Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electric Machines

Transcription

1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electric Machines Problem Set Solutions September 17, 5 Problem 1: First, note that λ 1 = L 1 i 1 + Mi cos ωt, so that v 1 = L 1 di 1 dt + M cos ωtdi dt ωmi sin ωt Also, we already know that T e = Mi 1 i sin θ So, for constant i == I, T e V 1 = MI 1 I sin ωt = ωmi sin ωt Problem 1, Constant Secondary Current 1 Torque, Nm Terminal 1 Voltage Time, s Figure 1: Part 1: Constant Current in Both Coils Now, if λ =, i = M L I 1 cos ωt 1

2 and Torque becomes: and voltage is di dt = ωmi 1 L sin ωt T e = M I M 1 sin ωt cos ωt = I1 L L sinωt v 1 = ω M I 1 cos ωt sin ωt + ω M I 1 sin ωt cos ωt = ω M I 1 sinωt L L L 1 Problem 1, Zero Secondary Flux.5 Torque, Nm Terminal 1 Voltage Time, s Figure : Part : Zero Flux in Coil Note that there is no average of voltage or torque and so no real power for these two parts. This makes sense as there is no place for dissipation. In the third part we do have dissipation. It is convenient, since this is a sinusoidal steady state problem, to use complex variables. Voltage on the secondary side can be written: v = Re {V e jωt} The complex amplitude of v is found by writing KVL: V = jωmi 1 + jωl I = RI So the complex amplitude of current in the secondary is:

3 I = jωmi 1 jωl + R The magnitude and angle of the secondary current are easily written: I = ωmi 1 (ωl ) + R I = π arctan ωl R = θ = π φ Torque is then calculated: T e = Mi 1 i sin ωt = MI 1 I sin ωt cos(ωt θ) { 1 = MI 1 I sinθ + 1 } sin(ωt θ) Voltage is, using the chain rule for differentiation: v 1 = dλ 1 dt These are plotted in Figure 3 = ωmi sin ωt + M cos ωt di dt = ωm I sin ωt cos(ωt θ) ωm I cos ωt sin(ωt θ) = ωm I sin(ωt θ) As a check, we might consider the average power input to the machine and the average power dissipated in the resistor, as these must be equal. Power into the machine through the shaft is easily estimated as speed times torque: { 1 P m = ωt e = ωmi 1 I sin θ + 1 } sin(ωt θ) Power dissipated in the resistor is P e = i R These are plotted in Figure 4. They appear to have the same average value, but we can prove it easily. See that average electrical power is, substituting for the magnitude of i : P e = 1 I R = 1 ω M I1 (ωl ) + R R Average mechanical power in is: 3

4 .5 Problem 1, Secondary shorted through resistor Torque, Nm Terminal 1 Voltage Time, s Figure 3: Part 3: Coil shorted through a resistor P m = ωt e = ω M (ωl ) + R sin θ Noting that θ = π + arctan ωl R, we see that sin θ = cos arctan ωl R = R (ωl ) + R And making the substitution for sinθ into the expression for P m we have proven that P e = P m. Problem : Area A = wd is.5 square meter, so force per unit area is So that required flux density is P = F A = 5N.5m = 1 6 Pa = B µ B = µ P = T Magentic field is H = B µ = A/m, And then required ampere-turns are NI = gh = Note that this implies a current density in the coil of 5 amperes per square centimeter: a value that would imply the need for good cooling. 4

5 4 Problem 1, Electrical and Mechanical Power 3 Mech Solid, El Dotted Time, s Figure 4: Power in and out for Part 3 Now, if the car is off center, we need to be a step more sophisticated in modeling. Note that the permeances of the two gaps will be: P 1 = µ ( ) D w g + x P = µ ( ) D w g x Total permeance is just that of the two gaps in series: P = P 1P = µ ( ) D w P 1 + P gw 4 x Inductance is then simply: L = µ N D wg ( w 4 x ) Vertical force is, since this is a linear, singly excited system: F y = I Thus required exciting ampere-turns are: L g = µ (NI) ( ) D w g 4 x (NI) = wg F µ D w 4 x where F is the required vertical force. This is plotted in Figure 5. 5

6 3 x Problem..5 Ampere Turns Lateral Position Figure 5: Ampere-Turns required to produce lift force Lateral force is F x = I L x = µ (NI) D x wg When we substitute for the frquired ampere-turns to produce the required vertical force, and doing a little algebra, we find: 4xg F x = F w 4 x This is plotted in Figure 6 for both constant current and for current that must be maintained to produce the correct lift. 6

7 3 x Problem. 1 Lateral Force Lateral Position Figure 6: Lateral Force: Constant Current and corrected for current variation 7

8 Matlab Script for Problem 1 % Problem Set, Problem 1 % Parameters L1 =.1; L =.1; M =.8; I1 = 5; R = ; om = *pi*6; t =-1/6:1/6:1/6; % Part 1: Constant Current I = 5; T_e1 = -M*I1*I.* sin(om.* t); V11 = -om*m*i.* sin(om.* t); figure(1) subplot 11 plot(t, T_e1) title( Problem 1, Constant Secondary Current ) ylabel( Torque, Nm ) subplot 1 plot(t, V11); ylabel( Terminal 1 Voltage ) xlabel( Time, s ) % Part : Rotor shorted T_e = (M^/(*L))*I1^.* sin(*om.*t); V1 = (om*m^*i1/l).* sin(*om.*t); figure() subplot 11 plot(t, T_e) title( Problem 1, Zero Secondary Flux ) ylabel( Torque, Nm ) subplot 1 plot(t, V1); ylabel( Terminal 1 Voltage ) xlabel( Time, s ) % Part 3: Rotor shorted through a resistor Im = om*m*i1/sqrt((om*l)^+r^); phi = atan(om*l/r); theta = pi/ + phi; 8

9 T_e3 = -(M*I1*Im/).* (sin(theta) + sin(*om.* t - theta)); V13 = -om*m*im.* sin(*om.* t - theta); figure(3) subplot 11 plot(t, T_e3) title( Problem 1, Secondary shorted through resistor ) ylabel( Torque, Nm ) subplot 1 plot(t, V13); ylabel( Terminal 1 Voltage ) xlabel( Time, s ) % part 4: Power for Part 3 I_ = Im.* cos(om.* t - theta); P = R.* I_.^; Pm = -om.* T_e3; Pmav = om*m*i1*im*sin(theta)/; Peav = R * Im^/; Pma = Pmav.* ones(size(t)); Pea = Peav.* ones(size(t)); figure(4) subplot 11 plot(t, P, t, Pea, -- ) title( Problem 1, Electrical and Mechanical Power ) ylabel( Dissipated in Resistor ) subplot 1 plot(t, Pm, t, Pma, -- ); ylabel( Mech Power In ) xlabel( Time, s ) figure(5) plot(t, P, --, t, Pm) title( Problem 1, Electrical and Mechanical Power ) ylabel( Mech Solid, El Dotted ) xlabel( Time, s ) 9

10 Matlab Script for Problem % Problem Set, Problem % Parameters D =.5; % length F = 5e4; % required force muzero = pi*4e-7; g =.1; % relative motion gap w =.1; % pole width, coil width and depth % First, do the centered case A = w*d; % area B = sqrt(*muzero*f/a) N_I = *g*b/muzero x = -.5*w:.5*w:.5*w; NIS = (*w*g^*f/(muzero*d))./ (w^/4 - x.^); NI = sqrt(nis); figure(1) plot(x, NI) axis([ e4]) title( Problem. ) ylabel( Ampere-Turns ) xlabel( Lateral Position ) % now lateral force % fixed current F_xf = -(*muzero*d*n_i^/(w*g)).* x; % accounting for curent F_xc = (-*muzero*d/(w*g)).* NIS.* x; F_chk = -(F*4*g).* x./((w^)/4 - x.^); figure() plot(x, F_xf, --, x, F_xc, x, F_chk) title( Problem. ) ylabel( Lateral Force ) xlabel( Lateral Position ) 1