lecture 9 fundamentals of ... - biomechatronics

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METE 3100UActuators and Power Electronics

Lecture 9Fundamentals of Electromechanical

Energy Conversion

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Outline of Lecture 9

By the end of today’s lecture, you should be able to

• Calculate the stored electromagnetic energy

• Understand the principles of energy conversion

• Model and analyse an electromagnet

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Applications

DC motors convert electrical energy into mechanical energy and vice-versa.How is the energy converted?

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Applications

The magnetic levitation train uses two sets of magnets, one set to repel andpush the train up off the track. How can them be used to propel the train?

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Applications

Electromagnets and solenoids are widely used in mechatronic systems. Howcan we calculate the force deployed by these actuators?

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Energy conversion process

Principle of conservation of energy

We = Wm + Wf + Wloss (1)

→ We : electrical energy

→ Wm: mechanical energy

→ Wf : field stored energy

→ Wloss : all energy losses→ Electrical: Joule’s loss i2R→ Mechanical: Friction b→ Magnetic: Eddy currents

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Energy conversion process

Neglecting losses, the instantaneous energy change during a time interval dt is

dWe = dWm + dWf + Wloss (2)

If g = 0, then Wm = 0 and thus

dWe = dWf (3)

The electric energy is

dWe = eidt (4)

What is the field energy?

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Energy conversion process

The differential field energy is

dWf = idλ (5)

Integrating both sides yields

Wf =ˆ λ

0idλ (6)

where λ is the coil flux linkage defined by

λ =ˆ

edt → e = dλdt (7)

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Energy conversion process

The electromotive force relates to the magnetic field intensity H as (Lecture 2):

Ni = Hc`c + Hg`g → i = Hc`c + Hg`g

N (8)

Thus

Wf =ˆ λ

0

Hc`c + Hg`g

N dλ (9)

(` is the the length of a magnetic path)

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Field energyAccording to Faraday’s law (Lecture 2):

Φ =ˆ

A

~B · d~A (10)

A is the cross-sectional area. For a coil with N turns and no leakage:

λ = NΦ = NBA→ dλ = NAdB (11)

which yields:

Wf =ˆ

Hc`c + Hg`g

N NAdB (12)

If the flux density B = φ/A is constant, for the air gap we have

Hg = Bµ0

(13)

Wf =ˆ

Hc A`c︸︷︷︸ dB +ˆ

Bµ0

A`g︸︷︷︸ dB (14)

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Field energy

Wf =ˆ

Hc A`c︸︷︷︸Vc

dB +ˆ

Bµ0

A`g︸︷︷︸Vg

dB

Wf =(ˆ

HcdB)

Vc +(

B2

21µ0

)Vg

Wf = Wfc + Wfg

→ wfc =´

HcdBc is the energy density in the magnetic core

→ wfg = B2

2µ0is the energy density in the air gap

→ Vg , Vc , are the volume of the air gap and magnetic core

→ Wfc is the energy in the magnetic core

→ Wfg is the energy store in the air gap

Typically: Wfg >> Wfc and thus

Wf ≈Wfg (15)

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Energy and coenergy

Wf =ˆ λ

0idλ, Wf =

ˆ i

0λdi (16)

For a linear system

W ′f = Wf (17)

→ Wf if the field energy

→ W ′f if the field coenergy

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Mechanical force

The core moves from x = x1 (point a) to x = x2 (point b), with g2 < g1

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Mechanical forceCase 1: The movement occurs quickly. What happens to the current?

If λ is constant, the work done is a decrease in the field energy

dWm = dWf (18)

If fm is the mechanical force causing a displacement dx

fmdx = dWf

fm = ∂Wf (λ, x)∂x

∣∣∣∣λ=cte

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Mechanical forceCase 2: The movement occurs slowly. What happens to the current?

If i is constant, the work done is an decrease in the field co-energy

dWm = dW ′f (19)

If fm is the mechanical force causing a displacement dx

fmdx = dW ′f

fm = ∂W ′f (i , x)∂x

∣∣∣∣i=cte

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Experiment - Hysteresis

An external magnetic field causes the atomic dipoles to align themselves with it

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Exercise 38

The magnetic core of the actuator shown is made of cast steel with a magneticpermeability of µ = 1.5× 10−3 Hm−1. The coil has 250 turns and the coilresistance is 5 Ω. For a fixed air gap of 5 mm, a DC supply is connected to thecoil to produce a flux density of 1 Tesla in the air gap.

(a) Calculate the required DC voltage

(b) Calculate the store field energy

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Exercise 38 - continued

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Exercise 38 - continued

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Exercise 38 - continued

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Exercise 39

The relationship between the magnetic flux linkage λ and the current i of anelectromagnet is given by

i =(

λ

0.09g)2

which is valid for the limits 0 < i ≤ 4 A and 3 < g < 10 cm. For a current of 3Amps and air gap length of g = 5 cm, find the mechanical force on the movingpart using the energy and co-energy of the field.

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Exercise 39 - continued

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Exercise 39 - continued

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Exercise 40

In a translational motion actuator, the λ− i relationship is given by

i = λ32 + 2.5λ(x − 1)2 (20)

for 0 < x < 1 m, where i is the current in the coil of the actuator. Determinethe force on the moving part at x = 0.6 m.

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Exercise 40 - continued

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Exercise 40 - continued

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Exercise 41

In a translational motion actuator, the λ− i relationship is given by

λ = 1.2 i 12

g (21)

where g is the air gap length. For a current i = 2 A and when g = 10 cm,determine the mechanical force on the moving part using

(a) The stored field energy

(b) The stored field coenergy

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Exercise 41 - continued

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Exercise 41 - continued

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Next class...

• Mechanical force in electromagnetic systems

Additional supporting materials for Lecture 9:

What is coenergy?: https://youtu.be/dDTnoQTeA24

Example of an induction motor: https://youtu.be/FHgw_l-Z_s0

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