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Project on Transformer Design | Electrical Machine Design

This document details the design process for a 65 kVA, 11 kV/400 V, 50 Hz, three-phase core type transformer. It outlines objectives, design requirements, calculations for dimensions, winding configurations, losses, and efficiency metrics, using specific formulas and materials such as silicon steel for the core. The final analysis indicates that additional cooling methods are necessary to maintain acceptable temperature levels during operation.

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Transformer Design Course on Electrical Machine Design Designed By Md. Jikrul sayeed Hossain (1503048) Tanbin islam rohan (1503099) Lubaina saffana siddiqua (1503022) Mamun sarker (1503106) Group: 25
Problem on Transformer Design Design a 65 KVA, 11 KV/400 V, 50 Hz, 3 Phase, core type, delta/star distribution Transformer, 5% tapping is used in the HV side. (Choose cooling system) i Objectives  To Learn About The Procedure of Transformer Designing  To Gather Knowledge on The Basic Requirements to Design a Transformer  To minimize the cost of Designing
Design Requirements Self Coolling HV 6.6KV DELTA LV 433V STAR 3 PHASE 50HZ 5% Tap 75 KVA 1
Core Design 2
Core Design The value of k is taken from the table, [7.2 in A course in Electrical Machine Design. AK. Sawhney] K=0.45 for 3-phase core type distribution transformer. Voltage per turn Et =K 𝑄 =0.45 75 =3.90 V Therefore, Flux in the core, Φm = 𝐸𝑡 4.44×𝑓 = 3.90 4.44×50 = 0.01757 Wb Hot rolled silicon steel grade 92 is used. The value of flux density Bm is assumed as 1.0 Wb/m2. Net iron Area Ai= 0.01757 1.0 = 0.01757 = 17.57×103 mm2 Using a cruciform core, Ai =0.56d2 Diameter of circumscribing circle, d = 17.57×103 0 . 56 =177.13 mm≈177 mm Reference widths of laminations: a=0.85d=0.85×177 =150.45 mm b=0.53d=0.53×177=93.81 mm The nearest standard dimensions are: a =151 mm and b =94 mm 3 Type K Single phase shell type 1-1.2 Single phase core type 0.75-0.85 Three phase shell type 1.3 Three phase core type(distribution) 0.45 Single phase shell type(power) 0.6 to 0.7 Area percentage of circumscribing circle Square Cruciform Three Stepped Four step Net core area,Ai 0.45 0.56 0.6 0.62
Window Dimensions 4
Window Dimensions The window space factor for the required rating transformer is Kw = 10/(30+KV)= = 10 30 + 6 . 6 = 0.273 ≈ 0.27 The current density in the windings is taken 2.5 A/mm2 Output of transformer: Q =3.33fBmKwδAwAi× 10−3 Therefore, Window Area, Aw= 75 3.33×50×1×0.273×2.5×106×0.0×10−3 =37.56 m2 =37.56× 103 mm2 Taking the ratio Height to Width of window as 2.7: Hw×Ww = 37.56 × 103 mm2 or, 2.7×Ww² = 37.56 × 103 therefore width of window, Ww=118 mm & Window height, Hw =318 mm Area of window provided, Aw =118 × 318 = 37524mm2 = 0.0375 m2 Distance between adjacent core center, D = Ww+d =118 + 177 =295 mm 5 For a distribution Transformer with rating 65KVA, Kw = 10/(30+KV) For distribution , large power transformers,self oil cooled type 𝜹 = 2.2 to 3.2 A/mm2
Yoke Design 6
Yoke Design The area of Yoke is taken as 1.2 times that of limb. Therefore, Flux density in Yoke = 1/1.2 = 0.833 Wb/m2 Net area of Yoke =1.2×17.57×103 =21.084×103 mm2 Gross area of Yoke =21.084×103 /0.9=23.426×103 mm2 Taking the section of Yoke as rectangular, Depth of Yoke, Dy=a=151mm Therefore, height of Yoke, Hy = 23.426×10 3 151 ≈ 155 mm 7
Overall Dimension of frame 8
Overall Dimension of Frame Height of frame, H=Hw+2Hy=318+2*155=628 mm Width of frame, W=2D+a=2×305 +151=761 mm Depth of frame, Dy=a=151 mm 9
Low voltage winding 10
Low Voltage Winding Continued… Secondary voltage = 433 V (Connection= Star) Secondary Phase Voltage ,Vs= 433 3 =250V Number of turns per phase, Ts = 𝑉𝑠 𝐸𝑡 = 250 3.90 =64.10≈64 Secondary Phase Current, Is= 75×1000 3×250 =100 A Current density 2.5 A/mm2 is Used. Area of Secondary Conductor as=100/2.5=40 mm2 Using a bare conductor of 13×3.2 mm, Area of bare conductor , as=41.6 mm2 Current density in secondary winding , δs=100/41.6=2.40 A/mm2 The conductor are paper covered. The increase in dimension on account of account of paper covering is 0.5 mm. Therefore Dimension of insulated conductor = 13×3.7 mm2 11
Low Voltage Winding Using 3 layers for the winding. Helical winding is used. Therefore, space has to be provided ( 𝑇𝑠 3 + 1)=( 64 3 + 1) =22.3= 23 (approx) turns along the axial depth. Axial depth of L.V winding, Lcs =23×13=299 mm The Height of window is 486 mm. This leaves a clearance of (318-299)/2=9.5 mm of each side of the windings. [Which fulfill the minimum requirement 6 mm] Radial depth of low voltage winding, bs=no. of layers x radial depth of conductor + insulator betn layers =3×3.2+2×0.5 = 10.6 mm Diameter of circumscribing circle, d =177 mm Using pressboard wraps 1.5 mm thick as insulation between l.v. winding and core. Inside diameter of l.v. winding=117+1.5×2=120 mm Outside diameter of l.v. winding= 120+10.6 =130.6 mm 12
High voltage winding 13
High Voltage WindingContinued… Primary Line voltage =6600V (Connection: Delta) Primary Phase Voltage, Vp= 6600 V Therefore, no. of turns per phase, Tp=Vp × 𝑇𝑠 𝑉𝑠 = 6600×64 250 =1689.6 = 1690 (approx) As ±5% tapping is to be provided, therefore the no. of turns is increased to Tp=1.05×1690=1775 The voltage per coil is about 1500 V, therefore using 15 coils, Voltage per coil= 6600/15=440 V Turns per coil= 1775 /15=118 Using 14 normal coils of 120 turns and one reinforced coil of 95 turns. Total H.V turns provided, Tp =14 × 120 + 1×95=1775 Taking 24 layers per coil. Turns/layer=120/24= 5 (approximately) Maximum voltage between layers = 2×5×3.9 =39 V H.V. winding phase current, Ip =(75×1000)/(3×6600)=3.78 A As the current is below 20A, cross-over coils are used H.V winding taking a current density of 2.1 A/mm2 Area of H.V. conductor, ap =3.78/2.1=1.8 mm2 14
High Voltage WindingContinued… Diameter of bare conductor = 4 𝜋 × 1.8 =1.51 mm Using paper covered conductors, the nearest standard conductor size: Bare diameter =1.5 mm insulated diameter=1.58 mm with fine covering. Modified area of the conductor, ap= 𝝅 𝟒 × 𝟏. 𝟓 𝟐 =1.76 mm2 Actual value of current density used, δp= 𝟑.𝟕𝟖 𝟏.𝟕𝟔 =2.13 A/mm2 Axial depth of one coil =5x 1.58 =7.9 mm The space betn adjacent coils are 5 mm in height. Axial length of H.V winding: Lcp =no. of coils x axial depth of each coil + depth of spacers =15×7.9 + 15×5=193.5 mm The height of window is 318 mm & therefore the space left betn winding & window is (318-193.5)/2 =62.25 mm. The clearance left on each side is 62.25 mm, which is sufficient for 6.6kV transformer. 15
High Voltage Winding The insulation used between layers is 0.3 mm thick paper Radial depth of H.V coil, bp=15 × 1.58 + 14×0.3=27.9 Thickness of insulation between h.v. and l.v. windings= 5+0.9kV=5+0.9*6.6=10.94 mm Inside diameter of H.V. winding =Outside diameter of L.V winding + 2 x thickness of insulation. =130.6+2×10.94≈152.48 mm Outside diameter of H.V. winding De = Inside diameter of H.V. winding + 2 x Radial depth of H.V coil =152.48+2×27.9=208.08 mm Clearance betn windings=D – Outside diameter of H.V. winding =295– 208.08=86.92 mm (Which is for enough) 16
Resistance & Leakage Reactance 17
Resistance & Leakage ReactanceContinued… Mean diameter of primary winding = 𝟏𝟓𝟐.𝟒𝟖+𝟐𝟎𝟖.𝟎𝟖 𝟐 = 180.28 mm Length of mean turn of primary winding, Lmtp=𝝅×180.28×10-3 =0.566 m Resistance of primary winding at 75℃, rp= 𝑻𝒑×⍴×𝐋 𝐦𝐭𝐩 𝒂 𝒑 = 𝟏𝟕𝟕𝟓×𝟎.𝟎𝟐𝟏×𝟎.𝟓𝟔 𝟏.𝟖 =11.6 Ω Mean Diameter of Secondary winding= 𝟏𝟐𝟎+𝟏𝟑𝟎.𝟔 𝟐 =125.3≈125mm Length of mean turn of Secondary winding, Lmts=𝝅×125×10-3 =0.3927 m Resistance of Secondary winding at 75℃, rs= 𝑻𝒔×⍴×𝐋 𝐦𝐭𝐬 𝒂 𝒔 = 𝟔𝟒×𝟎.𝟎𝟐𝟏×𝟎.𝟑𝟗𝟐𝟕 𝟒𝟏.𝟔 =0.01268Ω Therefore, Total Resistance referred to primary side , Rp=11.6+ 𝟏𝟕𝟕𝟓 𝟔𝟒 𝟐 × 𝟎. 𝟎𝟏𝟐𝟔𝟖=21.35 Ω P.U. resistance of transformer εr= 𝑰𝒑×𝑹𝒑 𝑽𝒑 = 𝟑.𝟕𝟖 ×𝟐𝟏.𝟑𝟓 𝟔𝟔𝟎𝟎 = 0.012 R e s i s t a n c e 18
Resistance & Leakage Reactance Mean diameter of windings = 120+208.08 2 =164.04 mm Length of mean turn, Lmt = 𝜋 ×164.04×10-3=0.515 m Height of winding, Lc =(Lcp +Lcs) /2 =(193.5 +299)/2 = 246.25 mm Therefore leakage Resistance referred to primary side , Xp=2×𝜋×50×4× 𝜋 ×10-7 ×17752× 0.515 0.2462 × 10.94 + 10.6+ 27.9 3 × 10−3 =61.85 Ω P.U. leakage resistance of transformer εΦ= 3.78 ×61.85 6600 =0.035 P.U Impedance εs = 0.012 2 + 0.035 2=0.037 R e a c t a n c e 19
Voltage Regulation 20
Voltage Regulation P.U. regulation ε= εr 𝐜𝐨𝐬𝚽 + εssinΦ Therefore, P.U. regulation at unity power factor ε=εr =0.012 P.U regulation at 0.8 p.f. lagging = 0.012*0.8 + 0.037*0.6 =0.0318 21
Losses 22
Losses LOSSES: I2R loss at 75℃ = 3 × 𝐼𝑝2 × 𝑅𝑝=3 × 3.782 × 21.35=915.17 W Taking stray loss 15% above. Total I2R loss, 𝑃𝑐=1.5 X 915.17 =1372.755 W CORE LOSS: Taking density laminations as 7.6X103 kg/m3 Weight of 3 limbs =3×0.3×0.01757×7.6×103=107.388 kg Corresponding to flux density 1 Wb/ m2, specific core loss 1.2 W/kg Core loss in limbs=107.388×1.2=128.8656 W Weight of two Yokes = 2X0.761X0.021084X7.6X103=243.88 kg Corresponding to flux density 0.833 Wb/m2, specific core loss 0.85 W/kg Core loss in Yoke=243.88X0.85=207.298 W Total core loss, Pi=128.87 + 207.3 =336.17 W 23
Efficiency 24
Efficiency Total losses at full load = 336.17 +1372.755 =1708.925 W Efficiency at full load unity P.f. = 𝟕𝟓𝟎𝟎𝟎 𝟕𝟓𝟎𝟎𝟎+𝟏𝟕𝟎𝟖.𝟗𝟐𝟓 × 𝟏𝟎𝟎% =97.77 % For maximum efficiency x= 𝑷𝒊 𝑷𝒄 = 𝟎. 𝟒𝟗4 Thus maximum efficiency occurs at 49.4% of full load. This is a good figure for distribution transformer. 25
No load current 26
No load Current Corresponding to flux densities of 1 Wb/m2 & 0.833 Wb/m2 in core & yoke respectively atc= 120 A/m & aty = 80 A/m. So, total magnetizing m.m.f. = 3×120×0.296+2×80×0.705 = 219.36 A So, magnetizing m.m.f. per phase, ATo = 219.36/3 = 73.12 Magnetizing current, Im = ATo/ 𝟐 TP = 𝟕𝟑.𝟏𝟐 𝟐×𝟏𝟔𝟗𝟎 = 30.59×10-3 A Loss component of no load current, IL = 𝟑𝟑𝟔.𝟏𝟕 𝟐×𝟔𝟔𝟎𝟎 = 36.02×10-3 A No load current, Io = 𝟑𝟎. 𝟓𝟗 × 𝟏𝟎 ^ − 𝟑 𝟐 + 𝟑𝟔. 𝟎𝟐 × 𝟏𝟎^ − 𝟑 𝟐 = 47.26×10-3 A No load current as a percentage of full load current = 𝟒𝟕.𝟐𝟔×𝟏𝟎 − 𝟑 𝟑.𝟕𝟖 ×100% = 1.25% Allowing for joints etc the no load current will be about 2.5% of full load current 27
Tank Design 28
Tank Design Height over Yoke, H = 628 mm Allowing 50 mm at the base & 150 mm for oil, Height of oil level =628+50+150=828 mm Allowing 200 mm Height for leads etc, Height of Tank Ht=828 + 200 =1028mm ≈ 1.028 m Assuming a clearance of 40 mm along the width on each side, Width of the Tank, Wt=2D+De+2l=2 × 295 + 208.08 + 2 × 40 =878.08 mm≈0.87 m Length of the tank, Lt=De+2b=208.08+2X50 =308.08 mm≈0.308 m Total loss dissipating surface of tank =2 0.87 + 0.308 × 1.028 = 2.422 m2 Specific loss dissipation due to radiation & convention is 12.5 w/m2℃ Temperature rise = 1708.925 2.422×12.5 = 56.44 ℃ This is not bellow 35℃ and therefore plain tank is not sufficient for cooling and tubes are required. 29
Tank With Tubes 30
Tank With Tube Area of tubes= x* 2.422= 2.422x Loss dissipating surface= (1+x)*St= 2.422*(1+x) Loss dissipated= 12.5+8.8𝑥 𝑥+1 w/m2 - ℃ Temperature rise with tubes, Ɵ= 𝑃𝑖+𝑃𝑐 𝑆𝑡(12.5+8.86) = 1708.925 2.422(12.5+8.86𝑥) Pi + Pc = 1708.925, x= 0.494 Therefore, Ɵ = 41.88 ℃ Total area of tubes = 1 8.8 ( 𝑃𝑖+𝑃𝑐 Ɵ − 12.5 𝑆𝑡) =1.2 m2 Let, dt = 0.05 m Lt = 1.00 m Hence, Number of tubes nt = 1 8.8𝜋∗𝑑𝑡 ∗ 𝐿𝑡 ( 𝑃𝑖+𝑃𝑐 Ɵ − 12.5 𝑆𝑡) = 7.62 But, It can’t be a fraction so nt = 8. 31
Design Sheet 32
Final Voltage Current Rating 33 Line Volage HV: 6.6 kV LV: 433 V Phase voltage HV: 6.6 kV LV: 250 V Line current HV: 6.55 A LV:100 A Phase current HV: 3.78 A LV: 100 A
Design Sheet (Continued…) 1 Material --- 0.35mm thick 92 grade 2 Output Constant K .45 3 Voltage per turn Et 3.90 V 4 Circumscribing circle diameter d 177 mm 5 No. of steps --- 2 6 Dimensions a 151 mm b 94 mm 7 Net iron area Ai 17.57 × 103 mm2 8 Flux density Bm 1.0 Wb/m2 9 Flux Φm 0.01757 Wb 10 Weight 107.388 kg 11 Specific iron loss 1.2 W/kg 12 Iron loss 128.8656 W CORE: 34
Design Sheet (Continued…) 1 Depth of Yoke Dy 151 mm 2 Height of Yoke Hy 155 mm 3 Net Yoke area 21.084x103 mm2 4 Flux density 0.833 Wb/m2 5 Flux 0.01757 Wb 6 Weight 244 kg 7 Specific iron loss 0.85 W/kg 8 Iron loss 207.298 W YOKE: WINDOWS: 1 Number 2 2 Window space factor Kw 0.27 3 Height of window Hw 318 mm 4 Width of window Ww 118 mm 5 Area of window Aw 0.0375 m2 35
Design Sheet (Continued…) 1 Distance betn adjacent limbs D 295 mm 2 Height of Frame H 628 mm 3 Width of Frame W 761 mm 4 Depth of window Dy 151 mm 1 Betn L.V. winding & Core Press board wraps 1.5mm 2 Betn L.V. winding & H.V. winding Bakelized paper 5mm 3 Width of duct betn L.V & H.V. 5mm INSULATION: FRAME: 36
Design Sheet (Continued…) Sl no. Properties L.V. H.V. 1 Type of winding Helical Cross-over 2 Connections Star Delta 3 Conductor Dimensions bare 13x3.2 mm2 Diameter=1.5 mm insulated 13x3.7 mm2 Diameter=1.58mm Area 41.6 mm2 1.76 mm2 No. in parallel None None 4 Current Density 2.5 A/mm2 2.13 A/mm2 5 Turns per phase 64 1690 (1775 at ±5% tapping) 6 Coils total number 3 3x15 per core leg 1 15 7 Turns Per coil 64 14 of 120 turns, 1 of 95 Per layer 22 5 8 Number of layers 3 24 9 Height of winding 318 mm 193.5 mm 10 Depth of winding 299 mm 27.9 mm 11 Insulation Betn layers 0.5 mm press board 0.3mm paper Betn coils 5mm spacers 12 Coil Diameters Inside 120 mm 152.48 mm Outside 130.6 mm 208.08 mm 13 Length of mean turn 0.566 m 0.3927 m 14 Resistance at 75℃ 11.6 Ω 0.01268 Ω WINDINGS: 37
Design Sheet TANK: 1. Dimensions Height Ht 1.028 m Length Lt 0.308 m Width Wt 0.87 m 1. Tubes 8 1. Temperature rise --- 56.44 ℃ 1. Impedance P.U. Resistance --- 0.035 P.U. Reactance --- 0.012 P.U. Impedance --- 0.037 1. Losses Total Core loss --- 336.17 W Total copper loss --- 1372.755 W Total losses at full load --- 1708.925 W Efficiency at full load & unity p.f. --- 97.77 % 38
39 Thank You all !

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Project on Transformer Design | Electrical Machine Design

  • 1.
    Transformer Design Course onElectrical Machine Design Designed By Md. Jikrul sayeed Hossain (1503048) Tanbin islam rohan (1503099) Lubaina saffana siddiqua (1503022) Mamun sarker (1503106) Group: 25
  • 2.
    Problem on TransformerDesign Design a 65 KVA, 11 KV/400 V, 50 Hz, 3 Phase, core type, delta/star distribution Transformer, 5% tapping is used in the HV side. (Choose cooling system) i Objectives  To Learn About The Procedure of Transformer Designing  To Gather Knowledge on The Basic Requirements to Design a Transformer  To minimize the cost of Designing
  • 3.
  • 4.
  • 5.
    Core Design The valueof k is taken from the table, [7.2 in A course in Electrical Machine Design. AK. Sawhney] K=0.45 for 3-phase core type distribution transformer. Voltage per turn Et =K 𝑄 =0.45 75 =3.90 V Therefore, Flux in the core, Φm = 𝐸𝑡 4.44×𝑓 = 3.90 4.44×50 = 0.01757 Wb Hot rolled silicon steel grade 92 is used. The value of flux density Bm is assumed as 1.0 Wb/m2. Net iron Area Ai= 0.01757 1.0 = 0.01757 = 17.57×103 mm2 Using a cruciform core, Ai =0.56d2 Diameter of circumscribing circle, d = 17.57×103 0 . 56 =177.13 mm≈177 mm Reference widths of laminations: a=0.85d=0.85×177 =150.45 mm b=0.53d=0.53×177=93.81 mm The nearest standard dimensions are: a =151 mm and b =94 mm 3 Type K Single phase shell type 1-1.2 Single phase core type 0.75-0.85 Three phase shell type 1.3 Three phase core type(distribution) 0.45 Single phase shell type(power) 0.6 to 0.7 Area percentage of circumscribing circle Square Cruciform Three Stepped Four step Net core area,Ai 0.45 0.56 0.6 0.62
  • 6.
  • 7.
    Window Dimensions The windowspace factor for the required rating transformer is Kw = 10/(30+KV)= = 10 30 + 6 . 6 = 0.273 ≈ 0.27 The current density in the windings is taken 2.5 A/mm2 Output of transformer: Q =3.33fBmKwδAwAi× 10−3 Therefore, Window Area, Aw= 75 3.33×50×1×0.273×2.5×106×0.0×10−3 =37.56 m2 =37.56× 103 mm2 Taking the ratio Height to Width of window as 2.7: Hw×Ww = 37.56 × 103 mm2 or, 2.7×Ww² = 37.56 × 103 therefore width of window, Ww=118 mm & Window height, Hw =318 mm Area of window provided, Aw =118 × 318 = 37524mm2 = 0.0375 m2 Distance between adjacent core center, D = Ww+d =118 + 177 =295 mm 5 For a distribution Transformer with rating 65KVA, Kw = 10/(30+KV) For distribution , large power transformers,self oil cooled type 𝜹 = 2.2 to 3.2 A/mm2
  • 8.
  • 9.
    Yoke Design The areaof Yoke is taken as 1.2 times that of limb. Therefore, Flux density in Yoke = 1/1.2 = 0.833 Wb/m2 Net area of Yoke =1.2×17.57×103 =21.084×103 mm2 Gross area of Yoke =21.084×103 /0.9=23.426×103 mm2 Taking the section of Yoke as rectangular, Depth of Yoke, Dy=a=151mm Therefore, height of Yoke, Hy = 23.426×10 3 151 ≈ 155 mm 7
  • 10.
  • 11.
    Overall Dimension ofFrame Height of frame, H=Hw+2Hy=318+2*155=628 mm Width of frame, W=2D+a=2×305 +151=761 mm Depth of frame, Dy=a=151 mm 9
  • 12.
  • 13.
    Low Voltage WindingContinued… Secondary voltage = 433 V (Connection= Star) Secondary Phase Voltage ,Vs= 433 3 =250V Number of turns per phase, Ts = 𝑉𝑠 𝐸𝑡 = 250 3.90 =64.10≈64 Secondary Phase Current, Is= 75×1000 3×250 =100 A Current density 2.5 A/mm2 is Used. Area of Secondary Conductor as=100/2.5=40 mm2 Using a bare conductor of 13×3.2 mm, Area of bare conductor , as=41.6 mm2 Current density in secondary winding , δs=100/41.6=2.40 A/mm2 The conductor are paper covered. The increase in dimension on account of account of paper covering is 0.5 mm. Therefore Dimension of insulated conductor = 13×3.7 mm2 11
  • 14.
    Low Voltage Winding Using3 layers for the winding. Helical winding is used. Therefore, space has to be provided ( 𝑇𝑠 3 + 1)=( 64 3 + 1) =22.3= 23 (approx) turns along the axial depth. Axial depth of L.V winding, Lcs =23×13=299 mm The Height of window is 486 mm. This leaves a clearance of (318-299)/2=9.5 mm of each side of the windings. [Which fulfill the minimum requirement 6 mm] Radial depth of low voltage winding, bs=no. of layers x radial depth of conductor + insulator betn layers =3×3.2+2×0.5 = 10.6 mm Diameter of circumscribing circle, d =177 mm Using pressboard wraps 1.5 mm thick as insulation between l.v. winding and core. Inside diameter of l.v. winding=117+1.5×2=120 mm Outside diameter of l.v. winding= 120+10.6 =130.6 mm 12
  • 15.
  • 16.
    High Voltage WindingContinued… PrimaryLine voltage =6600V (Connection: Delta) Primary Phase Voltage, Vp= 6600 V Therefore, no. of turns per phase, Tp=Vp × 𝑇𝑠 𝑉𝑠 = 6600×64 250 =1689.6 = 1690 (approx) As ±5% tapping is to be provided, therefore the no. of turns is increased to Tp=1.05×1690=1775 The voltage per coil is about 1500 V, therefore using 15 coils, Voltage per coil= 6600/15=440 V Turns per coil= 1775 /15=118 Using 14 normal coils of 120 turns and one reinforced coil of 95 turns. Total H.V turns provided, Tp =14 × 120 + 1×95=1775 Taking 24 layers per coil. Turns/layer=120/24= 5 (approximately) Maximum voltage between layers = 2×5×3.9 =39 V H.V. winding phase current, Ip =(75×1000)/(3×6600)=3.78 A As the current is below 20A, cross-over coils are used H.V winding taking a current density of 2.1 A/mm2 Area of H.V. conductor, ap =3.78/2.1=1.8 mm2 14
  • 17.
    High Voltage WindingContinued… Diameterof bare conductor = 4 𝜋 × 1.8 =1.51 mm Using paper covered conductors, the nearest standard conductor size: Bare diameter =1.5 mm insulated diameter=1.58 mm with fine covering. Modified area of the conductor, ap= 𝝅 𝟒 × 𝟏. 𝟓 𝟐 =1.76 mm2 Actual value of current density used, δp= 𝟑.𝟕𝟖 𝟏.𝟕𝟔 =2.13 A/mm2 Axial depth of one coil =5x 1.58 =7.9 mm The space betn adjacent coils are 5 mm in height. Axial length of H.V winding: Lcp =no. of coils x axial depth of each coil + depth of spacers =15×7.9 + 15×5=193.5 mm The height of window is 318 mm & therefore the space left betn winding & window is (318-193.5)/2 =62.25 mm. The clearance left on each side is 62.25 mm, which is sufficient for 6.6kV transformer. 15
  • 18.
    High Voltage Winding Theinsulation used between layers is 0.3 mm thick paper Radial depth of H.V coil, bp=15 × 1.58 + 14×0.3=27.9 Thickness of insulation between h.v. and l.v. windings= 5+0.9kV=5+0.9*6.6=10.94 mm Inside diameter of H.V. winding =Outside diameter of L.V winding + 2 x thickness of insulation. =130.6+2×10.94≈152.48 mm Outside diameter of H.V. winding De = Inside diameter of H.V. winding + 2 x Radial depth of H.V coil =152.48+2×27.9=208.08 mm Clearance betn windings=D – Outside diameter of H.V. winding =295– 208.08=86.92 mm (Which is for enough) 16
  • 19.
  • 20.
    Resistance & LeakageReactanceContinued… Mean diameter of primary winding = 𝟏𝟓𝟐.𝟒𝟖+𝟐𝟎𝟖.𝟎𝟖 𝟐 = 180.28 mm Length of mean turn of primary winding, Lmtp=𝝅×180.28×10-3 =0.566 m Resistance of primary winding at 75℃, rp= 𝑻𝒑×⍴×𝐋 𝐦𝐭𝐩 𝒂 𝒑 = 𝟏𝟕𝟕𝟓×𝟎.𝟎𝟐𝟏×𝟎.𝟓𝟔 𝟏.𝟖 =11.6 Ω Mean Diameter of Secondary winding= 𝟏𝟐𝟎+𝟏𝟑𝟎.𝟔 𝟐 =125.3≈125mm Length of mean turn of Secondary winding, Lmts=𝝅×125×10-3 =0.3927 m Resistance of Secondary winding at 75℃, rs= 𝑻𝒔×⍴×𝐋 𝐦𝐭𝐬 𝒂 𝒔 = 𝟔𝟒×𝟎.𝟎𝟐𝟏×𝟎.𝟑𝟗𝟐𝟕 𝟒𝟏.𝟔 =0.01268Ω Therefore, Total Resistance referred to primary side , Rp=11.6+ 𝟏𝟕𝟕𝟓 𝟔𝟒 𝟐 × 𝟎. 𝟎𝟏𝟐𝟔𝟖=21.35 Ω P.U. resistance of transformer εr= 𝑰𝒑×𝑹𝒑 𝑽𝒑 = 𝟑.𝟕𝟖 ×𝟐𝟏.𝟑𝟓 𝟔𝟔𝟎𝟎 = 0.012 R e s i s t a n c e 18
  • 21.
    Resistance & LeakageReactance Mean diameter of windings = 120+208.08 2 =164.04 mm Length of mean turn, Lmt = 𝜋 ×164.04×10-3=0.515 m Height of winding, Lc =(Lcp +Lcs) /2 =(193.5 +299)/2 = 246.25 mm Therefore leakage Resistance referred to primary side , Xp=2×𝜋×50×4× 𝜋 ×10-7 ×17752× 0.515 0.2462 × 10.94 + 10.6+ 27.9 3 × 10−3 =61.85 Ω P.U. leakage resistance of transformer εΦ= 3.78 ×61.85 6600 =0.035 P.U Impedance εs = 0.012 2 + 0.035 2=0.037 R e a c t a n c e 19
  • 22.
  • 23.
    Voltage Regulation P.U. regulationε= εr 𝐜𝐨𝐬𝚽 + εssinΦ Therefore, P.U. regulation at unity power factor ε=εr =0.012 P.U regulation at 0.8 p.f. lagging = 0.012*0.8 + 0.037*0.6 =0.0318 21
  • 24.
  • 25.
    Losses LOSSES: I2R loss at75℃ = 3 × 𝐼𝑝2 × 𝑅𝑝=3 × 3.782 × 21.35=915.17 W Taking stray loss 15% above. Total I2R loss, 𝑃𝑐=1.5 X 915.17 =1372.755 W CORE LOSS: Taking density laminations as 7.6X103 kg/m3 Weight of 3 limbs =3×0.3×0.01757×7.6×103=107.388 kg Corresponding to flux density 1 Wb/ m2, specific core loss 1.2 W/kg Core loss in limbs=107.388×1.2=128.8656 W Weight of two Yokes = 2X0.761X0.021084X7.6X103=243.88 kg Corresponding to flux density 0.833 Wb/m2, specific core loss 0.85 W/kg Core loss in Yoke=243.88X0.85=207.298 W Total core loss, Pi=128.87 + 207.3 =336.17 W 23
  • 26.
  • 27.
    Efficiency Total losses atfull load = 336.17 +1372.755 =1708.925 W Efficiency at full load unity P.f. = 𝟕𝟓𝟎𝟎𝟎 𝟕𝟓𝟎𝟎𝟎+𝟏𝟕𝟎𝟖.𝟗𝟐𝟓 × 𝟏𝟎𝟎% =97.77 % For maximum efficiency x= 𝑷𝒊 𝑷𝒄 = 𝟎. 𝟒𝟗4 Thus maximum efficiency occurs at 49.4% of full load. This is a good figure for distribution transformer. 25
  • 28.
  • 29.
    No load Current Correspondingto flux densities of 1 Wb/m2 & 0.833 Wb/m2 in core & yoke respectively atc= 120 A/m & aty = 80 A/m. So, total magnetizing m.m.f. = 3×120×0.296+2×80×0.705 = 219.36 A So, magnetizing m.m.f. per phase, ATo = 219.36/3 = 73.12 Magnetizing current, Im = ATo/ 𝟐 TP = 𝟕𝟑.𝟏𝟐 𝟐×𝟏𝟔𝟗𝟎 = 30.59×10-3 A Loss component of no load current, IL = 𝟑𝟑𝟔.𝟏𝟕 𝟐×𝟔𝟔𝟎𝟎 = 36.02×10-3 A No load current, Io = 𝟑𝟎. 𝟓𝟗 × 𝟏𝟎 ^ − 𝟑 𝟐 + 𝟑𝟔. 𝟎𝟐 × 𝟏𝟎^ − 𝟑 𝟐 = 47.26×10-3 A No load current as a percentage of full load current = 𝟒𝟕.𝟐𝟔×𝟏𝟎 − 𝟑 𝟑.𝟕𝟖 ×100% = 1.25% Allowing for joints etc the no load current will be about 2.5% of full load current 27
  • 30.
  • 31.
    Tank Design Height overYoke, H = 628 mm Allowing 50 mm at the base & 150 mm for oil, Height of oil level =628+50+150=828 mm Allowing 200 mm Height for leads etc, Height of Tank Ht=828 + 200 =1028mm ≈ 1.028 m Assuming a clearance of 40 mm along the width on each side, Width of the Tank, Wt=2D+De+2l=2 × 295 + 208.08 + 2 × 40 =878.08 mm≈0.87 m Length of the tank, Lt=De+2b=208.08+2X50 =308.08 mm≈0.308 m Total loss dissipating surface of tank =2 0.87 + 0.308 × 1.028 = 2.422 m2 Specific loss dissipation due to radiation & convention is 12.5 w/m2℃ Temperature rise = 1708.925 2.422×12.5 = 56.44 ℃ This is not bellow 35℃ and therefore plain tank is not sufficient for cooling and tubes are required. 29
  • 32.
  • 33.
    Tank With Tube Areaof tubes= x* 2.422= 2.422x Loss dissipating surface= (1+x)*St= 2.422*(1+x) Loss dissipated= 12.5+8.8𝑥 𝑥+1 w/m2 - ℃ Temperature rise with tubes, Ɵ= 𝑃𝑖+𝑃𝑐 𝑆𝑡(12.5+8.86) = 1708.925 2.422(12.5+8.86𝑥) Pi + Pc = 1708.925, x= 0.494 Therefore, Ɵ = 41.88 ℃ Total area of tubes = 1 8.8 ( 𝑃𝑖+𝑃𝑐 Ɵ − 12.5 𝑆𝑡) =1.2 m2 Let, dt = 0.05 m Lt = 1.00 m Hence, Number of tubes nt = 1 8.8𝜋∗𝑑𝑡 ∗ 𝐿𝑡 ( 𝑃𝑖+𝑃𝑐 Ɵ − 12.5 𝑆𝑡) = 7.62 But, It can’t be a fraction so nt = 8. 31
  • 34.
  • 35.
    Final Voltage CurrentRating 33 Line Volage HV: 6.6 kV LV: 433 V Phase voltage HV: 6.6 kV LV: 250 V Line current HV: 6.55 A LV:100 A Phase current HV: 3.78 A LV: 100 A
  • 36.
    Design Sheet (Continued…) 1Material --- 0.35mm thick 92 grade 2 Output Constant K .45 3 Voltage per turn Et 3.90 V 4 Circumscribing circle diameter d 177 mm 5 No. of steps --- 2 6 Dimensions a 151 mm b 94 mm 7 Net iron area Ai 17.57 × 103 mm2 8 Flux density Bm 1.0 Wb/m2 9 Flux Φm 0.01757 Wb 10 Weight 107.388 kg 11 Specific iron loss 1.2 W/kg 12 Iron loss 128.8656 W CORE: 34
  • 37.
    Design Sheet (Continued…) 1Depth of Yoke Dy 151 mm 2 Height of Yoke Hy 155 mm 3 Net Yoke area 21.084x103 mm2 4 Flux density 0.833 Wb/m2 5 Flux 0.01757 Wb 6 Weight 244 kg 7 Specific iron loss 0.85 W/kg 8 Iron loss 207.298 W YOKE: WINDOWS: 1 Number 2 2 Window space factor Kw 0.27 3 Height of window Hw 318 mm 4 Width of window Ww 118 mm 5 Area of window Aw 0.0375 m2 35
  • 38.
    Design Sheet (Continued…) 1Distance betn adjacent limbs D 295 mm 2 Height of Frame H 628 mm 3 Width of Frame W 761 mm 4 Depth of window Dy 151 mm 1 Betn L.V. winding & Core Press board wraps 1.5mm 2 Betn L.V. winding & H.V. winding Bakelized paper 5mm 3 Width of duct betn L.V & H.V. 5mm INSULATION: FRAME: 36
  • 39.
    Design Sheet (Continued…) Slno. Properties L.V. H.V. 1 Type of winding Helical Cross-over 2 Connections Star Delta 3 Conductor Dimensions bare 13x3.2 mm2 Diameter=1.5 mm insulated 13x3.7 mm2 Diameter=1.58mm Area 41.6 mm2 1.76 mm2 No. in parallel None None 4 Current Density 2.5 A/mm2 2.13 A/mm2 5 Turns per phase 64 1690 (1775 at ±5% tapping) 6 Coils total number 3 3x15 per core leg 1 15 7 Turns Per coil 64 14 of 120 turns, 1 of 95 Per layer 22 5 8 Number of layers 3 24 9 Height of winding 318 mm 193.5 mm 10 Depth of winding 299 mm 27.9 mm 11 Insulation Betn layers 0.5 mm press board 0.3mm paper Betn coils 5mm spacers 12 Coil Diameters Inside 120 mm 152.48 mm Outside 130.6 mm 208.08 mm 13 Length of mean turn 0.566 m 0.3927 m 14 Resistance at 75℃ 11.6 Ω 0.01268 Ω WINDINGS: 37
  • 40.
    Design Sheet TANK: 1. DimensionsHeight Ht 1.028 m Length Lt 0.308 m Width Wt 0.87 m 1. Tubes 8 1. Temperature rise --- 56.44 ℃ 1. Impedance P.U. Resistance --- 0.035 P.U. Reactance --- 0.012 P.U. Impedance --- 0.037 1. Losses Total Core loss --- 336.17 W Total copper loss --- 1372.755 W Total losses at full load --- 1708.925 W Efficiency at full load & unity p.f. --- 97.77 % 38
  • 41.