Question 6.6: (a) Determine the parameters of the motor of Example 6.5 sol......
(a) Determine the parameters of the motor of Example 6.5 solving for the leakage reactances using Eq. 6.59. (b) Assuming the motor to be operating from a 220-V, 60-Hz source at a speed of 1746 r/min, use MATLAB to calculate the output power for the two sets of parameters.
X_{\mathrm{bl}}=X_1 +X_2 \quad \quad \quad (6.59)
a. As found in Example 6.5
X_{\mathrm{nl}} = 21.8 Ω \quad X_{\mathrm{bl}} = 2.01 Ω\\ R_1 = 0.262 Ω \quad R_{\mathrm{bl}} = 0.652 Ω
Thus, from Eq. 6.42,
X_{\mathrm{nl}}=X_{11}= X_1 + X_m \quad \quad \quad (6.42)
X_1 + X_m =X_{\mathrm{nl}} = 21.8 Ω
and from Eq. 6.59
X_1 +X_2 = X_{\mathrm{bl}} = 2.01 Ω
From Table 6.1, X_1 = 0.3(X_1 + X_2) = 0.603 Ω \text{and thus} X_2 = 1.41 Ω \text{and} X_m = 21.2 Ω.
Finally, from Eq. 6.56,
R_2 = ( R_{\mathrm{bl}} – R_1 ) \left( \frac{X_2 +X_m}{X_m} \right) ^2 =0.444 Ω
Comparison with Example 6.5 shows the following
b. For the parameters of Example 6.6, P_{shaft} = 2467 [W] while for the parameters of part (a) of this example, P_{shaft} = 2497 [W]. Thus the approximation associated with Eq. 6.59 results in an error on the order of 1 percent from using the more exact expression of Eq. 6.54.
X_{\mathrm{bl}} = X_1 + X_2 \left( \frac{X_m}{X_2 + X_m} \right) \quad \quad \quad (6.54)
This is a typical result and hence this approximation appears to be justifiable in most cases.
Here is the MATLAB script:
| Table 6.1 Empirical distribution of leakage reactances in induction motors. | |||
| Fraction of \bf X_1 + X_2 |
|||
| Motor class | Description | \bf X_1 |
\bf X_2
|
| A | Normal starting torque, normal starting current | 0.5 | 0.5 |
| B | Normal starting torque, low starting current | 0.4 | 0.6 |
| C | High starting torque, low starting current | 0.3 | 0.7 |
| D | High starting torque, high slip | 0.5 | 0.5 |
| Wound rotor | Performance varies with rotor resistance | 0.5 | 0.5 |
| Source: IEEE Standard 112. | |||
| Parameter | Example 6.5 | Example 6.6 |
| R_1 | 0.262 Ω | 0.262 Ω |
| R_2 | 0.447 Ω | 0.444 Ω |
| X_1 | 0.633 Ω | 0.603 Ω |
| X_2 | 1.47 Ω | 1.41 Ω |
| X_m | 21.2 Ω | 21.2 Ω |
Script File
clc
clear
% Here are the two sets of parameters
% Set 1 corresponds to the exact solution
% Set 2 corresponds to the approximate solution
R1 (1) = 0.262 ; R1 (2) = 0.262 ;
R2 (1) = 0.447 ; R2 (2) : 0.444 ;
X1 (1) = 0.633 ; X1 (2) = 0.603 ;
X2 (1)= 1.47 ; X2 (2) : 1.41 ;
Xm (1)= 21.2 ; Xm (2) = 21.2 ;
nph = 3 ;
poles = 4 ;
Prot = 354 ;
%Here is the operating condition
V1 = 220/sqrt (3) ;
fe = 60 ;
rpm = 1746 ;
%Calculate the synchronous speed
ns = 120 * fe/poles ;
omegas = 4 * pi * fe/poles ;
slip = (ns-rpm) / ns ;
omegam = omegas * (1 - slip) ;
% Calculate stator Thevenin equivalent
% Loop over the two motors
for m = 1 : 2
Zgap = j * Xm (m) * (j * X2 (m)+R2 (m) / slip) / (R2 (m) / slip + j * (Xm (m) + X2 (m) ) ) ;
Zin = R1(m) + j * X1 (m) + Zgap ;
I1 = V1 / Zin ;
I2 = I1 * (j * Xm (m) ) / (R2 (m) / slip + j * (Xm (m) + X2 (m) ) ) ;
Tmech = nph * abs (I2)^2 * R2 (m) / (slip*omegas) ; %Electromechanical torque
Pmech = omegam*Tmech; %Electromechanical power
Pshaft = Pmech - Prot ;
if (m == 1)
fprintf (' \nExact solution:')
else
fprintf (' \nApproximate solution:')
end
fprintf (' \n Pmech= %.if [W], Pshaft = % . if [W]' , Pmech , Pshaft )
fprintf (' \n I1 = %.if [A] \n' , abs(I1) ) ;
end % end of "for m = 1 : 2" loop