ENERGY CONVERSION ONE (Course 25741) CHAPTER SIX (& S.G. parallel op. with Pow. Sys. of chapter 5)...

ENERGY CONVERSION ONE (Course 25741)

CHAPTER SIX (& S.G. parallel op. with Pow. Sys. of chapter 5)SYNCHRONOUS MOTORS

Operation of AC Generators in Parallelwith Large Power Systems

• Isolated synchronous generator supplying its own load is very rare (emergency generators)

• In general applications more than one generator operating in parallel to supply loads

• In Iran national grid hundreds of generators share the load on the system

• Advantages of generators operating in parallel: 1- several generators can supply a larger load 2- having many generators in parallel increase the reliability of power system 3- having many generators operating in parallel allows one or more of them to be removed for shutdown & preventive maintenance 4- if only one generator employed & not operating near full load, it

will be relatively inefficient


INFINITE BUS• When a Syn. Gen. connected to power system,

power sys. is so large that nothing operator of generator does, have much effect on pwr. sys.

• Example: connection of a single generator to a large power grid (i.e. Iran grid), no reasonable action on part of one generator can cause an observable change in overall grid frequency

• This idea belong to definition of “Infinite Bus” which is: a so large power system, that its voltage & frequency do not vary, (regardless of amount of real and reactive power load)


• When a syn. Gen. connected to a power system:

1-The real power versus frequency characteristic of such a system

2-And the reactive power-voltage characteristic


• Behavior of a generator connected to a large system

A generator connected in parallel with a large system as shown

• Frequency & voltage of all machines must be the same, their real power-frequency (& reactive power-voltage) characteristics plotted back to back


• Assume generator just been paralleled with infinite bus, generator will be “floating” on the line, supplying a small amount of real power and little or no reactive power

• Suppose generator paralleled, however its frequency being slightly lower than system’s operating frequency

At this frequency power supplied by generator is less than system’s operating frequency, generator will consume energy and runs as motor


• In order that a generator comes on line and supply power instead of consuming it, we should ensure that oncoming machine’s frequency is adjusted higher than running system’s frequency

• Many generators have “reverse-power trip” system

• And if such a generator ever starts to consume power it will be automatically disconnected from line

Synchronous Motors

• Synchronous machines employed to convert electric energy to mechanical energy

• To present the principles of Synchronous motor, a 2-pole synchronous motor considered

• It has the same basic speed, power, & torque equations as Syn. Gen.

ckwisecountercloBkB SRind

Synchronous MotorsEquivalent Circuit

• Syn. Motor is the same in all respects as Syn. Gen., except than direction of power flow

• Since the direction of power flow reversed, direction of current flow in stator of motor may also be reversed

• Therefore its equivalent circuit is exactly as Syn. Gen. equivalent circuit, except that the reference direction of IA is reversed


• 3 phase Eq. cct.

• Per phase Eq. cct.


• The related KVl equations: Vφ=EA+jXS IA + RAIA EA =Vφ-jXSIA –RAIA • Operation From Magnetic

Field Perspective • Considering a synchronous

generator connected to an infinite bus

• Generator has a prime mover turning its shaft, causing it to rotate (rotation in direction of Tapp)

FIGURE (1) for Generator


• The phasor diagram of syn. Motor, & magnetic field diagram

FIGURE (2)

Synchronous MotorsOperation

• Induced torue is given by:

Tind=kBR x Bnet (1)

Tind=kBR Bnet sinδ (2)

• Note: from magnetic field diagram, induced torque is clock wise, opposing direction of rotation in Generator related diagram

• In other words; induced torque in generator is a countertorque, opposing rotation caused by external applied torque Tapp


• Suppose, instead of turning shaft in direction of motion, prime mover lose power & starts to drag on machine’s shaft

• What happens to machine? Rotor slows down because of drag on its shaft and falls behind net magnetic field in machine BR slows down & falls behind Bnet , operation of machine suddenly changes

• Using Equation (1), when BR behind Bnet , torque’s direction reverses & become counterclockwise


• Now, machine’s torque is in direction of motion

• Machine is acting as a motor

• With gradual increase of torque angle δ, larger & larger torque develop in direction of rotation

until finally motor’s induced torque equals load torque on its shaft

• Then machine will operate at steady state & synchronous speed again, however as a motor

Synchronous MotorsSteady-state Operation

• Will study behavior of synchronous motors under varying conditions of load & field current , also its application to power-factor correction

• In discussions, armature resistance ignored for simplicity

- Torque-Speed Characteristic• Syn. Motors supply power to loads that are constant

speed devices• Usually connected to power system, and power

systems appear as infinite buses to motors• Means that terminal voltage & system frequency will

be constant regardless of amount of power drawn by motor


• Speed of rotation is locked to applied electrical frequency

• so speed of motor will be constant regardless of the load

• Resulting torque-speed characteristic curve is shown here

• S.S. speed of motor is constant from no-load up to max. torque that motor can supply (named : pullout torque) so speed regulation of motor is 0%


• Torque equation:• Tind=kBRBnet sinδ • Tind = 3 Vφ EA sinδ /(ωm XS) (chapter 5, T=P/ωm)

• Pullout torque occurs when δ=90◦

• Full load torque is much less than that, may typically be 1/3 of pullout torque

• When torque on shaft of syn. Motor exceeds pullout torque, rotor can not remain locked to stator & net magnetic fields. Instead rotor starts to slip behind them.


• As rotor slows down, stator magnetic field “laps” it repeatedly, and direction of induced torque in rotor reverses with each pass

• Resulting huge torque surges, (which change direction sequentially) cause whole motor to vibrate severely

• Loss of synchronization after pullout torque is exceeded known as “slipping poles”

• Maximum or pullout torque of motor is: Tmax=kBRBnet Tmax=3VφEA/(ωm XS)• From last equation, the larger the field current, larger

EA , the greater the torque of motor• Therefore there is a stability advantage in operating

motor with large field current or EA


• Effect of load changes on motor operation

• when load attached to shaft, syn. Motor develop enough torque to keep motor & its load turning at syn. Speed

• Now if load changed on syn. motor, let examine a syn. motor operating initially with a leading power factor


• If load on shaft increased, rotor will initially slow down• As it does, torque angle δ becomes larger & induced

torque increases• Increase in induced torque speeds the rotor back up,

& rotor again turns at syn. Speed but with a larger torque angle δ

• last figure show the phasor diagram before load increased

• Internal induced voltage EA=Kφω depends on field current & speed of machine

• Speed constrained to be constant by input power supply, and since no one changed field current it is also constant


• |EA| remain constant as load changes• Distances proportional to power increase (EA sinδ or IA

cosθ) while EA must remain constant• As load increases EA swings down as shown & jXSIA has to increase & Consequently IA Increase, Note: p.f. angle θ change too, causing less leading & gradually lagging


• Example: A 208 V, 45 kVA, 0.8 PF leading, Δ connected, 60 Hz synchronous machine has XS=2.5 Ω and a negligible RA. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially the shaft is supplying a 15 hp load, and motor’s power factor is 0.8 leading

(a) sketch phasor diagram of this motor, and find values of IA, IL, and EA

(b) assume that shaft load is now increased to 30 hp Sketch the behavior of the phasor diagram in response to this change (c ) Find IA, IL, and EA after load change. What is the new motor power factor


• Solution:

(a) Pout=(15 hp)(0.746 kW/hp)=11.19 kW

electric power supplied to machine:

Pin=Pout+Pmech loss+Pcore loss+Pcopper loss=

= 11.19+1.5+1.0+0=13.69 kW

Since motor’s PF is 0.8 leading, the resulting line current flow is:

IL=Pin/[√3 VT cos θ]=13.69/[√3 x 208 x 0.8]=47.5 A

IA=IL/√3 IA=27.4 /_36.78◦ A


• To find EA :

EA=Vφ-jXS IA =208 /_0◦ - (j2.5 Ω)(27.4/_36.78◦)=

208- 68/_126.87◦ =249.1 –j 54.8 V=255/_-12.4◦V


• (b) As load power on shaft increased to 30 hp, shaft slows momentarily, & internal voltage EA swings to a larger δ, maintaining its magnitude


• (c) After load changes, electric input power of machine is:

• Pin=Pout + Pmech loss+Pcore loss+Pcopper =

(30)(0.746)+1.5+1.0+0=24.88 kW

since P= 3 Vφ EA sinδ /XS

torque angle determined:

δ=arcsin [XSP/(3VφEA)]

=arcsin[(2.5x24.88)/(3x208x255)]=

arcsin(0.391)=23◦


• EA=355/_-23◦ V• IA= [Vφ-EA] / (jXS) = [208-255/_-23◦]/(j2.5)=

= [103/_105◦]/(j2.5) =41.2/_15◦ A

and IL :

IL= √3 IA =71.4 A

the final power factor is cos(-15)=0.966 leading


• Effect of Field current changes on a synchronous motor• It was shown how change in shaft load affects motor torque

angle and the supply current• Effect of field current change:

• Above phasor diagram shows a motor operating at a lagging p.f.

• Now increase its IF & see what happens to motor• This will increase EA, however don’t affect real mechanical

power supplied by motor. Since this power only changes when shaft load torque change


• Since change in IF does not affect shaft speed nm

and, since load attached to shaft is unchanged, real mechanical power supplied is unchanged

• VT is constant (by power source supply)• power is proportional to following parameters in

phasor diagram : EAsinδ & IAcosθ and must be constant

• When IF increased, EA must increase, however it can be done along line of constant power as shown in next slide

Synchronous Motors Effect of an increase in field current

• Note: as EA increases first IA decreases and then increases again

• At low EA armature current is lagging , and motor is an inductive load, consuming reactive power Q

• As field current increased IA lines up with Vφ motor like a resistor, & as IF increased further IA become leading and motor become a capacitive load (capacitor-resistor) supplying reactive power

Synchronous Motors Steady-state Operation

• A plot of IA versus IF for a synchronous motor shown below known as V curves in each curve

of constant P, minimum IA occur at unity P.F.

ENERGY CONVERSION ONE (Course 25741) CHAPTER SIX (& S.G. parallel op. with Pow. Sys. of chapter 5)...

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