Consider the motor of Fig. 7.2. Establish the following points (|Ia| and |Ef|) on the V curve corresponding to full-load power output: (a) Stability limit, δ = - 90 deg. (b) Power factor of 0.8 Igging. (c) Unity power factor. (d) Power factor of 0.8 leading.

Question

Consider the motor of Fig. 7.2. Establish the following points ([latex]\ (|I_{a}|[/latex] and [latex]\mid E_{f}\mid[/latex]) on the V curve corresponding to full-load power output: (a) Stability limit, δ = - 90 deg. (b) Power factor of 0.8 Igging. (c) Unity power factor. (d) Power factor of 0.8 leading.

Accepted Answer

We recall the following specifications from Example (7.2) on a per-phase
basis:

[latex]V_{t}={\frac{2300}{\sqrt{3}}}=1327.91\ \mathrm{V} \ P={\frac{764 imes10^{3}}{3}}\ \mathrm{W} \ X_{s}=1.9\;\Omega[/latex]

As a result, we have

[latex]{\frac{764 imes10^{3}}{3}}={\frac{-1327.91}{1.9}}|E_{f}|\sin\delta[/latex]

Thus

[latex]|E_{f}|\sin\delta=-364.38\ \mathrm{V}[/latex]

(a) For δ = -90 deg, we get

[latex]|E_{f}|=364.38\ \mathrm{V}[/latex]

We find [latex]I_{a}[/latex] by using

[latex]I_{a}={\frac{V_{t}-E_{f}}{j X_{s}}}={\frac{1327.91-364.38\angle-90\mathrm{~deg}}{j1.9}} \ =724.73\angle-74.66\ \mathrm{deg}\ \mathrm{A}[/latex]

(b) For [latex]\phi=-\cos^{-1}0.8\;=\;-\,36.87\;\mathrm{deg},[/latex]

[latex]I_{a}=\frac{764 imes10^{3}}{{\sqrt3}(2300)(0.8)}=239.73\angle-36.87\ \mathrm{deg}\ \mathrm{A}[/latex]

We find [latex]E_{f}[/latex] from

[latex]E_{f}=V_{t}-j I_{a}X_{s} \ =1327.91-(239.73\angle-36.87\ \mathrm{deg})(1.9)\angle90\ \mathrm{deg} \ =1115.793\angle-19.061\ \mathrm{deg}\ \mathrm{V}[/latex]

(c) For unity power factor

[latex]I_{a}=\frac{764 imes10^{3}}{2300\sqrt{3}}=191.78\angle0\,\mathrm{A}[/latex]

We can get the same result from

[latex]-|I_{a}|X_{s}=|E_{f}|\sin\delta=-364.38\,\mathrm{V}[/latex]

The value of [latex] extstyle E_{f}[/latex] is obtained from

[latex]E_{f}=1327.91-191.78(1.9)\angle90\ \mathrm{deg} \ =1377\angle-15.34\ \mathrm{deg}\,\mathrm{V}[/latex]

(d) For a power factor of 0.8 leading, we have

[latex]I_{a}=239.73\angle36.87\ \mathrm{deg}\ \mathrm{A} \ E_{f}=1327.91-(239.73\angle36.87\ \mathrm{deg})(1.9)\angle90\ \mathrm{deg} \ =1642.17\angle-12.82\ \mathrm{degV}[/latex]

This concludes this example. We now discuss methods for starting a synchronous motor.

Question 7.9: Consider the motor of Fig. 7.2. Establish the following poin......

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