Lecture 32 DC Motors Part II

Learning Objectives Analyze the circuit equivalent of a permanent magnetic DC motor that accounts for armature resistance, induced electromotive force (back EMF), developed electromagnetic torque, and applied (input) voltage. Define the power output of a permanent magnetic DC motor in terms of developed electromagnetic torque and angular velocity. Relate power output in terms of horse power. Determine the efficiency of a permanent magnetic DC motor using the given or calculated power in and power out.

Basic DC Motor Operations

Parts of a DC Motor SOURCE: Gears Educational Systems.

Rotary DC Motor Torque developed Td=Kv Ia Power Developed Pd = Td ω = Kv Ia ω Back EMF Ea = Kv ω Angular velocity ω = 2  (RPM/60)

DC motor power flow Pelec loss Pmech loss Pin Pout Pdev = Pout + Pmech loss 6

Electrical Power losses Electrical loss occurs due to the armature resistance and is expressed as Pelec loss = Ia2 Ra. Pelec loss 7

Torque losses (Tloss) Pmech loss represents losses due to the friction of mechanical parts, magnetic inefficiencies of the material, and losses coupling brushes and commutator and is expressed as Pmech loss = Tloss wm Pmech loss 8

Output Power Pd represents the power developed by the motor which includes power out and mechanical losses (Ploss). It is expressed Pd = Pmech loss + Pout = Tdev w = Kv IA wm Power out is the power that ultimately gets to the load and is expressed Pout = TLOADwm Pout Pd = Pout + Pmech loss 9

Power Conversion Diagram Electrical Mechanical

Motor Efficiency Developed power is: If we Ignore rotational losses, Pd=Pout, and machine efficiency can be calculated as:

Magnetic field Instead of permanent magnet, we could raise the field strength B with an electromagnet. The wires wrapped around a ferromagnetic core are known as field windings. The field windings are stationary and are part of the stator.

Magnetic field 2 poles

Magnetic poles Increasing the number of poles will increase and smooth the output torque. four-pole dc machine eight-pole dc machine

Example Problem 1 A 24 V DC motor is rated for 15 A. RA = 0.20 Ω Assumed no rotational losses. Determine: The input power The power loss due to the resistance of the armature The power developed The back EMF (EA) The efficiency assuming no mechanical power loss Draw a power conversion diagram and fill-in the values for power in, electrical power loss, power developed, mechanical power loss and power out.

Example Problem 2 A permanent magnet DC motor is rated for 120V, 17A and 1200 rpm. The machine is 90% efficient at rated conditions Tloss = 0.0334 N·m Find Ra and Kv and torque developed by the motor .

Example Problem 3 We wish to design a 1/4 hp, 28 V DC motor with an efficiency of 96%. What current can we expect to draw? If the machine constant is Kv= 0.2139 ν·s, determine Tout if we ignore mechanical losses. Calculate rated speed in rpm.

Why is there so much field wiring on the Practical Exercise? ARMATURE

PE 20 Diagram FIELD ARMATURE

Starting Circuit At startup, when the speed is zero, the back EMF is zero. The armature current is limited only by the armature resistance. Since this resistance is very low, a large armature current would result if full terminal voltage were applied directly to the armature. VDC=Ea Speed (RPM) Current (A)

Starting Circuit This large current could overheat the armature windings and cause damage to the motor. Additional resistance (called starting resistance) is placed in series with the armature at startup and removed after the rotor comes up to speed and back EMF is produced. Shorting switch shut

Motor Starting Circuit Larger motors will require a more complex starting circuit. This circuit has 4 starting resistors As the motor speeds up, the contact arm moves to sequentially remove starting resistors from the circuit The Emergency Propulsion Motor (EPM) on a submarine uses a starting circuit like this.

Motor Starting Circuit