- 2. Introduction Constituents of transformer: i. Magnetic Circuit ii. Electric Circuit iii. Dielectric Circuit iv. Other accessories 2
- 3. Classification or Types Transform ers Based on Core Core Type Shell Type Based on transform er Ratio Step up Step Down Based on Service Distributio n Transform er Power Transform er 3
- 8. Constructional Details The requirements of magnetic material are, High permeability Low reluctance High saturation flux density Smaller area under B-H curve For small transformers, the laminations are in the form of E,I, C and O as shown in figure The percentage of silicon in the steel is about 3.5. Above this value the steel becomes very brittle and also very hard to cut 8
- 9. Transformer Core Core type Construction Shell Type Construction 9
- 10. Stepped Core Construction Transformer Core 10
- 11. Shell Type Core Type Transformer Core Ww Hw 11
- 12. Transformer Winding Types of transformer windings are, Concentric Sandwich Disc Cross over Helical 12
- 15. Insulations Dry type Transformers: Varnish Enamel Large size Transformers: Unimpregnated paper Cloths Immersed in Transformer oil – insulation & coolent 15
- 16. Comparison between Core & Shell Type Description Core Type Shell Type Construction Easy to assemble & Dismantle Complex Mechanical Strength Low High Leakage reactance Higher Smaller Cooling Better cooling of Winding Better cooling of Core Repair Easy Hard Applications High Voltage & Low output Low Voltages & Large Output 16
- 17. Classification on Service Details Distribution transformer Power Transformer Capacity Upto 500kVA Above 500kVA Voltage rating 11,22,33kV/440V 400/33kV;220/11kV…etc., Connection Δ/Y, 3φ, 4 Wire Δ/Δ; Δ/Y, 3φ, 4 Wire Flux Density Upto 1.5 wb/m2 Upto 1.7 wb/m2 Current Density Upto 2.6 A/mm2 Upto 3.3 A/mm2 Load 100% for few Hrs, Part loadfor some time, No-load for few Hrs Nearly on Full load Ratio of Iron Loss to Cu loss 1:3 1:1 Regulation 4 to 9% 6 to 10% Cooling Self oil cooled Forced Oil Cooled 17
- 18. Output Equation of Transformer It relates the rated kVA output to the area of core & window The output kVA of a transformer depends on, Flux Density (B) – related to Core area Ampere Turns (AT) – related to Window area Window – Space inside the core – to accommodate primary & secondary winding Let, T- No. of turns in transformer winding f – Frequency of supply Induced EMF/Turn , Et=E/T=4.44fφm ……. (1) 18
- 19. Window in a 1φ transformer contains one primary & one secondary winding. Output Equation of Transformer ) 3 ( a I a I windings the both in same is ) ( Density Current ) 2 ( A K A window, in area Conductor A A factor,K Space Window Window of area Total Window in area Conductor factor,K Space Window s s p p w w c w c w w 19
- 20. Output Equation of Transformer s s p p I a ; I a If we neglect magnetizing MMF, then (AT)primary = (AT)secondary AT=IpTp=IsTs (4) Total Cu. Area in window, Ac=Cu.area of pry wdg + Cu.area of sec wdg ) 5 ( AT 2 AT AT 1 I T I T 1 I T I T a T a T a T a T conductor sec of section - X of area X turns sec of . No conductor pry of section - X of area X turns pry of . No s s p p s s p p s s p p s s p p 20
- 21. Therefore, equating (2) & (5), kVA rating of 1φ transformer is given by, Output Equation of Transformer ) 6 ( A K 2 1 AT AT 2 A K w w w w 3 w w m 3 w w m 3 t t 3 - p p p p -3 p p -3 p p 10 A K f 22 . 2 10 A K 2 1 . f 44 . 4 ) 6 ( 10 AT E T E E ), 1 ( from 10 I T T E 10 I E 10 I V Q 21
- 22. We know that, Three phase transformer: Each window has 2 primary & 2 Secondary windings. Total Cu. Area in the window is given by, Output Equation of Transformer i m m i m m A B and A B kVA 10 K A A B f 22 . 2 Q 3 w w i m Ww Hw 4 A K AT A K AT 4 ), 7 ( & ) 2 ( Compare ) 7 ( AT 4 A a T 2 a T 2 A w w w w c s s p p c 22
- 23. kVA rating of 3φ transformer, Output Equation of Transformer kVA 10 K A A B f 33 . 3 10 A K 4 1 . f 44 . 4 3 10 AT E 10 I T T E 3 10 I E 3 Q 3 w w i m 3 w w m 3 t 3 - p p p p 3 p p 23
- 24. EMF per Turn Design of Xmer starts with the section of EMF/turn. Let, f 44 . 4 10 r Q f 44 . 4 10 r Q 10 f 44 . 4 r Q 10 r f 44 . 4 10 ) AT ( f 44 . 4 10 I T f 44 . 4 10 I V Q AT r loading Electric to loading magnetic Specific of Ratio 3 m 3 3 2 m 3 m m 3 m 3 p p m 3 p p m 24
- 25. Q 10 r f 44 . 4 f 44 . 4 Q . 10 r f 44 . 4 f 44 . 4 f 44 . 4 10 r Q f 44 . 4 f 44 . 4 E , t . k . w 3 3 3 m t Q K Et EMF per Turn 3 10 r f 44 . 4 K , where K depends on the type, service condition & method of construction of transformer. 25
- 26. EMF per Turn Transformer Type Value of K 1φ Shell Type 1.0 to 1.2 1φ Core Type 0.75 to 0.85 3φ Shell Type 1.2 to 1.3 3φ Core Type Distribution Xmer 0.45 to 0.5 3φ Core Type Power Xmer 0.6 to 0.7 26
- 27. Optimum Design Transformer may be designed to make one of the following quantitites as minimum. i. Total Volume ii. Total Weight iii. Total Cost iv. Total Losses In general, these requirements are contradictory & it is normally possible to satisfy only one of them. All these quantities vary with AT r m 27
- 28. Optimum Design Design for Minimum Cost Let us consider a single phase transformer. kVA 10 K A A B f 22 . 2 Q 3 w w i m w w c 3 c i m A K A kVA 10 A A B f 22 . 2 Q Assuming that φ & B are constant, Ac.Ai – Constant Let, In optimum design, it aims to determining the minimum value of total cost. 2 A A K 2 1 AT A B AT r c w w i m m m 28 ) 1 ( M2 A A i c
- 29. Optimum Design Design for Minimum Cost c i m c i m A A B 2 2 A A B r ) 2 ( r B 2 A A B 2 A A r m c i m c i β is the function of r alone [δ & Bm – Constant] From (1) & (2), 29 M A & M A c i 2 M A A i c
- 30. 30 Optimum Design Design for Minimum Cost Let, Ct - Total cost of transformer active materials Ci – Cost of iron Cc – Cost of conductor pi – Loss in iron/kg (W) pc – Loss in Copper/kg (W) li – Mean length of flux path in iron(m) Lmt – Mean length of turn of transformer winding (m) Gi – Weight of active iron (kg) Gc – Weight of Copper (kg) gi – Weight/m3 of iron gc – Weight/m3 of Copper c c i i c i t G c G c C C C
- 31. 31 . ly respective copper and iron of ts cos specific c & c , where A g c A g c C c i c c c i i i t mt i L l Optimum Design Design for Minimum Cost M g c M g c C c c i i t mt i L l Differeniating Ct with respect to β, 2 / 3 c c 2 / 1 i i t M g c 2 1 ) ( M g c 2 1 C d d mt i L l For minimum cost, 0 C d d t 2 / 3 c c 2 / 1 i i M g c 2 1 ) ( M g c 2 1 mt i L l 1 c c i i g c g c mt i L l i c c c i i A A g c g c mt i L l
- 32. 32 c c c i i i A g c A g c mt i L l Optimum Design Design for Minimum Cost c i c c i i C C G c G c Hence for minimum cost, the cost of iron must be equal to the cost of copper. Similarly, For minimum volume of transformer, Volume of iron = Volume of Copper For minimum weight of transformer, Weight of iron = Weight of Conductor For minimum loss, Iron loss = I2R loss in conductor c i c i c c i i g g G G or g G g G c i G G c 2 i P P x
- 33. Optimum Design Design for Minimum Loss and Maximum Efficiency Total losses at full load = Pi+Pc At any fraction x of full load, total losses = Pi+x2Pc If output at a fraction of x of full load is xQ. Efficiency, Condition for maximum efficiency is, 33 c 2 i P x P xQ xQ x 0 d d x x c 2 i c 2 c 2 c 2 i c c 2 i c c 2 i c 2 i c c 2 i P x P P x P x xQ P x P xQ x 2xP Q P x P xQ xQ 2xP Q Q P x P xQ P x P xQ xQ 2xP Q - Q P x P xQ x 0 d d 2 x
- 34. Variable losses = Constant losses for maximum efficiency 34 Optimum Design Design for Minimum Loss and Maximum Efficiency i c 2 c i c c i i 2 c c i i c i p p x G G or G p G p x G p G p P P
- 35. Design of Core Core type transformer : Rectangular/Square /Stepped cross section Shell type transformer : Rectangular cross section 35
- 36. For core type distribution transformer & small power transformer for moderate & low voltages Rectangular coils are used. For shell type, 36 Design of Core Rectangular Core 2 4 . 1 to Width Depth 3 2 to Core of Depth limb Central of Width
- 37. Used when circular coils are required for high voltage distribution and power transformer. Circular coils are preferred for their better mechanical strength. Circle representing the inner surface of the tubular form carrying the windings (Circumscribing Circle) 37 Design of Core Square & Stepped Core
- 38. Dia of Circumscribing circle is larger in Square core than Stepped core with the same area of cross section. Thus the length of mean turn(Lmt) is reduced in stepped core and reduces the cost of copper and copper loss. However, with large number of steps, a large number of different sizes of laminations are used. 38 Design of Core Square & Stepped Core
- 39. Let, d - diameter of circumscribing circle a – side of square Diameter, Gross core area, Let the stacking factor, Sf=0.9. Net core area, 39 Design of Core Square Core 2 2 2 2 2 2 d a a a a a d 2 2 2 5 . 0 2 d A d a square of Area A gi gi 2 2 45 . 0 5 . 0 9 . 0 d d Ai
- 40. Gross core area includes insulation area Net core area excludes insulation area Area of Circumscribing circle is Ratio of net core area to Area of Circumscribing circle is Ratio of gross core area to Area of Circumscribing circle is 40 2 4 d 58 . 0 4 45 . 0 2 2 d d 64 . 0 4 5 . 0 2 2 d d Design of Core Square Core
- 41. Useful ratio in design – Core area factor, 41 Design of Core Square Core 45 . 0 d d 45 . 0 d A K 2 2 2 i C Circle bing Circumscri of Square area Core Net
- 42. Let,a – Length of the rectangle b – breadth of the rectangle d – diameter of the circumscribing circle and diagonal of the rectangle. θ – Angle b/w the diagonal and length of the rectangle. 42 Design of Core Stepped Core or Cruciform Core θ d b b a a (a-b)/2 (a-b)/2 The max. core area for a given ‘d’ is obtained by the max value of ‘θ’ For max value of ‘θ’, From the figure, 0 d dAgi sin d b d b sin cos d a d a cos
- 43. Two stepped core can be divided in to 3 rectangles. Referring to the fig shown, 43 Design of Core Stepped Core or Cruciform Core θ d b b a a (a-b)/2 (a-b)/2 2 2 b ab 2 b ab ab b 2 ) b a ( 2 ab b 2 b a b 2 b a ab gi A area, core Gross On substituting ‘a’ and ‘b’ in the above equations, 2 2 2 gi 2 2 2 gi 2 gi sin d 2 sin d A sin d sin cos d 2 A ) sin d ( ) sin d )( cos d ( 2 A For max value of ‘θ’, 0 d dAgi
- 44. i.e., Therefore, if the θ=31.720, the dimensions ‘a’ & ‘b’ will give maximum area of core for a specified ‘d’. 44 Design of Core Stepped Core or Cruciform Core 0 1 1 2 2 2 2 gi 72 . 31 2 tan 2 1 2 tan 2 2 2 tan 2 2 cos 2 sin 2 sin 2 cos 2 cos sin 2 d 2 cos 2 d 0 cos sin 2 d 2 cos 2 d d dA d 53 . 0 ) 72 . 31 sin( d b sin d b d b sin d 85 . 0 ) 72 . 31 cos( d a cos d a d a cos 0 0
- 45. Gross core area, The ratios, 45 2 2 2 gi 2 gi 2 gi d 56 . 0 d 618 . 0 9 . 0 , 9 . 0 d 618 . 0 A ) d 53 . 0 ( ) d 53 . 0 )( d 85 . 0 ( 2 A b ab 2 A i i f A area Core Gross X factor Stacking A area, core Net S factor, stacking Let Design of Core Stepped Core or Cruciform Core 79 . 0 d 4 d 618 . 0 71 . 0 d 4 d 56 . 0 2 2 2 2 circle bing Circumscri of Area area core Gross circle bing Circumscri of Area area core Net
- 46. Core area factor, Ratios of Multi-stepped Cores, 46 Design of Core Stepped Core or Cruciform Core 56 . 0 d d 56 . 0 d A K 2 2 2 i C Circle bing Circumscri of Square area Core Net Ratio Square Core Cruciform Core 3-Stepped Core 4-Stepped Core 0.64 0.79 0.84 0.87 0.58 0.71 0.75 0.78 Core area factor,KC 0.45 0.56 0.6 0.62 circle bing Circumscri of Area area core Net circle bing Circumscri of Area area core Gross
- 47. Flux density decides, Area of cross section of the core Core loss Choice of flux density depends on, Service condition (DT/PT) Material used Cold rolled Silicon Steel-Work with higher flux density Upto 132 kV : 1.55 wb/m2 132 to 275kV: 1.6 wb/m2 275 to 400kV: 1.7 wb/m2 Hot rolled Silicon Steel – Work with lower flux density Distribution transformer : 1.1 to 1.4 wb/m2 Power Transformer : 1.2 to 1.5 wb/m2 47 Design of Core Choice of Flux Density in Core
- 51. Design of Winding Transformer windings: HV winding & LV winding Winding Design involves: Determination of no. of turns: based on kVA rating & EMF per turn Area of cross section of conductor used: Based on rated current and Current density No. of turns of LV winding is estimated first using given data. Then, no. of turns of HV winding is calculated to the voltage rating. 51 winding LV of Current Rated I winding LV of voltage Rated V where, T winding, LV in turns of No. LV LV LV t LV LV I AT ) or ( E V
- 53. Cooling of transformers Losses in transformer-Converted in heat energy. Heat developed is transmitted by, Conduction Convection Radiation The paths of heat flow are, From internal hot spot to the outer surface(in contact with oil) From outer surface to the oil From the oil to the tank From tank to the cooling medium-Air or water.
- 54. Methods of cooling: 1. Air Natural (AN)-upto 1.5MVA 2. Air Blast (AB) 3. Oil natural (ON) – Upto 10 MVA 4. Oil Natural – Air Forced (ONAF) 5. Oil Forced– Air Natural (OFAN) – 30 MVA 6. Oil Forced– Air Forced (OFAF) 7. Oil Natural – Water Forced (ONWF) – Power plants 8. Oil Forced - Water Forced (OFWF) – Power plants Cooling of transformers
- 55. Specific heat dissipation due to convection is, The average working temperature of oil is 50-600C. For The value of the dissipation in air is 8 W/m2.0C. i.e, 10 times less than oil. Cooling of transformers Transformer Oil as Cooling Medium m surface, g dissipatin of Height oil, to relative surface the of difference e Temperatur - where, . / 3 . 40 0 0 2 4 1 H C C m W H conv , 1 5 . 0 & 200 m to H C . . W/m 100 to 80 0 2 C conv
- 56. Transformer wall dissipates heat in radiation & convection. For a temperature rise of 400C above the ambient temperature of 200C, the heat dissipations are as follows: Specific heat dissipation by radiation,rad=6 W/m2.0C Specific heat dissipation by convection, conv=6.5 W/m2.0C Total heat dissipation in plain wall 12.5 W/m2.0C The temperature rise, St – Heat dissipating surface Heat dissipating surface of tank : Total area of vertical sides+ One half area of top cover(Air cooled) (Full area of top cover for oil cooled) Cooling of transformers Temperature rise in plain walled tanks t c i S P P 5 . 12 tank of surface g dissipatin Heat n Dissipatio heat Specific losses Total
- 57. Design of tanks with cooling tubes Cooling tubes increases the heat dissipation Cooling tubes mounted on vertical sides of the transformer would not proportional to increase in area. Because, the tubes prevents the radiation from the tank in screened surfaces. But the cooling tubes increase circulation of oil and hence improve the convection Circulation is due to effective pressure heads Dissipation by convection is equal to that of 35% of tube surface area. i.e., 35% tube area is added to actual tube area.
- 58. Let, Dissipating surface of tank – St Dissipating surface of tubes – XSt Loss dissipated by surface of the tank by radiation and convection = Design of tanks with cooling tubes t t S S 5 . 12 5 . 6 6 ) 1 ( 8 . 8 5 . 12 8 . 8 5 . 12 tubes and by walls dissipated loss Total t t t S X XS S t t XS 8 . 8 XS 100 135 5 . 6 convection by tubes by dissipated Loss ) 1 ( S S S tubes and s tank wall of area total Actual t t t X X
- 59. Design of tanks with cooling tubes ) 2 ( ) 1 ( ) 8 . 8 5 . 12 ( ) 1 ( ) 8 . 8 5 . 12 ( surface g dissipatin of m per dissipated Loss area Total dissipated losses Total surface g dissipatin of m per dissipated Loss 2 2 X X X S X S t t 5 . 12 P P 8 . 8 P P ) 8 . 8 5 . 12 ( ) 8 . 8 5 . 12 ( P P have, we (3), and (1) From ) 3 ( P P P losses, Total Dissipated Loss loss Total tubes cooling r with Transforme in rise e Temperatur c i c i c i c i loss t t t S X S X X S 5 . 12 P P 8 . 8 1 c i t S X
- 60. Design of tanks with cooling tubes ) 6 ( 5 . 12 P P 8 . 8 1 each tube of Area tubes of area Total tubes, of number Total , ) 5 ( 5 . 12 P P 8 . 8 1 5 . 12 P P 8 . 8 1 tubes cooling of area Total c i c i c i t t t t t t t t t t t t S l d n n l d tubes of area Surface tubes of Diameter d tubes of Length l Let S S S Standard diameter of cooling tube is 50mm & length depends on the height of the tank. Centre to centre spacing is 75mm.
- 62. Dimensions of the tank: Let, C1 – Clearance b/w winding and tank along width C2 - Clearance b/w winding and tank along length C3 – Clearance b/w the transformer frame and tank at the bottom C4 - Clearance b/w the transformer frame and tank at the top Doc – Outer diameter of the coil. Width of the tank, WT=2D+ Doc +2 C1 (For 3 Transformer) = D+ Doc +2 C1 (For 1 Transformer) Length of the tank, LT= Doc +2 C2 Height of the tank, HT=H+C3+ C4 Design of tanks with cooling tubes
- 63. Clearance on the sides depends on the voltage & power ratings. Clearance at the top depends on the oil height above the assembled transformer & space for mounting the terminals and tap changer. Clearance at the bottom depends on the space required for mounting the frame. Design of tanks with cooling tubes
- 64. Voltage kVA Rating Clearance in mm C1 C2 C3 C4 Up to 11kV <1000kVA 40 50 75 375 Upto 11 kV 1000-5000kVA 70 90 100 400 11kV – 33kV <1000kVA 75 100 75 450 11kV – 33kV 1000-5000kVA 85 125 100 475 Design of tanks with cooling tubes
- 65. Estimation of No-load Current No-load Current of Transformer: Magnetizing Component Depends on MMF required to establish required flux Loss Component Depends on iron loss 65
- 66. Total Length of the core = 2lc Total Length of the yoke = 2ly Here, lc=Hw=Height of Window ly= Ww=Width of Window MMF for core=MMF per metre for max. flux density in core X Total length of Core = atc X 2lc= 2 atc lc MMF for yoke=MMF per metre for max. flux density in yoke X Total length of yoke = aty X 2ly= 2 aty ly Total Magnetizing MMF,AT0=MMF for Core+MMF for Yoke+MMF for joints = 2 atc lc +2 aty ly +MMF for joints The values of atc & aty are taken from B-H curve of transformer steel. 66 Estimation of No-load Current No-load current of Single phase Transformer lC lC ly ly
- 67. Max. value of magnetizing current=AT0/Tp If the magnetizing current is sinusoidal then, RMS value of magnetising current, Im=AT0/√2Tp If the magnetizing current is not sinusoidal, RMS value of magnetising current, Im=AT0/KpkTp The loss component of no-load current, Il=Pi/Vp Where, Pi – Iron loss in Watts Vp – Primary terminal voltage Iron losses are calculated by finding the weight of cores and yokes. Loss per kg is given by the manufacturer. No-load current, 67 Estimation of No-load Current No-load current of Single phase Transformer 2 2 m 0 I I I l
- 68. 68 lC lC lC ly ly Estimation of No-load Current No-load current of Three phase Transformer
- 69. 69
- 70. 70 1000 mm 500 mm 75 mm 50 mm 50 mm 25 mm Total width = 1000 mm D=50mm Tube dia 13*50=650 Gap = 12*25=300 975mm 1000-975=25mm 12.5 mm 12.5 mm 12.5 mm 12.5 mm 37.5 mm 37.5 mm 37.5 mm 37.5 mm Total length = 500 mm D=50mm Tube dia 6*50=300 Gap = 5*25=125 425 mm 500-425=75mm 1st row horizontal – 13*2 = 26 tubes 1st row vertical – 6*2 =12 tubes 2nd row horizontal 12*2 = 24 tubes
- 71. 71 St=2Ht(Wt+Lt) = 8.14 5 . 12 P P 1 . 9 1 c i t S X t t t t S l d n 5 . 12 P P 1 . 9 1 c i Ht=1.85m Lt=0.65m Wt=1.55m