the abc’s of synchronous motors
TRANSCRIPT
SYN
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The ABC’s of Synchronous Motors200-SYN-42
The ABC’s of Synchronous Motors
1. High efficiencySynchronous motors have a unique andmerited position as the most efficient electricaldrive in industry. They perform with greateconomy while converting electrical power tomechanical power and can be built with uniquephysical features. Synchronous motors can bedesigned to effectively operate over a widerange of speeds to provide the best drive for awide variety of loads.
2. Power factor correctionBecause they can operate at leading powerfactors, synchronous motors can help reduceyour power cost and improve efficiency of thepower system by correcting low power factor.In a few years, savings in power bills mayequal the original investment in the motor.
3. Reduced maintenanceSynchronous motors with brushless exciterspractically eliminate the need for motormaintenance other than routine inspection andcleaning.
4. Space savingsEngine-type synchronous motors can beconnected directly to the shaft and bearings ofthe driven equipment. This approach saves onfloor space, extra building costs, and makes fora simple installation.
5. Constant speed means better quality productsSynchronous motor speed is unaffected by lineor load conditions. In certain applications, e.g.paper mill pulp machines, constant speedoperation results in superior uniformity andquality of the product being produced.
6. Adjustable speed means lower power costsIn many cases it is much more economical tooperate driven equipment at reduced speeds.This can be accomplished using synchronousmotors in conjunction with magnetic drives, orthrough electronic methods, which can be usedto supply an adjustable frequency to themachine and control its operating speed.Typical applications for adjustable speedsinclude pumps and fans. Either approach canreduce power costs and provide a more cost-effective process.
The Greeks had a word for it —SYNCHRONOUS— “syn” a prefix meaning“with”, and “chronos” a word denoting time. Asynchronous motor literally operates “in time with”or “in sync with” the power supply.
The modern synchronous motor is a double-duty motor. It is a highly efficient means ofconverting alternating current electrical energy intomechanical power. Also it can operate at eitherleading, unity, or in rare cases at lagging powerfactor, providing the function of a synchronouscondenser for power factor correction.
In the synchronous motor, the basic magneticfield is obtained by direct current excitation ratherthan through the air-gap from the armature, as isthe case with induction motors. Comparativelylarge air-gaps are used, making practicable themanufacture, even in relatively low horsepowerratings of low speed synchronous motors. In alllow speed ratings and in large high speed ratings,synchronous motors are physically smaller and lesscostly to build than squirrel-cage induction motorsof equivalent horsepower.
A synchronous motor can be applied to anyload which can be successfully driven by a NEMADesign B squirrel-cage motor. However, there arecertain types of loads for which the synchronousmotor is especially well suited. The correctapplication of synchronous motors results insubstantial savings in several ways:
2Synchronous motor 1500 HP at 720 RPM with Paper MillDripproof™ enclosure driving a refiner in a paper mill.
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EM synchronous motors are designed and built to industry standards
ANSI, IEC, IEEE and NEMA standards helpboth manufacturer and user by providingdefinitions and suggested values. They arecontinually updated to incorporate the latesttrends in machine use. Although these standardsare gradually coming closer together, this is a longprocess. ANSI and/or NEMA provide thestandards for synchronous motors in the UnitedStates, whereas IEC standards provide theguidelines for European applications. Standardsare a great help in understanding the applicationof motors. Please refer to NEMA MGI part 21,Synchronous Motors, or ANSI C50.
The parts of a synchronous motor
FRAME ... supports and protects the motor; it isbuilt in horizontal and vertical types and invarious protective constructions for indoor oroutdoor service.
STATOR ... consists of the stationary magneticparts, including the core and winding whichoperate off the alternating current power supplyto provide a rotating magnetic field.
ROTOR ... consists of the rotating active parts,and includes the spider, the field-pole windingand the amortisseur winding. Field poles aremagnetized by direct current from the exciter ordirectly through collector rings and brushes; theyinterlock magnetically through the air-gap andrevolve in synchronism with the rotating magneticpoles of the stator.
EXCITER ... supplies magnetizing current to thefield winding. Modem exciters are of the rotatingbrushless type with no collector rings or brushes.
AMORTISSEUR WINDING ... makes asynchronous motor self-starting like a squirrel-cage induction motor. Depending on the type ofloads and the torque required, a variety of barmaterials and arrangements can be used toprovide adequate starting torque.
StatorFrame
Exciter
Amortisseur Winding
Rotor
Synchronous motors are“geared” to the power supply
GEARED
Figures 4 and 4a complete a one-quarter turnor 90° of a two-pole motor. Continuation of thiscompletes one revolution.
The rotor shown in the above figures is that ofa two-pole synchronous motor having definiteNorth and South poles developed by directcurrent flowing in the field winding. The currentflows from the observer in the left hand portion ofthe winding, towards the observer in the righthand portion. Using the “right-hand rule” thisdevelops a North pole at the bottom of the rotorand a South pole at the top.
The magnetic poles developed in the stator areexactly opposite. However, unlike poles attract,so the North pole of the rotor lines up under theSouth pole of the stator.
As the North pole of the stator shifts 30°clockwise, the rotor is pulled along with it, and(assuming no load on the rotor shaft and nolosses) the rotor and stator poles line up exactly asshown. Each rotor pole lags behind itscorresponding stator pole by 20 to 30 electricaldegrees at full load. This is called the “loadangle,” and it is a function of motor load, whereincreased torque requirements result in greatermagnetic pull and a greater displacement frommagnetic center.
A normal two pole stator differs in manyrespects from that shown in Figures 1 through 4.Instead of one slot per phase there are six or seven.The coil pitch is about 120° instead of 60° and willoverlap the coils of the various phases. Theconfiguration shown results in rotation by a seriesof jerks. The use of more closely spaced coils willresult in a smoother, more uniform turning action.
During operation, the stator of a polyphasealternating current motor has a rotating magneticfield. This is developed by the direction andamount of current flow in the stator coils.
Figure 1 represents a two-pole, three-phasestator, having one slot per pole per phase in whichthere are six stator coils. These coils are connectedto a three phase power supply having phaserotation A, B, C.
At the instant shown, the current in phase A inmaximum positive enters the motor at A+ andleaves at neutral where the three conductors meet.The “A” arrows in Figure 1 represent the magneticflux (lines of force) developed by this current. Atthis instant the current in phases B and C is negative,each equal to one-half the current in A. Themagnetic flux developed by B is shown by the “B”arrows, each equal to half the “A” arrows. The fluxdeveloped by C is represented by the “C” arrow,each equal to half of A. B and C also have horizontalcomponents but, as shown, they cancel out.
The result is a vertically polarized stator, witha North pole at the top and a South pole at thebottom. The arrows show the flux flowingdownward through the rotor.
One-twelfth revolution later (Figure 2), thepositive current in A has dropped in value and isequal to the negative value in C, which hasincreased in value. The current in B is zero. Equalfield strengths are developed by A and C and,canceling out the opposing components, we findthe total flux in Figure 2a to be represented by thetwo “A” and two “C” arrows, 30° from the formerposition.
At the 60° position, shown in Figure 3, thecurrent in phase B and the corresponding statorcoils are positive and equal to phase A. Thecurrent in phase C is maximum negative. Aftercanceling out the opposing components, we find aresultant field, as shown in Figure 3a composedprincipally of current in phase C, supplementedby smaller equal amounts in phases A and B.
4
Figures 1 through 4 AC Generator.Three-phase power supply for Synchronous Motor. Generatorvoltage is assumed equal and in phase with the current; thereis no load torque for friction so rotor and stator poles are freeto line up magnetically.
A
BC
Figure 1 See top left of page 5 for balance of this diagram, plus its caption.
5
0–
+
0°
90 180DEGREES
PHASE A PHASE B PHASE C
270 360
A'
A'
B'C'
CB
A+
N
N
S
S S
NB–
0–
+
30°
90 180DEGREES
A
B C
270 360
A'
A'
C'
C
A+
NZERO
N
S
SS
N
B–
0–
+
90 180DEGREES
A
B
C
270 360
60° A+
N
N
S
SS
N
B+A'
A'
B'C'
CB
A
0–
+
90 180DEGREES
B
A
C
270 360
90° A
N
N S
SS N
B+A'
A'
B'C'
CB
ZERO
Figure 1 Current relationship, and magnetic resultant (Figure 1a)of a two-pole synchronous motor with rotating magnetic field at 0°.SEE TOP LEFT OF PAGE 4 FOR BALANCE OF THIS DIAGRAM.
Figure 2 Current relationship, and magnetic resultant (Figure 2a) of a two-pole synchronous motor with rotatingmagnetic field at 30°.
Figure 1a
Figure 2a
Figure 3 Current relationship, and magnetic resultant (Figure 3a) of a two-pole synchronous motor with rotatingmagnetic field at 60°.
Figure 3a
Figure 4 Current relationship, and magnetic resultant (Figure 4a) of a two-pole synchronous motor with rotatingmagnetic field at 90°.
Figure 4a
How synchronous motorscontrol power factor
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Power factor is the factor by which apparentpower, or kVA, is multiplied to obtain actualpower, or kW, in an ac system. It is the ratio of thein-phase component of current to total current. Itis also the cosine of the angle by which the currentlags (or leads) the voltage.
The conversion of electrical energy tomechanical energy in a motor is accomplished bymagnetic fields. In the previous section, it wasexplained how magnetic poles are formed whencurrent flows in coils placed around the stator gapbore. As explained, these poles rotate around thestator. When voltage is applied to a motor, anarmature current flows to provide the necessarymagnetic push (mmf or ampere turns) to producea flux which, in turn, produces a voltage (backemf) that opposes the applied voltage. This mmfis a magnetizing current. It is loss-less, except forthe i2r in the winding and any core loss due to thechanging flux in the iron. The magnetic energy istransferred from the line to the motor and backagain each half cycle. The net power is zero, andthe power factor is zero.
The power factor of a synchronous motor iscontrollable within its design and load limits. Itmay operate at unity, leading, or in rare cases,lagging power factor and may be used to modifythe power factor of the system to which it isconnected. A simplified explanation of phasorrelationships will describe what takes place undervarious load and excitation conditions.
Phasors are vectors that represent therelationship between voltages or currents. Thephasor length represents the magnitude and, as itis allowed to rotate counterclockwise about theorigin, the projection on the x axis is theinstantaneous value.
Figures 5, 6 and 7 show the instantaneousvoltage, current and power (V times I) for unity,leading and lagging power factors. The magnitudeof the current and voltages are held constant for allcases. Also shown is the corresponding phasordiagram for each in Figure 5a, 6a and 7a. Note thatfor lagging power factor, the current phasor lagsthe voltage phasor and the instantaneous currentpasses through zero after the voltage. The powerfor unity power factor is always positive whileboth leading and lagging power factors have somenegative power values, and the peak is less thanfor unity. Thus, the average value of power is lessfor either leading or lagging power factor for thesame current and voltage.
Once the synchronous motor is synchronized,the field poles on the rotor are in line with therotating magnetic poles of the stator. If dc isapplied to the rotor pole windings, the rotor cansupply the necessary ampere turns to generate theflux which produces the internal motor voltage.Thus, the field current can replace part or all of themagnetizing current. In fact, if more dc fieldcurrent is supplied, the increased flux will try toincrease the line voltage. To increase the linevoltage, the motor will supply ac magnetizingcurrent to all “magnets” on the system to increasetheir magnetic flux. This is leading power factor.
The synchronous torque developed is roughlyproportional to the angle of lag (load angle) of themotor rotor with respect to the terminal voltage,which, at full load, is in the area of 20 to 30electrical degrees. If a restraining force is appliedto the motor shaft, it will momentarily slow downuntil a torque is developed equal to the appliedrestraint. The motor will then continue to operateat synchronous speed.
Figure 8 shows a somewhat simplified phasordiagram for a unity power factor motor connectedto a large system such that the terminal voltage isfixed. Since synchronous machines have polesand interpole spaces, a system of phasorsrepresent each of the “direct” (in line with thepole) and “quadrature” axes (interpole).
In drawing the diagram, the reactance drop issubtracted from the terminal voltage to obtain theinternal voltage and the direction of thequadrature and direct axes. The internal voltagesets the level of flux in the motor, which, in turn,sets the mmf required (ampere-turns ormagnetizing current). The current flowing in thearmature winding produces an mmf calledarmature reaction. The portion, in line with the
Direct Axis Internal Dropidxd
TerminalVoltage Et
QuadratureAxis Current
Quad
ratu
re A
xis
Inte
rnal
Dro
pi qx
q
i d iq
MMF due toInternal Volts
δLoad AngleQuadrature
Axis
Dire
ct A
xis
Curr
ent
Direct Axis
Armature Current
Internal Voltage Ed(voltage due to field excitation)
Two-Reaction Diagram(Unity Power Factor)
Figure 8
7
2.5
2
1.5
1
0.5
0
Average Power = 1.0
Peak e = √2
I = 1.25
V = 1.00
Figure 7a
P.F.∠
Average Power = 1= V x I x P.F.
Power returned to system–0.5
–1
–1.5
–2
Peak i = √2/0.8
2.5
2
1.5
1
0.5
0
Peak e = √2 ERMS
Peak i = 1.77 (√2 x 1.25)
V = 1.00
i = 1.25(unity P.F.)
Figure 5a
Average Power = 1.25= V x I x P.F.
–0.5
–1
–1.5
–2
Peak p = 2.5
p, i, e are instantaneous values
2.5
2
1.5
1
0.5
0
Average Power = 1.0
Peak voltage = √2 ERMS
I = 1.25
V = 1.00
Figure 6a
P.F.∠
Average Power = 1= V x I x P.F.
Power returned to system–0.5
–1
–1.5
–2
Figure 5 (top) 0.8 PF Motor at unity PF.Figure 6 (middle) 0.8 PF Leading.Figure 7 (bottom) 0.8 PF Lagging.
PF∠ stands for Power Factor Angle.
direct axis, must be added (or subtracted, forlagging P.F.) to the mmf, producing the internalvoltage. Since the field winding on the rotor polesis in line with the direct axis, dc in this windingcan supply the required mmf.
If the dc is increased, the internal voltage willincrease, and since the terminal voltage is fixed bythe connected system, the internal drop mustincrease. The result of this is shown in Figure 9
where the line current is increased andrepositioned to a leading power factor. Note thatthe current in phase with the voltage representsthe load and is unchanged, but a magnetizingcomponent has been added. This componentflows to the line to try to raise the system voltage.The motor has a leading power factor.
If, instead, the dc is decreased, the internalvoltage will try to reduce the system voltage,resulting in a magnetizing current drawn from thesystem. The motor then has a lagging power factor.
The fact that the field poles of a synchronousmotor are magnetically in line with the stator fluxallows dc current on the field to replace acmagnetizing current on the stator. Since powerfactor is determined by the magnetizing current,the power factor is adjustable by changing thefield current.
idxd iqxq
Et
Ed
iq
id
Direct Axis
Armature Current
Two-Reaction Diagram(Leading Power Factor)
QuadratureAxis
θ
δ
Figure 9
power factor. If excitation is increased from the1.0 power factor condition, the line current willagain increase, but now the motor will operateunder a leading power factor condition.
For any load, the power factor will be the ratioof minimum line amperes at that load to the lineamperes at the excitation value underconsideration. To operate at 1.0 power factor formaximum efficiency, set the excitation forminimum ac line current.
“V” CurvesIt is generally assumed that the line voltage
will be substantially constant, and it is apparentfrom the preceding discussion that load,excitation, line amperes, and power factor areclosely related. This relationship is readilyexpressed by a family of characteristic curvesknown from their shape as “V” curves. These arerepresented in Figure 10. Note that for each curvethe power is constant and the excitation is variedto give different magnetizing currents.
The minimum value of line amperes for eachload condition is at 1.0 or unity power factor. Asexcitation is decreased, the line current willincrease and the motor will operate at lagging
% O
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.8 P
.F. L
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AMPE
RES
% OF FULL LOAD .8 P.F. EXCITATION
FULL LOAD 1.0 P.F.
FULL LOAD .8 P.F.
FULLLOAD
HALFLOAD
NOLOAD
A160
80
40
00 25
LAG LEAD
50 75 100 125 150
120
B
C
O
D
UNIT
YPO
WER
FACT
ORLI
NE
LEADING
LAGGINGkVA
kW kVAR
Plant loading and power factor (induction motor proposed)Monthly Averages Calculated Transformer
Load MWH P.F. Hrs (Est) kW kWR kVA Load RatingT1 Office 77 0.92 194 400 -170 435 435 T1=1000T2 Pumps 587 0.72 335 1750 -1687 2431 2431 T2=2000T3 Air 235 0.82 335 700 -489 854 854 T3=1500T4 Mill 1347 0.88 464 2900 -1565 3295T4 Proposed 335 0.81 335 1000 -724 1235 4522 T4=4000Overall 0.824 6750 -4635 8188
Plant Loading and Power factor (synchronous motor proposed)Monthly Averages Calculated Transformer
Load MWH P.F. Hrs (Est) kW kWR kVA Load RatingT1 Office 77 0.92 194 400 -170 435 435 T1=1000T2 Pumps 587 0.72 335 1750 -1687 2431 2431 T2=2000T3 Air 235 0.82 335 700 -489 854 854 T3=1500T4 Mill 1347 0.88 464 2900 -1565 3295T4 Proposed 335 -.8 335 1000 750 1250 4522 T4=4000Overall 0.906 6750 -3161 7453
Figure 10For any given load the power factorat some given excitation value will
be the ratio of minimum lineamperes at that load, to the lineamperes at the excitation valueunder consideration. Assume a
steady load condition in which theline amperes, at various excitationvalues, lie along line A-O-B; withpoint O representing minimum acinput and point B representing the
input at the desired excitation value.The power factor at that excitation
is CO ÷ DB.
Figure 11Plant loading and power factor.
Figure 12For leading power factor synchronous
motors, leading kVARs indicates kVARs fedto the system, where as induction motors
always absorb kVARs from the system.
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System Power Factor CorrectionThe power usage and the power factor of a
plant should be periodically reviewed to avoidsurprises. A typical plant power system is shownin the form of a spreadsheet in Figure 11. Here itis assumed that you have monthly records for thevarious load centers. A proposed plant expansionemploying either an induction or a synchronousmotor is under consideration. T1, T2, etc.represent transformers or load centers.
From the MWH and the estimated operatinghours, the average kW is calculated. Next,calculate the kVAR and kVA using the powerfactor. Tabulation of results is shown in Figure 13.
Note that the terms “leading” and “lagging”refer to the current leading or lagging the voltage.The current to induction motors will lag, and themotor will draw magnetizing current from thesystem. Leading P.F. synchronous motors feedmagnetizing current to the system. As shown inFigure 12, lagging currents are treated as negative,leading as positive.
kVAR = kVA2-kW2
kVAR = kW x Tan (θ) (where θ is the angle in degrees, by whichthe current differs from the voltage.)
kW = kWH/HMotor rated kVA = 0.746 x HP
Eff. x P.F.
If the proposed expansion is an inductionmotor, the overall power factor will be less than0.9. This will result in additional charges by thepower company. Note that transformer T4 inFigure 11 will be overloaded.
If the addition is a 0.8 leading P.F.synchronous motor, the total kW will be the same,but the kVAR will be less, and the overall powerfactor will be over 0.9. The overload on T4 iseliminated. In calculating the load on T4, it isnecessary to first combine the kWs and kVARsseparately and then calculate the kVA as thesquare root of the sum of the squares.
Figure 14 shows the reactive capability oftypical synchronous motors. These curves plot thepercent kVAR vs. percent load. To use the curve,determine the kVAR based on the percent loadand rated power factor of the motor. The reactivekVAR is then the rated horsepower multiplied bythe percent kVAR determined. For example, a 250horsepower, 0.8 power factor motor operating at100% load gives a 0.6 factor. The kVAR is then 250X 0.6 = 150 kVAR supplied to the system.Excitation is maintained at the full load value.
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kVAR
IN %
OF
RATE
D HO
RSEP
OWER
% OF RATED LOAD ON MOTOR
100
80
60
40
20
00 20 40 60 80 100 120 140
0.7 P.F. MOTOR
0.8 P.F. MOTOR
0.9 P.F. MOTOR
1.0 P.F. MOTOR
Power Ratio Power Ratio Power RatioFactor kVAR/kW Factor kVAR/kW Factor kVAR/kW1.00 .000 .80 .750 .60 1.333.99 .143 .79 .776 .59 1.369.98 .203 .78 .802 .58 1.405.97 .251 .77 .829 .57 1.442.96 .292 .76 .855 .56 1.480.95 .329 .75 .882 .55 1.518.94 .363 .74 .909 .54 1.559.93 .395 .73 .936 .53 1.600.92 .426 .72 .964 .52 1.643.91 .456 .71 .992 .51 1.687.90 .484 .70 1.020 .50 1.732.89 .512 .69 1.049 .49 1.779.88 .540 .68 1.078 .48 1.828.87 .567 .67 1.108 .47 1.878.86 .593 .66 1.138 .46 1.930.85 .620 .65 1.169 .45 1.985.84 .646 .64 1.201 .44 2.041.83 .672 .63 1.233 .43 2.100.82 .698 .62 1.266 .42 2.161.81 .724 .61 1.299 .41 2.225
Figure 13 Power Factor Table
Figure 14 Leading reactive kVA in percent of motorhorsepower for synchronous motors at part load and at variouspower factor ratings. Solid lines are based on reduction inexcitation, at overload, to maintain rated full-load amperes;dotted lines represent values with rated excitation to maintainrated pull-out torque.
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Torque makes the wheels go ’round
TORQ
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Torque may represent either the turning effortdeveloped by a motor or the resistance to turningeffort exerted by the driven load. It is defined astangential pull at a radius of one unit from thecenter of rotation. In the United States it ismeasured in foot-pounds or inch-pounds. For agiven condition:
Torque = HP X 5250rpm
Torque = Tangential effort in pounds at onefoot radius
HP = Horsepower developedrpm = Revolutions per minute
In some cases motor torques are expressed infoot-pounds; however, they are usually expressedas a percentage of full load torque.
A synchronous motor has many importanttorques that determine the ability of the motor tostart, accelerate, pull its connected load into step,and operate it through anticipated peak loadswithin its design limits.
These torques may be described as:
1. Starting torque or breakaway torque.This is the torque developed at the instant ofstarting at zero speed.
2. Accelerating torque.This is the motor torque developed from stand-still to pull-in speed minus the load torque.
3. Pull-up torque.This is the minimum torque developedbetween stand-still and the pull-in.
4. Pull-in torque.This is the torque developed during thetransition from slip speed to synchronousspeed and generally is defined at 95% speed.
5. Synchronous torque.This is the steady state torque developedduring operation. It is load dependent.
6. Pull-out torque.This is defined as the maximum steady statetorque developed by the motor, for one minute, before it pulls out of step due tooverload.
Starting and accelerating torquesThe characteristics of these torques are closely
allied and will be discussed together. Fiveindividual torques contribute to the turning effort ofa synchronous motor during starting andaccelerating. They are produced by:
1. Amortisseur windings2. Field windings3. Hysteresis in pole faces4. Eddy currents in pole faces5. Variation in magnetic reluctance
Of the five torques, only the first two aresignificant during starting and accelerating, andonly their characteristics will be discussed here.
The amortisseur windingsThese windings develop most of the starting
and accelerating torque in motors of this type. Thiswinding consists of a partially distributed cagewinding, with bars imbedded in the pole faces andshort circuited at each end by end rings. Varyingthe number, location and resistance of these barswill have a substantial effect on the torque and onthe kVA input. In some cases a double cage,consisting of one row of shallow bars and one rowof deepset bars, is used. These may be separate, orthey may have common end rings as circumstanceswill dictate. Typical amortisseur windings areillustrated in Figure 15.
% O
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TORQ
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ND k
VA
% SYNCHRONOUS RPM
500
400
300
200
100
0250 50 75 100
kVA CURVES
TORQUE CURVES
4
3
2
1
4
321
11
As discussed previously, the application ofpolyphase power to the stator winding results inthe development of a rotating magnetic field.Magnetic lines of force developed by this rotatingmagnetic field cut across the cage bars andgenerate voltages, causing currents at slipfrequency to flow in them. The interactionbetween these currents and the rotating magneticfield develops torque, tending to turn the rotor inthe same direction as the rotating magnetic field.
Recommended starting torques range from40% to 200% of full load torque. See NEMArecommendations on torques for varioussynchronous motor applications.
Torque comes from the action of the fluxproduced by the stator. Getting more torquerequires more flux and, therefore, more iron.More torque also results in more inrush. For anygiven motor, the effective resistance and reactanceof the cage winding are the principal factors indetermining motor torques. Typical starting andaccelerating torques and their associated kVAvalues are shown in Figure 16.
The double cage motor was developed to giveessentially constant pull-up torque. The shallowcage has a high resistance and low reactance, whilethe deep cage has low resistance and highreactance. At standstill, slip frequency equals linefrequency, and the high reactance of the deep cageforces most of the induced current to flow throughthe shallow cage, high resistance bars, resulting inhigh starting torque.
0 50 100
TORQUE FROM UPPER CAGE
TORQUE RESULTANT - DOUBLE CAGE
TORQUE FROM LOWER CAGE
% SYNCHRONOUS SPEED
TORQ
UE
As the rotor comes up to speed, the slipfrequency decreases and, with it, the impedance ofthe deep cage. This permits more of the inducedcurrent to flow through the deep-set, low-resistance bars, resulting in high torques at speedsnear pull-in. Figure 17 illustrates the resultanttorques using this type of construction.
Ideal applications for double cage windingsare those where the load is uniformly high duringthe entire accelerating period. Disadvantagesinclude saturation, due to the number of cage barsin the pole head, and stress, due to the differentialthermal expansion in bars that are close together.Double cage motors are relatively uncommon.
Figure 15 Modern amortisseur windings use a variety of barmaterials and arrangements to secure the desired resistance
and reactance to produce the required torques.
Figure 16 (left) Curves show starting torque and startingkVA of synchronous motors using various types of amortisseurwinding constructions. For any given motor the effectiveresistance and reactance of the cage winding are the principalfactors in determining the shape of the curves, and the ratio oftorque to kVA.
Figure 17 (right) Curves show double-cage torque resultingfrom high resistance upper cage winding, and low resistance,high reactance lower cage winding.
Amortisseurwindings and
end ring
Field WindingsThe amortisseur winding imparts squirrel-cage
motor starting and accelerating characteristics tothe synchronous motor. The field windingdevelops a torque similar to that of a wound rotormotor with a single-phase secondary circuit.
The single-phase characteristic results in twotorque components, one rotating in the samedirection as the rotor and one rotating in theopposite direction. The second component ispositive in value to 50 percent speed and negativefrom that point to pull-in. Figure 18 illustrates thetorque components and the resultant torquederived from the field windings. If a motor wasbuilt with a relatively weak cage, the dip in torquedue to the effect of the negative rotating componentcould result in the motor not being able toaccelerate much past half-speed.
During starting and accelerating, the fieldwinding is short-circuited through a field dischargeresistor to limit the induced field voltage. The ratioof this resistance to the field resistance has asignificant effect on the starting torque, the torqueat pull-in and, to a lesser degree, on starting kVA.Figure 19 shows the effect of varying the fielddischarge resistance on one specific motor. It willaffect the negative rotating component of the fieldwinding torque, and a high value of resistance mayreduce it considerably. However, a high value offield discharge resistance results in a high inducedvoltage across the field winding. In extreme cases,as in the case of open-circuit field starting (infiniteresistance) the induced field voltage may exceed100 times the normal excitation voltage.
Pull-up TorquePull-up torque is the minimum torque
developed from stand-still to the pull-in point.The pull-up torque must exceed the load torque(torque required by the driven machine) byenough margin to maintain a satisfactory rate ofacceleration from stand-still to pull-in under theminimum expected voltage condition.
Net Accelerating TorqueNet accelerating torque is the margin by
which the motor torque exceeds the load torquefrom stand-still to the pull-in point.
In the case of high inertia loads, it is importantthat the starting time be determined so properrelaying action of overcurrent relays andamortisseur winding protective relays, etc., can beobtained.
Accelerating time from standstill to the pull-inpoint can be approximated by applying thefollowing relationship:
Accelerating time t = WK2 x ∆rpm (seconds)308T
WK2 = total inertia of the load and themotor in lb.ft.2
∆rpm = the change in speed (rpm2 - rpm1)308 = a constant
T = net accelerating torque, lb.ft. fromrpm1 to rpm2
t = the time increment to acceleratefrom rpm1 to rpm2
Note the motor torque as a function of speed,the load torque as a function of speed, and the
Figure 18 Torque components and resultant torque derivedfrom the field winding of a synchronous motor during starting.
Figure 19 Effect of ratio of field discharge resistance to field resistance on the starting and pull-in characteristics ofthe motor.
TORQ
UE
% SYNCHRONOUS SPEED
TORQ
UE
PERC
ENT
TORQ
UE
RESULTANT
PULL-IN TORQUE
STARTING TORQUE
NEGATIVEROTATING
COMPONENT
POSITIVEROTATING
COMPONENT
140
120
100
80
60
40
RATIO: % OF EXTERNAL RESISTOR% OF FIELD WINDING
0 50 100
0 10 20 30 40 50
12
13
motor and load inertias. The latter are obtainedfrom the manufacturers as constants; the former aremost frequently illustrated as speed-torque curves.
Refer to Figure 20, which shows curves for atypical 600 hp. 900 rpm motor and a throttledcentrifugal fan load. The total motor and load WK2
(inertia) is given as 29770 lb.ft.2.Net accelerating torque at any speed is
determined by subtracting the load torque fromthe motor torques at that speed. By taking valuesat 10% intervals, starting at 5% speed andcontinuing to 95% speed for example, a reasonableapproximation of acceleration time from standstillto pull-in speed can be calculated. Refer to Figure21 for a tabulation from this example.
During the starting period, the amortisseurwinding must absorb energy. In addition to theenergy required to overcome the load torque,energy must also be absorbed to accelerate theinertia of the rotating system. Since hightemperatures that affect material strength can berealized in a short period of time, it is important that a means be provided for transmitting as muchheat as possible into the pole body and also that the heat absorbed will not overheat the amortisseurwinding. During starting the temperature rise of theamortisseur winding may be as high as 150°C.Quantitatively, the energy absorbed in acceleratinga rotating mass is covered by this formula:
H = 0.231 x WK2 x (rpm)2
(1000)2
H = Energy in kW secondsWK2 = Total WK2
rpm = Speed in rpm0.231 = a constant
In the case of the motor and fan referred toabove, the energy absorbed in the rotor, due toinertia, in bringing the unit to 95% of synchronousspeed is:
H = 0.231 x 29,770 x (855)2 = 5020 kW seconds(1,000)2
H is expressed in per unit by dividing theabove by rated kW (rated horsepower x 0.746).Interestingly, two times H in per unit is theaccelerating time if the accelerating torque is equalto rated torque. Applying this to our example: H = 11.2 and the average accelerating torque is76%. Accelerating time (t) = 11.2/0.76 x 2 = 29.5seconds.
Some types of loads, such as ball mills, havevery low inertia values, but their load torque mayapproach that developed by the motor if the rate ofacceleration and the accelerating torque is low.Large fans may have both high inertia values andappreciable torque requirements. In such cases, itis desirable to determine the total energy input to
the rotor, both as a result of accelerating theinertia, and of overcoming the load torque. Sinceall of the torque developed by the motor musteither accelerate inertia or overcome load torque,and since slip loss is a function of this torque andof slip, it is readily possible to determine thisenergy input.
Referring again to the motor and loadcharacteristics covered by Figure 20 and Figure 21,we can proceed as follows:
Slip loss = 0.746 x TM x S x HP x t, in kilowatt seconds
TM = % motor torqueS = % slipt = Time at that torque
Note that slip loss is proportional to averageslip and to average total motor torque (not netaccelerating torque) for each 10% speed intervalconsidered. As shown in Figure 21, total loss inthe rotor is 6400 kW seconds.
% SYNCHRONOUS RPM
% F
ULL
LOAD
TOR
QUE
40
60
80
100
20
00 20 40 60 80 100
MOTOR TORQUE
NO LOADTORQUE OF FAN
% % Motor % Fan % Net Ft.Lb. Time inRPM Torque Torque Torque Torque Seconds
5 86 5 81 2840 1.4615 88 3 85 2980 2.7825 91 6 85 2980 2.7835 93 10 83 2920 2.8445 97 14 83 2920 2.8455 99 19 80 2800 2.9665 102 25 77 2700 3.0875 104 32 72 2520 3.3085 104 40 64 2240 3.8095 100 50 50 1750 4.75
30.59
Figure 21
Figure 20 Motor and load torques for centrifugal fan, illustrating method fordetermining accelerating time of the motor.
14
Since it was previously determined that theloss due to accelerating the inertia was 5020 kWseconds, we now determine the additional lossdue to the load is 6400 - 5020 = 1380 kW seconds.
In some cases the speed of the driven unit willdiffer from that of the driving motor, because ofbelting or gearing arrangements. All WK2 valuesused in determining accelerating time or energyabsorption must be effective values referring to themotor speed.
Effective WK12 = WK2
2 x(rpm2)2
(rpm1)2
Effective WK12 = WK2 referred to the
motor shaft rpmWK2
2 = Inertia of driven machinein lb.ft.2 @ rpm2
rpm2 = Speed of load shaft inrevolutions per minute
rpm1 = Motor shaft speed inrevolutions per minute
Pull-in TorqueThe pull-in point (when a synchronous motor
changes from induction to synchronouscharacteristics and operation) is usually the mostcritical period in the starting of a synchronousmotor. The torques developed by the amortisseurand field windings become zero at synchronousspeed. They cannot pull the motor into step, so,the reluctance torque and the synchronizingtorque (resulting from exciting the field windingswith direct current) take over.
Reluctance TorqueAny magnetic object tends to align itself in a
magnetic field so that the magnetic reluctance is ata minimum. There are as many positions of asalient pole rotor within the rotating statormagnetic field as there are poles. Belowsynchronous speed, a pulsating torque calledreluctance torque, is developed. It has a netaverage value of zero and a frequency equal totwice the slip frequency. This torque may lock alightly loaded rotor into step and developapproximately 30% pull-out torque.
Synchronizing TorqueWhen excitation is applied, definite polarity is
developed in the rotor poles. If the slip is lowenough, the attraction between unlike poles in therotor and the stator will lock the rotor into stepwith the rotating stator magnetic field. Neglectingthe influence of reluctance torque, the value of slipfrom which the motor will pull into step may beexpressed as follows:
S = C
xHP
rpm WK2
S = Slip in % of synchronous speedWK2 = Inertia of motor rotor plus load inertia
C = Constant taking into account suchfactor as efficiency, power factor, etc.
The relationship between slip and inertia in agiven motor is shown by the curve in Figure 22.The larger the inertia the higher the speedrequired to achieve synchronization.
Load InertiaSince load inertia is so important in
determining the speed to which the motor mustaccelerate before pull-in is possible, NEMA hasestablished normal load WK2 values. These arecovered by the formula:
Normal load WK2 =3.75 x HP1.15
(rpm/1000)2
Figure 23 illustrates the relationship betweenrated horsepower, speed and normal load WK2.
We may designate as “inertia factor” the ratioof actual load WK2 to normal load WK2. Thisvaries over a wide range. Typical values areshown in Figure 24.
NEMA defines pull-in torque as the maximumconstant torque under which the motor will pullits connected inertia load into synchronism atrated voltage and frequency, when excitation isapplied. This is sometimes referred to as “load”pull-in torque.
RATIO LOAD Wk2 TO NORMAL LOAD Wk2
% S
LIP
% S
YNCH
RONO
US R
PM
0 20 40 60 80 100
8
10
6
4
2
0
92
90
94
96
98
100
Figure 22 Relationship between load inertia and minimumslip for pulling into step of a specific synchronous motor.
TORQ
UE
15
From Figure 24, we note that the inertia factorof the connected load may vary from 1 to 100. Forthe specific motor covered by Figure 22, pull-incan be affected at 8% slip with normal load WK2.However, with an inertia factor of 100, the motormust reach 99.2% of synchronous speed on thecage winding before the motor will pull into stepon application of excitation. A motor will thereforehave a much higher “load” pull-in torque whendriving a load having a low inertia factor thanwith a load having a high inertia factor.
It is rarely possible for the motormanufacturer to shop test the pull-in torque of amotor under actual load conditions. As a practicalsubstitute, the motor torque developed at 95%speed (“nominal” pull-in torque) is used as adesign and test point to predetermine the pull-incapability of the motor. Since the torque at 95%speed is a steady state condition, and actual pull-in is transient, it is necessary to take into accountinertia, motor reactance and other factors.
To investigate the relationship of “load” pull-in torque and “nominal” pull-in torque, assume anapplication of a centrifugal fan having an inertiafactor of 20, and connected to a 1000 HP, 1200 rpmmotor. The fan load at synchronizing speed, withclosed discharge, is 50% of full load torque.Assume it is determined that the motor slip mustnot exceed 1.8% if the motor is to pull into step onapplication of excitation. In this case, the effect ofreluctance torque is neglected and it is assumedexcitation may be applied at the worst phase anglefor synchronizing.
In Figure 25, point A is the speed which mustbe attained on the amortisseur winding to permitsynchronizing. Assume that the torque developedby the amortisseur winding is directlyproportional to slip from zero slip (synchronousspeed) to 5% slip (95% speed). We can thenextend line 0A, as a straight line to B, at 95%speed. Point B will then determine the torquerequired at the 95% speed point (in this case135%). This value is the “nominal” pull-in torque.
If, instead of the centrifugal fan discussed above,a motor of this rating were connected to a centrifugalpump, again having a load a synchronizing speed of50% but having an inertia factor of 1, it would pullinto step, on application of excitation, from a speedof 96% or at a slip of 4%. This is indicated by pointC. The corresponding “nominal” pull-in torque atpoint D is 63%.
This illustrates two applications, each havingload pull-in torques of 50%, one requiring 135%“nominal” pull-in torque and the other 63%“nominal.” The difference is due to the connectedinertia load.
100
50
20
10
5
2
1
RATED HORSEPOWER0 500 1000 2000 3000 4000 5000
NORM
AL L
OAD
Wk2 IN
LB.
FT.
2 (THO
USAN
DS)
150 RPM
225 RPM
300 RPM
450 RPM
600 RPM
900 RPM
1200 RPM
1800 RPM
14
200
160
120
80
40
0
% SLIP
MOTOR TORQUE(FAN DRIVE)
MOTOR TORQUE(PUMP DRIVE)
12 10 8 6 4
B
DC A
2 0
86% SYNCHRONOUS RPM
% F
ULL
LOAD
TOR
QUE
88 90 92 94 96 98 100
NOMINAL PULL-INTORQUE LINE
LOADTORQUE
Figure 23 Curves show the relationship between rated horsepower,speed and normal load WK2 for synchronous motors.
Figure 25 Relation of load and “nominal” pull-in torque, which isthe torque required at 95% speed of pulling into step.
(Inertia Factor)Type of Load Load WK2/Normal Load2
Ball Mill 3Band Saw 100Centrifugal Fan 12-60Chipper 30-100Pump, Centrifugal 1Air Compressor 10Hammer Mill 25
Figure 24
16
Figure 26 shows various common loadshaving typical horsepower and speed values,typical inertia factors and corresponding values ofmaximum permissible slip for synchronizing, withtypical load pull-in torque values, andcorresponding nominal pull-in torques.
The load pull-in torque value (indicated incolumn 6) for chippers is merely friction andwindage as chippers must be started, accelerated,and synchronized unloaded.
In some cases, in order to obtain an adequaterate of acceleration, it will be desirable to designfor higher nominal pull-in torque values thanthose shown in column 7 of Figure 26.
Frequently, where it has been difficult for themotor to pull into step on application of excitation,raising the applied excitation voltage will facilitatepull-in.
During the starting interval, the field windingis short-circuited through the field dischargeresistor. As shown in Figure 27, the frequency ofthe field discharge current is line frequency (60Hertz on a 60 Hertz system) at standstill, decliningto 3 Hertz at 95% speed. The time constant of thefield winding is quite long. The excitation currentbuilds up relatively slowly when excitationvoltage is applied. As shown in Figure 28, if theexcitation is applied at the proper interval of thefield discharge current wave, it will readily buildup to a maximum value, being assisted by the fielddischarge current, thus pulling the rotor into steppromptly and with a minimum of linedisturbance. Improved synchronization can beobtained by proper timing of field application.
1 2 3 4 5 6 7Ball Mill 1000 180 4 4.3 100 115Fan 1000 1200 20 1.8 50 138Pump 1000 1200 1 4.0 50 63Centrifugal Compressor 2500 900 15 1.8 50 138Chipper 1500 277 50 1.0 10 50
1. Type of load2. Horsepower3. RPM4. Assumed inertia factor5. Maximum slip from which motor will pull-in, in percent6. Load torque required at pull-in7. Nominal pull-in torque
Figure 26
Figure 27 Frequency of the current induced in the fieldwinding of a synchronous motor declines as the motor comesup to speed.
Figure 28a Oscillogram shows synchronizing of the motor byapplication of dc excitation. If the point of application anticipatesthe interval before synchronizing as shown, the excitation willreadily build up to pull the rotor into step smoothly.
Figure 28b Stator Current. Rotor pulls into synchronism with minimum stator current. (Note: Pulsations of current aftersynchronism are oscillations due to very large flywheel effect of the load.)
Inducedfieldcurrent
Excitationappliedhere
2.0
Rotor pulls promptly intosynchronism without pole slippage
2.5
INDUCED FIELD CURRENT
TIME (seconds)
Excitation applied here
Rotor pulls promptlyinto synchronismwithout pole slippage
TORQ
UE
60 CYCLESFREQUENCY OF FIELD-DISCHARGE CURRENT MOTOR SYNCHRONIZED
CORRESPONDING MOTOR SPEED
30 CYCLES 6 CYCLES
0% 50% 90%
3 CYCLES
95% FIELD EXCITATIONAPPLIED
X
Synchronous TorqueThe unlike poles of the stator rotating
magnetic field and of the rotor lock together andoperate simultaneously. Figure 29 representsconditions of a typical motor at no load, rotating ina counter-clockwise direction. The South pole, S,of the rotor is directly opposite the North pole, N,of the stator. At that point all torques are zero.However, at any displacement from that position,as with the rotor pole dropping back to theposition S', a resultant torque along curve C willbe developed. This is the resultant of torquecurves A and B.
The torque curve A is due to magneticreluctance. It becomes zero at 90° displacementand is negative from that point to 180°. Torquecurve B is that due to definite polarity from theexcited poles. It becomes a maximum at 90° andzero at 180°.
The resultant C curve is the synchronoustorque. It reaches a maximum at approximately70° lag of the rotor behind the rotating statormagnetic field and is unstable from that point to
17
STATOR MAGNETIC FIELD ROTATION
CB
Ah
m
d0° 90° 180°
δ
32°
N
N
S
S S'
ROTOR
dotted line –unstableoperation
magneticflux inair-gap
electricaldegrees
180°. If the load suddenly increases, the rotor willpull back and oscillate at its natural frequencyabout the next load point. There will be a slightvariance in the speed for a moment.
Synchronizing power is measured in kilowattsalong the line d-m. The angle of lag is the angulardistance between S and S'. In this case we assumeit to be 0.56 radian (32°) at full load. Thensynchronizing power Pr is defined as:
Pr =kWd
Pr = Synchronizing power in kilowatts perradian at full load displacement
kW = Power measured at motor shaftd = Displacement in electrical radians
Pr is important because it is a factor of thestiffness of magnetic coupling between the statorand the rotor. It affects the pull-out torque andalso the neutral frequency. Pr is approximatelyHP x 1.35 for unity power factor, and HP x 1.8 for0.8 leading power factor motors.
Figure 29 Diagram shows conditions in a synchronous motor when operating in synchronism and at no load. When the motoris loaded the rotor will drop back along the red curve C, the curve of synchronous torque, sufficiently to develop the load torque. Cis the resultant of magnetic reluctance torque, A, and the definite polarity torque, B. The maximum synchronous torque is reachedat about 70 electrical degrees lag of the rotor.
Pull-out TorqueNEMA defines pull-out torque of a
synchronous motor as the maximum sustainedtorque which the motor will develop at synchronousspeed for one minute at rated frequency and normalexcitation. Normal pull-out torque is usually 150%of full load torque for unity power factor motorsand 200% to 225% for 0.8 leading power factormotors. It can be increased by increasing the fieldstrength, which generally means increasing the air-gap and flux density. This usually means increasingthe physical size of the motor.
As shown in Figure 29, any sudden increase inload is accompanied by an increased angle positionof the rotor and stator and a corresponding increasein line current. Due to transformer action, there is atransient increase in excitation and pull-out torque.This increased value disappears in a few cycles so itcan be effective only on instantaneous peaks.However, the change in excitation can be used to“trigger” an increase in excitation voltage, thustending to maintain the value of excitationestablished by transformer action.
The reactance of the motor windings willdetermine the amount of increase in excitationcurrent and pull-out torque for a typical motor.The effect is shown in Figure 30.
18
Pulsating TorqueSince the rotor of a salient pole synchronous
motor has poles and interpole spaces, its magneticcircuit varies from point to point around the rotor.The instantaneous starting or accelerating torqueis greater when the flux set up by the stator passesa rotor position where the poles provide a goodflux path. Less torque is developed when passingbetween poles. It is common to show a torquecurve as a smooth curve from zero to 95% speed;this is really the average torque. At a given speed,the torque varies from maximum to minimum attwice slip frequency. At standstill, the torquepulses at 120 Hz. As shown in Figure 31, thetorque at 95% speed has a pulsating component at2 x 60 x (1.00-0.95) = 6 Hz; one cycle every 0.167second. At 80% speed this is 2 x 60 x (1.00-0.80) =24 Hz, or 0.042 second per cycle. Figure 32demonstrates torque pulsation at 80% speed.
The pulsating component of torque needs to becarefully evaluated as to its effect on the combinedmotor driven equipment. A typical pulsatingtorque is approximately 20% of rated torque.Typically, we have the mass of the motor, the gearand the load (i.e. compressor) all connected by theshaft. Each individual part has a natural frequencyas well as the combinations. The responsibility toavoid a torsional natural frequency that is nearrunning speed is that of the system designer.Further, special consideration has to be given to thepulsating torque in certain applications. Forexample, large 6 pole motors driving compressorsfrequently have a resonance between 60 and 80%speed. Fortunately, it is only present duringstarting. If the start-up is fast, the resonance haslittle time to build up. The motor manufacturerprovides curves of the average and the pulsatingtorque so that an overall system analysis can bemade. Figure 33 is a representation of a torque vs.speed curve. Shaft sizes may be adjusted or thecoupling changed, but the pulsating torque willstill be there.
Figure 30 Transient pull-out torque of a synchronousmotor. This torque is influenced by reactance of the motorwindings.
Synchronous motor 1750 HP at 240 RPM with Paper MillDripproof™ enclosure and double shaft extensions drivingvacuum pumps in a paper mill.
The ABC’s of Synchronous
Motors
TORQ
UE
200
175
150
125
100
00 2 4 6 8 10 12 16
PULL
-OUT
TOR
QUE
TIME IN CYCLES
TRANSIENT TORQUE
STEADY STATE PULL-OUT TORQUE I.O. P.F. MOTOR
TRANSIENT TORQUE
STEADY STATE PULL-OUT TORQUE O.8 P.F. MOTOR
.167 sec.
3 cycles
TIME IN SECONDS
~20%
0.0 0.1 0.2 0.3 0.4 0.5
.042 sec.
12 cycles
TIME IN SECONDS
~20%
0.0 0.1 0.0 0.3 0.4 0.5
95
% SYNCHRONOUS SPEED0
0
MOTOR %TORQUE
100
25 50 75 100
19
Figure 31Pulsations in total torquedue to reluctancevariations in magneticcircuit at 95% speed; 6 Hz.or .167 sec./cycle.
Figure 32Pulsations in total torque due to reluctancevariations in magneticcircuit at 80% speed; 24 Hz. or .042 sec./cycle.
Figure 33Typical speed-torque curveshowing average torque.
Starting & excitation methods for synchronous motors
STAR
TIN
G&
EXCI
TATI
ON
20
StartingAs will be noted from the application section,
the starting (zero speed) torques required ofsalient-pole industrial synchronous motors rangefrom 40% to 200%.
The required pull-in torque depends upon theload torque and the total inertia. As covered in theprevious section, the nominal pull-in torque is theactual torque the motor develops at 95% speed.
The starting kVA depends on both the startingtorque and the nominal pull-in torque. For a givenstarting torque, the starting kVA of a synchronousmotor in percent of the full-load kVA will increasewith an increase in the ratio of nominal pull-intorque to starting torque. Figure 34 shows sometypical values.
The starting kVA of synchronous motors oftentaxes the capacity of the generating or distributing
system to which it is connected. Extreme caremust then be taken in analysis of the system’scapacity, torque requirements, starting kVA, andstarting method.
Full-voltage starting, due to its simplicity,should be used wherever possible. Most systemscan stand dips of up to 20%. System capacities areusually expressed as MVA short circuit capacity.A quick approximation of the voltage dip duringstarting can be made by converting the motorinrush to MVA as follows:
SCMVA = System short circuit capacity in MVA
MMVA = (% motor inrush/100) x(motor kVA/1000)
MMVA = motor starting kVAMotor kVA = HP x 0.746/(P.F. x Eff.)
% dip = MMVA x 100/(SCMVA + MMVA)
In some cases a reduction in starting kVA isnecessary to limit voltage dip. Line diagrams forvarious reduced kVA starting methods areillustrated in Figure 35.
Figure 36 shows motor voltage, line kVA andmotor torque relationships for various methods of
% Starting % Nominal % StartingTorque Pull-in Torque Kva
80 60 35580 80 41080 100 470
Figure 34
Motor Voltage Line kVA Motor TorqueType of in % of in % of in % of FullStarting Full Voltage Full Voltage kVA Voltage TorqueFull Voltage 100 100 100ReducedVoltage 78 60 60Reactor 60 60 36
Figure 36
FULL-VOLTAGE
Full-voltage starting should always beused unless (1) limited capacity ofpower system makes reduced kVAstarting necessary, or (2) torqueincrements are required in starting.
Line kVA is reduced approximately asthe square of the motor terminalvoltage, as is the motor torque.Closed transition auto-transformerstarters avoid the line surgesexperienced with earlier open-transition switching.
Line kVA is reduced approximatelydirectly as the motor terminal voltage,but torque is reduced as the squareof the voltage. The starter has theadvantage of simplicity and closedtransition switching.
AUTOTRANSFORMER REACTOR
Figure 35 Common starting methods for synchronous motors.
21
starting. Figure 37 shows full voltage startingkVAs for 1.0 P.F. motors at high and low speeds.
It is important that sufficient torque bedeveloped to start and accelerate the loadproperly. In some applications, incrementalstarting which employs part winding, reactor,resistor, or autotransformer is used. The motorneed not start on the first step, but additionaltorque increments are successively applied untilthe motor starts and accelerates. This methodeases the mechanical shock on driven equipmentand, on regulated systems, permits restoration ofnormal voltage before application of the nexttorque and kVA step.
The motor pulls into step on application ofexcitation to the rotor field winding. The startingmethod applies excitation at the proper speed andphase angle to assure proper synchronizing.
ExcitationAll synchronous motors need a source of
direct current for their field winding. Historically,this was supplied by an exciter, a specialized dcgenerator which may have been direct-connectedto the shaft of the motor so that if the motor isrunning, dc is available. In the past, conventional
APPROXIMATE FULL-VOLTAGE STARTING kVA OF UNITY POWER FACTOR* SYNCHRONOUS MOTORS(In % of full-load kVA of motor; for various starting, pull-in and pull-out torques; for 50 and 60 Hertz motors)
High-Speed Motors (500 to 1800 rpm) Low-Speed Motors (450 rpm and Lower)% Torques % Starting kVA % Torques % Starting kVA
Starting Pull-in** Pull-out at Full Voltage Starting Pull-in** Pull-out at Full Voltage50 50 150-175 375 40 40 150 250-32550 75 150-175 400 50 75 150 350-37575 75 150-175 400 75 110 150 50075 110 150-175 500 100 50 150 375-400110 110 150-175 500 160 110 150 575125 110 200-250 550-600 100 110 200-250 600-650150 110 150-175 550 125 110 200-250 600-650175 110 200-250 600-650 125 125 200-250 600-650
*For 0.8 power factor motors the percent starting kVA will be approximately 80 % of the values shown.** The above pull-in torques are based on normal load WK2.NOTE: The percent starting kVA in the tables above are approximate for estimating purposes only.
Specific values for particular motors can be furnished.
Figure 37
exciters were small dc generators withcommutators to convert the generated voltage todc. Brushes allow the current to come from theexciter to the stationary control center. The currentis then routed back through slip-rings and brushesto the motor field. This system works very wellexcept for a minor nuisance, the brushes. Brushescan last as long as a year, but there is the carbondust to clean up and continual operator visualinspection is required. Occasionally conditions willchange and brushes will wear rapidly. Theselection of the proper grade of brush depends oncurrent density, ring material, ring speed,humidity, brush pressure, and many other factors.It may take a long time to find the right brush. Putthis together with the brush and ring maintenanceand there’s an opportunity for a better method.
Collector ringand brushesare not requiredor used withbrushlessexcitation.
STAR
TIN
G&
EXCI
TATI
ONBrushless excitation is more reliable withno brush or collector ring maintenance
By the 1960s, solid state diodes and thyristorshad advanced to where they could carry thecurrent and block the voltages as required. It wasthen that Electric Machinery ManufacturingCompany developed the brushless exciter (seeFigure 38). The exciter is physically directconnected as before. The rotor has a three phaseac armature winding. The stationary fieldwinding is on poles on the stator and is connectedto a variac and rectifier or equivalent source ofvariable dc. The generated ac current is directlyconnected along the shaft to a rotating diodewheel, where it is rectified to dc before going tothe motor field. The magnitude of the motor fieldcurrent is adjusted by changing the current to thestationary exciter field.
The most amazing part of this design was themounting of the control on the rotor. Starting asynchronous motor requires shorting the fieldwith a discharge resistor and blocking the dccurrent until the rotor is near full speed. The dccurrent is then applied and the discharge resistoris removed. For the brush type machines this wasin a control cabinet the size of a four drawer filecabinet. Now it is a small package on the rotor.
To show what is required of the control,consider the starting of a synchronous motor. Theac breaker closes, applying three phase voltage tothe stator winding. The stator winding has beenwound to form a number of magnetic poles
depending on the rated speed of the motor. Thesepoles on the stator cause a magnetic field to rotateat rated speed. The rotor has not started to move,so the field winding and cage are swept by therotating magnetic field. This will induce a veryhigh voltage (thousands of volts) in the fieldwinding. To avoid this, the field is shorted by adischarge resistor. In addition, this resistor isdesigned to give additional pull-in torque. As therotor accelerates, the frequency of the dischargecurrent is proportional to the difference in speedbetween the stator flux and the rotor winding (slip).At synchronous speed there is no slip, so no voltageis induced and no torque is produced by the cage.Since there is some load torque, the speed neverreaches synchronous speed but reaches equilibriumwhere the decreasing cage torque matches the loadtorque. At this point, dc is applied to the field,creating strong magnetic North and South poles.These are attracted to the opposite magnetic poleson the stator. If the attraction is great enough, therotor with its inertia will be pulled up to synchronousspeed. The rotor is thereafter locked to the rotatingstator magnetic field.
Field Application SystemFunctions of Field Application System:
1. Provide a discharge path for the currentinduced in the field of the motor duringstarting, and open this circuit when excitationis applied.
2. Apply field excitation positively when themotor reaches an adequate speed. Thisexcitation should be applied with suchpolarity that maximum torque will beobtained at the time of pull-in.
3. Remove excitation and reapply the fielddischarge resistor immediately if the motorpulls out of step.
The circuit for accomplishing this is shown inFigure 39. The field discharge resistor protects themotor field winding from the high voltageinduced in starting and provides the voltagesource for the control circuit. The ac output of theexciter is converted to dc by the rotating rectifierdiodes. This output is switched on or off to themotor field winding by silicon controlled rectifierSCR-1, which is gated by the control circuit.
22
Figure 38 Shaft mounted brushless exciter.
Field discharge resistors
SCR-2 and D10
Control module
Exciter armature
Control circuitThe control circuit keeps the SCR-1 from firing
until the induced field current frequency is verylow, representing a close approach to synchronousspeed, and then fires the rectifier SCR-1 at theproper time and applies excitation to thesynchronous motor field. At the same time, thefield discharge resistor is removed from the circuit.This is done by the inherent operatingcharacteristics of silicon controlled rectifier SCR-2.This frequency sensitive part of the control circuitassures that field excitation is applied at the properpull-in speed for successful synchronizing and atthe proper polarity to give maximum pull-in torquewith minimum line disturbance.
The control circuit operates to remove excitationshould the motor pull out of step due to a voltagestep or excessive mechanical load. On the first halfcycle after pull-out, the induced field voltage causesthe net field current to pass through zero, turningSCR-1 off, automatically removing excitation. SCR-2operates to connect the field discharge resistor backin to the circuit. During this time the motoroperates as an induction motor. When conditionspermit, field is then re-applied as during starting.
In Figure 39, the voltage from the exciter-rectifier is blocked by SCR-1 until the point ofsynchronization. The field has an alternatingvoltage causing current to flow first through SCR-2and the discharge resistor. On the next half cycle,current flows through the diode and dischargeresistor. The control circuit waits until thefrequency drops to the preset value, indicating therotor is at an adequate speed. Then, after a Northpole on the stator is in the right position to beattracted to what will be a South pole on the rotor, ittriggers SCR-1 to apply excitation. If the rotor doesnot synchronize, it will slip a pole, the induced fieldvoltage will oppose the exciter voltage causing thecurrent to go to zero, turning SCR-1 off. SCR-2 is
23
DC TOEXCITER FIELD
3 PHASEAC
1 2 3FDR
SCR-2
D104 5
6
SCR-1
FIELD FIELDARMATURE
ROTATING EQUIPMENT
EXCITER ROTATING HEATSINK ASSEMBLY
SYNCHRONOUSMOTOR
ARMATURERECTIFIER FIELD APPLICATION CIRCUIT
“SYNC-RITE”™CONTROL
UNIT
Figure 39
turned on only at a voltage higher than the excitervoltage so it will not be on when SCR-1 is on.
Occasionally a lightly loaded motor willsynchronize without excitation being applied. Thisis due to the reluctance torque. Reluctance torqueresults from the magnetic circuit having lessreluctance when the poles line up with the statorflux. The EM design includes a “zero slip” circuit toapply excitation in these situations.
While this may seem complex, it has proven tobe a highly reliable system.
Features of the brushless exciter
1. No brushes, no carbon dust problems, nobrush maintenance.
2. No commutator or slip rings to resurface.
3. Completely automatic field application at thebest angle for sure synchronization.
4. Removal of excitation and application of fielddischarge resistor in the event of out of stepoperation and automatic resynchronization.
5. No sparking. Can be used in hazardous areas.
6. No field cubicle with FDR and field contactorto maintain.
Where synchronous motorsshould be applied
24
Synchronous motors can handle any loadwhich can be driven by a NEMA design Bsquirrel-cage motor. Whether they should be usedfor any specific application is a matter of someinvestigation. A rough rule of thumb is thatsynchronous motors are less expensive thansquirrel-cage motors if the rating exceeds 1 HP perrpm. However, that considers only initial costand does not take into account:
1. Higher efficiency of the synchronous motor.
2. Power factor improvement of the synchronousmotor.
These two factors become important at speedsbelow 500 rpm where induction motorcharacteristics leave much to be desired. On theother side of the balance are these:
3. Necessity for excitation source and field controlmeans for the synchronous motors.
4. Relatively low torque/kVA efficiency instarting.
5. Slightly greater maintenance cost, especiallywith motors with slip rings and brushes.
A few general rules may be established as apreliminary guide in selection of synchronous vs.induction motors, but specific local conditionsmust govern:
1. If especially low starting kVA, controllable torqueor adjustable speed are desired, a synchronousmotor and slip coupling may be used.
2. At 3600 rpm, synchronous motors aresometimes used from 10,000 - 20,000 HP.Above that range they are the only choice.
3. At 1800 rpm, motors show an advantage above1000 HP. From 2000 to 10,000 HP, synchronousmotors are used if power factor improvement isespecially important. Above 15,000 HP, they arethe only choice.
4. Induction motors for operation below 500 rpmhave lower efficiency and a lower power factor.Synchronous motors are built at unity orleading power factor and with good efficienciesat speeds as low as 72 rpm. For directconnection in ratings above 1000 HP and inspeeds below 500 rpm, the synchronous motorshould be the first choice for compressors,grinders, mixers, chippers, etc.
Figure 40 is a general sizing chart forsynchronous and induction motors.
Leading power factor motors are slightly lessefficient than unity power factor motors, but theydo double duty. Their application depends on thenecessity for power factor correction.
Induction motors require from 0.3 to 0.6reactive magnetizing kVA per HP of operatingload. 0.8 leading power factor synchronousmotors will deliver from 0.4 to 0.6 correctivemagnetizing kVA per HP depending on themechanical load carried. Equal amounts ofconnected HP in induction motors and 0.8 leadingpower factor synchronous motors will result inapproximate unity power factor.A
PPLIC
ATIO
NS
Synchronous motor 2500 HP at 327 RPM with WPII enclosuredriving a reciprocating compressor in a refinery.
25
Synchronous motors for air and gascompressors and vacuum pumps
Air and gas compressors and vacuum pumpsmay be reciprocating, rotary or centrifugal.Synchronous motors may be used with any ofthese types. Detailed application informationfollows:
Reciprocating compressors and vacuum pumpsMany more synchronous motors are direct
connected to reciprocating compressors than areapplied to all other types of loads combined.Factors contributing to this are:
1. Low starting and pull-in torque requirement ofunloaded reciprocating compressors.
2. Elimination of belt, chain and gear drives.
3. High efficiency of low speed synchronousmotors direct connected to compressors.
4. Power factor of low speed synchronous motorsdirect connected to compressors.
5. Minimum floor space requirement.
6. Low maintenance cost.
20,000
15,000
10,000
7,500
5,000
2,000
1,000
500
3600
HORS
EPOW
ER
RPM1800 1200 900 720 600 514 450 400 360 327 300
SYNCHRONOUS
INDUCTION
EITHERSYNCHRONOUSOR INDUCTION
NEMA Recommended Torques in % ofTorques for Full-load TorqueReciprocating Compressors Starting Pull-in Pull-outAir or Gas –Starting Unloaded 40 30 150Vacuum Pumps – Starting Unloaded 40 60 150
Figure 41
Figure 40 General areas of application of synchronous and induction motors.
Motors designed for direct connection toreciprocating compressors are usually engine typeor single bearing construction. The stator ismounted on soleplates set in the foundation. Aspecial case is the flange-mounted motor in whichthe stator frame is mounted on a flange on thecompressor frame. In either case, torquerequirements are usually within the range ofnormal design, low speed motors. See Figure 41.
The torque values may be exceeded, for certaincompressors and vacuum pumps, the manufacturershould be checked for specific values.
26
Current pulsationA reciprocating compressor has a varying
torque requirement per revolution, depending onnumber of cylinders, crank angle, etc. As discussedpreviously, the angle of rotor lag and the amount ofstator current vary with torque. There will be acyclic pulsation of the load current during eachrevolution. This in itself is not too objectionable,but excessive current pulsation may result inappreciable voltage variation and thus affect lightsor other devices sensitive to voltage change.
Present standards limit current pulsation to 66%of rated full-load current, corresponding to anangular deviation of approximately 5% from auniform rotative speed. The current pulsation is thedifference between maximum and minimum valuesexpressed in percent of full-load current. Figure 42shows an actual oscillogram of current input to asynchronous motor driving an ammonia compressor.
In some cases, step unloading of compressorsinvolving successive unloading of various cylinderends is used. This introduces additionalirregularity into the crank effort and usuallyincreases the current pulsation.
Natural frequencyAny system having mass in equilibrium and
having a force tending to return this mass to itsinitial position, if displaced, will tend to have a
natural period of oscillation. The commonpendulum is a good example.
The natural frequency of this oscillation in asynchronous motor is determined by the formula:
NF =35200 Pr x Hz
N WK2
NF = Natural frequency in oscillations per minuteN = Synchronous speed in revolutions per minutePr = Synchronizing powerHz = Line frequency (CPS)
WK2 = Flywheel effect in foot pounds squared
It is assumed there is no effective dampingand the motor is connected to an infinite system.This natural frequency should differ from anyforcing frequency by at least 20%
Compressor factorNeglecting inertia and damping forces, a
single cylinder, double acting or a two cylinder,single acting compressor will have a crank effortdiagram similar to Figure 43. A two cylinder,double acting or a four cylinder, single actingcrank effort diagram is illustrated in Figure 44.
Various factors may tend to disturb thesymmetry of the torque effort diagramsillustrated. For instance, on the two cylinder,
ONE REVOLUTION
DEGREES0 180 360
TORQ
UEIN
FOO
T PO
UNDS
BA
C
Figure 42Oscillogram of
synchronous motordriving a reciprocatingcompressor. Line A is
envelope of current wave,B-C is current pulsation.
B-C divided by rated full-load motor current is
percent pulsation = 55.2%.
Figure 43Crank effort diagram for a one
cylinder, double-actingcompressor, or two cylinder,
single-acting compressor.
����
APPLIC
ATIO
NS
NEMA Recommended Torques in % of Full-load TorqueTorques for Compressors Starting Pull-in Pull-out
Compressors, centrifugal – starting with:a. Inlet or discharge valve closed 30 40-60 150b. Inlet or discharge valve open 30 100 150
Compressors, Fuller Companya. Starting unloaded (by-pass open) 60 60 150b. Starting loaded (by-pass closed) 60 100 150
Compressors, Nash-Hytor – starting unloaded 40 60 150
Compressors, reciprocating – starting unloadeda. Air and gas 30 25 150b. Ammonia (discharge pressure 100-250 psi) 30 25 150c. Freon 30 40 150
Figure 46
27
double acting compressor with 90° cranks, a“galloping” effect due to acceleration of theunbalanced reciprocating parts is introduced onceper revolution.
Two-stage compressors usually do not haveexactly equal loading on all cylinder ends. In somecases, reciprication compressors operate on two ormore suction systems operating at differenttemperatures and at different pressures. Variousarrangements to secure part-load operation such aslifting suction valves, opening clearance pockets,cutting in partial by-pass systems, etc., mayintroduce various degrees of unbalance. Mosttypes of compressors are covered by a NEMA tablein which each common application is assigned anapplication number and “Compressor Factor.”Figure 45 illustrates percent current pulsation forNEMA application No. 5 covering a single stage,two cylinder, double acting compressor with 90°cranks, plotted against “Compressor Factor” “X” asthe abscissa. This curve shows that with “X”values of 2.0 to 6.0 or above 12.0 the currentpulsation will not exceed 66%.
Flywheel effectThe required flywheel effect to limit current
pulsation is proportional to the compressor factor“X” shown in this formula:
WK2 = X x Hz x Pr x GWK2 = Total flywheel effect in foot pounds
squaredX = Compressor factor
Hz = Line frequency in CPSPr = Synchronizing power
G = 1.34 x (100)4
rpm4
Flywheel effect values calculated by the aboveformula and using compressor factors establishedby NEMA will satisfy the condition that thenatural and forcing frequencies differ sufficientlyto meet allowable pulsation.
Centrifugal compressors and vacuum pumpsVarious types of centrifugal, positive
displacement compressors are used to compressair or various gas or to serve as vacuum pumps.The Hytor, or water seal type, is frequently usedas a vacuum pump in paper mills, and NEMArecommends special torque consideration. SeeFigure 46.
ONE REVOLUTION
DEGREES0 90 270180 360
TORQ
UEIN
FOO
T PO
UNDS
0
200
240
160
120
8066%
40
0
COMPRESSOR FACTOR “X”2 4 6 8 10 12 14 16
CURR
ENT
PULS
ATIO
N “Y
”
Figure 44Crank effort diagram for a twocylinder, double-actingcompressor, or four cylinder,single-acting compressor.
Figure 45 X-Y curve for determining flywheel requirements.
28
Centrifugal fans, blowers, compressors, and exhausters
Centrifugal fans, blowers and compressorsdiffer from reciprocating units in that the pressuredeveloped is a function of the density of the air orgas handled, the speed of the impeller, and ofrestriction to flow. The gas is accelerated inpassing through the rapidly rotating impeller, thisvelocity being converted into pressure in thevolute casing surrounding the impeller.
Units operating at pressures above 30 psi arecalled compressors. For pressures above 10 psi,multi-stage compression is commonly used. Theseunits are especially suited for high volume at lowpressure. NEMA recommended torques areshown in Figure 47.
Many of these applications have substantiallymore than normal load WK2. The actual valueshould be taken into account in applying anddesigning motors for such loads. Care should alsobe taken in applying motors to fans normallyhandling hot gases. The load, when handling coldair, as in starting, will be substantially higher.
Synchronous motors for centrifugal pumpsThe term “centrifugal” may be applied to
radial-flow, mixed-flow or axial-flow pumps. Inall cases, energy is added to the flowing liquid bymeans of pressure differences created by vanes ofa rotating impeller or propeller. The radial-flowpump discharges the liquid at right angles to theshaft, the mixed-flow discharges it at an angle, andthe axial-flow propels the liquid in an axialdirection.
Hydraulically, the difference is fundamentallyone of the “specific speed” used in the design.Mathematically, it is shown by this formula:
Ns =NQH3⁄4
Ns = Specific speedN = rpmQ = gpmH = Total head in feet where Q and H
correspond to conditions at whichmaximum efficiency is obtained
These differences are important in theapplication of synchronous motors, as can be seenin Figure 48.
The important factor is the torque required atfull speed and closed discharge. In the radial-flowcentrifugal pump this is usually about 55% of fullload torque. In the axial-flow it may reach 220%.Obviously, no attempt should be made tosynchronize a motor driving a mixed-flow oraxial-flow pump under shut-off conditions.However, the inertia of the water column alonesimulates a partially closed discharge in case ofattempted rapid acceleration, as in the case ofpulling into step. See Figure 49.
The pull-in torques given above areinadequate where mixed-flow or axial-flowpumps are concerned. In some cases it is possibleto start these pumps with partially empty casings,thus reducing pull-in requirements.
On vertical pump motors, additional thrustcapacity is usually required where rigid couplings orhollow shafts are used, and the motor thrust bearingmust support all rotating parts plus hydraulic load.Where thrust loads are unusually high, the break-away or starting torque required may beconsiderably in excess of values in Figure 49. In suchcases hydraulic lift bearings should be considered.
����
APPLIC
ATIO
NS
Type Specific Speed % Torque at Shut-off
Radial-flow 500 451000 502000 603000 70
Mixed-flow 5000 120Axial-flow 10000 220
Figure 48
NEMA Recommended Torques Torques in % of Full-load TorqueStarting Pull-in Pull-out
Blower, Centrifugal –Inlet or discharge valve closed 30 40-60 150Inlet and discharge valve open 30 100 150Compressor, Centrifugal Inlet or discharge valve closed 40 40-60 150Inlet and discharge valve open 40 100 150Fans, Centrifugal (Except Sintering)Inlet and discharge valve open 30 40-60 150Inlet and discharge valve open 30 100 150Fan, Sintering –Inlet gates open or closed 40 100 150Fan Propeller – Discharge open 30 100 150
Figure 47
29
Synchronous motors for crushers,grinders, and mills
Crushing, grinding or pulverizing is anecessary step in separation of metal from ore,preparation of crushed rock for the constructionindustry, preparation of agricultural limestone,and manufacture of portland cement.
Primary crushers, usually jaw, gyratory orhammermill type, are fed the rock or ore directlyafter blasting. The material next goes to thesecondary crusher where reduction to about 1/2"may be accomplished. Secondary mills may begyratory, cone, hammermill, or roll type.
Rod mills may also be used as secondarygrinders. A rod mill consists of a steel shell linedwith wear-resistant material and rotated about ahorizontal cylinder axis. The mill uses steel rods 2to 5 inches in diameter running the length of themill. As the mill rotates, the tumbling action of therods causes grinding.
Final grinding, possibly down to a materialwhich will pass through a 200 mesh screen orfiner, is usually done in a ball mill. This is similarto a rod mill except that steel balls tumble over thematerial to be crushed. Autogenous mills use thecrushed rock to grind itself to powder. See Figure50 for NEMA recommended torques for thevarious types of mills.
Synchronous motors for pulp and paper millsOne of the largest power consuming
industries is that of processing wood and woodproducts in the manufacture of paper and pulp.
Although alternative fibers may be used inrelatively small quantities in the manufacture ofpulp and paper, by far the largest source is wood.Paper may be made either by the groundwood orchemical process. The application table andrecommended NEMA torques shown in Figure 51cover the usual synchronous motor applications inpaper mills, most of which are common to bothprocesses.
Paper mills are large users of synchronousmotors for pumps, refiners, chippers and otherequipment.
NEMA Recommended Torques in % ofTorques Full-load Torque
Starting Pull-in Pull-outRefiners (unloaded) 50 50-100 150Conical (Jordan) disc 50 50 150Chippers – Empty (1) 60 50 250Grinders (unloaded)Magazine 50 40 150Pocket (unloaded) 40 30 150Vacuum Pumps (Hytor) 60 100 150
(1) These are high-inertia loads and motor torque requirements can not be determined from load values alone. WK2 values must be known to permit proper motor application.
Figure 51
Torques in % of Full-load TorqueApplication Starting Pull-in Pull-outGyratory (unloaded) 100 100 250Cone (unloaded) 100 100 250Hammer Mill (unloaded) 100 80 250Roll Crusher (unloaded) 150 100 250Rod Mill – Ore 160 120 175Ball Mill – Ore 150 110 175Ball Mill – Rock or Coal 150 110 175
Figure 50
NEMA Recommended Torques Torques in % of Full-load Torquefor Pumps Starting Pull-in Pull-outPumps, axial flow, adjustable blade – starting with:a. Casing dry 5-40 15 150b. Casing filled, blades feathered 5-40 40 150
Pumps, axial flow, fixed blade– starting with:a. Casing dry 5-40 15 150b. Casing filled, discharge closed 5-40 175-250 150c. Casing filled, discharge open 5-40 100 150
Pumps, centrifugal, Francis impeller– starting with:a. Casing dry 5-40 15 150b. Casing filled, discharge closed 5-40 60-80 150c. Casing filled, discharge open 5-40 100 150
Pumps, centrifugal, radial impeller– starting with:a. Casing dry 5-40 15 150b. Casing filled, discharge closed 5-40 40-60 150c. Casing filled, discharge open 5-40 100 150
Pumps, mixed flow – starting with:a. Casing dry 5-40 15 150b. Casing filled, discharge closed 5-40 82-125 150c. Casing filled, discharge open 5-40 100 150
Pumps, reciprocating – starting with:a. Cylinders dry 40 30 150b. By-pass open 40 40 150c. No by-pass (three cylinder) 150 100 150
Figure 49
Adjustable SpeedSPEED
30
Some applications have varying requirements.For example, a process may at times require morewater or air. The pump or compressor and motorare sized for the maximum, and a throttle is usedto reduce the flow. This does not reduce thepower used substantially, since the reducedquantity is offset by the greater head caused by thethrottle. See Figure 52.
To overcome this, a slip clutch (eddy currentdrive) is sometimes used. Electric Machinerymanufactures a magnetic drive for this use. Herethe motor runs at rated speed and the magneticdrive slips, allowing the compressor to run at alower speed. At the lower speed the output and thetorque required are reduced. Although the torqueis reduced, the motor still runs at rated speed, sothe HP is down in proportion to the torque. Thereis some slip loss in the slip clutch. This type ofvariable speed drive is declining in use.
Adjustable speed with synchronous motorsWhen one thinks of synchronous motors, the
idea of a motor running at an absolutely constantspeed comes to mind. While this mental picture iscorrect, it is so only because the power line frequencyis a constant, and synchronous motors are “lockedin” to the line frequency.
As technology has advanced over the latterthird of the twentieth century, the ability to changethe frequency of the incoming line power - and thus,to control of the speed of “constant speed” motors -has become a reality. This has led to adjustablespeed drives for pumps, fans, compressors andother loads, yielding higher operational efficiencies,lower wear rates, softer starting, lower startingcurrents, and even quieter operation as machinesrun at reduced speeds. See Figure 54 and compareit with Figure 53 to see potential efficiencyimprovements with adjustable frequency vs. eddycurrent drive.
A “tool” which has come out of developmentefforts and is finding application is the LoadCommutated Inverter, or LCI. The LCI isreminiscent of the “incher,” an array of sixcontactors used to switch direct current insequence, first positive polarity and then negative,to the three phases of a normally excitedsynchronous motor, thus enabling it to slowly“inch” its load around to position it formaintenance reasons. The difference is that theLCI can do this job at a wide variety offrequencies, including frequencies higher than linefrequency, thus even enabling supersynchronous(> synchronous speed) operation. The incher waslimited to one slow speed.
The LCI drive (also commonly referred to as“VFD,” or variable frequency drive) consists of anincoming section which supplies direct current viaa current-smoothing choke to an inverter section.The inverter supplies dc power to the synchronousmotor by switching it between phases at thedesired frequency corresponding to speed.Because of the functions being accomplished, theinverter and motor together are sometimes referredto as a brushless dc motor. Refer to Figure 55.
Fan head characteristicat constant speed
Required head, orsystem resistancecurve
Excess head, tobe dissipated
by throttling
0 20 40 60 80 100 120Volume flow rate or capacity
Syst
em re
sist
ance
or f
an h
ead
14
12
10
8
6
4
2
0
Figure 52 System resistance and fan head characteristics.
31
Typical fan powerinput curve,throttled
Eddy current driveinput power required
T x constant RPM(speed squared curve)
Throttlinglosses saved byusing magnetic
drive
% Rated Output
% R
ated
Sha
ft Ho
rsep
ower
100
75
50
25
00 25 50 75 100
Typical fan powerinput curve,throttled
VFDinput power
requiredT x variable RPM
(speed cubed curve)
Throttlinglosses saved by
using VFD
% Rated Output
% R
ated
Sha
ft Ho
rsep
ower
100
75
50
25
00 25 50 75 100
Figure 53 Power into eddy current drive is product of torquerequired and constant input rpm. With an unthrottled fan orpump, the torque requirement is nearly a direct function of speedsquared, thus, input power varies as the output speed squared.
Figure 55 Load commutated variable frequency drive (6-pulse)
Figure 54 Power into variable frequency drive is the product of thetorque required and the output rpm. With a load as in figure 53, theinput varies as the speed cubed.
CURRENT FLOW
TRANSFORMER
BRUSHLESS DC MOTOR
MOTOR
ELECTRONICCOMMUTATOR
DCSUPPLY
CONVERTER
POWERFLOW
ACBREAKER
HIGHVOLTAGESUPPLY
+
–
A smoother-acting system can be assembledby taking advantage of the 30º difference betweenthe secondary voltages in wye and delta-connected transformers. Using this motor andtransformer combination, a twelve-step invertercan be applied.
As with any thyristor circuit, once a thyristor isturned on, the current must be reduced to zerobefore the thyristor returns to the non-conductingstate. With an excited synchronous motor as theload, the commutation is accomplished using theopen circuited winding to generate opposingvoltages in the machine. (Figures 56 and 57.) Toprovide reliable operation at low speeds, encodersare sometimes used to indicate the instantaneousposition of the rotor to the control so thatsucceeding thyristors can be turned on at the properinstant. Variations in technologies also exist whichuse no encoders, but work by sensing the backvoltages from the machine. Whichever method isused, a rotating magnetic field is established.
The simplest type of LCI uses a six-stepapproach. Refer to Figure 59 and note theprogression of the resultant vector as dc is appliedin sequence and at alternating polarity to each phaseof the machine. The vector, representing themagnetic field, progresses around the stator in onecomplete cycle.
The six-step approach, while the simplest, isnot the smoothest. The pulsations associated withthe finite progression of magnetic fields around thestator are reflected in torque pulsations andmagnetic noise. Care must be exercised to avoidcritical frequencies in the motor shaft and otherelements which could respond to pulsations set upby the inverter action. In some cases, specialcouplings must be applied to avoid prematurefailures. The six-step approach is commonly usedfor horsepower in the 500-2500 range, sometimesup to 6000HP.
32
C-A
4 6 2
1 3 5
A
B
C
C-A
4 6 2
1 3 5
A
B
C
Figure 56 Load Current Path (C-A)
Figure 57 Load Current Path (C-B)
Figures 58 Back to backtest of 2-14200 HP 3200RPM 2 pole synchronousmotors for VFD application.The shaft extensions arecoupled together, and withtheir respective VFD's oneoperates as a motor and theother operates as agenerator, achievingessentially full loadoperation of each machine.In this type of test the localutility must only make upthe losses of the machinesand VFD's.
SPEED
Refer to Figure 60. In Figure 61, the progression ofthe vector can be followed as above, but the stepsnow are at 30º intervals rather than at 60º, as in thesix-step inverter. Another significant advantage isa reduction in harmonics imposed on theincoming line as well as torque harmonics to theload. Figures 62 and 63 further illustrate thedifferences between six and twelve-step inverteractions. The fifth, seventh, seventeenth,nineteenth, and many higher order harmonicscancel in the wye and delta secondaries; they donot appear in the incoming power to the twelve-step converter. The result is smoother operation,fewer concerns about torsional stresses, quieterrunning, lower harmonics on the incoming line,and a reduced output requirement from eachthyristor bridge.
While two transformers could be used to feedthe twelve-step converter, a three-windingtransformer is frequently used (note in Figure 60).Although this could be done with transformersbetween the inverter and the motor, the motor,rather, is frequently wound with a dual windingto isolate the outputs of the two inverter sections.The two separate windings are shifted 30 electricaldegrees. Each part of the motor stator actuallysees a six-step progression, but the rotor sees thetwelve-step sum of the two inputs to the machine.The twelve step approach is typically used forhorsepowers of 2000 and larger, even up to 50,000.
Current-limiting functions are often employedduring starting, keeping currents to valuesconsiderably under normal locked rotor values. Thecurrent limit is adjustable and is set as a function ofthe amount of torque required. Available startingtorque, when operating with an LCI, is more like themaximum operating torque of the motor, since it is
33
4 6 2
1 3 5
T1
Windings shifted by 30 electrical degrees
Winding example of 6 pulse machine Both windings used on 12 pulse machine
2 6 4
5 3 1
T2
T3
T1'
T2'
T3'
Υ
∆
1.
T3 T2
T1
T3'
T2'
T1'
-T2
T1T1-T2
2. -T2
T1T1-T2
3. -T3
T1 T1-T3
-T3
T1'
T1'-T3'
-T3
T1'
T1'-T3'
PLAIN
PLAIN WINDING PRIME WINDING
PRIME RESULTANT
T1'-T2'
T1'-T2'T1-T2
T1-T2
T1-T3
T1'-T2'
T1'-T3'
T1'-T3'
Figure 59 Single Winding 6 Pulse.
Figure 62 SCR firing sequence for 6 pulse and 12 pulse drive.
Figure 61 Dual Winding 12 Pulse. (Only 3 of the 12 pulses are shown)
Figure 60 Top: 6 Pulse Single Winding.Bottom: 12 Pulse Dual Winding.
1.
T3 T2
T1
-T2
T1T1-T2
2. -T3
T1T1-T3
3.
-T3T2
T2-T3
6.
T3-T2
T3-T2
5.
T3
-T1 T3 -T1
4.
T2
-T1T2-T1
1 2 3 4 5 6 1 2 3 4 5 6 CommentSingle Winding
60 X X T1 T2120 X X T1 T3 SCR’s180 X X T2 T3 Switch 240 X X T2 T1 Every300 X X T3 T1 60°360 X X T3 T2
30 X X T1 T2 X X T1' T2'60 X X T1 T2 X X T1' T3'90 X X T1 T3 X X T1' T3'120 X X T1 T3 X X T2' T3'150 X X T2 T3 X X T2' T3' SCR’s180 X X T2 T3 X X T2' T1' Switch 210 X X T2 T1 X X T2' T1' Every240 X X T2 T1 X X T3' T1' 30°270 X X T3 T1 X X T3' T1'300 X X T3 T1 X X T3' T2'330 X X T3 T2 X X T3' T2'360 X X T3 T2 X X T1' T2'
Degr
ees
12 P
ulse
Dua
l Win
ding
6 St
ep S
ingl
e W
indi
ng
Curr
ent I
nCu
rren
t Out
Curr
ent I
nCu
rren
t Out
SCR SCR'
Id
(a)
synchronized with the very low frequency source.Greater starting torques may be available at thesame time that lower starting currents aredemanded when compared with across-the-linestarting. Reduced speed operation is controlled at aconstant volts per hertz basis so the magneticstructure is not saturated.
Occasionally, a by-pass contactor or circuitbreaker will be supplied to connect the linedirectly to the machine. This will affect the cost ofthe installation, but it may also be desired by theuser as a backup system. It is also possible to startor operate multiple machines while using only oneLCI. For example, there may be two pumps; onemay be operating via the LCI, but its capacity hasbeen exceeded by the demand. This machine canbe transferred to across-the-line operation via theby-pass contactor and the LCI transferred to thesecond machine, which it can bring up tooperating speed, increasing the total capacity in avery smooth and bumpless manner. Refer toFigure 64.
Occasionally, it will be desired that voltagewill be increased to rated level at speeds over 80%(of line frequency rating). For this condition, themotor is built with enough iron so that magneticsaturation does not become problematic. Theresult is that the machine can run without a powerfactor penalty. Low voltage means delayed firingangles in the front (input) end of the converter,and that translates into a low power factor. Sinceit is common for large machines (forced draft andinduced draft fans) to run at speeds between 80%and 100%, it is desirable to operate at 100% voltageand realize the benefit of the improved powerfactor (as seen by the line) available from 80 to100% speed.
While unity or leading power factors aregenerally associated with synchronous motors,this feature does not hold true when the motor isoperated through an LCI. The motor will beoperating at a leading power factor, but the linedoes not see the motor; it sees the input section ofthe LCI, and that is always a lagging power factorload. The lower the speed (voltage) the morelagging the power factor, and the best powerfactor is still only about 90% when output voltageto the motor is at maximum.
Cooling the machine may be by use of anintegral fan if the load is variable torque and is notexpected to operate at speeds below about 50% forextended periods. If a constant torque load isbeing driven, however, and particularly if thespeed will be less than 50% rated, external coolingfans must be employed. Often, the motor will beoversized to lessen the effects of reduced coolingat lower speeds.
Sometimes present in the output supplied tothe motor from the LCI are common mode34
6-Pulse Converter With Delta/WYE TransformerFourier Series:
i = 2 3 Id (cos θ + 1⁄5 cos 5θ - 1⁄7 cos 7θ - 1⁄11 cos 11θ+ 1⁄13 cos 13θ + 1⁄17 cos 17θ - 1⁄19 cos 19θ —)
(b)
Id
6-Pulse Converter With Delta/Delta TransformerFourier Series:
i = 2 3 Id (cos θ - 1⁄5 cos 5θ + 1⁄7 cos 7θ - 1⁄11 cos 11θ+ 1⁄13 cos 13θ - 1⁄17 cos 17θ + 1⁄19 cos 19θ —)
(c)
Id
Current in Delta/Delta and Delta/WYE TransformersFourier Series:
Itotal = 2 3 Id (cos θ - 1⁄11 cos 11θ + 1⁄13 cos 13θ- 1⁄23 cos 23θ + 1⁄25 cos 25θ —)
��π
��π
��π
Figure 63 Sine wave simulation of VFD output.
SPEED
voltages which may impose higher-than-usualstresses to ground from the machine winding.Accordingly, it is necessary to provide groundwallinsulation with 1.3 to 1.8 times the normaldielectric strength compared to ordinary machineswhen the unit is to be operated from an inverter.
The exciter design must differ fromconventional designs in that conventional excitersextract energy from the rotating shaft, whereas,that source would be such a variable factor – evenapproaching zero at very low speeds – the energymust be transmitted to the rotor by another means.Exciter designs for these motors use three-phase acexcitation on the stator, applied so the rotatingfield is counter-rotating to the rotor. In a very realsense, it is a rotating transformer with an air gapin the magnetic circuit, and all of the powerdeveloped in the rotor comes from the excitationsource. The exciter may or may not have the samenumber of poles as the motor; however, thefrequency in the rotor circuit is not critical, since itis rectified anyhow.
An advantage of the LCI drive which has notbeen discussed is the ability to run atsupersynchronous speed (higher than thesynchronous speed of a machine with the samenumber of poles when connected to the line). Ifthis kind of operation is planned, it must be keptin mind that centrifugal forces are a function of thesquare of the speed and loads for centrifugalpumps and fans are related to the cube of thespeed. Realizing this, it soon becomes apparent
that a small increase in speed results in muchlarger changes in centrifugal stresses and inloading; thus, these expectations must be knownby the machine designer so that propercapabilities can be incorporated into the design.Critical frequencies which, conceivably, mightotherwise occur at speeds only slightly abovenormal synchronous speed could also be avoided.
We have seen that the motors used withinverters are much the same as those which areapplied only at line voltages and frequencies; themajor differences being in two-winding stators forthe twelve-step inverter applications, possibleheavier ground-wall insulation, cooling, and three-phase exciter stator windings, plus torsionalconsiderations and, possibly, provisions forsupersynchronous operation which affect themechanical facets of the design as well as theelectrical. However, these are but refinements tothe machine to adapt it to a special application.
Advances in technology, bringing about suchequipment as the Load Commutated Inverter,continue to open new doors of opportunity forapplication of that old, reliable workhorse ofindustry, the Synchronous Motor.
Synchronous motors fill an important place inindustry. They are usually associated with loads ofmajor importance. A corresponding degree of carein their application is well advised and worth theextra effort.
35
Figure 64 By-Pass and LCIContractor areinterlocked to bemutually exclusive.LCI
DRIVE
BY-PASSCONTACTOR 1
BY-PASSCONTACTOR 2
MOTOR 1 LCICONTACTOR 1
LCICONTACTOR 2
MOTOR 2
About the authorsAU
TH
ORS
200-SYN-42
Steve MollSteve Moll is presently Senior Marketing Representative with Electric Machinery.
He holds a Bachelor of Electrical Engineering degree from the University of Minnesota.His experience at EM includes Senior Design Engineer where he worked with
motor and generator controls, as well as high voltage switchgear and controls for gasturbine driven generators. In his present position, his area of specialty is applicationand marketing of 2-pole turbine driven generators. He has significant experience withsynchronous and induction motors as well.
Several of Steve’s articles have appeared in trade as well as company publications.
Jack ImbertsonJack Imberton holds an Electrical Engineering degree from the University of
Minnesota. While at EM, he held the positions of Design Engineer, Programmer,Manager of Synchronous Machines, and Manager of Turbo Generators. He has beenactive in IEEE and was a past chairman of the IAS section. He also served as amember of the ANSI C50 standards committee on synchronous machines.
Since retirement, he has taught electro-mechanics at the University of Minnesotaand does some consulting.
Ben ImbertsonBen Imbertson holds a Bachelors Degree in Electrical Engineering from the
University of Minnesota and is a registered Professional Engineer. While at EM, heworked as a Control Design Engineer, Control Engineering Section Head, andPrincipal Engineer-Control Products.
His EM experience included applications in synchronous and induction motorcontrol, generator control, as well as adjustable-speed drives. His post EM experiencehas been in similar areas.
Ben’s writing has appeared in several EM publications including theSynchronizer, “The ABC’s of Motor Control.”
Gerry OscarsonNow deceased, Gerry Oscarson was the original author of this special issue of
the EM Synchronizer. He held Electrical Engineering degrees from the University ofMinnesota. Specializing in electric power apparatus, Mr. Oscarson was directlyassociated with the industrial application of motors, generators, and controls. Formany years, he was Chief Application Engineer of the Electric Machinery Mfg.Company. The ABC series of EM Synchronizers are Gerry’s work. These remain asstandard references for industrial plants around the world, and are often used as textsby engineering schools.