Question 6.2: A 15-hp, 220-V, three-phase, 60-Hz, six-pole, Y-connected in...
A 15-hp, 220-V, three-phase, 60-Hz, six-pole, Y-connected induction motor has the following parameters per phase:
R_{1} =0.15 ohm
R_{2} =0.1 ohm
X_{T} =0.5 ohm
G_{c} =6\times 10^{-3}
B_{m} =0.15 S
The rotational losses are equal to the stator hysteresis and eddy-current losses. For a slip of 3 percent, find the following:
A. the line current and power factor;
B. the horsepower output;
C. the starting torque.
A. The voltage specified is line-to-line value as usual. Utilizing the approximate equivalent circuit of Figure 6.3, the rotor current can be seen to be given by
I_{r}=\frac{\frac{220}{\sqrt{3} } }{\left\lgroup0.15+\frac{0.1}{0.03} \right\rgroup+j0.5 }
=36.09\angle -8.17° A
The no-load current I_{\phi} is obtained as
I_{\phi } =\frac{220}{\sqrt{3} } \left(6\times 10^{-3} -j0.15\right)
=0.7621-j19.05 A
As a result, the line current (stator current) is
I_{s} =I_{r}+I_{\phi }
=43.772\angle -33.535°
Since V_{1} is taken as reference, we conclude that
\phi _{s}=33.535°
\cos \phi _{s} =0.8334
B. The air-gap power is given by
P_{g} =3I_{r} ^{2} \left\lgroup\frac{R_{2} }{s} \right\rgroup=3\left(36.09\right) ^{2} \left\lgroup\frac{0.1}{0.03} \right\rgroup = 13,024.881 WThe mechanical power to the shaft is
P_{m}=\left(1-s\right) P_{g} =12,634.135 WThe core losses are
P_{c} =3E_{1} ^{2}\left(G_{c} \right) =290.4 WThe rotational losses are thus
P_{rl}=290.4 wAs a result, the net output mechanical power is
P_{out} =P_{m}-P_{rl}
=12,343.735 W
Therefore, in terms of horsepower, we get
hp_{out}=\frac{12,343.735}{746} =16.547 hpC. At starting, s = 1:
\left|I_{r} \right| =\frac{\frac{220}{\sqrt{3} } }{\left(0.15+0.1\right) +j0.5} =227.215AP_{g}=3 \left(227.215\right) ^{2}\left(0.1\right) = 15,487.997 W
\omega _{s} =\frac{2\pi \left(60\right) }{3}=40\pi
T=\frac{P_{g} }{\omega _{s} } =\frac{15,487.997}{40\pi } =123.25 N.m
The following script implements Example 6.2 in MATLAB^{TM}:
% Example 6-2
%
V=220/3^.5;
s=0.03;
f=60;
R1=0.15;
R2=0.1;
Xt=0.5;
Gc=6*10^-3;
Bm=0.15;
Ir=V/((R1+R2/s)+i*Xt);
abs(Ir)
angle(Ir)*180/pi
Iphi=V*(Gc-i*Bm)
Is=Ir+Iphi;
abs(Is)
angle(Is)*180/pi
% V1 is taken as reference
phi_s=-angle(Is);
pf=cos(phi_s)
% B. The airgap power
Pg=3*(abs(Ir))^2*(R2/s)
% The mechanical power to the shaft
Pm=(1-s)*Pg
% The core loss
E1=V;
Pc=3*E1^2*Gc
% The rotational losses
Prl=Pc
% The net output mechanical power
Pout=Pm-Prl
hpout=Pout/746
MATLAB^{TM} con’t.
% At starting s=1
s=1;
Ir=V/((R1+R2/s)+i*Xt);
abs(Ir)
angle(Ir)*180/pi
Pg=3*(abs(Ir))^2*(R2/s)
omega_s=2*pi*f/3;
T=Pg/omega_s
The results obtained from MATLAB^{TM} are as follows:
EDU»
ans = 36.0943
ans = -8.1685
Iphi = 0.7621-19.0526i
ans = 43.7750
ans = -33.5313
pf = 0.8336
Pg = 1.3028e+004
Pm = 1.2637e+004
Pc = 290.4000
Prl = 290.4000
Pout = 1.2347e+004
hpout = 16.5506
ans = 227.2150
ans = -63.4349
Pg = 15488
T = 123.2496