Question 6.4: The full-load slip of a squirrel-cage induction motor is 0.0...
The full-load slip of a squirrel-cage induction motor is 0.05, and the starting current is five times the full-load current. Neglecting the stator core and copper losses as well as the rotational losses, obtain:
A. the ratio of starting torque (st) to the full-load torque (fld), and
B. the ratio of maximum (max) to full-load torque and the corresponding slip.
\left\lgroup\frac{I_{st} }{I_{fld} } \right\rgroup ^{2} =\frac{\left\lgroup\frac{R_{2} }{0.05} \right\rgroup ^{2}+X^{2}_{T} }{R^{2}_{2}+X^{2}_{T} } =\left(5\right) ^{2}
This gives
\frac{R_{2} }{X_{T} } =\sqrt{\frac{24}{375} }\cong 0.25A.
T=\frac{3I^{2}_{r}\left(R_{2} \right) }{s\omega _{2} }
\frac{T_{st} }{T_{fld} } =\frac{I^{2}_{st}}{I^{2}_{fld}}\left\lgroup\frac{S_{fld}}{S_{st}} \right\rgroup =\left(5\right) ^{2}\frac{0.05}{1} =1.25
B.
S_{\max _{T} } =\frac{R_{2} }{X_{T} } =0.25
\frac{T_{\max } }{T_{fld} }=\frac{I^{2}_{\max } }{I^{2}_{fld} } \left\lgroup\frac{S_{fld} }{S_{\max _{T} } } \right\rgroup
=\left\lgroup\frac{S_{fld} }{S_{\max _{T} } } \right\rgroup \frac{\left\lgroup\frac{R_{2} }{S_{fld} } \right\rgroup ^{2}+X^{2}_{T} }{(2X^{2}_{T} )}
=\frac{S_{fld}}{S_{\max _{T} }} \frac{\left\lgroup\frac{S_{\max _{T} } }{S_{fld} } \right\rgroup ^{2}+1}{2}
=\frac{0.05}{0.25}\left[\frac{\left(5\right) ^{2}+1 }{2} \right]
Thus,
\frac{T_{\max } }{T_{fld} }=2.6The following script implements Example 6.4 in MATLAB^{TM}:
% Example 6-4
% A scuirrel cage induction motor
sfld=0.05;
sst=1;
% Ist=5*Ifld;
% ratio1=Ist/Ifld=5
ratio1=5;
%
(ratio1)^2=((R2/sfld)^2+(XT)^2)/(R2^2+(
XT)^2)
% (R2/XT)^2*((1/sfld)^2-
ratio1^2)=ratio1^2-1
% ratio2=R2/XT
f=[((1/sfld)^2-ratio1^2) 0 -(ratio1^2-
1)]
ratio2=roots(f);
ratio2=ratio2(1)
% A. T=3*Ir^2*R2/(sfld*ws)
% ratio3=Tst/Tfld
ratio3=ratio1^2*(sfld/sst)
% B.
s_maxT=ratio2
%Tmax/Tfld=(Imax/Ifld)^2*(sfld/s_maxT)
%=(sfld/s_maxT)*((R2/sfld)^2+XT^2)/(2*X
T^2)
%
(Tmax/Tfld)=(sfld/s_maxT)*((s_maxT/sfld
)^2+1)/2
% ratio4=Tmax/Tfld
ratio4=(sfld/s_maxT)*((s_maxT/sfld)^2+1
)/2
The results obtained from MATLAB^{TM} are as follows:
EDU»
f = 375 0 -24
ratio2 = 0.2530
ratio3 = 1.2500
s_maxT = 0.2530
ratio4 = 2.6286