Model Answers

1.1 What is the magnetic circuit concept?

Solution: Similarly to electric circuits, there are magnetic circuits.

Instead of electromotive force (voltage) magneto motive force (mmf) is what drives magnetic circuits.

F = Ni

Direction of mmf is determined by RHR

Like the Ohm’s law, the Hopkinson’s Law:

F = φR

F – mmf, φ – magnetic flux, R – reluctance

1.2 What is magnetizing intensity? What is magnetic flux density? How are they related?

Solution: The Amperes law:

∫H .dl=Inet

Where H [A-turns/m] is the intensity of the magnetic field produced by the current Inet

For the ferromagnetic cores, almost all the magnetic field produced by the current remains inside the core, therefore the integration path would be lc and the current passes it N times.

Inet = Ni

H = N il

Magnetic flux density

B = μH= μN il

where μ is the magnetic permeability of a material.

μ0=4 π ×10−7 Hmis the permebilty of free space

r – the relative permeability

1.3 Why all cores exposed to ac flux variations laminated?

Solution: Voltages are generated within a ferromagnetic core by a time-changing magnetic flux same way as they are induced in a wire. These voltages cause currents flowing in the resistive material (ferromagnetic core) called eddy currents. Therefore, energy is dissipated by these currents in the form of heat

The amount of energy lost to eddy currents is proportional to the size of the paths they travel within the core. Therefore, ferromagnetic cores are frequently laminated: core consists of a set of tiny isolated strips. Eddy current losses are proportional to the square of the lamination thickness.

1.4 What is faradays law?

Solution: If a flux passes through a turn of a coil of wire, a voltage will be induced in that turn that is directly proportional to the rate of change in the flux with respect to time:

eind = -dφdt

Or, for a coil having N turns:

eind = -Ndφdt

eind – voltage induced in the coil

N – number of turns of wire in the coil

- magnetic flux passing through the coil

1.5 What is real, reactive, and apparent power? What the units are they measured in? How are they related?

Solution: The average value of the first term is the average or real power

P = VI cosθ Watt

The second term represents a reactive power: power that is transmitted to the load and then reflected back to the power source either through electric energy stored in a capacitor or through magnetic energy stored in an inductor.

The reactive power of a load is given by:

Q = VI sinθ

By the convention, Q is positive for inductive loads and negative for capacitive loads. Units are volt-amperes reactive (var).

The power that “appears” to be supplied to the load is called the apparent power:

S = VI (VA)

The real, reactive, and apparent power supplied to a load is related by the power triangle.

cos is usually called a power factor of the load:

1.6 Define the difference between LAP and WAVE windings in DC motors.

Solution: LAP winding: This type of winding is used in dc generators designed for high-current applications. The windings are connected to provide several parallel paths for current in the armature. For this reason, lap-wound armatures used in dc generators require several pairs of poles and brushes.

Wave windings: This type of winding is used in dc generators employed in high-voltage applications. Notice that the two ends of each coil are connected to commutator segments separated by the distance between poles. This configuration allows the series addition of the voltages in all the windings between brushes. This type of winding only requires one pair of brushes. In practice, a practical generator may have several pairs to improve commutation.

1.7 Define the turn ratio of an ideal transformer.

Solution: The ratio of the primary voltage to the secondary voltage both caused by the mutual flux is equal to the turns ratio of the transformer.

vp (t)vs (t)

=NpNs

=a

1.8 Define different types of losses in a real transformer accurately.

Solution: To model a real transformer accurately, we need to account for the following losses:

1. Copper losses – resistive heating in the windings: I2R.

2. Eddy current losses – resistive heating in the core: proportional to the square of voltage applied to the transformer.

3. Hysteresis losses – energy needed to rearrange magnetic domains in the core: nonlinear function of the voltage applied to the transformer.

4. Leakage flux – flux that escapes from the core and flux that passes through one winding only.

1.9 List and define three major losses in DC motors.

1.10 Compare between the three types of DC motors

3. A ferromagnetic core with a mean path length of 40 cm, an air gap of 0.05 cm, a cross-section 12 cm2, and r = 4000 has a coil of wire with 400 turns. Assume that fringing in the air gap increases the cross-sectional area of the gap by 5%, find

(a) The total reluctance of the system (core and gap),

(b) The current required to produce a flux density of 0.5 T in the gap.

(a) The reluctance of the core:

Rc= lcμAc

= lcμoμrAc

= 0.14000∗0.0012∗4 π∗10−7=66300 A−turns/wb

Since the effective area of the air gap is 1.05 x 12 = 12.6 cm2, its reluctance:

Ra= laμoAa

= 0.00050.00126∗4 π∗10−7=316000 A−turns/wb

The total reluctance:

Req = Rc + Ra = 66300 + 316000 = 382300 A-turns/wb The air gap contribute most of the reluctance!

(b) The mmf: F = φR = Ni = BARTherefore:

i= ¿N

=0.5∗0.00126∗382300400

=0.602 A ¿

Since the air gap flux was required, the effective area of the gap was used.

4. In a simplified rotor and stator motor, the mean path length of the stator is 50 cm, its cross-sectional area is 12 cm2, and r = 2000. The mean path length of the rotor is 5 cm and its cross-sectional area is also 12 cm2, and r = 2000. Each air gap is 0.05 cm wide, and the cross-section of each gap (including fringing) is 14 cm2. The coil has 200 turns of wire. If the current in the wire is 1A, what will the resulting flux density in the air gaps be?

Solution: The reluctance of the stator is:

Rc= lcμoμrAc

= 0.52000∗0.0012∗4 π∗10−7

=166000 A−turns /wb

The reluctance of the rotor is:

Rr= lrμoμrAr

= 0.052000∗0.0012∗4 π∗10−7=16600 A−turns/wb

The reluctance of each gap is:

Ra= laμoAa

= 0.00050.0014∗4 π∗10−7=284000 A−turns/wb

The total reluctance:

Req = Rc + Ra1 + Rr + Ra2 = 751000 A-turns/wb

Finally, the magnetic flux density in the gap is:

B= φA

= ¿RA

= 200∗1751000∗0.0014

=0.19T

6. We need to determine the equivalent circuit impedances of a 20 kVA, 8000/240 V, 60 Hz transformer. The open-circuit and short-circuit tests led to the following data:

VOC = 8000 V VSC = 489 V

IOC = 0.214 A ISC = 2.5 A

POC = 400 W PSC = 240 W

Solution: The power factor during the open-circuit test is

PF=cosθ= POCVOC∗IOC

= 4008000∗0.214

=0.234 lagging

The excitation admittance is

YE= IOCVOC

←cos−1PF=0.2148000

←cos−10.234=0.0000063− j 0.0000261= 1RC

− j 1XM

Therefore: RC= 10.0000063

=159 kΩ; XM= 10.0000261

=38.3 kΩ

The power factor during the short-circuit test is

PF=cosθ= PSCVSC∗ISC

= 240489∗2.5

=0.196 lagging

The series impedance is given by

ZSC=VSCISC

<cos−1PF=4892.5

<78.7 °=38.4+ j192Ω

Therefore: Req = 38.3 Ω; Xeq = 192Ω

The equivalent circuit

Solution:

Solution: