What Is the Formula of the Specific Electric Loading
How many factors are used to determine the specific magnetic charge? Before deriving an expression for power, a brief digression may be useful for those familiar with linear rather than rotary systems. In the SI system, the unit of work or energy is the joule (J). A joule represents the work done by a force of 1 Newton moving 1 meter in its own direction. Therefore, the work done (W) by a force F moving a distance d, given by With F in Newtons and d in meters, Wis is clearly in Newton meters (Nm), from where we see that a Newton meter is the same as a joule. In rotary systems, it is more convenient to work in terms of torque and angular distance than in force and linear distance, but these are closely related, as we can see when we consider what happens when a tangential force F is applied within a radius r of the center of rotation. The torque is simply indicated by T = F x r. Now suppose that the arm rotates through an angle U, so that the circumferential distance traveled by the force is r x U. The work performed by force is then given by We find that while in a linear system the work is force times distance, in terms of rotation the work is couple times angle. The units of torque are newton meters, and the angle is measured in radians (which is dimensionless), so the units of work performed are Nm or Joule as expected. (The fact that torque and work (or energy) are measured in the same units doesn`t seem to be taken for granted by this author!) In order to find the strength or pace of work, we divide the work done by the time required. In a linear system and assuming that the velocity remains constant, the power is therefore given, where v is the angular velocity (constant), in radians per second. We can now express the output power in terms of rotor dimensions and specific loads using equation 1.9, which gives equation 1.13, which emphasizes the importance of speed in determining the output power.
If we want a motor with some power for certain specific and magnetic loads, we can choose between a large (and therefore expensive) low-speed motor or a small (and cheaper) high-speed motor. The latter choice is preferred for most applications, although some form of speed reduction (with belts or gears) is required because the smaller motor is cheaper. Well-known examples are portable power tools, in which rotor speeds of 12,000 rpm or more make it possible to reach powers of hundreds of watts, and electric traction, in which in both cases the high speed of the motor for the final drive is reduced. In these examples, in which volume and weight are in the foreground, direct training would not be an option. The importance of speed is emphasized when we reorganize equation 1.13 to obtain an expression of the specific output power (power per unit volume of the rotor), Q, given by To obtain the highest possible specific power for given values of specific magnetic and electrical charges, we must clearly operate the motor at the highest possible speed. The obvious disadvantage of a small high-speed engine and gearbox is that the acoustic noise (both from the engine itself and from the power transmission) is higher than with a larger direct drive motor. Therefore, when noise needs to be minimized (e.g. B ceiling fans), a direct drive motor is preferred despite its larger size. In light of the previous discussion, we can obtain the entire tangential force by first looking at an area of the rotor surface of width w and length L. The axial current circulating in width w is given by I = wA, and on average this total current is exposed to the radial flux density B, so that the tangential force (from equation 1.2) is given by B x wA x L. The surface area is wL, so the force per unit area is B x A. We see that the product of the two specific loads expresses the average tangential stress on the rotor surface.
To obtain the total tangential force, we must multiply by the area of the curved surface of the rotor, and to obtain the total torque, we multiply the total force by the radius of the rotor. Therefore, for a rotor of diameter D and length L, the total torque is determined by this equation is extremely important. The term D2L is proportional to the volume of the rotor, so we see that for given values of specific magnetic and electrical charges, the torque of any motor is proportional to the volume of the rotor. We are free to choose a long, thin or short rotor, but once the rotor volume and specific loads are specified, we have effectively determined the torque. It should be noted that we did not focus on a specific type of engine, but we approached the issue of torque generation from a completely general point of view. Essentially, our conclusions reflect the fact that all engines are made of iron and copper and differ only in the way these materials are disposed of. We must also recognize that in practice the total volume of the motor is important and not the volume of the rotor. But again, we find that regardless of the type of motor, there is a fairly close relationship between the total volume and the rotor volume, for motors with similar torque. We can therefore say boldly but generally accurately that the total volume of an engine is determined by the torque it must generate. There are, of course, exceptions to this rule, but as a general guideline for engine selection, it is extremely useful. .