I am being asked to find the speed of a 16.5kW shunt DC motor that is being supplied with terminal DC voltage of 300V when operating at 50% rated load with an armature current of 30A. I need to use the armature current when solving this, and am wondering if it decreases due to the motor operating at just 50% load, or if I can use the 30A.
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\$\begingroup\$ 300V * 30A = 9 kw which is just over half the 16.5kw motor rating. If that rating covers 1.5kw for magnetisation and friction + 15kw output power, 30A might be its consumption at half load (7.5kw output) \$\endgroup\$– user16324Commented Dec 14, 2014 at 12:20
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1 Answer
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For a shunt DC motor, the field and armature coils are typically connected in parallel, therefore the voltage is the same across the coils, and their current is proportional to their resistance. As you increase the voltage, the current will increase on both coils. So, to double the power (full load), the voltage and current have to be increased by the square root of 2 $$ 2^{1/2} x \ 2^{1/2} = 2$$ In the event that the coils are not connected in parallel and the voltage is to remain the same, then the current on the armature has to double, to provide the necessary power (ignoring losses).
It should now be obvious that the armature current decreases as the load decreases, and that for the second case at full load, the current would have to be 60A.
