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Unformatted text preview: EE 4002: Project # 1 Dynamic Simulation of an Elevator Driven by a Separately Excited DC Motor by Means of a System of Pulley and Gears By Derek Scott October 20, 2004 1 Elevators are used as methods of raising and lowering people and cargo to different levels of a specific structure. With the invention of the electric elevator and cheap steel, Americans raced to construct buildings taller than previously imaginable. The new technology made it possible to reach the 103rd floor of the Empire State Building in New York City without being in great physical shape. Also the cargo carrying capability of the elevator made it possible to successfully lift heavy objects safely with the walls of the structure repeatedly. Elevators enabled Americans to built America. The design in question calls for an elevator shown schematically in Figure 1 to be driven by a separately excited DC motor with an available supply voltage of 110V. A system of gears and pulleys is used to reduce the input power delivered by the motor. The elevator cab at full load is kg M Cab 733 = . The counterweight of the elevator cab is kg M M Cab Ctw 25 . 183 25 . = = . The radius of the pulley is m R 2 . = . The gear ratio is defined as 10 = = M L G r r K , where r L and r M are the radii of the gears on the load side and the motor side, respectfully. The combined gear and pulley moment of inertia is 2 3 . 3 ) ( m kg J J G P • = + . The combined coefficient of friction of the gear and pulley was found to be s rad m N B • = 3 . 3 . A representation of the elevator system can be found in Figure 2. The motor is considered fully loaded when lifting the elevator at a rate of s m v Cab 5 . 1 = . In order to select the proper motor, the rated power of the motor needs will be calculated. The rated power is equal to ) ( M L M P P = , where P L(M) is the power of the load on the motor side of the gears. Given the elevator linear speed at steady state, the process of calculating P M will start with this value to find the speed of the rotor of the motor. Upon finding the rotary speed of the motor, the electromagnetic torque of the motor will be calculated. The product of these two values gives the input armature power of the motor. The power losses of the gears and pulley system will be neglected. The rating of the motor will be rounded up to next standard horsepower motor, so the losses will be taken into account. • Speed of the Lifting Rope s m v v Cab r . 3 5 . 1 2 2 = × = = , where v r is the speed of the lifting rope • Speed of the Pulley L r R v ϖ × = , where ω L is the speed of the pulley s rad R v r L 15 2 . . 3 = = = ϖ • Speed of the Rotor of the Motor s rad K G L M 150 10 15 = × = × = ϖ ϖ , where ω M is the speed of the rotor 2 Figure 1....
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This note was uploaded on 02/06/2012 for the course EE 4002 taught by Professor Scalzo during the Fall '06 term at LSU.
 Fall '06
 Scalzo

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