Equivalent Circuit of Induction Motors, No-load, locked-rotor tests

[@TheVcasf]

Hello sir. Thank you for this great video. I am helping a carpenter with his machines. His machines start at very high current. For example an old bandsaw. What is the simplest/cheapest way to reduce the starting current? Is it possible to use capacitors? Or is the starting motor the best method? thank you again my brother

[@erdemcanaz6394]

You have the option to utilize soft starting techniques. I can suggest various approaches for you to ensure a smooth motor start, ranging from simple to more complex methods. Implementing resistive starters: Consider a 3-phase switch with three operational modes: open circuit, resistor-added path, and shorted path. Connect the motor to the grid through this switch. During motor startup, use the resistor-added mode to decrease the current drawn. However, be mindful that the resistors may heat up quickly. Therefore, once the induction machine reaches its rated speed, promptly switch back to the shorted path, which is akin to directly connecting the induction machine to the grid. This is the easiest and most cost-effective solution as per your request. Utilizing Delta-Star switches: Note that if the motor is delta-connected, the voltage across its phases is higher compared to a star connection (approximately 1.7 times greater). Consequently, if you initiate the motor in a star configuration and subsequently switch to a delta connection, the initial current is nearly two times less than in the delta configuration. You can employ delta-y switches for this purpose. However, I do not recommend this method unless you possess the necessary qualifications to alter the phase connections of the induction motor or understand the motor's ratings. This method is also cost-effective. There are additional methods worth exploring, such as induction machine drivers that gradually increase the electrical frequency of the stator windings, thereby reducing the slip and minimizing the drawn current. Furthermore, certain power devices can modify the AC voltage applied to the motor terminals. By gradually increasing the voltage at the motor terminals, the drawn current remains low. However, these methods tend to be expensive.

[@learner1585]

Why you didn't consider stator resistance and stator leakage reactance in no load condition

[@erdemcanaz6394]

The short answer is for the sake of simplicity. In the equivalent model, there is a component called a "parallel branch," where Rc and Xm are connected in parallel. Let me explain the meaning of these components to enhance clarity in our discussion. *Rc: This represents the equivalent resistance that models the core losses. 1) Hysteresis: When using magnetically permeable materials to generate more flux, it's important to note that these materials are not ideal. They exhibit hysteresis characteristics. Imagine an electromagnet used in school projects. Even after you stop applying current, it remains magnetically attracted to other magnetic materials, such as bolts. This is a basic example of hysteresis. In AC circuits, hysteresis causes power losses. (To change the state of something, you have to convince it. :)) Additionally, when the material is saturated, the magnetizing current is incredibly high, resulting in more conduction losses in the windings. However, for now, let's ignore this aspect, assuming the motor is designed not to saturate. 2) Eddy current losses: The phenomena on which induction machines rely are also a significant cause of power losses. As the alternating magnetic flux travels through the core material, it induces voltages across the core body. Although the core material is designed with low conductivity and lamination, it still causes losses. The Rc in the parallel branch models the combined "core" losses. You might ask whether it is constant or not, which is a very good question. Both hysteresis and eddy current losses are functions of the voltages applied to the motor terminals. It's worth noting that the stator winding resistance and inductance are not significantly high, so we can disregard the voltage drop over them. In this case, the voltage drop over Rc is not significantly different whether it is placed on the right side (which would be more accurate) or moved to the input terminals. This simplifies our analysis, and we don't lose much accuracy. You can try solving this circuit in both ways to understand what I mean. *Xc: This represents the equivalent inductance that models the inductance of the core. Consider a simple case with transformer windings that have low resistances. Despite this, they do not blow up, and their input current remains low when connected to the grid. Thanks to the law of induction, when an AC input is applied, the voltage induced across the transformer terminals is equal to the voltage drop across the primary windings and the induced voltage across the coil (L(dI/dt)). In transformers, we keep L very high, so even a slight change in current with respect to time results in very high voltages that counteract the increase. In short, since the air gap in induction machines reduces L compared to transformers, the inductance is still relatively high compared to leakage inductance. Now let's address your question. Engineers designing these machines aim to optimize their products by ensuring that Rc and Xc are guaranteed to be very high compared to Ra and Xa. You will observe that if you: Do not consider the parallel branch on the input side and do not ignore the stator resistance and reactance, Consider the parallel branch on the input side and ignore the stator resistance and reactance, I assure you that the results obtained will match by more than 95%. :)

[@ababababababababababababababa]

@@erdemcanaz6394 hocam gerçekten çok güzel bir cevap, çok güzel bir açıklama. ellerinize sağlık.