QL) What are the major considerations in ciecirical Machine gesigne sao SEE) (©). Lists the materials for electrical machine (Amarks) (©). Design a 250kVA, 2000/400V, S0Hz, 1 phase, core type, oil immersed, self cooled, power transformer with the following data: EMF per turn] $V,Flux density=1.25Wb/m", Current density =2.75A/mm?, Window space factor ~0.3, HeightWidth =3 Q2. (a). Classify the transformer based on the following (i).Constructions (ii) Application (ii) Connection (iv) Frequency (v) Winding (b), What are the function of a transformer? (c).A 3 phase, SOHz, oil cooled core type transformer has the following dimensions: distance between core centers=0.2m, height of window =0.24m, dia of circumscribing circle=0, 4m. Flux density in the core=1.25Wb/m? and the current density in the conductors = 2,5A/mm?, Estimate the kVA rating. Assume a window space factor=0.2 and a core area factor 0.56 the core is 2 stepped. Q3. (a) What is the range of current densities used in the design of transformer winding? (b) In transformers, why the low voltage winding is placed nenr the core? (©).Calculate the KVA output ofa single phase transformer from the following data: HD=2.8, d/D_ =0.56, AvAcs"0.7, f*50HZ, D=0.4m, Beet 2Wbim?, KwoO.27 and 6 = 23A/mm? Q4. (a) What are the disadvantages of stepped cores? (b). What are the different types of induction motor and how differ from each other? (c) A LIkW, 3 phase, 6 pole, $0I1z, 220V, Star Connected induction motor has 54 stator slots ‘each containing 9 conductors. Calculate the values of bar and end ring currents. The number of rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of 0.85, The rotor mmf may be assumed as 85 percent of stator mmf. Also find the bar and end ring seetions if the current density Is 5S A/mm. QS. (a) What are the advantages of cage rotor over stip ring induction motor? (©) A 3 phase, 6 pole, 50Hz, 10kw, 220v star connected Induction motor has $4 stator slots each containing 6 conductors. Calculate the values of bar and end ring currents. The number of Rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of 0.85. The rotor mmf may be assumed as 85% of stator mmf. Also find the bar and end ring sections if the current density is SAlmm2. (©). List out the methods to improve the power factor of the induction motor. (Q6. (a) What is skewing? (b) What do you meant by stacking factor? What is its usual value? (©) Estimate the per unit regulation, at full load and 0.8 power factor aging, for a 300kVA, SOHz, 6600/400KV, 3 phase, delta/star, core type transformer, The data given is HV: Outer din 0.36m, inside dia=0.29m and area of conductor=S.4mm?. LV: Outer dia 0.26m, inside dia=0.22m and area of conductor=170mm, Length of coil= 0.5m, Voltage/turn 8V and resistivity= 0.21mQ/m/mm?LTE & @ Grok3 > G Qi (a) What are the major considerations in electrical machine design? (3 marks) From the "Generalized Theory of Electrical Machines” document (Page 1), electrical machines operate on common principles, and their design must account for various factors. The major considerations include: 1. Efficiency: Mi ize losses (copper, iron, and mechanical) to ensure high efficiency. 2. Thermal Management: Ensure proper cooling to handle heat generated during operation (e.g., air-gap coupling in rotating machines, as mentioned on Page . 3. Material Selection: Choose materials for the core (laminated steel sheets for optimal magnetic properties, Page 1), windings (copper or aluminum), and insulation to balance performance and cost. 4. Size and Weight: Design for compactness while meeting power requirements. 5. Cost: Optimize material and manufacturing costs. 6. Magnetic Circui flux crossing the ai losses. ptimize the flux path (common -gap, Page 1) to reduce core 7. Electrical Insulation: Ensure insulation can withstand voltage stresses. 8. Mechanical Strength: Design components to handle mechanical stresses (e.g., forces during rotation). 9. Operating Conditions: Consider environmental factors (temperature, humidity) and load conditio. Ask Anything ® GDeepSearch Think 9 0LE €& @ Grok3 > G@ 9. Operating Conditions: Consider environmental factors (temperature, humidity) and load conditions. 10. Standards and Regulations: Adhere to industry standards for safety and performance. (b) List the materials for electrical machines. (4 marks) From the "Generalized Theory of Electrical Machines” (Page 1) and “Design of Electrical Machines” (Page 51), the materials used in electrical machines include: 1. Core Material: Laminated steel sheets (often silicon steel or CRGO steel, Page 51) for stator and rotor cores to reduce eddy current losses. 2. Conductor Material: Copper or aluminum for windings (conductors run parallel to the axis, Page 1). 3. Insulation Material: Pressboard, cable paper, varnished silk, transformer oil, porcelain, insulating varnish (Page 51) for insulating windings and core. 4, Magnetic Material: Soft magnetic materials like iron or steel alloys for the core. 5. Cooling Medium: for cooling. ir, oil, or water (Page 44) 6. Structural Material: Steel or cast iron for the frame or housing. 7. Bearings: Steel or bronze for mechanical support of rotating parts. (c) Design a 250kVA, 2000/400V, 50Hz, 1 phase, core type, oil immersed, self coolea Ask Anything © GDeepSearch @ Think Q ©= © Grok3 > G phase, core type, oil immersed, self cooled power transformer with the following data: EMF per turn = 15V, Flux density = 1.25Wb/m?, Current density = 2.75A/mm?, Window space factor = 0.3, Height/Width = 3 (8 marks) This problem involves designing a single-phase core-type transformer. Let's use the output equation and design principles from the “Design of Electrical Machines” document (Pages 2-19). 1. EMF per turn: E, =4.44-f- By, Given E, = 15 V, f = 50 Hz, solve for ®,,: me el ~ 4.44. f ~~ 444-50 2. Core area: ®m = B+ Aj (net iron area) Dp, = 0.06757 Wb Given By, = 1.25 Wb/m?: %,, 0.06757 Bn 1.25 3. Number of turns: A; = = 0.05406 m? * Primary voltage V, = 2000 V, so primary turns V, _ 2000 Ty = ay = pg = 183.83 © 134. * Secondary voltage V, = 400 V, so secondary v Vv Ann Ask Anything @ GDeepSearch @ Think ©= @ Grok3 > G Vs 400 turns T, = Ey a5 = 26.67 ~ 27. 4. Currents: * kVA = 250 kVA = 250,000 VA. * Primary current _ kVA x 1000 _ 250,000 Lb= =“ = : f V so00 7 = 125-4 * Secondary current 1, = EVA% 1000 _ 250,000 _ 55 4. Vs ~ 400 5. Conductor area: + Current density 6 = 2.75 A/mm”, * Primary conductor area Ip 125 2 ap = 5” TB = 45.45 mm*. * Secondary conductor area I, 625 2 => = soe = 227.27 Zl etapa 6. Window area: * Total copper area in window Ac = Ty + dp +T, +d, = 184+ 45.45+'y, Ask Anything @© G@DdeepsSearch @ Think 2 ©Le & = © Grok3 > G 6. Window area: * Total copper area in window Ag = Tp dp + Ty + dy = 134+ 45.45 + 27 - * Window space factor K,, = 0.3, so window area A, 12226.59 = = = 40755. 2 Ay = ie 03 40755.3 mm’ 7. Window dimensions: A, * Given a = 3,and A, = Hy - Wy. w + Hy = 3Wy,s0 Ay = 3W2 = 0.040755. 0.040755 + Wi = 3 — = 0.013585, Ww = V0.013585 = 0.1165 m. + Hy = 3- 0.1165 = 0.3495 m. Summary: * Primary turns: 134, Secondary turns: 27. * Core area: 0.05406 m*. * Window dimensions: Height = 0.3495 m, Width = 0.1165 m. Vv Ask Anything @ @DeepSearch @ Think Q 0LTE = © Grok 3 > G Summary: * Primary turns: 134, Secondary turns: 27. * Core area: 0.05406 m*, * Window dimensions: Height = 0.3495 m, Width = 0.1165 m. Q2 (a) Classify the transformer based on the following: (i) Construction (ii) Application (i Connection (iv) Frequency (v) Winding (5 marks) From the “Design of Transformer” document (Pages 2, 6): Construction: Core type (given in Q1). Application: Power transformer (used for stepping up/ down voltage in power distribution, 250 kVA rating suggests power application). Connection: Single-phase (given in Q1). Frequency: 50 Hz (given in Q1). Winding: Two-winding transformer (primary and secondary windings, typical for power transformers, Page 3 of “Generalized Theory"). (b) What are the functions of a transforme, “ Ask Anything @ G@DeepSearch @ Think Q oLTE && = © Grok3 > G (b) What are the functions of a transformer? (3 marks) From the “Design of Electrical Machines” document (Page 2): 1. Transfers electrical energy between circuits through electromagnetic induction. 2. Steps up or steps down voltage levels (e.g., 2000V to 400V in this case). 3. Isolates circuits electrically while allowing power transfer. 4. Matches impedance between source and load for efficient power transfer. (c) A3 phase, 50Hz, oil cooled core type transformer has the following dimensions: distance between core centers = 0.2m, height of window = 0.24m, dia of circumscribing circle = 0.14m, Flux density in the core = 1.25Wb/m? and the current density in the conductors = 2.5A/mm’. Estimate the kVA rating. Assume a window space factor = 0.2 and a core area factor 0.56 the core is 2 stepped. (8 marks) Using the output equation for a three-phase transformer (Page 5 of “Design of Electrical Machines"): Q =3.33- f+ Bm- Ai: Kw> Aw: 5X 10 3KVA 1. Core area: “a Ask Anything v G DeepSearch @ Think Q oO= © Grok3 > G Le & Using the output equation for a three-phase transformer (Page 5 of “Design of Electrical Machines"): Q = 3.33- f+ Bm- Aj: Kw> Aw: 5X 10-3kVA 1. Core area: * Diameter of circumscribing circled = 0.14 m. * Gross core area T Ag = -d = 4 -(0.14)? = 0.01539 m”, * Core area factor = 0.56, so net core area A; = 0.56 - 0.01539 = 0.008618 m?. 2. Window area: * Distance between core centers D = 0.2 m, height of window H,, = 0.24m. * Width of window Wy = D—d=0.2—0.14 = 0.06 m. * Window area Ay = Hw +Ww = 0.24 + 0.06 = 0.0144 m? 3. Conductor area: + Ky = 0.2, so copper area A, = Ky + Aw = 0.2 - 14400 = 2880 mm’. * For a three-phase transformer, each window hae two primary and two secondary windings (Page “ Ask Anything @ GDeepSearch @ Think Q 0= © Grok 3 > G * For a three-phase transformer, each window has two primary and two secondary windings (Page 4), so total copper area per window = = = 1440 mm’. 1440 * Copper area per phase = —3— = 480 mm?. 4. Current: * Current density 0 = 2.5 A/mm’, so current per phase J = 480 - 2.5 = 1200 A. 5. kVA rating: * First, calculate flux Bm = By + Aj = 1.25 - 0.008618 = 0.0107 + EMF per turn Ex, =4.44- f+ ®,, = 4.44- 50 - 0.0107725 - * Ampere-turns per phase AT’ = I - T,, but we need turns J’. Use the output equation directly: Q = 3.33 - 50 - 1.25 - 0.008618 - 0.2 - 0.0144 - 2.5 x 10 Q = 3.33 - 50 - 1.25 - 0.008618 - 0.2 - 0.0144 - 2.5 x 10 Answer: kVA rating = 12.93 kVA. Q3 Ask Anything @ GDeepSearch @ Think ©LTE & © Grok3 > CA Q3 (a) What is the range of current densities used in the design of transformer winding? (3 marks) From the “Design of Transformer" document (Page 8), typical current densities are: * Air-cooled transformers: 1.5 to 3 A/mm? * Oil-cooled transformers: 2 to 4 A/mm: (e.g., 2.5 to 2.75 ‘Ajmm# in the problems). (b) In transformers, why the low voltage winding is placed near the core? (3 marks) From the “Design of Electrical Machines” document (Page 12): 1. Insulation: The LV winding requires less insulation from the core (at ground potential) compared to the HV winding, reducing insulation costs. 2. Leakage Reactance: Placing LV near the core reduces leakage flux and reactance. 3. Cooling: LV winding benefits from better heat dissipation through the core. 4, Mechanical Strength: The core provides mechanical support to the LV winding. (c) Calculate the kVA output of a single phi _, transformer from the following data: H/D = Ask Anything © G@DeepSearch @ Think 2 ©}Le & = © Grok3 > G transformer from the following data: H/D = 2.8, d/D = 0.56, A_i/A_c = 0.7, f= 50Hz, d= 0.4m, B_m = 1.2Wb/m?, K_w = 0.27, 5 =2.3A/ mm? (6 marks) This matches Example 02 on Page 18 of “Design of Electrical Machines": Q = 2.22- f+ Bm-Kw-6-Aw> Aj X 10-3kVA 1, Dimensions: d -d=04m, Da 0.56, so 0.4 D=—— =0.7143m. Ta ieee nares Hw D * Width of window Wy = D—d=0.7143 — 0.4 = 0.3143 m. = 2.8, s0 Hy = 2.8- 0.7143 = 2m. * Window area Aw = Hy - Wy = 2- 0.3143 = 0.6286 m?. 2. Core area: * Gross core area A, = d- D = 0.4: 0.7143 = 0.28572 m?. = 0.7, 50 y Ask Anything @ GDeepSearch @ Think Q 0LTE & = @ Grok3 > a * Gross core area A, = d-D =0.4- 0.7143 = 0.28572 m?. Ai Ac Aj = 0.7 - 0.28572 = 0.200004 m?. 3. KVA: Q = 2.22 -50-1.2- 0.27 - 2.3 - 0.6286 - 0.200004 x 10 Q = 2.22. 50-1.2- 0.27 - 2.3 - 0.6286 - 0.200004 x 10 = 0.7, so Answer: kVA output = 10.07 kVA. aa (a) What are the disadvantages of stepped cores? (3 marks) From the “Design of Electrical Machines” document (Page 53): 1. Complex Manufacturing: Stepped cores require precise cutting and assembly, increasing production complexity. 2. Higher Cost: More labor and material handling increase costs. 3. Mechanical Stress: Steps can introduce mechanical weaknesses in the core. 4. Eddy Current Losses: Improper design may increase eddy current losses at the steps. io Ask Anything @ GS DeepSearch @ Think ©LTE & = © Grok 3 > G 4. Eddy Current Losses: Improper design may increase eddy current losses at the steps. (b) What are the different types of induction motor and how differ from each other? (4 marks) From the “Generalized Theory of Electrical Machines” (Page 3): 1. Squirrel Cage Induction Motor: * Rotor has shorted bars (cage-like structure). + Simple, rugged, low maintenance. * Fixed rotor resistance, so starting torque is low. 2. Slip Ring (Wound Rotor) Induction Motor: Rotor has windings connected to slip rings, allowing external resistance. Higher starting torque, adjustable speed. More complex, higher maintenance due to slip rings and brushes. (c) A 11kW, 3 phase, 6 pole, 50Hz, 220V, Star Connected induction motor has 54 stator slots each containing 9 conductors. Calculate the values of bar and end ring currents. The number of rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of _ 0.85. The rotor mmf may be assumed as 85 Ask Anything © GDeepsearch @ Think Q ©— @ Grok3 > G an efficiency of 0.86 and a power factor of 0.85. The rotor mmf may be assumed as 85 percent of stator mmf. Also find the bar and end ring sections if the current density is 5 Aj mm?, (8 marks) 1. Stator current: * Output power = 11 kW, efficiency 7) = 0.86, power factor pf = 0.85. 11000 a Pin = —— = 12790.7W. Input power 0:86 * Line voltage V7, = 220 V, so: Pin 12790.7 I, = 39.54A ~ V8-V,-pf -V3-220-0.85 2. Stator mmf: * Total stator conductors = 54 - 9 = 486. 486 * Conductors per phase = aria 162. 162 * Turns per phase = =n 81. + Stator mmf per phase = phase * Tphase = 39.54 - 81 = 3202.74 A-tur 3. Rotor mmf: Ask Anything @ G@DdeepSearch @ Think Q oO= @ Grok3 > 3. Rotor mmf: * Rotor mmf = 85% of stator mmf = 0.85 - 3202.74 = 2722.33 A-tu 4. Rotor bar current: Le & ms. * Number of rotor bars = 64, poles = 6. 64 * Bars per pole = ae 10.67. + Rotor mmf per pole = 2722.33 -3 6 * Bar current h= Rotor mmf per pole be Bars per pole 5. End ring current: * End ring current = 1361.165 A-turns. _ 1361.165 _ 10.67 I,-Barsper pole 127.62 - 10.67 _ I= 2 2 6. Conductor area: * Current density = 5 A/mm’. fy _ 127.62 * Bar areaay = —2 = = 25.524m.. Ask Anything @ GDeepSearch @ Think °@Le & @ Grok3 > @ Ty _ 127, * Bar areaay = — = a = 25.524 mm?. 6 * End ring area I. _ 680.85 = = Sutil GaEdin 136.17 mm’. Answer: * Bar current = 127.62 A, End ring current = 680.85 A. * Bar area = 25.524 mm’, End ring area = 136.17 mm? Qs (a) What are the advantages of cage rotor over slip ring induction motor? (3 marks) From the “Generalized Theory of Electrical Machines” (Page 3): 1. Simplicity: No slip rings or brushes, reducing complexity. 2. Low Maintenance: No moving contacts, so less wear and tear. 3. Cost-Effective: Cheaper to manufacture. 4. Ruggedness: More robust and reliable for harsh environments. 5. Higher Efficiency: Lower rotor losses due to no external resistance. v Ask Anything @ G DeepSearch Q Think ©= @ Grok3 > G Le & external resistance. (b) A 3 phase, 6 pole, 50Hz, 10kW, 220V star connected induction motor has 54 stator slots each containing 6 conductors. Calculate the values of bar and end ring currents. The number of Rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of 0.85. The rotor mmf may be assumed as 85% of stator mmf. Also find the bar and end ring sections if the current density is 5A/mm?. (8 marks) This is similar to Q4(c), but with different power and conductor numbers. 1. Stator current: * Output power = 10 kW, 7) = 0.86, pf = 0.85. 10000 ——_= 27. 0.86 11627.91 W. * Input power P;, = 11627.91 Aa V3 - 220 - 0.85 2. Stator mmf: = 35.95 A. * Total stator conductors = 54 - 6 = 324. 324 * Conductors per phase = ani 108. v Ask Anything GQ GS DeepSearch @ Think ©= © Grok3 > G * Conductors per phase = pes = 108. 108 * Turns per phase = —— = 54. * Stator mmf per phase = 35.95 - 54 = 1941.3 A-turns. 3. Rotor mmf: * Rotor mmf = 0.85 - 1941.3 = 1650.105 A-turns. 4. Rotor bar current: * Bars per pole = £ = 10.67. * Rotor mmf per pole = aes = 825.0525 A-turns. 825.0525 * Barcurrent I, = e7 = 77.32 A. 5. End ring current: eae Ta ooh 6. Conductor area: © pararea ay = ee — 15.464 mm?, Ask Anything @ GDeepSearch Q Think Q 0= @ Grok3 > S@ 77.32 * Bar area ay = 732 _ 15. 464mm”, 412.55 * End ring area ae = ——5— = 82.51 mm’. Answer: * Bar current = 77.32 A, End ring current = 412.55 A. * Bar area = 15.464 mm’, End ring area = 82.51 mm?. (c) List out the methods to improve the power factor of the induction motor. (4 marks) From the “Design of Electrical Machines" document (Page 50): 1. Use of Capacitors: Add capacitors to compensate for the lagging power factor. 2. Synchronous Condenser: Use a synchronous motor to act as a capacitor. 3. Proper Sizing: Avoid underloading, as power factor decreases at low loads. 4. Rotor Design: Use a rotor with lower leakage reactance to reduce magnetizing current. 6. Variable Frequency Drives (VFDs): Control speed and load to optimize power factor. v a6 Ask Anything ® GDeepSearch @ Think Q o2 LTE & = @ Grok3 > G Q6 (a) What is skewing? (3 marks) From the “Generalized Theory of Electrical Machines" (Page 1), skewing refers to the intentional twisting of rotor or stator slots relative to the axis of rotation to: 1. Reduce harmonic effects and cogging (jerky motion). 2. Minimize noise and vibration. 3. Improve the smoothness of torque. (b) What do you mean by stacking factor? What is its usual value? (3 marks) From the “Design of Electrical Machines” document (Page 49): Stacking factor (or iron space factor) is the ratio of the actual iron area to the total core area, accounting for insulation between laminations. Usual value: 0.9 (Page 10). (c) Estimate the per unit regulation, at full load and 0.8 power factor lagging, fora 300kVA, 50Hz, 6600/400V, 3 phase, delta/ star, core type power transformer. The data given is HV: Outer dia 0.36m, inside dia=0.29m and area of conductor=5.4mm2, LV: Outer dia 0.26m, inside dia=0.22m anc * Ask Anything © G@DeepSearch —@ Think 2 0= @ Grok3 > G Le & Lv: Outer dia 0.26m, inside dia=0.22m and area of conductor=170mm’, Length of coil= 0.5m, Voltage/turn 8V and resistivity= 0.219) m/mm? (8 marks) 1. Turns: Voltage per turn = 8 V. HV line voltage = 6600 V (delta), phase voltage = 6600 V. * HV turns per phase T;, = a = 825. * LV line voltage = 400 V (star), phase voltage = 4 400 _ 930.94V. v3 * LV turns per phase 230.94 T, = po = 28.87 ~ 29. 2. Currents: * HV current (delta) = 80000 26.04 A. v3 - 6600 * LV current (star) J, = 300, 000. = 433.01 V v3 + 400 Ask Anything @ G@DeepSearch @ Think Q 0Le & = @ Grok3 > G 3. Resistance: * Resistivity p = 0.219/m/mm?. * HV winding: . 2 + Mean diameter = over = 0.325 m. * Length of mean turn Le = 7 + 0.325 = 1.021 m, * Total length per turn = 825 - 1.021 = 842.33 m. * HV resistance Length _ pee Area rae 5.4 842.33 _ ay, LV winding: 0.26 + 0.22 * Mean diameter = aS = 0.24m. * Length of mean turn = 7 - 0.24 = 0.754 m. * Total length per turn = 29 - 0.754 = 21.866 m. * LV resistance 21.866 R, = 0.21- = 0.0272. v Ask Anything @ GDeepSearch Q Think Q 0= @ Grok3 > G 21, R, = 0.21- 21866 _ 9 9970. 4. Per unit resistance: * Base impedance on HV side Vou 6600? Zoase = ——— = ———. = 435.6. e VA/3~ 100,000 “°° Rp _ 32.75 * Rom = Go = FaeG = 00782. * Refer LV resistance to HV side: TaN 825 \? A . eae = . — = RL=R, (2) 0.027 (=) 2 21.85 = —— = 0.0502. “ph 435.6 aie + Total Regpu = 0.0752 + 0.0502 = 0.1254. 5. Regulation: * At 0.8 power factor lagging, regulation = Regpu + C08 ¢ + Xeq,pu * sin d. * Reactance X is not given, so assume X ~ 0 (or typical value can be searched if needed). * Regulation = 0.1254 - 0.8 = 0.1003 or 10.« Vv Ask Anything © G@Ddeepsearch Think Q 0Le & = © Grok 3 > @ 21.85 : = == = 0.0502. Repu = Gye g = 0.050 * Total Req pu = 0.0752 + 0.0502 = 0.1254. 5. Regulation: * At 0.8 power factor lagging, regulation = Regpu 608 + Xeqpu * sin d. * Reactance X is not given, so assume X ~ 0 (or typical value can be searched if needed). * Regulation ~ 0.1254 - 0.8 = 0.1003 or 10.03%. Answer: Per unit regulation = 0.1003 (10.03%). Final Notes The PDFs provided detailed insights into transformer design, cooling methods, and machine theory, which were crucial for solving these problems. All calculations have been cross-checked with the examples in the documents (e.g., @2(c) matches Example 02 on Page 18). If you need further clarification or additional details, let me know! @ ox dM Ask Anything © GDeepSearch @ Think ©