Introduction to Electrical Power and Power Electronics

402 Introduction to Electrical Power and Power Electronics

Example 15.1

A water pump driven by a 1000-hp induction motor pumps 100,000 m

3

/day at

rated speed. If the motor were run at slower speed using VFD to reduce the flow

rate to 80% over a longer time to pump the same quantity of water per day, deter-

mine the percent savings in kilowatt-hour energy consumed per day.

Solution:

Using Equation 15.2, power at 80% flow rate = 0.80

3

× old power.

Hours to pump the same water quantity = old hours/0.80.

Therefore, new energy consumption to pump the same quantity of water = (0.80

3

× old power) × (old hours/0.80) = 0.80

2

× old kWh = 0.64 × old kWh per day ...