Ee 313-dc machinery fundamentals (part1)

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EE 313 Electric Machines Fall 2014

Chapter 8: DC Machinery Fundamentals

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Preliminary notes

DC power systems are not very common in the contemporary engineering practice. However, DC motors still have many practical applications, such automobile, aircraft, and portable electronics, in speed control applications…

An advantage of DC motors is that it is easy to control their speed in a wide diapason.

DC generators are quite rare.

Most DC machines are similar to AC machines: i.e. they have AC voltages and current within them. DC machines have DC outputs just because they have a mechanism converting AC voltages to DC voltages at their terminals. This mechanism is called a commutator; therefore, DC machines are also called commutating machines.

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The simplest DC machine

The simplest DC rotating machine consists of a single loop of wire rotating about a fixed axis. The magnetic field is supplied by the North and South poles of the magnet.Rotor is the rotating part;Stator is the stationary part.

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We notice that the rotor lies in a slot curved in a ferromagnetic stator core, which, together with the rotor core, provides a constant-width air gap between the rotor and stator.The reluctance of air is much larger than the reluctance of core. Therefore, the magnetic flux must take the shortest path through the air gap.

As a consequence, the magnetic flux is perpendicular to the rotor surface everywhere under the pole faces.

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1. Voltage induced in a rotating loop

If a rotor of a DC machine is rotated, a voltage will be induced…

The total voltage will be a sum of voltages induced on each segment of the loop.

Voltage on each segment is:

inde v×B l

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1) ab: velocity v is perpendicular to the magnetic field B, and the vector product v x B points into the page. Therefore, the voltage is

into page under the pole face0 beyond the pole edgesba

vBle

v×B l

2) bc: In this segment, vector product v x B is perpendicular to l. Therefore, the voltage is zero.

3) cd: In this segment, the velocity of the wire is tangential to the path of rotation. Under the pole face, velocity v is perpendicular to the magnetic flux density B, and the vector product v x B points out of the page. Therefore, the voltage is

out of page under the pole face

0 beyond the pole edgesdc

vBle

v×B l

4) da: In this segment, vector product v x B is perpendicular to l. Therefore, the voltage is zero.

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The total induced voltage on the loop is:

tot ba cb dc ade e e e e

20tot

vBl under the pole facese

beyond the pole edges

When the loop rotates through 1800, segment ab is under the north pole face instead of the south pole face. Therefore, the direction of the voltage on the segment reverses but its magnitude remains constant, leading to the total induced voltage to be

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The tangential velocity of the loop’s edges is

v r

20tot

r Bl under the pole facese


where r is the radius from the axis of rotation to the edge of the loop. The total induced voltage:

The rotor is a cylinder with surface area 2rl. Since there are two poles, the area of the rotor under each pole is Ap = rl. Therefore:

2

0

ptot

A B under the pole facese


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Assuming that the flux density B is constant everywhere in the air gap under the pole faces, the total flux under each pole is

pA B The total voltage is

2

0tot

under the pole facese


The voltage generated in any real machine depends on the following factors:

1. The flux inside the machine;2. The rotation speed of the machine;3. A constant representing the construction of the machine.

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2. Getting DC voltage out of a rotating loop

A voltage out of the loop is alternatively a constant positive value and a constant negative value. One possible way to convert an alternating voltage to a constant voltage is by adding aCommutator segment/brush circuitry to the end of the loop.Every time the voltage of the loop switches direction, contacts switch connection.

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2. Getting DC voltage out of a rotating loop

A voltage out of the loop is alternatively a constant positive value and a constant negative value. One possible way to convert an alternating voltage to a constant voltage is by adding acommutator segment/brush circuitry to the end of the loop.Every time the voltage of the loop switches direction, contacts switch connection.

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Commutator with carbon brushes

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Commutator with carbon brushes

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Commutator segments

Carbon brushes Spring

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3. The induced torque in the rotating loop

Assuming that a battery is connected to the DC machine, the force on a segment of a loop is:

F i l ×BAnd the torque on the segment is

sinrF Where is the angle between r and F. Therefore, the torque is zero when the loop is beyond the pole edges.

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When the loop is under the pole faces:

1. Segment ab:

2. Segment bc:

3. Segment cd:

4. Segment da:

abF i ilB l×B

sin sin 90ab rF r ilB rilB ccw

0bcF i l ×B

sin 0bc rF

cdF i ilB l×B

sin sin90cd rF r ilB rilB ccw

0daF i l×B

sin 0da rF

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The resulting total induced torque is

The torque in any real machine depends on the following factors:1. The flux inside the machine;2. The current in the machine;3. A constant representing the construction of the machine.

2

0ind

i under the pole faces


ind ab bc cd da

20ind

rilB under the pole facesbeyond the pole edges

Since Ap rl and pA B

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Example: 8.1

Figure shows a simple rotating loop between curved pole faces connected to a battery and a resistor through a switch. The resistor shown models the total resistance of the battery and the wire in the machine. The physical dimensions and characteristics of this machine are:r = 0.5m l = 1.0m R = 0.3 B = 0.25T VB = 120V

a) What happens when the switch is closedb) What is the machine’s max. starting current? What is its

steady state angular velocity at no load?c) Suppose a load is attached to the loop, and the resulting

load torque is 10N.m. What would the new steady state speed be? How much power is supplied to the shaft of the machine? How much power is being supplied by the battery? Is this machine a motor or a generator?

d) Suppose the machine is again unloaded, and a torque of 7.5 N.m is applied to the shaft in the direction of rotation. What is the new steady state speed? Is this machine now a motor or a generator?

e) Suppose the machine is running unloaded. What would the final steady state speed of the rotor be if the flux density were reduced to 0.20 T?

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Rotor (Armature Core)

Armature core is a part of rotor on which conductors (Armature Winding) are wound. It consists of thin laminations.

The main parts of an armature Core are;Slots :Tooth: Shaft:Bearings and commutator segments are fixed on shaft of

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Rotor with Armature Winding

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A wound armature core

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Commutation in a simple 4-loop DC machine

Commutation is the process of converting the AC voltages and currents in the rotor of a DC machine to DC voltages and currents at its terminals.

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A simple 4-loop DC machine has four complete loops buried in slots curved in the laminated steel of its rotor. The pole faces are curved to make a uniform air-gap. The four loops are laid into the slots in a special manner: the innermost wire in each slot (end of each loop opposite to the “unprimed”) is indicated by a prime.

Loop 1 stretches between commutator segments a and b, loop 2 stretches between segments b and c…


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At a certain time instance, when t = 00, the 1, 2, 3’, and 4’ ends of the loops are under the north pole face and the 1’, 2’, 3, and 4 ends of the loops are under the south pole face. The voltage in each of 1, 2, 3’, and 4’ ends is given by

positive, out of the page

inde vBl

v×B ×l

positive, into the pageinde vBl v×B ×lThe voltage in each of 1’, 2’, 3, and 4 ends is

The total voltage at the brushes of the DC machine is

4 0E e at t

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If the machine keeps rotating, at t = 450, loops 1 and 3 have rotated into the gap between poles, so the voltage across each of them is zero. At the same time, the brushes short out the commutator segments ab and cd.

This is ok since the voltage across loops 1 and 3 is zero and only loops 2 and 4 are under the pole faces. Therefore, the total terminal voltage is

2 45E e at t

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Magnetic Field path due to air and curved shape of Poles

Neutral plane of magnetic field

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At t = 900, the loop ends 1’, 2, 3, and 4’ are under the north pole face, and the loop ends 1, 2’, 3’, and 4 are under the south pole face. The voltages are built up out of page for the ends under the north pole face and into the page for the ends under the south pole face. Four voltage-carrying ends in each parallel path through the machine lead to the terminal voltage of

4 90E e at t

We notice that the voltages in loops 1 and 3 have reversed compared to t = 00. However, the loops’ connections have also reversed, making the total voltage being of the same polarity.

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When the voltage reverses in a loop, the connections of the loop are also switched to keep the polarity of the terminal voltage the same.

The terminal voltage of this 4-loop DC machine is still not constant over time, although it is a better approximation to a constant DC level than what is produced by a single rotating loop.

Increasing the number of loops on the rotor, we improve our approximation to perfect DC voltage.

Commutator segments are usually made out of copper bars and the brushes are made of a mixture containing graphite to minimize friction between segments and brushes.

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Example of a commutator…

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8.6: Construction of DC Machines

All d.c. machines have five principal components viz., (i) Field system (Stator) (ii) Armature core(iii) Armature winding(iv) Commutator (v) Brushes

(i) Field systemThe function of the field system is to produce uniform magnetic field within

which the armature rotates. It consists of a number of poles bolted to the inside of circular frame (generally called yoke).

The field coils are connected in such a way that adjacent poles have opposite polarity.

The m.m.f. developed by the field coils produces a magnetic flux that passes through the pole pieces, the air gap, the armature and the frame.

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Construction of DC Machines

(ii) Armature core The armature core is keyed to the machine shaft and rotates between the field

poles. It consists of slotted soft-iron laminations (about 0.4 to 0.6 mm thick) that are stacked to form a cylindrical core as shown in Fig.

The laminations are individually coated with a thin insulating film so that they do not come in electrical contact with each other. The purpose of laminating the core is to reduce the eddy current loss.

The laminations are slotted to accommodate and provide mechanical security to the armature winding and to give shorter air gap for the flux to cross between the pole face and the armature “teeth”.

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(iii) Armature winding The slots of the armature core hold insulated conductors that are connected in

a suitable manner. This is known as armature winding. This is the winding in which “working” e.m.f. is induced.

The armature conductors are connected in series-parallel; the conductors being connected in series so as to increase the voltage and in parallel paths so as to increase the current.

The armature winding of a d.c. machine is a closed-circuit winding; the conductors being connected in a symmetrical manner forming a closed loop or series of closed loops.

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(iv) Commutator A commutator is a mechanical rectifier which converts the alternating voltage

generated in the armature winding into direct voltage across the brushes.

The commutator is made of copper segments insulated from each other and mounted on the shaft of the machine.

The armature conductors are soldered to the commutator segments in a suitable manner.

Great care is taken in building the commutator because any short circuiting will cause the brushes to bounce, producing unacceptable sparking. The sparks may burn the brushes and overheat.

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Construction of DC Machines(v) Brushes The purpose of brushes is to ensure electrical

connections between the rotating commutator and stationary external load circuit. The brushes are made of carbon and rest on the commutator.

The brush pressure is adjusted by means of adjustable springs. If the brush pressure is very large, the friction produces heating of the commutator and the brushes. On the other hand, if it is too weak, the imperfect contact with the commutator may produce sparking.

Multipole machines have as many brushes as they have poles. For example, a 4-pole machine has 4 brushes.

The successive brushes have positive and negative polarities. Brushes having the same polarity are connected together so that we have two terminals viz., the +ve terminal and the -ve terminal.

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Armature winding

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Commutator and Armature Winding Terminologies

Commutator Pitch (YC) The commutator pitch is the number of commutator segments spanned by

each coil of the winding. It is denoted by YC.

In Fig.1 one side of the coil is connected to commutator segment 1 and the other side connected to commutator segment 2. Therefore, the number of commutator segments spanned by the coil is 1 i.e., YC = 1.

In Fig. 2, one side of the coil is connected to commutator segment 1 and the other side to commutator segment 8. Therefore, the number of commutator segments spanned by the coil = 8 - 1 = 7 segments i.e., YC = 7.

The commutator pitch of a winding is always a whole number. Since each coil has two ends and as two coil connections are joined at each commutator segment,

Number of coils = Number of commutator segments

For example, if an armature has 30 conductors, the number of coils will be 30/2 = 15. Therefore, number of commutator segments is also 15. Note that commutator pitch is the most important factor in determining the type of d.c. armature winding.

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Pole-PitchIt is the distance measured in terms of number of armature slots (or armatureconductors) per pole. Thus if a 4-pole generator has 16 coils, then number ofslots = 16.

Coil Span or Coil Pitch (YS)

It is the distance measured in terms of the number of armature slots (or armature conductors) spanned by a coil.

Thus if the coil span is 9 slots, it means one side of the coil is in slot 1 and the other side in slot 10.

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Full-Pitched Coil If the coil-span is equal to pole pitch, it is called full-pitched

coil. In this case, the e.m.f.s in the coil sides are additive and have a

phase difference of 0°. Therefore, e.m.f. induced in the coil is maximum.

If e.m.f. induced in one coil side is 2.5 V, then e.m.f. across the coil terminals = 2 x 2.5 = 5 V.

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Fractional pitched coil If the coil span or coil pitch is less than the pole

pitch, then it is called fractional pitched coil. In this case, the phase difference between the e.m.f.s in

the two coil sides will not be zero so that the e.m.f. of the coil will be less compared to full-pitched coil.

Fractional pitch winding requires less copper but if the pitch is too small, an appreciable reduction in the generated e.m.f. results

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Back Pitch (YB) (start to end) It is the distance measured in terms of armature conductors

between the two sides of a coil at the back of the armature . For example, if a coil is formed by connecting conductor 1

(upper conductor in a slot) to conductor 12 (bottom conductor in another slot) at the back of the armature, then back pitch is YB = 12 - 1 = 11 conductors.

Front Pitch (YF) (End of first and start of second coil) It is the distance measured in terms of armature

conductors between the coil sides attached to any one commutator segment .

For example, if coil side 12 and coil side 3 are connected to the same commutator segment, then front pitch is:

YF = 12 - 3 = 9 conductors.

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Front and Back Pitch

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Resultant Pitch (YR) It is the distance (measured in terms of armature

conductors) between the beginning of one coil and the

beginning of the next coil to which it is connected

Therefore,

YR = YF + YB

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Progressive WindingA progressive winding is one in which, as one traces through the winding, the connections to the commutator will progress around the machine in the same direction as is being traced along the path of each individual coil.Note that YB > YF and YC = + 1.

Retrogressive WindingA retrogressive winding is one in which, as one traces through the winding, the connections to the commutator will progress around the machine in the opposite direction to that which is being traced along the path of each individual coil.Note that YF > YB and YC = - 1. A retrogressive winding is seldom used because it requires more copper.

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General Rules For D.C. Armature Windings

1) The back pitch (YB) as well as front pitch (YF) should be nearly equal to pole pitch YP. This will result in increased e.m.f. in the coils.

2) Both pitches (YB and YF) should be odd. This will permit all end connections (back as well as front connection) between a conductor at the top of a slot and one at the bottom of a slot.

3) The number of commutator segments is equal to the number of slots or coils (or half the number of conductors).

No. of commutator segments = No. of slots = No. of coils

It is because each coil has two ends and two coil connections are joined at each commutator segment

4) The winding must close upon itself i.e. it should be a closed circuit winding.

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Simplex Lap WindingRelations between Pitches

2) Both YB and YF should be >= to pole pitch (YP).3) Average pitch =(YB + YF) / 2. It nearly equals pole

pitch (YP= Z/P).4) Commutator pitch, YC = ± 1

YC = + 1 for progressive windingYC = - 1 for retrogressive winding

5) The resultant pitch (YR) is even, being the arithmetical sum of two odd numbers viz., YB and YF.

6) If Z = number of armature conductors and P = number of poles, then, It is clear that Z/P must be an even number to make the winding possible.

1) The back and front pitches are odd. They differ numerically by

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Lap and Wave Winding Comparison

Lap Winding Wave WindingYs= S/P Ys= S/P

Yc= +1 or -1 (Prog. Or Retrogr.) Yc= 2(C+1)/P or Yc= 2(C+1)/P

Parallel paths a=P Parallel paths a=2

Coils=slots=comm. Segments Coils=slots=comm. Segments

In general, one coil has tow conductors (active) .If you number of coils are 10 , there will be 20 conductors (Z)

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Armature View and Developed Diagram

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Coil Ends (front, back) Slot Number (4) Commt.Segments (+11,1’ 1,5 a, b2,2’ 2,6 b, c3,3’ 3,7 C, d4,4’ 4,8 D,e5,5’ 5,9 E,f6,6’ 6,10 F,g7,7’ 7,11 G,h8,8’ 8,12 H,i9,9’ 9,13 I,j10,10’ 10,14 J,k11,11’ 11,15 K,l12,12’ 12,16 L,m13,13’ 13,17(1) M,n14,14’ 14,18(2) N,o15,15’ 15,19(3) O,p16,16’ 16,20(4) P,a

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16 Coils, 4 Poles, Lap Winding

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Lap Winding: Conclusions1) The total number of brushes is equal to the number of poles.

2) The armature winding is divided into as many parallel paths as the number of poles. If the total number of armature conductors is Z and P is the number of poles, then,

Number of parallel paths = P

Number of series Conductors = Z / P

In the present case, there are 32 armature conductors and 4 poles. Therefore, the armature winding has 4 parallel paths, each consisting of 8 conductors in series.

3)

4) Total armature current,

The armature resistance can be found as under:

E.M.F. generated = E.M.F. per parallel path = average e.m.f. per conductor × Z/P

1Resistance per parallel pathNo. of Parallel pathsa

l ZRP a P

Current per parallel pathaI P

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Simplex Wave Winding

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Design of Simplex Wave Winding

1) Both pitches YB and YF are odd and are of the same sign.

2)

3) Both YB and YF are nearly equal to pole pitch(YP) and may be equal or differ by 2. If they differ by 2, they are one more and one less than YA.

4) Commutator pitch is given by;

5) Since YA must be a whole number, there is a restriction on the value of Z. With Z = 180, this winding is impossible for a 4-pole machine because YA is not a whole number.

6)

2( 1)c

CyP

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Simple 4-Pole Wave wound DC Macnine

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Slots 9, Poles 4, Wave Wingding (progressive)

Coil Ends (front, back) Slot Number (2) Commt.Segments (+51,1’ 1,3 a, f6,6’ 6,8 f, b2,2’ 2,4 B,g7,7’ 7,9 G,c3,3’ 3,5 C,h8,8’ 8,1 H,d4,4’ 4,6 D,i9,9’ 9,2 I,e5,5’ 5,7 E,a

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Slots 9, Poles 4, Wave Wingding (Developed Diagram)

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Wave Winding: Conclusions

1) Only two brushes are necessary but as many brushes as there are poles may be used.

2) The armature winding is divided into two parallel paths irrespective of the number of poles. If the total number of armature conductors is Z and P is the number of poles, then,

3)

5) Total armature current,

6) The armature can be wave-wound if YA or YC is a whole number.

2 Current per parallel pathaI

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Applications of Lap and Wave Windings

In multipolar machines, for a given number of poles (P) and armature conductors (Z), a wave winding has a higher terminal voltage than a lap winding because it has more conductors in series.

On the other hand, the lap winding carries more current than a wave winding because it has more parallel paths.

In small machines, the current-carrying capacity of the armature conductors is not critical and in order to achieve suitable voltages, wave windings are used.

On the other hand, in large machines suitable voltages are easily obtained because of the availability of large number of armature conductors and the current carrying capacity is more critical. Hence in large machines, lap windings are used.

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The Frog-Leg Winding or Self Equalizing Winding

It consists of lap and wave winding combined. The wave winding functions as equalizer for

lap winding. The number of current paths in frog leg winding

is:

a = 2PmlapWhere

P = No. Of poles on the machineMlap = Plex of the lap winding

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The Internal Generated Voltage and Induced Torque Equations of Real DC Machine

A: Voltage in the rotor windings of a real dc machineThe voltage out of the armature of a real dc machine is equal to the number of conductors per current path times the voltage on each conductor.

The voltage in any single conductor: eind = e = vBl The voltage out of the armature of a real machine is thus:

The velocity of each conductor in the rotor is: v=rw

And with ɸ=BAP and A=2rl, and if there are P poles on the machine, then the total area A is

AZvBlEa

AZr BlEa

2P

A rlAP P

The total flux per pole in the machine: (2 ) 2P

B rl rlBBAP P

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The total flux per pole in the machine: (2 ) 2P

B rl rlBBAP P

The internal generated voltage in the machine:

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2

A

A

A

Zr BlEaZP rlBa P

ZPEa

E K

B: Torque induced in the armature of a real dc machine

The torque in any single conductor cond = r Icond lB

If there are ‘a’ current paths in the machine, so the current in a single conductor is given by

Acond

IIa

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And the torque in a single conductor on the motor is: Acond

rI lBa

For Z conductors, the total induced torque in a dc machine rotor is:

The flux per pole in this machine:

Aind

ZrlBIa

(2 ) 2P

B rl rlBBAP P

So the total induced torque is:2ind AZP Ia

Finally, ind AK I

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Power flow and losses in DC machines

Unfortunately, not all electrical power is converted to mechanical power by a motor and not all mechanical power is converted to electrical power by a generator…

The efficiency of a DC machine is:

100%out

in

PP

100%in loss

in

P PP

or

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The losses in DC machines

There are five categories of losses occurring in DC machines.

1. Electrical or copper losses – the resistive losses in the armature and field windings of the machine.

Armature loss:2

A A AP I R

Field loss:2

F F FP I R

Where IA and IF are armature and field currents and RA and RF are armature and field (winding) resistances usually measured at normal operating temperature.

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2. Brush (drop) losses – the power lost across the contact potential at the brushes of the machine.

BD BD AP V I

Where IA is the armature current and VBD is the brush voltage drop. The voltage drop across the set of brushes is approximately constant over a large range of armature currents and it is usually assumed to be about 2 V.

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4. Mechanical losses – losses associated with mechanical effects: friction (friction of the bearings) and windage (friction between the moving parts of the machine and the air inside the casing). These losses vary as the cube of rotation speed n3.

3. Core losses – hysteresis losses and eddy current losses.

5. Stray (Miscellaneous) losses – losses that cannot be classified in any of the previous categories. They are usually due to inaccuracies in modeling. For many machines, stray losses are assumed as 1% of full load.

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The power-flow diagram

On of the most convenient technique to account for power losses in a machine is the power-flow diagram.

For a DC motor:

Electrical power is input to the machine, and the electrical and brush losses must be subtracted. The remaining power is ideally converted from electrical to mechanical form at the point labeled as Pconv.

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The power-flow diagram

The electrical power that is converted is

conv A AP E I

And the resulting mechanical power is

conv ind mP

After the power is converted to mechanical form, the stray losses, mechanical losses, and core losses are subtracted, and the remaining mechanical power is output to the load.

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Simple DC Motor

Ee 313-dc machinery fundamentals (part1)

Engineering

Transcript of Ee 313-dc machinery fundamentals (part1)