Solar Panels on residential Homes

[mull982]

The other day I was driving through a neighborhood and noticed several averaged sized homes with solar panels covering their rooftops. I began to wonder weather or not these solar panels were enough to power the entire homes. I also wondered if they provided enough power to also supply additional power back to the grid.

My questions are:

1) Can solar panels on a rooftop be enough to power an entire home typically? I'm sure this depends on the number of panels but I'm curioius what the typical setup is capable of supplying.

2) Can typical resedential rooftop solar panels supply enough power to power the home as well as put excess power back onto the Utility grid?

3) Can the panels provide excess power back onto the grid at the same time they are supplying the loads in the house? How does this scheme work supplying power to home and back to grid simoultaneously?

Any links or additional information to this topic would be greatly appreciated.

 

[iwire]

1) Can solar panels on a rooftop be enough to power an entire home typically? I'm sure this depends on the number of panels but I'm curioius what the typical setup is capable of supplying.

No, not when the people are home and are typical power hogs.

Each panel is only about 200 to 300 watts.

2) Can typical resedential rooftop solar panels supply enough power to power the home as well as put excess power back onto the Utility grid?

Sure when the folks are at work and they do not leave heavy loads on.

3) Can the panels provide excess power back onto the grid at the same time they are supplying the loads in the house? How does this scheme work supplying power to home and back to grid simoultaneously?

Yes it can happen, the systems are connected to the grid via a 'Grid Interactive Inverter' and any power produced in excess of the homes needs feeds into the grid.

College of Engineering Southwest Technology Development Institute

College of Engineering Southwest Technology Development Institute Codes & Standards

http://www.nmsu.edu/~tdi/pdf-resources/pdf version divided PV:NEC/PV-NEC 1.91/PV-NEC-V-1.91-opt.pdf

The North American Board of Certified Energy Practitioners (NABCEP)

 

[gar]

100922-0829 EST

mull982:

In Michigan I am told by several sources that a 4 KW rated array in a fixed location south facing at an optimum inclination would have an annual production of about 4400 KWH. In others words multiply the W rating by 1.1 to get the yearly KWH. If a tracking array is used, then I believe the output is about 10% greater. A ball park figure is $30,000 for a 4 KW system. That is $6.82 per KWH if written off in one year, or at 0.13/KWH (our present rate) 52.4 years. Current incentives reduce this considerably.

However, personal property tax would probably be about $500 to $1000 per year. 4400 KWH costs $572 per year at $0.13 per KWH. So I would pay taxes to the city on the equipment about equal to what I would have paid directly to the power company for the electricity. It does not compute in favor of solar panels even with incentives.

In the future power company power will cost more because the government is forcing it higher in various ways. So where is the balance I do not know? I am not rushing to put a large solar array on my house. I would need over a 12 KW system to meet my average needs.

.

 

[mull982]

Yes it can happen, the systems are connected to the grid via a 'Grid Interactive Inverter' and any power produced in excess of the homes needs feeds into the grid.

College of Engineering Southwest Technology Development Institute

College of Engineering Southwest Technology Development Institute Codes & Standards

http://www.nmsu.edu/~tdi/pdf-resources/pdf version divided PV:NEC/PV-NEC 1.91/PV-NEC-V-1.91-opt.pdf

The North American Board of Certified Energy Practitioners (NABCEP)

Thanks for the info. I guess my question is in theory with the solar panels connected to the grid how does the resetential system know to push power out to the grid rather than pull it in from the grid? So if two sources (utility and solar panels) are both tied together then why does the home take all of its power from the solar and none from the grid if avalaible. How is it in theory that the excess power is able to be pushed out to the grid?

Does this have something to do with power system stability?

 

[iwire]

I guess my question is in theory with the solar panels connected to the grid how does the resetential system know to push power out to the grid rather than pull it in from the grid?

I have been involved in a few installations and have wired the inverters but as far as how it actually works is a bit beyond me.

I just accept that it does, perhaps Gar could explain the details.

 

[BretHeilig]

Thanks for the info. I guess my question is in theory with the solar panels connected to the grid how does the resetential system know to push power out to the grid rather than pull it in from the grid? So if two sources (utility and solar panels) are both tied together then why does the home take all of its power from the solar and none from the grid if avalaible. How is it in theory that the excess power is able to be pushed out to the grid?

Does this have something to do with power system stability?

It's just basic electrical theory. The inverter develops a voltage, and that voltage seeks a path to discharge itself. If there is no such path in the nearest load center (ie all switches in the house are off, etc.), the next nearest path is probably in the next-door neighbor's load center, which is connected in parallel through the service lateral. On the way, the flow of current (which is heading out of the house instead of into it) causes the meter to spin backwards.

 

[gar]

100922-1051 EST

mull982:

It is simple electrical theory.

Consider a car battery and automotive electrical system.

When the engine is off and generator or alternator is not supplying power, then there has to be a device, called a cutout in the old days, between the battery and the generator to prevent the generator from becoming a load on the battery when the generator voltage is below the battery voltage. As soon as the generator voltage is equal to or greater than the battery voltage then current is allowed to flow from the generator to the electrical system.

Actually it is somewhat below battery voltage when current starts to flow from the generator. This is a function of the internal impedances of the battery and the generator interacting with the load impedance.

Today the cutout function is performed by the rectifier diodes in the alternator. In the old generator days it was performed by a relay. It could have been performed by a diode, but no low voltage drop high current diodes existed. But with a generator you still would not use a diode because of the loss of power in the diode. Diodes are inherently a requirement in an alternator for rectification of the alternators AC output.

Once the generator voltage is high enough for current to flow out of the generator, then the internal impedances of the generator and battery determine the distribution of current from each source to the load.

Once the generator voltage becomes high enough that thru its internal impedance it can supply all of the load current, then the current from the battery becomes zero.

As you raise the generator voltage further, then current above that required for the load will flow to the battery charging it.

The same thing happens in a grid-tied solar or wind system.

.

 

[flatlander]

100922-0829 EST

mull982:

In Michigan I am told by several sources that a 4 KW rated array in a fixed location south facing at an optimum inclination would have an annual production of about 4400 KWH. In others words multiply the W rating by 1.1 to get the yearly KWH. If a tracking array is used, then I believe the output is about 10% greater. A ball park figure is $30,000 for a 4 KW system. That is $6.82 per KWH if written off in one year, or at 0.13/KWH (our present rate) 52.4 years. Current incentives reduce this considerably.

However, personal property tax would probably be about $500 to $1000 per year. 4400 KWH costs $572 per year at $0.13 per KWH. So I would pay taxes to the city on the equipment about equal to what I would have paid directly to the power company for the electricity. It does not compute in favor of solar panels even with incentives.

In the future power company power will cost more because the government is forcing it higher in various ways. So where is the balance I do not know? I am not rushing to put a large solar array on my house. I would need over a 12 KW system to meet my average needs.

.

Are you for real aboutpersonal property taxes

 

[qcroanoke]

Are you for real about personal property taxes

Yeah, really! Do they tax solar panels as personal property? Just like they would your vehicle?
In Va. vehicles are personal property and are taxed as such.

 

[flatlander]

Yeah, really! Do they tax solar panels as personal property? Just like they would your vehicle?
In Va. vehicles are personal property and are taxed as such.

Do they tax other items attached to the home?
Is this in addition to real property tax?

Wow this is pure insanity.

 

[gar]

100922-1310 EST

In the State of Michigan -- YES.

I just called the accessor to get a rough idea. Here the market value divided by two is the taxable valuation. Per year we pay about $46/1000 of taxable valuation in taxes. The only question is how much does a given amount of solar power addition increase the value of the property. Is it one to one or some other ratio?

If I were to install enough solar to offset my electrical usage it take at least $100,000 of investment. If the above ratio was one to one, then I would be taxed on $50,000 or $2300 per year. My electric bill for 2009 was $1791.21 for 13,792 KWH or an average of 37.8 KWH/day.

Under these conditions I would lose $500/year just on taxes without considering anything else.

If by adding solar to a house it did not change the market value, then there would be no tax penalty.

For Michigan I should not have used the term personal property but real property when it is attached to the home. Mounted on a trailer as a portable unit I believe I would have no tax on it in Michigan.

.

 

[qcroanoke]

100922-1310 EST

In the State of Michigan -- YES.

I just called the accessor to get a rough idea. Here the market value divided by two is the taxable valuation. Per year we pay about $46/1000 of taxable valuation in taxes. The only question is how much does a given amount of solar power addition increase the value of the property. Is it one to one or some other ratio?

If I were to install enough solar to offset my electrical usage it take at least $100,000 of investment. If the above ratio was one to one, then I would be taxed on $50,000 or $2300 per year. My electric bill for 2009 was $1791.21 for 13,792 KWH or an average of 37.8 KWH/day.

Under these conditions I would lose $500/year just on taxes without considering anything else.

If by adding solar to a house it did not change the market value, then there would be no tax penalty.

.

Ok, In Va. Real Estate taxes are your home and the land it is on.
Personal property is your vehicles including trailers, boats and motorcycles.
But we mustn't linger on this topic lest we be chastised for getting off topic!

 

[mull982]

mull982:

It is simple electrical theory.

Consider a car battery and automotive electrical system.

When the engine is off and generator or alternator is not supplying power, then there has to be a device, called a cutout in the old days, between the battery and the generator to prevent the generator from becoming a load on the battery when the generator voltage is below the battery voltage. As soon as the generator voltage is equal to or greater than the battery voltage then current is allowed to flow from the generator to the electrical system.

Actually it is somewhat below battery voltage when current starts to flow from the generator. This is a function of the internal impedances of the battery and the generator interacting with the load impedance.

Today the cutout function is performed by the rectifier diodes in the alternator. In the old generator days it was performed by a relay. It could have been performed by a diode, but no low voltage drop high current diodes existed. But with a generator you still would not use a diode because of the loss of power in the diode. Diodes are inherently a requirement in an alternator for rectification of the alternators AC output.

.

So is there a similar cutout switch installed in these solar panel to grid interfaces. I'm thinking there cant be due to the fact the solar panels alone might not be enough to deliver the required power.

Once the generator voltage is high enough for current to flow out of the generator, then the internal impedances of the generator and battery determine the distribution of current from each source to the load.

So do the two sources share the loads in proportion to their source impedances similar to two transformers in parallel with different impedance values? So if both the utility and solar panels have the same source impedance values then the loads in the house will split between these two sources?

Once the generator voltage becomes high enough that thru its internal impedance it can supply all of the load current, then the current from the battery becomes zero.

Can you explain this one further I'm not seeing it. Why if the loads split between the sources depending on their impedances would the contribution from the battery eventually become zero? Why does the alternator then completely take over?

As you raise the generator voltage further, then current above that required for the load will flow to the battery charging it.

How much higher does the generator have to go over the battery voltage? Would this be the same as putting the two sources in parallel and performing a mesh analysis between the load and battery in parallel with the generator?

How does this part translate to the solar panels feeding back to the grid?

 

[GrayHeadedMule]

I work for a minicipality. Every time someone secures a permit, that permit is sent electronically to the appraisal district. The appraisal district views all permits as improvements to your house and adjusts the appraised value accordingly every year.

 

[qcroanoke]

I work for a minicipality. Every time someone secures a permit, that permit is sent electronically to the appraisal district. The appraisal district views all permits as improvements to your house and adjusts the appraised value accordingly every year.

And that is why no homeowner wants to get a permit.
Just sayin.......

 

[gar]

100922-1708 EST

mull982:

Consider the following very simple circuit.

Two idea batteries. This means their voltage is absolutely constant independent of current flow and the direction of current flow. They have zero internal impedance. One is exactly 10 V the other is adjustable +/- some amount relative to 10 V. The first is V1 and the second V2.

We will put their internal resistance external to the battery. V1 will have 0.1 ohm as its internal resistance. V2 will have 1 ohm. These internal resistances are connected to the positive battery terminals. The other internal resistance ends are connected together and to the load of 9.9 ohms. Both battery negative terminals are connected together and to the other end of the load.

When V2 is equal to 9.9 V how much current flows from V1 and from V2?
Change V2 to 9.8 V what are the currents?
Change V2 to 10.0 V what are the currents?
Next V2 to 11.010101, same question?
And last V2 to 12 V, ?

.

 

[mull982]

100922-1708 EST

mull982:

Consider the following very simple circuit.

Two idea batteries. This means their voltage is absolutely constant independent of current flow and the direction of current flow. They have zero internal impedance. One is exactly 10 V the other is adjustable +/- some amount relative to 10 V. The first is V1 and the second V2.

We will put their internal resistance external to the battery. V1 will have 0.1 ohm as its internal resistance. V2 will have 1 ohm. These internal resistances are connected to the positive battery terminals. The other internal resistance ends are connected together and to the load of 9.9 ohms. Both battery negative terminals are connected together and to the other end of the load.

When V2 is equal to 9.9 V how much current flows from V1 and from V2?
Change V2 to 9.8 V what are the currents?
Change V2 to 10.0 V what are the currents?
Next V2 to 11.010101, same question?
And last V2 to 12 V, ?

.

Gar I'm a little rusty on my circuit analysis. I used (2) KVL loops with I1 and I2 in each loop. Solving for I1 and I2 I got 3.6A for I1 and 21.36A for I2. This was for the first scenario with V2=9.9V.

I'm not sure if I'm going about this right so I wanted to see if I was going about solving the circuit right first before I proceeded with the rest of the scenarios.

 

[gar]

100923-0900 EST

mull982:

Do a little intuitive analysis.

If I have a 10 V source with a 10 ohm load the load current is 1 A. If the load resistance is split into two resistors consisting of 0.1 and 9.9 ohms, then what is the voltage across the 9.9 ohm resistor? It is 9.9 V.

If I connect to this circuit a 9.9 V battery across the 9.9 ohm resistor and use the correct polarity, then how much current flows to or from the 9.9 V battery? If the wrong polarity, then what are the currents?

In my previous post in the first sentence "idea" should have been "ideal".

.

 

[DownRiverGUy]

Refer to this link for Property Tax Exemptions for solar by State:

http://solarpowerrocks.com/files/2010/08/property-tax-exemption-by-state-solar-2010.png

Looks like I'll be calling up my city before putting solar on a house... (I live in Michigan).

 

[mull982]

100923-0900 EST

mull982:

Do a little intuitive analysis.

If I have a 10 V source with a 10 ohm load the load current is 1 A. If the load resistance is split into two resistors consisting of 0.1 and 9.9 ohms, then what is the voltage across the 9.9 ohm resistor? It is 9.9 V.

If I connect to this circuit a 9.9 V battery across the 9.9 ohm resistor and use the correct polarity, then how much current flows to or from the 9.9 V battery? If the wrong polarity, then what are the currents?

In my previous post in the first sentence "idea" should have been "ideal".

.

Would there be 1A flowing from the 9.9V battery to the load with the correct polarity. I'm not sure about the wrong polarity. Would it be 0A?