Can you explain the details of how we arrive at this specifically for the Co-Ene

Question

Can you explain the details of how we arrive at this specifically for the Co-Energy? Finding the electromagnetic torque from there is simple enough.

From the Flux Linkage equations, what portions are contributing to the resultant Co-Energy (Wc), which are being discounted, why?

The flux linkage equations for the machine shown to the right can be expressed as: A, = (Ls + Lms)Isa-Lms sin().-Lms cos@rx, rb ra ra rSa msra COS Sa Sa ra where Lms-50 mH and 7s Determine the value of the electromagnetic torque Sa 6, -36.9° ra ra rb rb Sa sa

Explanation / Answer

given that

isa=8A,ira'=-4A,irb'=3A

Lms=50mH,Lls=Llr'=1mH,angle=36.9o

angular speed is given by

We=0.5Lmsisa2+0.5Lmsira2+0.5Lmsirb2-Lms(isaira-irbisa)

by substituting all the values we can get

We=0.826

torque=(2/p)J(dW/dt)

here p=2

by substituting all the values we can get

T=0.826J

here torque depends on inertia.

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