The synchronous machine
The Synchronous Machine
The workhorse for the generation of electricity is the synchronous machine. The bulk of electric energy is produced by three-phase synchronous generators. Synchronous generators with power ratings of several hundred MVA are com- mon; the biggest machines have a rating up to 1500 MVA. Under steady-state conditions, they operate at a speed fixed by the power system frequency, and therefore they are called synchronous machines. As generators, synchronous machines operate in parallel in the larger power stations. A rating of 600 MVA is then quite common.
In a power plant the shaft of the steam turbine is mounted to the shaft of the synchronous generator. It is in the generator that the conversion from mechanical energy into electrical energy takes place. The two basic parts of the synchronous machine are the rotor and the armature or stator. The iron rotor is equipped with a DC-excited winding, which acts as an electromagnet. When the rotor rotates and the rotor winding is excited, a rotating magnetic field is present in the air gap between the rotor and the armature. The armature has a three-phase winding in which a time-varying EMF is generated by the rotating magnetic field.
Synchronous machines are built with two types of rotors: cylindrical rotors that are driven by steam turbines at 3000–3600 RPM (see Figure 2.24 (a)) and salient-pole rotors that are usually driven by low-speed hydro turbines (see Figure 2.24 (b)). In the cylindrical rotor the field winding is placed in slots, cut axially along the rotor length. The diameter of the rotor is usually between 1 and 1.5 m, and this makes the machine suitable for operation at 3000 or 3600 RPM. These generators are named turbo generators. A turbo generator rotor has onek pair of poles. Salient-pole machines have usually more pair of poles, and to produce a 50 Hz or 60 Hz frequency, they can operate at a lower rotational speed.
The frequency of the EMF generated in the stator windings and the rotor speed are related by
- f the electrical frequency [Hz]
- n the speed [RPM]
- p the number of pairs of poles (i.e., the number of poles divided by two)
Hydraulic turbines in hydro plants rotate at a few hundred RPM, depending on the type, and therefore need many pole pairs to generate 50 or 60 Hz.
The efficiency of the generators is very important. Synchronous generators in power plants have an efficiency of 99%. This means that for a 600 MW generator 6 MW heat is produced, and therefore the machine has to be cooled.Large turbo generators are cooled with hydrogen or water. Hydrogen has 7 times the heat capacity of air and water 12 times. The hydrogen and/or water flows through the hollow stator windings. Cooling equalizes the temperature distribution in the generator, because temperature hot spots affect the life cycle of the electrical insulation. When the temperature gradient is small, the average temperature of the machine can be higher, and this means that the generator can be designed for a higher output. The evolution of big turbo generators has been determined by better materials, novel windings, and sophisticated k cooling techniques. Low-speed generators in hydro plants are always bigger than high-speed machines of equal power in thermal power plants, and a good air-cooling system with a heat exchanger usually does the job for low-speed machines.
Before a synchronous generator can be connected to the grid, four conditions must be satisfied. The generator voltage must:
- Have the same phase sequence as the grid voltages
- Have the same frequency as the grid
- Have the same amplitude at its terminals as the one of the grid voltage
- Be in phase with the grid voltage
When the generator is connected to a large grid, its output voltage and frequency are locked to the system values and cannot be changed by any action on the generator. We say that the generator is connected to an infinite bus: an ideal voltage source with a fixed voltage amplitude and frequency. The equivalent circuit of the synchronized generator is represented in Figure 2.25. The reactance X is called the synchronous reactance and is constant during normal steady-state conditions. The resistance of the stator coil is neglected in the equivalent circuit. Immediately after synchronization and coupling, the generator is neither feeding power to nor absorbing power from the grid. The steam going into the turbine is the same as before the coupling action and is just enough to drive the rotor and compensate for the losses of the turbine and the generator.
If more steam is fed into the turbine, one might expect the generator to speed up, but this is not possible because the generator is connected to an infinite bus. It is much like having two objects tied together via an elastic spring. The infinite bus is one end of the spring and moves with a constant speed, and the synchronous generator is connected to the other end of the spring as shown in Figure 2.26 (a) for the situation where there is no power exchange between the generator and the grid. What happens if more steam is fed into the turbine is shown in Figure 2.26 (b). The torque supplied to the generator axis by the prime mover tries to speed up the generator but can only extend the spring a bit as it is tied to the infinite bus, and all the extra energy is transformed into electrical k energy: the machine injects power into the grid. In relation to the extension of the spring, the internal EMF E now leads the terminal voltage; in a way, the internal EMF “drags along” the grid voltage as shown in Figure 2.27.
The amplitude of the generator internal EMF Eis a function of the field current If and is controlled by the operator of the generator. In Figure 2.27 E leads the system voltage Vby an angle δ, which is known as the power angle. As a consequence, current, and hence power, is fed into the grid. The expression for the current is
the current phasor:
the machine internal EMF phasor:
E = |E|∠δ[V] the system voltage phasor: V= |V |∠0 [V]the machine reactance [Ω]
The three-phase complex power supplied to the power system equals S=3VI∗ =P+jQg
Expressions for the active and reactive power can be found by either eliminating V or eliminating I from Equation 2.23.