A 15 kW, 415 V, 4-pole, 50 Hz delta-connected motor gave the following results on test (voltages and currents are in line values): No-load test 415 V 10.5 A 1,510 W Blocked-rotor test 105 V 28 A 2,040 W Using the approximate circuit model, determine: (a) the line current and power factor for

Question

A 15 kW, 415 V, 4-pole, 50 Hz delta-connected motor gave the following results on test (voltages and currents are in line values):

No-load test	415 V	10.5 A	1,510 W
Blocked-rotor test	105 V	28 A	2,040 W

Using the approximate circuit model, determine:

(a) the line current and power factor for rated output,

(b) the maximum torque, and

(c) the starting torque and line current if the motor is started with the stator star-connected.

Assume that the stator and rotor copper losses are equal at standstill.

Hint Part (a) is best attempted by means of a circle diagram. For proceeding computationally from the circuit model, we have to compute the complete output-slip curve and then read the slip for rated output.

Accepted Answer

[latex] O P_{0}=10.5 A[/latex]

[latex] \cos \phi_{0}=\frac{1,510}{\sqrt{3 imes 415 imes 10.5}}=0.2[/latex]

Or [latex] \phi_{0}=78.5^{\circ}[/latex]

[latex] O P_{S C}=\frac{28 imes 415}{105}=110.7 A[/latex] (at 415 V)

[latex] \cos \phi_{S C}=\frac{2,040}{\sqrt{3} imes 105 imes 28}=0.4[/latex]

[latex] \phi_{ SC }=66.4^{\circ}[/latex]

Rated output [latex] =15 kW =\sqrt{3} V(I \cos \phi)_{ ext {rated }}[/latex]

[latex] (I \cos \phi)_{ ext {rated }}=\frac{15,000}{\sqrt{3} imes 415}=20.87 A[/latex]

(a) The circle diagram is drawn in Fig. 9.30. [latex] P P^{\prime}[/latex] is drawn parallel to the output line [latex] P_{0} P_{S C}[/latex] at a vertical distance 20.87 A above [latex] P P_{S C}[/latex] The point P pertains to full load ([latex] P^{\prime}[/latex] is the second but unacceptable solution—too large a current):

Then,

[latex] I_{1}=O P=30.5 A[/latex]

[latex] pf =\cos \phi=\cos 30^{\circ}=0.866[/latex] lagging

(b) Torque line [latex] P_{0} F[/latex] is drawn by locating F as the midpoint of [latex] P_{S C} G [/latex] (equal stator and rotor, losses). The maximum torque point is located by drawing a tangent to the circle parallel to [latex] P_{0} F[/latex]. The maximum torque is given as

RS = 44 A

[latex] T_{\max }=3 imes \frac{145}{\sqrt{3}} imes 44=31,626 syn ext { watts }[/latex]

[latex] n_{s}=1,500 rpm , \quad \omega_{ s }=157.1 rad / s[/latex]

[latex] T_{\max }=\frac{31,626}{157.1}=201.3 Nm[/latex]

(c) If the motor is started to delta (this is the connection for which test data are given)

[latex] I_{s}( ext { delta })=O P_{S C}=110.7 A[/latex]

[latex] P_{S C} F=22 A[/latex]

[latex] T_{s}( ext { delta })=\frac{(\sqrt{3} imes 415 imes 22)}{157.1}=100.7 Nm[/latex]

Star connection [latex] I_{ s }( star )=\frac{100.7}{3}=33.6 A[/latex]

[latex] T_{ s }( star )=\frac{100.7}{3}=33.57 Nm[/latex]

Question 9.7: A 15 kW, 415 V, 4-pole, 50 Hz delta-connected motor gave the...

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