Figure 1.3-2 Alternating circuits (resistive).
The magnitude of the angle (or power factor) depends on how capacitive or inductive the load is. In a purely capacitive circuit, the current will lead the voltage by 90°, whereas in a purely inductive one, the current will lag the voltage by 90° (see Figure 1.3-3). Here, the sinusoidal voltage E is applied to a circuit comprised of resistive, capacitive, and inductive elements. The resulting angle between the current and the voltage depends on the value of the resistance, capacitance, and inductance of the load.
A circuit that has capacitive or inductive characteristics is referred to as being a reactive circuit. In such a circuit, the following parameters are defined:
Figure 1.3-3 Alternating circuits (resistive‐inductive‐capacitive).
| S: The apparent power | S = E × I, given in units of volt‐amperes or VA. |
| P: The active power | P = E × I × cos φ, where φ is the angle between the voltage and the current. P is given in units of watts. |
| Q: The reactive power | Q = E × I × sin φ, given in units of volt‐amperes‐reactive or VAR. |
The active power P of a circuit indicates a real energy flow. This is power that may be dissipated on a resistance as heat, or may be transformed into mechanical energy. However, the use of the word “power” in the definition of S and Q has been an unfortunate choice that has resulted in confounding most individuals without an electrical engineering background for many years. The fact is that apparent power and reactive power do not represent any measure of real energy. They do represent the reactive characteristic of a given load or circuit, and the resulting angle (power factor) between the current and voltage. This angle between voltage and current significantly affects the operation of an electric machine.
For the time being, let us define another element of AC circuit analysis: the power triangle. From the relationships shown above among S, P, Q, E, I, and φ, it can be readily shown that S, P, and Q form a triangle. By convention, Q is shown as positive (above the horizontal), when the circuit is inductive, and vice versa when capacitive (see Figure 1.3-4).
Figure 1.3-4 Definition of the “power triangle” in a reactive circuit.
To demonstrate the use of the power triangle within the context of large generators and their interaction with the power system, we need to consider a one‐line schematic that includes the generator, transmission system, and the connected load at the end (see Figure 1.3-5).
The voltage required at the load, so that it will operate correctly, is given as 1000 V. The transmission line resistance and reactance are provided and the line impedance calculated as shown, using the power triangle approach. If we now consider an actual load for the simple system of Figure 1.3-5, we can calculate the current drawn by the load and the voltage required from the generator source to compensate for all the line losses and voltage drop across the line. Two cases are provided to illustrate the effect of a purely resistive load versus a load with a reactive component include (see Figures 1.3-6 and 1.3-7). Working out the required voltage from the generator for the two different loads by the power triangle method shows how reactive loads greatly affect the power system operation and the generation requirements. Reactive power compensation is a large part of synchronous generator operation and affects generator design in a significant way, as will be discussed later on in Chapter 2. There is a delicate balance between generation and load that is clearly shown by the two cases presented and the comparison of operational results (see Table 1.3-1).
Figure 1.3-5 Schematic of a simple system in one‐line form.
Figure 1.3-6 Case 1. The load is purely resistive in this example, and the system is operating at the “unity” power factor.
Figure 1.3-7 Case 2. The load is resistive and inductive in this example, and the system is operating in the “lagging” power factor range.
Although the “real” power consumed is the same, the addition of the reactive component in Case 2 has caused an increase in current drawn from the generator, an increase in line losses, a higher volt drop across the line, and, therefore, a higher voltage required from the generator source.
TABLE 1.3-1 A comparison of Case 1 and Case 2
| Load | 100 kW | 100 kW and 50 kVAR |
|---|---|---|
| Power consumed by the load (kW) | 100 | 100 |
| Current (A) | 57.8 | 64.6 |
| Line losses (kW) | 33.4 | 41.6 |
| Voltage drop along line (V) | 817 | 913 |
| Required delivery voltage at generating end (V) | 1680 | 1892 |
The above examples show that there is a considerable demand placed on the generator to operate the various loads on a system. In reality, the generator terminal voltage Vs is constant, plus or minus 5% by design. As the load increases or decreases, the current from the generator changes significantly and the voltage drop on the system Vload requires compensation (Figure
