The following readings were obtained when no-load and blocked rotor tests were performed on a 3-phase, 400 V, 14.9 kW induction motor:
No-load test: 400 V, 1250 W, 9 A
Blocked rotor test: 150 V, 4000 W, 38 A
Find full-load current and power factor of the motor using circle diagram.
Using the data available, V = \frac{V_{L}}{\sqrt{3} } = \frac{400}{\sqrt{3} } = 231 V
\cos \phi _{0} = \frac{1250}{\sqrt{3} \times 400\times 9} = 0.2004 lagging\phi _{0} = \cos^{-1} 0.2004 = 78.5^{\circ } lagging
\cos \phi _{SC} = \frac{4000}{\sqrt{3} \times 150\times 38} = 0.405 lagging
\phi _{SC} = \cos^{-1} 0.405 = 66.1^{\circ } lagging
Short circuit current with normal voltage,
I _{SN} = I _{SN} = \frac{V_{SN}}{V_{SC}} \times I_{SC} = \frac{400 / \sqrt{3} }{150 / \sqrt{3} } \times 38 = 101.3 A
Power drawn with normal voltage would be
= \left\lgroup\frac{400 / \sqrt{3} }{150 / \sqrt{3} }\right\rgroup^{2} \times 4000 = 28440 W
Construction of Circle Diagram
Draw a vector OV along y-axis, as shown in Fig. 9.44.
Let the scale for current be 5 A = 1cm.
Draw vector OO′ i.e., I_{o} = 9 A \left(OO^{\prime } = \frac{9}{5} = 1.8 cm\right) lagging behind the voltage vector OV by an angle \phi _{0} \left(\phi _{0} = 78.5^{\circ }\right) .
Draw vector OA i.e., I_{SN} = 101.3 A \left(OA = \frac{101.39}{5} = 20.26 cm\right) lagging behind the voltage vector OV by an angle \phi _{SC} \left(\phi _{SC} = 66.1^{\circ }\right) .
Note: To adjust the figures on the page, the size of the figures are reduced.
Draw horizontal x-axis (OX) perpendicular to OY and a line O′B parallel to x-axis.
Draw a line O′A and its bisector which meets the line O′B at C. Draw the semicircle O′AB from centre C with radius CO′.
Draw a line AQ parallel to y-axis which represents 28440 W, as calculated above from blocked rotor test. Measure this line which comes out to be 8.1 cm.
Hence the scale for power is 1 cm = \frac{28440}{8.1} = 3511 W
Now full-load motor output = 14.9 kW = 14900 W14900 W = \frac{14900}{3511} = 4.245 cm
For locating full-load power point on the circle, extend QA to QA such that AA′ = 4.245 cm. From A′ draw a line parallel to output line O′A which intersects the semicircle at L. From L drop a perpendicular on x-axis which meets the line OX at M.
Line current = OL = 6 cm = 6 \times 5 = \mathbf{30 A}
Phasor OL makes an angle \phi with the voltage vector OV.
\phi = 30^{\circ } \hspace{30 pt} \hspace{30 pt} \left(by measurement\right)
Power factor, \cos \phi = \cos 30^{\circ } = \mathbf{0.886 \left(lagging\right) } \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \left(Also \cos \phi = \frac{OM}{OL} = \frac{5.3}{6} = 0.863\right)
