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Shunt motor pump: resistance for 800 rpm

Short answer

Shunt field constant, flux constant. At constant torque, armature current remains 48A. E_b at 1000 rpm = 195.2V, at 800 rpm = 156.16V. Required series resistance R = (200–156.16)/48 – 0.1 = 0.813Ω.

A 200 V DC shunt motor drives a centrifugal pump where constant torque is required. The motor draws a current of 50 A when running at 1000 rpm. What value of resistance must be inserted in the armature circuit to reduce the speed to 800 rpm at constant torque? Given that armature winding resistance, Ra = 0.1 Ω and field winding resistance, Rf = 100 Ω.

At constant torque and constant flux (shunt field is unchanged), the armature current stays the same. So the speed is reduced by inserting resistance, which lowers the back emf.

Equivalent circuit

 +--[Ra=0.1]--[R ?]--(Eb)--+
 |       Ia = 48 A          |
 +---- Rf = 100 ohm --------+--- V = 200 V

Step 1: initial state

  • I_sh = 200 / 100 = 2 A
  • I_a1 = 50 − 2 = 48 A
  • E_b1 = 200 − 48 × 0.1 = 195.2 V at 1000 rpm

Step 2: new state (constant torque)

T ∝ Φ I_a, Φ unchanged, so I_a2 = I_a1 = 48 A.

E_b2 = E_b1 × (800/1000) = 195.2 × 0.8 = 156.16 V

Step 3: required resistance

V = E_b2 + I_a (R_a + R)

200 = 156.16 + 48 (0.1 + R)

0.1 + R = 43.84 / 48 = 0.9133 Ω

R = 0.9133 − 0.1 = 0.813 Ω

Answer: a resistance of about 0.813 Ω must be inserted in series with the armature.

✓ Answer checked by Claude Opus 5.5 (high effort)

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