Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Problem Set 1 Solutions September 1, 5 Problem 1: If we assume, as suggested in the problem statement, that fields outside the coil can be ignored, magnetic field inside the coil is simply H = i z Ni and outside the coil magnetic field is zero. On the inner surface of the coil the normal vector is i r and r = µ H z so that traction on the surface is pressure pushing out: P r = µ ( ) Ni Hoop force per unit length is just pressure times radius: F h = RP r = µ ( ) Ni R To do this same problem using the principal of virtual work, see that co-energy is just coenergy per unit volume times volume, or W m = u ( ) Ni πr Hoop force is the first derivative of co-energy with respect to hoop circumference, which is C = πr: F h = W m C = 1 W m π R = 1 ( ( ) µ Ni πr ) = µ ( ) Ni R π R to get hoop force per unit length, divide by and we get the same answer. 1 A I T_r T_o T_r Time intervals 1 3 4 t Figure 1: Current in coil for Problem 1
Problem : I have re-drawn the current waveform to show my notation for time intervals (Figure 1). If we note H z as magnetic field within the conductive cylinder, current in that cylinder must be: K θ = σt s ( ) µ πri πr i t H H z z = T s t where t s and R i are shell thickness and radius, respectively. The shell time constant is T s = µ σt s R If K θ is azimuthal current in the shell, field inside the shell is: so that H z = K θ + Ni T s H z t + H z = Ni Now we have to patch together a solution. Note that if we note H z = Ni, the step response of our differential equation would be: ( ) Hz s = H z 1 e t Since the ramp which is the current waveform during interval 1 is simply the integral of a step, and noting that the response of an ODE to the integral of a waveform is the integral of the response to the waveform itself, the response during the first interval is H z1 = t H z (1 e t ) ( t dt = H z T ( ) ) s 1 e t At the end of this interval the field has risen to: ( H z (1) = H z 1 T ( )) s 1 e Tr During the second time interval the excitation is constant and we have the equivalent of a step response with an initial condition: ( H z = H z H z H (1) z ) e t where t is time from the start of interval. With a little bit of manipulation this is found to be: ( H z = H z 1 T ( ) ) s 1 e Tr e t In the third time interval we have an excitation which is the same as the steady state minus the same ramp as started the problem: ( t3 H z3 = H z H z T ( )) s 1 e t 3 Writing that out and evaluating at the end of the third time interval, when excitation current reaches zero, ( ( ) ( )) H z (3) = H z 1 e Tr 1 e To+Tr
and in the fourth time interval the system is homogeneous: H z4 = H (3) z e t 4 It is relatively easy in MATAB to build up the waveforms for H z by simply concatenating the time periods. Shown in Figure are the field inside of the cylinder and outside (which is just Ni ). 1 6.685 Problem 1. Axial Field Outside Shell, A/m 8 6 4.1..3.4.5.6.7.8.9.1 1 Inside Shell, A/m 8 6 4.1..3.4.5.6.7.8.9.1 Time, s Figure : Magnetic Fields Net pressure on the cylinder is, noting that the normal vector inside is just i r and outside is i r, P r = µ ( ( ) ) Ni Hz so that hoop force is F H = RP r This is shown in Figure 3. Note that it starts out negative as the field outside is greater than the field inside. To compute coil voltage we need: v = Ri + N dφ dt Now, the resistive term Ri is straightforward. Flux consists of two parts: Φ = µ A 1 H z + µ A Ni 3
3 6.685 Problem 1.: Hoop force per unit length 1 N/m 1 3 4.1..3.4.5.6.7.8.9.1 Time, s Figure 3: Net Pressure where A 1 = πr i (R i is the shell radius) and A = π(r o R i ) (R o is the coil radius). The rate of change of current is ± I or zero, depending on time interval and that can be pieced together in MATAB. Rate of change of field inside the cylinder is: dh z = 1 ( ) Ni dt T s H z The result is shown in Figure 4 To get dissipation in the cylinder we can simply find the current in the cylinder: and then K θ = Ni H z P d = πr i K θ σt s Cylinder current is shown in Figure 5 and resulting dissipation in Figure 6 Problem 3 We assume here that speed is very close to synchronous, so that: Required torque is simply: Ω = π 6 p T = P Ω 4
1 Problem 1.: Coil Voltages Inductive 5 5 1.1..3.4.5.6.7.8.9.1 1 Resistive 5 5 1.1..3.4.5.6.7.8.9.1 1 5 Volts 5 1.1..3.4.5.6.7.8.9.1 Time, s Figure 4: Voltage Induced in the Coil We established in class that: T = πr = V rotor τ where R and are rotor radius and length, and τ is peripheral shear stress. If length is twice diameter, V rotor = 4πR 3 I have written a simple script that carries out these calculations (appended) and here is the result (I have removed a bunch of blank lines) Om = 376.9911 188.4956 15.6637 T = 1.e+3 *.656 5.35 7.9577 Vol =.133.65.398 D =.36.566.937 =.473.5131.5874 5
4 6.685 Problem 1.: Shell Current 3 1 A/m 1 3 4.1..3.4.5.6.7.8.9.1 Time Figure 5: Azumuthal Current in Conductive Cylinder 14 6.685 Problem 1.: Dissipation 1 1 8 Watts 6 4.1..3.4.5.6.7.8.9.1 Time Figure 6: Dissipation in Cylinder 6
Matlab Code for Problem % Massachusetts Institute of Technology % 6.685, Fall Term, 5 % Problem set 1, Problem Solution % Parameters, etc. muzero = pi*4e-7; sig = 6e7; % conductivity of the cylinder ts =.1; % thickness of the cylinder Ri =.1; % radius of the cylinder Ro =.15; % radius of the coil = 1; % axial extent of the system N = 1; % turns I = 1; % peak current T_r =.1; % rise time T_o =.5; % steady current time T_f =.3; % end time (kinda arbitrary) dw =.; % coil wire diameter % minor side calculations R_c = N**pi*Ro/(sig*(pi/4)*dw^); % coil resistance T_s = muzero*sig*ts*ri/; % shell time constant % now build the times dt = T_r/; % basic time step t1 = :dt:t_r; % first interval ti = dt:dt:t_o; % constant current time t = ti + T_r; % so we can put things together t3i = dt:dt:t_r; % downgoing ramp te = t3i + T_o; % extension of interval t3 = t3i + T_r + T_o; % for assembly purposes t4i = dt:dt:t_f; % not sure how far to carry this... t4 = t4i + *T_r + T_o; t = [t1 t t3 t4]; % complete time line T_e = *T_r+T_o+T_f; % construct current waveform i1 = (I/T_r).* t1; i = I.* ones(size(ti)); i3 = I.* (1 - (1/T_r).* t3i); i4 = zeros(size(t4)); I = [i1 i i3 i4]; H_o = (N/).* I; % first interval % constant current interval % downward ramp % now construct magnetic field H1 = (N*I/).* (t1./ T_r - (T_s/T_r).* (1 - exp(-t1./ T_s))); H = (N*I/).* (1 - (T_s/T_r)*(1-exp(-T_r/T_s)).* exp(-ti./ T_s)); H3 = (N*I/).* (1 - (T_s/T_r)*(1-exp(-T_r/T_s)).* exp(-te./ T_s))... 7
- (N*I/).* (t3i./ T_r - (T_s/T_r).* (1 - exp(-t3i./ T_s))); H4 = (N*I/).* (T_s/T_r) * (1 - exp(-t_r/t_s)) * (1 - exp(-(t_o+t_r)/t_s)).* exp(-t4i./ T_s H = [H1 H H3 H4]; figure(1) subplot 11 plot(t, H_o) title( 6.685 Problem 1. Axial Field ) ylabel( Outside Shell, A/m ) subplot 1 plot(t, H) ylabel( Inside Shell, A/m ) xlabel( Time, s ) % hoop force is easily computed from the fields: % this is force per unit length F_h = (Ri*muzero/).* (H.^ - H_o.^); figure() plot(t,f_h) title( 6.685 Problem 1.: Hoop force per unit length ) ylabel( N/m ) xlabel( Time, s ) % Now to compute voltage dhzdt = (N*I/(*T_r)).* [ones(size(t1)) zeros(size(t)) -ones(size(t3)) zeros(size(t4))]; dhzdt = (1/T_s).* (H_o - H); A1 = pi*ri^; A = pi*(ro^ - Ri^); v_i = muzero*n.* (A1.* dhzdt + A.* dhzdt); v_r = R_c.* I; v = v_r + v_i; figure(3) subplot 311 plot(t, v_i) axis([ T_e -1 1]) title( Problem 1.: Coil Voltages ) ylabel( Inductive ) subplot 31 plot(t, v_r) axis([ T_e -1 1]) ylabel( Resistive ) subplot 313 plot(t, v) axis([ T_e -1 1]) ylabel( Volts ) xlabel( Time, s ) % and, finally current in the shell and then loss 8
Kth = H_o - H; % this is current in the shell P_d = (*pi*ri/(sig*ts)).* Kth.^ ; % dissipation figure(4) plot(t, Kth) title( 6.685 Problem 1.: Shell Current ) ylabel( A/m ) xlabel( Time ) figure(5) plot(t, P_d) title( 6.685 Problem 1.: Dissipation ) ylabel( Watts ) xlabel( Time ) Matlab Code for Problem 3 % 6.685 Problem Set 1, Problem 3 p = [1 3]; Om = *pi*6./ p T = 1e6./ Om Vol = T./ e5 R = (Vol./ (4*pi)).^ (1/3); D =.* R =.* D 9