0195136047.pdf

532 ELECTROMECHANICS

(c) (1) The excitations are direct currentsIsandIr. For the given conditions ofωs=ωr=

ωm=0 andα=0,

Te=−LIsIrsinθ 0

which is a constant. Hence,

(Te)av=−LIsIrsinθ 0

The machine operates as adc rotary actuator,developing a constant torque against

any displacementθ 0 produced by an external torque applied to the rotor shaft.

(2) Withωs=ωr, both excitations are alternating currents of the same frequency. For

the conditionsωs=ωrandωm=0,

Te=−

LIsIr

4

[sin( 2 ωst+α+θ 0 )+sin(− 2 ωst−α+θ 0 )+sin(−α+θ 0 )

+sin(α+θ 0 )]

The machine operates as anac rotary actuator,and the developed torque is fluctu-

ating. The average value of the torque is

(Te)av=−

LIsIr

2

sinθ 0 cosα

Note thatαbecomes zero if the two windings are connected in series, in which case

cosαbecomes unity.

(3) Withωr=0, the rotor excitation is a direct currentIr. For the conditionsωr=

0 ,ωs=ωm, andα=0,

Te=−

LIsIr

4

[sin( 2 ωst+θ 0 )+sinθ 0 +sin( 2 ωst+θ 0 )+sinθ 0 ]

or

Te=−

LIsIr

2

[sin( 2 ωst+θ 0 )+sinθ 0 )]

The device operates as an idealizedsingle-phase synchronous machine,and the

instantaneous torque is pulsating. The average value of the torque is

(Te)av=−

LIsIr

2

sinθ 0

since the average value of the double-frequency sine term is zero. If the machine

is brought up tosynchronousspeed (ωm=ωs), an average unidirectional torque is

established. Continuous energy conversion takes place at synchronous speed. Note

that the machine is not self-starting, since an average unidirectional torque is not

developed atωm=0 with the specified electrical excitations.

(4) Withωm=ωs−ωr, the instantaneous torque is given by

Te=−

LIsIr

4

[sin( 2 ωst+α+θ 0 )+sin(− 2 ωrt−α+θ 0 )

+sin( 2 ωst− 2 ωrt−α+θ 0 )+sin(α+θ 0 )]