ECE 411 Introduction to Electric Drives Fall Due Date: Thurs., Oct. 9. Required Homework #04 Lec : Induction Machine Performance

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1 ECE 4 Introduction to Electric Drives Fall 204 Due Date: Thurs., Oct. 9 Name: Required Homework #04 Lec. 0-: Induction Machine Performance Note: Please fill in the answers on these sheets to simplify grading but attach separate sheets that include all calculations, graphs and code used to complete your homework so that the grader can assign partial credit when appropriate. A wye-connected 50hp, 460Volt (line-to-line rms), 60Hz, three-phase, 4-pole induction machine has the following equivalent circuit per-unit parameters: r = x = 0. r 2 = The machine is operating at rated voltage and frequency. x m = 2.5 x = 0. R 2 a) For the per-unit system appropriate for this machine, determine the base quantities for power, voltage, current, impedance, torque, electrical frequency, and mechanical frequency. Using the base quantities, covert the equivalent circuit per-unit values for the 50hp machine into actual values. m Base Quantities P = kw B V B = V rms 2 Z B = T B = N-m 2 = elec. rad/s B = mech. rad/s m b) Determine the steady-state values of rated slip S R, rated rotor speed I R, and rated power factor cos R mb Machine Parameters r = r = L = H L = H L m = H R = rr, rated torque T er, rated stator current. Find the rotor speed, torque, and current in both pu and actual units. S R = rr = elec. rad/s rmr = mech. rad/s rr = rmr = pu T er = N-m T er = pu I = pu I R = A rms R cos R = c) Determine the stator losses, rotor losses, and efficiency R for operation at rated output power (part b). Stator winding losses = Watts = pu Rotor winding losses = Watts = pu Core losses = Watts = pu = % R

2 ECE 4 Introduction to Electric Drives Fall 204 d) Find the slip for maximum motoring torque S mt (S<), stator current I MT, and the maximum torque T emax in pu, and compare with the approximate values calculated using the simplified equivalent circuit consisting of only X and r / 2 S and Eq s & Exact Approx S mt = S mt = I MT = pu I MT = pu T emax = pu T emax = pu e) Calculate the starting stator current I start (slip=.0) and starting torque T e _ start expressed in pu and as a multiple of the corresponding quantities at rated slip. Ignore rotor skin effect. Compare with the approximate values using Eq s & -5. f) Find the slip S g (<0) rated mechanical power input as a generator (i.e., e corresponding stator current amplitude I, input torque T e, and rotor speed Look at Fig for braking operation at speeds above 800rpm.) r T T I start e _ start I start e _ start Exact = pu = pu Approx = pu = pu T = -pu = -50hp = -37.3kW) and the r in pu. Assume V = pu. (Hint: S g = I = pu T = pu e = pu r

3 Contents Machine Parameters Part a) Find base value for pu conversion and the real machine parameters Part b) Find steady state slip value and corresponding rated values Part c) calculate stator losses, rotor losses and efficiency for rated operation Part d) find the slip for maximum torque and corresponding values Part e) calculate the starting condition Part f) generating mode operation End %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % University of Wisconsin-Madison % % ECE 4 Introduction to Electric Drives % % Homework #04 Solution % % Copyright: 204 Fall by B. Ge % % Go Badgers! % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clc clear close all format short g Machine Parameters r_pu = 0.022; r_2pu = 0.026; x_pu = 0.; x_2pu = 0.; x_mpu = 2.5; R_mpu = 2; P_r = 50*746; f_e = 60; Page of 8

4 P = 4; V_llrms = 460; Part a) Find base value for pu conversion and the real machine parameters P_B = P_r; % rated power as base V_B = V_llrms/sqrt(3); % rated voltage as base, phase rms value I_B = P_B/(3*V_B); Z_B = V_B/I_B; omega_b = 2*pi*f_e; omega_mb = omega_b/(p/2); T_B = P_B/omega_mB; r_ = r_pu*z_b; r_2 = r_2pu*z_b; x_ = x_pu*z_b; L_ = x_/omega_b; x_2 = x_2pu*z_b; L_2 = x_2/omega_b; x_m = x_mpu*z_b; L_m = x_m/omega_b; R_m = R_mpu*Z_B; disp('part a solution:') Page 2 of 8

5 disp([p_b r_;v_b r_2;z_b L_;T_B L_2;omega_B L_m;omega_mB R_m]) part a solution: Part b) Find steady state slip value and corresponding rated values % iterative method ss = r_2pu; % initial guess V_pu =.0; % rated voltage P_rpu =.0; % rated output err = e-0; % error control while R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/ss); % equivalent impedance I_pu = V_pu/R_eqpu; I_2pu = I_pu*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/ss); T_epu = abs(i_2pu)^2*r_2pu/ss; P_pu = T_epu*( - ss); if abs(p_pu - P_rpu) < err % found break else % scaling initial guess ss = P_rpu*(ss/P_pu); Page 3 of 8

6 I_rpu = abs(i_pu); T_erpu = T_epu; omega_rr = ( - ss)*omega_b; omega_rmr = ( - ss)*omega_mb; T_eR = T_erpu*T_B; I_R = abs(i_rpu)*i_b; pf = cos(angle(i_pu)); % power factor disp('part b solution:') disp([nan NaN ss;omega_rr omega_rmr -ss;nan T_eR T_erpu;NaN I_R abs(i_rpu);nan NaN pf]) part b solution: NaN NaN NaN NaN NaN NaN Part c) calculate stator losses, rotor losses and efficiency for rated operation P_statpu = abs(i_pu)^2*r_pu; P_stat = P_statpu*P_B; P_rotopu = abs(i_2pu)^2*r_2pu; P_roto = P_rotopu*P_B; V_mpu = V_pu - I_pu*(r_pu + i*x_pu); % voltage across magnetizing branch P_corepu = abs(v_mpu)^2/r_mpu; P_core = P_corepu*P_B; eta = P_rpu/(P_rpu + P_statpu + P_rotopu + P_corepu); disp('part c solution:') disp([p_stat P_statpu;P_roto P_rotopu;P_core P_corepu;NaN eta]) part c solution: Page 4 of 8

7 NaN Part d) find the slip for maximum torque and corresponding values % Approximate T_emax_appr = V_pu^2/(2*(x_pu + x_2pu)); s_mt_appr = r_2pu/(x_pu + x_2pu); I_mt_appr = V_pu/(sqrt(2)*(x_pu + x_2pu)); % Exact, iterative method % please ref. to % other method like bisection search, interpolation are also acceptable sr = 0.0; sl =.0; err = e-0; % error control while sm = 0.68*sl *sr; st = 0.382*sl *sr; R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/sm); I_pu = V_pu/R_eqpu; I_2pu = I_pu*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/sm); T_epum = abs(i_2pu)^2*r_2pu/sm; R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/st); I_pu = V_pu/R_eqpu; I_2pu = I_pu*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/st); T_eput = abs(i_2pu)^2*r_2pu/st; if T_epum > T_eput sr = st; else sl = sm; if abs(sr - sl) < err Page 5 of 8

8 break s_mt_exac = (sr + sl)/2; R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/s_mt_exac); I_mt_exac = V_pu/R_eqpu; I_2mt_exac = I_mt_exac*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/s_mt_exac); T_emax_exac = abs(i_2mt_exac)^2*r_2pu/s_mt_exac; disp('part d solution:') disp([s_mt_exac s_mt_appr;abs(i_mt_exac) I_mt_appr;T_emax_exac T_emax_appr]) part d solution: Part e) calculate the starting condition % Approximate % please ref. to lecture notes I_start_appr = V_pu/(x_pu + x_2pu); T_estart_appr = V_pu^2/((x_pu + x_2pu)^2)*r_2pu; % Exact ss =.0; R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/ss); I_start_exac = V_pu/R_eqpu; I_2start_exac = I_start_exac*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/ss); T_estart_exac = abs(i_2start_exac)^2*r_2pu/ss; disp('part e solution:') disp([abs(i_start_exac) abs(i_start_exac)/i_rpu;t_estart_exac T_estart_exac/T_erpu;NaN I_start_appr;NaN T_estart_appr]) Page 6 of 8

9 part e solution: NaN NaN Part f) generating mode operation % same iterative method as part b) V_pu =.0; P_rpu = -.0; ss = -r_2pu; err = e-0; % error control while R_eqpu = r_pu + i*x_pu + par(par(i*x_mpu, R_mpu), i*x_2pu + r_2pu/ss); I_pu = V_pu/R_eqpu; I_2pu = I_pu*par(i*x_mpu, R_mpu)/(par(i*x_mpu, R_mpu) + i*x_2pu + r_2pu/ss); T_epu = abs(i_2pu)^2*r_2pu/ss; P_pu = T_epu*( - ss); if abs(p_pu - P_rpu) < err break else ss = P_rpu*(ss/P_pu); disp('part f solution:') disp([ss;-abs(i_pu);t_epu;-ss]) part f solution: Page 7 of 8

10 End format short Published with MATLAB R204a Page 8 of 8