###### Principles of Electric Machines with Power Electronic Applications

70 SOLVED PROBLEMS

Question: 7.9

## 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. ...

We recall the following specifications from Exampl...
Question: 8.A.3

## For the motor of Problem 8.A.2, assume that the stator impedance is given by Z1 = 1.86 +j2.56 Ω Find the internal mechanical power, output power, power factor, input power, developed torque, and efficiency, assuming that friction losses are 15 W. ...

We need the backward-field impedance Z_{b}=...
Question: 8.A.2

## The forward-field impedance of a one-quarter-hp four-pole 110-V 60-Hz single-phase induction motor for a slip of 0.05 is given by Zf = 12.4 + j16.98 Ω Assume that Xm = 53.5 Ω Find the values of the rotor resistance and reactance. ...

We have Z_{f}=\frac{j0.5X_{m}[0.5(R_{2}^{\p...
Question: 8.A.1

## A single-phase induction motor takes an input power of 490 W at a power factor of 0.57 lagging from a 110-V supply when running at a slip of 5%. Assume that the rotor resistance and reactance are 1.78 Ω and 1.28 Ω, respectively, and that the magnetizing reactance is 25 Ω. Find the resistance and ...

The equivalent circuit of the motor yields ...
Question: 8.2

## For the single-phase induction motor of Example 8.1, it is required that the power and torque output, and the efficiency when running at a slip of 5%, be found. Neglect core and rotational losses. ...

In Example 8.1, we obtained Z_{i}=26.841\an...
Question: 8.1

## The following parameters are available for a 60-Hz four-pole single-phase 110-V 1/2-hp induction motor. R1 = 1.5 Ω X1 = 2.4 Ω Xm = 73.4 Ω R’2 = 3 Ω X’2 = 2.4 Ω Calculate Zf, Zb, and the input impedance of the motor at a slip of 0.05. ...

Z_{f}=\frac{j\,36.7(30+j1.2)}{30+j37.9}=22....
Question: 7.13

## Asynchronous machine is supplied from a constant-voltage source. At no load, the motor armature current is found to be negligible when the excitation is 1.0 p.u. The p.u. motor constants are Xd = 1.0 and xq = 0.6. (a) If the machine loses synchronism when the angle between the quadrature axis and ...

(a) P=\frac{V E_{f}}{X_{d}}\,\mathrm{sin}\,...
Question: 7.12

## The machine of Example 7.11 is connected to an infinite bus through a link with reactance of 0.2 p.u. The excitation voltage is 1.3 p.u. and the infinite-bus voltage is maintained at 1 p.u. For a power angle of 25 deg, compute the active and reactive power supplied to the bus. ...

We have \begin{array}{l}{{X_{d}=0.95+0.2=1....
Question: 7.11

## The reactances Xd and xq of a salient-pole synchronous generator are 0.95 and 0.7 p.u., respectively. The armature resistance is negligible. The generator delivers rated kVA at unity PF and rated terminal voltage. Calculate the excitation voltage. ...

We apply E_{f}^{\prime}=V_{t}+j I_{a}x_{q}[...