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Shunt motor pump, torque ∝ speed²

Short answer

For T ∝ N², armature current ∝ N². Initial I_a=48A. At 800 rpm, I_a=30.72A, E_b=156.16V. Required series resistance R = (200–156.16)/30.72 – 0.1 ≈ 1.327Ω.

A 200 V dc shunt motor drives a centrifugal pump where torque is proportional to the square of speed. 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. Given: Armature resistance (Ra) = 0.1 Ω and field winding resistance (Rf) = 100 Ω.

Here the load torque is not constant: T ∝ N². Flux is constant (shunt motor), so T ∝ I_a. Therefore I_a ∝ N².

Equivalent circuit

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

Initial condition (1000 rpm)

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

At 800 rpm

  • I_a2 = I_a1 (N₂/N₁)² = 48 × (0.8)² = 48 × 0.64 = 30.72 A
  • E_b2 = E_b1 × (800/1000) = 156.16 V

Resistance

V = E_b2 + I_a2 (R_a + R)

200 = 156.16 + 30.72 (0.1 + R)

0.1 + R = 43.84 / 30.72 = 1.4271 Ω

R = 1.4271 − 0.1 = 1.327 Ω

CaseI_a (A)E_b (V)Series R (Ω)
Constant torque (for comparison)48156.160.813
T ∝ N²30.72156.161.327

Answer: R ≈ 1.327 Ω must be inserted in the armature circuit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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