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Electric Machines I DC Machines - DC Motors

1

Dr. Firas Obeidat

2

Table of Contents

1 • DC Motor Principle

2 • Types of DC Motors

3 • E.M.F. Equation of DC Motor

4 • Armature Torque of DC Motor

5 • Speed Regulation of DC Motor

6 • Total Losses in DC Motor

7 • Power Stages and Efficiency

8

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

DC Motor Principle

Constructionally, there is no basic difference between DC generator and

DC motor. The DC machine can be used as generator or as motor.

When field magnets are excited in multipolar DC motor, and its

armature conductors are supplied with current from the supply, they

experience a force tending to rotate the armature. Armature conductors

under N-pole are assumed to carry current downwards (crosses) and

those under S-poles to carry current upwards (dots).

By applying Fleming’s left hand rule, each conductor experiences a force

F which tends to rotate the armature in anticlockwise direction. These

forces collectively produce a driving torque which sets the armature

rotating.


Types of DC Motors

𝐼𝐴 = 𝐼𝐿

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴

𝑉𝐹 = 𝐼𝐹 𝑅𝐹

𝐸𝐴 = 𝑘ϕ𝜔𝑚

Where

IA: is the armature current

IL: is the load current

EA: is the internal generated voltage

VT: is the terminal voltage

IF: is the field current

VF: is the field voltage

RA: is the armature winding resistance

RF: is the field winding resistance

ϕ: is the flux

𝜔m: is the rotor angular speed

1- Separately Excited DC Motor

EA

-

+ RA

IA IL

-

+

VT

RF

LF

-

+

VF

IF

2- Shunt DC Motor

𝐼𝐿 = 𝐼𝐴 + 𝐼𝑓

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴 𝑅𝐴

𝑉𝐹 = 𝐼𝐹 𝑅𝐹

𝐸𝐴 = 𝑘ϕ𝜔𝑚 EA

-

+ RA

IA IL

-

+

VT

RF

LF

IF


Types of DC Motors

Speed control of separately excited and shunt DC motors

Adjusting the field resistance RF

• Increasing RF causes 𝑰𝑭 = 𝑽𝑻/𝑹𝑭 to decrease.

• Decreasing IF decreases ϕ.

• Decreasing ϕ, lowers 𝐸𝐴=𝑘ϕ𝜔𝑚 .

• Decreasing EA increases 𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨)/𝑹𝑨.

• Increasing IA increases Tind=𝑘ϕIA , with the change in IA dominant over the change in flux.

• Increasing Tind makes Tind>Tload and the speed 𝜔m increases.

• Increasing 𝜔m increases 𝐸𝐴=𝑘ϕ𝜔𝑚.

• Increasing EA decreases IA.

• Decreasing IA decreases Tind until Tind=Tload at a higher speed 𝜔𝑚.

Adjusting the terminal voltage

applied to the armature

• An increase in VT increases 𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨)/𝑹𝑨.

• Increasing IA increases Tind=𝑘ϕIA

• Increasing Tind makes Tind>Tload and the speed 𝜔m increases.

• Increasing 𝜔m increases 𝐸𝐴=𝑘ϕ𝜔𝑚.

• Increasing EA decreases 𝑰𝑨 = (𝑽𝑻 − 𝑬𝑨)/𝑹𝑨.

• Decreasing IA decreases Tind until Tind=Tload at a higher speed 𝜔𝑚

Inserting a resistor in series with the armature circuit

• resistor is inserted in series with the armature circuit, the effect is to drastically increase the slope of the motor's torque- speed characteristic, making it operate more slowly if loaded.

• The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted resistor are very large. For this reason, it is rarely used. It will be found only in applications in which the motor spends almost all its time operating at full speed or in applications too inexpensive to justify a better form of speed control.


Types of DC Motors

3- The Permanent Magnet DC Motor

A permanent-magnet DC (PMDC) motor is a DC motor whose poles are made of permanent magnets. Permanent-magnet dc motors offer a number of benefits compared with shunt dc motors in some applications. Since these motors do not require an external field circuit, they do not have the field circuit copper losses associated with shunt dc motors.

Because no field windings are required, they can be smaller than corresponding shunt DC motors. they are especially common in smaller fractional- and sub fractional-horsepower sizes.

PMDC motors are generally less expensive, smaller in size, simpler, and higher efficiency than corresponding DC motors with separate electromagnetic fields. This makes them a good choice in many DC motor applications. The armatures of PMDC motors are essentially identical to the armatures of motors with separate field circuits, so their costs are similar too. However, the elimination of separate electromagnets on the stator reduces the size of the stator, the cost of the stator, and the losses in the field circuits.

PMDC motors also have disadvantages. Permanent magnets cannot produce as high a flux density as an externally supplied shunt field, so a PMDC motor will have a lower induced torque Tind per ampere of armature current IA than a shunt motor of the same size and construction. In addition, PMDC motors run the risk of demagnetization.

A permanent-magnet DC motor is basically the same machine as a shunt dc motor, except that the flux of a PMDC motor is fixed. Therefore, it is not possible to control the speed of a PMDC motor by varying the field current or flux. The only methods of speed control available for a PMDC motor are armature voltage control and armature resistance control.

The techniques to analyze a PMDC motor are basically the same as the techniques to analyze a shunt dc motor with the field current held constant.


Types of DC Motors

𝐼𝐴 = 𝐼𝑆 = 𝐼𝐿

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴(𝑅𝐴 + 𝑅𝑆)


4- Series DC Motor

EA

-

+ RA

IA IL

-

+

VT

Rs Ls

Is

Speed control of series DC motors 1) Change the terminal voltage of the motor.

2) Insertion of a series resistor into the motor circuit, but this

technique is very wasteful of power and is used only for

intermittent periods during the start-up of some motors.


Types of DC Motors

5- Compound DC Motor

Speed control of Cumulatively compound DC motors 1) Change the field resistance RF.

2) Change the armature voltage VA.

3) Change the armature resistance RA.

EA

-

+ RA

IA IL

-

+

VT

Rs Ls RF

LF

IF

EA

-

+ RA

IA IL

-

+

VT

Rs LsRF

LF

IF

𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴(𝑅𝐴+𝑅𝑠)

For Long Shunt Cumulatively Compound DC Motor


For Short Shunt Cumulatively Compound DC Motor

𝑉𝑇 = 𝐼𝐹 𝑅𝐹

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴𝑅𝐴 +𝐼𝐿𝑅𝑠


𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹


Types of DC Motors

Example: A 220V DC shunt machine has an armature resistance of 0.5Ω. If

the full load armature current is 20A, find the induced emf when the machine

acts as (i) generator (ii) motor.

EA

-

+ RA

-

+ VT =

220V

RF

LF

IF

ILIA=20A

EA

-

+ RA

ILIA=20A

-

+ VT =

220V

RF

LF

IF

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴𝑅𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴

𝐸𝐴 = 220 + 20 × 0.5 = 230𝑉

𝐸𝐴 = 220 − 20 × 0.5 = 210𝑉

(i) As generator

(ii) As motor


Types of DC Motors

Example: A 440V shunt DC motor has an armature resistance of 0.8Ω and

field resistance of 200Ω. find the back emf when giving an output of 7.46kW at

85% efficiency.

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴 × 𝑅𝐴 = 440 − 17.7 × 0.8 = 425.84𝑉

EA

-

+ RA

ILIA

-

+

VT

RF

LF

IF

𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛× 100%

85% =7.46𝑘

𝑃𝑖𝑛× 100%

𝑃𝑖𝑛 =7.46𝑘

0.85= 8.7765𝑘𝑊

𝑃𝑖𝑛 = 8.7765𝑘 = 𝑉𝑇𝐼𝐿 = 440 × 𝐼𝐿

𝐼𝐿 =8.7765𝑘

440= 19.9𝐴

𝐼𝐹 =440

200= 2.2𝐴 𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 19.9 − 2.2 = 17.7𝐴


Types of DC Motors

Example: A 25kW, 250V DC shunt machine has an armature and field

resistances of 0.06Ω and 100Ω respectively. Determine the total armature

power developed when working (i) as generator delivering 25kW output and

(ii) as motor taking 25kW input.

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 250 + 0.06 × 102.5 = 256.15

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 250 − 97.5 × 0.06 = 244.15V

𝑃𝐴 = 𝐸𝐴𝐼𝐴 = 256.15 × 102.5 = 26255.375𝑊

(i) As generator

(ii) As motor

EA

-

+ RA

-

+

VT

RF

LF

IF

ILIA

EA

-

+ RA

ILIA

-

+

VT

RF

LF

IF

𝑃𝑜𝑢𝑡 = 𝐼𝐿𝑉𝑇 → 𝐼𝐿 =𝑃𝑜𝑢𝑡

𝑉𝑇=

25000

250= 100𝐴

𝐼𝐹 =250

100= 2.5𝐴 𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 100 + 2.5 = 102.5𝐴

𝑃𝑖𝑛 = 𝐼𝐿𝑉𝑇 → 𝐼𝐿 =𝑃𝑖𝑛

𝑉𝑇=

25000

250= 100𝐴

𝐼𝐹 =250

100= 2.5𝐴 𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 100 + 2.5 = 97.5𝐴

𝑃𝐴 = 𝐸𝐴𝐼𝐴 = 244.15 × 97.5 = 23804.625𝑊


E.M.F. Equation of DC Motor

Let

ϕ: flux/pole in weber.

Z: total number of armature conductors

Z=number of slots × number of conductors/slot

A: number of parallel paths in armature

N: armature rotation in rpm

E: emf induced in any parallel path in armature

Generated emf EA=emf generated in any one of the parallel paths

Average emf generated/conductor=dϕ/dt volt

Flux cut/conductor in one revolution dϕ=ϕP Wb

Number of revolutions /second=N/60

Time for one revolution dt=60/N second

E.M.F. generated/conductor= dϕ/dt= ϕPN/60 volt



For simplex wave-wound motor

Number of parallel paths=2

Number of conductors (in series) in one path=Z/2

𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =𝜙𝑃𝑁

60×

𝑍

2=

𝜙𝑃𝑍𝑁

120𝑣𝑜𝑙𝑡

For simplex lap-wound motor

Number of parallel paths=P

Number of conductors (in series) in one path=Z/P

𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =𝜙𝑃𝑁

60×

𝑍

𝑃=

𝜙𝑍𝑁

60𝑣𝑜𝑙𝑡

In general

𝐸𝐴 =𝜙𝑍𝑁

60×

𝑃

𝐴𝑣𝑜𝑙𝑡

where

A=2 for simplex wave-winding

A=P for simplex lap-winding

𝐸𝐴 =1

2π×

2π𝑁

60× 𝜙𝑍 ×

𝑃

𝐴=

𝑍𝑃

2π𝐴𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝜔𝑚 =

2π𝑁

60

For a given DC machine Z,P and A are constant

𝐸𝐴 = 𝑘𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝑘 =𝑍𝑃

2π𝐴



Example: A 4-pole, 32 conductor, lap-wound DC shunt generator with

terminal voltage of 200V delivering 12A to the load has RA=2 Ω and field

circuit resistance of 200Ω. It is driven at 1000rpm. Calculate the flux per pole

in the machine. If the machine has to be run as a motor with the same

terminal voltage and drawing 5A from the mains, maintaining the same

magnetic field, find the speed of the machine.

EA

-

+ RA

-

+VT =

200V

RF

LF

IF=1A

IL=12AIA=13A

Action as Generator

EA

-

+ RA

-

+RF

LF

IF=1A

VT =

200V

IL=5AIA=4A

Action as Motor

(i) As generator

𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 200 + 13 × 2 = 226𝑉

𝐼𝐹 =200

200= 1𝐴 𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 12 + 1 = 13𝐴

𝐸𝐴 = 226 =𝜙𝑍𝑁

60×

𝑃

𝐴 A=P for lap-winding

𝜙 =226 × 60

1000 × 32= 0.42375𝑤𝑏

(ii) As motor

𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 5 + 1 = 4𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 200 + 4 × 2 = 192 =𝜙𝑍𝑁

60×

𝑃

𝐴𝑉

𝑁 =192 × 60

0.42375 × 32= 850𝑟𝑝𝑚


Armature and Shaft Torque of DC Motor

Let Ta be the torque developed by the armature of a motor, then the power

developed

𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 = 𝑇𝑎 × 2𝜋𝑁/60 𝑤𝑎𝑡𝑡

The electrical power converted into mechanical power in the armature=EAIA

Equating the above two equations yields

𝑇𝑎 × 2𝜋𝑁 = 𝐸𝐴𝐼𝐴

𝑇𝑎 =𝐸𝐴𝐼𝐴

2𝜋𝑁/60=

60

2𝜋

𝐸𝐴𝐼𝐴

𝑁= 9.55

𝐸𝐴𝐼𝐴

𝑁N. m


60×

𝑃

𝐴𝑣𝑜𝑙𝑡

𝑇𝑎 =9.55

60𝜙𝑍𝐼𝐴 ×

𝑃

𝐴= 0.159𝜙𝑍𝐼𝐴 ×

𝑃

𝐴N. m

Or

𝑇𝑠ℎ = 9.55𝑀𝑜𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡

𝑁N. m

(𝑇𝑠ℎ − 𝑇𝑎) is known as lost torque and is due iron and friction losses of the motor

Armature Torque of DC Motor

Shaft Torque of DC Motor



Example: A DC motor takes an armature current of 110A at 480V. The

armature circuit resistance is 0.2Ω. The machine has 6 poles and the armature

is lap-connected with 864 conductors. The flux per pole is 0.05wb. Calculate

the speed and the gross torque developed by the armature.

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 480 − 110 × 0.2 = 458V


60×

𝑃

𝐴=

0.05 × 864 × 𝑁

60= 458 𝑁 = 636 𝑟𝑝𝑚

𝑇𝑎 = 9.55𝐸𝐴𝐼𝐴

𝑁= 9.55

458 × 110

636≈ 756N. m

𝑇𝑎 = 0.159 × 𝜙 × 𝑍 × 𝐼𝐴 = 0.159 × 0.05 × 864 × 110 ≈ 756N. m Or

Example: Determine armature torque and motor speed of 220V, 4-pole series

motor with 800 conductors wave connected supplying a load by taking 45A from

the mains. The flux per pole is 25mwb and its armature circuit resistance is 0.6Ω.

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 220 − 45 × 0.6 = 193V


60×

𝑃

𝐴=

0.025 × 800 × 𝑁

60×

4

2= 193 𝑁 = 579 𝑟𝑝𝑚

𝑇𝑎 = 0.159𝜙𝑍𝐼𝐴 ×𝑃

𝐴= 0.159 × 0.025 × 800 × 45 ×

4

2= 286.2N. m



Example: A 220V DC shunt motor runs at 500 rpm when the armature

current is 50A. Calculate the speed if the torque id doubled. Given that

RA=0.2Ω.

In shunt DC motor the flux is constant.

𝑇𝑎1 = 0.159𝜙𝑍𝐼𝐴1 ×𝑃

𝐴 𝑇𝑎2 = 2𝑇𝑎1 = 0.159𝜙𝑍𝐼𝐴2 ×

𝑃

𝐴

𝑇𝑎1

2𝑇𝑎1=

𝐼𝐴1

𝐼𝐴2 →

1

2=

50

𝐼𝐴2 → 𝐼𝐴2 = 2 × 50 = 100𝐴

𝐸𝐴1 =𝜙𝑍𝑁1

60×

𝑃

𝐴 𝐸𝐴2 =

𝜙𝑍𝑁2

60×

𝑃

𝐴

𝐸𝐴1

𝐸𝐴1=

𝑁1

𝑁2→

210

200=

500

𝑁2 → 𝑁2 =

500 × 200

210= 476𝑟𝑝𝑚

𝐸𝐴1 = 𝑉𝑇 − 𝐼𝐴1𝑅𝐴 = 220 − 50 × 0.2 = 210V

𝐸𝐴2 = 𝑉𝑇 − 𝐼𝐴2𝑅𝐴 = 220 − 100 × 0.2 = 200V



Example: A 500V, 37.3kW, 1000rpm DC shunt motor has on full load an

efficiency of 90%. Determine (i) full load line current (ii) full load armature

torque (neglect iron and friction losses).

𝑀𝑜𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 =𝑃𝑜𝑢𝑡

𝜂=

37300

0.9= 41444𝑊

𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑙𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 =41444

500= 82.9𝐴


𝑁≈ 9.55

𝑜𝑢𝑡𝑝𝑢𝑡

𝑁= N. m (neglect iron and friction losses)

𝑇𝑎 = 9.5537300

1000= 356N. m

(i)

(ii)

Example: Determine the torque established by the armature of a four-poles

DC motor having 774 conductors, two paths in parallel, 24 milli-webers of

pole-flux and the armature current is 50A.

𝑇𝑎 = 0.159 × 𝜙 × 𝑍 × 𝐼𝐴 = 0.159 × 0.024 × 774 × 50 × 4/2 = 295.36N. m


Speed Regulation of DC Motor

The Speed Regulation (SR) is defined as the change in speed when the load on

the motor is reduced from rated value to zero, expressed as percent of the

rated load speed.

The speed regulation refers to the change in speed of a motor with change in

applied load torque, other conditions remaining constant.

𝑆𝑅 =𝑁𝑁𝐿 − 𝑁𝐹𝐿

𝑁𝐹𝐿× 100%

Where

NNL: noload speed

NFL: full load speed

Example: A 4-pole series motor has 944 wave connected armature conductors.

At a certain load, the flux per pole is 34.6 mWb and the total mechanical

torque developed is 209 N.m. calculate the line current taken by the motor and

the speed at which it will run with an applied voltage of 500V. Total motor

resistance is 3 ohm.

𝑇𝑎 = 209 = 0.159 × 𝜙 × 𝑍 × 𝐼𝐴 ×𝑃

𝐴= 0.159 × 0.0346 × 944 × 𝐼𝐴 ×

4

2

𝐼𝐴 = 20.1𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 500 − 20.1 × 3 = 439.7V

𝐸𝐴 = 439.7 =𝜙𝑍𝑁

60×

𝑃

𝐴=

0.0346 × 944 × 𝑁

60×

4

2 → 𝑁 = 403.8𝑟𝑝𝑚


Speed Regulation of DC Motor

𝑆𝑅 =𝑁𝑁𝐿 − 𝑁𝐹𝐿

𝑁𝐹𝐿× 100% =

1200 − 1120

1120× 100% = 7.1%

Example: A 230V DC shunt motor has an armature resistance of 0.5Ω and

field resistance of 115Ω. At no load, the speed is 1200rpm and the armature

current 2.5A. On application of rated load, the speed drops to 1120 rpm.

Determine the speed regulation, line current and power input when the motor

delivers rated voltage.

𝑁1 = 1200𝑟𝑝𝑚 𝐸𝐴1 = 𝑉𝑇 − 𝐼𝐴1𝑅𝐴 = 230 − 2.5 × 0.5 = 228.75V

𝑁2 = 1120𝑟𝑝𝑚 𝐸𝐴2 = 𝑉𝑇 − 𝐼𝐴2𝑅𝐴 = 230 − 𝐼𝐴2 × 0.5

𝐸𝐴1

𝐸𝐴2=

𝑁1

𝑁2 →

228.75

𝐸𝐴2=

1200

1120 → 𝐸𝐴2 =

1120 × 228.75

1200= 213.5𝑉

𝐸𝐴2 = 213.5 = 230 − 𝐼𝐴2 × 0.5 → 𝐼𝐴2 =230 − 213.5

0.5= 33𝐴

𝐼𝐿 = 𝐼𝐴2 + 𝐼𝐹 = 33 +230

115= 35𝐴

𝑃𝑖𝑛 = 𝐼𝐿𝑉𝑇 = 35 × 230 = 8050W


Total Loss in a DC Motor

Tota

l L

oss

es

Copper Losses

Armature Cu Loss

Shunt Cu Loss

Series Cu Loss

Iron Losses

Hysteresis Loss

Eddy Current Loss

Mechanical Losses

Friction Loss

Air Friction or Windage Loss

Stray Losses

Iron and mechanical losses are collectively known as Stray (Rotational) losses.

Constant or Standing Losses

Field Cu losses is constant for shunt and compound generators. Stray losses

and shunt Cu loss are constant in their case. These losses are together known

as Constant or Standing Losses (Wc).


Power Stages and Efficiency

Mechanical Efficiency

𝜂𝑚 =𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠× 100% =

𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝐸𝐴𝐼𝐴× 100%

Electrical Efficiency

𝜂𝑒 =𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠× 100% =

𝐸𝐴𝐼𝐴

𝑉𝑇𝐼𝐿× 100%

Overall or Commercial Efficiency

𝜂𝑐 = 𝜂𝑚 × 𝜂𝑒 =𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠× 100% =

𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝑉𝑇𝐼𝐿× 100%


Example: The armature winding of a 4-pole, 250V DC shunt motor is lap

connected. There are 120 slots, each slot containing 8 conductors. The flux per

pole is 20mWb and current taken by the motor is 25A. The resistance of

armature and field circuit are 0.1Ω and 125Ω respectively. If the rotational

losses amount to be 810W. Find

(i) gross torque (Ta) (ii) useful torque (Tsh) (iii) efficiency.


𝐼𝐹 =𝑉𝑇

𝑅𝐹=

250

125= 2𝐴 𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 25 − 2 = 23𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 250 − 23 × 0.1 = 247.7𝑉

𝐸𝐴 = 247.7 =𝜙𝑍𝑁

60×

𝑃

𝐴=

0.02 × (120 × 8) × 𝑁

60×

4

4 → 𝑁 = 773𝑟𝑝𝑚


𝑁= 9.55

247.7 × 23

773= 70.4N. m

(i) gross torque (Ta)



Driving power in armature = 𝐸𝐴𝐼𝐴 = 247.7 × 23 = 5697.1𝑊

(ii) useful torque (Tsh)

Optput power = 𝐸𝐴𝐼𝐴 − 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 5697.1 − 810 = 4887.1𝑊


𝑁= 9.55

4887.1

773= 60.4N. m

(iii) efficiency

𝜂𝑚 =𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑊𝑎𝑡𝑡𝑠

𝐸𝐴𝐼𝐴× 100% =

4887.1

5697.1× 100% = 85.8%

𝜂𝑒 =𝐸𝐴𝐼𝐴

𝑉𝑇𝐼𝐿× 100% =

5697.1

250 × 25× 100% = 91%

𝜂𝑐 =4887.1

250 × 25× 100% = 78.2%


Example: A 20hp (14.92kW), 230V, 1150rpm, 4poles DC shunt motor has a

total of 620 conductors arranged in two parallel paths and yielding an

armature circuit resistance of 0.2Ω. When it delivers rated power at rated

speed, it draws a line current of 74.8A and a field current of 3A. Calculate

(i) the flux per pole (ii) armature torque (iii) the rotational losses (iv) total

losses expressed as a percentage of power.


𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 74.8 − 3 = 71.8𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 230 − 71.8 × 0.2 = 215.64𝑉

𝐸𝐴 = 215.64 =𝜙𝑍𝑁

60×

𝑃

𝐴=

𝜙 × 620 × 1150

60×

4

2 → 𝜙 = 9𝑚𝑊𝑏


𝑁= 9.55

215.64 × 71.8

1150= 128.6N. m

(i) the flux per pole

(ii) armature torque

Rotational losses=EAIA – output power =215.64×71.8 - 14920= 562.952W

(iii) the rotational losses

(iv) total losses expressed as a percentage of power

Total losses=input power (VTIL)-output power = 230×74.8 - 14920= 17204 - 14920= 2284W

total losses expressed as a percentage of power=2284/17204=13.3%


Example: A 7.46kW, 250V shunt motor takes a line current of 5A when

running light. Calculate the efficiency as a motor when delivering full load

output, if the armature resistance are 0.5Ω and 250 Ω respectively.


𝑃𝑖𝑛 = 𝑉𝑇𝐼𝐿 = 250 × 5 = 1250𝑊

𝐼𝐹 =𝑉𝑇

𝑅𝐹=

250

250= 1𝐴 𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 5 − 1 = 4𝐴

1- When loaded lightly

Field Cu loss = 𝑉𝑇𝐼𝐹 = 250 × 1 = 250W

Armature Cu loss = 𝐼𝐴2𝑅𝐴 = 42 × 0.5 = 8W

Iron and friction losses=input power-Armature Cu loss-Filed Cu loss

Iron and friction losses=1250-250-8=992W Iron and friction losses is constant

2- at full load

𝑉𝑇𝐼𝐴 = 𝐼𝐴2𝑅𝐴 + 𝐸𝐴𝐼𝐴 𝑉𝑇𝐼𝐴 = 250𝐼𝐴

𝐸𝐴𝐼𝐴 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 + 𝑖𝑟𝑜𝑛 𝑙𝑜𝑠𝑠 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑙𝑜𝑠𝑠 = 7460 + 992 + 0 = 8452𝑊

250𝐼𝐴 = 𝐼𝐴2 × 0.5 + 8452



0.5𝐼𝐴2 − 250𝐼𝐴 + 8452 = 0

Rearrange the above equation yields

𝐼𝐴 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎=

250 ± (−250)2−4 × 0.5 × 8452

2 × 0.5= 36.5 𝑜𝑟 463.5

𝐼𝐴 = 36.5𝐴

𝐼𝐿 = 𝐼𝐴 + 𝐼𝐹 = 36.5 − 1 = 37.5𝐴

𝑃𝑖𝑛 = 𝑉𝑇𝐼𝐿 = 250 × 37.5 = 9375𝑊

𝑃𝑜𝑢𝑡 = 7460𝑊

𝜂𝐹𝐿 =𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟× 100% =

7460

9375× 100% = 79.6%

250𝐼𝐴 = 𝐼𝐴2 × 0.5 + 8452


Example: A 6-pole, 500V, wave connected shunt motor has 1200 armature

conductors and useful flux/pole of 20mWb. The armature and field resistances

are 0.5Ω and 250Ω respectively. What will be the speed and torque developed

by the motor when it draws 20A from the supply mains? If magnetic and

mechanical losses amount to 900W, find (i) output in kW (ii) useful torque

(Tsh) (iii) efficiency (𝜂𝑐) at this load.


𝐼𝐹 =𝑉𝑇

𝑅𝐹=

500

250= 2𝐴

𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 20 − 2 = 18𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 500 − 18 × 0.5 = 491𝑉

𝐸𝐴 = 491 =𝜙𝑍𝑁

60×

𝑃

𝐴=

0.02 × 1200 × 𝑁

60×

6

2 → 𝑁 = 410𝑟𝑝𝑚


𝑁= 9.55

491 × 18

410= 205.9N. m

(i) output in kW

Field Cu loss = 𝑉𝑇𝐼𝐹 = 500 × 2 = 1000W

Armature Cu loss = 𝐼𝐴2𝑅𝐴 = 182 × 0.5 = 162W

Iron and friction losses=900W



Total losses=armature Cu loss + field Cu loss + iron and friction loss

Total losses=162+1000+900=2062W

Motor input power=VTIL=500×20=10000W

Motor output power=motor input power-motor losses=10000-2062=7938W


𝑁= 9.55

7938

410= 184.9N. m

(ii) useful torque (Tsh)

(iii) Efficiency (𝜂𝑐) at this load

𝜂𝑐 =𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟× 100% =

7938

10000× 100% = 79.38%


Characteristics and Applications of DC Motors

Type of Motor Characteristics Applications

Shunt Motor

Approximately constant

speed

Adjustable speed

Medium starting torque (up

to 1.5 F.L. torque)

For driving constant speed line

shafting

Lathes

Centrifugal pumps

Machine tools

Blowers and fans

Reciprocating pumps

Series Motor

Variable speed

Adjustable variable speed

High starting torque

For traction work i.e. Electric

locomotives

Rapid transit systems

Trolley, Cars etc

Cranes and hoists

conveyers

Cumulative

Compound

Motor

Variable speed

Adjustable variable speed

High starting torque

For intermittent high torque loads

For shears and punches

Elevators

Conveyers

Heavy planers

Rolling mills, Ice machine, printing

presses, Air compressors

electric machines i - philadelphia.edu.jop for simplex lap-winding ... example: a 4-pole, 32...

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