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Shunt motor driving pump: constant torque resistance

Short answer

Constant torque requires the same armature current of 48 A. At the reduced speed of 800 rpm, the back emf is 156.16 V, so the total armature circuit resistance must be (200 – 156.16) / 48 = 0.9133 Ω, yielding an external resistance of 0.8133 Ω.

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 ΩR_a = 0.1\ \Omega and field winding resistance Rf=100 ΩR_f = 100\ \Omega.

The field is across the supply, so the flux is constant. Constant torque means constant armature current. The speed is lowered by an external armature resistance RxR_x, so N∝EbN\propto E_b.

Initial condition:

If=200100=2 A,Ia=50−2=48 AI_f = \frac{200}{100} = 2\ \text{A}, \qquad I_a = 50 - 2 = 48\ \text{A} Eb1=200−48(0.1)=195.20 VE_{b1} = 200 - 48(0.1) = 195.20\ \text{V}

At 800 rpm:

Eb2=Eb1×8001000=195.20×0.8=156.16 VE_{b2} = E_{b1}\times\frac{800}{1000} = 195.20\times 0.8 = 156.16\ \text{V}

The torque is constant, so Ia=48I_a = 48 A:

V=Eb2+Ia(Ra+Rx)  ⇒  Ra+Rx=200−156.1648=0.9133 ΩV = E_{b2} + I_a(R_a + R_x) \;\Rightarrow\; R_a + R_x = \frac{200 - 156.16}{48} = 0.9133\ \Omega Rx=0.9133−0.1=0.8133 ΩR_x = 0.9133 - 0.1 = 0.8133\ \Omega

Answer: external armature resistance Rx=0.8133 ΩR_x = 0.8133\ \Omega.

✓ Answer checked by Claude Opus 5.5 (high effort)

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