Question 7.10: An 800-hp three-phase induction motor operated at 0.7 PF lag......
An 800-hp three-phase induction motor operated at 0.7 PF lagging from a 2300-V supply.
(a) Find the line current.
(b) The synchronous motor of Example 7.2 is connected in parallel with the induction motor and is operated as a capacitor (zero power factor). Find the necessary excitation voltage to improve the line power factor to 0.9 lagging.
(c) Repeat part (b) for an improved power factor of 0.95 lagging.
(a) The line current with the induction motor operating alone is
I_{L_{0}}={\frac{800\times746}{{\sqrt{3}}(2300)(0.7)}} \\ =214.01\angle-45.57\ \mathrm{deg}\ \mathrm{A}(b) To improve the line power factor to 0.9 PF lagging, we have the phasor diagram of Fig. 7.26. The armature current of the synchronous motor is denoted by I_{c} and the line current is I_{L_{1}} .Thus
I_{L_{1}}=I_{L_{0}}+I_{c}As a result,
|I_{L_{1}}|\angle-\mathrm{cos}^{-1}\,0.9=214.01\angle-45.57\ \mathrm{deg}+j|I_{c}|The real part of the foregoing yields
0.9|I_{L1}|=149.81Thus
|I_{L_{1}}|=166.46\,\mathrm{A}The imaginary part yields
0.44|I_{L_{1}}|= 152.84 – |I_{c}|Thus
|I_{c}|=80.28\,\mathrm{A}To obtain the required excitation voltage, we use
E_{f}=V_{t}-j I X_{s} \\ ={\frac{2300}{\sqrt{3}}}-j(80.28\angle90\ \mathrm{deg})(1.9)\\ ={1480.44}\,\mathrm{V}(c) To improve the power factor to 0.95 lagging, we repeat the foregoing to obtain
\begin{array}{c}{{|I_{L_{2}}|=\,157.69\,\mathrm{A}}}\\{{|I_{c}|=\,103.6\,\mathrm{A}}}\end{array}and
E_{f}=1524.74\,\mathrm{V}We note that the magnitude of the line current is reduced by inserting the synchronous capacitor. A higher improvement in power factor requires a larger field excitation.
