When no-load and blocked rotor tests were performed on a 3-phase, 400 V, 50 Hz, star connected induction motor, the following results were obtained:
No-load test: 400 V, 8.5 A, 1100 W
Blocked-rotor test: 180 V, 45 A, 5700 W
Draw the circle diagram and estimate the line current and power factor of the motor when operating at 4% slip. The stator resistance per phase is measured as 0.5 ohm.
From no-load test;
Phase voltage, V = \frac{V_{L}}{\sqrt{3} } = \frac{400}{\sqrt{3} } = 231 V; I_{o} = 8.5 A
\cos \phi_{o} = \frac{1100}{\sqrt{3} \times 400 \times 8.5} = 0.187 lagging; \phi_{o} = \cos^{-1} 0.187 = 79.2^{\circ } lagging
From blocked-rotor test:
\cos \phi_{SC} = \frac{5700}{\sqrt{3} \times 180 \times 45} = 0.4063 lagging; \phi_{SC} = \cos^{-1} 0.4063 = 66^{\circ } lagging
Short circuit current at rated voltage,
I_{SN} = \frac{V_{SN}}{V_{SC}} \times I_{SC} = \left\lgroup\frac{400 / \sqrt{3} }{180 / \sqrt{3} } \right\rgroup \times 45 = 100 A
Short-circuit power input at normal voltage would be
P_{SN} = \left\lgroup\frac{400 / \sqrt{3} }{180 / \sqrt{3} } \right\rgroup^{2} \times 5700 = 28150 W
Construction of circle diagram
Draw horizontal and vertical axis OX and OY respectively, as shown in Fig. 9.46.
Draw a phasor OV along y-axis.
Let scale for the current be 10A = 1cm
Draw OO′ i.e., I_{o} = 8.5 A \left(OO^{\prime } = \frac{8.5}{10} = 0.85 cm\right) lagging behind the phasor OV by an angle of \phi _{0} \left(\phi _{0} = 79.2^{\circ }\right) .
Draw phasor OA i.e., I_{SN} = 100 A \left(OA = \frac{100}{10} = 10 cm\right) lagging behind the phasor OV by an angle \phi _{SN} \left(\phi _{SN} = 66^{\circ }\right) .
From O′ draw a line O′X′ parallel to x- axis.
Draw a line O′A and its bisector which meets the line O′X at C. taking C as centre draw a semicircle O′AB having radius CO′, this line O′A is called output line.
Draw a line AQ parallel to y-axis which represents total losses at short circuit with normal voltage i.e., P_{SN} = 28150 W. Measure this line which comes out to be 4.05 cm.
Hence, scale for power is 1 cm = \frac{28150}{4.05} = 6950 W
Stator copper loss on short circuit = 3 I_{SN}^{2} R_{1} = 3\times \left(100\right)^{2} \times 0.5 = 15000 W\therefore EF = \frac{15000}{6950} = 2.16 cm
Hence, point E is located on the line AQ.
Draw a line O′E by joining point E with O′.
Now, slip, S = \frac{Rotor copper loss}{Rotor input} = 0.04 = \frac{NK}{LK}
Using this ratio \biggl(i.e. \frac{LK}{NK} = \frac{1}{0.04} = 25\biggr) locate point L on the circle by hit and trial method.
Thus, line current, I_{L} =OL = 2.42 cm = 10\times 2.42 = \mathbf{24.2 A}
Power factor, \cos \phi = \frac{OM}{OL} = \frac{2.1}{2.42} = \mathbf{0.866 lagging}
