I am considering installing a windmill generator to my main service panel, connecting to my utility using a bi-directional meter. I am a little confused with the concept. My windmill generator is located approx. 500' away from my service panel and the utility transformer is approx. 150' away from the panel. The windmill inverter will match freq/voltage so I can tie directly with the utility. My confusion is if there is higher line resistance between the windmill and panel than the utility transformer and panel, how much of a load is drawing from the windmill as opposed to the utility considering the difference in line resistance? I realize wire sizing is important. Is the load divided between the two sources of power? Any feedback is appreciated.
windmill power
- Thread starter Ken 6789
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dmagyar
Senior Member
- Location
- Rocklin, Ca.
Any transfer switch?
Any transfer switch?
My guess is that it won't matter if the distance between the panel and the windmill is farther than the distance from the panel to the utility drop. Whatever amount of power your windmill can supply is power that you won't have to take from the utility. What I'd be more concerned about though is if the power goes out, you'll have the ability to backfeed the utility lines and anyone working on the lines has a good chance of getting injured. You didn't mention if your inverter is designed like those in use for PV setups. It thats the case then if the power goes out, so does your potential for backfeeding the utility lines. If not then you'll need something to stop that from happening and it has to be an automatic fail open transfer or contactor.
Any transfer switch?
My guess is that it won't matter if the distance between the panel and the windmill is farther than the distance from the panel to the utility drop. Whatever amount of power your windmill can supply is power that you won't have to take from the utility. What I'd be more concerned about though is if the power goes out, you'll have the ability to backfeed the utility lines and anyone working on the lines has a good chance of getting injured. You didn't mention if your inverter is designed like those in use for PV setups. It thats the case then if the power goes out, so does your potential for backfeeding the utility lines. If not then you'll need something to stop that from happening and it has to be an automatic fail open transfer or contactor.
brian john
Senior Member
- Location
- Kilmarnock, Va
- Occupation
- Retired after 52 years in the trade.
But the controller is at the panel?
The windmill has a safety feature if it senses utility power is out, a mechanism disengages it preventing backfeed to the utility. There is no transfer switch because it's designed to sell back power through a bi-directional meter. Where I am confused is how does the panel loads recognize what is windmill power vs utility power and where to draw power from. The windmill connects to the inverter to match voltage/freq. and to a regular 2 pole circuit breaker in service panel.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100115-2126 EST
Ken 6789:
You need to study electrical theory a little.
Consider the following setup:
1. Voltage source V1 (the utility).
2. An internal impedance between V1 and the main panel.
3. Voltage source V2 (windmill).
4. An internal impedance between V2 and the main panel.
5. A load on the main panel.
If V2 is greater than V1 then some power will flow from V2 to the load and/or V1. If V2 has enough power available to more than supply the load, then all the power to the load will be from V2 and the excess will flow to V1.
There will be power loss in both internal impedances.
If V2 can not supply all the power required by the load, then some power will come from V1.
For V2 to deliver its maximum power it has to have the ability to adjust V2 to achieve this operating point.
I do not know how your system detects loss of V1.
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Ken 6789:
You need to study electrical theory a little.
Consider the following setup:
1. Voltage source V1 (the utility).
2. An internal impedance between V1 and the main panel.
3. Voltage source V2 (windmill).
4. An internal impedance between V2 and the main panel.
5. A load on the main panel.
If V2 is greater than V1 then some power will flow from V2 to the load and/or V1. If V2 has enough power available to more than supply the load, then all the power to the load will be from V2 and the excess will flow to V1.
There will be power loss in both internal impedances.
If V2 can not supply all the power required by the load, then some power will come from V1.
For V2 to deliver its maximum power it has to have the ability to adjust V2 to achieve this operating point.
I do not know how your system detects loss of V1.
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rcwilson
Senior Member
- Location
- Redmond, WA
How does the water in the pipes know whether it supposed to run to the sink or the shower? Of course, it doesn't. The water main or tank supplies the pressure and the water flows wherever it can.
Same with power. Kilowatts always flow to the load. If your windmill is making more watts than your house is using, the excess goes to the utility, spinning the meter backwards, paying you money. If the panel loads need more watts than the windmill can supply the utility supplies the rest.
Using the water analogy, say your windmill pumped water from a well to a tank on your roof that was high enough to put 10 psi pressure on your water pipes and the water main ran at 10 psi also. When your srinklers turn on, the tank starts draining so the pressure drops below 10 psi and drops water flows from the water main. When no water is being used the tank fills up higher making the pressure higher and the tank water flows into the water main so your neighbor can use it.
The windmill is the same. The different cable impedances of the windwill and utility feeds only affect the losses and not the flow of power.
(A problem with this analogy is it can lead to you thinking that pressure = voltage so if the windmill's volts are higher, power flows to the utility. Not true. Kwatts flow to the loads and vars flow from high voltage to low voltage. Your meter only reads watts, not amps or vars.)
Your windmill/inverter makes watts, if your loads don't suck them up, your neighbors loads will. Simple addition: Power to utility = (windmill power) - (panel loads). If loads are greater than windmill power, power to utility is negative meaning power is coming into your house.
Same with power. Kilowatts always flow to the load. If your windmill is making more watts than your house is using, the excess goes to the utility, spinning the meter backwards, paying you money. If the panel loads need more watts than the windmill can supply the utility supplies the rest.
Using the water analogy, say your windmill pumped water from a well to a tank on your roof that was high enough to put 10 psi pressure on your water pipes and the water main ran at 10 psi also. When your srinklers turn on, the tank starts draining so the pressure drops below 10 psi and drops water flows from the water main. When no water is being used the tank fills up higher making the pressure higher and the tank water flows into the water main so your neighbor can use it.
The windmill is the same. The different cable impedances of the windwill and utility feeds only affect the losses and not the flow of power.
(A problem with this analogy is it can lead to you thinking that pressure = voltage so if the windmill's volts are higher, power flows to the utility. Not true. Kwatts flow to the loads and vars flow from high voltage to low voltage. Your meter only reads watts, not amps or vars.)
Your windmill/inverter makes watts, if your loads don't suck them up, your neighbors loads will. Simple addition: Power to utility = (windmill power) - (panel loads). If loads are greater than windmill power, power to utility is negative meaning power is coming into your house.
rcwilson
Senior Member
- Location
- Redmond, WA
To answer the question about how the inverter detects loss of utility, it senses voltage and frequency. During normal operation the utiltiy keeps the windmill or generator lcoked onto 60 Hz and 240 V. When the incoming utility is lost, the frequency starts changing.
If windmill is > than load, the frequency goes up, if freqeuncy is less than load, the frequency drops. Inverters and generators are required to disconnect if frequency or voltage goes outside of a tight band. Time delays prevent nuisance trips.
If windmill is > than load, the frequency goes up, if freqeuncy is less than load, the frequency drops. Inverters and generators are required to disconnect if frequency or voltage goes outside of a tight band. Time delays prevent nuisance trips.
size matters
size matters
how big do you want to go?
Vestas V82 1.8 MW, school has one
vindkraftverk will draw power 5kw approx until sustained wind speed has reached minimum, rotor is at speed, then PFC system Power Factor Correction system essentially turns on allowing produced power to match grid performance
600V sent down tower to PFC and then transformer
GE's system features pdf
your power goes to the grid, and you draw off the grid at the same time
consider your PPA options Power Purchase Agreement
spinning the meter backwards, net billing is great for them, lousy for you
you want to sell your produced power at retail or better rates
others are avoided cost, and wholesale rates
wholesale rates may drop to nil if there's surplus of wind and deficit of demand
next time, please indicate what size and type you want
size matters
how big do you want to go?
Vestas V82 1.8 MW, school has one
vindkraftverk will draw power 5kw approx until sustained wind speed has reached minimum, rotor is at speed, then PFC system Power Factor Correction system essentially turns on allowing produced power to match grid performance
600V sent down tower to PFC and then transformer
GE's system features pdf
your power goes to the grid, and you draw off the grid at the same time
consider your PPA options Power Purchase Agreement
spinning the meter backwards, net billing is great for them, lousy for you
you want to sell your produced power at retail or better rates
others are avoided cost, and wholesale rates
wholesale rates may drop to nil if there's surplus of wind and deficit of demand
next time, please indicate what size and type you want
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100116-1056 EST
Denis:
I would not describe the power from the windmill going to the grid and then coming back under the following condition:
1. The windmill can not supply all the power required by the load.
All power from the windmill goes to the load plus additional power from the grid goes to the load. Analyze the V1 and V2 discussion in my previous post.
Suppose your load power is zero at night from 0000 to 0800, and your windmill produces X KWH fed into the grid. Then during the day from 0800 to 2400 the windmill produces 0 KWH output, but you consume X KWH. In this case you supplied energy to the grid at night and during the day you used energy from the grid. But the energy you supplied at night is not the energy you used during the day.
Now suppose you have a battery bank that is charged during the night and you draw energy from the battery during the day. Then the energy you generated at night is the energy you used during the day.
If the windmill can produce more power than the load, then the excess power from the windmill goes to the grid, and all the load power is supplied from the windmill.
If the windmill connects to the grid thru its own transformer, then all the windmill power goes to the grid. Any power required by the load will be obtained from the grid thru a different transformer than the windmill transformer.
Probably the real demarcation point is to which side of the load meter is the windmill connected.
A somewhat different subject. Why the power company wants to pay you less for power you supply to them than power you buy from them. Very roughly speaking the energy you buy from the power company is worth about about 1/3 of the bill. Another 1/3 is the distribution system. And another 1/3 is overhead. Obviously this breakdown is an overall system approximation.
If I am a very large industrial plant, like the Ford Rouge Plant, right next door to the generating plant, then distribution costs are very low. In the case of the Rouge Plant they previously had their own power plant. Now they have contracted with DTE to build a generation plant adjacent to the Rouge for their power supply. I believe that Dow Chemical did the same thing with Consumers Power.
.
Denis:
I would not describe the power from the windmill going to the grid and then coming back under the following condition:
1. The windmill can not supply all the power required by the load.
All power from the windmill goes to the load plus additional power from the grid goes to the load. Analyze the V1 and V2 discussion in my previous post.
Suppose your load power is zero at night from 0000 to 0800, and your windmill produces X KWH fed into the grid. Then during the day from 0800 to 2400 the windmill produces 0 KWH output, but you consume X KWH. In this case you supplied energy to the grid at night and during the day you used energy from the grid. But the energy you supplied at night is not the energy you used during the day.
Now suppose you have a battery bank that is charged during the night and you draw energy from the battery during the day. Then the energy you generated at night is the energy you used during the day.
If the windmill can produce more power than the load, then the excess power from the windmill goes to the grid, and all the load power is supplied from the windmill.
If the windmill connects to the grid thru its own transformer, then all the windmill power goes to the grid. Any power required by the load will be obtained from the grid thru a different transformer than the windmill transformer.
Probably the real demarcation point is to which side of the load meter is the windmill connected.
A somewhat different subject. Why the power company wants to pay you less for power you supply to them than power you buy from them. Very roughly speaking the energy you buy from the power company is worth about about 1/3 of the bill. Another 1/3 is the distribution system. And another 1/3 is overhead. Obviously this breakdown is an overall system approximation.
If I am a very large industrial plant, like the Ford Rouge Plant, right next door to the generating plant, then distribution costs are very low. In the case of the Rouge Plant they previously had their own power plant. Now they have contracted with DTE to build a generation plant adjacent to the Rouge for their power supply. I believe that Dow Chemical did the same thing with Consumers Power.
.
cadpoint
Senior Member
- Location
- Durham, NC
Respectfully if someone said anything that I'm about, I missed it... Sorry :\
What does the POCO have to say about this and their requirements!
Google, Images; Using, "electrical diagram wind mill" will give you plenty of show and tell of what your thinking about doing.
See my first, two comments.
There's a whole lot of what if's there! IMO. I think of a wind mill in the simple sense first as you keep the power till you can deposit it to the POCO.
Don't get me wrong it can be straight dumping of power, but the POCO is going to make you match their requirements placed on the distrubtion equipment.
I'd all so to suggest more search engine research on *.edu sites, and books addressing your exact application and repeating a big shout out to the POCO.
Sounds great, I wish I had the land and ungreen as it sounds less trees... I can bearly go Solar due to former, and the ratio payout...
But then again you haven't seen the back yard, nor I! LOL!
What does the POCO have to say about this and their requirements!
Google, Images; Using, "electrical diagram wind mill" will give you plenty of show and tell of what your thinking about doing.
See my first, two comments.
There's a whole lot of what if's there! IMO. I think of a wind mill in the simple sense first as you keep the power till you can deposit it to the POCO.
Don't get me wrong it can be straight dumping of power, but the POCO is going to make you match their requirements placed on the distrubtion equipment.
Thats what we do, day in and day out, worry about the Wire.
I'd all so to suggest more search engine research on *.edu sites, and books addressing your exact application and repeating a big shout out to the POCO.
Sounds great, I wish I had the land and ungreen as it sounds less trees... I can bearly go Solar due to former, and the ratio payout...
But then again you haven't seen the back yard, nor I! LOL!
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mivey
Senior Member
A great point that is often missed by the customer.
Another thing they fail to get is that the bill most get has an average energy cost. A lot of times the energy the customer wants to pump onto the grid at night has less economic value than the energy they want to use during the day or on average across the month.
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I was involved with water powered generators and the utility had very strict requirements regarding the system protection plan. The amount they paid for KWH's delivered to the grid was the amount it cost them to produce a KWH.
So if they were charging 0.10 per KWH and they produced it for 0.04 per KWH, they paid the customer 0.04 cents.
I think your installation may have a problem pumping KWH's back into the grid.
Looking back at the generator I see a higher impedance than the utility system. The utility system is usually very stiff and maintains the voltage level.
Unless you can provide at least the same voltage level as the utility does under load, your system will not supply many KWH's to your home.
Some one correct me if I have misspoken.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100116-1918 EST
bob:
I disagree.
The grid is quite stiff, but the pole transformer may not be. My pole transformer has a much greater impedance than the wires from the transformer to the main panel.
Suppose the wind generator can supply more power at the end of its supply line to the main panel than is consumed by the load on the panel. This means that the voltage at the main panel from the wind generator is greater than or equal to the open circuit voltage of the transformer. If the wind supplied voltage is greater, then some power flows to the grid.
Even if the grid is very stiff and a low impedance coupling exists you can supply power to the grid.
..
bob:
I disagree.
The grid is quite stiff, but the pole transformer may not be. My pole transformer has a much greater impedance than the wires from the transformer to the main panel.
Suppose the wind generator can supply more power at the end of its supply line to the main panel than is consumed by the load on the panel. This means that the voltage at the main panel from the wind generator is greater than or equal to the open circuit voltage of the transformer. If the wind supplied voltage is greater, then some power flows to the grid.
Even if the grid is very stiff and a low impedance coupling exists you can supply power to the grid.
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