prob uni 1 Q&A

May 27, 2018 | Author: Aravind Ck | Category: Transmission (Mechanics), Torque, Machines, Mechanical Engineering, Manufactured Goods


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Electric Drives and Control(EEE 402) Introduction to Electric Drives1) A motor drives two loads, one has rotational motion and another has translational motion. The moment of inertia of the motor is 1.2 kg-m2. The motor runs at a speed of 1000 rpm. Following are the details about the two loads: Load Type of motion Speed Inertia / Mass Torque / Force ---------------------------------------------------------------------------------------------------- 1. Rotational 200 rpm 7 kg-m2 10 N-m 2. Translational 10 m/s 10 kg 20 N Calculate the equivalent inertia and torque of the system referred to the motor shaft assuming the gear ratio a=0.1, 80% efficiency in both rotational and transmission systems. Ans: 1.3, 3.9 2. A motor drives two loads. One has rotational motion, it is coupled to the motor through a reduction gear with a =0.1 and efficiency 90%. The load has moment of inertia of 10kg-m2 and a torque of 10 N-m. Other load has translational motion and consists of 1000 kg weight to be lifted at a uniform speed of 1.5 m/s. coupling between this load and the motor has an efficiency of 85%. Motor has inertia of 0.2 kg-m2 and runs at a constant speed of 1420 rpm. Determine equivalent inertia referred to the motor shaft and power delivered by the motor. Ans: J=0.4, Tl=117.53 3) (a) A drive has the following parameters: T = 150-0.1N Nm, where N is the speed in rpm. Load torque T1 = 100Nm Initially the drive is operating in steady state. The characteristics of the load torque are changed to T1 = -100Nm. Calculate initial and final equilibrium speeds. (b) A motor is used to drive a hoist. Motor characteristics are given by Dr Umashankar, Asso Prof, SELECT-VIT When hoist is loaded.1N_ ( _100) = 0 50_0.1N = 0 _ 0. TL= _100 Nm (changed) Initially the drive is operating in steady-state. T_TL = 0 150_ 0.1N = _ 50 N1= 500 rpm Final equilibrium speed.2N. Obtain the equilibrium speeds for operation in all the four quadrants.2N Nm Quadrants II.1N. Nm TL = 100 Nm (load) TL = _80 Nm (unloaded) Equilibrium speeds = ? T= 200_ 0.2N.Nm.1N_100 = 0 50_0. load Torque TL = 100 Nm.Electric Drives and Control(EEE 402) Introduction to Electric Drives Quadrants I.2N.2N. SELECT-VIT .2N Nm where N is the speed in rpm. net load torque T1 = -80Nm. T_TL = 0 150_ 0. Nm Quadrants ІІ. Nm Dr Umashankar.1N = _ 250 N2 = 2500 rpm  initial equilibrium speed N1 = 500 rpm final equilibrium speed N1 = 2500 rpm (b) Quadrants І.1N + 100 = 0 _ 0. ІІ and ІV : T= 200_ 0. Nm ІІ І TL= _80 Nm (unload) TL=100 Nm (loaded) ІІІ ІV T= _200_ 0.II and IV: T = 200-0. (a) T= 150_ 0. ІІІ and ІV : T= _200_ 0. T_TL = 0 Initial equilibrium speed. III and IV: T = -200-0. Asso Prof. the net load torque T1 = 100Nm and when it is unloaded. TL= 100 Nm 200_0. Nm.2N = _280 N2= 1400 rpm In Quadrant ІІІ.2N.2N = 300 N4 = _1500 rpm Dr Umashankar. TL= _80 Nm (unloaded) T_TL = 0 200_0.2N = 0 _0.2N = _100 N1= 500 rpm In Quadrant ІІ.2N = 0 _0. T = _200_0.2N _ ( _80) = 0 200 + 80 _ 0. T_TL = 0 T= 200_0.2N.2N. T = _200_0.2N _ 100 = 0 _300 _ 0. TL=100 Nm (loaded) T_TL = 0 _200_0.2N = _120 N3 = _600 rpm In Quadrant ІV. Nm. SELECT-VIT .2N = 0 _0.2N = 0 _0. Nm.Electric Drives and Control(EEE 402) Introduction to Electric Drives In equilibrium steady-state condition.2N _( _80) = 0 _200 + 80 _ 0. Asso Prof.2N.2N _100 = 0 100 _ 0. T = 200_0. Nm. T_TL = 0 In Quadrant І. TL= _80 Nm (unloaded) T_TL = 0 _200_0. SELECT-VIT .Electric Drives and Control(EEE 402) Introduction to Electric Drives Dr Umashankar. Asso Prof.
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