Chapter two introduction to Machines Transformers

Chapter two introduction to Machines, Static Machines

1. Introduction to Electrical Machines ECEg3131 Chapter-2 Transformer College of Engineering & Technology Department of Electrical and Computer Engineering

2. INTRODUCTION 2/26/2024 2 transformer is a static device that transfers electrical energy from one electrical circuit to another electrical circuit through the medium of magnetic field and without a change in the frequency. The electric circuit which receives energy from the supply mains is called primary winding and the other circuit which delivers electrical energy to the load is called secondary winding. Actually the transformer is an electric energy conversion device, since the energy received by the primary is converted to useful electrical energy in the other circuits (secondary winding circuit). If the secondary winding has more turns than the primary winding, then the secondary voltage is higher than the primary voltage and the transformer is called a step-up transformer .

3. Contu …. 2/26/2024 3 When the secondary winding has less turns than the primary windings then the secondary voltage is lower than the primary voltage and the transformer is called step down transformer . Note that a step-up transformer can be used as a step-down transformer, in which the secondary of step-up transformer becomes the primary of the step-down transformer. Actually a transformer can be termed a step-up or step-down transformer only after it has been put into service. The most important tasks performed by transformers are:- 1. Changing voltage and current levels in electrical power systems 2. Matching source and load impedances for maximum power transfer in electronic and control circuit and 3. Electrical isolation (isolating one circuit from another ) Transformers are used extensively in ac power systems. AC electrical power can be generated at one central location, its voltage stepped up for transmission over long distances at very low losses and its voltage stepped down again for final use.

4. Construction of Transformer 2/26/2024 4 Core and copper(winding) are the main parts of transformer. Based on their construction there are basically two types of transformer, the core-type and the shell-type. The two types differ from each other by the manner in which the windings are wound around the magnetic core. The magnetic core is a stock of thin silicon-steel laminations about 0.35mm thick for 50Hz transformers . In order to reduce the eddy current losses, these laminations are insulated from one another by thin layer of varnish . The vertical portions of the core are usually called limbs or legs and the top and bottom portions are called yoke.

5. Core type Txr Shell type Txr 2/26/2024 Consists of simple rectangular laminated piece of steel with the transformer windings wrapped around two sides of the rectangle. T he windings surround a considerable part of steel core. For a given output and voltage rating, requires less iron but more conductor material. The type of winding employed are concentric coils. For single phase it is two legged. Primary and secondary winding are wound over different legs. Consists of a three-legged laminated core with the windings wrapped around the center leg. T he steel core surrounds a major part of the windings. For a given output and voltage rating, requires less conductor but more iron material. I nterleaved (or sandwiched) coils for shell- type transformers. It is three legged. Primary and secondary windings are wound over the same leg. 5 Contu ….

6. Contu …. 2/26/2024 6 In core-type transformers, most of the flux is confined to high permeability core. However , some of the flux leaks through the core legs and non-magnetic material surrounding the core. The flux called leakage flux , links one winding and not the other. As reduction in this leakage flux improves the transformer performance considerably hence, an effort is always made to reduce it by placing half of the low voltage (LV) winding over one leg and the other half over the second leg or limb and for the high voltage (HV) winding also , half of the winding is over one leg and the other half over the second leg. Low voltage winding is placed adjacent to the steel core and high voltage winding outside, in order to minimize the amount of insulation required . In shell-type transformer the low voltage and high voltage windings are wound over the central limb and are interleaved or sandwiched . In core-type transformer, the flux has a single path around the legs or yokes where as in shell-type transformer, the flux in the central limb divides equally and returns through the outer two legs.

7. Contu …. 2/26/2024 7 In both core and shell-type transformers, the individual laminations are cut in the form of long strips of L's, E’s and I's. Figure long strips of E’s, L’s and I’s laminations In order to avoid high reluctance at the joints where the laminations are butted against each other, the alternative layers are stacked. During the transformer construction first the primary and secondary winding are wound, then the laminations are pushed through the coil openings, layer by layer and the steel core is placed. The laminations are then tightened by means of clamps and bolts. Low-power transformers are air cooled whereas larger power transformers are immersed in oil for better cooling. In oil-cooled transformer, the oil serves as a coolant and also as an insulation medium.

8. Principle of Transformer Action 2/26/2024 8 The primary winding P is connected to an alternating voltage source, therefore, an alternating current I m starts flowing through N 1 turns. The alternating mmf N 1 I m sets up an alternating flux  which is confined to the high permeability iron path as indicated in Figure . The alternating flux induces voltage E 1 in the primary P and E 2 in secondary S. If a load is connected across the secondary, load current starts flowing. Schematic diagram of a two-winding transformer

9. Ideal two-Winding Transformer 2/26/2024 9 For a transformer to be an ideal one, the various assumptions are as follows. Winding resistances are negligible. All the flux set up by the primary links the secondary windings i.e. all of the flux is confined to the magnetic core. The core losses (hysteresis and eddy current losses) are negligible. The core has constant permeability, i.e. the magnetization curve for the core is linear. An ideal transformer is a lossless device with an input winding and an output winding. EMF Equation of a Transformer When a sinusoidal( sine wave) voltage V 1 is applied to the primary , then sinusoidal current I m starts to flow and, therefore, the flux  will flow with the variations of I m . That is, the flux  is in time phase with the current I m and varies sinusoidally . Let sinusoidal variation of flux  be expressed as

10. Contu …. 2/26/2024 10 Where  m is maximum of the magnetic flux in Weber and  = 2  f is the angular frequency in rad /sec and f is the supply frequency in Hz. The emf e 1 in volt, induced in the primary of N 1 turns by the alternating flux is given by: Its maximum value, E 1max occurs when is equal to 1.  And  The RMS value of the induced emf E 1 in the primary winding is

11. Contu …. 2/26/2024 11 Since the primary winding resistance is negligible hence e 1 , at every instant, must be equal and opposite of V 1 . That is, The emf induced in the secondary is  Rms value of emf E 2 induced in secondary winding is given by

12. Contu …. 2/26/2024 12 Voltage Transformation Ratio(K) From emf Eqs . we get The ratio is known as voltage transformation ratio. If N 2 > N 1 i.e., K>1 , then the transformer is called a step-up transformer. If N 2 < N 1 i.e., K<1 , then the transformer is known as a step-down transformer. Again in an ideal transformer Hence, the currents are in the inverse ratio of the (voltage) transformation ratio also, the ratio of and this shows that the emf per turn in each of the windings is the same. Example: A single phase transformer has 350 primary and 1050 secondary turns. The net cross-sectional area of the core is 55 cm 2 . If the primary winding be connected to a 400 V, 50 Hz single phase supply, calculate ( i ) the maximum value of flux density in the core and (ii) the voltage induced in the secondary winding.

13. Contu …. 2/26/2024 13 Solution Voltage applied to the primary = 400 V Induced emf in the primary, E 1  voltage applied to the primary, V 1 = 400 V Number of turns in the primary N 1 = 350 Net cross-sectional area A i = 55 cm 2 = 55  10 -4 m 2 Frequency of the supply f = 50 Hz Induced emf in the primary is given by Maximum value of flux density in the core, Number of turns in the secondary winding, N 2 = 1050 For an ideal transformer, Voltage induced in the secondary winding ,

14. Contu …. 2/26/2024 14 Example : The required no-load voltage ratio in a single phase 50 Hz, core type transformer is 6600/500. Find the number of turns in each winding, if the flux is to be 0.06 Wb . Solution No-load voltage ratio = No-load voltage of low voltage winding = 500 V Flux  = 0.06 Wb Frequency f = 50Hz Induced emf in the low voltage winding (secondary) of the transformer is given by, Number turns in the low voltage, Winding in the core type transformer is accommodated on both the limbs. i.e. half number of turns of each winding on one limb. As such the number of turns in each winding should be even. Considering these facts, the number of turns in low voltage winding, N 2 = 38 Number of turns in high voltage winding Considering all the factors mentioned above, the number of turns in the high voltage winding N 1 =500. because the high voltage winding will be split up into a number of coils. With 250 turns on each limb, high voltage winding on one limb can be split into 5 coils of 50 turns each.

15. Equivalent Circuit of A Transformer 2/26/2024 15 The equivalent circuit for any electrical engineering devices can be drawn if the equations describing its behavior are known. The equivalent circuit for electromagnetic devices consists of a combination of resistances, inductances, capacitances, voltages etc. Equivalent circuit is simply a circuit representation of the equations describing the performance of the device . In the equivalent circuit of transformer ( r l +jx 1 ) and (r 2 + jx 2 ) are the leakage impedances of the primary and secondary windings respectively. The voltage is treated as a voltage drop in the direction of I 1 and it’s magnitude does not change appreciably from no load to full load in large transformers. It’s magnitude depends on f , N1 and  m , since .

16. Contu …. 2/26/2024 16 The primary current I 1 consists of two components. One component is the load component and counteracts the secondary mmf . I 2 N 2 completely. The other component is exciting current I e which is composed of I c and I m . The current I c is in phase with V1 ` and product of V1 ` I c gives core loss. The resistance R c parallel with V1 ` represents the core loss P c , such that. The current I m lags V1 ` by 90° and this can, therefore , be represented in the equivalent circuit by a reactance X m , such that R c and X m shown in the exact equivalent circuit of a transformer are called core-loss resistance and magnetizing reactance, respectively. For minor changes in supply voltage and frequency, which is common under normal operation, R c and X m are treated constant.

17. Contu …. 2/26/2024 17 In transformer analysis, it is usual to transfer the secondary quantities to primary side or primary quantities to secondary side. Secondary resistance drop I 2 r 2 when transferred to primary side must be multiplied by the turns ratio N 1 /N 2 . Secondary resistance drop, when transferred to primary =

18. Contu …. 2/26/2024 18 When the secondary resistance is referred to primary If resistance is placed in the primary circuit, then the relation between voltage V 1 and V 2 is unaffected. This resistance is called the secondary resistance referred to primary . Therefore, the total resistance in the primary circuit is Hence r e1 is called the transformer equivalent (or total) resistance referred to primary winding. Secondary leakage reactance referred to primary and total primary leakage reactance is

19. Contu …. 2/26/2024 19 Likewise, the equivalent or total leakage reactance referred to secondary is The equivalent (or total) leakage impedance referred to primary is The equivalent (or total) leakage impedance referred to secondary is Following the above procedure, it can be shown that

20. Contu …. 2/26/2024 20 In general, when values are referred to either circuit the following conditions should be kept in mind The energy condition (i.e. the active and reactive) power should be remain unchanged The phase angle between voltage and current i.e. power factor, should be remain the same and The referring factor must be the same for all values of the same type. Since the exciting current is only about 2 to 6 percent of the rated winding current in power and distribution transformers, the error introduced by neglecting I e ( r 1 +jx 1 ) or including I e (r 2 + jx 2 ) is insignificant. Therefore the following approximate equivalent circuits can be used

21. Contu …. 2/26/2024 21 Transformer approximate equivalent neglecting shunt branch parameters. Transformer phasor diagram Assume that the secondary load voltage V 2 load current I 2 and angle  2 , by which I 2 lags V 2 are known. First of all draw I 2 lagging V 2 by an angle  2 and then add I 2 (r 2 + jx 2 ) to V 2 to obtain E 2 . It is obvious from figure that current I m due to voltage E 2 , must lag it by 90° and further I c must be in phase with E 2 .

22. Contu …. 2/26/2024 22 The phasor sum of I c and I m gives I e and phasor sum of I 2 and I e gives I 1 . The voltage drop I 1 (r 1 + jx 1 ) is now added to E 2 to obtain V 1 as shown in Figure . The secondary p.f . is cos  2 lagging and the primary p.f . is cos  1 lagging . The voltage drops I 1 (r 1 + jx 1 ) and I 2 (r 2 + jx 2 ) have been drawn to a much larger scale, in comparison-with V 1 or V 2 for the sake of clarity.

23. OPEN-CIRCUIT AND SHORT-CIRCUIT TESTS 2/26/2024 23 These two tests on a transformer help to determine The parameters of the equivalent circuit . The voltage regulation and Efficiency The equivalent circuit parameters can also be obtained from the physical dimensions of the transformer core and its winding details. The power required during these two tests is equal to the appropriate power loss occurring in the transformer. Open Circuit (or No-Load) Test To perform open circuit test on a single phase transformer a voltmeter, wattmeter and an ammeter are connected on the low voltage side of the transformer. The high voltage side is left open circuited.

24. Contu …. 2/26/2024 24 The rated frequency voltage applied to the primary, i.e. low voltage side, is varied with the help of a variable ratio auto-transformer. When the voltmeter reading is equal to the rated voltage of the L.V. winding , all three instrument readings are recorded. (a) Circuit diagram for open-circuit test on a transformer and (b) approximate equivalent circuit at no load The ammeter records the no-load current or exciting current I e . Since I e is quite small (2 to 6%) of rated current), the primary leakage impedance drop is almost negligible, and for all practical purposes, the applied voltage V 1 is equal to the induced emf E 1 .

25. Contu …. 2/26/2024 25 The input power given by the wattmeter reading consists of core loss and ohmic loss. The exciting current being about 2 to 6 percent of the full load current, the ohmic loss in the primary varies from 0.04 percent to 0.36 percent of the full-load primary ohmic loss . In view of this fact, the ohmic loss during open circuit test is negligible in comparison with the normal core loss (approximately proportional to the square of the applied voltage). Hence the wattmeter reading can be taken as equal to transformer core loss . V 1 = Applied rated voltage on L.V. side, I e = exciting current ( or no-load current) and Pc = core loss Then

26. Contu …. 2/26/2024 26  No load p.f . From phase diagram , it follows that Core loss resistance  Magnetizing reactance It must be kept in mind that the values of R c and X m , in general, refer to the side, in which the instruments are placed (the L.V. side in the present case).

27. Contu …. 2/26/2024 27 A voltmeter is sometimes, used at the open-circuited secondary terminals, in order to determine the turns ratio. Thus the open-circuit test gives the following information: ( i ) core loss at rated voltage and frequency, (ii) the shunt branch parameters of the equivalent circuit, i.e. R c and X m and (iii) turns ratio of the transformer. Short-Circuit Test The low voltage-side of the transformer is short-circuited and the instruments are placed on the high voltage side.

28. Contu …. 2/26/2024 28 The applied voltage is adjusted by auto-transformer, to circulate rated current in the high voltage side. In a transformer, the primary m.m.f . is almost equal to the secondary m.m.f ., therefore, a rated current in the H.V. winding causes rated current to flow in the L.V. winding. connection diagram for short circuit test on a transformer A primary voltage of 2 to 12% of its rated value is sufficient to circulate rated currents in both primary and secondary windings. The wattmeter, in short circuit test, records the core loss and the ohmic loss in both windings .

29. Contu …. 2/26/2024 29 Since the core loss has been proved to be almost negligible in comparison with the rated voltage core loss, the wattmeter can be taken to register only the ohmic losses in both windings. At rated-voltage, the exciting Current is 2 to 6% of full load current. When the voltage across the exciting branch is 1 to 6% of rated voltage, the exciting current may be 0.02 percent to 0.36% percent of its full-load current and can, therefore, be safely ignored. Let V SC , I SC and P SC be the voltmeter, ammeter and wattmeter readings; then equivalent leakage impedance referred to H.V. side, Equivalent resistance and equivalent leakage reactance referred to H.V. side, are

30. Contu …. 2/26/2024 30 In the analysis of transformer equivalent circuit, the values of equivalent resistance and equivalent leakage reactance referred to either side are used. However, if the leakage impedance parameters for both primary and secondary are required separately, then it is usual to take r 1 = r 2 =½ r e ) and x 1 = x 2 =½ x e , referred to the same side. Thus, the short-circuit test gives the following information: ( i ) ohmic loss at rated current and frequency and (ii) the equivalent resistance and equivalent leakage reactance. Voltage regulation of a transformer can be determined from the data obtained from short-circuit test. Data of both open-circuit and short-circuit tests is necessary ( i ) for, obtaining all the parameters of exact equivalent circuit and (ii) for calculating the transformer efficiency

31. Contu …. 2/26/2024 31 Example : A 20 kVA , 2500/250 V, 50 Hz, single-phase transformer gave the following test result Open-circuit test (on L.V. side):250 V, 1.4 A, 105 W. Short-circuit test (on H.V. side): 104 V, 8 A, 320 watts. Compute the parameters of the approximate equivalent circuit referred to high-voltage and low-voltage sides. Also draw the exact equivalent circuit referred to the low -voltage side. Solution From open-circuit test: No-load power factor,  = 72.55  and sin  =0.954 I c = I e cos  = 1.4  0.3 = 0.42 A and I m = I e sin  = 1.4  0.954 = 1.336 A  

32. Contu …. 2/26/2024 32 hence, Alternatively , the value of R cL and X mL can be determined as follows:

33. Contu …. 2/26/2024 33 From short circuit test: Equivalent circuit parameters referred to L.V. side are: R cL = 595  X mL = 187   

34. Contu …. 2/26/2024 34 This equivalent circuit is shown in Figure below (a) (a) approximate equivalent circuit referred to L.V. side and (b)exact equivalent circuit referred to L.V. side. Equivalent circuit parameters referred to H.V. side are: r eH = 5  ; x eH = 12 

35. Contu …. 2/26/2024 35 Exact equivalent circuit parameters referred to L.V. side are: R cL = 595  and X mL = 187 

36. Transformer regulation, loss and Efficiency 2/26/2024 36 VOLTAGE REGULATION Constant voltage is the characteristics of most domestic, commercial and industrial loads. It is therefore, necessary that the output voltage of a transformer must remain within narrow limits as the load and its power factor vary. This requirement is more stringent in distribution transformers as these directly feed the load centers. The voltage regulation is defined as voltage in secondary terminal voltage, expressed as a percentage (per unit) of secondary rated voltage i.e. Where E 2 = Secondary terminal voltage at no load V 2 = Secondary terminal voltage at any load It is stipulated that the secondary rated voltage of a transformer is equal to the secondary terminal voltage at no load, i.e. E 2 . 

37. Contu …. 2/26/2024 37 At no-load, the primary leakage impedance drop is almost negligible, therefore, the secondary no-load voltage . The expression for voltage regulation can also be written as Here V1 is the primary applied voltage. The change in secondary terminal voltage with load current is due to the primary and secondary leakage impedances of the transformer. The magnitude of this change depends on the load power factor , load current , total resistance and leakage reactance of a transformer. A distribution transformer should have a small value of voltage regulation (i.e. good voltage regulation) so that the terminal voltage at the consumers does not vary widely as the load changes. For a transformer of large voltage regulation (i.e. poor voltage regulation), the voltage at the consumers' terminals will fall appreciably with increase in load.

38. Contu …. 2/26/2024 38 Thus distribution transformer should be designed to have a low value of leakage impedances. The voltage regulation of a transformer can be obtained form its approximate equivalent circuit referred to primary or secondary. From phasor diagram Thus the voltage drop in the secondary terminal voltage The change in secondary terminal voltage is equal to the magnitude of E 2 minus the magnitude of V 2 . Per unit voltage regulation for any load current I 2 is

39. Contu …. 2/26/2024 39 In case I 2 , is rated current, then Similarly, for rated current I 2 , The per unit voltage regulation at rated current is given by Percentage voltage regulation at rated load

40. Contu …. 2/26/2024 40 For leading power factor loads, the phasor diagram of Figure reveals that Therefore , secondary terminal voltage drop, for any load current I 2 , is p.u . voltage regulation at any load current I 2 is given by In case I2 is the rated ( or full-load)current, then p.u . voltage regulation is given by Condition for zero voltage regulation If load power factor is varied with constant values of load current and secondary emf , then zero voltage regulation will occur when

41. Contu …. 2/26/2024 41 From the above equation Magnitude of the load p.f . The negative value of tan  2 indicates a leading power factor. Therefore, zero voltage regulation occurs when load power factor is leading . For leading p.f.s . greater than , the voltage regulation will be negative, i.e. the voltage will rise from its no load value, as the transformer load is increased. Condition for maximum voltage regulation The condition for maximum voltage regulation is obtained by differentiating the above expression with respect to  2 and equating the results to zero. Here again the load current and secondary emf are assumed to remain constant.

42. Contu …. 2/26/2024 42 Here tan  2 is positive, therefore, maximum voltage regulation occurs at lagging load p.f . equal to . In other words, maximum voltage regulation occurs when load power-factor angle  2 is equal to the leakage impedance angle  of the transformer. The magnitude of maximum voltage regulation is: Thus the magnitude of maximum voltage regulation is equal to the p.u value equivalent leakage impedance of the transformer. Example : A 6600/440 V, single-phase transformer has an equivalent resistance of 0.02 p.u . and an equivalent reactance of 0.05 p.u . Find the full-load voltage regulation at 0.8 pf lagging, if the primary voltage is 6600 V. Find also the secondary terminal voltage at full load.

43. Contu …. 2/26/2024 43 Solution  For a primary voltage of 6600 V, the secondary no load voltage E 2 is 440 V.  The change in the secondary terminal voltage E 2  V 2 = 440 (0.046) = 20.25 V and secondary terminal voltage V 2 = 440 - 20.25 V

44. Contu …. 2/26/2024 44 Example: A short-circuit test, when performed on the H.V. side of a 10 kVA , 2000/400 V, single-phase transformer gave the following data: 60 V, 4 A, 100 W If the L.V. side is delivering full load (or rated) current at 0.8 p.f . lag and at 400 V, find the voltage applied to H.V. side. Solution From short circuit data For the L.V. side, the parameters are Full load secondary current

45. Contu …. 2/26/2024 45 Now For V 2 = 400 V, E 2 = 400 + 13.16 = 413.16 V  The voltage applied to the primary is

46. Transformer Losses and Efficiency 2/26/2024 46 Equipment is desired to operate at a high efficiency. Fortunately, losses in transformers are small. Because the transformer is a static device , there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer the efficiency can be as high as 99%. Transformer Losses There are mainly two kinds of losses in a transformer, namely Core loss and ohmic loss Core loss: The core loss P c occurring in the transformer iron, consists of two components, hysteresis loss P h and eddy current loss P e i.e. P c = P h + P e

47. Contu …. 2/26/2024 47 The hysteresis and eddy current losses in the core can be expressed by Where K h = proportionality constant which depends upon the volume and quality of the core material and units used. K e = Proportionality constant whose value depends on the volume and resistivity of the core material, thickness of laminations and the units employed B m = maximum flux density in the core and f = frequency of the alternating flux The value of the exponent x (called Steinmetz’s constant) varies from 1.5 to 2.5 depending upon the magnetic properties of the core material. Therefore, the total core loss is P c = K h fB 1.6 m + K e f 2 B 2 m

48. Contu …. 2/26/2024 48 Ohmic Loss When a transformer is loaded, ohmic loss (I 2 R) occurs in both the primary and secondary winding resistances. Since the standard operating temperature of electrical machines is 75 C. The ohmic loss should be calculated at 75 C. Transformer Efficiency The efficiency of a transformer ( or any other device) is defined as the ratio of the output power to input power. Thus Where P c = total core loss I 2 2 R = total ohmic losses V 2 I 2 = output VA Cos  2 = load power factor

49. Contu …. 2/26/2024 49 Condition for Maximum Efficiency . P c is constant and the load voltage V 2 remains practically constant. A specified values of load p.f . cos  2 , the efficiency will be maximum when . Hence the maximum efficiency occurs when the variable ohmic loss is equal to the fixed core loss Pc. The load current I 2 at which maximum efficiency occurs is given by

50. Contu …. 2/26/2024 50 If both sides of above equation are multiplied by , we get  kVA load for maximum Thus the maximum efficiency, for a constant load current, occurs at unity power factor (i.e. at purely resistive load). The load current at which maximum efficiency occurs does not depend upon the load power factor because P c and r e2 are almost unaffected by a variation in the load power factor. A reduction in the load power factor reduces the transformer output and therefore the transformer efficiency is also reduced accordingly .

51. Contu …. 2/26/2024 51 Note that transformer efficiency is maximum at the same load current regardless of variation in the load power factor. Effect power factor on efficiency Example : A 100 kVA , 1000/10000 V, 50 Hz, single phase transformer has an iron loss of 1100 W. The copper loss with 5 A in the high voltage winding is 400 W. Calculate the efficiencies at ( i ) 25 %, (ii) 50 % and (iii) 100 % of normal load for power factors of (a) 1.0 and (b) 0.8. The output terminal voltage being maintained at 10000 V. Find also the load for maximum efficiency at both power factors.

52. Contu …. 2/26/2024 52 Solution Efficiency at 25% of normal load, unity pf : Iron losses = 1100 W Copper losses with 5 A in secondary = 400 W Secondary full load current, I 2 Current in the secondary at 25 % full load = ¼  10 = 2.5 A Copper losses at 25% full load = = 100 W Output at 25% full load = 0.25  100  1000  1 = 25 000 W Efficiency at 25 % load, unity pf

53. Contu …. 2/26/2024 53 Efficiency at 25 % full load, 0.8 pf : Efficiency at 50 % full load, unity pf : Copper losses at 25% full load = 400 W Output at 50 % full load, unity pf = 0.5  100  1000  1 = 50 000 W Iron losses = 1100 W Efficiency at 50 % full load, 0.8 pf :

54. Contu …. 2/26/2024 54 Efficiency at 100 % full load, unity pf : Copper losses at 100% full load = = 1600 W Output = 100  1000  1 = 100 000 W Iron losses = 1100 W Efficiency at full load, 0.8 pf :

55. Contu …. 2/26/2024 55 Load for maximum efficiency at unity pf : Maximum efficiency occurs when the iron losses equal the copper losses. Let the maximum efficiency occur at x per cent of the full load. Copper losses at x % of full load = x 2  1600 Thus x 2  1600 = 1100 x = 0.829 Load for maximum efficiency = 0.829  100 = 82.9 kVA Load for maximum efficiency at 0.8 pf : Load for maximum efficiency will remain the same irrespective of power factor Thus load for maximum efficiency = 82.9 kVA Example : A single phase transformer working at unity power factor has an efficiency of 90% at both one half load and at the full load of 500 W. Determine the efficiency at 75 % of full load.

56. Contu …. 2/26/2024 56 Solution Efficiency of the transformer at full-load = 0.9 Output at full load = 500 W Let the iron losses of the transformer be = x watts and the total copper losses at full load be = y watts Then, the total losses at full load = x + y Hence , Or 0.9 x + 0.9 y = 50 Efficiency of the transformer at half of full load = 0.9 Total copper losses at half of full load = Output of the transformer = ½  500 = 250 W Thus, 0.9 x + 0.225 y = 25

57. Contu …. 2/26/2024 57 Solving the above equations y = 37 W and x = 18.53 W i.e. total copper losses at full load = 37 W and Iron losses = 18.53 W Output of the transformer at 75 per cent of full load = 0.75 x 500 = 375 W Total copper losses at 75 per cent of full load = (0.75) 2 x 37 = 20.8 W Efficiency at 75 percent of full load

58. Polarity test and Parallel Operation of 1-phase Transformers 2/26/2024 58 Polarity test On the primary side of a two-winding transformer, one terminal is positive with respect to the other terminal at any one instant. At the same instant, one terminal of the secondary winding is positive with respect to the other terminal. These relative polarities of the primary and secondary terminals at any instant must be known if the transformers are to be operated in parallel or are to be used in a polyphase circuit.

59. Contu …. 2/26/2024 59 Terminals A 1 and A 2 marked plus and minus arbitrarily. Now terminal A 1 is connected to one end of the secondary winding and a voltmeter is connected between A 2 and the other end of the secondary winding. A voltage of suitable value is now applied to the H.V. winding. Let E 1 and E 2 be the e.m.fs induced on H.V. and L.V. sides respectively. If the voltmeter reading is equal to E 1 –E 2 then secondary terminal connected to A 1 is positive and is marked a 1 , the L.V. terminal connected to A 2 through the voltmeter is negative and is marked a 2 . If voltmeter reading is equal to E 1 +E 2 , then the terminals connected to A 1 and A 2 are negative and positive and are marked a 2 and a 1 respectively. The subscript numbers 1,2 on the H.V. and L.V. windings are so arranged that when A 2 is negative with respect to A 1 . a 2 is also negative with respect to a 1 at the same instant. In other words, if the instantaneous emf is directed from A 2 to A 1 in H.V. winding, it is at the same time directed from a 2 to a 1 in the L.V. winding

60. Contu …. 2/26/2024 60 When the voltmeter reads the difference E 1 –E 2 , the transformer is said to possess a subtractive polarity and when voltmeter reads E 1 +E 2 the transformer has additive polarity . In subtractive polarity, the voltage between A 2 and a 2 (or A 1 and a 1 ) is reduced. The leads connected to these terminals and the two windings are, therefore, not subjected to high voltage stress. In additive polarity the windings and the leads connected to A 1 , A 2 , a 1 and a 2 are subjected to high voltage stresses. On account of these reasons, subtractive polarity is preferable to additive polarity.

61. Contu …. 2/26/2024 61 PARALLEL OPERATION OF SINGLE-PHASE TRANSFORMERS When electric power is supplied to a locality a single transformer, capable of handling the required power demand, is installed. Though two or more transformers may be expensive than one large unit, In some cases, it may be preferable to install two or more transformers in parallel, instead of one large unit due to the following advantages .  With two or more transformers, power system becomes more reliable . For instance if one transformer develops a fault, it can be removed and the other transformers can maintain the flow of power , though at reduced load.  Transformers can be switched on or off , depending upon the power demand. In this manner, the transformer losses decrease and the system becomes more economical and efficient in operation .  The cost of standby (or spare) unit is much less when two or more transformers are installed.  Electric power demand may become more than rated KVA capacity of already existing transformer or transformers. Under such circumstances, the need for extra transformer arises; the extra unit must be connected in parallel

62. Contu …. 2/26/2024 62 Note that the parallel operation of transformers requires that their primary windings as well as secondary windings are connected in parallel .   The various conditions which must be fulfilled for the satisfactory parallel operation of two or more single-phase transformers are as follows:  A)The transformer must have the same voltage ratios, i.e with the primaries connected to the same voltage sources, the secondary voltage of all transformers should be equal in magnitude.  B)The equivalent leakage impedance in ohms must be inversely proportional to their respective KVA ratings., In other words, per unit ( pu ) leakage impedance of transformers based on their KVA rating must be equal.  C)The ratio of equivalent leakage reactance to equivalent resistance i.e. X e /r e should be the same for all transformers . 

63. Contu …. 2/26/2024 63 D)The transformer must be connected properly as far as their polarities are concerned.  Out of the conditions listed above, condition(d) must be strictly fulfilled. If the secondary terminals are connected with wrong polarities, large circulating currents will flow and the transformers may get damaged. Condition (a) should be satisfied as accurately as possible ; since different secondary voltages would give rise to undesired circulating currents. For conductions (b) and (c), some deviation is permissible. Thus the fulfillment of condition (d) is essential whereas the fulfillment of the other conditions is desirable.

64. Contu …. 2/26/2024 64 Two single-phase transformers in parallel   The Figure shows single-phase transformers in parallel, connected to some voltage source on the primary side. Zero voltmeter reading indicates proper polarities. If the voltmeter reads the sum of two secondary voltages, the polarities are improper and can be corrected by reversing the secondary terminals of any one of the transformers.

65. THREE-PHASE TRANSFORMERS 2/26/2024 65 Generation, transmission and distribution of electric energy is invariably done through the use of three-phase systems. A large number of three-phase transformers are inducted in a 3-phase energy system for stepping-up or stepping – down the voltage as required. For 3-phase up or down transformation, three units of 1-phase transformers or one unit of 3-phase transformer may be used. When three identical units of 1-phase transformers are used the arrangement is usually called a bank of three transformers or a 3-phase transformer bank. A single 3-phase transformer unit may employ 3–phase core-type construction or three phase shell type construction. A single-unit 3-phase core-type transformer uses a three-limbed core, one limb for each phase winding.

66. Contu …. 2/26/2024 66 A 3-phase core-type transformer costs about 15% less than a bank of three 1-phase transformers. Also, a single unit occupies less floor space than a bank. (a)Three-phase transformer bank, both windings in star;(b) three-phase core-type transformer

67. Three-Phase Transformer Connections 2/26/2024 67 Three-phase transformers may have the following four standard connections (a) Star-Delta ( Y-  ) (b) Delta-Star (  -Y) (c) Delta-Delta (  -  ) (d) Star-Star (Y-Y) Let V and I are taken as input line voltage and line current respectively. Primary and secondary windings of one phase are drawn parallel to each other. With phase turns ratio from primary to secondary as N 1 /N 2 = a, the voltages and current in the windings and lines can be calculated. Star-delta ( Y-  ) Connection This connection is commonly used for stepping down the voltage from a high level to a medium or low level. The insulation on the h.v . side of the transformer is stressed only to 57.74%

68. Contu …. 2/26/2024 68 For per-phase m.m.f . balance, I 2 N 2 =I 1 N 1 Here primary phase current, I 1 = primary line current I Also, voltage per turn on primary = voltage per turn on secondary Secondary phase voltage, Secondary line voltage = secondary phase voltage =

69. Contu …. 2/26/2024 69 Input VA = 3 I = output VA = 3. Phase and line values for voltages and currents on both primary and secondary sides of star-delta transformer are shown in Figure. (a) Star-delta connection and (b) delta-star connection of 3-phase transformers Delta-Star (  -Y) connection This type of connection is used for stepping up the voltage to a high level. For example, these are used in the beginning of h.v . transmission lines so that insulation is stressed to about 57.74% of line voltage.

70. Contu …. 2/26/2024 70 Delta-star transformers are also generally used as distribution transformers for providing mixed line to line voltage to high-power equipment and line to neutral voltage to 1-phase low-power equipment. Delta-star distribution transformer is used to distribute power to consumers by 3-phase four-wire system. Three-phase high–power equipment is connected to 400V, three line wires, whereas 1-phase low-power equipment is energized from 231 V line to neutral circuits.

71. Contu …. 2/26/2024 71

72. Contu …. 2/26/2024 72 Delta-Delta (  -  ) Connection This scheme of connections is used for large 1.v transformers. It is because a delta-connected winding handles line voltage, so it requires more turns per phase but of smaller cross-sectional area. The absence of star point may be a disadvantage in some applications. In case a bank of three transformers is used, then one transformer can be removed for maintenance purposes while the remaining two transformers (called an open-delta or V-connection) can still deliver 58% of the power delivered by the original 3-phase transformer bank. For per phase mmf balance, I 2 N 2 = I 1 N 1 :

73. Contu …. 2/26/2024 73

74. Contu …. 2/26/2024 74 Phase and line value for voltages and currents on both primary and secondary sides of a 3-phase delta-delta transformer are shown in Figure . Figure (a) Delta-delta connection and (b) Star-star connection of three-phase transformers. Star-Star (Y-Y) Connection This connection is used for small h.v transformers . With star connection, turns per phase are minimum and the winding insulation is stressed to 57.74% of line voltage. Star-star connection is rarely used in practice because of oscillatory neutral problems.

75. Contu …. 2/26/2024 75

76. Contu …. 2/26/2024 76 Example : A 3-phase transformer is used to step-down the voltage of a 3-phase, 11kV feeder line. Per-phase turns ratio is 12. For a primary line current of 20A, calculate the secondary line voltage, line current and output KVA for the following Connections: (a)star-delta (b) delta-star (c) delta-delta (d) star-star. Neglect losses. Solution a) Three-phase transformer with star-delta connection.

77. Contu …. 2/26/2024 77 (b) Delta-star connection of 3-phase transformer. (c) Delta-delta connection of 3-phase transformer.

78. Contu …. 2/26/2024 78 3-phase transformer with star-star connection. Example : An 11000/415V, delta-star transformer feeds power to a 30 kW, 415V, 3-phase induction motor having an efficiency of 90% and full-load pf 0.833. Calculate the transformer rating and phase and line currents on both high and low voltage sides.

79. Contu …. 2/26/2024 79 Solution For star connected 1.v. winding, phase current in 1.v. winding = line current on 1.v side = 55.65A. Line current on HV, side of transformer = For delta connected HV winding, phase current in HV winding: = ( line current on h.v . side) = x 2.1 = 1.212A

80. AUTOTRANSFORMERS 2/26/2024 80 In principle and in general construction, the autotransformer does not differ from the conventional two-winding transformer so far discussed. It does differ from it, however, in the way the primary and secondary windings are interrelated. When the primary and secondary voltage are derived from the same winding , the transformer is called an autotransformer . An ordinary two-winding transformer may also be used as an autotransformer by connecting the two windings in series and applying the impressed voltage across the two, or merely to one of the windings. It depends on whether it is desired to step the voltage down or up, respectively.

81. Contu …. 2/26/2024 81 Figure Autotransformers: (a) step-down; (b) step-up . The input voltage V 1 is connected to the complete winding (a-c) and the load R L is connected across a portion of the winding, that is, (b-c). The voltage V 2 is related to V 1 as in the conventional two-winding transformer, that is,

82. Contu …. 2/26/2024 82 Where N bc and N ac are the number of turns on the respective windings. The ratio of voltage transformation in an autotransformer is the same as that for an ordinary ­transformer, namely, with k > 1 for step-down. Assuming a resistive load for convenience, then, Assume that the transformer is 100% efficient. The power output is Note that I 1 flows in the portion of winding ab , whereas the current (I 2 – I 1 ) flows in the remaining portion bc .

83. Contu …. 2/26/2024 83 The resulting current flowing in the winding bc is always the arithmetic difference between I 1 and I 2 , since they are always in opposite sense. Remember that the induced voltage in the primary opposes the ­primary voltage. As a result ,the current caused by the induced voltage flows opposite to the input current. In an autotransformer, the secondary current is this induced current, that is, The power required by the load is determined by

84. Contu …. 2/26/2024 84 The load power consists of two parts. The first part is The second part is Most of the power to the load is directly conducted by winding ab where as, the remaining power is transferred by the common winding bc .

85. Contu …. 2/26/2024 85 Example : A standard 5-kVA 2300/230-V distribution transformer is connected as an autotransformer to step down the voltage from 2530 V to 2300 V. The transformer connection is as shown in slide81 Figure (a). The 230-V winding is section ab , the 2300-V winding is bc . Compare the kVA rating of the autotrans­former with that of the original two-winding transformer. Also calculate P c , P tr , and the currents. Solution The rated current in the 230-V winding (or in ab ) is The rated current in the 2300-V winding (or in bc ) is Therefore ,

86. Contu …. 2/26/2024 86 The secondary current I 2 can also be calculated from Since the transformation ratio The conducted power is and that transformed is An autotransformer of given physical dimensions can handle much more load power than an equivalent two-winding transformer. An autotransformer is not suitable for large percentage voltage reductions as is a distribution transformer. This is due to the required turns ratio becoming too large; hence the power-handling advantage would be minimal.

87. Contu …. 2/26/2024 87 In situations where autotransformers can be used to their full advantage, it will be found that they are cheaper than a conventional two-winding transformer of similar rating. They also have better regulation (i.e., the voltage does not drop so much for the same load), and they operate at higher efficiency. In all applications using autotransformer it should be realized that the primary and secondary circuits are not electrically isolated, since one input terminal is common with one output terminal.

88. END 2/26/2024 88 Thank you!!! Any Question