Compared with constant torque (0.8133 ohm), Rx here is:

Larger, 1.3271 ohm, as current is lower

Asked 2/21
2068 Bhadra · 8 marks

Shunt motor driving pump: torque proportional to speed squared

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 ( $R_a$ ) = $0.1\ \Omega$ and field winding resistance ( $R_f$ ) = $100\ \Omega$ .

For a fan or centrifugal pump, torque is proportional to the square of speed: $T\propto N^2$ . The flux is constant, so $T\propto I_a$ and therefore $I_a\propto N^2$ . The speed is lowered by an external armature resistance $R_x$ , and $N\propto E_b$ .

Initial condition (1000 rpm):

I_f = \frac{200}{100} = 2\ \text{A}, \qquad I_{a1} = 50 - 2 = 48\ \text{A}

E_{b1} = 200 - 48(0.1) = 195.20\ \text{V}

At 800 rpm:

I_{a2} = I_{a1}\left(\frac{800}{1000}\right)^2 = 48\times 0.64 = 30.72\ \text{A}

E_{b2} = E_{b1}\times\frac{800}{1000} = 156.16\ \text{V}

R_a + R_x = \frac{V - E_{b2}}{I_{a2}} = \frac{200 - 156.16}{30.72} = 1.4271\ \Omega

R_x = 1.4271 - 0.1 = 1.3271\ \Omega

Answer: external armature resistance $R_x = 1.3271\ \Omega$ (compare with $0.8133\ \Omega$ for a constant-torque load).

✓ Answer checked by Claude Opus 5.5 (high effort)

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