(a) A three-phase, 5 kW, 440 V, 6 pole, star connected synchronous motor having negligible stator resistance and synchronous…

(a) Synchronous Motor: Torque Angle, Pull-out Torque, Half-Load Operation Assume supply frequency f = 50 Hz and neglect rotational losses. Per-phase voltage: Vph = 440/√3 = 254.03 V. At full load, P = 5000 W, cosφ = 0.8, sinφ = 0.6.

Armature current: I = P/(√3 VL cosφ) = 5000/(√3 × 440 × 0.8) = 8.201 A.

For a synchronous motor with Ra = 0: Vph = Ef + jXsIa, so Ef = Vph − jXsIa. Since Ia lags, Ia = I(0.8 − j0.6). Thus jXsIa = j × 6 × 8.201(0.8 − j0.6) = 29.52 + j39.37 V. Hence Ef = (254.03 − 29.52) − j39.37 = 224.51 − j39.37 V.

|Ef| = √(224.51² + 39.37²) = 227.94 V.

(i) Torque angle: δ = arctan(39.37/224.51) = 9.94°. The excitation emf Ef lags Vph by δ = 9.94°.

(ii) Pull-out occurs at δ = 90°: Pmax = 3VphEf/Xs = 3 × 254.03 × 227.94/6 = 28.95 kW. Synchronous speed: Ns = 120f/P = 120 × 50/6 = 1000 rpm. ωs = 2πNs/60 = 104.72 rad/s. Pull-out torque: Tpo = 28951.6/104.72 = 276.5 N m.

(iii) At half rated torque, P = 2500 W, assuming Ef remains unchanged. P = 3VphEf sinδh/Xs sinδh = 2500 × 6/(3 × 254.03 × 227.94) = 0.08635. δh = 4.95°, cosδh = 0.9963.

Armature current components: Ia = (Ef sinδh/Xs) − j(Vph − Ef cosδh)/Xs = 3.28 − j4.49 A.

|Ia| = √(3.28² + 4.49²) = 5.56 A. Power factor angle φh = arctan(4.49/3.28) = 53.86°. So pf = cos53.86° = 0.590 lagging.

(b) PDF of Z = 3X + 4Y (b)(i) For fY(y) = A e^(−3y), 0 ≤ y < ∞: ∫₀^∞ A e^(−3y) dy = A/3 = 1, hence A = 3.

(ii) Let U = 3X and V = 4Y. Since X is uniform on [−2,2], U is uniform on [−6,6]: fU(u) = 1/12, −6 ≤ u ≤ 6. Since fY(y) = 3e^(−3y), for V = 4Y: fV(v) = (3/4)e^(−3v/4), v ≥ 0.

For Z = U + V, use convolution: fZ(z) = ∫ fU(u) fV(z − u) du. The integrand is nonzero for −6 ≤ u ≤ 6 and z − u ≥ 0, i.e. u ≤ z.

Thus:

• For z < −6, fZ(z) = 0.
• For −6 ≤ z ≤ 6, fZ(z) = ∫₋₆^z (1/12)(3/4)e^(−3(z−u)/4) du = (1/12)(1 − e^(−3(z+6)/4)).
• For z ≥ 6, fZ(z) = ∫₋₆^6 (1/12)(3/4)e^(−3(z−u)/4) du = (1/12)(e^(−3(z−6)/4) − e^(−3(z+6)/4)).

Therefore, fZ(z) = 0, z < −6 fZ(z) = (1/12)(1 − e^(−3(z+6)/4)), −6 ≤ z ≤ 6 fZ(z) = (1/12)(e^(−3(z−6)/4) − e^(−3(z+6)/4)), z ≥ 6.

(c) Stokes Theorem Verification For cylindrical coordinates, Hρ = 2ρz, Hφ = 3z sinφ, Hz = −4ρ cosφ.

Curl H: (∇×H)ρ = (1/ρ)∂Hz/∂φ − ∂Hφ/∂z = sinφ. (∇×H)φ = ∂Hρ/∂z − ∂Hz/∂ρ = 2ρ + 4cosφ. (∇×H)z = (1/ρ)[∂(ρHφ)/∂ρ − ∂Hρ/∂φ] = 3z sinφ/ρ.

On z = 1, taking normal as az, so dS = ρ dρ dφ az. Surface integral: ∫(∇×H)·dS = ∫₀^(π/4) ∫₀² (3 sinφ/ρ) ρ dρ dφ = 6∫₀^(π/4) sinφ dφ = 6(1 − 1/√2) = 6 − 3√2 A = 1.757 A.

Line integral, taking positive normal az: C1: φ = 0, ρ = 0 → 2. ∫ H·dl = ∫₀² 2ρ dρ = 4 A. C2: ρ = 2, φ = 0 → π/4. ∫ H·dl = ∫₀^(π/4) 6 sinφ dφ = 6 − 3√2 A. C3: φ = π/4, ρ = 2 → 0. ∫ H·dl = ∫₂⁰ 2ρ dρ = −4 A. Contribution at ρ = 0 is zero.

Total: ∮H·dl = 4 + (6 − 3√2) − 4 = 6 − 3√2 A = 1.757 A.

Both sides of Stokes’ theorem are equal: 6 − 3√2 A.