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47 questions By Tony R. Kuphaldt
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Question 46 of 47
A very large 3-phase alternator at a hydroelectric dam has the following continuous full-power ratings:
- 600 MW power output
- 15 kV line voltage
- 23.686 kA line current
Calculate the continuous full-load apparent power for this alternator (in MVA), its continuous full-load reactive power (in MVAR), and its power factor (in percent).
Reveal answerS = 615.38 MVA Q = 136.72 MVAR P.F. = 97.5%
Notes:These figures came from a real hydroelectric generator that I saw once on a tour. Needless to say, this alternator was very large!
- Maximum continuous ratings
- 615.38 MVA apparent power output
- 600 MW true power output
- 15 kV line voltage
- 23.686 kA line current
- 97.5% power factor
- 365 volts exciter voltage
- 3,425 amps exciter current
- Maximum continuous overload ratings
- 707.69 MVA apparent power output
- 690 MW true power output
- 15 kV line voltage
- 27.239 kA line current
- 97.5% power factor
- 410 volts exciter voltage
- 3,640 amps exciter current
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Question 47 of 47
Large power distribution centers are often equipped with capacitors to correct for lagging (inductive) power factor of many industrial loads. There is never any one value for capacitance that will precisely correct for the power factor, though, because load conditions constantly change. At first it may seem that a variable capacitor would be the answer (adjustable to compensate for any value of lagging power factor), but variable capacitors with the ratings necessary for power line compensation would be prohibitively large and expensive.
One solution to this problem of variable capacitance uses a set of electromechanical relays with fixed-value capacitors:
Explain how a circuit such as this provides a step-variable capacitance, and determine the range of capacitance it can provide.
Reveal answerCapacitors may be selected in combination to provide anywhere from 0 μF to 15 μF, in 1 μF steps.
Notes:Although semiconductor-based static VAR compensator circuits are now the method of choice for modern power systems, this technique is still valid and is easy enough for beginning students to comprehend. A circuit such as this is a great application of the binary number system, too!
