Synchronous Machine Design ( Dr. R. C. Goel & Nafees Ahmed )

Transcription

1 Synchronous Machine Design ( Dr. R. C. Goel & Nafees Ahmed ) y Nafees Ahmed Asstt. rof. Department of Electrical Engineering DIT, University, Dehradun, Uttarakhand References:. Notes by Dr. R. C. Goel. Electrical Machine Design by A.K. Sawhney. rinciples of Electrical Machine Design by R.K Agarwal 4. VTU e-learning repared by: Nafees Ahmed

2 Synchronous Machines Design OUTUT EQUATION: - It gives the relationship between electrical rating and physical dimensions (Quantities) The rating is given by Q V I KVA () h h K f N I KVA () Q 4.44 pd h h Vh.Kd fn h Where K pd =.955 For a short pitched (chorded) distributed winding N f Frequency of output voltage N = Speed in rpm D L D = Inner diameter of stator L = Length of the machine Total electrical loading 6N h I h ac D _ ac D N hi h 6 utting the values of K d, f, & N h I h in equation () we get N D ac D Q L 6 Or Where Q (. Q CD LN 5 KVA ac ) D LN 5 C Output Co efficient. Note : For Induction Motor Q CD LN KW -5 Where C 7.4 ac cos ac cos -5. ac cos CHOICE OF MAGNETIC LOADING ( ): is maximum value of fundamental flux density in the air gap) ( Range of =.8 to Tesla ac KVA repared by: Nafees Ahmed

3 Fundamental flux distribution Average value Average Flux Density D _.5 to. 65 CHOICE OF SECIFIC ELECTRIC LOADING ac :. Water wheel alternator (Hydro Alternator): Salient pole machines ac =, to 4, A/m. Turbo Alternator: Non-Salient pole machines ac =5, to 75, A/m MINI AND MAXI VALUE OF C: 5 We know C. ac. Water wheel alternator ( to 4.5) 5 C..8 min max 4.5. Turbo alternator (5 to 8) C. C min max ERIHERAL SEED: - D N V peri m / s 6. Water Wheel alternator to 8 m/s Maxi limit 4 m/s C. Turbo alternator Maxi limit 5 m/s 75 m/s RUNAWAY SEED: - If an alternator is delivering rated load and the load is suddenly through-off, the speed of the runner becomes abnormally high, this repared by: Nafees Ahmed

4 abnormal high speed is called as runaway speed.. Water Wheel alternator elton wheel turbine:.8 times of rated speed Frances Turbine: to. times of rated speed Kaplan Turbine:.5 to.8 times of rated speed. Turbo alternator.5 times of rated speed ESTIMATION OF MAIN DIMENSIONS (D, L): Q We know D L () CN. Water Wheel alternator: It is of two types a. Round ole or Circular ole: / b = ole arc or width of pole shoe = Length of the pole () = L (for round pole with square pole shoe)- L b. Long ole: to 5 b b p b p N b p S Round or circular ole Round or circular ole (With square pole shoe) 4 repared by: Nafees Ahmed

5 Solving equation () & () we can find out D & L. Finally check the peripheral speed.. Turbo alternator: For large rating machines 5 MVA we use peripheral speed as limit D N V peri 5m / s () From above equation taking N= rpm 5 6 D m So for large rating machines dia of alternator is determined by maximum permissible peripheral speed. Thus for peripheral speed of 5 m/s, the maximum permissible dia D = meter Now put D = in equation () and solve for L Note: Output/meter of length Q CD LN KVA = 8 x x x KVA for L = meter = 4 MVA/meter Hence we can say For Q = 48 MVA M/C L = m D = m For Q = 96 MVA M/C L = 4 m D = m DETERMINATION OF AIR GA LENGTH: - Where. to.5. to.5 D Water wheel Turbo ESTIMATION OF NO OF STATOR SLOTS (S): - 5 sg mm 4 mm 6 mm No of stator slots D S sg S q = Slots per pole per phase May be an integer or fractional (say yx ) Up to KV Up to 6.6 KV Up to 6 KV For symmetrical winding, no of pole pairs should be divisible by y x 7 x (I.e. if =4 no of pole pairs =, so change ) y y repared by: Nafees Ahmed

6 Hence find out the corrected value of q q corrected and S corrected = q corrected EFFECTIVE, OVERALL & GROSS IRON LENGTH OF MACHINE: (Note: - Same as for induction motor) Generally l = l = l =. = l n l l l l n L Let n v =No of ventilating ducts b v = Width of one ventilating duct (Generally for every cm of core length there used to be cm ventilating duct) Gross Iron length l = l + l + l +. +l n Actual Iron length l i =K i *l b v Where K i =Stacking factor =.9 to.9 Overall length L = l + n v *b v Effective length ' L L n b e v v ' 5 Where bv bv =Effective width of ventilating duct (< b b v due to fringing) v 5 ESTIMATION OF NO OF TURNS N h, TOTAL NO CONDUCTORS Z, CONDUCTORS ER SLOTS N C, FUNDAMENTAL FLUX AND MAXIMUM VALUE OF FUNDAMENTAL FLUX DENSITY : - So Vh.K d f N h N () pd h Vh () 4.44 K f D l K pd =.955 e () Z = 6 N h (4) 6 repared by: Nafees Ahmed

7 Z N c (5) S It should be an integer, if not make an integer and divisible by for double layer winding. So find N c,corrected Z corrected using equation (5) N h,corrected using equation (4) using equation (),corrected,corrected using equation () ESTIMATION OF FLUX DENSITY DISTRIUTION CURVE IN THE AIR GA: - Shape of the voltage generated depends on shape of air gap flux. If air gap flux is sinusoidal then generated emf will be sinusoidal. Hence aim is to get sinusoidal flux in the air gap. Synchronous machines may be. Salient pole rotor : rojected oles a. Sinusoidal air gap b. Constant air gap. Cylindrical rotor x m 9 (x).. D (a) Sinusoidal air gap For sinusoidal air gap x cos θ is taken from centre of pole,at pole centre θ.. D (b) Constant air gap & at interpolar axis θ 9. Estimation of air gap flux for salient pole: Sinusoidal Air gap There are three methods a. Wiesemann method of curvilinear square: b. Simplified method (Developed in Germany): c. Carter s fringe curve method: 7 repared by: Nafees Ahmed

8 a. Wiesemann method of curvilinear square: Consider half pole pitch of sinusoidal air gap The flux path in the air gap under the pole may be divided into tubes of flux as shown in following figure. Tubes of force(flux) a x Curvilinear Squares Equipotential lines Interpolar Axis ole Axis These two Lengths are equal Curvilinear Squares Assumptions are i. The line of force (flux) and equipotential line intersects at right angle and form curvilinear square. ii. Flux enters and emerges at perpendicular to the equipotential surface. It is possible if permeability of iron is infinity. iii. Stator & rotor surface are assumed at equipotential. iv. Each tube of force caries equal number of flux lines. v. Each tube of force is divided into equal number of curvilinear squares. vi. The depth of each tube being unity in a direction parallel to shaft. The flux density plot shown in above figure can be used to determine i. ermeance of air gap ii. Form factor iii. ercentage of harmonics. This then help in shaping the pole correctly to obtain sinusoidal flux distribution in air gap and reducing the harmonics and eliminating the more pronounced ones when necessary. 8 repared by: Nafees Ahmed

9 9 ermeance of each square A Width of square Unit depth of pole (Machine) μ μ L Length along air gap path Here width and length of each square are made equal as far as possible so that each square will have same permeance i.e ermeance ermeance in series m= no of squares along the air gap =6 ermeance in parallel n= no of squares along the pole pitch =5x (x for full pole pitch) Thus the permeance of air gap path under each pole m ermeance / pole n For flux density distribution curve Let W x =mean width of a flux tube l gx =mean length of flux tube Then permeance of this tube considering unit depth Wx Wx μ μ l l Flux gs x AT f gs MMF ermeance Flux density at armature surface where width of tube is a x, x Wx x ATf μ () a a l x x gs repared by: Nafees Ahmed W l x gx (AT f = Filed MMF/ ole) Now at the centre of the pole: a x =W x & l g =l gx So flux density at the pole centre μ m ATf l () () % () g W l x g x m μ () a xlgs Calculate flux density at various points using equation () each or 5 apart on armature surface and draw the curve. rms Form factor av b. Simplified method (Developed in Germany): Assumption: i. ermeability of iron is infinity MMF / ole We know o H o Air gap length So flux density at pole centre MMF m o () o Flux density at distance x from the pole centre

10 () / () MMF x o () x x m () x Inter-polar Axis Direct Axis Apparent Flux density curve Actual Flux density curve Error Triangle K L M b O s G H I N s J r D E F r x x r A C b Note:. ole Tip is rounded with a radius of.. b= distance between point G & I. b= distance between point A & C For all the points on this surface, ratio b/b = always Flux Density Distribution Curve Under the pole flux density can be find out by using the equation () but between pole tip and inter-polar axis we have to apply another method to find out flux density. Steps to find out flux density between pole tip to Interpolar axis: (i) Assume three points A, & C on the curve portion of the pole tip. (ii) Draw tangents AD, E & CF from points A, & C respectively. (iii) With D, E & F as centers, draw arcs of circle with radius of AD, E & CF respectively. They intersect horizontal line at points G, H & I. (iv) Join H &, let Chord H = s Radius E = r (v) So flux density corresponding to point H x at point H HK m m x k s Where k is a constant and its value can be find out from graph (See reference notes figure 6 page 6.) repared by: Nafees Ahmed

11 lx k s r/s.5 b /b (vi) (vii) (viii) (ix) (x) (ix) Similarly we can find out flux density at each and every point on the pole tip. Draw the flux density curve. It is known as apparent flux density curve. The value of flux density at inter-polar axis must be zero but it is not zero in above calculated graph. So it is not the actual flux density curve. To get actual flux density curve. Draw triangle MJO, known as error triangle. For point K Calculated flux density (Apparent flux density) = HK Error = HN Actual flux density HL = HK-HN. Similarly we can find out point of actual flux density curve after subtracting the error value. The actual flux density distribution curve so obtained has a fundamental as well as odd harmonics. Actual flux density distribution curve: Divide half of the pole pitch into 6 equal s parts as shown bellow. Fundamental flux Density distribution Actual flux Density Distribution m u u u u4 u5 u6 D.86u.67u.6u.89u 4.u u.u.6u.89u 4.6u u u 6 6 repared by: Nafees Ahmed

12 Note: Equation for actual flux density m Sin Sin 5Sin 5 7Sin 7 It Average value av Its RMS value av Ratio K m Flux per pole l e Actual flux per pole l p av e av K m 5 7 m K K. & are used for calculation in electrical circuit design and used in magnetic circuit calculation.. ole tip is rounded by an arc of radios.... m &. Estimation of flux density distribution curve for cylindrical rotor: In cylindrical rotor synchronous machine / rd portion of rotor is left unslotted. Length of air gap is constant Flux is assumed to be trapezoidal are Actual flux density Assumed flux density (Trapezoidal) N L +I f -I f repared by: Nafees Ahmed

13 m o MMF / ole Air gap length N TOOTH AND SLOT DESIGN: S Cylindrical Rotor In alternators generally open slots are used. D Slot pitch sg Flux through a slot pitch m l We know So sg K m m K sg Flux through the tooth (we assume that entire flux per slot pitch passes through tooth). b.. l sg t tt e ermissible value of flux density in tooth t =.6 to.8 T If flux density t >.7 T in stator tooth, magnetic unloading will take place: A part of the flux enters through the slot. e So width of slot Height of slot So b b tt o s sg.. l b s t e sg h ( to 5) b b o tt repared by: Nafees Ahmed

14 Reactance Gap Wedge Wall Insulation (4 mm) Mica Tape (.5 to.75 mm) Mica Separation (.5 mm) Mica (.5 mm) Cu Strips Slot for Water Wheel Alternator ( KV-rick Construction) Slot for Turbo Alternator ( KV) Thickness of insulation With mica or leatheroid as a wall insulation KV mm With improved insulation (Semica Therm) KV mm Thickness =.5 KV +.7 mm Advantages of Semica Therm: (a) Much better heat is dissipated for higher rating machines due to less thickness of wall insulations. (b) Insulation occupies little less space in the slot. ESTIMATION OF SECTIONAL AREA OF STATOR CONDUCTOR (F C ): er phase stator current Q I h V h I h So FC mm Where Current density 4 5 A / mm STATOR YOKE DESIGN: Flux through stator yoke is half of the flux per pole y hy K il Where y = flux density in yoke =. to.5 T / So hy K l OUTER DIA OF MACHINE (D O ): 4 y i repared by: Nafees Ahmed

15 D D h h o s y ESTIMATION OF STATOR RESISTANCE, STATOR CUR LOSS AND WEIGHT OF COER: Length of mean turns Lmt \ L D Where ole pitch DC Resistance of stator winding per phase at 75 C L Coil Span < ole pitch R., h dc, 75 C AC Resistance of stator winding per phase at 75 C 6 L F mt c N h One Turns.5 to. Rh, dc, 75 C R R h ac C h,, 75 Copper loss in the stator winding I R h h Weight of cu used in stator winding =89*(*L mt *N h *F C ) ESTIMATION OF STATOR IRON LOSS: (Same as for Induction Motor) We find out flux density in the stator tooth at Slot pitch at sg ht rd of tooth height from narrow end D hs S rd rd of tooth height from narrow end Width of the tooth at of tooth height from narrow end b b t ht sg ht s rd Area of one stator tooth at of tooth height from narrow end b Kil (Where l i = k i l=actual iron length) t ht Area of all the stator teeth under one pole A t h t Area of one tooth No of teeth S per pole 5 repared by: Nafees Ahmed

16 b t ht S Kil D h S s S bs Kil So mean flux density in teeth t ht A t ht Corresponding to flux density in tooth graph given on page 9, fig 8. find out iron loss per Kg from the So t h t t ht p it W / Kg Iron loss in teeth = p it * density * volume of iron in teeth = p it * 76 * volume of iron in teeth Corresponding to flux density in yoke find out iron loss per Kg from the graph given on page 9, fig 8. So y piy W / Kg Iron loss in yoke = p iy * density * volume of iron in yoke = p iy * 76 * volume of iron in yoke Yoke volume hy K i l ( D hy ) Total Stator iron losses i = Iron loss in teeth + Iron loss in yoke y 6 repared by: Nafees Ahmed

17 ROTOR DESIGN:. Sectional area of pole core(f C ): l e =Useful flux per pole (average value of fundamental flux) l = Actual flux per pole (average value of actual flux) p p av. e total (Take % as leakage flux) total FC C Where C = ermissible value of flux density in pole core =.5 to.7 T Diameter corresponding to this area F C d 4 d 4F C r =.5L d L d Round ole Long (Rectangular) ole 7 repared by: Nafees Ahmed

18 . Sectional are of pole yoke(f Y ): total FY Y Where Y = ermissible value of flux density in pole yoke =. to. T. Shaft diameter: KW d shaft. m RM 4. Damper winding design: Sectional area of copper in damper bars per pole ac F D. ole D (As a thumb rule we provide % of stator cu in damper bars) Where D = ermissible current density = to 4 A/mm No of damper bars Damper bar slot pitch.8 time the stator slot pitch ole Arc N d.8 Stator Slot itch 5. Estimation of OCC from design data: It involves calculation of magnetic circuit (mmf) per pole pair under rated terminal voltage on NO load. Magnetic circuit for a pole pair consists of air-gaps length stator teeth height stator yoke length rotor pole core and pole shoe length rotor yoke length MMF required for two air-gaps AT ' : Effective length of air gap ' K C Where sg K C = Cartor s gap coefficient r b r 5 b We know H So AT ' ' sg ht MMF required for teeth heights AT : 8 repared by: Nafees Ahmed

19 D Slot pitch at tooth top sgt D h Slot pitch at tooth middle t sgm D ht Slot pitch at tooth bottom sgb So Slot width at tooth top btt sgt b Slot width at tooth middle btm sgm b Slot width at tooth bottom btt sgb b Flux through a slot pitch m l sg sg e So Flux density at tooth top sg tt attt b K l ( AT / m) tt i sg Flux density at tooth middle tm attm ( AT / m) b K l sg Flux density at tooth bottom tb attb ( AT / m) btbkil Note: For calculating the ampere turns per meter corresponding to flux density in stator tooth use figure 4 if t. 7 and figure 5 if t. 7 of given reference notes. Average mmf/m for the tooth is given by Simpson s rule attt 4attm attb atht 6 Hence MMF required for teeth heights AT ht atht ht MMF required for stator yoke (AT Y ): So AT Where y at y y at l y y tm i 9 repared by: Nafees Ahmed

20 D h D ht hy h ly or MMF required for pole core and pole shoe (AT CS ): at CS CS atcs So AT h Where h h CS C h CS h h C MMF required for rotor yoke (AT Y ): Y at Y So ATY aty ly Where D hcs hy ly S.No art Length of path Flux density at (AT/m) Stator Yoke l y y at y AT y AT pole-pair Stator Tooth h t at ht AT ht t h t Air Gap ' at ' AT ' 4 Rotor Tooth h t at ht AT ht t h t 5 Rotor Yoke l ry ry at ry AT ry AT pole-pair = AT = Hence MMF (AT)/ole air MMF (AT)/ole ATs AT AT ht ATY AT ' CS ATs AT AT AT AT AT AT Y ' ht Y CS Y OCC is graph between terminal voltage at no load and filed mmf per pole i.e. V ( or) Vs AT / pole or V OCC AT/ole repared by: Nafees Ahmed

21 6. Estimation of SCC: Apply otier s concept The excitation required under short circuit donation to circulate rated current consists of two parts (i) Excitation A required for leakage reactance drop IX a. (ii) Excitation (A) required to balance the effect of armature reactance Total excitation required = A A A is given by A.9K NI Where So excitation A A condition or V pd N = No of turns for all phases per pole N h N will circulate rated current under the short circuit Rated No load Voltage E M OCC SCC N Air gap line IX a A A A AT/ole Q 7. Estimation of voltage regulation: E V V. R. V % V I hasor Diagram (Lagging F) 8. Design of filed winding: The excitation voltage (V dc ) is usually to 6 volts dc. About 5% drop is taken in leads and also to provide excess excitation when required. So.85 I R () Where I f f Vdc f f 75 C Filed current F () f Cf Current density in filed winding( 4A/mm E IXa F cf = Sectional area of field conductor IXs ) repared by: Nafees Ahmed

22 d D h =d D L mtf R f,75 C = Filed resistance ole and filed Winding L 6 mtf f () L mtf = Length of mean turn for field winding N f = Total number of field turns Using equation () & (), equation () will become F N Cf.85V N Filed current f dc f F Cf.85V.. dc 6 f L mtf 6 L mtf F N Cf f Where I f AT N ole f N f = No of field turns per pole Sectional area of filed conductor I f FC f Note:. Excitation power required 5 KVA to KVA: - 5% of the machine rating 5 MVA to 5 MVA:.5% to.5% of the machine rating. Losses and efficiency: a. Efficiency of > MVA synchronous machine: up to 98.5% repared by: Nafees Ahmed

23 b. Efficiency of >5 MVA synchronous machine: 9% or up to 95% 9. SCR: Short Circuit Ratio: V (p.u.) I (p.u) C OCC SCC D O I fo A The short circuit ration (SCR) of a synchronous machine is defined as the ratio of filed current required to produce rated voltage on open circuit to filed current required to circulate rated current at short circuit E I fsc I SCR I fo fsc OA OE Consider similar triangles OA and OED OA A A SCR OE ED AC AC A SCR pu voltage on open circuit corresponding pu current on short circuit SCR X S Thus the SCR is the reciprocal of synchronous reactance. Range of SCR SCR=.5 to.9 for turbo alternator SCR=.9 to. for water wheel alternator Effect of SCR on performance of alternator:. Stability: ower developed EV d, cylindrical sin X d, salient s EV V X d X q sin sin X X X s A machine which can develop more power is more stable. So lower the value of X s (or X d ), higher will the stability SCR. arallel Operation: Synchronizing power X s So for higher synchronizing power X s should be small SCR d q repared by: Nafees Ahmed

24 . Short Circuit Current: V I sc X s So to limit short circuit current X s should be high SCR 4. Voltage Regulation: E V IR a jx S E V VR V For VR to be low, X s should be high SCR 5. Self Excitation: Lower the value of Xs, higher will be the current in the filed path due to residual voltage. SCR Hence We see that if SCR is high the machine will have higher stability, better parallel operation, good voltage regulation and more suitable for self excitation. The value of X s is controlled by the length of air gap. y increasing the air gap length we can reduce armature reaction and so decrease synchronous reactance and increase SCR. If we increase the air gap, more filed MMF is required and it will result in more cost for the field system. For salient pole rotor X d = X s X q = (.55 to.65) X d 4 repared by: Nafees Ahmed