hw7_solution

hw7_solution - EE321 Spring 2008, HW #7 Problem 30 ra :=...

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EE321 Spring 2008, HW #7 Problem 30 r a 0.2 := L af 200 10 3 := r f 4 := T estar 2 := Now, we want to control the field current, so as to minimize lass P loss r a i a 2 r f i f 2 + = T e L af i a i f = So i a T e L af i f = So P loss r a T e L af i f 2 r f i f 2 + = Taking the derivative with respect to field current and setting it to zero we find the extrema of power loss (the minimum) 02 r a T e L af 2 i f 3 2r f i f + = i f r a r f 0.25 T e L af 0.5 = Our current command are thus i f r a r f 0.25 T estar L af 0.5 := i f 1.495 = i a T estar i f L af := i a 6.687 =
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i f P loss r a i a 2 r f i f 2 + := P loss 17.889 = Problem 31 The steady-state impedance looking in the machine is Zr a r f r L af + () j ω e L aa L ff + + = The rms magnitude of the terminal current (armature or field) may be expressed I t V s Z = Recall the torque may be expressed T e L af i t i t = where i sub t is the terminal current (armature or field).
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hw7_solution - EE321 Spring 2008, HW #7 Problem 30 ra :=...

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