Small wind turbine in parallel with mains 3

[lyledunn]

May I bother you with a technical query slightly outside my capability? Consider a 1Kva wind turbine being fed on to a common single phase busbar also supplied by the grid via a standard 50Kva tx. Such a situation would be found in a typical domestic installation that has installed the micro wind turbine of the sort now widely available in retail stores. Lets suppose the load is resistive and rarely exceeds 1Kw for long periods. Can you advise if simple proportionality can be applied to assess the current drawn from the two sources (even as a rough guide) or is information required on the internal impedance of the respective sources? The kernel of my query lies in the claims made by the suppliers made about pay back period.

Regards,

Lyledunn

 

[davidbeach]

Assuming that the wind turbine is connected on the customer side of the meter, and all the interconnect requirements have been complied with, the load will use all of the power produced by the wind turbine up to 100% of the load. Any output from the wind turbine in excess of the available load will go out onto the utility.

 

[alehman]

I think some can be configured to operate in different modes, e.g. utilize all power locally, charge batteries or export.

You could talk with one of the manufacturers of such products. Some units are provided with specialized electronic controllers that handle the interconnection and load balancing.

 

[lyledunn]

Thanks for the replies. There is no utility buy back and no batteries to charge. The point is that the load in my home rarely exceeds 1Kw thus with 2 available emf sources connected in parallel and if we suppose each had the same internal impedance then the load would be shared equally between the two sources. Consequently my utility meter would still register for as long as the situation prevailed. The pay back period being quoted by the turbine supplier seems to disregard the fact that the utility supply is constantly present.

Regards,

Lyledunn

 

[dpc]

In an ac system, it's mainly the phase angle that causes power transfer, not the voltage or impedance. The wind turbine doesn't know about power sales agreements with the utility and it doesn't even know whether you are connected to the utility or not. If the wind blows hard enough, the wind turbine power output will equal your consumption and there will be no net power flow. If the winds blows harder, power will flow back to utility, unless there is relaying installed at the interface point to shutdown the wind turbine or reduce its output. You just won't be getting paid for it.

It is possible to install a tie controller that could regulate the turbine output to avoid net power export, but I suspect it won't catch it all.

Also, if you have an induction generator, you may have real power going in one direction and reactive power going in the other.

 

[itsmoked]

I believe running the two as you're describing in parallel is fraught with problems. Motorizing, wrong sharing, phase locking, rat-you-out-meters, legality, etc.

You should maybe consider running the wind generator into a battery bank and then running a large,(now relatively inexpensive), inverter out. Then you will decouple the turbine from the grid and gain some backup capability. Selectively run appropriate circuits in your house with circuit sized transfer switches.


Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[lyledunn]

DPC many thanks for you reply. I am sorry to be thick but are you saying that if say my demand was 1Kw and sufficient wind was available for the turbine to output 1 kw, my utility meter would not be registering energy demand?

Regards,

Lyledunn

 

[itsmoked]

No he's saying it depends on what the phase relationship is between the utility and the turbine. Whichever is leading by some infinitesimal amount will be doing most or all of the work. (running the loads) If the phase relationship exceeds a certain point then you could be running the turbine with the utility.. Not much savings a-what? This would be; you not needing any power, but the meter showing 1kW because you are running a fan on a tower.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[davidbeach]

Yes, the wind turbine would supply the load. If the mechanical side is putting 1kW of power into the shaft of the generator, the generator will run at a phase angle sufficiently ahead of the utility for 1kW to flow from the generator. The wind turbine/generator will produce the power it produces based on the wind conditions and that power will enter the electrical system to be utilized by the load or to flow further out onto the utility system.

 

[dpc]

Yes, that's basically what I'm saying. If your turbine is putting out 1000 watts and you are tied to the grid, that power is going to go somewhere. Does it go into your house or does it go to the utility? That depends on a lot of things, but as far as your power bill is concerned, it makes no difference. The *net* power measured at your meter will decrease by whatever power the windmill can make.

If the windmill is producing more than you are using, the power flows back to the utility, regardless of what your contract says.

 

[waross]

It may be easier to describe the action of a synchronous diesel generator.
We are going to run a small diesel generator in parallel with the grid.
First we run it up to rated speed, say 1800 rpm.
Then we match the voltage.
Then we watch the synchroscope and close the breaker at the correct time.
Now the governor/throttle is set to provide enough fuel to maintain the set at 1800 RPM.
The generator is not supplying or drawing any power.
However the generator is now locked to the utility power. If we reduce the fuel the engine will not slow down but the utility will supply enough power to keep the set running at 1800 rpm.
If we increase the fuel feed the set will not speed up but will supply power to the grid.
The power out of a generator in parallel with the grid depends only on the throttle setting. The more power the engine produces the more power will be supplied to the grid.
dpc mentioned phase angle. This is correct. The generator shaft moves a few degrees (or in some cases only part of a degree) ahead of the synchronous position, and when this is converted into electrical degrees, this is the phase angle.
When the set is being motored the generator shaft position is a little behind the synchronous position. When the set is producing power the generator shaft is a little ahead of the synchronous position.
Power out depends only on power in (plus losses).
The exact small number of degrees is academic and is only considered in the rare cases when someone wants to run two synchronous frequency convertors in parallel, run two generators on one shaft and synchronize them or something similar.
What does this mean to a wind generator?
The amount of power depends still on the power in. The harder the wind blows, the more power the wind generator will put out.
An induction generator does not lock on like a synchronous generator does but slips. Now instead of a few degrees of mechanical displacement we are talking about several RPM of mechanical slip. However, the phase displacement of the resulting power will be the similar few degrees of leading phase angle.
The harder the wind blows, the more power your wind machine will produce and a corespondingly less amount of power will flow through your power meter.
Similar to a diesel generator a wind turbine may be motored.
If the wind does not keep the wind turbine above synchronous speed, the utility will supply the power to keep it turning at near synchronous speed. As itsmoked pointed out, you now have a large fan that is quite expensive to run.
itsmoked makes some good points about other issues.
remember, the harder the wind blows the more power your set puts out.
The farther you pull the throttle on a diesel set, the more electrical power it puts out.
respectfully

 

[davidbeach]

Nice explanation waross.

 

[LionelHutz]

If your turbine is putting out 1000 watts and you are tied to the grid, that power is going to go somewhere.

This describes it in one sentence. If you don't use it in the house then your meter runs backwards. Then, if you use more power than your turbine produces your meter runs forwards again. You should never just unload a wind turbine so you can't just "keep" the power inside your house.

At that power level I take it you're talking about a permanent magnet DC windmill with a grid-tied inverter. That seems to be the standard in the 1kW size.

The biggest determinator in the payback period will be figuring out your on-site wind speeds and power outputs. You typically calculate the (power for that wind speed) times (# of hours of that wind speed per year) to get the kWh of production in a year at each speed. Do it for each wind speed from the cut-in speed to max output speed, ie 2-4m/s, 4-6m/s etc etc and then add up to get the total kWh per year. Subtract some losses for cables and the grid-tied inverter to get it more accurate. Multiply by the going utility rate and you get your $$ per year figure.

 

[tinfoil]

I am aware of some utilities that will not or legally can not purchase from small generators. So they install revenue meters that only measure the 'first quadrant' observed by the revenue meter - in other words, while your windmill output is less than your house load, you will be paying for the energy difference, but while your windmill output is greater than your house load, your revenue meter would ignore the 'export' and call it zero. (It's still a cheaper arrangement for you than putting in equipment to prevent exports, and the utulity gets a bit of free power).

It depends on your local regulatory environment.

Some jurisdictions require a 'generator's license' just to be able to hook one up.

 

[vtpower]

Every state has different net metering rules (assuming your in the US) Here, we allow customers to install a unit at their residence, such as a windturbine up to 15KW,and run in parallel with the system. As everyone has mentioned, if you are generating more energy than the house is consuming, you will push back to the utility. In our case the meter actually does spin backwards, and registers this as negative energy flow, or essentially banks it and allows you to apply it back to your home when you are consuming from the utility. At the end of the year the slate is wiped clean and the customer starts at 0 again.

 

[Freelance00]

warcross did a very good job answering many of my questions on this topic. I would like to concentrate now on the apparent voltage difference and phase angle when any two ac generators of the same output voltage are connected in parallel with each other and also in parallel with a load. It seems to me that the two generator outputs at their common connection point (the load) must have identical phases and voltages. Otherwise we would know that they are not truly in parallel. Just by monitoring the common voltage point, we are not able to tell if a selected generator is sinking or sourcing power. Internal to both generators there is undoubtedly some equivalent output impedance that allows the "internal" ideal generator of one generator to be slightly different from that of the other "internal" ideal generator because the internal impedances allow for a voltage drop and a phase angle difference between the two "internal" generators. If by measurement of the output currents we determine that Gen. A is sourcing current and that Gen B is sinking current, we surmise that the driving force to Gen. A is greater than the driving force to Gen. B for identical generators. In that case, Gen. B may be acting as a motor rather than as a generator. Also note that the circulating current between Gen. A and Gen B may be far greater than the current to the parallel load since it is limited only by the two internal impedances and the voltage difference between the two internal generators. So what bothers me is that I don't have a good explanation for how a nation-wide Electric Grid works. Who (or what) is in charge of the amplitude and phase of the the Grid system? I can only envision the potential (hah!) for very heavy circulating currents and instability on the Grid unless there is some very tight control. Who does the controlling? Can someone explain this phenomina to me in simple terms? I am certainly not an expert here. For a reference, see: "What's Wrong with the Electric Grid?" at

http://www.aip.org/tip/INPHFA/vol-9/iss-5/p8.pdf.

 

[davidbeach]

Power flows from a point of leading phase angle to lagging phase angle. If I run my generator and apply mechanical torque into the shaft of the generator, that torque will cause the phase angle of the generator to increase (become more leading) to the point that the phase angle difference between the generator and the surrounding system is such that the generator can supply an amount of power to the system equal to amount of power being applied to its shaft, less losses. Internal impedance only matters if you are starting with the internal (air gap) voltage of the generator; you can get close enough for understanding the concept strictly using terminal voltage and ignoring internal impedance - the internal impedance is much more important in fault calculations.

The surrounding system sets the general voltage level and frequency, but your generator will raise the local voltage level some. Most voltage control is very distributed, load tap changers on transformers and excitation controls on generators being run in voltage control mode - which a small generator run in parallel with the utility is not - with some supervisory control. Properly set up generator control systems allow generators to run in parallel with each other and in parallel with the outside world without any circulating current. Bad setup will have mismatched power flows and circulating or oscillating VAr flows, but it need not be that way.

 

[waross]

A standard component of a control scheme for a generator that will be run in parallel with other generators is a reverse power relay. If the input power or torque driving a generator drops below the level needed to sutain the no-load losses and the generator starts to motor, the reverse power relay opens the breaker to disconnect the generator and in many installations stops the prime mover. Don't worry about generators motoring.
When generators are run in parallel the governors control the speed and thus the frequency. A characteristic of governors called droop allows the generators to share the load even if the load is changing.
The excitation setting on one set determines how many VARs the set will produce and to a much smaller extent the system voltage. You adjust the excitation on one set to adjust the power factor of that individual set. You adjust the excitation of all sets to adjust the system voltage.
In the case of a small set feeding into the grid, the set will not be able to affect the voltage of the grid and the excitation adjustment will determine the power factor of the generator only.
Note, with a single generator, the load determines the power factor. The load will demand a number of watts and a number of VARs. The generator will have no choice but to supply both watts and VARs.
When generators are in parallel the watts may be supplied by one generator and the VARs may be supplied by the other generator.
A single generator must run at a power factor determened by the load.
Parallel generators may run at different power factors but the system power factor will still be determined by the load.
In a small plant with generators operated in parallel there is a circuit that is sometimes called a quadrature circuit that automatically balances the excitation so as to balance the VAR output of each governor.
To shift the watts from one generator to the other, adjust the power input by adjusting the governors.
To shift the VARs from one generator to the other adjust the excitation.
Measuring the Watts output of a generator is the accepted way of determining the load.
I agree completely with David's post. David has given a good technical explanation. I have tried to give an explanation from the perspective of the operator on the floor of a small generating plant.
I hope both explanations are helpful.
respectfully

 

[davidbeach]

One small addition/clarification/amplification to waross's excellent post - The operational mode for generator(s) islanded with a load are quite different from the operational mode for generator(s) running parallel with the utility.

Standalone: The governor will control speed directly and real power indirectly - it takes enough power to maintain speed. The excitation system will control voltage directly and reactive power indirectly - the generator will supply however much reactive power it needs to to maintain voltage.

Parallel with utility: The governor has no control over speed (close enough to true for this discussion) and directly controls the real power output. The excitation system has essentially no control of output voltage and directly controls the reactive power output/input.

Multiple small generators - standalone: Two independent controls balance real power (governors) and reactive power (excitation) using various techniques, droop is the most common.

Multiple small generators - parallel with utility: Same as single small generator parallel with utility.

Big utility power plant generators are run in a a mode very similar to the "multiple small generators - standalone." Here they can have some influence on system frequency (speed) and voltage, but a small generator (say under 20MW, but YMMV) paralleled to the utility will not move the voltage or the frequency.

 

[resqcapt19]

David,

Big utility power plant generators are run in a a mode very similar to the "multiple small generators - standalone." Here they can have some influence on system frequency (speed) and voltage, but a small generator (say under 20MW, but YMMV) paralleled to the utility will not move the voltage or the frequency.

I don't know much about the workings of the US power transmission grid, but it is my understanding that there are 3 "interconnects" and the frequency within the interconnect is the same at all points within that interconnect. If my understanding is correct, then to change the frequency, the generator would have to output enough power to change raise the frequency of the interconnect. What am I missing? Thanks.
Don