Hawkins Electrical Guide v. 07 (of 10) / Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications

by N. Hawkins

The method of procedure can best be explained by reference to a concrete case. Assume a load of 450 kw. at .65 power factor. It is desired to raise the power factor to .9. What will be the rating of the condenser?

Fig. 2,488.—Diagram for synchronous condenser calculations.

Referring to the diagram, fig. 2,488, it is necessary to start with 450 kw. At .65 power factor, or 692 kva., this has a wattless lagging component of √(692² - 450²) = 525 kva. With the load unchanged and the power factor raised to .9, there will be 500 apparent kva., which will have a wattless component of √(500² - 450²) = 218 kva.

It is obvious that the condenser must supply the difference between 525 kva. and 218 kva., or 307 kva. A 300 kva. condenser would, therefore, meet the requirements.

If it be desired to drive some energy load with the condenser and still bring the total power factor to .9, proceed as indicated in fig. 2,489. Assume a total load of 150 kw. on the motor. As before, 450 kw. at .65 power factor, or 692 kva., with a wattless component of 525 kva.

The energy load will be increased from 450 to 600 kw. as indicated, and with the power factor raised to .9 there will be a kva. of 667 with a wattless component of √(667² - 600²) = 291.

There must be supplied 525 - 291 = 234 in leading kva.

The synchronous motor then must supply 150 kw. energy and 234 kva. wattless, which would give it a rating of √(150² + 234²) = 278 kva. at .68 power factor.