Question
The three-phase,
230-V, 60-Hz, 12-kW, four-pole induction motor R2 = 0.2 Ω is to be operated
from a variable-frequency, constant-volts-per-hertz motor drive whose terminal
voltage is 230 V at 60 Hz. The motor is driving a load whose power can be
assumed to vary as:
P_load=10.5
(n/1800)^3 Kw
Where n is the load
speed in r/min. Motor rotational losses can be assumed to be negligible. Write
a MATLAB script to find the line-to-line terminal voltage, the motor speed in
r/min, the slip and the motor load in kW for (a) a source frequency of 60 Hz
and (b) a source frequency of 40 Hz.
Answer
(a)
clc
clear all
close all
V1 = 230/sqrt(3)
;
N = 3;
poles = 4 ;
f1 = 60 ;
R1 = 0.095;
R2 = 0.2;
X1 = 0.680;
X2 = 0.672;
Xm = 18.7;
f2 = 60 ;
X3 = X1*(f2/f1)
;
X4 = X2*(f2/f1)
;
Xm1 = Xm*
(f2/f1) ;
V2 = V1*(f2/f1)
;
ws = 4*pi*f2/poles;
ns = 120*f1/poles;
V3 = abs(V2*1i*Xm/(R1 +
1i*(X1+Xm)) ) ;
Z = 1i*Xm*(R1+1i*X1)/(R1 +
1i*(X1+Xm)) ;
R3 = real(Z)
;
X5 = imag(Z)
;
slip = 0.0000000001 ;
error =512;
while error >=
0;
slip = slip
+ 0.00001;
nm = ns*(1-slip)
;
wm = ws*
(1-slip) ;
tm = (1/ws)*N*V3^2
*(R2/slip) ;
tm1 = tm/((R1+R2/slip)^2+(X3+X4)^2) ;
pm = tm1*wm;
pl = (10.5e3*((nm/1800)^3));
error=pl-pm;
end
pload=pl/1000;
fprintf('\nFor fe : %g [Hz] : ',f2)
fprintf('\n Terminal voltage = %g [volt] ',V2*sqrt(3))
fprintf('\n nm = %g',nm)
fprintf('\n slip = %g [unitless] ',slip)
fprintf('\n pload : %g [kw] ',pload)
fprintf ( ' \n\n' )
The Result
For fe : 60 [Hz] :
Terminal voltage = 230 [volt]
nm = 1720.67
slip = 0.04407 [unitless]
pload : 9.17207 [kw]
(b)
For fe : 60 [Hz] :
Terminal voltage = 230 [volt]
nm = 1720.67
slip = 0.04407 [unitless]
pload : 9.17207 [kw]
clc
clear all
close all
V1 = 230/sqrt(3)
;
N = 3;
poles = 4 ;
f1 = 60 ;
R1 = 0.095;
R2 = 0.2;
X1 = 0.680;
X2 = 0.672;
Xm = 18.7;
f2 = 40 ;
X3 = X1*(f2/f1)
;
X4 = X2*(f2/f1)
;
Xm1 = Xm*
(f2/f1) ;
V2 = V1*(f2/f1)
;
ws = 4*pi*f2/poles;
ns = 120*f2/poles;
V3 = abs(V2*1i*Xm/(R1 +
1i*(X1+Xm)) ) ;
Z = 1i*Xm*(R1+1i*X1)/(R1 +
1i*(X1+Xm)) ;
R3 = real(Z)
;
X5 = imag(Z)
;
slip = 0.0000000001 ;
error =512;
while error >=
0;
slip = slip
+ 0.00001;
nm = ns*(1-slip)
;
wm = ws* (1-slip)
;
tm = (1/ws)*N*V3^2
*(R2/slip) ;
tm1 = tm/((R1+R2/slip)^2+(X3+X4)^2) ;
pm = tm1*wm;
pl = (10.5e3*((nm/1800)^3));
error=pl-pm;
end
pload=pl/1000;
fprintf('\nFor fe : %g [Hz] : ',f2)
fprintf('\n Terminal voltage = %g [volt] ',V2*sqrt(3))
fprintf('\n nm = %g',nm)
fprintf('\n slip = %g [unitless] ',slip)
fprintf('\n pload : %g [kw] ',pload)
fprintf ( ' \n\n' )
The Result
For fe : 40 [Hz] :
Terminal voltage = 153.333 [volt]
nm = 1166.39
slip = 0.02801 [unitless]
pload : 2.85694 [kw]