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Power output control on large generator

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ps000205

Electrical
Jan 18, 2006
3
I have an ongoing discussion with several people as to how the power output on a generator is changed. The generator is connected to a steam turbine and is rated at 450 MW. Some say that by changing the excitation current you create a larger or smaller magnetic field, therefore changing generator power output. Isn't changing excitation current the primary means for generator voltage regulation and MVARS control not total power output? Others have suggested that by adding more steam flow you create more torque on the generator rotor therby increasing power output. I'm not sure how this would even be possible. Maybe all the different ideas just have me confused and I'm overlooking a simple answer. Any help is appreciated.
 
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More steam, more power output. Yep.

The standard visualization is: The generator is hooked to a massive network that cannot be materially altered by one generator. The generator is essentially trying to drag the network forward. More steam harder dragging.

Picture a gear train. More steam would cause more torque to turn the gear train.
 
Right on itsmoked
A single set, the power is determined by the load. More steam, it goes faster.
A single set, the more exitation, the more voltage.
Put it on a grid and that all changes.
More steam more power out, and somewhere else, one or more sets will put out a little less power.
Increase the exitation and power factor changes.
Same answer in different words.
When the plant is synchronized, the steam valve will be almost closed. It's just passing enough steam to spin the turbine and generator up to speed with no load. Once it's on the line, the steam valve is opened to pick up the load. If you're ever in the control room as the set is going online you will see the watt meter going up scale as the steam valve is opened.
 
The prime mover, your steam turbine in this case, can affect real power and speed. The exciter can affect voltage and reactive power.

When running stand alone, the load determines the amount of power, both real and reactive, necessary so opening the throttle enough on the prime mover (steam valve in your case) to maintain speed causes the necessary real power to be provided. The excitation system providing the desired terminal voltage causes the necessary reactive power to be provided.

When running against the utility, the utility sets speed (frequency) and voltage, so the throttle controls the amount of power into the prime mover, which is transmitted along the shaft to the generator, and out. Input power, less losses, equals output power. More in results in more out. With the voltage fixed by the utility, the excitation system controls the amount of reactive power produced, or absorbed, by the generator.
 
Thanks for the help, it makes things much clearer. The only thing I don't quite understand is just exactly how the mechanical energy in the steam turbine is transferred to electrical energy just by opening up the steam valve. Again, any help is greatly appeciated.
 
I hope this help to supplement the above explanations.
Generator operation works in a balance power in and out the generator. Load require from the generator active/real power (kW) and reactive power (kvar) to satisfy the demand. Here is a simplified interrelation with the prime mover and exciter:
o Active power (MW): is provided by the prime mover (i.e. steam turbine)
o Reactive power (mvar): Somehow the exciter is responsiblen without virtually any additional active power input. Two basic operation mode are:
1. Supplying vars: Overexcited mode with lagging power factor
2. Absorbing kvar: Underexited mode with leading power factor. Generator may work as synchronous condenser.
[blue]NOTE: Although generator could work in a wide region of their capability curve, most generator and particularly those own by IPP, are intend to work providing mainly kW. [/blue]

For graphical details, see the info below.
Gen_Reactive_Capability_Limits_2.jpg
 
Check for in a power plant book, handbook or just search for steam turbine "how things work"
Here is a sketch self-explanatory get from the net.

kombidampf_e.gif
 
""The only thing I don't quite understand is just exactly how the mechanical energy in the steam turbine is transferred to electrical energy just by opening up the steam valve. Again, any help is greatly appeciated.""

I want to throw this explanation out for comment.

Remember, There is a magnetic coupling between the power source and the electrical power output and it is controlled by excitation of the field.

With an islanded generator, increasing the mechanical power input is impossible if speed is held constant and field excitation is not increased to increase the magnetic flux coupling across the air gap and induce more current in the stator. Torque is directly related to current, and rpm controls frequency which does not vary (much), so to transmit more power means to transmit more torque. To get more current requires increased magnetic coupling to transmit the increased torque. Increased excitation of an islanded generator beyond that needed to maintain the current required by a constant load at the setpoint voltage would result in increased terminal voltage, decreased excitation would result in decreased terminal voltage if the load was constant.

In parallel operation (with the power grid for example) increasing mechanical power will not result in more electrical power output unless the excitation is increased to maintain the field poles synchronization with the stator field. Instead if underexcited severely or not excited at all torque may exceed magnetic coupling, and the generator might slip poles and loose syncrony or operate as an induction generator. Slight underexcitation increases var demand and overexcitation will result in vars exported since the voltage is controlled by the grid and the one generator cannot raise or lower the grid voltage by itself.

 
I think the point that hasn't been stated explicitly is that we are talking about AC generators on a a grid and not DC generators. All the rotors in all the generators on the grid are synchronized to the grid. If you open the throttle on one generator it will try to speed-up. The rotor will advance in phase angle compared to the grid by a fraction of a degree which will result in it pushing more power into the grid until the mechanical power put into the generator equals the electrical power going into the grid. I would use the analogy that all the rotors are connected to the grid with stiff magnetic springs.
 
I get this question a lot-

This is a system where such a large generator is attached to the main power grid. Basicly, the speed can't change. You are always fixed at 60hz (speed), so when you add more power to the shaft, you get power flow out of the generator windings.

Think of a car with a set cruise control at 60mph. You come to a hill, but to maintain speed you have to add more power. The car is moving the same speed, but you have added more shaft HP to keep it going.

You can basicly ignore the field in the rotor as actually doing anything with Watts, or true power. It takes power to excite the rotor, but you can ignore it for basic power flow. You vary the excitation only to vary the var flow.

Was that what you were asking?
JTK
 
I use the cruise control analogy alot; going uphill (more laod) or downhill (less load), the car/engine speed doesn't change, but the power output does. KW (KVA x pf) equates directly to horsepower and is therefore controlled by the turbine. KVAR (imaginary or reactive power) is contolled by the field. They work simultaneously to maintain setpoint in an islanded set and stability when hooked into the grid.
 
Thanks alot everyone! Really good information.
 
Another tip-

When people ask about watts, vars, VAs I like to use the glass of beer analogy- watts is the beer that does the work. Vars is the foam- you want some, but not too much. VAs is the glass.

The operators who run small plants seem to understand it better that way.

JTK
 
My favorite analogy is to take the Generator's Reactive Capability Curve like Cuky showed us and imagine pasting it on an Etch-A-Sketch. (Kid’s toy with two knobs that draws pictures on a screen. Left knob is up-down motion, right knob is left-right motion.)

Label the left knob "Excitation" and the right knob "Throttle". Now you can move the cursor to any point inside the operating curve by adjusting the throttle (KW) and excitation (KVARs).

This analogy works best when trying to explain problems with power factor control because you need to draw an angled line (constant pf) on the Etch-A-Sketch.

(We won't get into how KVAR's are like managers, they take up room on the system, increase losses, produce no useful work but are required to get the real work to the end of the line. My company's power factor is dropping.)
 
rcwilson,

I love your analogy!


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