Torque Equation of the Synchronous Motor

TORQUE EQUATION OF THE SYNCHRONOUS MOTOR

OL = Supply voltage/phase

I = Armature current

LM = Back emf at a load angle of δ

OM = Resultant voltage, E_R

E_r = IZ_s(or I X_s if R_a is negligible)

I lags/leads V by an angle ϕ and lags behind E_r, by an angle Ɵ.

Ꮎ = tan^-1 (X_s /R_a)

✓ Line NS is drawn at angle ϕ to LM

✓  LN and QS are perpendicular to NS

✓ Mechanical power developed per phase in the rotor,

P_mech =E_bI cos Ψ

Δ OMS, MS = I Z_s cos Ψ

MS = NS - NM = LQ - NM

I Z_s cos Ψ = Vcos (Ɵ - δ) - E_b cos Ɵ

 (or) I cos Ψ = V/Z_s cos (Ɵ - δ) - E_b/Z_s cos Ɵ

P_mech/phase = E_b [V/Z_s cos (Ɵ - δ) - E_b/Z_s cos Ɵ]

P_mech/phase = E_bV/Z_s cos (Ɵ - δ) - E_b²/Z_s cos Ɵ]

✓ This is the expression for the mechanical power developed in terms of load angle (α) and the internal angle of the motor for a constant voltage V & E_b.

Maximum power developed:

✓ Condition for maximum power developed can be found by differentiating the above expression with respect to load angle and then equating it is zero.

sin (Ɵ - δ) = 0 (or) Ɵ = δ

Value of maximum power,

This shows that the maximum power and hence torque depends on V and E_b that is excitation.

Maximum value of Ꮎ and hence α is 90° for all values of V and E_b, this limiting value of α is the same but maximum torque will be proportional to the maximum power developed.

If R_a is neglected, then Z_s = X_s and Ɵ = 90°.

COS Ɵ = 0

This gives the value of mechanical power developed in terms of δ - the basic variable of a synchronous machine.

P_mech(max) = E_bV/X_s when (δ = 90°)

This corresponds to the 'pull out' torque.

To determine the value of exitation or induced emf E_b to give maximum power developed possible, differentiate with respect to E_b and equal to zero.

Where

R_a → eflective resistance of the motor

Hence (P _mech)_max = V²/4R_a