ECE 35 Electric Energy Sytem Component 6- Three-Phae Induction Motor Intructor: Kai Sun Fall 015 1
Content (Material are from Chapter 13-15) Component and baic principle Selection and application Equivalent circuit
Undertanding the Nameplate 3-phae, 60Hz, AC induction motor Rated power: 30.0 hp (=.4kW) Service factor: 1.15 (i.e. 115% of rated power for hort-term ue) Rated full-load current: 35.0A Rated voltage: 460V Rated peed: 1765rpm Continuou duty at ambient 40 o C Full-load nominal efficiency: 93.0% Frame ize: haft height of 8/4=7 inche; body length of 6 Inulation cla: F (155 o C for 0,000 hour lifetime) NEMA deign: B (normal tarting torque combined with a low current) Start inruh kva: G (5.60-6.9kVA) http://www.pdhonline.org/coure/e156/e156content.pdf 3
Principle component Stator (tacked lamination) Rotor (tacked lamination) Wound rotor Squirrel-cage rotor Air gap (0.4mm-4mm) 4
Stator Hollow, cylindrical core made up of tacked lamination. Stator winding i placed in the evenly paced lot punched out of the internal circumference of the lamination. 5
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Wound Rotor Ha a 3-phae winding (uually in Y-connection) uniformly ditributed in rotor lot, imilar to the tator winding Rotor winding terminal are connected to three lip-ring which revolve with the rotor. 3 tationary bruhe connect the rotor winding to external reitor during the tart-up period and are hort-circuited during normal operation. 7
Squirrel-Cage Rotor More adopted in induction motor Intead of a winding, copper bar are puhed into the lot in the lamination Welded to two copper end-ring, all bar are hort-circuited at two end, o a to reemble a quirrel cage In mall and medium motor, the cage (bar and end-ring) are made of aluminum. 8
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Principle of Operation: Faraday Law and Lorentz Force A magnet moving above a conducting ladder induce a voltage E=Blv and a current I in the conductor underneath according to Faraday law (the right-hand rule). Cut by the moving flux, the current-carrying conductor experience a Lorentz force (the left-hand rule) alway acting in a direction to drag it along the moving magnet When the ytem reache a teady tate, the ladder move in the ame direction a the magnet but ha a lower peed <v (why?) Roll up the ladder into a cylindrical quirrel-cage rotor and replace the magnet by a tator an induction motor Parallel conductor are hort-circuited by end-bar A and B Conductor are hort-circuited by end-ring A and B) 10
Rotating magnetic field Conider a tator having 6 alient pole with Y-connected tator winding carrying balanced 3-phae alternating current. i ( t) 10co 10co( t) a i ( t) 10co( 10 ) 10co( t 10 ) b i ( t) 10co( 10 ) 10co( t 10 ) c Poitive current: alway flowing into terminal A, B and C 11
Rotating Magnetic Field: Intant 1 =0: 6 pole together produce a magnetic field having eentially one broad N pole and one broad S pole 1
Rotating Magnetic Field: Intant 60 o : the magnetic field move CW by 60 o ; it angular peed equal 13
Rotating Magnetic Field: Intant 3 10 o : the magnetic field move CW by 10 o ; it angular peed equal 14
Rotating Magnetic Field For the tator producing a magnetic field having one pair of N-S pole The field rotate 360 o during one cycle of the tator current. The peed of the rotating field i necearily ynchronized with the frequency f of the ource, o it i called ynchronou peed =60f (r/min) With the phae equence A-B-C, the magnetic field rotate CW. If we interchange any two of the three line connected to the tator, the new phae equence will be A-C-B and the magnetic field will rotate CCW at the ynchronou peed 15
Phae group In practice, intead of uing a ingle coil per pole, the coil i ub-divided into or more coil lodged and taggered in adjacent lot connected in erie, i.e. a phae group Each phae group produce one N/S pole, o uing more phae group allow u to increae the number of pole (denoted by p) No. of group = No. of phae No. of pole = 3 p 16
Number of N-S pole and ynchronou peed Synchronou peed n =10f / p [r/min] f : frequency of the ource [Hz] p: number of pole If f=60hz, n =10f / 4 =30f =1800r/min n =10f / 8 =15f=900r/min 17
Starting of an induction motor 1. The magnetic field produced by the tator current rotate CW at ynchronou peed n.. Relative to the magnetic field, the rotor rotate CCW. 3. According to Faraday Law (right-hand rule), the rotor winding (bar) have induced voltage in the direction a indicated. 4. Large circulating current are created in rotor winding (bar) by the induced voltage 5. The current have the maximum value at the taring point becaue the rotor ha the maximum peed relative to the magnetic field. 6. The rotor winding rotate CW ubjected to Lorentz force (left-hand rule) in the direction a indicated 7. The rotor will accelerate until the total Lorentz force equal the friction 8. Once the rotor tart rotating CW, it peed relative to the filed will decreae, o it winding current and Lorentz force will decreae Can the rotor reach n? If it did, then it current and Lorentz force would be zero and friction would low the rotor down, o the anwer i: No, the rotor peed n<n at a teady tate. Since friction i very mall, the rotor peed n n at no load condition 18
Slip Slip: i the difference between ynchronou peed n and rotor peed n, expreed a a per unit or percent of ynchronou peed. n n r f (pu) n f f: frequency of the ource connected to the tator [Hz] f : frequency of the voltage and current induced in the rotor [Hz] n =10f / p: ynchronou peed [r/min] n: rotor peed [r/min] r =n/30, =n /30 : correponding angular peed of n and n [rad/] 19
Example 13-3 A 0.5 hp, 6-pole induction motor i excited by a 3-phae, 60 Hz ource. Calculate the lip, and the frequency of the rotor current under the following condition: a. At tandtill b. Motor turning at 500r/min in the ame direction a the revolving field c. Motor turning at 500r/min in the oppoite direction to the revolving field d. Motor turning at 000r/min in the ame direction a the revolving field Solution: n =10f/p=1060/6 = 100 r/min a. = (n -n)/n =(100-0)/100 =1 pu f =f=60hz b. = (100-500)/100 = 0.583 pu f =f=0.583 60Hz = 35 Hz c. = [100- (-500)]/100 =1.417 pu f =f=1.417 60Hz = 85 Hz d. = [100-000)]/100 = -0.667 pu f =f=-0.667 60Hz = -40 Hz Slip =1 0<<1 >1 <0 Rotor peed & freq n=0 f =f 0<n<n 0<f <f n<0 f >f n>n f <0 Operating mode Tranformer (tandtill) Motor Brake Generator (revered phae equence A-C-B) 0
Equivalent circuit: wound-rotor induction motor 1. When the motor i at tandtill (i.e. n=0, =1), it act exactly like a conventional tranformer. Aume a Y-connection for both the tator and rotor, and a turn ratio of 1:1. Conider the per-phae equivalent circuit: E g : ource voltage, line to neutral, x 1, r 1 : tator leakage reactance and winding reitance x, r : rotor leakage reactance and winding reitance X m, R m : magnetizing reactance and reitance modeling loe of iron, R x : windage and friction external reitance, connecting one lip-ring to the neutral of the rotor Note: I o may reach 40% of I p due to the air gap between the tator and rotor (much bigger than the air gap between the P and S winding of a tranformer), o we cannot eliminate the magnetizing branch. 1
For motor maller than hp R =r +R X R X For motor exceeding hp R =r +R X R X
. When the motor run at a lip, i.e. n=(1-)n R =r +R X R X 1 : 1 f f E = E 1, I = I 1, jx jx r and R X are the ame I E E 0 = E arctan x R j x R j x R ( x ) R 1 1 Note: the phaor on the primary ide (E 1 and I 1 ) and the econdary ide (E and I ) cannot be drawn in one phaor diagram becaue they have different frequencie 3
Phaor diagram on the rotor and tator I 1 and I have the ame effective value even though they have different frequencie I 1 = I E 1 and E have the ame effective value E 1 = E E 1 lead I 1 and E lead I both by 4
Simplified equivalent circuit: referred to the tator ide I I 1 R E1 jx I 1 E E R / jx Z 1 1 def Z R / jx E I 1 1 jx =jx 1 +jx E 1 Note: The value of R / will vary from R to a the motor goe from tart-up (=1) to n (=0) 5
I p r 1 jx 1 jx R E g jx m I o R m I 1 jx R 1 R I m I f Total power aborbed by the motor: Eg Eg R S j I1 ( r1 jx ) R X m Eg P I r I R m Stator iron loe P E R Stator copper loe P I r j 1 1 m f g / m R 1 1 1 Active power upplied to the rotor: P I R / r 1 Total I R loe in the rotor circuit P I R P jr 1 r Mechanical power developed by the motor 1 Pm Pr Pjr Pr (1 ) I1 R Torque developed by the motor 9.55Pm 9.55 Pr(1 ) 9.55Pr T n n (1 ) n Note: The torque only depend on P r 6
Active power flow I p r 1 jx 1 jx R P I r j 1 1 P E R f g / m E g I m jx m I o R m I f I 1 1 R P I R jr 1 1 Pm I R P / P 1 L e 7
Phaor diagram of the induction motor S=P+jQ I p r 1 jx 1 jx R I o I 1 jx E g jx m R m R 1 R I m I f Z 1 =r 1 +jx =acco(p/ S ) =actan(x/r 1 ) 8
Breakdown (maximum) Torque Eg I1 R co E g IZ 1 1 co Eg I1 b Z1 co R b Z 1 When Z 1 = R /, P r and torque T both reach their maximum value E g T b 9.55Pr 9.55 I1 R / n n 9.55 E 4 n Z1 co g Z 1 I 1 I 1 R / I 1 9
Example 13-5 A 3-phae induction motor having a ynchronou peed of 100 r/min draw 80 kw from a 3-phae feeder. The copper loe and iron loe in the tator amount to 5 kw. If the motor run at 115 r/min, calculate a. The active power tranmitted to the rotor P r =P e P j P f =80 5 =75 kw b. The rotor I R loe, i.e. P jr =(n -n)/n =(100-115)/100=0.04 P jr =P r =0.0475=3 kw c. The mechanical power developed P m =P r P jr =75 3 =7 kw d. The mechanical power delivered to the load, knowing that the windage and friction loe equal to kw P L =P m - P V =7 =70 kw e. The efficiency of the motor =P L /P e =70/80=87.5% f. The torque developed by the motor T m =9.55 P r /n = 9.5575000/100=597 Nm 30
Two-wattmeter method to meaure 3-phae active power 3-phae load in Y or connection P 1 = V ac I a co( - 30 o )=E L I L co( - 30 o ) P = V bc I b co( + 30 o )=E L I L co( + 30 o ) S 3 P3 jq3 P1 P j 3( P1 P ) Proof: S3 ELI L co( 30 ) co( 30 ) j 3ELI L co( 30 ) co( 30 ) E I co co30 j 3E I in in 30 L L L L 3E L I L c o j 3E L I L in 31
Example 13-7 a. Power upplied to the rotor P e =P 1 +P =70 kw b. Rotor I R loe P jr r 1 =0.34/=0.17 P j =3I 1 r 1 =378 0.17=3.1 kw P r =P e P j P f =70 3.1 =64.9 kw =(n -n)/n =(1800-1763)/1800=0.005 P jr =P r =0.00564.9=1.33 kw Two-wattmeter method 1763 r/min A 3-phae induction motor having a nominal rating of 100 hp (~75 kw) and a ynchronou peed of 1800 r/min i connected to a 3-phae 600 V ource. Reitance between two tator terminal =0.34. Calculate c. Mechanical power upplied to the load P m =P r P jr =64.9 1.33 =63.5 kw P L =P m - P V = 63.5 1. =6.3 kw =6.3 1.34 = 83.5 hp d. Efficiency =P L /P e =6.3/70=89% e. Torque developed at 1763 r/m T m =9.55 P r /n = 9.55649000/1800 =344 Nm 3
Two practical quirrel-cage induction motor 33
Torque-Speed Curve n n R R, Pr I1, I1 Eg / ( Z1 ) n 9.55Pr T n 9.55 E Rn g R Z1 n n n n ( ) Stable operating Region (n, T) no-load torque The curve i nearly linear between no-load and full-load becaue i mall and R / i big 9.55 Eg 9.55 Eg I1 Eg / R T ( n ) n R n R n 34
The 5 hp induction motor Torque-Speed Curve E g =440/1.73 V R =1. Z 1 =1.5+6j n =1800 r/min T 9.55 Eg Rn Z1 ( n n) n n R Torque-Slip Curve Current-Speed Curve T 9.55 E 1 g R Z R / n I 1 Z 1 Eg Rn n n 35
Torque-Speed Curve: 5 hp motor I 1 P r T n When the motor i talled, i.e. locked-rotor condition, the current i 5-6 time the full-load current, making I R loe 5-36 time higher than normal, o the rotor mut never remain locked for more than a few econd Small motor (15 hp and le) develop their breakdown torque at about 80% of n 36
Torque-Speed Curve: 5000 hp motor Big motor (1500+ hp) : Relatively low tarting (locked-rotor) torque Breakdown torque at about 98% of n Rated n i cloe to n 37
Effect of rotor reitance R Rated Torque =15Nm Current at the rated torque.5r Rated torque =15Nm Current at the rated torque 10R Rated torque =15Nm Current at the rated torque 38
T LR Effect of rotor reitance 9.55 E R 9.55 E Z g g 1 R n Z1 n R I 1, LR Z 1 E g R P jr I 1 R T b 9.55 E 4 n Z1 co g When R increae, tarting (locked-rotor) torque T LR increae, tarting current I 1,LR decreae, and breakdown torque T b remain the ame Pro & Con with a high rotor reitance R It produce a high tarting torque T LR and a relatively low tarting current I 1,LR However, at the rated torque, it produce a rapid fall-off in peed with increaing load becaue the torque-peed curve become flat, and the motor ha high copper loe and low efficiency and tend to overheat Solution For a quirrel-cage induction motor, deign the rotor bar in a pecial way o that the rotor reitance R i high at tarting and low under normal operation If the rotor reitance need to be varied over a wide range, a wound-rotor motor need to be ued. 39
Effect of the terminal voltage: Problem 13-33 A 3-phae, 300kW, 300V 60Hz, 1780 r/min induction motor i ued to drive a compreor. The motor ha a full-load efficiency of 9% and power factor of 86%. If the terminal voltage rie by 0% while the motor operate at full load, how doe each of the following change? P m, T n FL Same I 1,LR T b 0% 44% 9.55 E T ( n n ) g FL FL Rn I 1,FL PF (Q) Average motor temperature Flux per pole About the ame (iron winding ) 0% T Q= I 1 x+ E g /X m LR 9.55 Eg Z n 1 R T LR About the ame (P f P j P jr ) 44% Exciting current Iron loe >0% (conidering aturation) T b I 1, LR 9.55 E 4 n Z1 co Z E g g R 1 40
Aynchronou generator Connect the 5 hp, 1800 r/min, 60Hz motor to a 440 V, 3-phae line and drive it at a peed of 1845 r/min =(n -n)/n =(1800-1845)/1800= -0.05 R /=1./(-0.05)=-48 <0 The negative reitance indicate that power i flowing from the rotor to the tator (aynchronou generator mode) Power flow from the rotor to the tator: E = 440/1.73=54 V I 1 = E / -48+1.5+ j6 =54/46.88= 5.4A P r = I 1 R /= -1410 W <0 1410W flow from the rotor to the tator Mechanical power and torque input to it haft: P jr = I 1 R =35. W P m =P r +P jr =1410 + 35.=1445W T=9.55 P m /n=.3 Nm Total active power delivered to the line: P j = I 1 r 1 =44.1 W P f = E /R m =71.1 W P e =P r -P j -P f = 1410-44.1-71.7=194 W P 3 =3P e =388W Reactive power aborbed from the line: Q 3 =( I 1 x+ E /X m ) 3 =(176+586) 3 =86 var Complex power delivered to the line: S 3 =P 3 - Q 3 =388 - j86 VA co=86.% Efficiency of thi aynchronou generator =P e /P m =194/1445=89.5% 41
Tet to determine the equivalent circuit I p r 1 jx 1 jx r / R / I o I 1 jx E g jx m R m R X / I m I f Etimate r 1, r, X m, R m and x (note: r +R X =R where R X i the external reitance) 1. No-load tet. Locked-rotor tet 4
No-load tet At no-load, lip 0 R / i high, I 1 <<I o Step: 1. Meaure tator reitance R LL between any two terminal (auming a Y connection) r 1 =R LL /. Run the motor at no-load uing rated line-to-line voltage E NL. Meaure no-load current I NL and 3-phae active power P NL S 3E I NL NL NL Q S P NL NL NL R m P P 3I r P P 3I r NL f NL 1 f NL NL 1 ( E / 3) E ( P P) / 3 P 3I r NL NL f NL NL 1 X m ( ENL / 3) ( Q ) / 3 NL (Ignoring P V ) E Q NL 43 NL
Locked-rotor tet When the rotor i locked, lip =1, I P >>I o r =r /R /, neglect the magnetizing branch Step: 1. Apply reduced 3-phae voltage E LR to the tator o that the tator current I p the rated value. Meaure line-to-line voltage E LR, current I LR and 3-phae active power P LR S 3E I LR LR LR Q S P LR LR LR x Q 3I LR LR P 3I r 3 I r r P / (3 I ) r LR LR 1 LR LR LR 1 44
Example 15-1 45
Homework Aignment #5 Read Chapter 13.0-13.16, 15.0-15.9 Quetion: 13-16, 13-3, 13-4, 13-8, 15-, 15-3, 15-4, 15-5 Due date: hand in your olution to Deni at MK 05 or by email (doipov@vol.utk.edu) before the end of 11/4 (Wed) 46