Solar Panels on residential Homes

[al hildenbrand]

I'm not sure about the wrong polarity. Would it be 0A?

Think about what two "D" cell batteries would do, if you paralleled two of them, + to + and - to -, and then reversed one of the "D" cells.

With the polarity reversed, will there be zero current or a large current limited by the internal resistances of the cells and any other parallel resistances?

 

[mull982]

Think about what two "D" cell batteries would do, if you paralleled two of them, + to + and - to -, and then reversed one of the "D" cells.

With the polarity reversed, will there be zero current or a large current limited by the internal resistances of the cells and any other parallel resistances?

I guess if you connected them in series you would have 24V across the combined internal resistance's of the 2 batteries. Would this resistance be small leading to a large amount of current?

 

[mull982]

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?

If there is 1A flowing from the 9.9V battery, is this one less A that the 10V battery needs to supply. I am just having a hard time visualsing this simple circuit.

 

[gar]

100923-1253 EST

mull982:

I tried to lead you into the answer. A new try with a slightly different incremental approach.

With an ideal 10 V battery connected to the series combination of 0.1 ohms and 9.9 ohms I have exactly 1 A thru those resistors and thus a 9.9 V drop across the 9.9 ohm resistor.

Now connect the negative end of the 9.9 V battery to the negative end of the 10 V battery. What is the voltage difference between the plus end of the 9.9 V battery and the node between the two resistors?

Next put a variable resistor between the 9.9 V battery and the node and measure the current as you adjust this resistor from infinity to 0. What change of current do you observe vs resistance, and why?

.

 

[mull982]

Now connect the negative end of the 9.9 V battery to the negative end of the 10 V battery. What is the voltage difference between the plus end of the 9.9 V battery and the node between the two resistors?
.

I believe you'd get 0V since they both had a 9.9V potential referenced to the - terminal

Next put a variable resistor between the 9.9 V battery and the node and measure the current as you adjust this resistor from infinity to 0. What change of current do you observe vs resistance, and why?

With 0V between the two adding a resistor there would be no current flow no matter how much you changed the resistor. Unless I answered the first part wrong and the potential between the node and positive terminal is 18.8V instead of 0V.

 

[mull982]

I believe I understand what happens if (2) different batteries are put in parallel. Lets say I have a 10V battery and a 5V battery in parallel and each battery has a 2.5ohm impedance (just for simplicity sake)

Now with these two batteries connected in parallel you will have a potential different of 5V between the two + battery terminals, and this 5V potential will drop across the 5ohm series internal resistance of the batteris giving a current flow of 1 A through the circuit.

If this is correct then I understand this far, and just need to visualize what happens when you add a load resistance in parallel.

Or is adding this load resistance just putting a resistance in series with the internal impedance of each battery.

 

[al hildenbrand]

Would this resistance be small leading to a large amount of current?

Yes.

This is the basic idea for the "reversed polarity" you mentioned above.

 

[gar]

100923-1727 EST

mull982:

Your answers of 0 V and 0 current are correct.

Now with the original voltages and resistances I presented to you what direction will current flow to the variable battery (V2) when V2's voltage is less than 9.9 V? The only question here is direction.

A different question. At What voltage for V2 will the current to V1 go to 0?

.

 

[dereckbc]

Guys i can help you out with the math as I have designed quite a few solar systems over the last 3 years or so. It is so simple a high school student can do it to some degree. :grin:

There are two kinds of systems:

• Grid Interactive aka Grid Tied.
• Off-Grid Stand Alone or Battery systems.

The two systems are similar but have very different principles. The easiest to grasp is Grid Tied. The first step is to completely remodel your home with all your appliance upgrades, doors, windows, insulation, ect... As this will give the most bang for your buck and on average for every dollar spent in upgrades will save you $3 on your solar installation.

Next step is to determine your daily average Kwh usage, and then decide what portion of that you want to replace with solar power. So for the example lets say it is 25 Kwh per day on average through out the year in Kansas City MO with Net 0 usage 100% solar. You then go to this website and download the Solar Insolation tables for your state, then look up your city to find the yearly average SUN HOURS,. This is where you learn real fast that location means everything. In this example for Kansas City the yearly average is 4.9 Hours. If you live in Gloomy Seattle it is down to 3.3, and Sunny Tucson is a whopping 5.6 Hours..

Ok you are now set to design the system. To find the solar panel wattage take the daily Kwh and multiply by 1.2 or divide by .8 to account for 80% efficiency factor. 25 Kwh / .8 = 30 Kwh. This is how much energy on average the panels must generate to net 0 usage over the course of a year. Now to find the solar panel wattage all you have to do is factor out the time element which is what you looked up wh / h = w = 30 Kwh / 4.9 h = 7.143 Kwh. Round that up to the nearest 100 and you need a 7200 watt solar panel array and an inverter to match it. In Tuscon you need a 4.76 Kw and Seattle you need 9.1 Kw ouch!

Battery systems are really ugly because we have to design for worse case and have to use two calculations, one for summer, and one for winter and use the larger of the two plus several tons of batteries. Staying with are above example using 25 Kwh we now Know right up front winter is the worse case with December being 3.3 Sun Hours. Next gotcha is at the very best a battery system is only 66% efficient. So we need to account for that for both solar panel wattage and battery calculations. So first take 25 Kwh and divide by .66 efficiency. 25 Kwh / .66 = 37.9 Kwh. Now take KCY December Sun Hours and factor out the wattage 37.9 Kwh / 3.3 h = 11.48 Kw. Think that is bad, Seattle gets a whopping 1.3 Sun Hours so you would need 29.15 Kw. Our Sunny Tucson friends get 5.6 Sun hours and only require 6.76 Kw. Location matters huh?

Now it really gets ugly, batteries. In order to get at least 5 years out of a battery you cannot discharge more than 20% capacity in one single day, and never discharge more than 50% ever or you only get a year or so going below 50%. Also you have to have some capacity to carry you through a cloudy day or two. So minimum capacity is 5 days which gives you 2.5 days to the 50% factor before you need to fire up your generator. So using the minimum of 5 days 37.9 Kwh x 5 = 189.5 Kwh reserve time. To find the amp hour requirement divide by the battery voltage which I will ignore for now because it is not important for this example. Here is the killer For each Kwh of battery requires 55 pounds and $130. So you are staring at 10,420 pounds of hazardous material which requires EPA permits and yearly FD inspection, spill containment, and $25,000 every 5 years just fore the batteries and no labor included. What a deal huh? If you select a battery system you just decided to pay 10 to 15 times more for electricity for the rest of your life.

Oh our KCY battery system would cost around $65,000 plus your labor and materials. Our Tuscon friend get off cheap at $50,000, and are gloomy friends in Seattle get stuck with a $100,000+ bill plus labor and materials for added insult and injury.. Location, location, location.

 

[Article 90.1]

I'll make it even simpler. If you have a grid tied system all you are is a generator supplying power to the grid. Yes, some electrons may flow into your building, but you really don;t care where they go, because all you need to do is log your generated power on whatever metering setup your local utility specifies.

Who cares what the load of your building is with a grid tied system? You are just selling power back and it does not matter how much power you use. If you generate 30 kWh for a month you will get paid the pre-agreed upon amount for that energy, regardless of how much you use. Of course, your bill will be more expensive if you consume more power, but that is irrelevant in a grid tied system in regards to the system paying for itself.

To make it even simpler, remove your building from the equation and just place the solar panels out on a piece of land you own, all you are is a co-gen facility.

 

[mull982]

Now with the original voltages and resistances I presented to you what direction will current flow to the variable battery (V2) when V2's voltage is less than 9.9 V? The only question here is direction.

When V2 drops below 9.9V to lets say 9.8V as you posed in your origonal example then I believe there would now be a .1V difference across the 1ohm source impedance of V2 and therefore a .1V/1ohm = .1A current would flow into the positive terminal of V2. Is this correct?

A different question. At What voltage for V2 will the current to V1 go to 0?

I'm i'm thinking about this right I try to determine what value of V2 will drop 10V across the 9.9ohm resistor when V1 is disconnected. Then when reconnecting V1 the potential between V1 and the node between the 0.1ohm and 9.9ohm resistor will be the same and no current will flow.

To calculate V2 I use a voltage divider to come up with 10V across 9.9ohm resistor in series with 1ohm V2 source impedance. Doing this I come up with a value of 11.01V for V2. Is this correct?

 

[gar]

mull982:

The following comment is not meant to help solve the previous question I asked you. That question you can do by inspection.

In the circuit I previously described you can replace the 10 V battery and the 0.1 and 9.9 ohm resistors with an equivalent circuit of a 9.9 V battery with an internal resistance of 1/(1/0.1 + 1/9.9) = 1/ 10.10101 ...... = 0.99000 ohms.

.

 

[mull982]

mull982:

The following comment is not meant to help solve the previous question I asked you. That question you can do by inspection.

In the circuit I previously described you can replace the 10 V battery and the 0.1 and 9.9 ohm resistors with an equivalent circuit of a 9.9 V battery with an internal resistance of 1/(1/0.1 + 1/9.9) = 1/ 10.10101 ...... = 0.99000 ohms.

.

Yes this I understand is Thevinin's theorm.

 

[gar]

100924-1117 EST

mull982:

Mistake, typo, on my post #32. The equivalent internal resistance should read 0.099 ohms.

Thus, when the variable voltage battery, V2, is at 9.8 V the positive current flow into V2 should be (9.9 - 9.8) / (1 + 0.099) = 0.1/1.099 = 0.090991811 A.


The second part of your post is the correct thinking and the correct answer, 11.010101 ..... V.

.

 

[gar]

100924-1220 EST

Locally we have some locations with solar panels. These are really experimental with Federal or State Grants purchasing the equipment.

Two installations are in Ypsilanti, Michigan with on line data output. The Y-axis is a screwy set of units. In some cases it is labeled watt-hr and others KWH. That part is OK, but it is for the amount of energy in 5 minute packets. You have to dig to find out that it is the energy per 5 minutes. The curve is drawn as a continuous curve, which it is not. As best that I can determine it should be plotted as spikes each 5 minutes. Then the Y-axis label would be tolerable, but hard to interpret.

The power company provides a meter that outputs a pulse every revolution of the disk. The 5 minute time of counting pulses is probably satisfactory, especially if interpolation is used.

When you are looking at a time varying function like this that is not the cumulative energy, then from a thinking perspective power would be a better variable. If I convert the curve to power based on 5 minute intervals, then I need to multiple the value by 12 and now I should have the approximate short time average power.

I believe this facility presently has on a rating basis about 2280 W worth of panels. 12 panels of different ratings. I would judge that this would mean the peak power from the system would not exceed 2280 W, and probably less.

At http://solar.ypsi.com/documents/YpsiFoodCoop_store/ProgressReports/PSC_08_31_Report_2010_04.pdf
is some data.

If you look at the first plot, this is for the food coop, they have a peak generated value of 160 watt-hr for 5 minutes. By my conversion this equates to about 160 * 12 = 1920 W peak power. This is 84% of the rating.

Next look at the plot Monthly Usage and the table below. For this type of plot the Y-axis units are correct. But they failed to label the plot and table as for the food coop. Adding the values in the table and assuming 110 for April I get a year total of 1643 KWH. Also note this installation has cost of maybe $40,000. 1643 / 2280 = 0.72 is a long way from the 1.1 ratio that I have been told is typical for Michigan.

We have a very cloudy day today, but you can look at current production at http://solar.ypsi.com/index.php?siteDetails=foodcoop
Sun must have broken thru for one short period.

At our farmers market they installed a 10 KW array. I am trying to find out if they have on-line data.

.

 

[gar]

100924-1359 EST

We are getting some intermittent bursts of sunshine, but no 5 minute periods.

I would much prefer that they had a power meter and provided a plot of instantaneous power generated. One second increments would be fine or maybe averaged over 5 second periods. Then maybe average power at 15 minute intervals. Along with this a cumulative energy production from midnight.

Note: my comments on the real time display only apply for today. The data will be different tomorrow.

.

 

[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, ?

.

Having gained a better understanding of your example here are the answers I come up with to your origonal problem.

When V2= 9.9V no current flows from V1 to V2

When V2 = 9.8V aprox .1A flows from V1 to V2

When V2 = 10V **I was not sure how to sove this one and could use a hint

When V2 = 11.010101 about 9.17E-9A flows from V1 to V2

When V2=12V .89A flows from V2 to V1

Other than the V2=10V problem I think I'm getting the hang of this. Are my answers on the right track?

 

[gar]

100927-0859 EST

Some more comments.

I should point out that I do believe we should be searching for economical means to obtain energy from renewable sources. However, I see a lot of hype with what I believe are unsupportable claims. Some are ignored unintended consequences. I go to our various energy meetings to learn what I can from real world experiences. However, most persons going to these get-togethers are really looking for government grants or money.

At a recent meeting I met a former CEO of the Dana Corporation, a company with which I have done a lot of business, who now heads a small company developing an energy recovery and reuse system for the auto and light truck industry. This is mostly a non-electrical and lower cost system for improving gas mileage in vehicles like postal or UPS trucks. In postal trucks they have improved mileage from about 5 MPG to maybe 12 MPG and are working to exceed 20 MPG. This is a hydraulic-pneumatic, and engine start-stop system. This type of system has great potential. Less weight and cost than an electrical hybrid system. Also safer because there is no high voltage high energy component that presents problems in accidents.

We have energy meetings this Thursday and the following week for which I am priming myself with questions to ask. Thus, part of the reason for my current probing for information that I have presented here.

Looking over the Ypsilanti material some more I found this report:
http://solar.ypsi.com/documents/City_Hall/20070120-CityHallSolarDesign.pdf
Within this report are calculations on expected payback. I think these are excessively optimistic. However, there is useful technical information on panels and inverters near the end of the report.

.

 

[SegDog]

modules or collectors? not always what ya think...

modules or collectors? not always what ya think...

When I first became interested in solar, I would notice more and more systems as I travelled.

I passed-by a development in New Jersey one day, and returned later to look at the systems up close. It turns out that they were solar collectors for heating water and were installed in the late seventies, and early eighties (after Arab Oil embargo). A number of these system were still operational many years later, if maintained.

Solar water heating collectors have been around for over a hundred years.

I've had more than one person ask me if I was talking about water or electric when I mentioned solar...

 

[gar]

100927-1229 EST

Mull982:

When V2= 9.9V no current flows from V1 to V2

Good.

When V2 = 9.8V aprox .1A flows from V1 to V2

OK. But I would prefer the answer had a little more precision because that would imply that you knew how to calculate the value.

When V2 = 10V **I was not sure how to solve this one and could use a hint

Here the equivalent circuit for the 10 V, 0.1 and 9.9 ohm resistors becomes useful. It is 9.9 V, call it V1', with an internal resistance of exactly 0.099 ohms.

Then the calculated current from V2 toward the load is the result of (V2 - V1')/(0.099+1.0) or (10.0-9.9)/1.099 = 0.090991811 A.

The voltage across the load is 10.0-0.090991811*1 = 9.909008189 V.

What is the current thru the load? Iload = 9.909008189/9.9 = 1.000909918 A. Thus, the current from V1 is Iv1 = 1.000909918-0.090991811 = 0.909918107 A. This current is clearly less than if the circuit to V2 was open. As you raise the voltage of V2 the amount of current from V2 to the load increases.

See if my calculations are correct.

When V2 = 11.010101 about 9.17E-9A flows from V1 to V2

This was a semi-trick question. What I wanted for an answer was zero. I have not checked your answer, but it is close to what I expected.

When V2=12V .89A flows from V2 to V1

Using the equivalent circuit method I get 2.1/1.099 = 1.910828025 A flowing from V2. The load voltage is 12.0-1.910828025 = 10.089171975 V. Thus, the current into V1 is about 0.089171975/0.1 = 0.89171975 A, and this checks your answer.

You are doing well.

,