Example 29.7. A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on its armature. It has armature and series field resistance of 0.75 ohm. The motor takes a current of 40 A. Estimateits speed and gross torque developed if it has a flux per pole of 25 mWb. Solution. Eb=ΦZN (P/A) Now,Eb=V−IaRa = 50 − 40 × 0.75 = 220 V ∴220 = 25 × 10−3× 782 ×N × 0.75 = 220 V ∴220 = 0.159 ΦZIa (P/A) = 0.159 × 25 × 10−3× 782 × 40 × (4/2) = 249 Nm Example 29.8. A d.c. shunt machine develops an a.c. e.m.f. of 250 V at 1500 r.p.m. Find its torque and mechanical power developed for an armature current of 50 A. State the simplifying assumptions. Solution. A given d.c. machine develops the same e.m.f. in its armature conductors whether running as a generator or as a motor. Only difference is that this armature e.m.f. is known as back e.m.f. when the machine is running as a motor. Mechanical power developed in the arm = EbIa = 250 × 50 = 12,500 W Ta = 9.55 EbIa/N = 9.55 × 250 × 50/1500 = 79.6 N-m. Example 29.9. Determine developed torque and shaft torque of 220-V, 4-pole series motor with 800 conductors wave-connected supplying a load of 8.2 kW by taking 45 A from the mains. The flux per pole is 25 mWb and its armature circuit resistance is 0.6 Ω Solution. Developed torque or gross torque is the same thing as armature torque. ∴Ta= 0.159 ΦZA (P/A) = 0.159× 25 × 10−3× 800 × 45 (4/2) = 286.2 N-m Eb=V− Ia Ra = 220 − 45 × 0.6 = 193 V Now,Eb=ΦZN (P/A) or 193 = 25 × 10−3× 800 ×Nπ× (4/2) ∴N= 4.825 r.p.s. Also,2πN Tsh = output or 2π× 4.825 Tsh = 8200 ∴Tsh = 270.5 N-m Example 29.10. A 220-V d.c. shunt motor runs at 500 r.p.m. when the armature current is 50 A. Calculate the speed if the torque is doubled. Given that Ra = 0.2 Ω Solution. As seen from Art 27.7, Ta∝ ΦIa . Since Φ is constant, Ta∝ Ia ∴Ta1∝Ia1 and Ta2∝Ia2 ∴Ta2/Ta1 = Ia2/Ia1 ∴2 = Ia2/50 or Ia2 = 100 A Now, N2/N1 = Eb2/Eb1 − since Φ remains constant. Fb1 = 220 − (50 × 0.2) = 210 V Eb2 = 220 − (100 × 0.2) = 200 V ∴N2/500 = 200/210 ∴N2 = 476 r.p.m.