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

Operation of AC Generators in Parallel with 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

Operation of AC Generators in Parallel with Large Power Systems 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)

Operation of AC Generators in Parallel with Large Power Systems 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

Operation of AC Generators in Parallel with Large Power Systems 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 

Operation of AC Generators in Parallel with Large Power Systems 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

Operation of AC Generators in Parallel with Large Power Systems 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.

Synchronous Motors Equivalent 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 I A is reversed

Synchronous Motors Equivalent Circuit 3 phase Eq. cct.  Per phase Eq. cct.

Synchronous Motors Equivalent Circuit The related KVl equations: V φ =E A +jX S I A + R A I A E A =V φ -jX S I A –R A I A 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 T app ) FIGURE (1)  for Generator

Synchronous Motors Equivalent Circuit The phasor diagram of syn. Motor, & magnetic field diagram FIGURE (2)

Synchronous Motors Operation Induced torue is given by: T ind =kB R x B net (1) T ind =kB R B net 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 T app

Synchronous Motors Operation 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  B R slows down & falls behind B net, operation of machine suddenly changes Using Equation (1), when B R behind B net, torque’s direction reverses & become counterclockwise

Synchronous Motors Operation 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 Motors Steady-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

Synchronous Motors Steady-state Operation 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%

Synchronous Motors Steady-state Operation Torque equation: T ind =kB R B net sinδ T ind = 3 V φ E A sinδ /(ω m X S ) (chapter 5, T=P/ω m ) 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.

Synchronous Motors Steady-state Operation 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: T max =kB R B net T max =3V φ E A /(ω m X S ) From last equation, the larger the field current, larger E A, the greater the torque of motor Therefore there is a stability advantage in operating motor with large field current or E A

Synchronous Motors Steady-state Operation 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

Synchronous Motors Steady-state Operation 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 E A =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

Synchronous Motors Steady-state Operation |E A | remain constant as load changes Distances proportional to power increase (E A sinδ or I A cosθ) while E A must remain constant As load increases E A swings down as shown & jX S I A has to increase & Consequently I A Increase, Note: p.f. angle θ change too, causing less leading & gradually lagging

Synchronous Motors Steady-state Operation Example: A 208 V, 45 kVA, 0.8 PF leading, Δ connected, 60 Hz synchronous machine has X S =2.5 Ω and a negligible R A. 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 I A, I L, and E A (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 I A, I L, and E A after load change. What is the new motor power factor

Synchronous Motors Steady-state Operation Solution: (a) P out =(15 hp)(0.746 kW/hp)=11.19 kW electric power supplied to machine: P in =P out +P mech loss +P core loss +P copper loss = = =13.69 kW Since motor’s PF is 0.8 leading, the resulting line current flow is: I L =P in /[√3 V T cos θ]=13.69/[√3 x 208 x 0.8]=47.5 A I A =I L /√3  I A =27.4 /_36.78◦ A

Synchronous Motors Steady-state Operation To find E A : E A =V φ -jX S I A =208 /_0◦ - (j2.5 Ω)(27.4/_36.78◦)= /_126.87◦ =249.1 –j 54.8 V=255/_-12.4◦V

Synchronous Motors Steady-state Operation (b) As load power on shaft increased to 30 hp, shaft slows momentarily, & internal voltage E A swings to a larger δ, maintaining its magnitude

Synchronous Motors Steady-state Operation (c) After load changes, electric input power of machine is: P in =P out + P mech loss +P core loss +P copper = (30)(0.746) =24.88 kW since P= 3 V φ E A sinδ /X S torque angle determined: δ=arcsin [X S P/(3V φ E A )] =arcsin[(2.5x24.88)/(3x208x255)]= arcsin(0.391)=23◦

Synchronous Motors Steady-state Operation E A =355/_-23◦ V I A = [V φ -E A ] / (jX S ) = [ /_-23◦]/(j2.5)= = [103/_105◦]/(j2.5) =41.2/_15◦ A and I L : I L = √3 I A =71.4 A the final power factor is cos(-15)=0.966 leading

Synchronous Motors Steady-state Operation 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 I F & see what happens to motor This will increase E A, however don’t affect real mechanical power supplied by motor. Since this power only changes when shaft load torque change

Synchronous Motors Steady-state Operation Since change in I F does not affect shaft speed n m and, since load attached to shaft is unchanged, real mechanical power supplied is unchanged V T is constant (by power source supply) power is proportional to following parameters in phasor diagram : E A sinδ & I A cosθ and must be constant When I F increased, E A 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 E A increases first I A decreases and then increases again At low E A armature current is lagging, and motor is an inductive load, consuming reactive power Q As field current increased I A lines up with V φ motor like a resistor, & as I F increased further I A become leading and motor become a capacitive load (capacitor-resistor) supplying reactive power

Synchronous Motors Steady-state Operation A plot of I A versus I F for a synchronous motor shown below known as V curves in each curve of constant P, minimum I A occur at unity P.F.