Small Solar Chargers

H. E. Taylor wrote:

I haven't had time to go searching for further material but the articles I have seen were not clear. It sounded something like your hot-dog example above, but the 40% claim seems to muddy the waters of standard measurements. If for example, the concentrator accoutrements were removed, what would be the native efficiency of the photovoltaic material under an AM1.5 sun?

As I recall, the efficiency was measured by NREL. While I don't have specific knowledge of their concentrator measurement procedures, I have more than passing familiarity with their measurement procedures in general and I'm virtually certain that they measured this one under the AM1.5 *spectrum*, though not the *intensity* (both of which are specified by the ASTM and ISO AM1.5 standards).
As for the other part of your query, if you put a concentrator cell under un-concentrated sunlight (i.e., 1 sun, as opposed to the hundreds of suns this one was probably measured under) you will measure a lower efficiency. This comes about purely through the physical design of the solar cell.
In *any* cell, voltage increases with the logarithm of the incident light intensity while current increases linearly. On the other hand, resistive losses increase with the square of the current. Both the higher intensity and higher resistive losses increase the cell's temperature, which depresses the voltage. All of this means that as you increase light intensity from zero you will typically see increasing efficiency up to a certain intensity, after which the losses will overwhelm the gains. In other words, efficiency reaches a maximum at a particular light intensity and is lower at any other intensity.
Typical solar cells like you buy in a standard PV module are designed with 1 sun in mind and probably have a peak efficiency at an intensity in the range of 0.75 to 2 suns (though the difference is a few tenths of a percent at most). Cells designed for concentration are doped more heavily, have thicker conductors on the front and back surfaces, and a few other features to push the peak efficiency up to very high light intensities. At 1 sun, these features work *against* the design: thicker conductors block incoming light, heavier doping increases dark saturation current and recombination (which depresses voltage), etc. I can't estimate what the 1-sun efficiency of this particular cell would be -- it's entirely dependent upon details of the design -- but I *can* say that it is definitely lower than a cell of similar structure, but designed for 1 sun, would be.

In article , R.H. Allen wrote:

H. E. Taylor wrote:
I haven't had time to go searching for further material but the articles I have seen were not clear. It sounded something like your hot-dog example above, but the 40% claim seems to muddy the waters of standard measurements. If for example, the concentrator accoutrements were removed, what would be the native efficiency of the photovoltaic material under an AM1.5 sun?
As I recall, the efficiency was measured by NREL. While I don't have specific knowledge of their concentrator measurement procedures, I have more than passing familiarity with their measurement procedures in general and I'm virtually certain that they measured this one under the AM1.5 *spectrum*, though not the *intensity* (both of which are specified by the ASTM and ISO AM1.5 standards).
As for the other part of your query, if you put a concentrator cell under un-concentrated sunlight (i.e., 1 sun, as opposed to the hundreds of suns this one was probably measured under) you will measure a lower efficiency. This comes about purely through the physical design of the solar cell.

I came across another article just today that mentions 244 suns...
<http://www.upi.com/Energy/view.php?StoryID=20061214-093927-4718r> 2006/12/14: UPI: Solar World: An efficiency milestone [Spectrolab]

[...]

<regards> -het
-- "The universe is full of magical things patiently waiting for our wits to grow sharper." -Eden Phillpotts
PV FAQ: http://www.autobahn.mb.ca/~het/energy/pv_faq.html H.E. Taylor http://www.autobahn.mb.ca/~het/

Hi,
So I suppose 1 sqr meter will generate about 300 W using these 40% efficiancy cells?
Let me know if I am wrong.
Parag
William.Mook@gmail.com wrote:

darthpup wrote: This is without doubt the greatest piece of circular reasoning posted on the net to date.
Nonsense. Circular reasoning must make reference to the conclusion at the outset. I do not. So, you're starting off on the wrong foot! Have you even read what I wrote? haha..
Capital costs must be divided by percent utilization.
I've done that skippy, iwhen I said I place the equipment in a location where there is 1800 hours per year of sunlight. haha.. You didn't catch that is equivalent of utilization!
For solar that number is 0.25.
Depends on location - utilization can vary. At my site in Northern Nevada, its 1800 hours per year.
So, take your cost of capital and divide by 0.25 and get a reasonable figure for COE..
COE meaning cost of equipment. You're reading a freakin' pamphlet and think you know something and then lecturing me as if I know nothing. What an asshole.
I've already accounted for utilization when I said how much sunlight fell on the site I was using - and you missed it because you don't understand the basis of the pamphlet you read..
Here, I'll make it easy on you. 1800 hours of sunlight per year means 1800 hours of utilization in a year. this is a location in northern Nevada where I control some land. To figure utilization know there are 8766 hours in a year. 1800 hours divided by 8766 hours equals 20.53% - slightly lower than the 25.00% utilization factor you give.
But I'm using real-world numbers based on real site and real equipment, whereas you are carping over a theoretical form you read in a pamphlet on solar energy that you understand so little of, you don't see what I'm doing and because the form I present my data is just slightly different, you miss it! lol.
And then you have such an attitude! You're a trip dude.
Try using a standard payout analysis to get reasonable energy costs, at least
Whatever! Such attitude from a complete moron! You talk as if payout analysis is important to figuring costs. You've got it bass ackwards! haha.. Equipment cost is important in figuring ultimate value, since you have to subtract that cost from the present value of the revenue stream to obtain a net value. But standard payout analysis doesn't give you your equipment cost. Sheez.
Look, a standard payout analysis is just another way of saying the present value of a stream of cash minus the cost of capital to generate that stream. Since I didn't talk about the stream of cash, and how I might create the stream of cash with this equipment your point doesn't even talk about what I wrote. Earning you the label - moron!
The cost figures stand despite your absolutely stupid statement.
Ultimately, the value created is the difference between the value of the stream of cash the equipment produces minus its cost. This is standard payout analysis. I was just talking about cost here - so your point doesn't even make sense relative to what I was saying. .
Now the value one obtains from standard payout analysis depends on how you cash in on the process. I wasn't discussing that. I was discussing costs. Just because I didn't look at the value of the stream of cash I MIGHT produce, doesn't mean the costs are wrong.
Okay,
Where I place my panels there are 1800 hours of operation in a year - that's a utilization of 20.53% if that's important to you. If you know enough to know how many hours are in a year.
Each watt, produces 1800 watt-hour - that's 1.8 kWh - of electrical output in a year. This cost is less than $0.10 per peak watt and includes balance of system cost for creating hydrogen.
This lets me store energy and sell it wherever there's a buyer for it. I can use CO2 and the Sabatier process to convert the hydrogen to methane, I can use coal and the Bergius process to convert the hydrogen to liquid fuels like jet fuel, diesel fuel, and gasoline. or I can sell the hydrogen directly.
The value of the hydrogen can be computed if you want to do your precious payout analysis. Although I bet you will use it to make more senseless arguments. haha..
A ton of hydrogen has the heat value of 23 barrels of oil. At $50 per barrel energy equivalent, that's $1,150 per ton of hydrogen value. Using this as the value we can do a payout analysis.
Using my process this requires 50 MWh of solar electrical energy to make a ton of hydrogen from 9 tons fo water. A MW of solar collectors with associated balance of system costs - is $100,000 - when made in GW quantities. in Northern Nevada this equipment would produce 1800 MWh of solar electricity in a year, which results in 36 tons of hydrogen per year being produced. This hydrogen has the heat equivalent of 828 barrels of oil. At $50 per barrel value - that's $41,400 per year from $100,000 worth of equipment.
Maintenance and operating costs of $5,000 per year for this equipment, reduces the value to $36,400 per year from $100,000 worth of equipment. Apply your own discount rates on the capital tied up, and a 20 year lifespan with a declining output of 2% per year - to obtain your vaunted payout analysis.
Vary the value of the hydrogen as you like based on what synfuel process you want to support with it - if you want to do that.
Any fool can see I'm radically undervaluing the electricity produced. At $80 per MWh for example, the elctricity alone is worth $144,000 per year so some will argue that I'm making less money than I otherwise might, especially when you see the hydrogen production component is half the cost!
But, this avoids the point that I can make synfuels from sunlight and sell them at market rates and make a huge return on investment. I'll sell electricity certainly, when the sun is shining, but the ability of utilities to absorb electricity at those times limit my output. So, clearly as the size of the installation grows electricity sales will only be a small part of the total. . But in the end, NONE of this affects the equipment cost which is what I spoke of originally.

One square meter of sunlight contains 850 watts of direct energy per square meter at Earth's surface. At 40% efficiency we have 340 watts electrical. In my application there are lenses, and so, this is reduced by another 10% - to about 300 watts overall per square meter.
The panels themselves are about $21 per square meter, and each square meter requires another $9 of attached systems to efficiently produce hydrogen from them. This cost is broken down as follows for 1 square meter;
Hot press molded PET sheets (2) 200 um thick: $0.10 Water filling the molded sheets ultrapure: $0.01 Hot press molded ABS base sheet 400 um: $0.10 0.775 sq in UTJ cells @ $20/sq in: $15.50 copper foil circuit connecting UTJ cells: $5.15 expanded styrene structural foam 3 in by 1 m2: $0.14
TOTAL/m2: $21.00
ELECTRICAL OUTPUT: 300 watts COST PER WATT: $0.07
Now, this is just PEAK POWER. The cost of energy is given by the average power which is much less. First off, through the day, the sun rises and sets, and so you have cosine losses over 12 hours of sunlight per day. This means only 70.7% of the light is available even under ideal conditions. You also have atmospheric effects. Light travelling through many miles more of air at sunrise and sunset is softer than light travelling through less air at noon. This reduces total light available to an optimally oriented flat plate to about 50% of what it might otherwise be. THEN, you have seasonal effects due to the panel being situated at 25 degrees north latitude - Southern Texas or Southern Florida to the Northern States 49 degrees north latitude - which when combined with the 23 degree tilt of the Earth's axis relative to its orbit around the Sun, further reduces light levels. Panels can be optimally tilted South to face the Sun in the North (or any latitude) but the variation of +/- 11.5 degrees around the center line extracts another 2% from the total. And seasonal variations can cause daylight in North America to vary from 8 hours to 16 hours depending on season and latitude
http://en.wikipedia.org/wiki/Image:Day_length.jpeg
And then there is cloud cover!
All these effects mean that we have 4 kWh/m2/da to 7 kWh/m2/da when averaged over the year in the continental US
http://en.wikipedia.org/wiki/Image:Us_pv_annual_may2004.jpg
So, at 40% efficiency, this means that a square meter produces 1.6 kWh/m2/da to 2.8 kWh/m2/da in the continental US
We have optioned land from mining interests in Northern Nevada. This has 5.5 kWh/m2/da and at 40% efficiency this translates to 2.2 kWh/m2/da.
There are 365.25 days in a year - so, we can produce 803.55 kWh per m2 per year. This square meter costs $21 as outlined. And with a discount rate of 8% per year and a 10 year life span we have the cost of capital of $3.13 per square meter and a cost of ENERGY of 0.387 cents per kWh!!!
The cost of the hydrogen conversion (and peak power matching) equipment is $9 for every square meter installed. Here's the complete breakdown per square meter for mysystem
Desc Panel Electrolyzer Cost $21.00 $9.00 Discount 8% 8% Life 10 year 10 years Cost/yr $3.13 $1.34 kWh/yr 808.55 808.55 Cost/kWh $0.00387 $0.00166 Cost/kg $0.19 $0.08 TOTAL Cost/ton $193.53 $82.94 $276.48
This consists of a plastic cylindrical tank filled half way with water and possessing 'D' shaped stainless steel electrodes riding on a plastic tube. The stainless steel is thin coated onto a molded PET sheet forming part of the tube arrangement. DC voltage is applied to either end of a 'stack' of sheets with the only electrical connection from one end of the tank to the other, being the electrodes and the water containing potassium hydroxide electrolyte in low concentration.
The top of the tank has a molded PET header that collects hydrogen and oxygen off the electrodes. By monitoring the current and voltage of the panels, a servo motor is rotated to bring the D shaped electrodes into and out of solution so that optimal current density is maintained as lighting conditions vary.
Water is admitted through a float valve arrangement in the stationary tanks at high pressure and oxygen is released through a turbine arrangement to the atmosphere connecting several 'tubs' with high pressure lines. The turbo arrangement runs an air compressor that pressurizes the water supply, which has a further assist from a DC electric motor.
The high pressure hydrogen is blown down through several tubs to be stored underground in a spent gas well nearby. 100 days supply is stored underground at high pressure. Hydrogen is retrieved and sent to a hydrogenation reactor where coal is converted to liquid fuels, gasoline, diesel fuel and jet fuel.
Physically we operate more than 1 sq meter at a time of course. Each panel we make is 3.0 sq meters (nearly 8ft x 4ft) and 3 inches thick. We string 1,100 panels together into 110 separate circuits and connect to 55 separate electrolyzer tanks at each end of the string (think of Christmas tree light production). Each string is 4,400 feet long and 8 feet wide, and at peak produces 1,122,000 watts of electricity. Over the course of a year a string produces 2,651,715 kWh of energy. A strong of 1100 panels when z-folded together, it conveniently on a 52' x 12' x 16' truck trailer. In the field, one end is staked down and a special tractor plants the string in less than 30 minutes. Two connections at either end, connect the tubs to the circuit. Water and gas lines connect strings at either end, to collectors at the edge of each 'square'
It takes about 50 kWh of energy to produce a kg of hydrogen from 9 kg of water. This is 53 tons of hydrogen per string. 660 strings are needed to cover a mile, and this square mile (with access on either end for pipelines, etc) produce 35,000 tons of hydrogen per year.
Each ton of hydrogen has the heat equivalent of 23 barrels of oil and 5.26 tons of high grade coal.
The hydrogen can be sold directly for this heat value and burned in coal fired generators. So, we arrange with electric utilities to exchange hydrogen for coal, at zero cost. They install hydrogen burners instead of stack cleaners to eliminate pollution.
We obtain the carbon credits and sell those credits in addition to getting the coal.
Each ton of coal is worth $35 and each ton of coal produces about 3.6 tons of CO2 and a ton of avoided carbon creates a credit worth $18. So, that's $66 per ton of avoided carbon emissions, and when added to the value of the coal its $101 per ton of coal equivalent hydrogen heat value.
Each ton of hydrogen has 142 MJ in it. Each ton of coal has 27 GJ. So each ton of solar hydrogen has the capacity to replace 5.26 tons of coal and avoid 18.9 tons of CO2 emissions.
Each ton of hydrogen used this way is worth $530 - so each 'square' described above produces $18.5 million per year producing 35,000 tons of hydrogen replacing 184,000 tons of coal and avoiding 662,000 tons of CO2.
Since the US uses about 1.1 billion tons of coal for electrical generation this means that a total of 5,975 squares are needed to displace all of that along with the pipelines (placed along the Union Pacific Rights of Way) to deliver the 209.12 million tons of hydrogen all over the continental US from our sites in Nevada.
Each 8'x4'x3" panel costs $63. A string costs $69,300. And the tanks on each end cost $14,850 each. Thus a completed string, installed is $99,000 - delivered on two flat bed trucks and installed by a crew of three.
660 strings in a square cost a total of $65.3 million and generates $18.5 million per year, so even with an 8% per year cost of capital and a 10 year life span, a substantial operating profit can be made!!!
Now, a ton of hydrogen can convert 8 tons of coal into 48 barrels of liquid fuels. So, to convert the 1.1 billion tons of coal into 6.6 billion barrels of liquid fuels (about what the US imports each year) requires the addition of 137.5 million tons of hydrogen which in turn requires 3,928 additional squares. Liquid fuels wholesale as commodities for about $63 per barrel. Thus, the 6.6 billion barrels of liquid fuels can generate $415 billion per year. If the same $530 per ton is paid for the solar hydrogen as the utilities pay, then $72.8 billion pays for the additional solar panels. The balance is profit to the oil operation and helps pay for high pressure hydrogenation reactors and fractional distillation units.
This ratio of 3,928 to 5,975 is the ratio of oil to electricity hydrogen production. But the high cost of oil favors its production, while electricity hydrogen sales make a lesser profit, but provide the coal that is already pre-purchased!
A preliminary setup in Nevada is being planned involving 100 squares With 40 squares making hydrogen for oil production and 60 squares making hydrogen for direct sale to nearby utilities. The 60 squares produce the equivalent of 11 million tons of coal each year, enough to power a 3 MW generator nonstop. The coal being exchanged for the hydrogen,is converted to liquid fuels, jet fuel, diesel fuel and gasoline, with the remaining 1.4 million tons of hydrogen, to create 66 million barrels of liquid fuels each year.
Nearly half the energy contained in the fuels come from sunlight, thus we call these fuels 'sunfuels' (C)
The 180,000 b/d facility is worth over $35 billion yet costs only $7 billion to build. Thus the value of the completed facility can be leveraged through bank financing to expand 100x over the next 12 years - to free America of its reliance on foreign energy, and end the stranglehold energy concerns have on our economy and economies around the world. This will free mid East oil to be sold to China and India, accelerating those economies, and making China and India consumer nations creating new markets for US goods and services.
In the end, Northern Nevada will be plated with 10,000 squares of solar collectors and provide copious hydrogen throughout the US to displace all fossil fuels and nuclear fuels in stationary electrical energy production. Hydrogen will also be used to convert the coal thus saved on long term contracts, to oil which will displace foreign oil purchases. Coal and oil production within the US will not be affected. A ton of hydrogen has the heat value of 131.2 mcf of natural gas. At $530 per ton this is equivalent to $4 per mcf natural gas. This is less than natural gas at the well head currently. And so, hydrogen will augment dwindling gas supplies and actually expand the market for direct use of these gases in heating and industrial processes. In all, domestic US energy companies will all benefit. Coal companies oil companies and natural gas companies will benefit from increased demand for their product, the US consumer will benefit from lowered costs and increased availability. Overseas, producers of oil will be free of US interference in their affairs, and increasing demand for oil products in Asia will take up the slack. Increased economic activity in Asia will create demand for US products and services and improve the US economy further. As oil supplies diminish replication of this process overseas will augment dwindling supplies, and direct use of hydrogen will rise.
In the end, within the next 28 years the entire world will enjoy an industrial lifestyle better than the average US citizen today, including US citizens, and the industrial processes linked to solar energy rather than extracted energy, will be some 11x what today's rate of consumption is, while the pollution rates will be some 1/20th of today's level. A total of 250,000 squares will be in place throughout the world.

wrote:

One square meter of sunlight contains 850 watts of direct energy per square meter at Earth's surface...

That's power, not energy.
Nick

Power is a type of measurement of energy. It ain't soup or dirt.
wrote in message

William.Mook@gmail.com> wrote:
One square meter of sunlight contains 850 watts of direct energy per square meter at Earth's surface...
That's power, not energy.
Nick

On Wed, 20 Dec 2006 17:37:30 -0500, "Solar Flare" wrote:

Power is a type of measurement of energy. It ain't soup or dirt.

That is like saying that horsepower is a type of measurement of gasoline...

Solar Flare wrote:

Power is a type of measurement of energy...

Of course. And miles per hour is a measure of distance :-)
Nick

You think electrical power is soup? I think electrical power is a form of energy, not soup.
Miles per hour is a measurement of vehicles.
wrote in message

Solar Flare wrote:
Power is a type of measurement of energy...
Of course. And miles per hour is a measure of distance :-)
Nick

Correct, I mis-spoke slightly trying to explain something more complex! lol. Where solar energy was coming from!
You are absolutely right. Power is the time rate of energy. Its the rate at which energy flows into or out of a system.
I was talking a little about the direct energy from the sun. That is the energy coming from the solar disk itself.
There is of course indirect energy from the sun. That is the energy bouncing off of other things - principally the air, making the sky blue - and lighting things up.
I guess if I were to re-write the sentence I would add two words;
"One square meter of sunlight contains 850 watts of diret energy per square meter per second at Earth's surface'
Cheers
nicksanspam@ece.villanova.edu wrote:

William.Mook@gmail.com> wrote:
One square meter of sunlight contains 850 watts of direct energy per square meter at Earth's surface...
That's power, not energy.
Nick

Wow - haha..
I mis-spoke trying to explain that not all the energy that arrives at Earth's surface if available for solar power applications and someone pointed out that I mis-spoke.
Solar energy of course arrives at a particular RATE that's fixed by the lighting conditions. That rate of energy arrival is power - measured in watts. The total amount of energy arriving in any particular time period is energy - and that can be measured in watt-seconds (aka joules) or watt-hours or kilowatt-hours among other things.
Horsepower is an older unit of measuring power. And it can be related to fuel economy though. The greater the horsepower the faster an engine burns fuel all things being equal - and the faster the car moves.
Miles per hour are not miles certainly. One is a velocity the other distance. The velocity tells you how much time it will take to travel a given distance all things being equal. Velocity is the time rate change of distance.
In an automobile you can relate the rate of energy use in the engine to the rate of fuel consumption and the rate of progress down the road - to get fuel economy - miles per gallon!
An auto engine produces a certain amount of power so you can go somewhere quickly.
Apply a force over a distance and you've used energy. The speed with which you move something against a force means you're using energy at a certain rate - using power. For example lifting a pint of water from a shopping bag to a shelf in your fridge - say 1 foot - you've applied a foot-pound of energy to the water since a pint of pure water weighs a pound. If you lifted the water in a second of time you've expended 1/550th of a horse power aka foot-pound/second.
So, if you know that an engine produces 65 hp while barreling down the road 65 miles per hour you know that it is producing 35,750 ft-lb/sec to travel at 95.3 ft/sec since you know that there are 550 ft-lb/sec in a HP and 3,600 seonds in an hour.
Now, divide ft-lb/sec by ft/sec and obtain the result that the automobile is fighting 375.1 pounds of drag travelling down this road at this speed!
If you know that a gallon of gas has 120 MJ of energy in it and you also know that the engine gets 18 miles per gallon at 65 mph,on this road from the car described above, then you know 3.61 gallons per hour are used by the engine because 65 mph/18 mpg = 3.61 gph
Divide this by 3600 the number of seconds in an hour to obtain 0.001 gallons per second. Mulitply this by 120 MJ per gallon to get 120.3 kJ/second aka 120.3 kW. Now since you know there are 0.746 kW per HP you can find out how much chemical power is being consumed to run the engine - dividing by kW by kW/HP obtains 161.4 HP.
But the engine makes only 65 HP of mehanical energy in the example above. So, we can see that at this speed this engine converts the energy in the fuel with an efficiency of 65/161.4 = 0.402 or 40.2% !!!
This leaves 59.8% wasted as heat. Heat is a form of energy and the rate at which heat is dumped overboard is ALSO a measure of power. The radiator at this speed must dump 96.4 HP of HEAT into the air. Let's say the radiator heats up 5 square feet of air as it travels down the road. So, in a second it travels 95.3 feet. So, a volume of air 95.3 x 5 = 476.5 cubic feet is raised to some temperature each second.
Knowing the heat capacity of air we can figure out what that temperature is knowing a few things.
For air the specific heat capacity is about 1 kJ/kg/K and at sea level the density of air is 1.2 kg/m3.
Well, we know the cubic feet per second = 476.5 and horsepower to be sunk as heat 96.4 HP.
So, let's convert everything to consistent units.
476.5 ft3/s = 13.5 m3/s and 96.4 HP = 129.2 kW. = 129.2 kJ/s
This means that 13.5 m3/s x 1.2 kg/m3 = 16.2 kg/s of air is passing over the radiator.
Divding 129.2 kJ/s by 16.2 kg/s = 7.98 kJ/kg -
which given the specific heat capacity of air means that the temperature of the air is raised by 7.98 kelvin, which is also 7.98 degrees centigrade which is 14.4 degrees farenheit.
And that's about all I have to say on this subjet! lol.

Wow - haha..
I mis-spoke trying to explain that not all the energy that arrives at Earth's surface if available for solar power applications and someone pointed out that I mis-spoke.
Solar energy of course arrives at a particular RATE that's fixed by the lighting conditions. That rate of energy arrival is power - measured in watts. The total amount of energy arriving in any particular time period is energy - and that can be measured in watt-seconds (aka joules) or watt-hours or kilowatt-hours among other things.
Horsepower is an older unit of measuring power. And it can be related to fuel economy though. The greater the horsepower the faster an engine burns fuel all things being equal - and the faster the car moves.
Miles per hour are not miles certainly. One is a velocity the other distance. The velocity tells you how much time it will take to travel a given distance all things being equal. Velocity is the time rate change of distance.
In an automobile you can relate the rate of energy use in the engine to the rate of fuel consumption and the rate of progress down the road - to get fuel economy - miles per gallon!
An auto engine produces a certain amount of power so you can go somewhere quickly.
Apply a force over a distance and you've used energy. The speed with which you move something against a force means you're using energy at a certain rate - using power. For example lifting a pint of water from a shopping bag to a shelf in your fridge - say 1 foot - you've applied a foot-pound of energy to the water since a pint of pure water weighs a pound. If you lifted the water in a second of time you've expended 1/550th of a horse power aka foot-pound/second.
So, if you know that an engine produces 65 hp while barreling down the road 65 miles per hour you know that it is producing 35,750 ft-lb/sec to travel at 95.3 ft/sec since you know that there are 550 ft-lb/sec in a HP and 3,600 seonds in an hour.
Now, divide ft-lb/sec by ft/sec and obtain the result that the automobile is fighting 375.1 pounds of drag travelling down this road at this speed!
If you know that a gallon of gas has 120 MJ of energy in it and you also know that the engine gets 18 miles per gallon at 65 mph,on this road from the car described above, then you know 3.61 gallons per hour are used by the engine because 65 mph/18 mpg = 3.61 gph
Divide this by 3600 the number of seconds in an hour to obtain 0.001 gallons per second. Mulitply this by 120 MJ per gallon to get 120.3 kJ/second aka 120.3 kW. Now since you know there are 0.746 kW per HP you can find out how much chemical power is being consumed to run the engine - dividing by kW by kW/HP obtains 161.4 HP.
But the engine makes only 65 HP of mehanical energy in the example above. So, we can see that at this speed this engine converts the energy in the fuel with an efficiency of 65/161.4 = 0.402 or 40.2% !!!
This leaves 59.8% wasted as heat. Heat is a form of energy and the rate at which heat is dumped overboard is ALSO a measure of power. The radiator at this speed must dump 96.4 HP of HEAT into the air. Let's say the radiator heats up 5 square feet of air as it travels down the road. So, in a second it travels 95.3 feet. So, a volume of air 95.3 x 5 = 476.5 cubic feet is raised to some temperature each second.
Knowing the heat capacity of air we can figure out what that temperature is knowing a few things.
For air the specific heat capacity is about 1 kJ/kg/K and at sea level the density of air is 1.2 kg/m3.
Well, we know the cubic feet per second = 476.5 and horsepower to be sunk as heat 96.4 HP.
So, let's convert everything to consistent units.
476.5 ft3/s = 13.5 m3/s and 96.4 HP = 129.2 kW. = 129.2 kJ/s
This means that 13.5 m3/s x 1.2 kg/m3 = 16.2 kg/s of air is passing over the radiator.
Divding 129.2 kJ/s by 16.2 kg/s = 7.98 kJ/kg -
which given the specific heat capacity of air means that the temperature of the air is raised by 7.98 kelvin, which is also 7.98 degrees centigrade which is 14.4 degrees farenheit.
And that's about all I have to say on this subjet! lol.

wrote:

I guess if I were to re-write the sentence I would add two words;
"One square meter of sunlight contains 850 watts of diret energy per square meter per second at Earth's surface'

This might work, with the word joules vs watts.
Nick

nicksanspam@ece.villanova.edu wrote:

William.Mook@gmail.com> wrote:
I guess if I were to re-write the sentence I would add two words;
"One square meter of sunlight contains 850 watts of diret energy per square meter per second at Earth's surface'
This might work, with the word joules vs watts.
Nick

Yes - quite right! haha - or add the word 'seconds' after watt - same difference.

If the U.S. stops burning coal and oil, it will make coal and oil far cheaper for the rest of the world to consume. The rest of the world will then use as much as the U.S. is using today if not more. So while the U.S. can feel better about itself, I doubt it will change any planetary dynamics.
But then I am quite the pessimist.
-mike
"stcfarms" wrote in message

PV is fine as far as it goes but it is inefficient as only a narrow band of wavelengths from the sun are turned into electricity, the rest of the wavelengths are wasted producing heat. A 5 terawatt geostationary orbit solar furnace could produce enough hydrogen to fuel the world. If the output of the furnace was aimed at a diffraction grating the wavelengths could be separated. Each wavelength could be used where it is most effective, some would even go to PV cells.
T.Keating wrote: First off, let me premise this discussion in that it will only consider ONLY the most up to date current and future trends. (Inefficient, wasteful, and obsolete technologies have no place in our energy future.)
Modern PV cell tech uses silicon wafers with thicknesses under 200um. Density of Si is around 2.33g/cm^3
The standard PV cell dimensions are 125mm *125mm * thickness..
Thus a kilogram of silicon yields a square block approximately 12.5cm (W) * 12.5cm(L)* 2.75cm(H).. "
some prelim calc figures.. or 27.5 mm (H) / 0.3mm == 91.6 cells (10.9g per cell) or 27.5 mm (H) / 0.2mm == 137.5 cells (7.3g per cell)*** or 27.5 mm (H) / 0.15mm == 183.3 cells (5.45g per cell)
Notes: 12.5cm^2 solar cell is exposed to 15.6 watts at std isolation(1000W/m^2 ) *.15(15% conversion eff) == 2.35 watts.. Ribbon process doesn't involve any wafer splitting. Ignot source requires wafer cutting, but wafers thickness is down to 150 to 200um range, with cutting process consuming 70 to 100um of Si material(recycled).
***Using mid range tech of 0.20mm cell thickness, (recycling waste Si) 2.35watts per cell * 137.5 == 323watts per kg.
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Average flat panel solar flux for the US lower 48 angled at latitude is 4.6 kWh/M^2/day.
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/colorgifs/208.GIF (Flat Plate Tilted South at latitude.)
Only a fool would construct them in a location with below average solar flux or mount them in horizontal flat orientation. (Covering maximum surface area is NOT a requirement NOR is it desirable).
The following chart/map when compared to the others demonstrates how much energy production one would loose by making a substandard mounting decision.
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/colorgifs/325.GIF (Flat plate oriented horizontally with the earth.. no tilt)
A significantly better location for large scale PV energy production is the desert SW with 9 to 10kWh/M^2/day of tracked solar flux(annual). (323 watts of PV @ 9kWh/M^2/day == 2.9kWh day/kg of Sit.
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/colorgifs/169.GIF (Flat plate with dual axis tracker.) (Notice the large area in the SW with 8 to 10 kWh/M^2/day production.)
In the end.. the energy payback for the initial SiO2 ->mgSi step using tracked PV in desert SW is somewhat less than a week.. (3 to 5 days depending on tech used)...
Wasteful, ultra pure IC grade silicon used for PV production is being phased out for cheaper alternatives. (It's really not needed anymore).
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As for carbon consumption..
To replace the 3883 billion kWh (2003) of electricity generated in US(2003) would require a ONE TIME investment of 3883e+9 kWh/ 365 / (2.9 kWh/kg) == 3.6 Million metric tons of refined Si..
Carbon (coal) needed to refine this amount of SI from SiO2 would be 3.6e+9 kg /28 * 12 == 1.5e+9 kg C or 1.5Million metric tons..
1.5 Million metric tons of C represents..(0.026%) or less than ONE THOUSANDTH of the US's annual carbon consumption.
http://www-cta.ornl.gov/data/tedb25/Spreadsheets/Table11_04.xls "U.S. Carbon Emissions from Fossil Energy Consumption" 2003..... 5,781.4 Million Metric tons of Carbon.
Imagine that... Just ONE day's worth of carbon usage dedicated towards PV manufacturing could chop US annual carbon emissions by more than 40%. A little conservation(*) and transitioning to EV's could chop off another 40%. (Using 15% of 3883B kWh PV output would propel 150 Million EV's for ~20K miles per year. (@~194wH per mile) ).
Notes:
Huge amounts of raw materials & energy, (consumed by the existing fossil fuel industries), would be conserved.. (ships, mines, ports, drilling rigs, pipelines, trucks, tankers, railroads, fueling stations, and other infrastructure.. mostly gone.. reduced by 90%.. and with any luck.. recycled).
* We will still have other forms of renewable energy production (wind, hydro), plus backup storage in the forms of, anticipatory load management, excess EV battery capacity, and combined cycle gas turbine plants fueled by H2.
H2 can be efficiently produced using a two step process, which combines a high yielding solar thermal energy component (60%), with a greatly reduced electrical energy input(~40% 0.6v) per unit of energy embodied in H2.
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Lastly... Long term PV production is a based on a renewable energy source. Very little mass of a large scale PV plant (>10GW peak) is consumed(lost) during a PV panel's/trackers operational lifetime.. Some erosion, some grease for the gears on the tracking mechanism, and some makeup oil for hydraulic pump & piston.
At the end of PV panel lifespan, ALL of it can all be recycled, on site, making new panels and trackers.
Collection and transportation energy costs near zero ! Separation energy costs near zero.. Clear PV grade Glass. ~80% energy savings.. Aluminum. ~95% energy savings.. (verses ~15kWh/kg for new Al production.) Silicon.. ~90% energy savings.. etc..
I.E. Higher initial energy investment (*), but long term energy production sustainability (>100 years) shifts dramatically towards the side of renewables. EROEI greater than 200 to 1....
* -- (Depends where the materials come from.. Each year the US misplaces 48% or 0.633 million metric tons of Aluminum used to make soda cans because someone was too lazy to recycle them.) http://www.aluminum.org/Content/NavigationMenu/News_and_Stats/Statistics_Reports/Facts_At_A_Glance/factsataglance05.pdf
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In summary.. PV as useful energy source that is progressing at a rapid rate..
Ultimately, It may not take on the exact characteristics as described in this post. But, I suspect, it will take a similar form... The long term EROEI aspects of this solution(>100years) most likely precludes day to day profit driven entities(mega corps) from being part of the solution. Thus it is up to the citizens of the world to force our governments to act on our behalf for the COMMON GOOD and our future generations.