Engineering (DC motor derive ω expression with respect time

[tcuay]

Homework Statement

I try to derive the expression of ω with respect to t,
but the problem is the armature current Ia or the flux is unknown. I don't know how to solve it.
Anyone could give me ideas? Thanks a lot

Homework Equations

τ=J(ω/t)
V_s=I_aR_a+E_a
T=K_EøI_a
E_a=K_Eøω

 

[Simon Bridge]

Either (a) you are mistaken, and there is some way to find Ia, or (b) you can make a substitution for Ia so the term no longer appears in the equation - like you do with simultaneous equations.

 

[tcuay]

Either (a) you are mistaken, and there is some way to find Ia, or (b) you can make a substitution for Ia so the term no longer appears in the equation - like you do with simultaneous equations.

Thanks for your reply.
I did in the following steps:
1.J(ω/t)=KøIa;
2. and then substitute Ia from Vs=IaRa+Ea into 1;
finally, i got:
ω=Vs/(Kø+JR_a/Køt)

But it turns out the term ø cannot be eliminated and it is unknown.
Where am i wrong?

 

[Simon Bridge]

But it turns out the term ø cannot be eliminated and it is unknown.

Really?
What physical property does "ø" represent?

τ=J(ω/t)

... isn't that supposed to be: $$\tau=J\frac{d\omega}{dt}$$...??

ω=Vs/(Kø+JRa/Køt)

recheck your algebra there...
how did you get this from what went before?

The fact they give you a bunch of initial values suggests to me you should look for a DE.

 

[tcuay]

ø is the magnetic flux.
i have double checked for it, the deriving process should be right.
and it seems reasonable if t approaches infinity
Vs=Køω quite similar to Ea=Køω

 

[Simon Bridge]

Derivation:

i have double checked for it, the deriving process should be right.

... So long as you are sure.

You don't appear to have used the fact that the armature resistance is "negligible".

What does that do to the derivation?