Can a generators resistance be variable to maximize efficiency?

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181127-1004 EST

Noswad4:

Quickly rereading this whole thread I think your question is:

Can I control an electrical generator, AC or DC, to dump energy into the electrical grid based upon some signal, your turbine output pressure. The answer is probably yes.

How can you control electrical power flow in a system? By changing a voltage or current.

From your various comments and questions it appears you want to dump electrical power into the electrical grid system. To make things simple consider a DC system. To a pair of electrical terminals on the grid those terminals looking toward the grid appear to have the characteristics of a very large battery. This means the grid looks like an ideal constant voltage source with some small internal impedance, electrical resistance.

I can remove power from the grid by placing a load electrical resistance on the two terminals. So by my choice I will call this a positive current flow. This current flows from the voltage source thru its internal resistance to my external resistive load. There is a voltage drop across the internal resistance, and a power dissipation in that internal resistance, and externally there is a voltage drop across my load resistance, and its power loss. One wants to make that internal power loss low compared to the load power. The sum of the two voltage drops has to equal the voltage of the ideal constant voltage source.

Keep in mine that the grid is very large compared to what I do at the load terminals.

Next shift gears and create an energy source external to the grid. This source will consist of a variable voltage source and its own fixed internal electrical resistance.

Connect the external energy source to the two grid terminals.

If the external voltage is less than the grid voltage, then the external energy source is really a load, and just like the resistive load is receiving energy from the grid. Under these conditions it is not an energy source.

Adjust the external voltage source to equal the grid voltage. No current flows. Thus, no energy transfer.

Adjust the external voltage to a value higher than the grid voltage. Now current flows into the grid as does energy. The amount of current flow is the voltage difference between the two voltage sources divided by the sum of the two internal impedances (resistances).

The output power from the external source is its internal voltage times said current. The power into the grid is the grid internal voltage times said current. The inefficiency is in the power loss in the two internal resistances.

By adjusting the internal voltage of the external power source you change the amount of power flow to the grid from the external source.

To control power flow you do not change the electrical resistance of the generator, but rather you change its internal source voltage or current. With any reasonable set of conditions you can use electronic circuitry to do this.

.
 
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gar said:
181127-1004 EST

Noswad4:

Quickly rereading this whole thread I think your question is:

Can I control an electrical generator, AC or DC, to dump energy into the electrical grid based upon some signal, your turbine output pressure. The answer is probably yes.

How can you control electrical power flow in a system? By changing a voltage or current.

From your various comments and questions it appears you want to dump electrical power into the electrical grid system. To make things simple consider a DC system. To a pair of electrical terminals on the grid those terminals looking toward the grid appear to have the characteristics of a very large battery. This means the grid looks like an ideal constant voltage source with some small internal impedance, electrical resistance.

I can remove power from the grid by placing a load electrical resistance on the two terminals. So by my choice I will call this a positive current flow. This current flows from the voltage source thru its internal resistance to my external resistive load. There is a voltage drop across the internal resistance, and a power dissipation in that internal resistance, and externally there is a voltage drop across my load resistance, and its power loss. One wants to make that internal power loss low compared to the load power. The sum of the two voltage drops has to equal the voltage of the ideal constant voltage source.

Keep in mine that the grid is very large compared to what I do at the load terminals.

Next shift gears and create an energy source external to the grid. This source will consist of a variable voltage source and its own fixed internal electrical resistance.

Connect the external energy source to the two grid terminals.

If the external voltage is less than the grid voltage, then the external energy source is really a load, and just like the resistive load is receiving energy from the grid. Under these conditions it is not an energy source.

Adjust the external voltage source to equal the grid voltage. No current flows. Thus, no energy transfer.

Adjust the external voltage to a value higher than the grid voltage. Now current flows into the grid as does energy. The amount of current flow is the voltage difference between the two voltage sources divided by the sum of the two internal impedances (resistances).

The output power from the external source is its internal voltage times said current. The power into the grid is the grid internal voltage times said current. The inefficiency is in the power loss in the two internal resistances.

By adjusting the internal voltage of the external power source you change the amount of power flow to the grid from the external source.

To control power flow you do not change the electrical resistance of the generator, but rather you change its internal source voltage or current. With any reasonable set of conditions you can use electronic circuitry to do this.

.
Click to expand...

:thumbsup:

I found this post to be very helpful in understanding how power is shared.

Would it be basically the same for AC, with the phases having to be matched? Does frequency come into the load sharing?
 
181127-1334 EST

JFletcher:

Works the same way with AC. For maximum power transfer into the AC grid you want the current going in to be in phase with the voltage. Maximum power transfer (real power) as I used it here is mean't to describe how the phase angle of the current is related to the grid voltage where current magnitude is a constant.

In an AC power distribution system you want the current to be as much in phase with the voltage as possible to reduce wasted power in the I^2*R distribution losses. Any out of phase current is heating wires, etc, without doing useful work in the destination load.

I would not consider frequency an issue other than you need matching and synchronized waveforms.

.
 
gar said:
181127-1334 EST

JFletcher:

Works the same way with AC. For maximum power transfer into the AC grid you want the current going in to be in phase with the voltage. Maximum power transfer (real power) as I used it here is mean't to describe how the phase angle of the current is related to the grid voltage where current magnitude is a constant.

In an AC power distribution system you want the current to be as much in phase with the voltage as possible to reduce wasted power in the I^2*R distribution losses. Any out of phase current is heating wires, etc, without doing useful work in the destination load.

I would not consider frequency an issue other than you need matching and synchronized waveforms.

.
Click to expand...

Thank you, gar. My experiences with Industrial generators is fairly limited, however I remember when we sync up our 2 megawatt 20 cylinder EMD 710 to the grid, the synchronizer dial rotated in a clockwise direction, and thr switch had to be manually operated at exactly 12 o'clock. I thought that the generator frequency had to be slightly higher than the grid to pick up the load, voltages were the same, 4160

Is that incorrect? Or does the generator run a bit faster because you're dropping a 2 megawatt dead load on it, which is going to slow down even a 4500 horsepower diesel engine?

It seems obvious to me now that all generators would have to run at the same frequency once they are sharing a load, yes?
 
181127-2441 EST

JFletcher:

I don't have direct experience tying a large alternator to the grid. But even a few megawatt unit is peanuts compared to the grid. The impedance into the grid is more of a factor. So your connection to the grid won't do much to the grid. Its frequency will stay the same, as will its internal equivalent voltage.

Other than for some form of small AC generator your generator can be considered a DC field excited alternator/synchronous motor. Whether it is a generator or motor is a function of the direction of energy flow. Once connected the motor/generator will run at line synchronous speed. For example exactly 1800 RPM on 60 Hz. Other possibilities at 60 Hz are 3600, 1200, etc.

If you have two voltage sources that are exactly the same. and connect them in parallel, then no current flows. If they are not the same voltage, then a large current can flow limited only by the impedance between them and the voltage difference.

When connecting one voltage source in parallel with another you want them matched to avoid large mechanical and electrical disturbances.

The alternator must run at AC line frequency. Mechanical and electrical phase angle might lead or lag, and internal equivalent voltage must be higher to force energy into the grid.

.
 
gar said:
181127-2441 EST

JFletcher:

I don't have direct experience tying a large alternator to the grid. But even a few megawatt unit is peanuts compared to the grid. The impedance into the grid is more of a factor. So your connection to the grid won't do much to the grid. Its frequency will stay the same, as will its internal equivalent voltage.

Other than for some form of small AC generator your generator can be considered a DC field excited alternator/synchronous motor. Whether it is a generator or motor is a function of the direction of energy flow. Once connected the motor/generator will run at line synchronous speed. For example exactly 1800 RPM on 60 Hz. Other possibilities at 60 Hz are 3600, 1200, etc.

If you have two voltage sources that are exactly the same. and connect them in parallel, then no current flows. If they are not the same voltage, then a large current can flow limited only by the impedance between them and the voltage difference.

When connecting one voltage source in parallel with another you want them matched to avoid large mechanical and electrical disturbances.

The alternator must run at AC line frequency. Mechanical and electrical phase angle might lead or lag, and internal equivalent voltage must be higher to force energy into the grid.

.
Click to expand...

The problem that I see is that to do its job the turbine will have to be able to run at a range of different speeds and flow rates. But to drive the grid requires a rigidly controlled frequency.
To be able to deliver all of the power than needs to be dropped by the turbine at any given moment into the grid requires either that the generator be a writable pole type where the output frequency is not tied directly to rotation speed or that the "generator" is actually a combination of a DC generator and a grid-tie-capable inverter.
I do not think that a variable speed transmission between turbine and generator would be a practical solution, as the mechanical losses would be far too great.
 
181128-0650 EST

GoldDigger:

Here is my view of a synchronous motor/generator.

When connected to the grid it runs at synchronous speed independent of whether it is a motor or generator. Assume unloaded it has a shaft phase angle relative to the AC line voltage of 0 degrees. This phase angle will change with loading. Assume for convenience + as a generator, and - as a motor. But it will always be at synchronous speed as long as it is not overloaded either as a motor or generator.

The only item that changes with loading is shaft torque, and relative shaft angle. As a motor the output load torque determines the electrical input power. As a generator the input torque determines the output electrical power when connected to the huge energy sink of the grid.

For this application the motor/generator only operates in the generator mode, and it is at a constant speed as defined by its connection to the grid. Let that be 1800 RPM.

To drive this alternator we have an air powered turbine. This turbine produces an output torque at any given speed that has some proportional relationship to the pressure drop across the turbine. Is there a speed or speeds of this turbine that can produce positive torque over the range of pressure change specified by Noswad4? If yes, then a fixed ratio gear box can connect the turbine to the alternator. The choice of that ratio then becomes the factor in determining maximum efficiency.

If constant RPM is not feasible with the turbine to achieve the above results, then one uses a DC generator to drive an electronic inverter to connect to the grid.

.
 
181128-0920 EST

I would like to see two plots of torque vs speed, one for the minimum pressure difference (200-50), and the other the maximum pressure difference (500-50), on a single sheet of graph paper for a sample turbine under consideration.

.
 
181128-1117 EST

To the plot I suggested in post #29 I would like two more curves added. These are efficiency vs speed for the two different pressure differentials. Then draw a straight line from the two maximum efficiency points and pick the midpoint of said straight line. The midpoint is the RPM to run the turbine at. Then pick a gear box to produce 1800 RPM at the generator.

.
 
Other than frequency, there are necessary conditions that must be met for a successful synchronization of generator to the grid.

Voltage magnitude of the generator must be equal to the grid and we all know that. Meaning the sinusoidal voltage of the grid.

Sustained voltage differential of the two systems could create havoc on the generator —being that the grid has more power to go around with.

If the generator voltage is too high the generator will put out large MVAR (mega volt amp reactive.) which could cause generator to get overexcited and therefore overheating occurs.

Inversely, when generator voltage magnitude is too low, the MVAR will be absorbed and the generator becomes underexcited.

Now, since we are mostly focused on the frequency, this is by no means a small fry to ignore.

An elephant in the room if you will.

The sinusoidal frequency of the generator must be the same as the grid.
As one poster had alluded to an instrument that shows whether the generator and the grid are in the process of synchronizing themselves, this is called synchroscope. When the two systems are not synchronized the synchroscope will keep spinning.

When the two systems reach synchronicity it (synchroscope) will stop at 12:00 oclock that is also zero. This is the moment when you throw the switch on.

Throwing the switch before they get synchronized will cause the generator to suffer from being out of step and subsequently behave like a motor—not to mention the stator and rotor would be slipping. A potential destructive force.

An old-school way of checking sychronicity is using a light bulb. When there is voltage differential the bulb is lit—when the bulb goes dim synchronicity is achieved.

The same destructive force would be enforced upon the generator when it is running too fast. When this happens the grid will try to slow it down.
On the other hand running the generator too slow will prompt the grid to speed it up with the same destructive result.

Phase angle and phase sequence are also important considerations when synchronizing generator to the the grid. Details of which would be for another discussion.

These problems had been overcome with the introduction of mercury-arc valves (a glorified term for rectifier) which are now obsolete with the arrival of solid state GTOs (gate turn off thyristors) and IGBTs for high voltage inverters.
Generator voltage is first rectified to DC and then inverted to AC for ease of synchronization.


We are using this strategy in our California PACIFIC INTERTIE that supplies power to the largest municipal owned utility company (LADWP).
 
When I hear generator I think essential/standby. A wind farm control system is going to incorporate a lot more solutions. Tip speed, rotor rotation, tip stall, etc. DC/investor, perhaps unloaded windmills whose only job is to get motored.
 
gar said:
181127-1004 EST

Noswad4:

Quickly rereading this whole thread I think your question is:

Can I control an electrical generator, AC or DC, to dump energy into the electrical grid based upon some signal, your turbine output pressure. The answer is probably yes.

How can you control electrical power flow in a system? By changing a voltage or current.

From your various comments and questions it appears you want to dump electrical power into the electrical grid system. To make things simple consider a DC system. To a pair of electrical terminals on the grid those terminals looking toward the grid appear to have the characteristics of a very large battery. This means the grid looks like an ideal constant voltage source with some small internal impedance, electrical resistance.

I can remove power from the grid by placing a load electrical resistance on the two terminals. So by my choice I will call this a positive current flow. This current flows from the voltage source thru its internal resistance to my external resistive load. There is a voltage drop across the internal resistance, and a power dissipation in that internal resistance, and externally there is a voltage drop across my load resistance, and its power loss. One wants to make that internal power loss low compared to the load power. The sum of the two voltage drops has to equal the voltage of the ideal constant voltage source.

Keep in mine that the grid is very large compared to what I do at the load terminals.

Next shift gears and create an energy source external to the grid. This source will consist of a variable voltage source and its own fixed internal electrical resistance.

Connect the external energy source to the two grid terminals.

If the external voltage is less than the grid voltage, then the external energy source is really a load, and just like the resistive load is receiving energy from the grid. Under these conditions it is not an energy source.

Adjust the external voltage source to equal the grid voltage. No current flows. Thus, no energy transfer.

Adjust the external voltage to a value higher than the grid voltage. Now current flows into the grid as does energy. The amount of current flow is the voltage difference between the two voltage sources divided by the sum of the two internal impedances (resistances).

The output power from the external source is its internal voltage times said current. The power into the grid is the grid internal voltage times said current. The inefficiency is in the power loss in the two internal resistances.

By adjusting the internal voltage of the external power source you change the amount of power flow to the grid from the external source.

To control power flow you do not change the electrical resistance of the generator, but rather you change its internal source voltage or current. With any reasonable set of conditions you can use electronic circuitry to do this.

.
Click to expand...

Gar:

Thanks for clarifying. You are correct. I would also add that I would like to try and control the pressure drop (signal). Maybe more correctly control the torque load on the turbine.

I was under the impression that a generator needs to produce the same or slightly more voltage than the grid it is feeding into.

Am I correct in assuming that through electronic circuitry one can manipulate torque load on the generator and in turn the turbine. The larger the torque load the more power that can be produced and sent to the grid?
 
Last edited:
myspark said:
Other than frequency, there are necessary conditions that must be met for a successful synchronization of generator to the grid.

Voltage magnitude of the generator must be equal to the grid and we all know that. Meaning the sinusoidal voltage of the grid.

Sustained voltage differential of the two systems could create havoc on the generator —being that the grid has more power to go around with.

If the generator voltage is too high the generator will put out large MVAR (mega volt amp reactive.) which could cause generator to get overexcited and therefore overheating occurs.

Inversely, when generator voltage magnitude is too low, the MVAR will be absorbed and the generator becomes underexcited.

Now, since we are mostly focused on the frequency, this is by no means a small fry to ignore.

An elephant in the room if you will.

The sinusoidal frequency of the generator must be the same as the grid.
As one poster had alluded to an instrument that shows whether the generator and the grid are in the process of synchronizing themselves, this is called synchroscope. When the two systems are not synchronized the synchroscope will keep spinning.

When the two systems reach synchronicity it (synchroscope) will stop at 12:00 oclock that is also zero. This is the moment when you throw the switch on.

Throwing the switch before they get synchronized will cause the generator to suffer from being out of step and subsequently behave like a motor—not to mention the stator and rotor would be slipping. A potential destructive force.

An old-school way of checking sychronicity is using a light bulb. When there is voltage differential the bulb is lit—when the bulb goes dim synchronicity is achieved.

The same destructive force would be enforced upon the generator when it is running too fast. When this happens the grid will try to slow it down.
On the other hand running the generator too slow will prompt the grid to speed it up with the same destructive result.

Phase angle and phase sequence are also important considerations when synchronizing generator to the the grid. Details of which would be for another discussion.

These problems had been overcome with the introduction of mercury-arc valves (a glorified term for rectifier) which are now obsolete with the arrival of solid state GTOs (gate turn off thyristors) and IGBTs for high voltage inverters.
Generator voltage is first rectified to DC and then inverted to AC for ease of synchronization.


We are using this strategy in our California PACIFIC INTERTIE that supplies power to the largest municipal owned utility company (LADWP).
Click to expand...

Myspark:

Could you elaborate on how wind turbines solve this issue when the blades spin at different speeds?
 
Most frequently the wind turbine will either directly produce DC or will produce "wild" (varying frequency and voltage) AC which is rectified to DC.
In either case, the DC then drives a grid-interactive inverter (or possibly a motor-generator) which drives current into the grid at the existing grid frequency. A grid interactive inverter can relatively easily be designed to convert all available DC power into AC power fed back into the grid. When less power is available it simply sources less current into the relatively constant grid voltage. In a way, that could be considered as varying its output resistance. Or at least its optimal output voltage and output current combination.
 
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181202-2500 EST

Noswad4:

By manual or automatic control you can change the mechanical load a generator applies to its driving mechanical source. That is what a dynamometer does for engine testing or other source. Simplest to visualize with a DC generator.

To drive energy into the grid, or battery, or whatever the generator voltage will need to be greater than some voltage source in the load. When sending energy to the grid the grid is the load.

If you want to control the torque load on the turbine, then you would use some torque measurement device as the signal source to your servo system.

Greater torque load from the generator to its source will send more power to the grid.

Speed times torque times a constant is generator power input which is the same as turbine power output. Generator power input minus losses is generator power output.

.
 
Noswad4 said:
Myspark:

Could you elaborate on how wind turbines solve this issue when the blades spin at different speeds?
Click to expand...

Noswad4

There are two ways to control the output of the generator that is driven by wind turbine. By this, I'm referring to those turbine whose blades resemble airplane propellers. . . those huge turbines that you see in Palm Desert Ca.

These huge blades can be positioned to vary the pitch-- according to the changing velocity of the wind that hits the blades. Think helicopter blades.

Generally the air speed of 15m/s could provide optimum output of a wind turbine.

This varying pitch also has impact on the load carried by the aircraft.

OK, so let's set that aside for the moment and get back to your question on how to solve this issue involving the nuances caused by differing speeds of the turbine.

The two ways of control that can be applied in turbine generators are Closed Loop Control and the Open Loop Control Systems.

Wind is something that we can't control but we can harness (and control) the power derived from the energy of the wind.
The way turbines are made-- consist of a shaft that is coupled to the input shaft of the generator. The coupling between turbine and generator has a brake that can be either friction type or electro-magnetic.

These components are housed in an enclosure called nacelle.

For this closed loop scheme, the output of the generator is used as feedback signal to control the magnetic coupling. If the wind is too strong above the 15m/s air speed, a signal from the generator is used to adjust the electro magnetic force-- thus allowing the coupling to slip and slow down the input shaft of the generator. This scheme can also be applied in changing the pitch of the blade if such feature is equipped.

I won't cover the details how the pitch of the blades called "angle of attack" can affect the torque and speed of the turbine. One page of this post won't be enough. And besides it will be boring for a Chemical Engineer like you. :)

This signal is sent to the nacelle. This is done automatically.

In all through this event the wind remains the way it is. Recall that we can't control mother nature. (well up to a certain extent.)

The Open Loop Control scheme is really just someone (in the breathing human form) watching the instruments and adjusting dials and flipping switches to make make sure the generator output is held constant to acceptable levels.

In this type of control-- things are done manually.

No signal is being fed back from the generator other than performance and behavior of the machine when in operation-- and thus reported to the guy at the control station.
 
Thanks for the input Gold, Myspark, and Gar

Putting it in terms of a wind turbine to help me explain. Instead of using the blade pitch or a brake on the coupling it would be possible to control the speed at which the blades spin by having the generator connected to a dynamometer.

What I am still unclear about is if instead of a dynamometer you can use the grid interactive inverter that GoldDigger mentioned? This would allow the DC generator to feed more power into the grid the faster it spun, but it doesn't sound like it has the ability to control the torque load on the turbine, which in turn could control the speed of the blades.

Just to try and make myself clear. My goal is to control the mechanical load a generator applies to its driving force and send more power to the grid as the mechanical load applied is increased.

Thanks again for working with me. I know I am not doing a great job of describing.
 
181203-2213 EST

Noswad4:

A dynamometer is the combination of a generator and an adjustable load resistor.

When you connect an electrical generator to a mechanical power source, a steam engine for example, the torque from the generator to the mechanical power source is a function of the electrical load on the generator. If that electrical load on the generator can be adjusted, then you can vary the torque load from the generator to the mechanical power source. This is what is done in an engine dynamometer.

On the other hand if the generator electrical load is a relatively constant voltage of some value with some internal impedance, the electrical grid, then what we do to change the input torque, mechanical power input to the generator, is change the electrical loading on the generator by changing the input rpm to the generator, or by changing magnetic field intensity in the generator.

To transfer electrical power from the generator to the grid the internal generated voltage of the generator has to be greater than the grid voltage. The power transferred is determined by the voltage of the grid times the current flowing to the grid. The current flowing is determined by the difference voltage between the grid and the generator divided by the internal impedances.

To adjust this power flow one adjusts the generator speed which will change the generated voltage, or if speed is held constant change the magnetic field excitation which will change the generated voltage. Both can be changed.

How can you run an experiment to see what happens? Get an automotive DC generator from the junkyard. Connect this to a motor, find a battery to represent the grid. Get or make a variable DC source to provide a means to adjust the generator field excitation. You also want some meters to make measurements. Torque is harder to measure, but you could measure power input to the motor as an estimate of torque. Current measurement to the motor is not a good measure of torque. With a 1/2 HP motor you could use a Kill-A-Watt EZ from home depot for power measurement.

.
 
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