2015 MACOSH Meeting Tampa OSHA Area Office Overview
This document provides details about the 2015 MACOSH meeting held at the OSHA Area Office in Tampa. It covers the current staffing, programs, and inspections related to maritime activities in the region. Information on safety specialists, industrial hygienists, compliance assistance, investigations, and more is included.
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Lecture # 8 Lecture # 8 SHUNT DC MOTOR
SHUNT DC MOTOR o Shunt motor, its field winding is in shunt (in parallel) with the armature. o The line voltage (Vt) supplies the field excitation and the current tothe armature. o The field current is constant since the line voltage (Vt) is in parallel withthe field circuit. Thus, the flux is constant. o T = ka ? Ia, so T is directly proportional to the armature current (Ia) only since the other quantities are constant. o As we apply voltage to a shunt motor, current will flow through the field circuit which magnetizes the motor. The magnetic field will have a specific polarity within the motor. o Shunt DC motor
SHUNT DC MOTOR - DIRECTION OFROTATION The direction of motor rotation will be determined depending on thepolarities of the magnetic field (?) and armature current(Ia). o To reverse the direction of the motor rotation, change the connection of either the armature or shunt field (i.e. interchange F1 & F2 see circuit). o If you reverse the connection of the terminal voltage Vt only, the direction of rotation remains unchanged since both field and armature currentreverse. o Shunt DC motor
EFFECT OF LOAD ON THE SHUNT DC MOTOR Torque o As the load increases, so does the torque until itis equal to the new load. Rated Torque o The Ia is proportional to the motor torque as well as the load torque. T = ka Ia o Remember that Eg = ka ? and Eg = Vt IaRa, so the angular speed Rated Current Armature current Speed (rad/s) =Vt IaRa Vt Ra ka Rated Speed = Ia ka = Vt IaRa ka ka Rated Current o Only ? and Ia vary since the other quantities are constant. o As load increases, Ia increases, the speeddecreases. I (A) a Shunt DC Motors has excellent speed regulation. 20
EFFECT OF FLUX ON THE SHUNT DC MOTOR o Assume Ia Ra is too small compared to Vt, so Extremely high speed at no-field (If 0) Vt V I R t ka a a Speed = ka o Therefore, is inversely proportionalto. o The purpose of the rheostat in the field circuit is to control the field current which affects theflux. Field current,If o If the field current decreases, the flux reduces, and the speed increases. But this presents a very dangerous situation; that s, the motor will be in a runaway condition which could damage the motor if the field circuit isan open circuit (i.e. If =0). Open circuit means the flux ? is almost zero which causes the speed to be very high.
SHUNT DC MOTOR Example 5-6: A 230 V, 1.5 kW shunt motor draws 8.2 A at full load. If Rf = 200 , Ra = 1.1 , and the rated speed is 156 rad/s, find: a) the developed power , b) the developed torque, c) the shaft torque, d) the stray power loss and e) theefficiency. Sol. a) The developed power If = Vt/Rf = 230/200 = 1.15A Ia = IL If = 8.2 - 1.15 = 7.05A Eg = Vt Ia Ra = 230 7.05 (1.1) Eg = 222.25V b) The developedtorque Tdev =Pdev =1567 = 10.04N.m. 156 Pdev = Eg Ia = 222.25 (7.05) = 1567 W
SHUNT DC MOTOR Example 5-6: A 230 V, 1.5 kW shunt motor draws 8.2 A at full load. If Rf = 200 , Ra = 1.1 , and the rated speed is 156 rad/s, find: a) the developed power , b) the developed torque, c) the shaft torque, d) the stray power loss and e) theefficiency. Sol. Pdev=EgIa Pi Po c) The shaft (output) torque To =Po Pcu Pst e) The efficiency T =1500 = 9.62N.m. o 156 Pi = VtIL Pi = 230 x 8.2 = 1886W d) The stray power loss Po Pix100 Pst = Pdev Po Pst = 1567 1500 = 67W % = % =1.5 x103 x 100 = 79.5% 1886 23
SHUNT DC MOTOR Example 5-7: Armature of a 6-pole, 6-circuits D.C. shunt motor takes 400Aat a speed of 350 rpm. The flux per pole is 80 mWb, the number of armature turns is 600,and 3% of the developed torque is lost in windage, friction and iron-loss. Calculatethe brake-horse-power. Pdev=EgIa Pi Po Sol. Pcu Pst NP 600 6 80 10 3 350 2 60 = 560 V Eg = a Eg= 6 Pst = 3%Pdev Po = 97 %Pdev =Pdev =Eg Ia =560 400= 6111.5 N m 350 2 dev T dev T 60 To= 0.97 Tdev= 5928.2 N m Po= To= 350 2 5928.2 = 217.28 kW or 291.3hp 60
SERIES DC MOTOR o Series motor, its field winding is in series with thearmature: IL = Is =Ia o Since they are in series, the flux is proportional Ia. o As Ia changes with the load, the flux will be affected as well. o Series motors are well-suited for high-torque loads like portable electric tools, cranes, winches, hoists, and automobile starters because of their high torque production and compact size. T = ka Ia = k I2 a
SERIES DC MOTOR - DIRECTION OFROTATION o To change the direction of rotation of the series motor, eitherchange the connection of either the series field or thearmature. Torque T=k'I2 a Armature current (Ia)
EFFECT OF LOAD ON A SERIES DC MOTOR o Applying KVL to the circuit, Vt Ia Rs+ Ra Vt Ia Rs + Ra k Ia Eg = Vt Ia Rs IaRa = = ka o When Ia is very small at no-load condition, is very high and the motoris running at very dangerous speed. 1 Ia o Therefore, a load is always placed on a series motor before startingit. Extremely high speed at no-load (Ia 0) o As before, the speed is inversely proportional to the flux. But if the field circuit is open, the series motor will notrun. Speed Series motors lack good speed regulation. Armature current (Ia) 27
