ELEC 3105 Basic EM and Power Engineering

Rotating DC MotorPART 2 Electrical

Motor / Generator Action

BrVemf

2

Equivalent circuit

emfVbat

V

R

terminalv

Expression of Vemf

Loop

Slide extracted from linear motor and modified for loop motor.

I

emfbatVIRV

BrRrB

Vbat

22

rBI2

Motor / Generator Action

BrRrB

Vbat 22

Linear relation betweenspeed and torque

rBVbat

loadno 2

RVrB bat2

0

Current flows in a direction to charge the battery.

Motor

Slide extracted from linear motor and modified for loop motor.

Stall torque

GeneratorLink

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

E = motor voltageRa = armature resistanceLa = armature inductanceV = Applied motor voltageIa = armature current

= magnetic fluxRf = field resistanceLf = field inductanceVf = Field voltageIf = field current

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

In steady state operation: FIELD SIDE

𝐼 𝑓=𝑉 𝑓

𝑅𝑓Φ=

2𝑛𝐼 𝑓𝔑reluctance

Magnetic circuit

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

In steady state operation: ARMATURE SIDE

𝐸=𝑘𝐸Φ𝜔𝑚

Motor constant

Back emf

Motor / Generator Action

BrVemf

2

Equivalent circuit

emfVbat

V

R

terminalv

Expression of Vemf

Loop

Slide extracted from linear motor and modified for loop motor.

I

emfbatVIRV

BrRrB

Vbat

22

rBI2

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

In steady state operation: ARMATURE SIDE

𝐸=𝑘𝐸Φ𝜔𝑚

Motor constant

Back emf

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

In steady state operation: ARMATURE SIDE

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼𝑎

Motor constant

Developed torque

𝑘𝐸=𝑘𝑇 Same motor constant in emf and developed torque

Motor / Generator Action

BrVemf

2

Equivalent circuit

emfVbat

V

R

terminalv

Expression of Vemf

Loop

Slide extracted from linear motor and modified for loop motor.

I

emfbatVIRV

BrRrB

Vbat

22

rBI2

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

In steady state operation: ARMATURE SIDE

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼 𝑎

Motor constant

Developed torque

𝑘𝐸=𝑘𝑇 Same motor constant in emf and developed torque

𝐸=𝑘𝐸Φ𝜔𝑚

Motor constant

Back emf

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

Power flow: ARMATURE SIDE; Conservation of energy

𝑉=𝐼𝑎𝑅𝑎+𝐸KVL

POWER 𝑉 𝐼 𝑎= 𝐼𝑎2 𝑅𝑎+𝐼 𝑎𝐸

Power in

Armature copper loss

Power developed

Motor / Generator Action

BrVemf

2

Equivalent circuit

emfVbat

V

R

terminalv

Expression of Vemf

Loop

Slide extracted from linear motor and modified for loop motor.

I

emfbatVIRV

BrRrB

Vbat

22

rBI2

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

Power flow: ARMATURE SIDE; Conservation of energy

𝑃𝑑𝑒𝑣= 𝐼𝑎𝐸=𝜔𝑚𝑇𝑑𝑒𝑣

Power developed

𝑃 𝑙𝑜𝑠𝑠𝑃 𝑖𝑛

𝑃𝑑𝑒𝑣

Electrical Mechanicalcopper

𝑉 𝐼 𝑎= 𝐼𝑎2 𝑅𝑎+𝐼 𝑎𝐸

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

Power flow: ARMATURE SIDE 𝑃𝑜𝑢𝑡=𝑃𝑑𝑒𝑣−𝑃𝑟𝑜𝑡

𝑃 𝑙𝑜𝑠𝑠𝑃 𝑖𝑛

𝑃𝑑𝑒𝑣

Electrical Mechanical

𝑃𝑟𝑜𝑡

𝑃𝑜𝑢𝑡

Rotational lossCopper

Electrical Equivalent

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

Motor sequence 𝑃𝑜𝑢𝑡=𝑃𝑑𝑒𝑣−𝑃𝑟𝑜𝑡

𝜔𝑚𝑉 𝑇 𝑑𝑒𝑣

Speed of rotation limiting loop

𝐸𝐼𝑎 𝑃𝑑𝑒𝑣

Shunt Connected Field

V

aI𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣𝐿 𝑓

𝑅 𝑓R

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼𝑎

𝑉=𝐼𝑎𝑅𝑎+𝐸

𝐸=𝑘𝐸Φ𝜔𝑚

𝑇 𝑑𝑒𝑣=𝑘Φ𝑉𝑅𝑎

− 𝑘2Φ2

𝑅𝑎𝜔𝑚

Developed torque Rotation rate

Shunt Connected Field

V

aI𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣𝐿 𝑓

𝑅 𝑓R

𝑇 𝑑𝑒𝑣=𝑘Φ𝑉𝑅𝑎

− 𝑘2Φ2

𝑅𝑎𝜔𝑚

Similar type of graph

Shunt Connected Field

V

aI𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣𝐿 𝑓

𝑅 𝑓R

𝑇 𝑑𝑒𝑣=𝑘Φ𝑉𝑅𝑎

− 𝑘2Φ2

𝑅𝑎𝜔𝑚

𝑇 𝑑𝑒𝑣

𝜔𝑚

𝐾Φ𝑉𝑅𝑎

𝑉𝑘Φ

MotorGenerator

Force motor to spin backwards

Generator

Force motor to spin to fast

Series Connected Field

V

aI𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓𝑅 𝑓

Universal motor design: works for D.C. and for A.C.

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼 𝑎

𝑉=𝐼 𝑎(𝑅¿¿𝑎+𝑅𝑓 )+𝐸 ¿

𝐸=𝑘𝐸Φ𝜔𝑚

𝑇 𝑑𝑒𝑣=𝑘′ 𝐼𝑎2 Since Φ∝ 𝐼𝑎

Series Connected Field

V

aI𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓𝑅 𝑓

Universal motor design: works for D.C. and for A.C.

𝑇 𝑑𝑒𝑣=𝑘′ [ 𝑉𝑅𝑎+𝑅 𝑓+𝑘′𝜔𝑚 ]

2

𝑇 𝑑𝑒𝑣

𝜔𝑚

1𝜔𝑚

2

Maximum Power Transfer

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓𝑉=𝐼 𝑎𝑅𝑎+𝐸

Power developed in the motor 𝑃𝑑𝑒𝑣=𝐸 𝐼𝑎

𝑃𝑑𝑒𝑣=𝐸 (𝑉 −𝐸 )/𝑅𝑎

𝑃𝑑𝑒𝑣=𝐸𝑉 −𝐸2

𝑅𝑎

Find maximum with respect to the motor voltage

Maximum Power Transfer

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓𝑉=𝐼 𝑎𝑅𝑎+𝐸

For extremes of a function, take derivatives and set to zero

𝑃𝑑𝑒𝑣=𝑚𝑎𝑥𝑖𝑚𝑢𝑚 h𝑤 𝑒𝑛𝐸=𝑉2

𝑃𝑑𝑒𝑣 𝑚𝑎𝑥 = 𝑣2

4 𝑅𝑎

Calculation example

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

A 120 volt dc motor has an armature resistance of 0.70 Ω. At no-load, it requires 1.1 A armature current and runs at 1000 rpm. Find the output power and torque at 952 rpm output speed. Assume constant flux.

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼𝑎𝑉=𝐼𝑎𝑅𝑎+𝐸

𝐸=𝑘𝐸Φ𝜔𝑚

Solution provided in class

Calculation example

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

A permanent magnet dc motor has the following information: 50 hp, 200 V, 200 A, 1200 rpm and armature resistance of 0.05 Ω. Determine the output power if the voltage is lowered to 150 V and the current is 200 A. Assume rotational losses are proportional to speed. Determine the rotational loss, armature resistance, no-load rpm, machine constant, efficiency?

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼𝑎𝑉=𝐼𝑎𝑅𝑎+𝐸

𝐸=𝑘𝐸Φ𝜔𝑚

Solution provided in class

Calculation example

V

fV

aI 𝑅𝑎𝐿𝑎

E

𝜔𝑚 ,𝑇 𝑑𝑒𝑣

𝐿 𝑓

𝑅 𝑓

An 80 V dc motor has constant field flux, separately excited, and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 A and the no-load current is 0.5 A. Assume constant rotational losses.

𝑇 𝑑𝑒𝑣=𝑘𝑇Φ 𝐼𝑎𝑉=𝐼𝑎𝑅𝑎+𝐸

𝐸=𝑘𝐸Φ𝜔𝑚

Solution provided in class