Question 5.1: A 60-Hz, three-phase synchronous motor is observed to have a......

a. Using the motor reference direction for the current and neglecting the armature resistance, the generated voltage can be found from the equivalent circuit of Fig. 5.3a or Eq. 5.23 as

\hat{V}_a=R_a\hat{I}_a+  jX_s \hat{I}_a+\hat{E}_{af}            (5.23)

\hat{E}_{af} = \hat{V}_a  –  jX_s\hat{I}_a

We will choose the terminal voltage as our phase reference. Because this is a line-to- neutral equivalent, the terminal voltage V_a must be expressed as a line-to-neutral voltage

\hat{V}_a =\frac{460}{\sqrt{3}}= 265.6  V, line-to-neutral

 A lagging power factor of 0.95 corresponds to a power factor angle θ = – \cos ^{-1} (0.95) = – 18.2°. Thus, the phase-a current is

\hat{I}_a = 120  e ^{-j18.2°}  A

Thus

\begin{aligned}\hat{E}_{af} &= 265.6  –  j 1.68(120  e ^{-j18.2°}) \\&= 278.8  e^{-j43.4°}  V, \text{line-to-neutral}\end{aligned}

and hence, the generated voltage E_{af} is equal to 278.8 V rms, line-to-neutral.

b. The field-to-armature mutual inductance can be found from Eq. 5.21. With ω_e = 120π,

E_{af}=\frac{ω_eL_{af}I_f}{\sqrt{2}}            (5.21)

L_{af}= \frac{\sqrt{2}E_{af}}{ω_e I_f}= \frac{\sqrt{2}×279}{120π×47}=22.3  mH

c. The three-phase power input to the motor P_{in} can be found as three times the power input to phase a. Hence,

\begin{aligned}P_{in}&=3V_aI_a \text{(power factor)} = 3 × 265.6 × 120 × 0.95\\& = 90.8  kW = 122  hp\end{aligned}