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Transcript
TEMPUS ENERGY: GENERATORS IN WIND TURBINES: EXERCISES
1: ASYNCHRONOUS GENERATOR WITH CAGE ROTOR
Consider a wind turbine where the rotor blades (1), using a gearbox (3), drive an asynchronous
generator (4) with cage rotor. Using a transformer (6), the generated power is injected into a high
voltage grid (7). Notice the capacitors (5) generate reactive power which avoids the grid has to supply
the reactive power required by the generator (or reduces the reactive power supplied by the grid).
The generator generates a 50 Hz voltage having a line voltage of 690 V. The generated active power
equals 750 kW and the power factor equals 0.86.
Questions part 1:
1) Calculate the apparent power of the generator.
2) Calculate the RMS value of the current provided by the generator.
3) Calculate the capacitor values C needed to obtain a power factor equal to 1.
Suppose the gearbox has 1.2 % losses. Suppose the generator has an efficiency πœ‚ = 95 % and the
transformer has an efficiency πœ‚ = 99 %.
Questions part 2:
1) Calculate the active power injected into the grid.
2) Calculate the overall efficiency of the entire installation.
Solutions part 1:
1) Since 𝑃 = 750 π‘˜π‘Š and π‘π‘œπ‘ πœ‘ = 0.86, 𝑆 = 872 π‘˜π‘‰π΄.
2) Since 𝑆 = √3 π‘ˆπ‘™π‘–π‘›π‘’ 𝐼𝑙𝑖𝑛𝑒 and π‘ˆπ‘™π‘–π‘›π‘’ = 690 𝑉, 𝐼𝑙𝑖𝑛𝑒 = 730 𝐴.
3) Since 𝑄 = 𝑆 π‘ π‘–π‘›πœ‘, 𝑄 = 445 π‘˜π‘‰π΄π‘…. One single capacitor, having a phase voltage of 400 𝑉,
has to generator 148 π‘˜π‘‰π΄π‘…. Since
𝑄 = π‘ˆ 2 πœ” 𝐢,
a 𝐢 = 3000 πœ‡πΉ is needed.
Solutions part 2:
1) The generator generates 750 π‘˜π‘Š, since πœ‚π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘Ÿ = 0.99 an active power of 742.5 π‘˜π‘Š is
injected into the grid.
2) The mechanical power at the output of the gearbox equals 750 π‘˜π‘Šβ„πœ‚π‘”π‘’π‘› = 789.5 π‘˜π‘Š. The
mechanical power at the input of the gearbox provided by the rotor blades equals
789.5 π‘˜π‘Šβ„πœ‚π‘”π‘’π‘Žπ‘Ÿπ‘π‘œπ‘₯ = 800 π‘˜π‘Š. This implies the overall efficiency equals πœ‚π‘‘π‘œπ‘‘ =
742.5 π‘˜π‘Šβ„800 π‘˜π‘Š = 93 %. Notice πœ‚π‘‘π‘œπ‘‘ = πœ‚π‘”π‘’π‘Žπ‘Ÿπ‘π‘œπ‘₯ πœ‚π‘”π‘’π‘› πœ‚π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘Ÿ .
2: DOUBLY-FED INDUCTION MOTOR
Consider a wind turbine equipped with a doubly-fed induction machine. The turbine has a nominal
power of 3 MW. The speed of rotation of the rotor blades varies between 13 revolutions per minute
and 20 revolutions per minute. The gearbox has a speed ration of 1:90.
The generator has four poles and the stator is connected with the 50 Hz public grid. The nominal
power of 3 MW is supplied in cased the speed of the rotor blades equals 20 revolutions per minute.
All losses are neglected.
Questions:
1)
2)
3)
4)
5)
6)
7)
8)
9)
Calculate the synchronous speed of the generator.
Calculate the speed of rotation of the rotor of the generator.
Calculate the slip speed.
Calculate the slip.
Calculate the power supplied by the stator (in MW).
Calculate the power supplied by the rotor (in MW).
Calculate the power injected into the grid by the frequency converter.
Calculate the rotor frequency.
Suppose the rotor frequency equals 7.6 Hz and the generator has a sub synchronous speed.
Calculate the speed of rotation of the rotor blades.
Solutions:
1) The stator of the generator, having four poles, is connected with a 50 Hz grid implying a
synchronous speed of 1500 π‘Ÿπ‘π‘š.
2) The speed of rotation of the rotor blades equals 20 π‘Ÿπ‘π‘š, due to the speed ratio 1⁄90, the rotor
of the generator has a speed of rotation which equals 1800 π‘Ÿπ‘π‘š. An super synchronous speed
is obtained.
3) The slip speed equals 300 π‘Ÿπ‘π‘š.
4) The slip equals βˆ’0.2.
5) Since
𝑃𝑅 β‰… 𝑠 𝑃𝑆
and 𝑠 < 0 in combination with 𝑃𝑆 < 0, 𝑃𝑅 > 0. This means active power is extracted from the
rotor windings and injected into the grid by the frequency converter. The total power injected
into the grid equals
βˆ’π‘ƒπ‘† + 𝑃𝑅 = βˆ’(1 βˆ’ 𝑠) 𝑃𝑆 = 3 π‘€π‘Š.
This implies 𝑃𝑆 = βˆ’2.5 π‘€π‘Š (the stator injects 2.5 π‘€π‘Š into the grid).
6) The rotor (using the frequency converter) injects 𝑃𝑅 = 𝑠 𝑃𝑆 = 0.5 π‘€π‘Š in the grid.
7) The losses in the frequency converter are neglected implying the frequency converter injects
0.5 π‘€π‘Š into the grid.
8) The frequency of the rotor currents and voltages equals
𝑓2 = 𝑠 𝑓1 = 10 𝐻𝑧 .
9) Since 𝑓2 = 𝑠 𝑓1 = 7.6 𝐻𝑧 with 𝑓1 = 50 𝐻𝑧, the slip 𝑠 = 0.152 implying a speed of rotation of
the rotor which equals
𝑁𝑅 = (1 βˆ’ 𝑠)𝑁𝑆 = 1272 π‘Ÿπ‘π‘š.
Due to the speed ratio 1⁄90 of the gearbox, the rotor blades have a speed of 14.1 π‘Ÿπ‘π‘š.