- #1

Numbskull

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- 1

- Homework Statement
- A 20kW DC shunt motor supplied with 500V, draws a full-load current of 45A and rotates at 600rpm. The no-load current is 5A. The field resistance is 220 Ohms and the armature resistance is 0.3 Ohms. Calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full-load value, assuming that flux per-pole does not change. These are ALL of the given values.

- Relevant Equations
- V = E + IaRa, P = I^2R

First some basic figures which are very rounded as I'm interested in the approach to the problem rather than accuracy of the answer at this stage:

The field current will be 500/220 = 2.2727 Amps, and so power is a fixed loss at 1136.36 Watts

The armature current is 45 - 2.2727 = 42.72 Amps. At full load, the power loss in the armature is 547 Watts. At no-load it's negligible at a couple of watts.

The question is exactly as it is posed, so I'm not sure if this suggests that half-load is half of the armature power or current, or half of the total power consumed by the machine under full load (which is 22.5kW and thus 11.25kW). I've made further calculations but is there something I've overlooked?

The field current will be 500/220 = 2.2727 Amps, and so power is a fixed loss at 1136.36 Watts

The armature current is 45 - 2.2727 = 42.72 Amps. At full load, the power loss in the armature is 547 Watts. At no-load it's negligible at a couple of watts.

**Calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full-load value, assuming that flux per-pole does not change.**The question is exactly as it is posed, so I'm not sure if this suggests that half-load is half of the armature power or current, or half of the total power consumed by the machine under full load (which is 22.5kW and thus 11.25kW). I've made further calculations but is there something I've overlooked?