Three Phase Induction Motor Design (Electrical Machine Design)
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The document presents a design project for a 0.746 kW, 400V, 3-phase squirrel cage induction motor, outlining the design parameters including dimensions, rotor specifications, and magnetic considerations. Key design elements include a star-delta starter, efficiency of 90%, and power factor of 0.8, alongside the calculations for core dimensions and winding configurations. The study concludes with detailed specifications including rotor and stator characteristics essential for effective performance.
![6
Main Dimension: Continued…….
KVA input, Q =
𝐾𝑊
𝞰×𝑝𝑓
=
0.746
0.9 ×0.8
= 1.036 𝐾𝑉𝐴
Volume Of Machine, 𝐃𝟐𝐋 =
𝑄
𝐶𝑜𝑛𝑠
=
1.036
34.67×25
= 1.2 × 10−3m3
We Know,
𝐿
𝜏
= 1 [Good overall Design]
∴
𝐿
𝜋
𝐷
4
= 1 [𝝉= 𝝅
𝑫
𝒑
]
⇒ 𝐋 = 0.79𝐷
AS , 𝑫𝟐𝑳= 𝟏. 𝟐 × 𝟏𝟎−𝟑m3
⇒ 𝐷2 × 0.79𝐷= 1.2× 10-3 m3
∴ 0.79𝐷3= 1.2 × 10−3 m3
Design
Feature
Ratio
𝑳
𝝉
Minimum Cost 1.5 to 2
Good Power Factor 1 to 1.25
Good Efficiency 1.5
Good Overall Design 1](https://image.slidesharecdn.com/group-02inductionmotordesignprojectelectricalmachinedesign-210705161618/85/Three-Phase-Induction-Motor-Design-Electrical-Machine-Design-6-320.jpg)
![10
Stator Winding:Continued……
Mush winding in tapered Semi-enclosed Slots is used here.
Coil Span,Cs=
𝑆𝑙𝑜𝑡𝑠
𝑃𝑜𝑙𝑒
=
24
4
= 6
But Coin Span Should not been Even integer In case of mush
windings.Therefore,a Coil Span Of 5 Slots is used.
Thus the Coils are Chorded by One Slot Pitch.
Angle Of Chording,𝜶 =
1
6
× 180° = 30°
Slots per pole per phase,m=2 [Taken]
𝜷 =
180
6
= 30°
Distribution Factor,Kd=
𝑆𝑖𝑛[
2×30°
2
]
2×sin[
30°
2
]
=0.966
Pitch Factor.Kp= cos(
30°
2
)=0.966
Stator Winding Factor,Kws=Kd×Kp=0.966×0.966=0.933
Stator Current Per Phase,IS=
𝑘𝑣𝑎
3×𝐸𝑠
=
0.746×103
3×400×0.9×0.8
= 0.863 𝐴](https://image.slidesharecdn.com/group-02inductionmotordesignprojectelectricalmachinedesign-210705161618/85/Three-Phase-Induction-Motor-Design-Electrical-Machine-Design-10-320.jpg)
![25
Short Circuit Current
Leakage Reactance:
Rotor Slot Leakage: For a Parallel Sided Slot,
Specific Slot Permeance,
𝜆𝑠 = 𝜇𝑜
ℎ1
3𝑊𝑠
+
ℎ2
𝑊𝑠
+
2ℎ3
𝑊𝑠
+𝑊0
+
ℎ4
𝑊𝑜
= 4𝜋 × 10-7[
7.0
3×5
+
2×1
5+2
+
1
2
]
= 1.57 × 10-6
This Referred To Stator Side,𝜆𝑠𝑟′ =
𝜆𝑠𝑟 ×
𝑘𝑤𝑠
2
×𝑆𝑆
𝑘𝑤𝑟
2
×𝑆𝑟
= 1.57 × 10-6×
0.933 2
×24
1 2
×21
=1.562× 10−6](https://image.slidesharecdn.com/group-02inductionmotordesignprojectelectricalmachinedesign-210705161618/85/Three-Phase-Induction-Motor-Design-Electrical-Machine-Design-25-320.jpg)
Three Phase Induction Motor Design (Electrical Machine Design)
- 1. EE-3220 ELECTRICAL MACHINE DESIGN PROJECT ON INDUCTION MOTOR DESIGN SUBMITTED TO, DR.MD.HABIBULLAH PROFESSOR, DEPARTMENT OF EEE,KUET ABU SAYED MD. JANNATUL ISLAM ASSISTANT PROFESSOR, DEPATMENT OF EEE,KUET SUBMITTED BY, GROUP 02 SAKIB HOSSAIN MD. SAJJAD HOSSAIN ROLL: 1703011 ROLL: 1703022 RAIAN HABIB HIMEL MUHAMMED ZUBAIR RAHMAN ROLL: 1703025 ROLL: 1703053
- 2. 2 Design problem: DESIGN THE MAIN DIMENSION AND ROTOR OF A 0.746KW, 400V, 3‐PHASE, 50HZ, 1432 RPM, SQUIRREL CAGE INDUCTION MOTOR. THE MACHINE IS TO BE STARTED BY A STAR‐DELTA STARTER. THE EFFICIENCY IS 90% AND POWER FACTOR IS 0.8 AT FULL‐LOAD.
- 3. 3
- 5. 5 The Speed Of motor is 1432 r.p.m. And The nearest Synchronous Speed corresponding to 50 HZ is 1500 r.p.m. Synchronous Speed,ns= 1500 60 r. p. s. = 25 r. p. s. Number Of poles,P = 2 𝑓 𝑛𝑠 = 2×50 25 =4 Assuming that, Specific Magnetic Loading,𝑩𝒂𝒗 = 0.33 𝑊𝑏 𝑚2 , Winding factor,𝐊𝐰 = 0.955 & Specific Electric Loading, 𝒂𝒄 = 10000 𝐴/𝑚 Output Coefficient, 𝑪𝒐 = 11 𝑩𝒂𝒗 𝑲𝒘 𝒂𝒄 × 10−3 = 11 × 0.955 × 0.33 × 10000 × 10−3 = 34.67 Main Dimension
- 6. 6 Main Dimension: Continued……. KVA input, Q = 𝐾𝑊 𝞰×𝑝𝑓 = 0.746 0.9 ×0.8 = 1.036 𝐾𝑉𝐴 Volume Of Machine, 𝐃𝟐𝐋 = 𝑄 𝐶𝑜𝑛𝑠 = 1.036 34.67×25 = 1.2 × 10−3m3 We Know, 𝐿 𝜏 = 1 [Good overall Design] ∴ 𝐿 𝜋 𝐷 4 = 1 [𝝉= 𝝅 𝑫 𝒑 ] ⇒ 𝐋 = 0.79𝐷 AS , 𝑫𝟐𝑳= 𝟏. 𝟐 × 𝟏𝟎−𝟑m3 ⇒ 𝐷2 × 0.79𝐷= 1.2× 10-3 m3 ∴ 0.79𝐷3= 1.2 × 10−3 m3 Design Feature Ratio 𝑳 𝝉 Minimum Cost 1.5 to 2 Good Power Factor 1 to 1.25 Good Efficiency 1.5 Good Overall Design 1
- 7. 7 Main Dimension: Continued……. ⇒ Stator Bore, 𝐃 = 3 1.2 × 10−3 0.79 = 115 𝑚𝑚 Stator Core Length,𝐋 = 0.79 × 𝐷 𝑚m Stator Core Length,𝐋 = 0.79 × 115 = 90.85 𝑚𝑚 = 0.09085 𝑚 Pole Pitch,𝛕 = 𝐋 = 90.85 𝑚𝑚 = 0.09085 𝑚 As Length of the Core is 90.85 mm<125 mm So there is no need Of providing any radial ventilating duct. Gross Iron length,Ls= L=90.85 mm Net Iron length, 𝐋𝐢 = 𝑠𝑡𝑎𝑐𝑘𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 × 𝐿s = 0.9 × 0.09085 = 0.082 𝑚 = 82 𝑚𝑚 Stator will be Provided With radial ventilating Duct if the core length exceeds 125 mm.The width of each duct will be then 8 to 10 mm.
- 9. 9 Stator Winding The machine is to be designed for delta connection as it is started by a star-delta starter. The stator voltage per phase, ES = 400V Flux per pole, 𝝓𝒎 = B av τ L = 0.33 × 0.09085 × 0.09085 = 2.724 × 10−3Wb Stator turns per phase, TS = 400 4.44×50×2.724×10−3×0.955 = 692.62 ≈ 693 Taking Stator Slots per pole per phase,qs=2 Stator Slots,SS=No of Phases×Poles×qs =3 ×4 ×2=24 Stator Slot Pitch,yss= 𝜋𝐷 𝑆𝑆 = 𝜋×115 24 = 15.05 𝑚𝑚 So It is Suitable for double layer Winding. Total Stator Conductors,Zs=6× 693 = 4158 Number of Stator Conductor Per Slot,Zss= 4158 24 = 173.25 ≈ 174
- 10. 10 Stator Winding:Continued…… Mush winding in tapered Semi-enclosed Slots is used here. Coil Span,Cs= 𝑆𝑙𝑜𝑡𝑠 𝑃𝑜𝑙𝑒 = 24 4 = 6 But Coin Span Should not been Even integer In case of mush windings.Therefore,a Coil Span Of 5 Slots is used. Thus the Coils are Chorded by One Slot Pitch. Angle Of Chording,𝜶 = 1 6 × 180° = 30° Slots per pole per phase,m=2 [Taken] 𝜷 = 180 6 = 30° Distribution Factor,Kd= 𝑆𝑖𝑛[ 2×30° 2 ] 2×sin[ 30° 2 ] =0.966 Pitch Factor.Kp= cos( 30° 2 )=0.966 Stator Winding Factor,Kws=Kd×Kp=0.966×0.966=0.933 Stator Current Per Phase,IS= 𝑘𝑣𝑎 3×𝐸𝑠 = 0.746×103 3×400×0.9×0.8 = 0.863 𝐴
- 11. 11 Stator Core Flux in stator core = 𝜙𝑚 2 = 2.724×10−3 2 = 1.362 × 10−3 Wb Assuming a flux density of 1.2 Wb/m2, Area Of Stator Core,ACS = 1.362×10−3 1.2 = 1.135 × 10−3 m2 Depth of Stator core, 𝒅𝒄𝒔 = 𝐴𝑐𝑠 𝐿𝑖 = 1.135×10−3 0.082 = 13.84 × 10−3 m=13.84 mm Taking the depth of core, 𝒅𝒄𝒔 𝑎𝑠 14 𝑚𝑚 Flux density in stator core, 𝑩𝒄𝒔 = 13.84×1.2 14 Wb/m2 =1.186 Wb/m2 Flux Desnity Of Stator Core Should be <1.5 Wb/m2 Stator Core
- 14. 14 Air Gap Length of Air Gap,𝒍𝒈 = 0.2 + 2 𝐷𝐿 = 0.2 + 2 0.115 × 0.09085 𝑚𝑚 = 0.404 𝑚𝑚 Air Gap is a mere Clearance between Rotor and Stator. It is made Smaller than the given equation if roller and ball bearings are used. To avoid large magnetizing current, a shorter gap need to be installed.Referring to table 10.2, Length of Air Gap,𝒍𝒈 = 0.3 𝑚𝑚 Diameter of rotor, 𝑫𝒓 = 𝐷 − 2 × 𝑙𝑔 mm = 115 − 2 × 0.3 mm = 114.4 𝑚𝑚
- 15. 15 Rotors Slots Since (Ss-Sr) Should not be equal to 0,±𝒑, ±𝟐𝒑, ±𝟑𝒑, ±𝟓𝒑, ±𝟏, ±𝟐, ± 𝒑 + 𝟏 , ±(𝒑 + 𝟐) As, No Of Poles,p=4 (Ss-Sr) Should not be equal to 0,±4, ±8, ±12, ±20, ±1, ±2, ± 5 , ±(6) So, (Ss-Sr) Can be ±3 𝑜𝑟 ± 7 No. Rotor Slots Is also 15% 𝑡𝑜 30% Smaller or larger than No. of Stator Slots. The Stator Slots, Ss=24 The Rotor Slots, Sr=24-3=21 Rotor slot pitch at the air gap, 𝒚𝒔𝒓 = 𝜋 𝐷𝑟 𝑆𝑟 = 3.14 × 0.1144 21 = 0.0171 𝑚𝑚 = 17.1 𝑚𝑚
- 16. 16 Rotors Bars Rotor bar current, 𝑰𝒃 = 2𝑚𝑠𝐾𝑤𝑠𝑇𝑠𝐼𝑠𝑐𝑜𝑠𝜑 𝑆𝑟 = 2×3×0.933×693×0.863×0.8 21 , ms=No of Phase = 127.54 𝐴 Bar Current Density, 𝜹𝒃 = 4.7 𝐴 𝑚𝑚2 Area of each rotor bar, 𝒂𝒃 = 127.54 4.7 = 27.14 𝑚𝑚21.1𝑚𝑚2 Referring to the table 23.1, the dimension of conductor to be used are 7.0× 4.0 𝑚𝑚2. Area Of bar used, ab = 27.1 mm2
- 17. 17 Rotors Bars:Continued……….. Width of rotor slot, 𝑾𝒔𝒓 = 4.0 + 0.5 + 0.5 = 5 𝑚𝑚 Depth of rotor slot, 𝒅𝒔𝒓 = 7.0 + 1 + 1 + 0.5 + 0.5 mm = 10 mm Slot pitch at the bottom of slots = 𝜋(𝐷𝑟−2×𝑑𝑠𝑟) 𝑆𝑟 = 3.14(114..4−2×10) 21 = 14.12 mm Tooth width at the root, 𝑾𝒕 = 14.12 − 𝑊 𝑠𝑟 = 14.12 − 5 mm = 9.12 mm Flux density at the root of rotor teeth = 𝝓𝒎 𝑆𝑟 𝑝 ×𝐿𝑖 ×𝑊𝑡 = 2.724×10−3 21 4 ×0.082×9.12×10−3 = 0.694 𝑊𝑏/𝑚2 The flux density is within limits. Semi Closed Rotor Slot All Dimensions is in mm
- 18. 18 Rotors Bars:Continued……….. The bars are extended by 15 mm beyond the core on each side and taking 15 mm as increment in length due to skewing. Length of each bar, 𝑳𝒃 = 𝐿 + 2 × 15 + 15 𝑚𝑚 = 90.85 + 30 + 15 𝑚𝑚 = 135.85 𝑚𝑚 Resistance of each bar, 𝒓𝒃 = 𝜌 ×𝐿𝑏 𝐴𝑏 = 0.021×135.85×10−3 27.1 , 𝜌 = 𝑟𝑒𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = 1.053 × 10−4 𝛺 Total Copper loss in bars = 𝑆 × 𝐼𝑏 2 × 𝑟𝑏 = 21 × 127.54 2 × 1.053 × 10−4 = 35.97 𝑊
- 19. 19 End Rings End Ring Current, 𝑰𝒄 = 𝑆𝑟𝐼𝑏 𝜋𝑝 = 21×127.54 3.1416×4 = 213.14 𝐴 Taking current density in end ring, 𝜹𝒆 = 5 𝐴/𝑚𝑚2 Area of the end ring, 𝒂𝒆 = 213.14 5 = 42.628 𝑚𝑚2 Using a ring of depth, 𝒅𝒆 = 𝟏𝟎 𝒎𝒎 ,thickness of the ring, 𝑡𝑒 = 𝟒. 𝟑 𝑚𝑚 Thus Area becomes, 𝒂𝒆 = 43 𝑚𝑚2 Outer diameter of end ring= Diameter Of Rotor – 2×depth of rotor slot = 𝐷𝑟 − 2 × 𝑑𝑠𝑟 = 114.4 − 2 × 10 = 94.4 𝑚𝑚 Inner diameter of end ring= Outer diameter – 2×depth of ring = 94.4 − 2 × 10 = 74.4 𝑚𝑚 Mean Diameter Of End Ring=(Outer diameter +Inner diameter)/2 = 94.4+74.4 2 = 84.4 𝑚𝑚
- 20. 20 End Rings Resistance of each ring, 𝒓𝒆 = 𝜌𝜋𝐷𝑒 𝑎𝑒 = 0.021 ×𝜋×84.4×10−3 43 = 1.2945 × 10−4 Ω Copper loss in two end rings = 2 × 𝐼𝑏 2 × 𝑟𝑒 = 2 × (213.14)2× 1.2945 × 10−4 = 11.76 𝑊 Total Copper Loss = 35.97 + 11.76 𝑊 = 47.73 𝑊 Now, 𝑅𝑜𝑡𝑜𝑟 𝐶𝑢 𝑙𝑜𝑠𝑠 𝑅𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑠 1−𝑠 = 47.73 0.746×103 = 0.064 ⇒ 𝑠 = 0.0602 Thus, full load slip, 𝒔 = 0.0602 = 6.02% Here, 𝑆𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑆𝑝𝑒𝑒𝑑, 𝑛𝑠 = 25 𝑟. 𝑝. 𝑠 𝒏𝒓 = 1 − 𝑠 × 𝑛𝑠 = 1 − 0.0602 × 25 = 23.495 Rotor Speed,𝑵𝒓 = 𝑛𝑟 × 60 = 23.495 × 60 = 𝟏𝟒10 𝑟. 𝑝. 𝑚. Which is nearly Equal to 1432 r.p.m.
- 21. 21 Rotor Core The Value Of Depth Of Rotor Core Is Taken equal to that Of Stator Core. ∴Depth Of Rotor Core,dcr= 14 mm Flux density in the Rotor Core,Bcr= 1.186 Wb/m2 Inner Diameter Of Rotor Laminations, Di=Dr-2dsr-2dcr =114.4-2× 10 − 2 × 14 =66.4 mm
- 22. 22 No Load Current Magnetizing Current: The MMF Required for Various Parts Of Magnetic Circuit (Only Rotor) are calculated below: (i) Air gap: For rotor slots, The ratio (slot opening/air gap length)= 2 0.3 = 6.67 Wo=2 mm Corresponding to a ratio 6.67,Carters Coefficient , 𝑲𝒄𝒔 =0.68 Gap Construction For Rotor Slots,Kgsr= 𝑦𝑠𝑟 𝑦𝑠𝑟 −𝑘𝑐𝑠 𝑊0 = 17.1 17.1−0.68×2 =1.086 Gap contraction factor for rotor slots,𝑲𝒈𝒔=𝑲𝒈𝒔𝒔 × 𝑲𝒈𝒔𝒓
- 23. 23 No Load Current (ii) Rotor Teeth: Width of rotor teeth at 1/3 height of narrow end, 𝑊𝑡𝑠 3 = 𝜋×(𝐷𝑟− 4×𝑑𝑠𝑟 3 ) 𝑆𝑟 − 𝑊 𝑠𝑟 = 𝜋(114.4−4× 10 3 ) 21 − 5 𝑚𝑚 = 10.12 𝑚𝑚 Area of teeth per pole at 1/3 height of narrow end 𝐴𝑡𝑟 = 21 4 × 0.082 × 10.12 × 10−3 = 4.356 × 10−3 𝑚2 Flux density in rotor teeth at 1/3 height,𝐵𝑡𝑟 3 = 0.61 𝑊𝑏/𝑚2 ∴ 𝐵𝑡𝑟66= 1.36 × 0.61 = 0.83 𝑊𝑏 𝑚2 In correspondence to this value,flux density, 𝑎𝑡𝑟 = 80 𝐴 𝑚 ∴MMF required for rotor teeth, 𝐴𝑇𝑡𝑟 = 80 × 10 × 10−3 = 0.8 𝐴. Turns
- 24. 24 No Load Current (iii)Rotor Core: Rotor Core area, 𝑨𝒄𝒓 = 𝐿𝑖𝑑𝑐𝑟 = 0.082 × 14 × 10−3 = 1.148 × 10−3 𝑚2 Flux density in the core = 1. 186 𝑊𝑏/𝑚2 Corresponding to this flux density, 𝒂𝒕𝒄𝒓 = 280 𝐴 𝑚 Length of flux path in rotor core, 𝒍𝒄𝒓 = 𝜋×(𝐷𝑟−2×𝑑𝑠𝑟−2×𝑑𝑐𝑟) 3×𝑝 = 𝜋(114.4−2×10−2×14) 3×4 = 17.38 × 10−3𝑚 = 17.38 mm MMF for rotor core, 𝐴𝑇𝑐𝑟 = 280 × 17.38 × 10−3 = 4.8664 𝐴. 𝑇𝑢𝑟𝑛𝑠
- 25. 25 Short Circuit Current Leakage Reactance: Rotor Slot Leakage: For a Parallel Sided Slot, Specific Slot Permeance, 𝜆𝑠 = 𝜇𝑜 ℎ1 3𝑊𝑠 + ℎ2 𝑊𝑠 + 2ℎ3 𝑊𝑠 +𝑊0 + ℎ4 𝑊𝑜 = 4𝜋 × 10-7[ 7.0 3×5 + 2×1 5+2 + 1 2 ] = 1.57 × 10-6 This Referred To Stator Side,𝜆𝑠𝑟′ = 𝜆𝑠𝑟 × 𝑘𝑤𝑠 2 ×𝑆𝑆 𝑘𝑤𝑟 2 ×𝑆𝑟 = 1.57 × 10-6× 0.933 2 ×24 1 2 ×21 =1.562× 10−6
- 26. DESIGN SHEET 26 KW:0.746 Phase:3 FREQUENCY:50HZ VOLTAGE:400V SLIP:6.02% TYPE:SQUIRREL CAGE CONNECTION: DELTA
- 27. 27 Rating 01 Full Load Output 0.746KW 02 Line Voltage 400V 03 Frequency f 50HZ 04 Phases 3 05 Efficiency 𝞰 90% 06 Power factor cos∅ 0.8 07 Number Of Poles 4 08 Synchronous r.p.s. ns 25 09 KVA input 1.036 10 Full Load Line Current 3 × 0.863 = 1.5 A
- 28. 28 Loading 01 Specific Magnetic Loading Bav 0.33 Wb/m2 02 Specific Electric Loading ac 10000 A/m 03 Output Coefficient C0 34.67 04 D2L 1.2 × 10-3 m3 Main Dimentions 01 Stator Bore D 115 mm 02 Stator Core Length 𝑳 90.85 mm 03 Ducts nd Nill 04 Gross Iron Length Ls 90.85 mm 05 Net Iron Length Li 82 mm 06 Pole Pitch 𝛕 90.85 mm
- 29. 29 Rotor 01 Length Of Air Gap lg 0.3 mm 02 Diameter of Rotor Dr 114.4 mm 03 Type Of Winding Squirrel Cage 04 Number Of Slots Sr 21 05 Slots Per Pole Per Phase qr 1.75 06 Conductors Per Slot Zsr 1 07 Winding Factor Kwr 1 08 Slot Pitch ysr 17.1 mm 09 Rotor Bar Current Ib 127.54 A 10 Rotor Bar Cross Section 7.0× 4.0 𝑚𝑚2 area ab 27.1 mm2 Length Lb 135.85 mm Current density 𝜹𝒃 4.7 A/mm2
- 30. 30 11 Resistance Of each bar rb 1.053 × 10−4 𝛺 12 Copper Loss In Bars SrIb 2rb 35.97 W 13 End Ring Current Ie 213.14 A 14 End Ring : Cross Section 10× 4.3 𝑚𝑚2 area ae 43 mm2 Mean diameter De 84.4 mm Current density 𝜹𝒆 5 A/mm2 15 Resistance Of each ring re 1. 2945 × 10−4 Ω 16 Copper Loss in End ring 2Ie 2re 11.76 W 17 Total rotor Copper Loss 47.73 W 18 Depth Of Rotor Core dcr 14 mm