Viability of Solar Power

[mmalluck]

How viable is solar power? I was asking myself this question and here's the numbers I came up with.

 

In 2001 the USA used 96275 trillion BTUs of energy that year. This comes to 3.22 trillion watts.

 

Now there are about 295 million people in the US, so this comes to about 11Kw per person at any given time.

 

This means each person uses is responsible for 262 Kwh of power per day.

 

Now lets say that square meter of sunlight provides 1 kw of energy on average, the average area gets 5 good hours of sunlight per day, and the typical solar panel is about 30% efficient. This means that for every square meter of solar panel would render 1.5 KwH every day.

 

This means that each man woman and child would need 174 square meters of panel to be responsible for all the energy made and used in their name!

 

If every person in the united states of America put up solar panels. We would have over 51 billion square meters of panel, that's close to 20,000 square miles of panel or the equivalent of covering most of West Virgina over in panels.

 

Now these numbers account for all energy used both domestic, industrial, and exported. Also these numbers do not account for the added or lost efficiency of converting systems over to pure electrical power as opposed to chemical energy processes like the internal combustion engine.

 

I left the links to my math in just incase I botched anything.

[swansont]

The required area is likely higher since you need to look at the area projection - the sun isn't always directly overhead. The solar power numbers assume an area perpendicular to the sun's rays.

 

It's interesting to put it in terms of the area, though, since the US could never be self-sustaining in the long term if the area required were larger than the country's total area (some fraction of it, really). All of our energy, other than nuclear, comes from the sun in one way or another.

[calbiterol]

There's an interesting technology out now that uses stirling cycle engines and mirrors to produce electrical power. It concentrates the thermal energy of the sun using a parabolic dish. It is highly efficient as far as solar power goes, and is the most efficient solar power available. The company who has developed the concept is called Stirling Solar, or something similar. This type of solar power is viable - a relatively small field of the solar dishes (only a few acres, I believe) rival the power of a nuclear reactor.

 

 

 

Also, the figure you used for BTU's in the US - does that include energy in the form of oil (cars, diesel trains, airplanes, etc), or just electrical energy?

[mmalluck]

If you look at the link I included, you'll see it includes all power used in the industrial, comercial,domestic, and transportation markets. Basically everything.

 

If the panels were 90% efficient and you could get 8 hours of sunlight out of them, you'd get 7.2Kwh out of every meter of panel per day..... Each person whould need 25 square meters of panel to break even and the total area of all the panels combined would be 2847 square miles. That's about the half the size of Connecticut.

 

That's still no small area.... I wonder how efficient plants are at converting light into biomass?

[calbiterol]

The dish I was talking about is made by Stiling Energy Systems. The 38-foot-diameter dish powers a 25-kilowatt generator, and the entire assembly (1 dish) costs about $250,000. It does not use conventional solar cells!!! It uses heat from cencentrated sunlight to drive a stirling cycle engine.

 

More info can be found at:

 

http://www.popsci.com/popsci/science/article/0,20967,1018934,00.html

http://www.stirlingenergy.com/solar_overview.htm

[mmalluck]

Still, you can't get more than 1kW of energy per square meter of sunlight.

 

Lets look at the area the 38-foot-diameter dish takes up.

 

(19 feet)^2*pi = 105.362726 m^2...... 1 kW of energy falls on each square meter, so in theory this generator should be able to make 105 kW, but it's pegged at 25kW.

 

25kW/105kW yeilds an efficentcy of 24%. Hmm. Looks like you'd be doing better with the 30% efficent solar panels.

 

Okay, neat fact time. The surface of the earth 5.1*10^14 meters squared. If the 1Kw / square energy density was true for every place on earth (it's not, but you try and calculate it all over the globe), then the total incident power that falls on the earth is around 510 trillion kWs.

[calbiterol]

Still' date=' you can't get more than 1kW of energy per square meter of sunlight.

 

Lets look at the area the 38-foot-diameter dish takes up.

 

(19 feet)^2*pi = 105.362726 m^2...... 1 kW of energy falls on each square meter, so in theory this generator should be able to make 105 kW, but it's pegged at 25kW.[/quote']

 

First, it is neither a circle nor a hemisphere. The dish is a parabola. If it were a circular mirror surface, then it would do no good - the rays from the sun would be reflected on areas other than the generator. Calculating the surface area of a parabola is rather extensive. It involves finding the arc length of the 2-d parabola and multiplying it by the circumference of the circle of the outer edge of the 3-d surface. If that made any sense at all, it would become readily apparent that it is impossible to calculate the surface area, because in order to know the equation of the parabola, you must know more than just the diameter of the final circle, and in order to calculate the surface area, you must know the parabolic equation.

 

Long story short, there is simply no way to calculate the surface area. Additionally, check your figure on kW/m^2. I'm not sure it's correct. Watts/day/meter can be found at http://www.stirlingenergy.com/incl_docs/images/Solar_2.jpg. This would suggest that under the best case scenario, 7.5 kWh per day per square meter is all that the sun puts out. This translates to (7.5*kW*h)/(24h*m^2), which is simplified to (7.5 kW)/(24m^2) and results in 312.5 W per m^2. Using your approximation of 105 m^2, the dish would recieve (312.5 W) / (105 m^2), or 32.8125 kW. The efficiency then becomes 25 kW / 32.8125 kw = 0.761904761904... In other words, assuming the figure of 25 kWh created is using the best possible conditions (which it may not be, and if it weren't, the efficiency below would only increase), the efficiency with an approximate area of 105 m^2 is 76%. That's more than twice that of the average traditional solar cell.

[swansont]

Okay, neat fact time. The surface of the earth 5.1*10^14 meters squared. If the 1Kw / square energy density was true for every place on earth (it's not, but you try and calculate it all over the globe), then the total incident power that falls on the earth is around 510 trillion kWs.

 

You haven't noticed that it's dark on half of the surface?

 

You don't want to use the surface area of a sphere. You want the area it projects, i.e. the area of the circle, which is 1/4 the size. Then you don't have to worry about the projection angle - it's already accounted for.

[mmalluck]

The projection area... that's what what I was using with the parabolic dish. The surface area doesn't matter, only the area across the apature of the dish. That's the most sunlight you can collect.

 

Calbiterol, your calculations look good for the most part. The "312.5 W per m^2" figure can be a little misleading, as this is the daily average power, and not the true power density for a given square meter of sunlight.

 

Lets say you get 10 hours of sunlight down there in sunny California.

(312.5 W *24h)/ 10h would give you a power density of 750W / m^2.

 

Going back to the equations:

750W/m^2 * 105 m^2 yeilds 78.7kW.

25kW/78.7kW gives us an efficiency of 31.7%. This is just barely edging out traditional solar panels. True, they may not have run this in "the" best area, which could make the efficiency rise.

 

I like the chart you included. It'll give me some better numbers to play with for more fun later.

[calbiterol]

The projection area... that's what what I was using with the parabolic dish. The surface area doesn't matter, only the area across the apature of the dish. That's the most sunlight you can collect.

 

Aha! That explains a lot. Just one more thing that little old me had absolutely no knowledge of... I guess if I would have thought about it, that might have come to mind. It's a very shallow parabola, anyways.

 

The "312.5 W per m^2" figure can be a little misleading, as this is the daily average power, and not the true power density for a given square meter of sunlight.

 

True. But an idea just occurred to me: we can ignore the whole daylight issue, because the chart is sunlight energy for the whole day... So, from the chart, the best case scenario is 7.5 kWh of sunlight per day. We know from our specs that the dish is pegged at 25 kWh. So, take the aperature area of the dish - about 105 meters squared - and divide it by 25 kWh. The result is 4.2 kWh per meter on the dish. 4.2 kWh divided by 7.5 kWh equals an efficiency of 56%. Unless I'm messing up my math somewhere (or my logic), this figure should be about correct.

 

I like the chart you included. It'll give me some better numbers to play with for more fun later.

 

Thanks!

[mmalluck]

But an idea just occurred to me: we can ignore the whole daylight issue' date=' because the chart is sunlight energy for the whole day...

[/quote']

 

I read through this and it looks pretty good, but I stumbled across a problem.

 

We're comparing 7.5Kwh/m^2/day to 25kWh/105m^2. The problem is we don't know what the daily average output of our dish is. I know it can't be 25kWh/105m^2/day because the would mean the dish runs all day and all night.

 

I guess we're stuck with making some kind of assumption as to how much daily power this dish can make, which is dependant on the day/night cycle.

[calbiterol]

I guess we're stuck with making some kind of assumption as to how much daily power this dish can make, which is dependant on the day/night cycle.

 

Poetic justice at its best. Ohh, the irony. I'd try and think about the rest of what you just said, but I'm not really in any condition to *think* about anything... All I can say is, my brain is tired... Zzzzzzzzzzzzzzzzzzzzzzzzzzzz (goes to get pillow)

[atinymonkey]

All this is based on the assumption that power consumption will remain a constant. Market values dictate that the unit price of electricity will increase as capacity to produce is reduced, which will decrease power use on a sliding scale.

 

Something else to consider is that only 5 > 7 % of the total power consumed in the US goes to domestic propertys.

 

There is also a huge wastage of power across the national grid, it is a very inefficent method of transferring power due to the consumption fluctuating from one second to the next. The effect of the fluctuation is that the national grid is forced to always produce excess to cope with the demand.

[Sayonara]

If every person in the united states of America put up solar panels. We would have over 51 billion square meters of panel, that's close to 20,000 square miles of panel or the equivalent of covering most of West Virgina[/url'] over in panels.

What's the surface area of all the roofs and sun-facing walls in the USA?

[calbiterol]

What's the surface area of all the roofs and sun-facing walls in the USA?

 

That's a very good point. Unfortunately, not all types of solar power are wall-mountable...

[Sayonara]

Then let's not use those models.

 

I am about to go to bed so I'm not about to check, but take a look at http://www.solarcentury.com and see if their cells are more efficient than you allowed for (per unit area). I seem to recall they do not suffer much in low light conditions.

 

 

[edit]

 

Also, don't forget that a lot of the domestic and industrial power requirements (i.e. heating etc) can be accommodated using those solar fluid tubes, which are cheaper, simpler and less technically demanding than PV cells.

[calbiterol]

Plus... Think about it. The basic technology behind the solar stuff I mentioned above was using a stirling cycle motor to drive a generator. What if you made the sides of a house and/or the roof a part of the gas expansion (heating) chanmber in a stirling cycle engine? If you could put a thin layer of sealed gas and it could readily absorb solar heat, then wouldn't it serve the same purpose in the end? You wouldn't get quite the efficiency of the mirror dish, but I would think it'd still be more efficient than normal solar power.

 

Just a thought.

[drz]

hey, you guys are pretty good with math. Atleast, lost me about 2nd post. Anyhow, interesting stuff.

 

You mentioned using the house itself to heat the sterling, have you heard of solar passive? A friend of mine's father built his house with this concept.

 

Solar passive

 

You got to click around a bit, but that site kinda explains it. Basicaly though, using a structure built like this, perhaps use a wood stove/boiler for extra heat, you could have a warm area.

 

I don't know what temp a sterling engine requires to operate, isn't it temperature difference??

[calbiterol]

I don't know what temp a sterling engine requires to operate' date=' isn't it temperature difference??[/quote']

 

Yep, it's temperature difference. Probably the best explanation of how stirling engines work is available here. If you wanted to do it on a house, you'd just have a MASSIVE expansion chamber - the area where the gas is heated, if you're looking at the HowStuffWorks page - and then you'd have the rest of the engine exposed to the outside air. A bonus of doing it like this that I hadn't thought of: in the winter, when less solar power reaches the ground, the air is cooler. Since Stirlings work on the difference in temperature, it'd remain at about the same power output throughout the year (don't forget that less sunlight equals less heat... the cooler air just balances it out). I don't know if it would do quite as well in winter as in summer, but it would definitely be less of a falloff than traditional solar panels would. Plus, you could make the whole thing out of materials that are cheaper and easier to come by than solar panels are.

 

Don't forget that using solar tubing to heat things, assuming you never store the energy, will mean that if the sun isn't out, you're cold!

[Guest bewick16]

I am currently researching the viabilty of a solar powered kiln for the production of Lime Roof Tiles in Bangladesh for the purpose of rainwater harvesting, the required temperature of the Kiln is 900C. I have looked at various souces and come up with the following calculations are they correct/viable?

 

Solar insolation average for Bangladesh = 4.78kWh/m^2/day (http://www.sdnbd.org)

Average Sunshine Hours = 7.55

 

4.78/7.55 = 0.63311

Solar Innsolation per hour = 633Whm^2

 

Solar Refelection systems max efficiency 30%, a figure of 10% is used.

Energy Reflected = 633 x 0.10 = 63.3J/hm^2

 

Energy Required to Heat 0.5m^3 of air to 900C.

Q = mc(change in T)

m = 0.5kg

c(Specific Heat Capacity) = 1.005

Change in T = 900 - 20 = 880

Q = 0.5 x 1.005 x 880

Q = 442J

 

Therefore the surface area of the solar reflector to focus enough energy to heat 0.5m^3 of air to 900C

= 442/63.3

=6.982622

=7m^2.

 

Is this correct? Do I need to take other variables into consideration such as the angle of incidence of the sun etc? Any help would be much appreciated.

[calbiterol]

Solar insolation average for Bangladesh = 4.78kWh/m^2/day (http://www.sdnbd.org)

Average Sunshine Hours = 7.55

 

4.78/7.55 = 0.63311

Solar Innsolation per hour = 633Whm^2

 

Solar Refelection systems max efficiency 30%' date=' a figure of 10% is used.

Energy Reflected = 633 x 0.10 = 63.3J/hm^2[/quote'] I'm not sure if this is correct - but I think it is. The meters squared would be the problem, if there is one. I have no idea where this efficiency fiure is coming from, though. I can't help you out any there.

 

Energy Required to Heat 0.5m^3 of air to 900C.

Q = mc(change in T)

m = 0.5kg

c(Specific Heat Capacity) = 1.005

Change in T = 900 - 20 = 880

Q = 0.5 x 1.005 x 880

Q = 442J

First off, it's C_p. Do that using [sub ][/sub ] tags (without the spaces).

 

Second, double-check your conversions. If your specific heat is in J/(g*c) then you have a problem - a kG is a thousand grams, a meter is 100 centimeters, and neither the jules or the centigrade numbers are increased. So, for simplicity and security, convert the meters cubed to centimeters cubed (and kilos to grams). Note that this results in 50^3 = 125000 centimeters cubed, because it is converting cubed units - meaning you need to cube the conversion rate. kG, however, is straight-up: a half-kilogram is 500 grams. That would give you 500g * 1.005 J/(g*C) * 880 or 500g* 1.005 J/(g*K) * 880, which is your number times one thousand. That would seem to be a MUCH more reasonable figure. But that's the only problem I see with your math.

 

 

Therefore the surface area of the solar reflector to focus enough energy to heat 0.5m^3 of air to 900C

= 442/63.3

=6.982622

=7m^2.

Remember that this is not the surface area of the reflector, it is the area of the aperature of the reflector. Double-check your maths and conversions.

 

Is this correct? Do I need to take other variables into consideration such as the angle of incidence of the sun etc? Any help would be much appreciated.

Well, it depends on conversions and the figures you got. Angle of incidence should be taken into consideration.

[Guest bewick16]

Thanks for the help, checked all the conversions and figures I had used, they all seemed correct, and the result made sense. I have obtained data about the path of sun in Bangladesh in 9 different locations over a 10 year period, plotted the differnernt paths, calculated the average and plotted that. From here I have calculated the average number of degress the tracking system would be out if it were set to follow the average path. This calculates to 7degress.

 

Is it possible to then calculate the energy loss due to this angle error, ie. the angle of incedence would be either 83degress of 97degress, i think??????

[JohnB]

Interesting. There was a debate recently in the Aussie Parliment about future energy production. It was stated that Oz would need around 2,500 sqare km of Solar Panels to meet current needs.

Each person whould need 25 square meters of panel to break even and the total area of all the panels combined would be 2847 square miles. That's about the half the size of Connecticut.

mmalluck, are the US states really that tiny? 2500 square miles is only a block 50 miles by 50 miles. We could stick a dozen of them in western Queensland and then lose the bloody things.

 

Another use for Solar energy is the Solar Tower. Information on the tower here.

 

I find this to be a very good combination of wind and solar energy.

[Sayonara]

Each person whould need 25 square meters of panel to break even and the total area of all the panels combined would be 2847 square miles. That's about the half the size of Connecticut.

They must be thinking of using pretty crappy panels

[Gayathri]

Actually, this is a great project regarding energy. Furthermore it is valuable solution for protects energy sources.

 

............................

GAYATHRI

 

http://www.goinggreenbuzz.com