Electric Motors M = Km⋅Φ⋅I from electrcal overview Lorentz force... Ref: Chapter 9 Km = constant_for_given_motor M = torque N⋅ m (ref: 2.93) Φ = magnetic_flux Wb = 1 weber I = current A = 1 amp (9.1) 1Wb⋅ 1A = 1 N⋅ when rotating, electromotive force induced in rotor given by .. from electrical overview Faraday's force ... E = KE⋅Φ⋅n (ref: 2.96) KE = constant_for_given_motor E = induced_electromotive_force V = 1 volt Φ = magnetic_flux Wb = 1 weber rpm = 6.283 n = rotation_speed 1 (9.2) Wb⋅ rpm = 0.105 V min model motor as resistance in series with EMF generator (opposing applied voltage) U = E + I⋅ R M = Km⋅Φ⋅I with ... I := n collect , Φ → (9.1) M KE⋅Φ − M ⋅ R n= 2 Km KE⋅ Φ (9.4) I E = KE⋅Φ⋅n and ... E := (U − I⋅ R) Km⋅Φ U 2 P = E⋅ I + I ⋅R (9.3) n := U KE⋅Φ (9.2) RA Um E E KE⋅Φ separately excited field M⋅ R − 2 KE⋅Km⋅Φ (9.5) XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX to see an example of DC motor behavior assume a set of reasonable parameters. Not all are independent. for fixed magnetic field Φ and rpm at maximum power , maximum current Im and maximum torque Mm M = Km⋅Φ⋅I set Φ, n, R and applied voltage U Φ := 1Wb derived n := 100rpm R := 2Ω U = E + I⋅ R E E⋅ Im = M m⋅n⋅ 2⋅ π M M m = Km⋅Φ⋅Im Km E = KE⋅Φ⋅n KE maximum current Um := 400V Im := 10A E := Um − Im⋅R E = 380 V M m := Km := KE := E⋅ Im M m = 57.753 N⋅ m assuming EMF*I converted into mechanical power n ⋅ 2⋅ π 2 Mm Km = 5.775 Φ⋅Im E 2 P = U⋅ I = E⋅ I + I ⋅R = M ⋅ n⋅ 2⋅ π + I ⋅R KE = 36.287 Φ⋅n XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 1 11/15/2006 M = U⋅ a := Km⋅Φ R Km⋅Φ − n⋅ KE⋅Km⋅Φ Um⋅ R b := R 2 Km KE⋅Φ R − n⋅ KE⋅Km⋅Φ 2 R = 57.753 N⋅ m 2 M ( U , n ) := ( U⋅ a − b ⋅ n) R nn( U , M ) := M 0 := 0N⋅ m Km⋅Φ M m = 57.753 N⋅ m ( calculate M when U and n known ... U⋅ a − M calculate n when U and M known useful at ends of torque range 0 - M m b ) nn Um , M m = 100 rpm derived check ( ) nn Um , 0 = 105.26316 rpm for example at U = 0.25 U m, calculate n at 0 and maximum torque ( ( ) ) n11 := nn Um⋅0.25, M m n11 = 21.053 rpm n12 := nn Um⋅0.25, M 0 n12 = 26.316 rpm maximum torque 0 torque n1 := n11 , n11 + 0.1rpm .. n12 Mm plot M vs n for U = 0.25*U m ( ) N⋅ m 50 M U m⋅0.25, n1 N⋅ m and if develop similar data for 0.5 * U m, 0.75 * Um and Um 0 obtain the following plot 0 10 20 30 n1 rpm plot data torque N*m 60 0.25 * Um 0.5 * Um 0.75 * Um 1.0 * Um Mm max torque 40 20 0 0 20 40 60 80 100 120 140 rpm 2 11/15/2006 additional operating envelope is available beyond design rpm by reducing the field strength Φ. But the region is limited by the maximum power available 3 Pmax := M m⋅ 2⋅ π⋅n beyond a M limited by Pmax Pmax = 3.8 × 10 W above base sped and torque with power consatant at P max, torque is limited inversely with rpm M max⋅nn⋅ 2⋅ π = Pmax Pmax M max_n( nn) := nn⋅ 2⋅ π nn1 := n , n + 1rpm .. 2 ⋅ n 60 Mm using .. again M max_n(nn1) 40 N⋅ m n= 20 100 150 200 U M⋅ R − KE⋅Φ KE⋅Km⋅Φ 2 (9.5) 250 nn1 rpm Pmax n= U KE⋅Φ M⋅ R − nn = KE⋅Km⋅Φ 2 U KE⋅Φ nn⋅ 2⋅ π − Pmax ⋅R U 2 nn = KE⋅Km⋅Φ KE⋅Φ 2 ⋅R 2⋅π ⋅nn − KE⋅Km⋅Φ 2 1 solved symbolicly on blank sheet nnmax( Φ ) := 2 1 4⋅π⋅KE⋅Km ⋅ 2 2 ⋅ Um⋅π⋅Km + 2 ⋅ ⎛ Um ⋅π ⋅Km − 2⋅π⋅KE⋅Km⋅ Pmax⋅R⎞ ⎝ ⎠ 2 2 Φ calculate n when U and M known - useful at ends of torque range 0 - M m and in this application a and b are functions of Φ a( Φ ) := Km⋅Φ R b ( Φ ) := M ( U , n) = U⋅ a − b ⋅ n Km KE⋅Φ 2 R calculate M when U and n known ... so with Φ = 0.75 Φ base n51 := nnmax( Φ⋅0.75) ( n52 := nn 0.75⋅Φ , M 0 nn( Φ , M ) := ) ( b(Φ ) MM( Φ , n ) := Um⋅ a( Φ ) − b ( Φ ) ⋅n ) and Φ = 0.6 of Φ base n51 = 133.333 rpm n52 = 140.351 rpm n5 := n51 , n51 + 1rpm .. n52 MM( Φ⋅0.75, n51 ) = 43.315 N⋅ m MM( Φ⋅0.75, n52 ) = 0 N⋅ m Um⋅a( Φ ) − M n61 := nnmax( Φ⋅0.6) n61 = 166.667 rpm ( n62 = 175.439 rpm n62 := nn 0.6⋅Φ , M 0 ) n6 := n61 , n61 + 0.1rpm .. n62 MM( Φ⋅0.6, n61 ) = 34.652 N⋅ m MM( Φ⋅0.6, n62 ) = 0 N⋅ m 3 11/15/2006 60 0.25 * Um 0.5 * Um 0.75 * Um 1.0 * Um Mm FI = 0.75 FI = 0.6 Pmax 50 torque N*m 40 30 20 10 0 0 50 100 150 200 250 rpm now if we plot this data in terms of power, nrpm ⎞ load( nrpm) := Pmax⋅ ⎛⎜ ⎟ 1.5⋅ n ⎝ power = torque⋅ rpm⋅ 2 ⋅ π and superimpose a cubic load curve reaching max power at 1.5 base rpm 0 ⎞ ⎛ 0 ⎜ Pmax ⎟ Pmax_plot := ⎜ n ⎟ ⎝ rpm W ⎠ 3 ⎠ nrpm := 0rpm , 1rpm .. 1.5⋅ n 4000 0.25 * Um 0.5 * Um 0.75 * Um 1.0 * Um M max FI = 0.75 FI = 0.6 P max load power (W) 3000 2000 1000 0 0 20 40 60 80 100 120 140 160 180 200 220 rpm So ... as observed in the text: "The operational envelopes show that a DC motor is very suited to drive a propeller for ship propulsion." 4 11/15/2006 practical aspects ... IA shunt motor: field in parallel with armature IF operates same as separate excitation ... field excitation constant RA shunt field Um E I series motor: field in series with armature now the field current and armature current is the same using the same relationships from above ... M := Km⋅Φ⋅I (9.1) E = KE⋅Φ⋅n (9.2) U = E + I⋅ R (9.3) Φ := KF⋅II Um I := U− E 2 M := Km⋅KF⋅I I → R ( I⋅ R = U − KE⋅KF⋅I⋅n 2 M → Km⋅KF⋅I E := KE⋅Φ ⋅n M := Km⋅KF⋅I E → KE⋅ KF⋅ I⋅ n E := KE⋅ KF⋅ I⋅ n U − KE⋅KF⋅I⋅ n ) I= U − KE⋅ KF⋅ I⋅ n I = R I⋅ R + KE⋅KF⋅n = U R U I := R + KE⋅KF⋅n 2 M → Km⋅ KF⋅ U R + KE⋅KF⋅n 2 U (R + KE⋅KF⋅n) (9.7) I = current 2 M := Km⋅Φ⋅I U = E + I⋅ R series field E KF = constant_for_given_motor eliminate I ... RA 2 M := Km⋅ KF⋅ U (R + KE⋅KF⋅n) (9.8) 2 some numerical values for a plot ... Um := 100V KE := 1 Km := 1 2 M ( U , n ) := Km⋅KF⋅ U (R + KE⋅KF⋅n) 2 Wb KF := 1 A R := 4Ω ( ) M Um , 100rpm = 47.747 N⋅ m Imax := 9A 2 M max := Km⋅KF⋅Imax M max = 81 N⋅ n := 1rpm , 2rpm .. 200rpm 5 11/15/2006 100 Mmax Torque (N*m) 80 N⋅ m 60 40 20 0 Um 0.5*Um 0 50 100 150 200 250 rpm motor suitable for traction purposes - high torque at low rpm Induction motors (AC) IF := 1 ω := 1 s t := FRAME⋅ 4⋅π 100 ⋅sec t to go from 0 to 4*π in 100 steps current sinusoidal (cos) with time π⎞ ⎛ Ib ( t) := IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟ 3 ⎠ ⎝ Ia( t) := IF⋅ cos( ω⋅t) π⎞ ⎛ Ic( t) := IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟ 3 ⎠ ⎝ field vector displaced by 2*π/3 and 4*p/3, and current at appropriate phase shift applied IF⋅ cos( ω⋅t) = 1 Bza( t) := IF⋅ cos( ω⋅t) π⎞ ⎛ ⎛ π⎞ ⎛ ⎛ π⎞ ⎞ Bzb ( t) := IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟ ⋅ ⎜ cos⎜ 2 ⋅ ⎟ + sin⎜ 2 ⋅ ⎟ ⋅i⎟ 3 3 ⎝ ⎠⎝ ⎝ ⎠ ⎝ 3⎠ ⎠ π⎞ ⎛ IF⋅ cos⎜ ω⋅t − 2 ⋅ ⎟ = −0.5 3 ⎠ ⎝ π⎞ ⎛ ⎛ π⎞ ⎛ ⎛ π⎞ ⎞ Bzc( t) := IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟ ⋅ ⎜ cos⎜ 4 ⋅ ⎟ + sin⎜ 4 ⋅ ⎟ ⋅i⎟ 3⎠ ⎝ ⎝ 3⎠ ⎝ ⎝ 3⎠ ⎠ ( ) ( ) ( Re_sum ( t) := Re Bza( t) + Re Bzb ( t) + Re Bzc( t) ) π⎞ ⎛ IF⋅ cos⎜ ω⋅t − 4 ⋅ ⎟ = −0.5 3 ⎠ ⎝ ( ) ( ) ( Im_sum( ) := Im Bza( t) + Im Bzb ( t) + Im Bzc( t) Re_sum ( t) = 1.5 Im_sum( t) = 0 2 ) 2 Re_sum( t) + Im_sum( t) = 1.5 6 11/15/2006 1.5 0 ⎛ ⎞ ⎜Im Bz ( t) ⎟ ⎝ ( a ) ⎠ 1 0 ⎛ ⎞ ⎜Im Bz ( t) ⎟ ⎝ ( b ) ⎠ 0.5 0 0 ⎛ ⎞ ⎜Im Bz ( t) ⎟ ⎝ ( c ) ⎠ 0.5 0 ⎞ ⎛ ⎜ ⎟ ⎝ Im_sum( t) ⎠ 1 1.5 1.5 1 0.5 0 0.5 1 1.5 0 0 0 ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎛ 0 ⎞ ⎟ ⎜Re Bz ( t) ⎟ , ⎜Re Bz ( t) ⎟ , ⎜Re Bz ( t) ⎟ , ⎜ ⎝ ( a ) ⎠ ⎝ ( b ) ⎠ ⎝ ( c ) ⎠ ⎝ Re_sum( t) ⎠ t1 := π 3 IF⋅ cos( ω⋅t1) = 0.5 sec ( ) ( ) ( ) Re Bza( t1) π⎞ ⎛ IF⋅ cos⎜ ω⋅t1 − 2 ⋅ ⎟ = 0.5 3 ⎠ ⎝ Re Bzb ( t1) imaginary part of field vector π⎞ ⎛ IF⋅ cos⎜ ω⋅t1 − 4 ⋅ ⎟ = −1 3 ⎠ ⎝ Re Bzc( t1) 2 ( ) 2 = 0.5 ( ) 2 = 0.5 ( ) 2 =1 + Im Bza( t1) 2 + Im Bzb ( t1) 2 + Im Bzc( t1) 1 0 1 1 0 1 real part of field vector 7 11/15/2006 speed of rotation of this machine = frequency of the supplied AC. as shown, there are two poles (one pair) N-S with multiple pairs the speed of rotation is reduced proportional to the number of poles ns = f = with AC frequency ω and two poles ω n s = rotation_speed 2⋅π f = frequency rpm Hz 1 ω = frequency for p poles ns = 2⋅ f Hz = 1 1 s Hz = 9.549 rpm s Hz ( 9.10 ) assumes radians 2⋅π⋅Hz = 60 rpm p one stator winding ... d E = −N⋅ Φ dt d Φ = −constant⋅Φ⋅f dt (2.95) small ... U = IF⋅R + E R<1 U=E= Φ⋅f E= => Φ⋅f KF U Φ = KF⋅ f => KF KF = constant_for_given_motor now consider the rotor,if it is turning at the same speed as the rotating magnetic field of the stator, there is no EMF the current induced in the rotor is strongly dependent on the relative speed define ... at low slip s= 0% n s − n (9.14) n s to .. 10% s = slip n s = rotation_speed_stator fR_EMF = n s − n n = rotation_speed_rotor fR_EMF = frequency_of_rotor_EMF (9.15) because the rotor induced current will be at slip frequency, EMF is low so reactance (L) will be low. I A depends primarily (only) on rotor (armature) resistance R A and is in phase with the flux pattern. The net result is torque is ~ directly proportional to slip M = K⋅ s (9.16) 8 11/15/2006 some typical curves from text for discussion M,I pull out torque I M increasing resistance load curve 0 n s load curve ns 0 0 n ns 0 s 1 fig. 9.20A Torque - speed curve varying rotor resistance from Woud 1 fig. 9.18 Torque - speed curve induction motor from Woud Um M I increasing resistance 0.75*Um 0.5*Um load curve load curve 0.25*Um 0 n s ns 0 0 1 fig. 9.20B Current - speed curve varying rotor resistance from Woud n s ns 0 1 fig. 9.21 Torque - speed curve induction motor varying supply voltage, from Woud starting has challenges and text reviews some alternatives There is much more to this subject and text covers quite well. Next lecture will review ship applications. One comment regarding opinion in text regarding DC motor drive. Dc drives not typical as commutation brushes require significant maintenance. DDX motor has innovative new brush technology. see: http://www.globalsecurity.org/military/library/report/2002/mil-02-04-wavelengths02.htm 9 11/15/2006