A 100-hp 440-V 0.8-PF-leading Δ-connected synchronous motor has an armature resistance of 0.3 Ω and a synchronous reactance of 4.0 Ω. Its efficiency at full load is 96 percent. (a) What is the input power to the motor at rated conditions? (b) What is the line current of the motor at rated

Question

A 100-hp 440-V 0.8-PF-leading [latex] \Delta [/latex] -connected synchronous motor has an armature resistance of 0.3 Ω and a synchronous reactance of 4.0 Ω . Its efficiency at full load is 96 percent.

(a) What is the input power to the motor at rated conditions?

(b) What is the line current of the motor at rated conditions? What is the phase current of the motor at rated conditions?

(c) What is the reactive power consumed by or supplied by the motor at rated conditions?

(d) What is the internal generated voltage [latex] \mathbf{ E }_{ A } [/latex] of this motor at rated conditions?

(e) What are the stator copper losses in the motor at rated conditions?

(f) What is [latex] { P }_{ conv } [/latex] at rated conditions?

(g) If [latex] { E }_{ A } [/latex] is decreased by 10 percent, how much reactive power will be consumed by or supplied by the motor?

Accepted Answer

(a) The input power to the motor at rated conditions is

[latex]P_{\mathrm{IN}}=\frac{P_{ ext {OUT }}}{\eta}=\frac{(100 \mathrm{\ hp})(746 \mathrm{~W} / \mathrm{hp})}{0.96}=77.7 \mathrm{~kW}[/latex]

(b) The line current to the motor at rated conditions is

[latex]I_L=\frac{P}{\sqrt{3} V_T \mathrm{\ PF}}=\frac{77.7 \mathrm{~kW}}{\sqrt{3}(440 \mathrm{~V})(0.85)}=120 \mathrm{~A}[/latex]

The phase current to the motor at rated conditions is

[latex]I_\phi=\frac{I_L}{\sqrt{3}}=\frac{120 \mathrm{~A}}{\sqrt{3}}=69.3 \mathrm{~A}[/latex]

(c) The reactive power supplied by this motor to the power system at rated conditions is

[latex]Q_{ ext {nated }}=3 V_\phi I_A \sin heta=3(440 \mathrm{~V})(69.3 \mathrm{~A}) \sin 36.87^{\circ}=54.9 \mathrm{\ kVAR}[/latex]

(d) The internal generated voltage at rated conditions is

[latex]\begin{aligned}& \mathbf{E}_A=\mathbf{V}_\phi-R_A \mathbf{I}_A-j X_S \mathbf{I}_A \& \mathbf{E}_A=440 \angle 0^{\circ} \mathrm{V}-(0.3 \Omega)\left(69.3 \angle 36.87^{\circ} \mathrm{A} ight)-j(4.0 \Omega)\left(69.3 \angle 36.87^{\circ} \mathrm{A} ight) \& \mathbf{E}_A=634 \angle-21.7^{\circ} \mathrm{V}\end{aligned}[/latex]

(e) The stator copper losses at rated conditions are

[latex]P_{\mathrm{CU}}=3 I_A{ }^2 R_A=3(69.3 \mathrm{~A})^2(0.3 \ \Omega)=4.3 \mathrm{~kW}[/latex]

(f) [latex]P_{ ext {conv }}[/latex] at rated conditions is

[latex]P_{ ext {conv }}=P_{\mathrm{IN}}-P_{\mathrm{CU}}=77.7 \mathrm{~kW}-4.3 \mathrm{~kW}=73.4 \mathrm{~kW}[/latex]

(g) If [latex]E_A[/latex] is decreased by 10%, the new value if [latex]E_A[/latex] = (0.9)(634 V) = 571 V. To simplify this part of the problem, we will ignore [latex]R_A[/latex] . Then the quantity [latex]E_A \sin \delta[/latex] will be constant as [latex]E_A[/latex] changes. Therefore,

[latex]\delta_2=\sin ^{-1}\left\lgroup\frac{E_{A 1}}{E_{A 2}} \sin \delta_1 ight group=\sin ^{-1}\left\lgroup\frac{634 \mathrm{~V}}{571 \mathrm{~V}} \sin \left(-21.7^{\circ} ight) ight group=-24.2^{\circ}[/latex]

Therefore,

[latex]\mathbf{I}_A=\frac{\mathbf{V}_\phi-\mathbf{E}_A}{j X_S}=\frac{440 \angle 0^{\circ} \mathrm{V}-571 \angle-24.2^{\circ}}{j 4.0}=61.9 \angle 19^{\circ} \mathrm{A}[/latex]

and the reactive power supplied by the motor to the power system will be

[latex]Q=3 V_\phi I_A \sin heta=3(440 \mathrm{~V})(61.9 \mathrm{~A}) \sin \left(19^{\circ} ight)=26.6 \mathrm{\ kVAR}[/latex]

Question 5.19: A 100-hp 440-V 0.8-PF-leading Δ-connected synchronous motor ...

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