Handbook of Small Electric Motors Episode 6 ppsx

To calculate theperformance for this motor, one must predict the air gap flux and calculate the no-load speed and current and the stall torque and current.. You can determine the flux in

Flashover and ring fire may be observed on starting some motors yet have verylittle effect on their life because it is intermittent and not seen under normal opera-tion.

It is obvious from the formulas that the best way to control flashover on motors

with variable loads is to limit Tpc This usually means more armature slots and largercommutators with more bars For a more detailed explanation of flashover, seePuchstein (1961) and Gray (1926)

4.4 PMDC MOTOR PERFORMANCE

This section is intended to give the PMDC motor designer a method to calculatePMDC motor performance given the material magnetic and electrical propertiesand physical dimensions

The basic construction of a PMDC motor is as shown in Fig 4.78 To calculate theperformance for this motor, one must predict the air gap flux and calculate the no-load speed and current and the stall torque and current A straight line drawn

between the no-load speed and the stall torque represents the speed-torque curve of

the motor A straight line drawn between the no-load current and the stall current

represents the current-torque performance curve Examples of such curves are

shown in Figs 4.79 and 4.80

FIGURE 4.78 PMDC motor construction.

4.4.1 Predicting Air Gap

A typical approach to predicting air gap flux is described here The magnets are erally attached to a magnetically soft steel housing When they are charged, they set

gen-up a nearly constant flux in the air gap between the magnets and the armature Inorder to determine the motor performance, you need to know the amount of air gapflux linking the armature conductors, the number of conductors, the number of

FIGURE 4.79 PMDC motor performance curves.

FIGURE 4.80 PMDC motor performance varying with voltage.

poles, and the current in the armature Figure 4.81 shows the direction of the flux due

to the magnets You can determine the flux in the air gap by the following procedure:

● Finding the permeance coefficient of the magnetic circuit

● Determining the flux density in the magnet

● Finding the total flux

● Factoring out the leakage flux

FIGURE 4.81 Flux due to magnets.

The remaining flux interacts with the armature conductors and produces the motortorque

The permeance coefficient is determined by the geometry of the cross section ofthe motor:

where σ =flux leakage factor (typically 1.05 to 1.15)

R f=reluctance factor (typically 1.1 to 1.3)

Lmr=radial length (thickness) of the magnet

L g=length of the air gap in the radial direction

A m=area of the magnet

A =area of the air gap

The area of the magnet is not necessarily the same as the area of the air gap,

because the magnets typically overhang the armature A g is usually smaller than A m

The permeance coefficient determines the load line of the magnet on its normaldemagnetization curve These curves are commonly supplied by the magnet manu-facturer A typical curve is shown in Fig 4.82; it is the upper left quadrant of the hys-

teresis loop shown in other chapters of this handbook The H and B axes must be

appropriately scaled for this technique to be accurate

FIGURE 4.82 Finding magnet load point.

To plot the load line, take the arctangent of the permeance coefficient, calculatethe angle ψ =tan−1P c , and plot the line as shown in Fig 4.82 The flux density in the magnet B mcan be found by finding the intersection of the load line and the normalcurve and reading the induction from the right vertical axis The flux in the magnet

is found by multiplying the flux density by the area of the magnet:

φm

σ

The effects of magnet overhang are predicted by calculating the permeance ficient of each section of the overhang using methods developed by Roters (1941).

coef-The magnetic air gap length Lglis determined by Carter’s method, but first youmust calculate the circumferential width of the armature slot, as follows

Given the following dimensions (Figs 4.83 and 4.84):

R a=outside radius of the armature lamination

Wast=width of the armature slot top in straight-line distance

Nat=number of armature teeth

Rpole=radius of the permanent-magnet pole face

FIGURE 4.83 PMDC motor cross-section.

To find the tooth pitch angle, in degrees, between two consecutive tooth centers:

360°



Nat

Width of the armature slot along the circumference, in inches:

lowing is a mathematical method for determining Carter’s coefficient K C , which

accounts for this fringing

per-Now that you know the magnetic air gap length, determine the magnetic areas ofthe air gap between the armature and the magnet

t p

p− λWasc

W2 asc

Geometric mean radius of the magnet, in inches:

where Rhiis the radius of the inside of the housing

The effective magnet area is the arc distance of the pole at the geometric meanradius times the mechanical stack length, in square inches:

where θpole=magnet pole arc, degrees

Lstk=length of armature lamination stack

The average air gap radius is the distance from the center of the armature to thecenter of the air gap, in inches:

Ags

Lgl

π

180

Rpole+R a

π

180

Permeance factor of the gap at the corner:

Area of the magnet at the corner, in square inches:

The flux at the end of the magnet segment will now be accounted for The

over-hang length per end Lmoecontributes flux to the armature stack depending on thelengths of the armature radius and the air gap

π

180

Agc



Lgc

FIGURE 4.85 Permeances for corner and end.

FIGURE 4.86 Permeances of corner and mean flux path.

Lmoe= (4.396)

If the overhang per end is equal to or greater than the armature radius R aplus the

air gap length L g , the mean flux path radius Rmf, in (Fig 4.85) is

To determine the area of the gap at the end, you must find the length of magnet

overhang which produces the flux you are accounting for Figure 4.85 shows Lom, in:

Ple=(0.0181511)Kmlθp Rmm (4.406)

where Kmlis the magnetic leakage constant based on the material type and

orienta-tion This constant accounts for energy product BHmaxof the magnet and its ability

to hold orientation

Age

Lme

π

180

π

180

π

180

Lma−Lstk



2

The permeance factor for the leakage along the axial length of the magnetaccounts for the flux which leaks off the sides of the straight edges of the magnet andinto the housing:

where M xis the slope of the demagnetization curve

The flux over the stack is the flux density in gauss converted to lines per squareinch times the area of the magnet over the stack

Flux over the stack, in lines:

In a similar manner to the method for determining flux over the stack, you mine the useful flux supplied by the corner segment of the magnet to the armature.The permeance coefficient for the corner is

Flux density of the magnet segment at the corner, in gauss:

where M xis the slope of the demagnetization curve

The useful flux supplied by the magnet at the corner is the flux density in gaussconverted to lines per square inch times the area of the magnet over the corner Fluxover the corner, in lines:

φme=(6.4516)BpeAme (4.417)The total flux supplied by the magnet is the sum of the individual sections Consid-ering two overhanging ends per magnet, the total flux supplied to the armature, inlines, is

φT= φms+2(φmc+ φme) (4.418)The air gap flux, in lines, is the total flux divided by the leakage factor:

4.4.2 Armature Calculation

Now that you know the air gap flux, you must solve the magnetic circuit by mining the flux densities in the sections of the armature and shell and then the mmfdrops First, analyze the geometry of the armature laminations This analysis givesyou the following:

deter-Gross slot area. This tells you how much cross-sectional area is available toaccommodate the armature windings (before adding insulation and slot pegs)

Net slot area. This is the net available area to accommodate the copper wireafter the slot insulators and pegs are subtracted from the gross slot area

Magnetic path length. For the armature, this is the distance that the flux willtravel and over which the mmf will be dropped

Magnetic path area. This is the cross-sectional area of the magnetic path as seenfrom the flux’s point of view The smaller the path area, the higher the flux den-sity, and the more mmf will be dropped along the path

Weights of steel. This must be known to calculate the inertias of the armatureassembly

Inertia. This is used in determining start-up acceleration or load matching.Using trigonometry, you break the armature lamination up into several sections

in order to calculate gross slot area The radii used in the calculations are readilyavailable from most lamination drawings supplied by the manufacturer

4.4.3 Armature Slot Calculations

To start the calculations, first determine the following from a drawing of a tion similar to that shown in Fig 4.87 or 4.88:

lamina-R a=outside radius of the armature lamination

R al=inside radius of the armature teeth

R s=radius of the shaft

Wast=width of the armature slot top in straight-line distance

R a2=radius of the circle on which the arc segments of the bottoms of the slotsare centered

φT

σ

Nat=number of armature teeth

Lstk=length of the armature stack

θpole=angle of pole arc

To find the tooth pitch angle, in degrees, between two consecutive tooth centers:

Tooth pitch, the circumferential distance between two tooth centers, in inches:

4.4.4 Magnetic Circuit

Special consideration is taken to account for saturation and for the air gap.The air gap is the area between the pole and the tooth tips on the armature The

mechanical L g and magnetic Lglair gap lengths were calculated earlier

The magnetic width of the air gap is the length of the arc along the center of themagnetic air gap, in inches:

Magnetic area of the air gap, in square inches:

Magnetic width of the housing, in inches (Fig 4.89):

where Dho=outside diameter of housing

Dhi=inside diameter of housing

The magnetic area of the housing is the magnetic width of the housing times the

axial length of the magnet Lma Magnetic area of the housing, in inches:

Flux is required for an electric motor to produce torque In order to produce flux,mmf must be supplied In the case of a PMDC motor, the mmf is supplied by themagnet In order to determine the mmf drop across a specific area, the flux density

must be determined From the BH curve for the material being used, the magnetic

field intensity in ampere-turns per inch can be determined The mmf drop inampere-turns for each specific area can be determined by multiplying the fieldintensity by the magnetic length for each specific area

For the previously determined flux in the air gap φg , kline, the flux densities for

the various paths are as follows:

Bgap, kline/in2, in the air gap:

In order to calculate the total mmf drops in the magnetic circuit, you need to read

the magnetic field intensity H (A  turn)/in, from the BH curve for each section.The

mmf drop for the section is the intensity times the length of the section

Field intensity Hgap, (A  turn)/in, in the air gap:

The last mmf drop to caculate for the armature is that of the armature yoke ay,

A  turn:

The mmf drop in the housing is calculated by reading the field intensity for the

housing H h from the appropriate BH curve Note that it is common to have the

hous-ing of different material than the armature Make sure that the proper curve is behous-ingused The mmf drop in the housing is taken over a distance of the arc of 30°using theradius to the center of the housing material

Magnetic length of the housing, in inches:

where P=number of poles

a=number of parallel paths in armature

Iline=line current, A

Zae=effective number of armature conductors (calculated later)

Kdl=empirically determined constant whose value depends on materials;use 0.15 if no data is available

Field distortion due to brush shift, in ampere-turns:

where α =brush shift angle, degrees

These values are mmf drops which are added to the total mmf drops in the steel,which were calculated previously

To continue the analysis of the mmf drops, the sum of the mmf drops, in turns, is

ampere-total=gap+att+at+ay+h+dl+bl (4.429)Now you know the total mmf drops in the magnetic circuit Calculate the mmfsupplied by the magnet

Total magnet area, in square inches:

Amt= π θpoleRmmLma (4.430)180

Average flux density in the magnet, in gauss:

Magnetizing intensity due to the magnet, using M xas the slope of the magnet curve,

in ampere-turns per inch:

to armature reaction and brush shift

Once the total mmf drops equal the supplied mmf, calculate the average ance coefficient

Or, to account for the fact that the conductors are not spanning the entire flux path,

two distribution factors are introduced The winding distribution factor K w1accountsfor the spread of the winding as compared to a full 180°

90θpoleP

360

Effective turns are also a function of the number of coils shorted by the brushes.

These formulas for K wtake that effect into account in many small motors

Zaemay be substituted in the following equations for Z a

The locked rotor current is found as follows:

where E t=terminal voltage

Rat=terminal resistance of the armature

R b=brush resistance

Torque developed by the motor at locked rotor, in oz  in:

Tdev=(2.256 ×10−7)φg ZaeIlock (4.441)

Since φgis relatively constant for a PMDC motor, the only variable in this torque

equation is the current I The developed torque can be described by defining a torque constant K t , (oz  in)/A:

K t=(2.256 ×10−7)φg Zae (4.442)The developed torque equation now becomes

where Iloadis the armature current at the particular load

The actual output locked-rotor torque is less the internal motor friction

(deter-mined in Fig 4.90) Internal motor friction Tfiis determined from test data on a ilar motor

sim-Locked rotor torque, in oz  in:

The true no-load speed can be calculated using the preceding values for terminalvoltage, number of conductors, and air gap flux.

True no-load speed, in rpm:

The true no-load speed is the speed the motor would run at if it had no internal tion or windage It is the speed at which the generated voltage of the armatureequals the terminal voltage

fric-The no-load current is determined by dividing the friction and windage torque bythe torque constant

No-load current, in amps:

where K f , (oz  in)/rpm, is the open circuit damping coefficient which was previouslydetermined from a similar motor, as shown in Fig 4.90 It is determined by drivingthe motor and measuring the reaction torque

The actual no-load speed Snlcan be calculated as follows:

times the armature resistance

Power lost in the windings as heat, in watts:

P L,copper=I2

Pout

Pin

Power lost in the brushes, in watts:

P L,brush=I2

Friction and windage losses, in watts:

The loss which has not been accounted for, the stray loss, is the input power −

(output power +losses)

amperes per square inch Using the diameter of the bare wire Daw,barethe currentdensity in the armature wire is found as follows:

where Tload=motor output torque, oz  in

Tfe=external friction torque, oz  in

J a=armature inertia, oz  in  s2

Jload=load inertia, oz  in  s2

The torque constant K thas already been determined

Back-emf constant Kbe, Volts/krpm (English units):

Back-emf constant Kbm, Volts/(rad  s) (metric units):

Kbm=(7.06 ×10−3)K t (4.459)Mechanical time constant, in seconds:

Zero-impedance (leads short-circuited) damping coefficient, in ounce-inches perradian per second:

where φg is in lines M should be greater than 50 for good commutation.

4.5 SERIES DC AND UNIVERSAL AC

PERFORMANCE*

The process is the same for both for series dc and universal ac performance, exceptthat in direct current there is no core loss in the field The reactance values go to 0.Transformer voltages go to 0, and the power factor goes to 1 The following describesthe ac method

Procedure

1 From the dimensions of a known lamination set, calculate the magnetic circuit

areas, lengths, and volumes

2 Plot a flux f versus magnetomotive force (mmf) curve for the lamination set on a

per-unit-length basis, including leakage flux

3 Assume a field winding and an armature winding.

4 Select a line voltage.

5 Select a fundamental current Ifundfor the motor and build a matrix of Ifundversusthe following:

Iload=load current

Fdl=field distortion mmf

Fbl=field distortion due to brush shift

Fnet=net mmf drop

ftemp=air gap flux per unit length

fact=actual air gap flux

Creal=real (or resistive) components of the phasor diagram less back-emf

Etf=field transformer voltage

Vim=imaginary (or reactive) components of the phasor diagram

qpf=power factor angle

PF =power factor

Vreal=real components of phasor diagram

E g=back-emf or generated voltage

Sload=load speed

Tdev=developed torque

Tfw=friction and windage torque

TCL=torque loss due to core loss

Tload=output torque at calculated load speed

Pout=output power, Watts

Pin=input power, Watts

Eff =motor efficiency

4.5.1 Performance Calculations

Using the lamination set of Fig 4.91, and knowing the stack length Lstk, we calculatethe magnetic circuit lengths, areas, and volumes

4.5.2 Air Gap Flux Versus MMF Drop

We will create an f versus mmf curve per inch of stack, so for these calculations we

will assume a 1.0-in stack height Therefore, the magnetic area of the air gap insquare inches is

Bmax=maximum flux density, line/in2

A T,atm=total armature tooth area per pole, in2

Starting with f g=0 and iterating toward fmax, we can now create our graph See

Figs 4.92 and 4.93 for an illustration of the flux paths For any value of f g , the

fol-lowing formulas apply

Bgap, kline/in2, in the air gap:

FIGURE 4.91 Laminations of sample motor.

Bat, kline/in2, in the armature teeth:

4.5.3 Calculation of Leakage Factor σ

The leakage factor is calculated based on several permeances See Fig 4.94 for a

graphical representation of the key variables involved First, calculate Rms, in, themean stator radius:

The permeance for the area between the magnetizing field coil and the armaturestack is the ratio of the area of the air gap over the stack to the magnetic air gaplength, in inches:

where µ0is a constant defined as the permeability of free space, the value of which isconstant for all of the permeance calculations and will cancel out when we calculatethe leakage factor

Flux leakage at the ends and sides of the stator:

le=1.04  θp  Rmm 0 (4.470)The permeance for the leakage along the axial length of the stator accounts for theflux which leaks off the sides of the straight edges and in the housing:

lsa=1.04  Lstk 0 (4.471)Permeance for the leakage from the stator pole back onto the stator yoke:

The leakage factor is the ratio of the sum of all permeances to the sum of the usefulflux permeance:

Note that this value for the leakage factor works well at about half the maximum

flux fmax There will be no leakage when there is no flux (or f g=0) and more leakagewhen the flux increases The actual leakage factor will increase as flux increases The

method used by the author is to use linear interpolation from s=0 to s1as f ggoes

from 0 to fmax/2 A maximum leakage factor smaxmust then be assumed smaxcan bedetermined from finite element analysis or from approximation Normally, one issafe to assume a maximum leakage factor of 1.2 to 1.3 Linear interpolation can be

used so that s =s1to s =smaxas f g increases from fmax/2 to fmax

To calculate the magnetic circuit variables in the stator, we take each section ofthe stator as follows:

Bsy, kline/in2, in the stator yoke:

where s is the leakage factor at that flux.

B s1, kline/in2, in stator section 1:

B s5, kline/in2, in stator section 5:

Bpole, kline/in2, in the stator pole:

In order to calculate the total mmf drops in the magnetic circuit, we need to read

the magnetic field intensity H, (A  turn)/in, from the BH curve for each section.The

mmf drop for the section is the intensity times the length of the section

Field intensity Hgap, (A  turn)/in, in the air gap:

MMF drop across the air gap, in ampere-turns:

For the following equations, the value of H is read from the BH curve for each ferent flux density B.

dif-MMF drop across the armature teeth:

MMF drop across the armature tooth tips:

MMF drop across the armature yoke, in ampere-turns:

Sum of the mmf drops, in ampere-turns:

total=gap+att+at+ay+sy+s1+s2+s3+s4+s5+pole (4.492)Calculated results for the sample motor are shown in Table 4.1, and the curve isshown in Fig 4.95

Bgap

3.19

We can now calculate the armature and field resistances based on the actual stackheight of the motor.

Resistance per armature coil, in ohms:

Rac=Lcucρa  Fwsa (4.493)where Rac=resistance per armature coil

Lcuc=total armature coil length, ft

ρa=resistivity of the armature wire,Ω/ft

Fwsa=armature wire stretch factor

The total armature resistance is the product of the number of active coils times theresistance per armature coil, or:

The field resistance R fis calculated by the following formula:

R f=2  Lfcρf  Fwsf (4.495)where R f=field resistance,Ω

Lfc=total field coil length, ft

ρf=resistivity of the field wire,Ω

Fwsf=field wire stretch factor

The total winding resistance is the sum of the armature plus the field resistance:

Air gap flux, kline

Stator yoke

Flux density 15.21 30.72 46.52 62.63 79.03 95.19 112.78 119.95 130.86 134.53 138.23 141.94Field intensity 1.79 2.71 3.51 5.90 10.45 30.57 180.75 363.61 922.71 1788.8 2947.1 4111.5MMF drop 0.69 1.04 1.35 2.26 4.01 11.72 69.32 139.44 353.86 686.03 1130.2 1576.7Stator section 1

Flux density 7.94 16.03 24.29 32.69 41.26 49.69 58.87 62.62 68.31 70.23 72.16 74.09

Air gap flux, kline

K w1=sin   (4.498)where span =number of armature teeth spanned by one coil

θp=pole arc, degrees

So:

Zeff=Ztot K w1  K w2 (4.499)where Zeff=effective number of conductors

Ztot=total number of conductors

We will start by assuming a fundamental current Ifund at which all of our load

points will be calculated Ifundwill then be incremented to give a complete torque-current curve

where Fdl=mmf drop due to field distortion

Fbl=mmf drop due to brush contact

θB=brush shift angle, degrees

P=number of poles

P a=number of paths

Fnet=Ifund TPCf−Fdl−Fbl (4.503)

where Fnetis the net MMF drop The air gap flux f gcan then be obtained from the

plot of f versus F (see Fig 4.96), using Fnet, and must be scaled according to stackheight

90 span  P

at

Stator field transformer voltage Etf:

where Etf=field transformer voltage

frequency =line-current frequency

TPCf=turns per coil of the field

The imaginary components of the phasor diagram consist of transformer voltage

and reactance voltages Now, let Cimbe the sum of the imaginary components of thephasor diagram

Cim=Etf+Ifund Xtot (4.508)

where Xtotis the total of the reactances of the motor

We can now calculate the power factor angle and power factor

θPF=sin−1

where Vrealis the total voltage drop in the real direction

We can now calculate the back-emf voltage E g:

Total developed torque, in ounce-inches:

The output torque is the difference between the total developed torque and the

fric-tion and windage torque Tfwwhere:

Tfw=Sload Kfe+Tfi (4.515)where Tfw=torque due to friction and windage, oz  in

Kfe=slope of the friction and windage torque versus speed line, (oz in)/rpm

Tfi=intercept of the friction and windage torque versus speed line, oz  inTorque due to core loss must be calculated for each section of the stator andarmature In general, the formula for core loss is as follows:

CL =Volsteel/sectionρsteel (4.516)

where: CL =core loss, W

Volsteel/section=volume of steel of each section

ρsteel=density of the lamination steel

W/lb =core loss per pound from the steel manufacturers’ core losscurve

Since core loss is dependent on frequency, that core loss in the armature will depend

on the load speed and the following formula:

For each section of the stator, core loss must be calculated at the line-current quency Torque loss, oz  in, due to core loss can then be calculated from CLtot, W, thesum of all core losses

4.5.4 Friction and Windage

Next, we must calculate the torque loss due to friction and windage Friction andwindage is more complicated to determine for a universal motor than it is for aPMDC or brushless dc (BLDC) motor In order to calculate the friction andwindage constants for a universal motor, the motor must be driven with anothermotor, usually a PMDC or BLDC motor To calculate the friction and windage con-stants of the driver motor, the motor must be disconnected from any load sources,such as a dynamometer Then, by recording speed versus current, the friction andwindage constants can be determined as follows:

1 Multiply current, A, by the torque constant Kt , (oz  in)/A, to get friction andwindage torque, oz  in

2 Plot speed, rpm, versus friction and windage torque, oz  in.

W

lb

1351.68 [(Ztot P)/(60  Pa)] φgact Ifund



108

3 The slope of the best line fit going through this curve will be Kfe, (oz  in)/rpm If

desired, K f , oz  in  s, can be calculated by multiplying by 60/2π

4 The intercept of the line is Tfi, oz  in

To determine the friction and windage constants for the universal motor, connectthe driver motor’s shaft to the test motor’s shaft with a coupling Repeat the test toobtain a curve like that shown in Fig 4.97, record the speed versus current, and cal-culate the constants as follows:

FIGURE 4.97 Friction and windage torque.

1 Multiply current, A, by the driver motor’s torque constant to get the total friction

and windage torque, oz  in

2 Calculate the driver motor’s friction and windage torque from the following

equation:

T f+w=speed  Kfe+Tfi (4.519)

3 Subtract the driver motor’s friction and windage torque from the total friction

and windage torque to get the test motor’s friction and windage torque

4 The slope of the best line fit going through this curve will be Kfe, (oz  in)/rpm, for

the test motor If desired, K f , oz  in  s, can be calculated by multiplying by 60/2π

5 The intercept of the line is Tfi, oz  in, for the test motor

6 Alternately, a polynomial may be used to determine the curve.

The load torque is then calculated by subtracting the friction and windage torqueand the torque loss due to core loss from the total developed torque:

Power, in watts, can be calculated from the following two formulas:

Pin=Vline PF  Ifund (4.522)The efficiency can be calculated from the following formula:

To calculate the performance of this motor, one first needs to calculate the fieldflux This is done by calculating the shunt winding resistance and using it to calculatethe shunt current Then, from the known ampere-turns one can calculate the fluxand determine the mmf in the magnetic circuit This procedure is the same as for thePMDC motor except that the magnet is replaced by a coil which produces the fieldmmf.

The motor geometry and magnetic paths for the field and armature are lated in the same manner as for the series-wound dc motor

calcu-Windings are selected such that the ampere-turns in the armature are about 75percent of the ampere-turns in the field at full load

The required no-load flux φnlis determined by the desired no-load speed Snland

is calculated as follows:

φnl= E g P a(60×108) (4.524)

PZ a Snl

FIGURE 4.99 Shunt-wound dc motor field.

FIGURE 4.100 Shunt-connected dc motor schematic diagram.

where E g= armature generated voltage or cemf

P= number of poles

P a= number of parallel armature paths

Z a= active armature conductors

where V L= line voltage

Ianl= estimated armature no-load current

Rat= armature terminal resistance

It is obvious that this type of motor limits the maximum no-load speed Snl The

constant flux causes the speed to stop increasing when E g=V L

When designing the field winding one must add in enough full-load ampere-turns

to account for the ampere-turns lost as a result of field distortion caused by ture reaction and brush shift

arma-According to Puchstein (1961), field distortion Fdl, A  turn, is calculated as lows:

Where: ψp= pole arc ÷pole pitch

Z a= active armature conductors

I a= current per armature path

speed-As the load increases, the armature

reaction causes Fdlto increase to a pointwhere the soft-iron pole tips are demag-netized This reduces the main pole flux,which results in a reduction of torque.This field distortion can be seen in thefinite element shown in Fig 4.102

To avoid commutation problems,shunt motors should be operated wellabove the point on the speed-torquecurve where field distortion becomessevere

FIGURE 4.101 Shunt-connected dc motor

field performance characteristics.

4.7 COMPOUND-WOUND DC

MOTOR CALCULATIONS

Compound-wound dc motor construction is similar to series-wound dc motor struction, as shown in Fig 4.103 The major difference is that this motor has both aseries and a shunt field The shunt field limits the maximum motor speed, in a man-ner similar to that in the PMDC and shunt-wound motors The series windingincreases starting torque and limits the starting current of the motor The inductance

con-of the series winding also helps to reduce flashover under rapid load changes

There are two common types of compound motor connections The long-shunt connection has the shunt winding across the power source, as shown in Fig 4.104.

Here the shunt field flux is a function of the line voltage and the shunt field tance

resis-The short-shunt connection is shown in Fig 4.105 In this configuration, the shunt

winding is across the armature but in series with the series winding Here the shuntfield current is a function of the voltage across the armature and the voltage drop inthe series winding At stall there is no generated voltage in the armature, so the

shunt winding sees a voltage Vsh

Vline=Vse+Vsh +Ra (4.528)

Vline=IseRse+Ise  (4.529)

This is a function of the series winding resistance Rseand parallel combination of the

shunt winding resistance Rshand the armature resistance R a

As the speed increases, the generated voltage component is added, which

effec-tively increases R a , causing the equivalent parallel resistance to increase This results

in an increase in voltage across the shunt winding This causes an increase in the field

RshR a



Rsh+R a

FIGURE 4.102 Magnetic field of shunt-connected dc motor with

field and armature energized.

flux At the same time, the current through the series winding is decreasing, whichresults in a decrease in field flux.

The shunt winding turn counts for the short and long connections would have to

be different to get the same resultant and field flux because of the different voltagesacross them

There are two different types of compound motors in common use They are the

cumulative compound motor and the differential compound motor In the tive compound motor, the field produced by the series winding aids the field pro-

cumula-FIGURE 4.103 Compound-wound dc motor field.

FIGURE 4.104 Compound motor long-shunt connection diagram.