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abet_solution

# abet_solution - EE321 Spring 2004 ABET Exam Solution...

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This is in A, or At Ni 210.085 = Ni Φ max R total := Φ max B max A := B max 1.1 := R total R center R edge + R center_air + R edge_air + := R edge_air R center_air := R center_air g µ 0 A := R edge R center := R center d s µ 0 µ r A := A π r c 2 := Let's begin by computing the cross section area (center post or outside) Problem 1 - Part B. t e 1.623 10 3 × = t e r c w s 2 r c 2 2 r c w s + w s 2 + 1 2 + := π r c 2 π r c w s + t e + ( ) 2 r c w s + ( ) 2 = In order to achieve the same areas Problem 1 - Part A. µ r 1000 := g .1mm := w s 2cm := µ 0 4 π 10 7 := d s 2cm := r c 1cm := mm 0.001 := cm 0.01 := Problem 1 EE321 Spring 2004 ABET Exam Solution

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Problem 1, Part C. The volume of copper will be: V cu d s π r c w s + ( ) 2 r c 2 pf = The area of the copper will be A cu d s w s pf N = The length of the copper will be L cu V cu A cu = L cu d s π r c w s + ( ) 2 r c 2 pf d s w s pf N = L cu π 2 r c w s + ( ) N = The resistance of the copper will be R cu ρ L cu A cu = R cu ρ π 2 r c w s + ( ) N d s w s pf N = R cu ρ π 2 r c w s + ( ) N 2 d s w s pf =
P loss r a T estar L af i f 2 r f i f 2 + = So the loss becomes i a T estar L af i f = To achieve the desired torque P loss r a i a 2 r f i f 2 + = Now the loss is given by T e L af i a i f = We know Problem 3 T e 2 x 4 3 λ as 3 2 λ as λ bs + 7 2 λ bs + = And so W f 4 x 2 + ( ) 4 3 λ as 3 2 λ as λ bs + 7 2 λ bs + = Thus W f2 4 x 2 + ( ) 7 2 λ bs 2 2 λ as λ bs + = Bring up the second flux linkage W f1 4 x 2 + ( ) 4 3 λ as 3 =

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