ECEN 4517 1Experiment No. 5The Synchronous MachineSynchronous ac machines find application as motors in constant speedapplications and, when interfaced to the power source with a variable-frequencyconverter system, in variable-speed applications. High-performance variable-speed motordrives are often constructed using a permanent-magnet synchronous motor. Anothercommon application is the alternator that generates the power for the automobileelectrical system. In addition, practically all electrical energy produced commercially isgenerated in rotating synchronous machines that are connected to the utility or otherpower systems. Most power systems are supplied by a number of such machinesoperating in parallel, and the systems themselves are interconnected to form power gridsof tremendous energy capacity. In comparing the capability of any single generator to theoverall capacity of the system to which it is connected, it can be seen that the individualmachine is relatively insignificant, even though its own power rating may be in theneighborhood of one million kilowatts. In other words, it can be thought of as beingconnected to a voltage bus of infinite capacity, so that regardless of how much energy itdelivers to (or receives from) the system, the voltage and frequency remain constant.The objectives of this experiment are (1) to develop a basic model for thesynchronous machine, (2) to apply the model to understand the characteristics of themachine when connected to an infinite bus, and (3) to synchronize an ac generator to alarge power system.1. An equivalent circuit model for the synchronous machineA basic synchronous machine is sketched in Fig. 1. The stator contains a three-phasearmature winding. When a source of three-phase ac is connected to this winding, amagnetic field of constant amplitude is produced within the machine; in a two-polemachine, this field rotates at frequency equal to the frequency of the applied ac. The rotorcontains a field winding that is excited by dc; this winding behaves as an electromagnet,producing a field of strength proportional to the applied field current that is aligned withthe axis of the field winding. Alternatively, the field winding may be replaced by apermanent magnet. Torque is produced by the two magnetic fields attempting to align,according to the formula T = FsFrsin δ(1)2where Fs and Fr arethe magnitudes of thestator and rotor fields,respectively. Theangle δ is the anglebetween the statorand rotor fields,commonly called thetorque angle.Theequivalent circuit ofthe synchronousmachi ne close lyresembles the dcmachine model usedin Experiment 1. The keydifference is the absence ofcommutator brushes, such thatthe armature voltage and currentare ac. Also, the field winding isnormally placed on the rotorwith the armature winding on thestator.The magnitude of themagnetic field produced by therotor field winding, Fr, isproportional to the applied fieldcurrent If: Fr=NfIfℜ(2)where Nf is the number of turnsin the field winding, and ℜ is thetotal reluctance in the magneticpath of the rotor flux.When the shaft turns, therotor field induces a voltage inFig. 1 A synchronous machine. Three-phase stator (armature) windingsare not explicitly shown. Torque is produced when the magneticfield of the dc rotor winding (aligned with the rotor axis)attempts to align with the rotating magnetic field of the three-phase ac stator winding.Fig. 2 Equivalent circuit model of the synchronousmachine.3the armature (stator) winding. This induced voltage is similar to the back EMF of the dcmachine, except that it is an ac voltage having a magnitude E and phase δ with respect tothe applied stator voltage. The magnitude E is proportional to the rotor field strength Frand the shaft speed ω: E = KFrω(3)where K is a constant of proportionality that depends on the machine and windinggeometry. ω is the angular shaft frequency. The stator winding model therefore includes avoltage source E∠δ (phasor notation) as in Fig. 2. In addition, a Thevenin-equivalentmodel of this winding contains a series impedance consisting of a series inductance,known as the synchronousinductance Ls, and a seriesresistance Rs that models theresistance of the wire.2. Solution of the equivalentcircuit, connected to aninfinite busThe circuit of Fig. 2 models one(line-to-neutral) phase of thethree-phase machine. The appliedvoltage at the terminals of thestator winding is taken to be V∠0.In a balanced three-phase system,all three phases have symmetricalwaveforms. The mechanicaloutput power Tω is then equal tothree times the average per-phaseelectrical power flowing into thevoltage source E: Tω = P = 3Re EIs*(4)In Eq. (4), E is the phasor E∠δ, Isis the phasor representing thestator current, Is* is the complexconjugate of the stator current,Fig. 3 A typical generator application. Field winding isexcited by dc current source If. Armature isconnected to three-phase ac infinite bus, modeledby voltage source (one phase shown). Shaft isconnected to a source of mechanical power, calledthe “prime mover.”4and Re(EIs*) is the average power flowing into the voltage source E.Figure 3 consists of the equivalent circuit of Fig. 2, connected to an infinite bushaving voltage V = V∠0. For simplicity, the stator winding resistance Rs has beenneglected. When the synchronous machine is connected to infinite bus V, the rotor mustturn at angular frequency ω equal to the angular frequency of the infinite bus. However,the rotor can be shifted in phase (by angle δ). Let us determine the stator winding currentIs and the average power 3Re(EIs*). Solution of the equivalent circuit of Fig. 3 to find Isleads to Is=E – VjωLs(5)Substitution of V = V∠0 and E = E∠δ leads to Is=E cos (δ)+ jE sin (δ)–VjωLs(6)The average power is then P = 3Re EIs*= 3Re E cos (δ)+ jE sin (δ)E cos (δ)– jE sin (δ)–V– jωLs=3VE sin (δ)ωLs(7)The generator loads the mechanical shaft with torque T=Pω=3VE sin (δ)ω2Ls(8)This equation is plottedin Fig. 4. For givenvalues of V and E, thereis a maximum torqueTmax that the machinecan produce, whichoccurs at δ = 90˚. Asthe power and torque