The no-load and blocked rotor tests were performed on a 3-phase, 200 V, 4 kW, 50 Hz, 4 pole, star connected induction motor and follows results were obtained:
No-load test: 200 V, 5 A, 350 W
Blocked-rotor test: 100 V, 26 A, 1700 W
Draw the circle diagram from the given data and estimate from the diagram line current and power factor at full load. Also estimate the maximum torque in terms of full-load torque considering that rotor copper loss at standstill is half the total copper loss.
Using data available from no-load test;
V = V =\frac{V_{L}}{\sqrt{3} } = \frac{200}{\sqrt{3} } = 115.5 V ; I_{o} = 5 A\cos \phi _{0} = \frac{350}{\sqrt{3} \times 200 \times 5} = 0.202 lagging;
\phi _{0} = \cos^{-1} 0.202 = 78.34^{\circ } lagging
Using data available from blocked-rotor test,
\cos \phi _{SC} = \frac{1700}{\sqrt{3} \times 100 \times 26} = 0.378 lagging ; \phi _{SC} = \cos^{-1} 0.378 = 67.82^{\circ } lagging
Short circuit current with normal voltage
I _{SN} = \frac{V_{SN}}{V_{SC}} \times I_{SC} = \frac{200 / \sqrt{3} }{100 / \sqrt{3} } \times 26 = 52 A
Power drawn with normal voltage at short circuit would be
= \left\lgroup\frac{200 / \sqrt{3} }{100 / \sqrt{3} }\right\rgroup^{2} \times 1700 = 6800 W
Construction of Circle Diagram
Draw line OX and OY as horizontal (x-axis) and vertical (y-axis) axis respectively. Draw a vector OV along y-axis, as shown in Fig. 9.45.
Let scale for the current be 2 A = 1 cm
Draw vector OO′ i.e., I_{0} = 5 A \left(OO^{\prime } = \frac{5}{2} = 2.5 cm\right) lagging behind the voltage phasor OV by an angle \phi _{0} \left(\phi _{0} = 67.82^{\circ }\right)
Draw phasor OA i.e., I_{SN} = 52 A \left(OA = \frac{52}{2} = 26 cm\right) lagging behind the voltage phasor OV by an angle \phi _{SC} \left(\phi _{SC} = 67.82^{\circ }\right)
Note: To adjust the figure on the page, the size of the figure has been reduced.
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. Then draw a semicircle O′AB taking centre C and radius CO′. This line O′A is called the output line.
Draw a line AQ parallel to y-axis which represents 6800 W, as calculated above from blocked rotor test, represents total losses. Measure this line which comes out to be 9.8 cm.
Hence scale for power is 1 cm = \frac{6800}{9.8} = 694 W
Out of total loss AQ, the portion QF represents no-load loss i.e., 350 W, whereas, the remaining portion AF represents total copper loss.
Find the centre point of line AF at E since at blocked-rotor, the stator copper loss is equal to rotor copper loss i.e., AE = EF
Join O′ and E, the line O′E represents torque line.
Now motor output = 4 kW = 4000 W= \frac{4000}{694} = 5.76 cm
Thus, locate the point L as follows:
Extend line FA to A′ so that AA′ = 5.76 cm and from A′ draw a line parallel to output line O′A which cuts the circle at L. From L drop a perpendicular on OX which meets OX at M. Join point L with origin O.
To locate point U which corresponds to maximum torque, proceed as follows:
From centre of the circle C drop a perpendicular on torque line O′E and extend it to meet the circle at U. This point U can also be located by drawing a parallel line with torque line O′E from A′.
From point U draw a line parallel to y-axis which meets the torque line at W, The section UW represents maximum torque.
Now in the circle diagram, OL represents full load current,
OL = 6.72 cm = 6.72 \times 2 = \mathbf{13.44 A}
Power factor at full load, \cos \phi = \frac{ML}{OL} = \frac{5.78}{6.72} = \mathbf{0.86 lagging}
\frac{Maximum torque}{Full-load torque} = \frac{UW}{LK} = \frac{10.2}{5.67} = \mathbf{1.8}
i.e., Maximum torque = 180 of full-load torque.
