*7.2. Static Model*

The standard-frame velocity in a static-source framework is that of the stator, i.e., *ω<sup>s</sup>* = *ω<sup>r</sup>* = 0. Such a standard frame is chosen when potential asymmetrical differences

in the stator are interrupted and the potential differences in the rotor are symmetrical. The potential difference equations of an induction machine in a static reference frame are determined by substituting *ω<sup>r</sup>* = 0 into (3). The resultant solution is the static design of an induction machine in which the rotor is removed. As a result, the machinery can be described as static.

It is noticeable that there is a major relationship between the stator and the rotor of an induction motor. As the subscripts, *r* and s indicate the stator (s) and rotor (*r*). The electromagnetic subscripts are *i, v*, and *λ*. The resistance is *r*, the leakage inductance is *Ll*, and the mutual inductance is *Lm*. The phase voltages are represented by *a*, *b,* and *c*.

The actual values of the induction motor's parameters can be derived from the model specified in Section 4 [26]. To determine the parameters, a no-load test and a load test must be performed, so the parameters can be determined as follows.

The no-load test is performed by supplying the voltage *VS* at a rated frequency. The motor will rotate close to a synchronous speed, resulting in a close-to-zero slip [22].

Assuming that *RS*, Ω, and *LSH* are much lower than the magnetising inductance *LmH*, the following equation is derived [1].

$$L\_m = \frac{V\_S}{2\pi f\_s I} \tag{20}$$

where *VS* is the applied phase voltage in the stator, *I* is the current supplied to the stator, and *fs* is the stator frequency.

Now that the magnetising inductance has been determined, the other parameters are extracted with the locked rotor test, and the resulting equivalent circuit is shown in Figure 11.

$$
\begin{bmatrix} v\_{ds} \\ v\_{qs} \\ v\_{dr} \\ v\_{qr} \end{bmatrix} = \begin{bmatrix} R\_S + sL\_s & 0 & sL\_m & 0 \\ 0 & R\_s + sL\_s & 0 & sL\_m \\ L\_m & R\_s + sL\_s & R\_r + sL\_r & \omega\_r L\_r \\ -\omega\_r L\_m & sL\_m & -\omega\_r L\_r & R\_s + sL\_r \end{bmatrix} \begin{bmatrix} i\_{ds} \\ i\_{qs} \\ i\_{dr} \\ i\_{qr} \end{bmatrix} \tag{21}
$$

**Figure 11.** Equivalent circuit of the locked rotor.

Should *ω<sup>r</sup>* = *ω<sup>s</sup>* = 0, Equation (21) becomes:


where *RS* and *Rr* are the resistances of the stator and rotor, *LS* and *Lr* are the stator's and rotor's self-inductance and the stator's and rotor's speeds.

Figure 12 illustrates a static model of an induction motor.

**Figure 12.** A static model of an induction motor.

#### **8. Results**

The following parameters were acquired from experimental work with a three-phase induction motor. These settings were then utilised to simulate and examine the behaviour of the induction motor using Matlab/Simulink.

*8.1. The Machine's Steady-State Performance Behaviour When Loaded from 0 to 125% of the Rated Load, Shown in Both Tabular and Graphical Form with Current, Power Factor, Real Power, Reactive Power, Speed, Efficiency, and Power Factor versus the Percentage or Per-Unit Loading*

The parameters indicated in Table 1 were used to simulate the induction machine using Matlab/Simulink were:


**Table 1.** Parameters used for induction motor simulation.

The rated torque was not provided; therefore, before we can calculate the rated torque, the synchronous speed of the machine must first be calculated. The synchronous speed is calculated as follows:

$$N\_5 = \frac{120 \times f}{p} = 1800 \, rp$$

Knowing the synchronous speed of a four-pole, 60 Hz machine, the rated torque may now be calculated.

$$T = \frac{P\_{\text{rated}} \times 9.5493}{N\_{\text{s}}} = 39.78875 \text{ N.m}$$

To simplify, 40 N.m was used as the rated torque value. Table 2 provides the results obtained from the induction machine.

As seen in Figure 13a, the higher the torque load is, the higher the current will be. In Figure 13b, we can see that the active power drawn by the induction machine is almost linear. Thus, the active power is proportional to the percentage of the loading.

As shown in Figure 14a, the efficiency of the induction motor is poor when the machine is lightly loaded. Theoretically, the optimal point must be at 100% loading; however, in this case, the optimal point is at 80–90%. This is mainly due to additional power losses (theoretical vs practical).

Figure 14b illustrates that the reactive power initially decreases. This is because the power factor is poor with no load and increases with the increase in the load; however, the reactive power drawn will also increase due to the increase in the load.


**Figure 13.** (**a**) Current vs torque load; (**b**) active power vs torque load.

**Figure 14.** (**a**) Efficiency vs torque load; (**b**) reactive power vs torque load.

As shown in Figure 15a, the power factor is similar to the efficiency. The lower the load is, the lower the power factor will be, and a very good power factor is reached at the rated torque.

As shown in Figure 15b, the induction machine's speed decreases as the torque load increases, starting from a value that is very close to the synchronous speed. Because the machine acts as an induction motor, it will always run below the synchronous speed.

**Figure 15.** Power factor vs torque load (**a**); speed vs torque load (**b**).

*8.2. Transient Behaviour of the Current and Torque Versus Speed during Starting, Assuming That (i) the Machine Is Unloaded, (ii) the Machine Is 50% Loaded, and (iii) the Machine Is 100% Loaded*

In Figure 16b, we see the three-phase current. One characteristic of an induction machine is that it has a very high starting current in the transient state.

As seen in Figure 17a, the electromagnetic torque oscillates during the transient state until the oscillation stops and moves towards the torque load.

In Figure 17b, we see the same oscillation behaviour in the transient state; however, the machine reaches an optimal point that is close to the synchronous speed due to lack of a load.

In Figure 18, we see a small disturbance in the speed. This is due to the oscillation of the electromagnetic torque. The motor reaches a steady-state close to the synchronous speed due to the lack of a load.
