Electrical Drive 4 th Class

Transcription

1 University Of Technology Electricl nd Electronics Deprtment Dr Nofl ohmmed Ther Al Kyt

2 A drive consist of three min prts : prime mover; energy trnsmitting device nd ctul pprtus (lod), hich perform the desired job The function of first to prts is to import motion nd operte the third one In electricl drives, the prime mover is n electric motor or n electromgnet A block digrm of n electricl drive my be s shon: Electricl otor Energy Trnsmitting Device Apprtus (lod) Control Equipments The electromgnetic forces or torque developed by the driving motor tend to propgte motion of the drive system This motion my be uniform trnsltionl or rottionl motion or non-uniform s in cse of strting, breking or chnging lod In uniform motion the motor torque must be in direction opposite to tht of the lod nd equl to it, tken on the sme shft (motor or lod shft) the stedy-stte condition hs uniform motion T m =, T m = 0 W m = Const Electricl otor W m T m For uniform motion T m, therefore the system ill either in ccelertion if T m > or declrtion hen T m < nd dynmic eqution relting them is given by: 1

3 T T = J, here J = moment of inerti of ll rotting prt m L d dt trnsformed to the motor shft Types of the lod: there re to min lods type: Active lod: hich provides ctive torque (grvittionl), deformtion in elstic bodies, springs forces, compressed ir, ) these lod my cuse motion of the system Pssive lods : re tht lods hich hve torque ll the times opposing the motion such s Frictionl lod nd Shering lods Also, lods my be subdivided into: Constnt lods hich re unchnged ith time or ith vrition of speed Liner vrying lods = + b here, b re constnt Second vrying lod (order) = + b + c here, b, c re constnt nd is the speed The frictionl lod my be considered constnt, hich pump lod consider liner nd compressor lods my consider second order Electricl drive my operte in one of three min modes op opertion: 1 Continuous mode: by hich the motor is strted nd operte for enough time to rech stedy-stte temperture then hen stopped, it must be left for enough to rech the initil temperture (room temperture) Temperture (θ) Stedy-stte θ θ θ z ON-Time OFF-Time (t) Time In such opertion nd to void therml stresses: rted torque of the motor T s > for the norml motor opertion T s = strting torque

4 Interruptive opertion: in such opertion the motor opertes nd stops for pproximtely equl intervls by hich t ny intervl, the motor ill not rech stedy-stte temperture hen strted nd not return to initil temperture hen stopped In such opertion nd ccording to the ON nd OFF intervls < T rted of the motor (depending on the number of ON nd OFF intervls) Temp (θ) Θ x limit θ 3 θ θ 1 ON OFF ON OFF ON Time (t) Interruptive Opertion For ech intervl T s > to insure strting The number of intervls is limited by the temperture reched hich must not exceed the designed temperture limit θ mx such opertion of drives mybe found in control system 3 Short time opertion: in such opertion the motor is strted nd operte for smll intervl of time, then sitch-off for very long period Temp (θ) θ ON OFF Time (t) Short Time Opertion 3

5 θ my rech θ x but the long OFF time ill be sufficient to return the temperture to its initil vlue t ON << t OFF The motor my be overloded during the ON time so the temperture θ my rech θ x t smll times, but the long OFF period ill return it bck to its initil vlue In such drive > T rted motor so multi-successive ON (strting) my dmged the motor Such derive mybe found in crs strting system For qudrnts opertion of drives: By the effect of ctive nd pssive lods, nd ccording to the reltionship nd directions of, T m nd the electricl drive my operte in ny qudrnt of 4-qudrnts digrm hose xis is ± T nd ± + T gb T m Genertor Breking otor II I -T +T T m T b III Referencing Breking I - The first qudrnt (I) represent motor opertion by hich the motor torque nd speed re in the sme direction (+ve) nd opposing the lod torque The speed of the motor nd it's torque is positive (counter clockise direction) The speed my be chnged from 0 to o here o = no-lod 4

6 ngulr speed nd the motor torque chnge from 0 t no-lod to T s t =0 (strting torque) For norml motor opertion the folloing reltion must be stisfied: Tr TL 0 For norml therml operting condition 0 < L o According to the selective method of speed vrition T > For ll motor opertion to insure strting s T r = rted torque of the motor (N) = lod torque of the motor shft (N) T s = strting torque of the motor (N) The norml stedy-stte opertion is hen T m = hich mke the dynmic d torque (T m - )=0 hich mens = 0 or = constnt = operting speed dt d Tm TL = J dt If for ny reson the motor speed exceed o, then the motor bck emf (E) ill E be genertor then the supply voltge () nd I = ill be negtive (return to the supply) nd produce torque opposing the speed (brek) I I F E φ W T Therefore moving the second qudrnt ill chnge the opertion from motor to genertor breking At genertor breking is positive but grter thn o 5

7 nd the motor torque oppose the motion (the lod torque in the sme direction s the motion) At no-lod speed o, E=, I = 0, T m = 0 (Y-xis) the 3 rd qudrnt represent lso motor opertion but in the reverse direction ith respect to tht in the first-qudrnt (ie rottion in the clockise direction nd T m lso in the clockise direction nd opposing ) The 4 th qudrnt represent brek opertion by hich the motor still tken current from the supply nd producing torque (+ve) but opposing the motion The motion is strted by the effect of the lod hich is grter thn T s (ctive) nd hence the motor move opposite to it's electricl direction cusing E to chnge it's direction E + E I = I = since E = -ve And hence T m increse s increse until t certin : T m = nd the lod ill brek t speed = o > > 0 Torque-speed chrcteristics (mechnicl chrcteristics) nd speed vrition methods: In mny pplictions the rottionl speed of the motor mybe much higher thn the required speed or vise verse, s ell s, in other pplictions multi-speeds re required To chnge the lod speed there re mechnicl nd electricl methods s ell s their combintion In mechnicl method, the use of pullies, ger my mke the lod speed equl to the required speed from given motor rottionl speed N 1 N r r 1 otor T m, m A B L Lod, L 6

8 A simple to gers my hve rdius r 1 nd r nd number of tooth N 1 nd N correspondingly my give the folloing reltionship: Since teeth pitch is the sme for both gers: hen ger A moves one revolution, (πr 1 ), ger B must move the sme length (πr 1 ) ccording to the rdius ie 1 πr1 = i(πr ) here lod speed L i = = motor speed m r1 N1 i = = r N, L = i m i > 1 if r 1 > r or N 1 > N If the gers hve the efficiency of 100%: P in = P out or T m m = L m Tm TL = Tm = L i And if the efficiency 100% (losses) 1 > ι > 0: Pout LTL P in = or mtm = ι ι m 1 ιtm TL = ι Tm = ι Tm = L i i If there re multi-stges of gers ith Ger rtio = i 1, i, i 3, L = ( i1 i i 3 ) m Tm And TL = i1 i i 3 for idel cse ( ι1 ι ι 3 ) Tm And TL = for ctul cse hen ι n represent the i1 i i 3 efficiency of stge n When the lod is trnsferred to the motor shft, nd hence it's speed ill be m the lod torque t the motor shft ill be: P L = L TL = m TL L = TL = i TL for idel cse otor m T \ i T = L for ctul cse ι m T m The equivlent system ill be s shon: 7

9 stedy-stte opertion hen: i TL T m = TL = ι Electricl ethods: there re mny electricl methods hich re used to mke the lod operte t the required speed by chnging m To select ny method the folloing points re tken into considertion: 1 The direction of speed vrition: UP; DOWN; UP nd DOWN ith respect to operting speed t nturl chrcteristics The dynmic rnge of speed vrition, ie the possible rtio of mximum to minimum speed chieved m D = dynmic rnge of speed = min min is tken to be t lest 10% of rted speed 3 The stiffness coefficient of mechnicl chrcteristics (β ) hich is equl to: dt β = This fctor represents the sensitivity of the motor to the lod d vrition Idel motor must keep constnt speed hen it's lod chnges, ie β = nd s this fctor decrese it mens the motor ill be much sensitive to lod vrition 4 The vlue of the lod torque 5 The efficiency, simplicity, cost, Different motors hve different electricl method of speed vrition, therefore dc nd c motors ill tken ech lone ith their possible speed vrition methods DC otors Prmetric ethods of Speed rition: A) Seprte excited or shunt motor The min eqution for seprte excited dc motor re: E I = Amp (1) E = k ϕ olt () T = k ϕ N (3) I I E T m m φ I F 8

10 ZP k = const = π Where: Z = Conductors / Armture P = No of poles = No of prllel pths = for ve nd = P for lp The mechnicl chrcteristics (=f(t)) cn be found from equtions 1, nd 3 s: W = T (4) ( ) For seprte or shunt motor interpdes nd compensting indings φ cn be considered = constnt independent of lod for nturl chrcteristics φ = constnt = φ nom The mechnicl chrcteristics ill be stright line since, k, φ, re constnts If = nom, φ = φ nom nd no dditionl elements re inserted the mechnicl chrcteristics is nturl chrcteristics At no-lod (T = I = 0) = = no-lod speed W o ; E = k ϕ At = 0, T = T so = strting condition Tso = k ϕ N ; E = 0 R W rd/sec A L r W o C B Nturl Chrcteristics T r T so T N If the motor is operte directly ith the lod t its nturl chrcteristics Tr TL 0, 0 L r for ny operting lod torque, the operting point ill ly on the mechnicl chrcteristics beteen points A nd B, for exmple t, L (point C): dt ( ) The stiffness coefficient β = = d R 9

11 -1- Adding resistnce to the rmture: R d I F I nom Φ nom m T m The mechnicl chrcteristics ill be represented by: nom + R d Wi = T hich give different speed i k ϕnom ( k ϕnom ) At different dded R d for give lod torque T nom At T = 0, i = = o = constnt independent of R d nom W rd/sec o Lo L1 L L3 Nturl Chrcteristics R d3 R d R d1 T S3 T S T S1 T so T N At = 0, T T si s R d si = k ϕ nom ( k φ ) R nom + R dt nom β = = s R d d + R d If 0 < R d1 < R d < R d3 d 10

12 T so > T s1 > T s > T s3 Lo > L1 > L > L3 β o > β 1 > β < β 3 The method cn decrese the speed only don Lo nd the stiffness coefficient ill be ore s R d increse The method is simple ith lo efficiency (high losses) due to R d ( I R d ), therefore it cn be used only for smll motors Any operting condition cn be chieved by certin R d hich mke the developed chrcteristics pss through tht points For exmple, it is required to operte lod of t speed L Given nturl chrcteristics such motor cn drive such lod W rd/sec o L R d? R d = 0 T S T N If T m rted Pr Pr (tt) Tr = = N πn r r 60 If the speed is given in RP, the nturl chrcteristics is not suitble if: nom = T L is grter thn L required k ϕnom ( k ϕnom ) To find R d such tht the lod operte t L the chrcteristics must hve R d so tht it pss through the point (, L ) R d cn be found if the point in the 1 st qudrnt (motor opertion) either by using the eqution: nom + R d L = T L k ϕnom ( k ϕnom ) R d cn be fount if the other prmeters re knon T so Or by similrity: o o L = T T s L otl Ts =, o L 11

13 nom Q Ts = k ϕ nom R d is found R + R d + o -T R R d =0 d R d1 -T so -T s1 -T s - o T s T s1 T so +T - For the motor to operte in the four qudrnt, ech developed chrcteristics cn be extended to the other qudrnt for brek opertion L ill be negtive ill is +ve For genertor breking L is positive > o but is negtive For reverse opertion: R d L = ( TL ) k ϕ k ϕ ( ) -- Chnging the rmture supply voltge: if the rmture supply voltge cn vry nom > 0 ill give set of mechnicl chrcteristics given by: = T I L k ϕ I k ϕ F nom ( ) nom At ny selected voltge, the chrcteristics is stright line prllel to ech other given by: Φ nom L 1

14 i i oi =, Tsi = nom k ϕnom As i, oi nd T si dt ( ) β = = = const ( The mechnicl chrcteristics is prllel to d ech other) -T o L + o L1 o1 L 1 = nom T s T s1 T so +T - nom > 1 > o > o1 > o T so > T s1 > T s At constnt lod : L > L1 > L The speed vrition only don By the use of poer electronic, the rmture supply voltge cn be chnger The efficiency of this method is better thn tht ith R d nd the stiffness coefficient remin constnt 13

15 -3- Chnging the field flux: By dding resistnce to the field circuit or by inserting series inding, the flux of the motor cn be chnged over nd under φ nom of single inding o1 o o + φ 1 < φ nom φ = φ nom -T T s1 T so φ > φ nom T s +T i = As φ i, nom k ϕ oi i For φ 1 < φ nom o1 > o T s1 > T so R ( k ϕ ) i i T L nom = nd k ϕ nom T si = R - And for φ > φ nom o < o T s > T so According to the vlue of incresing or decresing of the flux my give different results of speed (uncertinty), for exmple decresing the flux from φ 1 to φ ill give incresing of speed if the lod torque is T 1 from 1 to If the lod torque is T the chnge from the flux from φ 1 to φ ill not chnge the speed 14

16 At lod torque T 3 the chnging from the flux from φ 1 to φ ill decrese the speed from 3 to 4 s shon in the digrm φ φ 1 T 1 T T 3 T ( k ϕ ) The method ffect the stiffness coefficient β = s φ i but the method is more efficient thn tht ith R d in rmture Exmple: A seprte excited, 3KW, 150 olt, 1600 RP, dc motor hs n rmture resistnce = 1 Ω is to be used in drive through reducer of N m = 10 If the lod torque is 150 N find the required R d dded to the N L rmture to operte the lod t 1 rd/sec? Wht voltge reduction is used for such opertion if no R d is dded? Solution: π nr π 1600 r = = = rd/sec T \ L = = 15 N 10 \ L = 1 10 = 10 rd/sec T 3 mr = r Pr 3 10 = = 179 N (such motor cn drive the lod 15 N) i 15

17 For nturl chrcteristics: r = T r k ϕ ( k ϕ) = 179 k ϕ k ϕ ( ) ( ) 150( ) = ± ± 104 = = Either k ϕ = = or k ϕ = = = o, either o = = rd/sec or 150 o = = 1056 rd/sec 014 By neglect very fr from r k ϕ = 0753 To find R d required 1 + R 10 = 199 d 15 ( 0753) ( 1 + R d ) 645 = = R d = = 7153 R d = 6 153Ω For R d = 0 10 = ( 0753) 1 33 = = = = 74 olt 133 The reduction of supply voltge = = 16 olt 16

18 B Series otor: For series motor the vlue of R F is dded to rmture totl resistnce nd the chnge of lod (torque) ill chnge lso φ φ φ is no longer constnt but function of the rmture current for exmple I R F = AI + BI DI + C L Where A, B, C, D re constnt hose vlues gives liner reltionship t smll I nd constnt kφ t very high I (sturtion) (I = I F if no diverter only) At no-lod idelly φ = 0 nd o =, therefore no-lod or high lod opertion of series motor is not lloed since the resultnt speed is dngerously high For the sme reson movement from 1 st qudrnt (genertor brking) is not lloed lso The mechnicl chrcteristics is given by the sme eqution = T ith φ chnging ith T ( ) Adding the resistnce to the rmture or chnging the supply voltge ill chnge the φ lso, so the chrcteristics is no-longer stright line T 17

19 Exmple: A series 100 olt, KW, 800 RP motor hs = 05 Ω nd R F = 03 Ω The field is linerly proportionl to rmture current Find the required dditionl rmture resistnce to drive lod of 0 N t speed of 50 RP Solution: Let kφ = AI here A = Constnt 3 3 Pr Tr = = = = 387 N π 800 r T = I = AI + R F r = T r ( ) 100 ( ) 838 = ( ) ( ) 100( ) = ± ± 5998 = = kφ = 0954 or kφ = 038 (Neglect since T s T r ) kφ = 0954 = A I Tr = AI = 387 Tr = I = = 5 Amp, nd A = = To find R d, = 0 = A I =0038 I 0 I = = 563 I = 94 Amp 0038 = AI = = π L = = 618 rd/sec 60 + R F + R d L = T L ( ) 100 ( R ) 618 = d ( ) 18

20 618 = 1149 ( 08 + R d ) 6 4 ( 08 + R d ) 64 = = R d = = R d = = 56Ω Electricl Breking of DC otors: In order to stop the motor quickly nd to bsorb the kinetic energy in the rotting prts the motor ill produce torque hich oppose the motion (breking) by certin process nd connection clled electricl breking of dc motor When the supply is sitched OFF, the drive ill come to rest under the effect of frictionl force (torque) only hich my tke long time especilly for lrge inerti motors To increse tht time s it is required in control system electricl breking method re used Seprte excited motor electricl breking: Seprte excited motor cn be breking by one of the folloing methods: 1 Genertor breking Supply reverse breking 3 Dynmic breking 1 Genertor Breking: When the speed of seprte excited motor exceed it's no-lod speed, the motor ill exert torque oppose the motion nd the direction of the rmture current is reversed The current ill move from the mchine to the supply (the mchine ill operte s genertor) nd the produced torque is opposing the motion, hence, it clled genertor breking or regenertive breking L o o ne Operting Chrcteristics reduced = 1 T gb T s T 19

21 Suppose motor drive lod t speed L ccording to the mechnicl chrcteristics s shon in the figure If the voltge is reduced or φ is increse such tht the ne mechnicl chrcteristics hs o < L s shon Then t t = 0 hen is chnged the motor ill produced torque = T gb opposing the motion As reduced from L to ( o ne ) this torque ill chnge from T gb to zero For the operting chrcteristics o > L nd the mchine operte s motor ith E = (kφ L ) is less thn, the rmture current is E v I = = from the supply to the motor producing torque = R r kφi = k ϕ in the sme direction s tht of the speed (motor opertion) 1 When the voltge is chnged to 1, 1 < such tht o = < ne L 1 E 1 L I = = = -ve since kφ ne o ne = 1 Ie the current ill return to the supply nd produced torque ill be T gb = kφi ne = kφ(-i ne ) = -ve torque ie oppose the motion = brek Similr results hen φ increse to give o ne < L Notices tht the flux is remin constnt in vlue nd direction If the motors re shunt motor similr opertion cn be ctivted Series motor cn never be operte in such mode of opertion since its speed ill be dngerously high Supply Reversing Breking: If seprte excited dc motor operte t certin mechnicl chrcteristics t point (, L ) s shon By chnging the polrity of the rmture supply voltge the motor ill produce very lrge brek torque I t = 0 E A 1 A L I F I F φ E A 1 φ I b A 0 T b

22 B o L A Breking Chrcteristics Operting Chrcteristics -T s T b C -T s - 0 T s T D - L - o - L F Suppose point A is the operting point on the selected chrcteristics A(, L ) The rmture current ill be from the supply to the mchine (motor) nd E equl to I = E = brek emf = kφ L, > E When the supply polrity is reversed (φ kept constnt in mplitude nd direction) (in the sme direction s E hich remin the sme t t = 0) E + E I = = pssing through the rmture from the supply, but in opposite direction to tht for motor, (For motor from A 1 to A hile for supply reversing from A to A 1 ) E + The results torque = kφi = = ve R hich men opposing the motion (Brek) 1

23 + E The resultnt brek torque = T b = k ϕ t t = 0 nd s this brek ill reduce the speed quickly to zero E = kφ ill reduce nd the brek torque through breking process ill reduce too The mximum possible brek torque in such method is the cse of breking by supply reversing t no-lod ie if the motor is t no-lod, then = E, I = 0 hen is reversed then I = = = Is R R T b = I = I s = T ( -ve mens brek) s If the operting point t (0, o ) the brek torque t t = 0 = T s nd the rmture current = I s T T T depending on the position of the operting point A s b s At t = 0 the corresponding point on the brek chrcteristic is B nd the corresponding brek torque is T b As L reduced E, T b nd the point ill move on the brek chrcteristics from B to C At point C the lod speed is zero, therefore if the rmture supply is sitched OFF the system ill stop nd the time tken for stopping is very short (ith respect to the pervious method nd hich stopped due to frictionl force only) If the supply is not sitched OFF, then the motor ill hve torque = - T s k ϕ, therefore the motor ill be strt to operte in reverse direction The finl operting point such cse ill be point D(-, - L ) if the lod torque is pssive torque (frictionl), otherise the operting point ill be point (F), F(, - L ) nd L > (genertor breking) An o exmple of such cse is breking lift move up ord by supply reversing If supply voltge reversing is pplied to shunt or series motor F 1 F 1 F F A 1 F L A A F A 1 L

24 F F 1 F F 1 F A 1 F A 1 A L A L The motor ill not brked becuse the supply polrity reversing cuse the rmture current nd the field current both to chnge their direction then φ ill chnge to φ nd I ill chnge lso to I hich give: T = I k( ϕ)( I ) = I ie the shunt nd series motor ill not ffected by supply reversing Therefore, for shunt nd series motor supply reversing breking cn be done only if either the direction of rmture current or field current is kept the sme (Unchnged) This cn be done by the use of to similr field inding one ound opposite the other nd used lterntively hen the supply polrity reversing I F F A 1 A 1 F 1 F 1 F F 1 F F 1 I R A A L T b In such cse the current of the motor opertion pss through the rmture from A 1 to A nd through the upper field inding from F 1 to F cusing motor torque in sme direction s L When the supply polrity is reversed the upper field inding is disconnected nd the loer field inding is connected The rmture current direction ill be reversed (from A to A 1 ) hich mens I ith respect to the first cse but the field current in the loer field inding is still from F 1 to F hich men φ is not chnged T = I T = I b ( ) 3

25 3 Dynmic Breking: For the seprte excited motor operte ith lod t speed L t point A of the selective mechnicl chrcteristics cn be breking by sitching OFF the supply rmture voltge nd closing the rmture terminl through short circuit or through R d s shon ith field kept constnt (unchnged) For motor opertion: E I = ; E = R L I E I F φ And T L = I When = 0 nd the terminl of the rmture is short circuited: Either L 0 E E I b = = in the reverse direction R R I b I F Since φ is kept the sme (constnt) E φ T b = I b = E R As L, E nd T b OR T b For terminting the rmture through R d : 0 E E Ib = = in the opposite + R d + R d direction but less thn tht ith short circuit R d E I b I F φ E Tb = I b = R + R d As, I, T b 4

26 R d = 0 R d1 R d o A R d > R d > 0 -T s T b R d =0 T b R d1 T b R d T s T At = 0, I = 0, nd T b = 0 The mximum possible I b = t no-lod since E =, nd the brek torque is (T s ), s R d, I, nd T b If the lod is pssive lod, the finl position is the origin but if the lod is ctive lod the finl operting point ill move to the 4 th qudrnt t (,- L ) I F F OFF F I F A Rely A R d F E A I F 1 R d Rely B Contct B L TL Contct A 5

27 Operting in position (1) represented motor opertion in certin direction, hile operting in position () represented lso motor opertion in the reverse direction Chnging from position (1) to position () or vise vers through opertion ill represent supply reversing breking Chnging from position (1) to position (4) or from () to (3) ill represent dynmic breking through R d hich cn be selected s required Exmple: A seprte excited 150 olt, 3 KW, 960 RP dc motor hs totl rmture resistnce of 1 Ω is used to lift lod of 08T r t speed of 10% of it's no-lod speed Find: ) The required dditionl resistnce for such opertion? b) If the supply is reverse breking is used ht ill be the initil brek torque nd ht ill be the finl opertion? c) If the supply is sitched OFF nd the rmture is terminted through resistnce = 3 Ω ht ill be the initil brek torque nd ht ill be the finl position? Solution: πnr π 960 r = = = rd/sec Pr 3 10 L = 10% o, Tr = = = 9 84 N r = 08 T r = = 387 N From nturl chrcteristics: r = T r ( ) = = ( ) ( ) ( ) 150( ) = ± = 150 ± Either kφ = 15 Or kφ = 036 (Neglect given very high o nd T s T r ) 6

28 150 o = = = 10 rd/sec, L = 1 rd/sec 15 Supply Reversing o = 10 R d = 6Ω -T s T b B T s D T b 1 = L = 387 A T s T - o Dynmic Breking - L Supply Rev L DyB C E 7

29 To find R d : + R d L = T L ( ) R 1 = d ( 15) 10 1 = ( 1 + R d ) R d = = 7Ω 157 = 7 1 = 6Ω R d For supply reversing the initil breking is represented by point B (T b ) E Ib = = = 357 (In opposite direction to tht for t 7 motor opertion) Tb = I b = = 946N Since the lod is the potentil lod, therefore, finl operting position ill represented by point C (Generl breking for reverse opertion) t 7 L = + TL = = 693 rd/sec 15 ( ) ( ) For dynmic breking through 3 Ω: 0 E 15 1 Ib = = = 15 Amp in the opposite direction to motor t opertion T = lϕi = N (Point D) b b = The finl operting position ill lso be genertor breking (pot lod) t L = + TL = = 77 rd/sec Pint (E) 15 ( ) ( ) 8