Reducing the motor speed of a constant-torque pump reduces the energy consumption of the motor. When the motor is operated at 80-percent speed, what is the approximate motor energy consumption expressed as a percent of the full-speed motor energy consumption?
A. 80 percent energy
B. 65 percent energy
C. 50 percent energy
D. None of the above
Commentary
Variable-speed drives are increasingly applied in many applications to reduce energy consumption. Reducing the speed of a centrifugal pump reduces the pump discharge pressure by the square of its speed and its energy consumption by the cube of its speed. Even a small speed reduction, such as 5 percent, can save almost 15 percent of the energy consumed at full speed.
Such is not the case with constant-torque loads, such as positive-displacement pumps and positive-displacement compressors. The capacity and energy consumption of constant torque equipment are both approximately proportional to operating speed. Therefore, at 80-percent speed, the motor will consume approximately 80 percent of its full-speed energy consumption. Answer A is correct.
At 80-percent speed, the load will operate at approximately 80 percent of its capacity while consuming 80 percent of its full-speed energy. These relatively simple relationships for positive-displacement equipment make the operating speed and energy consumption considerably easier to predict as compared to centrifugal equipment.
Many variable-speed drive applications and installations require attention to detail to ensure that the motor and its mechanical equipment can function properly in operation. In addition, failure to attend to certain details of lower speed operation can damage the equipment – especially when variable-speed drives are applied to positive-displacement equipment.
David W. Spitzer is a regular contributor to Flow Control with more than 35
years of experience in specifying, building, installing, startup,
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has written over 10 books and 150 technical articles about instrumentation
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