Transformer design 2072 batch IOE - TU
Download as DOCX, PDF1 like1,492 views
The document is a lab report on the design of a 1000 KVA, 11/66 kv, 50 Hz, three-phase, core type distribution transformer. It provides details on the core design, window design, winding designs for the high voltage and low voltage coils, resistance and reactance calculations, efficiency calculations, regulation calculations, loss calculations, and tank design including the number of cooling tubes required. The transformer is designed to have a maximum temperature rise of 40°C and tappings of ±2.5% and ±5% on the high voltage winding.
![line voltage [
𝐻. 𝑉 660000𝑉
𝐿. 𝑉 11000𝑉
] phase voltage[
𝐻. 𝑉 660000𝑉
𝐿. 𝑉 6350𝑉
]
line current [
𝐻. 𝑉 3.5𝐴
𝐿. 𝑉 21𝐴
] phase current [
𝐻. 𝑉 2.02𝐴
𝐿. 𝑉 21𝐴
]
Type – Core Type of cooling: Oil Natural
Core Symbols Quantity
1 Material 0.35 mm thick 92 Grade
2 Output constant k 0.25
3 Voltage constant Et 9V
4 Circumscribing circle diameter d 235mm
5 Number of steps 2
6 Dimensions a= 200 mm, b= 125mm
7 Net iron area Ai 30.1185 × 10 3mm2
8 flux density Bm 1.3Wb/m2
9 flux ɸm 0.04054054Wb
10 Weight 494.42Kg
11 Specific iron loss 2.03 W/kg
12 Iron loss 1003.37W
Yoke
1 Depth of yoke Dy 200mm
2 Height of yoke Hy 200mm
3 Net yoke area 37.422×103 mm2
4 Flux density 1.08 Wb/m2
5 Flux 0.04054054Wb
6 Weight 711 kg
7 Specific iron area 1.2W/kg
8 Iron loss 853.2 W
Windows
1 Number 2
2 Window space factor Kw 1.25
3 Height of window Hw 720mm
4 Width of window Ww 290mm
5 Window area Aw 206. 11 × 103 mm2
Frame
1 Distance between adjacent limbs D 525 mm
2 Height of frame H 1120mm
3 Width of frame W 1250mm](https://image.slidesharecdn.com/transformerdesign2072bel05-190921182512/85/Transformer-design-2072-batch-IOE-TU-7-320.jpg)
More Related Content
Similar to Transformer design 2072 batch IOE - TU (20)
Transformer design 2072 batch IOE - TU
- 1. INSTITUTE OF ENGINEERING PURWANCHAL CAMPUS DHARAN-8, SUNSARI NEPAL A LAB REPORT ON TRANSFORMER DESIGN Submitted by: Submitted to: Gyan Kafle Bikash Gyawali (072/BEL/05) Asst. Lecturer Chet Raj Kandel (072/BEL/09) Department of Bishop Poudel (072/BEL/07) Electrical Engineering Akash Pandey (072/BEL/02) Purwanchal Campus Dharan
- 2. Designa 1000 KVA, 11/66 kv,50 HZ, 3-phase , Δ/Y , core type, oil immersed natural cooleddistribution transformer .The transformer is provided with tappings ±𝟐 𝟐 𝟏 % , ±𝟓 𝟐 𝟏 % on the HV winding. Maximum temperature rise not to exceed40℃. CORE DESIGN: For 3-𝜙 core type distribution transformer We use K=0.45 Voltage per turn Et =k√ 𝑄 =0.45√400 = 9V ∴ Flux in the core 𝜙 𝑚 = Et 4.44×𝑓 = 9 4.44×50 =0.04054054Wb Hot rolled silicon steel grade 92 is used so, the Bm is assumed to be 1.3Wb/m2 Net iron area Ai = 0.04054054 1.3 = 30.1185 × 103mm2 Using cruciform core, Ai = 0.56d2 Diameter of circumscribing circle d = √ 30.1185 ×10 3 0.56 = 235mm a=0.85d = 0.85×235=200mm b= 0.53d = 0.53 × 235 =125mm Window Design: The window space factor for 400 KVA transformer is given by formula kw = 12 30 +𝐾𝑉 = 12 30+66 = 0.125 Current density (𝛿 ) =1.5 Wb/mm2 (assumed) Q= 3.33fBmkw 𝛿AwAi× 10 -3 Aw = Q 3.33f 𝐵 𝑚 𝑘 𝑤 𝛿𝐴𝑖×10 −3 = Q 3.33 ×50×1×0.125×1.5 ×10 6× 0.04054054 ×10 −3 = 0.20611 m2 Aw =206. 11 × 103 mm2 Hw×Ww = 206. 11 × 103 2.5 W 𝑤 2 = 206. 11 × 103 Ww = 290mm Hw = 720 mm Aw = Hw×Ww = 290× 720 = 208×103 mm3 D= Ww+d = 290+235 = 525 mm Yoke Design The area of yoke is taken as 1.2 times that of limb Flux density in yoke = 1.3/1.2= 1.08 Wb/m2 Net area of yoke = 1.2×31.1185 ×103 mm2 = 37.422×103 mm2 Gross area of the yoke = 37.422×103 0.9 = 41.58 ×103 mm2
- 3. Taking the yoke as Dy = a= 200mm Hy = 41.58×103 𝑎 = 200mm Overall Dimensions of frame Height of frame (H) = Hw + 2Hy = 720 + 2×200 = 1120mm Width of frame W = 2D +a =2× 525 + 200 = 1250mm Depth of frame = a= 200mm LV Winding Star connection Secondary line voltage = 11kv Secondary phase voltage Vph = 11 ×1000 √3 = 6350V Number of turns per phase =Vph/Et =6350/9 = 705 Secondary phase current = 400 ×1000 3×6350 = 21 A A current density= 2.3 A/mm2 Area of secondary conductor, as = 21 2.3 = 9.13mm2 From the table (IS 1897-1962) using a bare conductor = 6.0 × 1.9 mm2 Area of bare conductor as = 9.29mm2 𝛿s = 21/9.29 = 2.26 A/mm2 The dimensions of insulated conductor = 6.5 × 2.4 mm2 Using 7 layers of helical winding. The no of turns along axial depth =100+1=101 Axial depth of L.V = 101× 6.5= 656.5mm Height of window= 720mm Clearance = (720-656.5)/2 = 31.75mm Radial depth bs= no of layers × radial depth of conductor + insulation between layers = 7× 2.4 +6×0.5 = 19.8 mm Inside diameter of L.V winding= d+2× insulation = 235+ 2× 1.5 = 238mm Outside diameter of L.V winding = 238+ 2× 19.8 = 277.6 mm H.V Winding: Primary voltage (line) = phase voltage = 66000kV (in delta) No of turns Tp = 66000/9 = 7334 For 5.5% tapping Tp max= 1.055 × 7334 = 7737 Using 20 coils, Turns per coil = 7737/20 = 387 Using 19 normal coils of 392 turns and one reinforced coil of 289 turns
- 4. Tp=19× 392+ 289= 7737 Taking 24 layers per coil, we have Turns per layer = 16 H.V winding phase current Ip = 400×1000 3×6600 = 2.02 A It is below 21 A so cross over coils are used for H.V winding Taking current density of 2.4 A/mm2 ap = 2.02/2.4 = 0.841 mm2 Diameter = ( 4 𝜋 × 0.841)1/2 Diameter of bare conductor =1.034 mm Using paper covered conductor of (IS: 1897 -1962) Bare diameter = 1.06mm and insulated diameter =1.26mm with fine covering Modified area of conductor= 𝜋 4 × (1.26)2 = 1.25mm2 Actual current density is = 2.02/1.25= 1.616 A/mm2 Axial depth of one coil = 16 × 1.26 = 20.16mm The spacers used between adjacent coils are 5mm in height. Axial length of H.V winding = no of coils × axial depth of each coil + depth of spacers =20× 20.16 + 20× 5 = 503.2 mm The height of window = 720 mm The clearance = (720- 503.2)/2 = 108.4 mm Radial depth of H.V coil bp = 24 × 1.26 + 23 × 0.5 = 41.74mm The thickness of insulation between H.V and L.V is given by = 5+ 0.9KV = 5+0.9× 66 = 64.4 mm Inside diameter of H.V winding = outer diameter of L.V + 2× thickness of insulation between L.V and H.V = 277 + 2× 64.4 = 406 mm Outside diameter of H.V winding = 406 + 2× 41.74 = 490mm Clearance between windings of two adjacent limbs = 525mm- 490mm=35mm Resistance Mean diameter of primary winding = 406 +490 2 = 448mm Lmtp = 𝜋 × 448 × 10-3 m = 1.41m Resistance of primary winding at 75℃ rp = 𝑇𝑝 𝜌×𝐿 𝑚𝑡𝑝 𝑎 𝑝 = 7334×0.021×1.41 1.246 = 174.29 Ω Mean diameter of secondary winding = 238 +277 .6 2 = 257.8mm Lmts = 𝜋 × 257.8 × 10-3 m = 0.81m Resistance of secondary winding at 75℃ rs = 705×0.021 ×0.81 9.29 = 1.29 Ω Total resistance referred to primary side Rp = 174.29 + ( 7334 705 )2 × 1.29 = 313.9Ω P.U resistance of transformer 𝜀r = 𝐼 𝑝 𝑅 𝑝 𝑉𝑝 = 1.29×313 .9 66000 = 0.00613 Ω
- 5. Leakage Reactance Mean diameter of windings = 238+490 2 = 364mm Length of mean turn Lmt = 𝜋 × 364 × 10-3 m = 1.143 m Height of winding Lc = 𝐿 𝑐𝑝+ 𝐿 𝐶𝑠 2 = 503 .2+656.5 2 = 579.85mm Width of duct, a = (364- 277.6)/2 = 43.2 mm Leakage reactance of transformer referred to primary side Xp = 2𝜋f µ0 Tp 2 𝐿 𝑚𝑡 𝐿 𝑐 (a+ 𝑏 𝑝+ 𝑏 𝑠 3 ) = 2𝜋 ×50×4 𝜋 × 10−7 × (7334)2 × 1.206 0.579 (a + 𝑏 𝑝+ 𝑏 𝑠 3 ) = 2𝜋 ×50×4 𝜋 × 10−7 × (7334)2 × 1.206 0.579 (43 + 41.74+ 19.8 3 ) × 10-3 = 2808 Ω P.U leakage reactance 𝜀x = (2.02 × 2808)/66000 = 0.086 P.U impedance 𝜀s = √0.006132 + 0.0862 = 0.08621 Regulation P.U regulation 𝜀 = 𝜀r cos ∅ + 𝜀x sin ∅ Per unit regulation at unity power factor 𝜀 = 𝜀r = 0.00613 At zero p.f lagging 𝜀 = 𝜀x = 0.086 At 0.8 pf lagging 𝜀 = 𝜀r cos ∅ + 𝜀x sin ∅ = 0.00613 × 0.8 + 0.086× 0.6 = 0.056504 Losses Copper loss: At 75℃ = 3I2 Rp = 3× 2.02 2 × 313.9 = 3842 watt Taking 15% stray loss = 1.15 × 3842 = 4420 watt Core loss Taking the density of laminations = 7.6 × 103 kg/m3 Weight of 3 limbs = 3× Hw × Ai × 7.6 × 103 = 3× 0.72 × 0.0301185 ×7.6 × 103 = 494.42kg The flux in the limbs is 1.3 Wb/mm2 From the specific core loss is 2.03 W/kg Therefore core loss in limbs = 494.42 × 2.03 = 1003.67 Watt Weight of two yoke = 2 × length of yoke × area of the cross section × density of laminations = 2 × 1.250 × 0.037422× 7.6 × 103 = 711 kg Corresponding to 1.08 Wb/m2 ≈ 1 .0 Wb/ m2 , specific core loss =1.2 W/kg Therefore core loss in yoke = 711 × 1.2 = 853.2 Watt Total core loss Pi = 1003.7+ 853.3 =1856.9 ≈ 1857 watt Efficiency: Total losses at full load = Pc + Pi = 4420 + 1857 = 6277 watt Efficiency at full load and unity power factor = 400×1000 400000+6277 =98.45% For maximum efficiency, x2 PC = Pi
- 6. ∴ x =√ 1857 4420 = 64.81% Tank Design Height of frame H =1120 mm Allowing for 50 mm for the base and 200 mm for oil Oil level = 1120 + 50+ 200 = 1370 mm Allowing another 250 mm for leads etc Height of tank Ht = 1370 + 250 = 1620 mm The height of tank is taken as 1.65m Allowing a clearance of 75 mm along the length Length of the tank Lt = 2D + De + 2l= 2×525 +490 + 2×75 =1690 mm The length of tank is taken as 1.7m The width of the tank Wt = De + 2× clearance = 490 + 2× 90 = 670mm ≈ 0.68m Loss dissipating surface of the tank St = 2(length + width) × height = 2(1.7 + 0.68) × 1.65=7.854m2 Temperature rise = total loss 12.5 ×𝑆𝑢𝑟𝑓𝑎𝑐𝑒 = 6277 12.5 ×7.854 = 64℃ So we need to design tank with tubes TANK WITH TUBES We have St be the dissipating surface of the tank The total heat dissipating by tank walls only = 12.5St W/℃ Let the area of tubes = xSt Total area of tank walls and tubes = St + xSt = St(1+x) Loss dissipated = 12.5+8.8x 𝑥+1 W/m2 -℃ Now x= 1 8.8 ( 𝑃𝑖 +𝑃𝑐 𝑆𝑡 𝜃 − 12.5) = 1 8.8 ( 6277 7.854× 35 − 12.5) = 1.175 Area of tubes = xSt = 1.175 × 7.854 = 9.23 m2 Using 50mm diameter tubes spaced 70 mm apart. The average length of tubes is assumed as 1.4m Dissipating area of each tubes =𝜋 ×0.05 × 1.4 = 0.22m2 Number of tubes to be provided = 9.23/0.22 = 42 Arrangement of tubes: Along length – 2 rows 11 and 10 tubes Design Sheet KVA rating = 400KVA, Phase = 3-phase, Frequency = 50HZ, Delta/star
- 7. line voltage [ 𝐻. 𝑉 660000𝑉 𝐿. 𝑉 11000𝑉 ] phase voltage[ 𝐻. 𝑉 660000𝑉 𝐿. 𝑉 6350𝑉 ] line current [ 𝐻. 𝑉 3.5𝐴 𝐿. 𝑉 21𝐴 ] phase current [ 𝐻. 𝑉 2.02𝐴 𝐿. 𝑉 21𝐴 ] Type – Core Type of cooling: Oil Natural Core Symbols Quantity 1 Material 0.35 mm thick 92 Grade 2 Output constant k 0.25 3 Voltage constant Et 9V 4 Circumscribing circle diameter d 235mm 5 Number of steps 2 6 Dimensions a= 200 mm, b= 125mm 7 Net iron area Ai 30.1185 × 10 3mm2 8 flux density Bm 1.3Wb/m2 9 flux ɸm 0.04054054Wb 10 Weight 494.42Kg 11 Specific iron loss 2.03 W/kg 12 Iron loss 1003.37W Yoke 1 Depth of yoke Dy 200mm 2 Height of yoke Hy 200mm 3 Net yoke area 37.422×103 mm2 4 Flux density 1.08 Wb/m2 5 Flux 0.04054054Wb 6 Weight 711 kg 7 Specific iron area 1.2W/kg 8 Iron loss 853.2 W Windows 1 Number 2 2 Window space factor Kw 1.25 3 Height of window Hw 720mm 4 Width of window Ww 290mm 5 Window area Aw 206. 11 × 103 mm2 Frame 1 Distance between adjacent limbs D 525 mm 2 Height of frame H 1120mm 3 Width of frame W 1250mm
- 8. 4 Depth of frame Dy 200mm Figure: Overall dimensions of 400KVA, 66/11KV 3 phase, 50Hz, coretype, distribution transformer
- 9. Windings L.V H.V 1 Type of winding Helical Cross over 2 Connection Star Delta 3 Conductor Dimensions-bare 6.0 × 1.9 mm2 1.06mm insulated 6.5 × 2.4 mm2 1.26mm Area 9.29mm2 1.25mm2 Number in parallel 4 Current density 2.26 A/mm2 1.616 A/mm2 5 Turns per phase 705 7334(7735)at 5.5% tap 6 Coils total number 3 3× 20 per core leg 1 20 7 Turns per coil 705 19 of 392 turns, 1 0f 289 per layer 101 16 8 Number of layers 7 24 9 Height of winding 656.8mm 503.2 10 Width of winding 19.8mm 41.74mm 11 Insulation between layers 0.5mm press board 0.5mm paper between coils 5.0 spacer 12 Coil diameter inside 238mm 406mm Outside 277.6mm 490mm 13 Length of mean turn 0.81m 1.41m 14 Resistance at 75°C 1.29 Ω 174.29Ω Insulation 1 Between L.V winding and core 2 Between L.V winding and H.V winding 3 Width of duct between L.V and H.V Tank 1 Dimensions- Height Ht 1.65m Length Lt 1.7m Width Wt 0.68m
- 10. 2 Oil level 1.37m 3 Tubes 42 tubes (11 and 10 in parallel) 4 Temperature 64℃ Impedance 1 P.U Resistance 0.00613 2 P.U Reactance 0.086 3 P.U impedance 0.08621 Losses 1 Total core loss 1857W 2 Total copper loss 4420W 3 Total losses at full load 6277W 4 Efficiency at full load and unity p.f 98.45%