# Document (5) ```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 &times; 0.75 = 220 V
∴220 = 25 &times; 10−3&times; 782 &times;N &times; 0.75 = 220 V
∴220 = 0.159 ΦZIa (P/A) = 0.159 &times; 25 &times; 10−3&times; 782 &times; 40 &times; (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 &times; 50 = 12,500 W
Ta = 9.55 EbIa/N = 9.55 &times; 250 &times; 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&times; 25 &times; 10−3&times; 800 &times; 45 (4/2) = 286.2 N-m
Eb=V− Ia Ra = 220 − 45 &times; 0.6 = 193 V
Now,Eb=ΦZN (P/A) or 193 = 25 &times; 10−3&times; 800 &times;Nπ&times; (4/2)
∴N= 4.825 r.p.s.
Also,2πN Tsh = output or 2π&times; 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 &times; 0.2) = 210 V
Eb2 = 220 − (100 &times; 0.2) = 200 V
∴N2/500 = 200/210
∴N2 = 476 r.p.m.
```