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

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Geoffrey Byrd
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1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 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 t Figure 1: Current in coil for Problem 1

2 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

3 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 ) Problem 1. Axial Field Outside Shell, A/m Inside Shell, A/m 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

4 Problem 1.: Hoop force per unit length 1 N/m 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

5 1 Problem 1.: Coil Voltages Inductive Resistive Volts 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 = T = 1.e+3 * Vol = D = =

6 Problem 1.: Shell Current 3 1 A/m Time Figure 5: Azumuthal Current in Conductive Cylinder Problem 1.: Dissipation Watts Time Figure 6: Dissipation in Cylinder 6

7 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

8 - (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( 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( 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

9 Kth = H_o - H; % this is current in the shell P_d = (*pi*ri/(sig*ts)).* Kth.^ ; % dissipation figure(4) plot(t, Kth) title( Problem 1.: Shell Current ) ylabel( A/m ) xlabel( Time ) figure(5) plot(t, P_d) title( Problem 1.: Dissipation ) ylabel( Watts ) xlabel( Time ) Matlab Code for Problem 3 % 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

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

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Problem Set Solutions September 17, 5 Problem 1: First, note that λ 1 = L 1 i 1 +