivalent circuit parameters of the transformer referred to the high side.(ii) Draw the equivalent circuit parameters of the transformer referred to the low side.3. The coil possesses 400 turns and links an ac flux having peak value of 2mWb. If thefrequency is 60Hz, calculate the induced voltage E.4. A transformer having 90 turns on the primary and 2250 turns on the secondary isconnected to a 120V, 60Hz source. Calculate:(i) The effective voltage for the secondary side.(ii) The peak voltage for the secondary side.5. An ideal-transformer having 90 turns on the primary and 2250 turns on the secondaryis connected to a 200V, 50 Hz source. The load across the secondary draws a current of2A at a power factor of 0.8 lagging. Calculate: (i) The effective value of the primary current.(ii) The primary current if the secondary current is 100mA.(iii) The peak flux linked by the secondary winding.6. The secondary winding of a transformer has a terminal voltage ofVs (t ) = 282.8 sin 377t V. The turn ratio of the transformer is 50/200. If the secondarycurrent of the transformer is is (t ) = 7.07 sin(377t − 36.87 ) A, calculate: 0(i) the primary current of the transformer(ii) voltage regulation 47
- 32. The impedances of this transformer referred to the primary side are Req = 0.05Ω Rc = 75Ω X eq = 0.225Ω X m = 20Ω7. A 20kVA 8000/277V distribution transformer has the following equivalent circuitparameter: R p = 32Ω Rs = 0.05Ω X p = 45Ω X s = 0.06Ω Rc = 250kΩ X m = 30kΩ(i) Draw the equivalent circuit of this transformer referred to the high voltage side.(ii) Calculate the voltage regulation if:- (a) the transformer is supplying rated load at 277V and 0.8 pf lagging, (b) the transformer is supplying load at 200V,100W and 0.85 pf lagging. (c) the transformer is supplying load at 18kVA, 250V and 0.8 pf leading.8. Short-circuit and open-circuit tests were performed on a 1000VA transformer, 50 Hz,rated at 230V/115V, and the results are listed as follows: Vsc = 13.2 V Voc = 115 V Isc = 4.35 A I oc = 0.45 A Psc = 20.1 W Poc = 30 W(i) Draw the equivalent circuit with the necessary parameters of the transformer referredto the low voltage side.(ii) Find the input voltage of the transformer if the transformer is connected to a 140Var,110V at 0.75 pf lagging 48

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