Electrical Machines Problems And Solutions

This book contains problems in Electrical Machines & Powert Systems (Problems with Solutions). I have used these and other problems in the class room for many years. In most of the solutions I have deliberately avoided giving theoretical explantions, because an average student should know the theyr well before attempting to solve any proble. SOLVED PROBLEMS Example 1: A source which can be represented by a voltage source of 8 V rms in series with an internal resistance of 2 kΩ is connected to a 50-Ω load resistance through an ideal. The problems cover every area of the electrical circuits, from basic modules to complex multi-phase circuits, port-based networks, and the use of Laplace transforms. Go directly to the answers and charts you need through a detailed index and reference.

SOLVED PROBLEMS

Example 1:

A source which can be represented by a voltage source of 8 V rms in series with an internal resistance of 2 kΩ is connected to a 50-Ω load resistance through an ideal transformer. Calculate the value of turns ratio for which maximum power is supplied to the load and the corresponding load power? Using MATLAB, plot the the power in milliwatts supplied to the load as a function of the transformer ratio, covering ratios from 1.0 to 10.0.

Solution:

For maximum power transfer, the load resistance (referred to the primary) must be equal to the source resistance.

Example 2

Problems And Solutions In Electrical Machines And Transformers Pdf

A 460-V:2400-V transformer has a series leakage reactance of 37.2 Ω as referred to the high-voltage side. A load connected to the low-voltage side is observed to be absorbing 25 kW, unity power factor, and the voltage is measured to be 450 V. Calculate the corresponding voltage and power factor as measured at the high-voltage terminals.

Power factor at primary terminals: cos(9.58 ) = 0.9861 lagging

Example 3:

The resistances and leakage reactances of a 30-kVA, 60-Hz, 2400-V:240-V distribution transformer are

R₁ = 0.68 Ω

R₂ = 0.0068 Ω

X_l1 = 7.8 Ω

X_l2 = 0.0780 Ω

where subscript 1 denotes the 2400-V winding and subscript 2 denotes the 240-V winding. Each quantity is referred to its own side of the transformer.

a.Draw the equivalent circuit referred to (i) the high- and (ii) the low-voltage sides. Label the impedances numerically.

b.Consider the transformer to deliver its rated kVA to a load on the low-voltage side with 230 V across the load. (i) Find the high-side terminal voltage for a load power factor of 0.85 lagging. (ii) Find the high-side terminal voltage for a load power factor of 0.85 leading.

c.Consider a rated-kVA load connected at the low-voltage terminals operating at 240V. Use MATLAB to plot the high-side terminal voltage as a function of the power-factor angle as the load power factor varies from 0.6 leading through unity power factor to 0.6 pf lagging.

Electrical Machines And Drives

Example 4:

A single-phase load is supplied through a 35-kV feeder whose impedance is 95 + j360 Ω and a 35-kV:2400-V transformer whose equivalent impedance is (0.23 + j1.27) Ω referred to its low-voltage side. The load is 160 kW at 0.89 leading power factor and 2340 V.

a.Compute the voltage at the high-voltage terminals of the transformer.

b.Compute the voltage at the sending end of the feeder.

Compute the power and reactive power input at the sending end of the feeder.

Example 5:

The following data were obtained for a 20-kVA, 60-Hz, 2400:240-V distribution transformer tested at 60 Hz:

a.Compute the efficiency at full-load current and the rated terminal voltage at 0.8 power factor.

b.Assume that the load power factor is varied while the load current and secondary terminal voltage are held constant. Use a phasor diagram to determine the load power factor for which the regulation is greatest. What is this regulation?

Solution:

(a)Rated current on the HV side = 20 kVA / 2400 = 8.33 A. Therefore, total power loss at full load current:

Example 6:

A three-phase generator step-up transformer is rated 26-kV:345-kV, 850 MVA and has a series impedance of 0.0035 + j0.087 per unit on this base. It is connected to a 26-kV, 800-MVA generator, which can be represented as a voltage source in series with a reactance of j1.57 per unit on the generator base.

(a)Convert the per unit generator reactance to the step-up transformer base.

(b)The unit is supplying 700 MW at 345 kV and 0.95 power factor lagging to the system at the transformer high-voltage terminals.

(i)Calculate the transformer low-side voltage and the generator internal voltage behind its reactance in kV.

(ii)Find the generator output power in MW and the power factor.

Solution:

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Solution Manual for Analysis of Electric Machinery and Drive Systems – 2nd and 3rd Edition

Author(s): Paul C. Krause, Oleg Wasynczuk, Scott D. Sudhoff, Steven Pekarek

Solution manual for 2nd edition includes all problems(From chapter 1 to chapter 15). Most of problems are answered. List of solved problems (Exercise, Discussion Question and …) exist in following.

Download Sample for 2nd edition

Solution manual for 3rd edition includes all problems(From chapter 1 to chapter 15). Most of problems are answered. List of solved problems (Exercise, Discussion Question and …) exist in following.

Download Sample for 3rd edition

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Total Number of Solved Problems	81
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