The Lathe Motor That Slows Under Cut
A workshop lathe is driven by a 230 V DC shunt motor. The armature resistance is 1.2 Ω and the field resistance is 230 Ω. At rated load the armature current is 10 A and the speed is 1500 rpm. During a heavy cut, the armature current rises to 18 A, the speed falls noticeably, and sparking appears at the brushes. The operator proposes to increase the field current to restore the speed. A second operator suggests that the quickest test is to disconnect the field circuit briefly and see whether the motor speeds up. The workshop has a multimeter, a clamp meter and a tachometer, but no spare motor. The supply voltage remains at 230 V throughout.
1. Calculate the field current, the back emf at rated load, and the back emf and approximate speed at 18 A. Explain why the speed falls when the load increases.
Read the worked answer
Field current I_f = V/R_f = 230/230 = 1.0 A. At rated load, back emf E_b = V − I_aR_a = 230 − (10 × 1.2) = 218 V. At 18 A, E_b = 230 − (18 × 1.2) = 208.4 V. For a shunt motor with constant flux, speed is proportional to back emf, so N₂/N₁ = 208.4/218 = 0.956, giving N₂ ≈ 1434 rpm. The speed falls because the heavier cut demands more torque. The motor responds by drawing more armature current, and the increased I_aR_a drop reduces the back emf. A lower back emf means less opposition to the supply, so current rises further until motor torque balances load torque. The speed settles at the value where the reduced back emf produces the current needed for the new torque.
2. Evaluate the operator's proposal to increase the field current to restore speed. What actually happens to speed and torque when field current is increased, and what is the correct action?
Read the worked answer
For a shunt motor, flux is approximately proportional to field current below saturation, and speed N ∝ E_b/Φ. Increasing the field current strengthens the flux, which increases torque for a given armature current but reduces speed. It cannot restore the speed lost under a heavy cut; it will slow the motor further and may increase field copper loss and saturation. The correct action is to reduce the load, that is, reduce the depth of cut or the feed rate, so the motor can return to its normal operating point. If the motor is driving a constant-torque load, field weakening is used to raise speed, but that reduces torque per ampere and can cause excessive armature current or stalling on a heavy load. Increasing the field is therefore the wrong direction for this symptom. The sparking should also be investigated: overload raises brush current density and armature reaction, so reducing the cut and checking brush pressure and commutator condition are appropriate.
3. Explain why the second operator's suggestion, briefly disconnecting the field circuit, is dangerous. Estimate the immediate armature current if the field is opened at rated load, stating your assumptions.
Read the worked answer
If the field circuit is opened, the field flux collapses to the small residual value. Back emf is proportional to flux and speed, so E_b falls almost to zero. The armature then sees nearly the full supply voltage across its small resistance. Ignoring the residual emf, the immediate armature current would be I_a = V/R_a = 230/1.2 = 192 A, compared with the rated 10 A. Even if the supply impedance and brush contact drop limit the current somewhat, it is many times the rating. The motor would also overspeed because, with almost no flux, the speed needed to generate even a small back emf is very high. The likely result is a flashover, destroyed commutator, burnt armature and a mechanical burst from overspeed. A shunt motor must never be run with its field open; a field-failure relay or undervoltage release is used so that the supply is tripped if field current is lost.