Electrical Engineering is a free introductory textbook to the basics of electrical engineering. See the editorial for more information....


Author: E.E. Kimberly

  Problem Answer
1 In the plant of Example 25-1, a 300-hp synchronous motor is installed to run at full load 576 hours per month with an efficiency of 92% and at a power factor of 0.8 leading. Calculate the new demand charge.
2 The motor of Problem 1 cost $1850.00. A similar motor capable of only 1.0 power factor could have been bought for $1775.00. What price per leading reactive kv-a at full load and 0.80 power factor was paid to install the motor with a power factor of 0.8 in preference to the motor with a power factor of 1.0? $0.41
3 A factory load consists of 400 kw of load with a power factor of 71% lagging. Because of the power-factor penalty clause provided in the power and energy contract when the power factor is less than 85% lagging, it is desirable to improve the over-all power factor just enough to avoid the penalty. How many kv-a must be provided by static condensers to effect the required correction? 150 kv-a
4 If in Problem 3 the correction were accomplished by adding synchronous motors at 0.80 power factor, how many horsepower of motors running at full load and 88% efficiency must be added to the existing load to avoid the penalty imposed for low power factor? 120 hp. Probably a 125-hp motor

Last Update: 2011-01-18