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R1 = 0105
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In the no-load test, the line voltage is 208 V, so the phase voltage is 120 V Therefore,
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X1 + X M =
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120 V = 5455 @ 60 Hz 220 A
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In the locked-rotor test, the line voltage is 246 V, so the phase voltage is 142 V From the locked-rotor test at 15 Hz,
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Z LR = RLR + jX LR =
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Therefore,
V I A,LR
142 V = 02202 645 A
PLR 2200 W = cos 1 3 (246 V )(645 A ) = 3682 S LR
RLR = Z LR cos LR = (02202 ) cos 3682 = 0176
R1 + R2 = 0176 R2 = 0071
X LR = Z LR sin LR = (02202 ) sin 3682 = 0132
At a frequency of 60 Hz,
60 Hz X LR = X LR = 0528 15 Hz
For a Design Class B motor, the split is X 1 = 0211 and X 2 = 0317 Therefore,
X M = 5455 - 0211 = 5244
The resulting equivalent circuit is shown below:
0105
j0211 j5244
j0317
0071
I2 1 s R2 s
A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:
% M-file: prob7_18m % M-file create a plot of the torque-speed curve of the % induction motor of Problem 7-18 % First, initialize the values needed in this program r1 = 0105; % Stator resistance x1 = 0211; % Stator reactance r2 = 0071; % Rotor resistance x2 = 0317; % Rotor reactance xm = 5244; % Magnetization branch reactance v_phase = 208 / sqrt(3); % Phase voltage n_sync = 1200; % Synchronous speed (r/min) w_sync = 1257; % Synchronous speed (rad/s) % Calculate the Thevenin voltage and impedance from Equations % 7-38 and 7-41
v_th z_th r_th x_th
= = = =
v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) ); ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm)); real(z_th); imag(z_th);
% Now calculate the torque-speed characteristic for many % slips between 0 and 1 Note that the first slip value % is set to 0001 instead of exactly 0 to avoid divide% by-zero problems s = (0:1:50) / 50; % Slip s(1) = 0001; nm = (1 - s) * n_sync; % Mechanical speed % Calculate torque versus speed for ii = 1:51 t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / (w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) ); end % Plot the torque-speed curve figure(1); plot(nm,t_ind,'b-','LineWidth',20); xlabel('\bf\itn_{m}'); ylabel('\bf\tau_{ind}'); title ('\bfInduction Motor Torque-Speed Characteristic'); grid on;
The resulting plot is shown below:
7-19
A 208-V four-pole 10-hp 60-Hz Y-connected three-phase induction motor develops its full-load induced torque at 38 percent slip when operating at 60 Hz and 208 V The per-phase circuit model impedances of the motor are
R1 = 033 X 1 = 042
X M = 16 X 2 = 042
Mechanical, core, and stray losses may be neglected in this problem (a) Find the value of the rotor resistance R2 (b) Find max , smax , and the rotor speed at maximum torque for this motor (c) Find the starting torque of this motor (d) What code letter factor should be assigned to this motor SOLUTION The equivalent circuit for this motor is
033
j042 j16
j042
1 s R2 s
The Thevenin equivalent of the input circuit is:
Z TH =
jX M ( R1 + jX 1 ) ( j16 )( 033 + j 042 ) = 0313 + j 0416 = 0520 53 = R1 + j ( X 1 + X M ) 033 + j ( 042 + 16 )
VTH =
( j16 ) jX M (120 0 V ) = 1169 12 V V = R1 + j ( X 1 + X M ) 033 + j (042 + 16 )
nm = (1 0038)(1800 r/min ) = 1732 r/min
(a) If losses are neglected, the induced torque in a motor is equal to its load torque At full load, the output power of this motor is 10 hp and its slip is 38%, so the induced torque is
ind = load =
(10 hp)( 746 W/hp) 2 rad 60 s (1732 r/min )
1r 1 min
2 3VTH R2 / s
= 411 N m
The induced torque is given by the equation
ind =
2 2 sync ( RTH + R2 / s ) + ( X TH + X 2 )
Substituting known values and solving for R2 / s yields
411 N m =
(1885 rad/s) ( 0313 + R2 / s ) 2 + ( 0416 + 042) 2
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