Parallel Operation of GovernorsIf Unit A had been operating alone carrying full load, thesystem frequency would be 60 cycles as shown by Line"A" at 60 cycles and 100% load. Now, if the frequencyof Unit B is adjusted by the engine speed control to beequal to that of Unit A and the circuit breaker of Unit B isclosed, the system would be operating under thefollowing conditions:1. Unit A is at 60 cycles and 100% load.2. Unit B is at 60 cycles and 0% load. Thecharacteristic of the governor of Unit B at this time isshown by the dotted line B.In order for Unit B to carry load, it is necessary toadvance the speed setting of the governor. If it isadvanced to the full load position, the governorcharacteristic of B will coincide with the characteristic ofA. Since the load on the system was 100% of one unit,no change was made in total load, and the availablecapacity is now 200%, the system will operate at 50%load on each unit, and 60.9 cycles for two units of equalcapacity (Point E). For units of unequal capacity, theload will be divided in proportion to the ratio of thecapacity of each to the total capacity. The systemfrequency will be determined by the points on thegovernor characteristics corresponding to these loads.The frequency will be the same for both units sinceparalleled alternators must operate at the same speed.If Unit A had been a 60 KW unit fully loaded and a 100KW Unit B was paralleled with it and the governoradjusted to the full load position, the final load divisionand frequency would be determined as follows:a. System load - 60 KWb. System capacity - 160 KWSystem LoadKW rating for unitc. The 60 KW unit will carry 60 x 60 KW = 22.5 KW160System capacityKW load for unitd. 100 KW unit will carry 60 x 100 KW = 37.5 KW160e. The system frequency can be determined readilyfrom step c or step d. The load carried by eachunit is 22.5 or 37.5 which figures out to be6010037.% of the capacity of either unit Again, using Figure 1for the governor characteristic of the 100 KW unitand reading up from the value of load (37.5% to point F),we find the system frequency to be 61.125 cycles.Figure 2-Paralleling two units, one with a 3% governorand the other with a hydraulic governor set forisochronous operation.Example 2-One 3% Governor and one 0%(Isochronous) Governor.The characteristics of the 3% governor (Unit A) is shownby line A of Figure 2 and the characteristics of theisochronous governor (Unit B) is shown by line B. Onlyat full load, 60 cycles, do the frequencies of the unitshave the same value. It is customary to operate asystem of this type with a system load greater than thecapacity of Unit A. In this way Unit A carries its full loadat 60 cycles and the additional load and load swings arehandled by Unit B, also at 60 cycles. The system canmaintain constant frequency by this method of operation.The system described in example 1 cannot maintainconstant frequency with load changes because of thespeed droop characteristics of the governors.In the system described in this example, if the load isless than the capacity of Unit A (which has the 3%governor) and can be carried by Unit B, the governorsetting of Unit A can be reduced to give a governorcharacteristic such as A, so Unit A will still carry thesteady part of the load and Unit B will carry the loadswings. If the system load is reduced to the point whereUnit A is not operating at the 60 cycle point of itsgovernor characteristic, Unit A will try to motor Unit Band the system frequency may be greater than 60cycles. The reason for using an isochronous governor ina power system is to maintain constant frequency aboveapproximately 40% load.18
Integrated Publishing, Inc. - A (SDVOSB) Service Disabled Veteran Owned Small Business