DC motor lifting a mass — Conversion de puissance 3 : Conversion électro-magnéto-mécanique

A DC machine of coupling constant

Φ_{0}

, of internal resistance

𝑟

and of inductance

𝐿

is used to lift a mass

𝑀

. The axis of the rotor is connected to a pulley of radius

𝑎

. On the pulley is fixed a string on which a mass is attached.

A friction couple

Γ_{𝑓} = - 𝑓 Ω

apply on the rotor. The moment of inertia of the system {rotor + pulley} is

𝐽

. Hysteresis and Foucault currents are neglected.

1/ Ascertain the current

𝑖

and the voltage

𝐸_{0}

delivered by the generator as a function of

𝑀

𝑔

𝑎

Φ_{0}

𝑓

𝑟

𝑟_{0}

Coup de pouce 1

Écrire une équation électrique et une équation mécanique et utiliser les relations entre grandeurs mécaniques et électriques pour une machine à courant continu.

Coup de pouce 2

Pour l’équation électrique, écrire la loi des mailles dans le circuit électrique équivalent de l’induit.

Coup de pouce 3

Pour l’équation mécanique, écrire le théorème du moment cinétique au système masse + câble + poulie + rotor.

Corrigé

Since the mass is lifted at a constant speed, every values are constant over time. The inductance $𝐿$ will have no influence and is not represented on the electrical schematic.

The electrical equation is obtained by applying the mesh law to the electrical circuit:

𝐸_{0} = 𝐸 + 𝑟 𝑖 + 𝑟_{0} 𝑖

Moreover $𝐸 = Φ_{0} Ω$ .

The pulley of radius $𝑎$ is linked to the rotor, so $𝑣_{0} = 𝑎 Ω$ .

The mechanical equation is obtained by applying the theorem of angular momentum to the system {pulley + rotor}:

0 = 𝐽 \frac{d Ω}{d 𝑡} = Φ_{0} 𝑖 + - 𝑓 Ω - 𝑀 𝑔 𝑎

From these equations, we obtain:

𝑖 = \frac{𝑀 𝑔 𝑎 Φ_{0} + 𝑓 𝐸_{0}}{𝑓 (𝑟_{0} + 𝑟) + Φ_{0}^{2}}

𝐸_{0} = \frac{(𝑟 + 𝑟_{0}) (𝑀 𝑔 𝑎 Φ_{0} + 𝑓 𝐸_{0})}{𝑓 (𝑟_{0} + 𝑟) + Φ_{0}^{2}} + \frac{Φ_{0} 𝑣_{0}}{𝑎}

DC motor lifting a mass ★★