Electromechanical Systems

These create mechanical motion by electromagnetic force…

Magnetic Coupling

Definition

For a conductor with length $L$, and current $i$ pass through it is in a magnetic field with a flux density $B$, a force $F=BLi$ is exerted on it.

We can apply the Lorentz force which states that $F=mg=Bli$ and find directions using Fleming’s left hand rule:

• Thumb is the force, index finger is the magnetic field, middle finger is the current

See DC Machines

$f=BLi$ is valid for $\{\begin{array}{l}1-\text{Straight conductors perp to a magnetic field}\\ 2-\text{Circular conductors in a radial field}\end{array}$

Reminder: If the conductor moves $\perp$ to the plane of a field-conductor then from DC Machines we know that $V_b=vBL$

DC Motors

Some Definitions

$v_a=\text{The voltage that is input to the armature}$ and $v_b=\text{The back e.m.f generated by the motion of the armature opposing the current}$

Mechanically…
The rotor rotates with angular velocity $\omega$ from generated torque $T$ which is opposed by damping $c$. There is also an opposing torque of the load being applied $T_L$.

$f=Bl{i}_a$ and therefore… $T=fr=nBl{i}_a=\left(nBlr\right)i_a$

Where we can group the following $K_T=nBlr$ such that we have $T=K_T{i}_a$

We also have $v_b=nBlv=nBlr\omega =K_B\omega$

Using these relationships, we can draw the conclusion that $K_B=K_T$

Mechanical power is given as $P_{mech}=Fv=\left(Bli\right)v=mgv$

Armature Circuit Relationships

Definitions

Note that here $\Omega$ is the Laplace variable for $\omega$, not resistance…

We can set up the following ODE’s for an armature circuit…
$v_a=R_a{i}_a-L_a{\frac{d{i}_a}{dt}}_{}-v_b=0$
$v_a-R_a{i}_a-L_a\frac{d{i}_a}{dt}-k_b\omega =0$

$I\frac{d\omega }{dt}=K_T{i}_a-c\omega -T_L$

Our outputs in this case are $\omega \left(t\right)$ and $i_a\left(t\right)$

$$\{\begin{array}{l}{V}_a-R_a{I}_a-L_{a}s{I}_a-k_b\Omega (s)=0\\ Is\Omega (s)=K_T{I}_a(s)-c\Omega (s)-T_L(s)\end{array}$$ $$\{\begin{array}{l}{I}_a(R_a+s{L}_a)=V_b-K_b\Omega \\ \Omega (c+Is)=K_t{I}_a-T_L\end{array}$$ $$\{\begin{array}{l}{I}_a=\frac{1}{R_a+s{L}_a}(V_a-k_b\Omega )\\ \Omega =\frac{1}{c+sI}(K_T{I}_a-T_L)\end{array}$$

Where the term in front of the voltages and currents are our mechanical or electrical impedances.

We can now create a block diagram using these relationships knowing that out input is $V_a$ and that our output is $I_a$ and $\Omega$.
Using the block diagram to derive the transfer function, you can put the denominator in a nice form to see the impedances where the denominator is the characteristic equation of the system and its roots are the poles of the system. You can find the rest of the transfer functions in a similar way.

Electrical power is given as $P=V_{1}i$

---

Primitive Motor w/ a Load (Stalled)

• Stationary condition w/ a load
  • $F=mg=Bli=k_{m}i$
  • Motor force is correlated with the current
  • If the resistance is R, then $V_1=iR$
  • Total power supplied is $i^{2}R$ but total mechanical power is 0
  • $i^{2}R$ is dissipated as heat so this is lost power

Primitive Motor w/ a Load (Stalled)

• Conductor moving at a constant speed with the same current
• Mechanical power here is $Fv=Bli\cdot v$
• Total power is $V_{1}i+Bliv$ which must equal the supply power $V_2-V_1=Blv=k_{b}v=E$ (emf)
• Motor velocity is correlated with voltage
• Motor efficiency is $\eta =\frac{\left(V_2-V_1\right)i}{V_{2}i}=\frac{Blv}{V_2}=\frac{E}{V_2}$ or ${\frac{P_{mech}}{P_{elec}}}_{}$

---

Field Generation

• EM flux is proportional to current and the following principle is relevant to electric motors when the magnetic flux is generated by a field coil rather than a permanent magnet (shunt and series DC motors) $B=\frac{\Phi }{A}=\frac{{\mu }_{0}Ni}{g}$