# A three phase alternator

EXERCISE: THREE PHASE ALTERNATOR

In this exercise, we propose to study a three-phase alternator with smooth poles and a wound rotor. The three phases are star-connected. We measured its electromotive force E as a function of the excitation current Ie at the speed of 3000rpm. The measurement of E (Ie) is shown in Table bellow:

 Ie(A) 0 2 4 6 8 10 12 14 16 18 20 22 24 Ie(A)(V) 0 50 100 148 190 227 260 283 300 305 310 312 314

The alternator has a nominal apparent rated power of 250kVA and a nominal phase-neutral voltage 230V.
1- Represent the connection scheme of the three phases corresponding to the star coupling of the alternator. In addition, draw the equivalent circuit of the synchronous machine (for one phase) according to Behn-Eschenburg.
2- The frequency of the phase voltages is 50Hz. Specify the number of poles of the alternator.
3- Calculate the nominal current value: In.
4- At short-circuit, the current of the alternator reaches the calculated nominal value In for a value of the excitation current: Ie = 6 A. Calculate the value of the synchronous reactance Xs if we neglect the resistance of the windings that constitute the phases.
5- The alternator is now connected to a set of unit power factor loads. These loads are three-phase balanced and star wired to the alternator. What is the value of the excitation current Ie allowing supplying 150kW to all the loads under a voltage between two phases of 400 V?
(Make a Fresnel diagram of the magnitudes of the equivalent single-phase circuit before starting any calculation.)

CORRECTION

1) FIG. 1 shows the star coupling of the alternator and the equivalent single-phase diagram of the machine according to Behn-Eschenburg

2) N =60f/p ⇒ p = 60×50/3000 = 1 pair of poles means 2 poles.

3) In = Sn/3.Vn = 250×103/(3×230) = 362A

4) If we neglect the resistance of the windings that constitute the phases, the short circuit current Isc is just limited by Xs. Thus, for Ie = 6 A, we read : E = 148 V.

Hence:  Xs = E/ISC(=In) = 148/362 = 0.4A

5) If we consider a set of unit power factor loads, that are three-phase balanced and star wired to the alternator, the Fresnel diagram of the magnitudes of the equivalent single-phase circuit
is as shown in figure 2:

Moreover, the current called by the load of 250 kW is in this case: I = P/ 3VnIn = 150×103/3×230 = 217A

We deduce the value of the electromotive force E by writing:  V2 + (Xs.I)2 = E2   so  E = [(V2 + (Xs.I)2)]½ = 245.8V

By linear regression of the values of table 1, we obtain: Ie = 7.2A