MDU RANK LIST IS OUT

Section III   Synchronous Machine Model The schematic diagram of a synchronous generator is shown in Fig. 1.15. This contains three stator windings that are spatially distributed. It is assumed that the windings are wye-connected. The winding currents are denoted by ia , ib and ic. The rotor contains the field winding the current through which is denoted by if . The field winding is aligned with the so-called direct ( d ) axis. We also define a quadrature ( q ) axis that leads the d -axis by 90°. The angle between the d-axis and the a-phase of the stator winding is denoted by θd. Fig. 1.15 Schematic diagram of a synchronous generator. Let the self-inductance of the stator windings be denoted by Laa, Lbb, Lcc such that (1.80) and the mutual inductance between the windings be denoted as (1.81) The mutual inductances between the field coil and the stator windings vary as a function of θd and are given by (1.82) (1.83) (1.84) The self-inductance of the field coil is denoted by Lff. The flux linkage equations are then given by (1.85) (1.86) (1.87) (1.88) For balanced operation we have Hence the flux linkage equations for the stator windings (1.85) to (1.87) can be modified as (1.89) (1.90) (1.91) For steady state operation we can assume (1.92) Also assuming that the rotor rotates at synchronous speed ωs we obtain the following two equations (1.93) (1.94) where θd0 is the initial position of the field winding with respect to the phase-a of the stator winding at time t = 0. The mutual inductance of the field winding with all the three stator windings will vary as a function of θd, i.e., (1.95) (1.96) (1.97) Substituting (1.92), (1.94), (1.95), (1.96) and (1.97) in (1.89) to (1.91) we get (1.98) (1.99) (1.100) Since we assume balanced operation, we need to treat only one phase. Let the armature resistance of the generator be R . The generator terminal voltage is given by (1.101) where the negative sign is used for generating mode of operation in which the current leaves the terminal. Substituting (1.98) in (1.101) we get (1.102) The last term of (1.102) is the internal emf ea that is given by (1.103) where the rms magnitude ‌ Ei ‌ is proportional to the field current (1.104) Since θd0 is the position of the d -axis at time t = 0, we define the position of the q -axis at that instant as (1.105) Therefore (1.94) can be rewritten as (1.106) Substituting (1.105) in (1.103) we get (1.107) Hence (1.102) can be written as (1.108)  .

	Chapter 1: Modelling Power System Components
...Contd from previous slide... The equivalent circuit is shown in Fig. 1.16. Let the current ia lag the internal emf ea by θa . The stator currents are then (1.109) (1.110) (1.111) Fig. 1.16 Three-phase equivalent circuit of a synchronous generator. The single-phase equivalent circuit is shown in Fig. 1.17. The phase angle θabetween eaand ia is rather difficult to measure under load as ea is the no load voltage. To avoid this, we define the phase angle between νa and ia to be θ . We assume that ea leads νaby δ . Therefore we can write (1.112) Then the voltages and currents shown in Fig. 1.17 are given as (1.113) (1.114) (1.115) Equations (1.113) to (1.115) imply that (1.116) The synchronous impedance is then defined as (1.117) The terminal voltage equation is then (1.118) Fig. 1.17 Single-phase equivalent circuit of a synchronous generator.