HW5 sol - spring 2019 hw 5 PDF

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spring 2019 hw 5...

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1. Problem 4.1 a. With the permeability of the core assumed to be infinite, the equivalent magnetic circuit is as shown in Fig. 1, where the reluctances are defined to be g µ0 W 2 x(t) Rx (t) = µ0 W 2 Rg =

and the total reluctance seen by the magnetomotive force is R(t) = 2Rx (t)||Rg =

2gx(t) µ0 (g + 2x(t))W 2

The flux linkage resulted from the impressed magnetomotive force is λ(t) = N φ(t) = N

N i(t) 1 1 = N 2 µ0 W 2 ( + )i(t) R(t) 2x(t) g

Figure 1: Problem 4.1

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b. The voltage at the terminal can be determined by taking the time-derivative of the flux linkage: dλ(t) dt ∂λ(t) di(t) ∂λ(t) dx(t) + = ∂i(t) dt ∂x(t) dt 1 di(t) i(t) dx(t) 1 + ) − 2 = N 2 µ0 W 2 [( ] 2x (t) dt 2x(t) g dt

v(t) =

2

2. Problem 4.2 a. The inductance of the reluctance motor is given to be L(θ) = L1 + L2 cos(2θ). In order to determine the inductance values L1 and L2 , at least two rotor positions where the total inductance is known must be chosen. For the generalized case, the inductance of the magnetomotive force is L(θ) =

N2 λ(θ) = R(θ) i

where R(θ) represents the total magnetic reluctance seen by the source. When θ = 0, the reluctance is a minimum, R(0) =

2g0 µ0 Ag0

causing the inductance to be a maximum: L(0) =

N 2 µ0 Ag0 = L1 + L2 2g0

When θ = 90, the reluctance is a maximum, R(90) =

2g1 µ0 Ag1

causing the inductance to be a minimum: L(90) =

N 2 µ0 Ag1 = L1 − L2 2g1

As such: µ0 N 2 Ag0 Ag1 + ) ( g1 g0 4 µ0 N 2 Ag0 Ag1 − ) L2 = ( g1 g0 4

L1 =

3

3. Special Problem 1 a. As the permeability of the iron core tends to infinity, the total magnetic reluctance reduces to the contribution from the air-gap: R(t) =

7.96 × 104 2g = x(t) µ0 Dx(t)

b. The flux linkage resulted from the impressed magnetomotive force is λ(t) = N φ(t) = N

N i(t) x(t)i(t) = R(t) 7.96

c. The voltage at the terminal can be determined by taking the time-derivative of the flux linkage: dλ(t) dt ∂λ(t) di(t) ∂λ(t) dx(t) + = i(t)R + ∂i(t) dt ∂x(t) dt x(t) di(t) i(t) dx(t) = i(t)R + + 7.96 dt 7.96 dt

v(t) = i(t)R +

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4. Special Problem 2 [The computation may be over challenging here, but it is necessary to understand the basic concepts tested in this problem] With the permeability of the core assumed to be infinite, the equivalent magnetic circuit has the reluctances defined to be g µ0 Ag x(t) Rx (t) = µ0 Ax Rg =

and the total reluctance seen by the magnetomotive force is R(t) = Rx (t)||Rg + Rg =

R x ( t) R g + Rg Rx (t) + Rg

The flux linkage resulted from the impressed magnetomotive force is λ(t) = N φ(t) = N

N i(t) N 2 µ0 Ag (gAx + x(t)Ag ) i(t) = gAx + 2x(t)Ag R(t) g

The voltage at the terminal can be determined by taking the time-derivative of the flux linkage: v(t) dλ(t) = dt ∂λ(t) di(t) ∂λ(t) dx(t) + = ∂i(t) dt ∂x(t) dt 2 dx(t) (gAx + 2x(t)Ag )Ag − 2Ag (gAx + x(t)Ag ) N µ0 Ag (gAx + x(t)Ag ) di(t) + Ag ] i(t) [ = 2 dt dt gAx + 2x(t)Ag g (gAx + 2x(t)Ag ) gAg2Ax dx(t) N 2 µ0 Ag (gAx + x(t)Ag ) di(t) i(t) [ − = ] 2 gAx + 2x(t)Ag (gAx + 2x(t)Ag ) g dt dt

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