Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric Machines Problem Set 1 Solutions September 6, 2013 Problem 1: If the coil is thin, Inductance is found easily to be: L = µ0 N 2 πR2 h If the coil is not thin, the magnetic ﬁeld inside will be as shown in Figure 1. H z w Ni h R Figure 1: Field Intensity vs. Radius Flux linked by a coil ﬁlament at radius r is then: Ni π (R − w)2 + φ(r) = µ0 h l r µ0 R−w � R − r′ 2πr ′ dr ′ w � In principle, this could be evaluated and then averaged over the coil radius to get inductance. However, it appears (to this calculus duﬀer) to re-parameterize the problem by assigning x = R − r, the distance from the outside of the coil. Then: Hz = = Ni x h w Ni h xw In which case, ﬂux can be found for the region of the coil: µ0 N i φ(x) = π(R − w)2 + h [ l w x x′ ′ 2π R − x dx w ( ′) After some manipulation, this becomes: φ(x) = µ0 πN i 1 Rx2 2 x3 + R2 − Rw + w2 − h 3 w 3w 1 ] then inductance is found by averaging φ over the extent of the coil, and with a few steps of grunge left out: L = = = 1 wN φ(x)dx i 0 w " # l µ0 πN 2 W 1 2 Rx2 2 x3 2 R − Rw + w − + dx 3 w 3w hw 0 l " 4w 1 µ0 πN 2 R2 1− + h 3R 2 � w R � 2 # Problem 2: With a steady DC current, ﬁeld along the centerline is simply: H0 = Ni h In the event of a transient, Faraday’s Law can be used: - · d-ℓ = − d E dt ll - · -nda B Taking this integral within the conductive cylinder at radius Rc , this is: 2πRc d (Hi − H0 ) + πRc2 µ0 Hi = 0 σ dt where Hi is the ﬁeld inside and H0 = Ni h is the ﬁeld outside the shell. The shell time constant is identiﬁed: Ts = µ0 σts Rc 2 and the diﬀerential equation for ﬁeld is: Ts Driven by a step of Ho = d Hi + Hi = Ho dt Ni h u(t), Hi = t � Ni � 1 − e− Ts h This is shown in Figure 2 To ﬁnd the response of the ﬁeld inside the shell, see that the symmetrical triangle wave can be expressed as the sum of two waveforms, each consisting of a ramp and a constant value, as shown in Figure 3. This observation allows us to calculate the magnetic ﬁeld once and then add the same solution, inverted in sign and delayed. First, we solve the ﬁeld problem for the ﬁrst excitation, consisting of a ramp of length T and NI I a slopt of N hT , followed by a constant value of h . During the ramp, since the ramp is the integral of the step, the response is the integral of the step response, and for the period after the end of the ramp, the solution follows a simple ﬁrst-order response to the steady state ﬁeld: 2 4 10 Problem 1.2, part 2a x 10 9 8 7 A/m 6 5 4 3 2 1 0 0 0.005 0.01 0.015 0.02 0.025 0.03 time, sec 0.035 0.04 0.045 0.05 Figure 2: Step Response Figure 3: Addition of two waves to form the triangle H(t) = H0 [ [ t τ − � τ T H(t) = H0 1 − τ T � t 1 − e− τ � T �] 1 − e− τ e− 0T τ It is straightforward to combine this solution with its inverted and shifted version, and now there are three time regions: H(t) = H0 [ H(t) = H0 1 − H(t) = τ T � � T [ t τ − 1 − e− τ e− � τ T ] t−T τ � t 1 − e− τ − H0 � T 2 H0 Tτ 1 − e− τ [ �] t−T T e− 0