DIRECT CURRENT MOTORS

The Electric Motor- It is a machine that converts electrical energy (or power) into mechanical energy (or power).

DC electric motor works on the principle that whenever a current carrying conductor is placed in a magnetic field, a force acts on a conductor causing it to move.

Speed Characteristics of a DC Motor

where:N = speed (rpm)a = number of armature current pathsP = number of polesZ = number of conductorsΦ = flux per pole (weber)Eb = back emf or counter emf (volt)

PZ

aEN b60

After the machine has been assembled, “PZ and a” are constant

where:k = proportionality constantSubscript 1 = for condition 1Subscript 2 = for condition 2

bEkN

1

2

2

1

2

1

b

b

E

E

N

N

Torque developed in the Armature

where:T = torque developed (newton-meter)Ia = armature current (ampere)

N = speed of armature rotation (rpm)a = number of armature current pathsP = number of polesZ = number of conductorsΦ = flux per pole (weber)

a

IPZT a

283.6

After the machine has been assembled, “PZ and a” are constant

akIT2

1

2

1

2

1

a

a

I

I

T

T

Example:

A simplex lap wound armature has 600 conductors and carries a current of 50 amperes per armature current path. If the flux per pole is 30 mWb, determine the electromagnetic torque developed by the armature.

m

IZ

mP

IPZ

a

IPZT aaa

283.6283.6283.6

mN

xT

24.1431283.6

501030600 3

Power Developed in the Armature

where:Pd = power developed in the armature (watt)

Ia = armature current (ampere)

Eb = back emf or counter emf (volt)

abd IEP

Mechanical Power Output in the Shaft

where:Hp = mechanical power output in the shaft (horsepower)N = speed of shaft rotation (rpm)T = torque developed

mNTrpmNNT

Hp

ftlbTrpmNNT

Hp

;760,44

2

;000,33

2

Types of DC Motor

SHUNT MOTOR- The armature and the field coils are connected

in parallel.

Note:In a shunt motor, unless otherwise specified, the flux is assumed to be constant.

where:Vs = supply voltage (volt)

Eb = back emf (volt)

Ra = armature equivalent resistance (ohm)

Rsh = shunt field winding resistance (ohm)

Im = current drawn from supply (ampere)

Ia = armature current (ampere)

Ish = shunt field current (ampere)

aasb RIVE sham III

sh

ssh R

VI

Example 1:

A shunt motor is taking 72 A at 120 V while developing an output of 10 bhp. Armature resistance is 0.05 ohm and shunt field resistance of 60 ohms. Determine

a. counter emfb. total copper lossesc. iron and friction losses

wattsP

RIRIP

B

voltsRIVE

AIII

AR

VI

A

copperloss

shshaacopperloss

aasb

shma

sh

ssh

48560205.070

.

5.11605.070120

70272

260

120

.

22

22

wattsP

P

PPP

wattsPPP

wattsP

wattsIVP

C

frictioniron

frictioniron

frictionironcopperlosses

outputinputlosses

output

msinput

695

4851180

118074608640

746074610

864072120

.

&

&

&

Series Motor- The armature and the field coils are connected in

series.

where:Rse = series field winding resistance (ohm)

ma II seaasb RRIVE

Example:

A 200V series motor has a field resistance of 0.2 ohm and an armature resistance of 0.1 ohm. The motor takes 30A of current at 900 rpm while developing full load torque. What is the motor speed when this motor develops 70% of full load torque?

Note:In a series motor, since the flux is proportional to the armature current, therefore, the torque is directly proportional to the square of the armature current.

VE

RRIVE

VE

RRIVE

AT

TI

T

TII

I

I

T

T

b

seaasb

b

seaasb

a

aaa

a

47.192

2.01.01.25200

191

2.01.030200

1.257.0

30

2

22

1

11

2

12

2

1

1

1

1

2

2

2

1

rpm

IE

INEN

IN

IN

E

E

IkkNE

ab

ab

a

a

b

b

a

10841.25191

3090047.192

'

21

12

2

1

2

1

12

2

1

Thus, E = k”NIa

LONG SHUNT COMPOUND MOTOR- The series field coil is connected in series with the armature coil while the shunt field coil is connected across the series combination.

sh

ssh R

VI

sham III

seaasb RRIVE

Example:

A long shunt compound motor draws a line current of 42A from a 230V DC source. The armature resistance is 0.1 ohm while the series and shunt field resistances are 0.2 ohm and 50 ohm respectively. If the iron and friction losses amount to 500W, determine the overall efficiency of the machine.

WVIP

WRIP

WRIP

AIII

AR

VI

sshsh

sease

aaa

shma

sh

ssh

10582306.4

752.2792.04.37

876.1391.04.37

4.376.442

6.450

230

22

22

%53.79

%1009660

628.19779660

966042230

628.1977

5001058752.279876.139&

x

P

PP

P

P

WIVP

WP

P

PPPPP

input

lossesinput

input

output

msinput

losses

losses

frictionironshsealosses

SHORT SHUNT COMPOUND GENERATOR- The series field coil is connected in series with the supply voltage while the shunt field coil is connected in parallel with the armature coil.

aasemsb RIRIVE

sham III

sh

semssh R

RIVI

Example:

A 220V short shunt compound motor has an armature resistance of 0.4 ohm, a shunt field resistance of 110 ohms and a series field resistance of 0.6 ohms. If this motor draws an armature current of 50A at rated load, determine

a. the counter emfb. horsepower developed in the

armature

hp

IEP

B

VE

RIRIVE

AI

II

III

II

R

RIVI

A

abd

b

aasemsb

m

mm

sham

mm

sh

semssh

325.11746

5097.168

746

.

97.1684.0506.0718.51220

718.51

005454.0250

005454.02110

6.0220

.

PONY BRAKE TEST OF A MOTOR- This test is used to determine the output

horsepower of a motor.

where:T = torque exerted by the motor during the testN = speed of the motor shaft during the test (rpm)Hp = output horsepower of the motor

T = (scale reading – dead weight) (length of arm)

Example 1:

The pony brake test of an elevator door drive shunt motor, the ammeter and voltmeter measuring the input read 34A and 220V. The speed of the motor is found to be 910 rpm and the balance on a 2 ft brake arm read 27.2 pounds. The tare weight of the arm is found to be positive 2.3 pounds. Determine the efficiency of the motor at this load.

T = (scale reading – dead weight) (length of arm)

%056.86

%100026.10

628.8

026.10746

34220

746

628.833000

8.499102

33000

2

8.4923.22.27

xP

P

hpIV

HP

hpNT

HP

ftlbT

in

out

msinput

output

Example 2:

Calculate the force that will be exerted on the scale in a pony brake test when a 20 hp, 1400 rpm motor is operating at full load. The length of the brake arm is 3 ft and the tare weight of the brake is 3.75 lbs.

Let: F = force exerted on the scale

T = (scale reading – dead weight) (length of arm)

lbsF

F

NTHP

FT

76.28

000,33

375.31400220

000,33

2

375.3

Substitute:

Assignment:

1. The shaft power of a shunt motor is 7.8 hp. It draws 50 A from 120 V. The field winding draws 1.2 A. What is the efficiency of the motor? Ans: 96.98%

2. A shunt motor was tested by means of a pony brake having a length arm of 3.5 feet and a tare weight of 5.7 lbs. The current drawn by the machine from a 240 V line was 50.9 A when the scale reading was 24 lbs and the speed of the motor was 1215 rpm. Calculate the rotational losses of the motor. The armature and shunt field resistances of the machine are 0.25 Ω and 240Ω, respectively. Ans: 673.5 W