From no-load test: V= \frac{V_{L}}{\sqrt{3} } = \frac{200}{\sqrt{3} } = 115.5 V; I_{o} = 7.7 A
\cos \phi _{o} = \frac{874}{\sqrt{3} \times 200 \times 0.77} = 0.327; \phi _{o} = \cos ^{-1} 0.327 = 78.75^{\circ } lagging
From blocked-rotor test:
\cos \underline{\phi} _{SC} = \frac{3743}{\sqrt{3} \times 100 \times 39.36} = 0.549; \phi _{SC} = \cos ^{-1} 0.549 = 63^{\circ } lagging
Short circuit current with normal voltage
I_{SN} = \frac{V_{SN}}{V_{SC}} \times I_{SC}= \left\lgroup\frac{200 / \sqrt{3} }{100 / \sqrt{3} } \right\rgroup \times 39.36 = 78.72 A
Power drawn with normal voltage at short circuit would be
P_{SN} = \left\lgroup\frac{200 / \sqrt{3} }{100 / \sqrt{3} } \right\rgroup^{2} \times 3743 = 14972 W
Let scale for the current be 5 A = 1 cm.
Construct the circle diagram as usual, as shown in Fig. 9.47.
Here, no-load current, I_{o}= 7.7 A;
OO^{\prime } = \frac{7.7}{5} = 1.54 cm lagging behind OV by 78.75^{\circ }
Phasor OA = I_{SN} = \frac{78.72}{5} = 15. 74 cm lagging behind phasor OV by 63^{\circ }
The vertical line AQ measures the power input on short circuit with normal voltage i.e.,14972 W and its measurement is 8.7 cm.
∴ Power scale is 1 cm = \frac{14972}{8.7} = 1721 W
Total copper loss, AF = 14972 – 874 = 14098 = 8.2 cm
Ratio of rotor copper loss and total copper loss i.e.,
\frac{AE}{AF} = \frac{Rotor resistance}{Rotor + stator resistance} = \frac{0.24}{0.24 + 0.38}
Or AE= \frac{0.24}{0.62} \times 8.2 = 3.18 cm
( i) Starting torque = AE = 3.18\times 1721 = \mathbf{5473 synch. watt}
( ii) From centre C draw a perpendicular to torque line O′E and extent it to meet the circle at U. from U draw a line parallel to y-axis which meets O′E at W, then UW represents maximum torque.
UW = 7.15 cm = 7.15 \times 1721 = \mathbf{12305 synch. watt}
(iii) For maximum power output, drop a perpendicular on output line O′A from centre C and extend it to meet the circle at R. Draw a line from R parallel to y-axis which meets the line O′A at J.
Then RJ represents maximum power output.
Maximum power output, P_{\max } = RJ = 5.9 cm = 5.9 \times 1721 = \mathbf{10154 W}
(iv) The slip for the maximum torque = \frac{WW^{\prime }}{WU} = \frac{1.4}{7.15} = 0.195
( v) For maximum power factor, draw a tangent on circle from O which touches the circle at P.
Then maximum power factor = \cos \angle POV = \cos 28.5^{\circ } = \mathbf{0.879 lagging}
