ECE 4501 Power Systems Laboratory Manual Rev 1.05.0 DC MOTORS5.1 DC SHUNT MOTOR 5.1.1 OBJECTIVETo study the torque vs. speed characteristic of a shunt wound DC motor and calculate its efficiency.5.1.2 DISCUSSIONThe speed of any DC motor depends directly on its armature voltage and the strength of its magnetic field. The field winding in a shunt motor is in parallel with the armature winding and the DC supply. If the DC line voltage is constant, the armature voltage will be constant and thus the magnetic field strength will be constant. This consistency leads to a reasonably constant speedof operation.The speed does tend to drop with increasing load on the motor. This drop in speed is a result of resistive losses in the armature winding. Shunt motors with low armature winding resistance tend to have nearly constant speed of operation.As with any energy conversion device, the DC shunt motor is not 100% efficient. Not all of the electric energy supplied to the motor is converted into useful work (mechanical power). The difference between electrical power supplied and mechanical power available at the shaft is lost in the form of heat inside the motor. Losses occur in the DC resistance of the field and armature windings, in the magnetic circuit that couples field and armature windings, in the friction and windage of the rotating armature and in the resistance of the brush contacts on the commutator. Losses increase as the load on the motor increases, resulting in significant heating of the motor at full load.5.1.3 INSTRUMENTS AND COMPONENTSPower Supply Module EMS 8821DC Metering Module EMS 8412DC Motor/Generator Module EMS 8211Electrodynamometer Module EMS 8911Hand Tachometer EMS 89205.1.4 PROCEDURECAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply after every measurement.1) Connect the following circuit. DO NOT APPLY POWER AT THIS TIME.- 1 -ECE 4501 Power Systems Laboratory Manual Rev 1.02) Set the shunt field rheostat control knob at its full clockwise position, for maximum field excitation. Make sure the brushes are in their neutral position (90 and 0).3) Set the Electrodynamometer control knob (or Prime Mover/Dynamometer control knob) at its full counterclockwise position (minimum load). Note that the Dynamometer will require a power source.4) Turn on the power supply and adjust the voltage control to 120 V DC. Note the direction of rotation. If it is not clockwise, turn OFF the power supply and swap the connections across the shunt field. Then turn on the power supply.5) Adjust the field rheostat counterclockwise for a no load motor speed of 1800 rpm as indicatedby the tachometer. Double check the voltmeter to ensure the source voltage is 120 V DC. Once the source voltage and no load speed are set, DO NOT change the field rheostat setting for the remainder of this experiment section.6) Measure and record the line current as indicated by the ammeter for the no load condition at 1800 rpm.Source (Volts) Line Current (Amps) Speed (RPM) Torque (N-M, Lbf-In)120 1800 0 n-m, 0 lbf-in120 0.35 n-m, 3 lbf-in120 0.7 n-m, 6 lbf-in120 1.05 n-m, 9 lbf-in120 1.4 n-m, 12 lbf-in- 2 -ECE 4501 Power Systems Laboratory Manual Rev 1.07) Apply a load to the motor by turning the dynamometer control clockwise until the torque reading is 3 pound – inches (lbf-in) or 0.35 Newton-meters. Adjust the DC voltage control to maintain 120 V as necessary. 8) Measure and record the line current and motor speed for the 3 lbf-in load condition.9) Increase the load to 6, 9 and 12 lbf-in, taking speed and current measurements at each point. Record them in the table provided above.10) Return the voltage control to zero percent and turn OFF the power supply.11) Plot the recorded points on the graph below and connect them with a smooth curve.The completed graph represents a “speed – torque curve” for the DC shunt motor.12) Calculate the speed regulation of the motor using the following equation: RPM (No Load) - RPM (Full Load)Speed Reg. = -------------------------------------------- x 100 % RPM (Full Load)Note: Full load for this motor is 9 Lbf – in. Speed Regulation = _____________ %- 3 -ECE 4501 Power Systems Laboratory Manual Rev 1.013) Now set the dynamometer control to its full clockwise position to maximize the starting load for the motor. Do NOT adjust the field rheostat.14) Turn on the DC power supply and increase the voltage control until the motor draws 3 amps of starting current. The motor will turn very slowly or not at all.15) Measure and record the DC voltage and the torque developed.V = __________ Volts DC Torque = __________ Lbf – in or N-m.16) Return the voltage control to zero percent and turn off the power supply.17) The line current drawn by the motor in step 14) above is limited only by the equivalent DC resistance of the armature winding and field windings. Calculate the value of the starting current drawn by the motor if full line voltage, 120 V DC, were applied:Starting Current at 120 V DC = __________ Amps5.1.5 CONCLUSIONS5.1.5.1 Calculate the horsepower, HP, developed by the shunt wound DC motor when the load torque is 9 Lbf – in (1.05 N-m):HP = (RPM)(Lbf – in)(1.59) / 100,000 OR HP = (RPM)(N-m)(14.07) / 100,000______________________________________________________________________________________________________________________________________________5.1.5.2 Based on the result above and knowing that 1 HP equals 746 Watts, what is the power developed by the motor in watts?______________________________________________________________________________________________________________________________________________5.1.5.3 What is the input power to the motor, in watts, as calculated using the voltage and current from the table for 9 Lbf – in of load?______________________________________________________________________________________________________________________________________________- 4 -ECE 4501 Power Systems Laboratory Manual Rev 1.05.1.5.4 Using the input and output power in watts, calculate the efficiency of the motor at full load.Efficiency,  = 100% (P-out) / (P-in)______________________________________________________________________________________________________________________________________________5.1.5.5 What are the losses, in watts, for the motor at full load? 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