lecture 9 fundamentals of ... - biomechatronics
TRANSCRIPT
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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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