Question

A 250 v shunt motor having an armature resistance of 0.2 ohm draws from the mains a current of 50 amp on half full load. The speed is to be increased to twice half full load speed. If the torque of the motor is of constant magnitude, determine the percentage change in flux required

Answer 1

The percentage change is 45.59%.

• The problem can be resolved by assuming that the armature current is equal to the supplied 40 A line current. It is true, as stated in a condensed response, that field flux must be lowered by one-third of its present value. The flux must be increased by $\frac{2}{3}$ times its present value or decreased by $\frac{1}{3}$ times its present value in order to reach the speed necessary for constant torque. The fundamental speed and torque equations of a DC shunt motor can be used to quickly determine this condition.
• Even though the cut-off of the membrane is orders of magnitudes bigger than the size of the solute molecules, the flux of ultrafiltration membranes may be significantly lowered when handling low-molecular-weight hydrophobic solutes. In this study, octanoic acid was used as a model drug to correlate the flow reduction to the membrane pore size. The liquid-liquid displacement porometry method and measuring the retention of a dextran solution were both used to measure the pore size for comparison.

Hence, the percentage change is 45.59%.

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