A 3-phase, 460 V, 6-pole, 60 Hz cylindrical rotor synchronous motor has a synchronous reactance of 2.5 Ω and negligible armature resistance. The load torque, proportional to the square of the speed, is 398 N.m at 1200 rpm. Unity power factor is maintained by excitation control. Keeping the v/f constant, the frequency is reduced to 36 Hz. The torque angle δ is

Concept:

T ∝ sinδ

\(P = \frac{{{V_s}{V_R}}}{X}sin\delta\)

\(N = \frac{{120f}}{P}\)

Calculation:

Given that N₁ = 1200 rpm

f₁ = 60 Hz

T₁ = 398 N-m

At f₂ = 36 Hz

\({N_2} = \frac{{120f}}{P} = \frac{{120 \times 36}}{6} = 720\;rpm\)

T ∝ N²

\(\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{N_1}}}{{{N_2}}}} \right)^2}\)

\(\Rightarrow \frac{{398}}{{{T_2}}} = {\left( {\frac{{1200}}{{720}}} \right)^2}\)

⇒ T₂ = 143.28 N-m

Now,

\({P_1} = \frac{{2\pi {N_1}}}{{60}} \times {T_1} = 50.014\;KW\)

\({P_1} = \frac{{{V_S}{V_R}}}{X}sin{\delta _1} = \frac{{{V^2}}}{X}sin{\delta _1}\)

\(\Rightarrow sin{\delta _1} = \frac{{2.5 \times 50.014 \times {{10}^3}}}{{{{460}^2}}}\)

⇒ sinδ₁ = 0.59

\(\frac{{{T_1}}}{{{T_2}}} = \frac{{sin{\delta _1}\;}}{{sin{\delta _2}}}\)

\(\Rightarrow \frac{{398}}{{143.28}} = \frac{{0.59}}{{sin{\delta _2}}}\)

⇒ sinδ₂ = 0.2127

⇒ δ₂ = 12.5°