Electrical Machines - Section 3
Electromagnetic Force and Torque
This section explains how stored magnetic-field energy is converted into mechanical energy in an electromechanical system. When a current-carrying coil interacts with a magnetic field, an electromagnetic force or torque can be produced and cause mechanical movement. The section develops the relationship between field energy, co-energy, mechanical movement, inductance, force, and torque. It also extends the analysis from a simple linear system to two coupled coils and rotary motion.
Q1. What is Simple Electromechanical System?
A1. A simple electromechanical system consists of an electrical system coupled to a mechanical system through a magnetic field.
When current flows through the coil, magnetic energy is stored. If the mechanical part is allowed to move, some of the energy can be converted into mechanical work.
For a slowly moving system, the electrical energy increment is: dWelec = e.i.dt
Since: e = dλ/dt
we obtain: dWelec = i.dλ
For a coil with turns: dWelec = i.N.dΦ
Since: F = N.i
then: dWelec = F.dΦ `
These relationships connect electrical input energy with magnetic-field energy and mechanical output.
Q2. What is Spring Energy in an Electromechanical System?
A2. A mechanical system may contain a spring connected to the moving part.
The energy stored in the spring is: Ws = 1/2*ks.x²
where:
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Ws = spring energy
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ks = spring constant
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x = displacement
The spring force is: Fs = ks.x
Therefore, as the electromagnetic system moves, energy can be transferred between:
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Electrical system
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Magnetic field
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Mechanical system
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Spring
Q3. What is Force Produced by the Magnetic Field?
A3. The magnetic field can produce mechanical force on a movable component.
For the simple system discussed in the chapter, the force can be obtained from the change in field energy with displacement:
F = - (dWfld / dx)
The negative sign depends on the displacement and force sign convention used in the chapter.
Using co-energy, the corresponding relationship is: F = dW'fld / dx
Thus, field energy and co-energy provide two methods for calculating electromagnetic force.
Q4. What is Force in Terms of Inductance?
A4. For a linear magnetic system: Wfld = 1/2*L.i²
Therefore, when inductance changes with position: F = - 1/2*i².dl/dx
This is an important equation in electromechanical energy conversion.
Important observation
If L depends on position X, then: dl/dx ≠ 0 and an electromagnetic force can be produced.
If inductance does not change with position, this particular mechanism does not produce force through the position-dependent inductance term.
Q5. What is Two Coupled-Coil System?
A5. Consider two magnetically coupled coils carrying currents and .
The total magnetic-field energy is: Wfld = 1/2*L11.i1²+1/2*L22.i2² + L22.i1.i2
where:
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L11= self-inductance of coil 1
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L22= self-inductance of coil 2
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L12= mutual inductance
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i1,i2= coil currents
If the inductances depend on displacement , the electromagnetic force is:
F = i1².dL11/dx+1/2*i2².dL22/dx+i1.i2.dL12/dx
Only those inductance terms that vary with position contribute to the force.
Q6. What is Conversion from Linear Motion to Rotary Motion?
A6. The same energy-conversion principle can be applied to a rotating machine.
For rotary systems: X➔θ and F➔T
where:
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X= linear displacement
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θ= angular displacement
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F= force
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T= torque
The corresponding torque equation is: T=∂Wfld/∂θ
Q7. What is Simple Rotary Electromechanical System?
A7. A simple rotary system consists of:
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Two stator coils, a and b
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One rotor coil r
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Stator coils separated by 90° electrical
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Rotor position represented by θ
If the rotor rotates at angular velocity ωr: θ = ωr.t
The mutual inductances vary with rotor position.
For the rotor-to-a coil: Lra = Lra(max).cosθ
or: Lra = Lra(max).cos(ωr.t)
For the rotor-to-b coil: Lrb = Lra(max).sinθ
or: Lrb = Lra(max).cos(ωr.t - 90°)
8. What is Energy of the Rotary System?
The total magnetic energy is composed of self-energy and mutual-energy terms.
The individual terms are:
W₁ = 1/2 Lₐₐi²ₐ
W₂ = 1/2 Lbbi²b
W₃ = 1/2 Lrri²r
The mutual terms are:
W₄ = Lrairiacosθ
W₅ = Lrairibsinθ
Therefore:
W = W₁ + W₂ + W₃ + W₄ + W₅
9. What is Torque in the Rotary System?
Torque is obtained by differentiating the magnetic-field energy with respect to angular displacement:
T=dW/dθ
For constant currents, the self-energy terms do not contribute to the torque because their inductances are constant with respect to θ
The chapter obtains:
T= Lrair [-ia sinθ +ib cosθ]
This equation shows that torque depends on:
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Rotor current
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Stator currents and
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Mutual inductance
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Rotor angular position
10. What are the Important Special Cases?
Case 1: iₐ = ib = i₅
Then:
T= Lrair [-ia sinθ +ib cosθ]
The chapter notes that the torque becomes zero at and reaches maximum at .
Case 2: iₐ = - iₛ, ib = + iₛ
Then:
T= Lrair [-ia sinθ +ib cosθ]
The maximum torque occurs at .
Important Equations of Section 3
Electrical energy
dWelec = ei dt = i dλ
Spring energy:
Wₛ = 1/2 kₛx²
Spring force:
Fₛ = kₛx
Force from field energy:
F = dWfld/dx
Force from co-energy:
F =-1/2i²dL/dx
Force using inductance:
F =-1/2i²dL/dx
Two-coil field energy:
Wfld = 1/2L₁₁i₁²+ 1/2L₂₂i₂²+ L₁₂i₁i₂
Two-coil force:
F = 1/2i₁²dL₁₁/dx+1/2i₂²dL₂₂/dc+i₁i₂dL₁₂/dx
Rotary torque
T=dw/dθ
Rotary-system torque
T = Lᵣₐiᵣ[-iₐsin sin θ + ib cos cos θ ]
10 MCQ Problems — Section 3
1. The electromagnetic force in a linear system can be obtained from:
A. F = Wfldx
B. F = -dWfld/dx
C. F = Wfld/x
D. F = dWfld dx
2. Force using magnetic co-energy is given by:
A. F =-dWfld/dx
B. F = dWfld/dx
C. F = Wfldx
D. F = Wfld/x
3. The energy stored in a spring is:
A. kx
B. kx²
C. 1/2kx²
D. 1/2kx
Answer: C
4. For a linear magnetic system, the force in terms of inductance is:
A. F =-1/2i² dL/dx
B. F = 1/2 i dL/dx
C. F = i²L
D. F = dL/dx
5. Which quantity must vary with position to produce force through the inductance mechanism?
A. Resistance only
B. Inductance
C. Frequency only
D. Temperature only
6. For rotary motion, linear displacement is replaced by:
A. Current i
B. Flux Φ
C. Angular displacement θ
D. Resistance R
7. The torque obtained from magnetic-field energy is:
A. T = dW/dθ
B. T= dW/di
C. T = Wθ
D. T = W/θ
8. In the simple rotary system, the mutual inductance between the rotor and stator coil varies as:
A. Lra = Lra(max) sin sin θ
B. Lra = Lra(max) cos cos θ
C. Lra = Lra(max) θ
D. Lra = 0
9. The two stator coils in the rotary system are separated by:
A. 30°electrical
B. 45°electrical
C. 90°electrical
D. 180°electrical
10. In a two-coupled-coil system, which terms contribute to force when their inductances vary with position?
A. Only resistance terms
B. Only current terms
C. Position-dependent inductance terms
D. Voltage terms only
Answer: 1(B), 2(B), 3(C), 4(A), 5(B), 6(C), 7(A), 8(B), 9(C), 10(C).
Calculation Problems —
Problem 1 — Force Using Inductance
A linear electromagnetic actuator has an inductance given by:
L = 0.5 + 2x
where L is in henry and x is in metre. The current is constant at 4A .
Calculate the electromagnetic force at any position x.
Use:
F =-1/2i² dL/dx
Problem 2 — Spring Force
A movable armature is connected to a spring having a spring constant of:
kₛ = 500N/m
If the armature moves by:
x = 0.02m
calculate the spring force.
Use:
Fₛ= kₛx
Problem 3 — Energy Stored in Two Coupled Coils
Two coupled coils have:
L₁₁ = 0.5H, L₂₂ = 0.2H, L₁₂ = 0.1H
and carry currents:
i₁ = 4A, i₂ = 3A
Calculate the total magnetic-field energy.
Use:
Wfld = 1/2L₁₁i₁² + 1/2L₂₂i₂²+ L₁₂i₁i₂
Problem 4 — Torque in a Rotary System
A simple rotary system has:
Lra=0.5H, ir=4A, ia=2A, ib=3A
at:
θ=30°.
Calculate the electromagnetic torque using:
T = Lrair [-ia sin sin θ + ib cos cos θ ]
Problem 5 — Force in a Two-Coil System
For a two-coil system:
i₁ = 5A, i₂ =2A
and at a particular position:
dL₁₁/dx = 0.4H/m, dL₂₂/dx = 0.2h/m, dL₁₂/dx = 0.1H/m.
Calculate the electromagnetic force.
Use:
F = 1/2i₁² dL₁₁/dx + 1/2i₂² dL₂₂/dx +i₁i₂ dL₁₂/dx