The system will work.
How much rubbish you can compact how many times depends on
- How much sun the panel gets.
- How big the panels are (Watts peak power in full sun)
- What technology they use (crystalline silicon, amorphous silicon, CdTe, ...) - And energy required for trash compaction (bin volume, rubbish type, rubbish volume)
Added long after:
Big Belly technical specification in my dropbox and on their site
In some places below I've added [Actual: xxx] figures next to my prior assumptions.
Based on calculations below it appears that
At a bad inner city location location (sun wise) in Chicago
in midwinter on an average December or January day
using a 20 W mono-crystalline silicon PV panel [Actual: 30 Watt]
and lead acid battery,
You'd probably get 10-15 compactions/day.
For all except 2.5 months of the year you'd get 2+ x that.
That sounds useful.
See below for the derivation of that result and the "assumptions" that it is based on.
As explained at the end, "assumptions" are initial conditions established based on best known information and past experience. They are stated clearly at the start so that the limitations of the system can be understood and so that they can be easily changed if other conditions apply.
The table below is from the wonderful www.gaisma.com site - provides solar and insolation & wind & more information from a vast range of sites worldwide.
The first line = insolation = equivalent hours of full sun daily on average, month by month.
Peak is 6.04 "sunshine hours" per day average in VII = July and lowest is 1.50 sunshine hours per day in December.
A "sunshine hour" will deliver 1000 Watts per square meter.
So an eg 50 Watt panel subject to 1.5 sunshine hours will deliver 50 x 1.5 = 75 Watt hours of energy if the sun is either full on or full off.
For reduced sunshine levels (cloud, shadow, rain, dawn/dusk, fog, ...) the light level will (of course) be lower.
As light levels reduce the best a cell can do is produce proportionately less power, but some cells are better at providing output at low levels than others.
The output of Silicon Monocrystalline cells reduces about in proportion to light level and they have the best output per sunshine hour of any technology commonly used.
Efficiencies vary with what you want to spend but the best cells have over 20% efficiencies and whole panel efficiencies of 17% but a target of 15% - 15% all up is reasonable. At 15% 1 m^2 = 150 Watts and 1 foot^2 = about 14 Watts. Lets assume a 20 W panel for now.
Gaisma Chicago
A 20 W panel in December gives an average of 30 W.h /day and in June - July it gives 120 Wh/day. By the time that gets stored and then used to power the compactor you'll get maybe 50%.
In winter 30 W.h x 50% = 15 W.h.
Assume the compactor uses a 1/4 HP or ~= 200 Watt motor [Actual: 1/6 HP, 130 Watt]
and that a compaction cycle takes 10 seconds (both of which I'd think should be very very adequate for a single garbage bin.).
200W x 10s x 1/3600 s/hr = 0.555 Wh/compaction - say 0.5 Wh.
So with 15 Wh available you get 15/.5 = 30 compactions/day.
BUT that's with a fully illuminated panel that gets what Chicago Winter sun it can when its available. At reduced light levels you get less.
I'll put compactions/day in [[square brackets]] in the following assuming 30 on a good winters day with panel pointed at sun.
A bright clouded sky when you can't see the sun location but it almost hurts your eyes to look at can approach 0.5 suns (50,000 lux)[[15]]. A good bright clouded sky, sun not obvious and not dazzling bright can be 20% of a sun = 20,000 lux[[6]]. Dimly overcast and in deep shaded skyscraper valleys etc can go from 10% / 10,000 lux on down [[<= 3]].
I think the 200 Watts/10 seconds per compaction is probably rather higher than needed. 200 W = 20kg.metre/second. At say 50% electrical to mechanical that's say 10 kg force over 10 metres or 100kg over 1 metre with 10 seconds of operation. You'd have to have some rather stroppy rubbish and a big bin to need this - so you may be able to get say 3 to 5 x as many compactions/day as above.
ie 10 - 15 compactions on an average winter day in a rather unfavourable location.
The above was based on a 20 Watt panel. Resize as required.
I said 50% panel to output via storage.
Battery has current storage efficiency - say 85% for lead acid, and
and voltage conversion efficiency = Vbattery out / Vpanel_rated.
- Using a lead acid battery (most usual) a panel rated at 18V (usually) delivers output from the battery at ~~ 12V so that's 2/3 efficient to start and current charge efficiency of LA is good but not 100% so say 2/3 x 85% =~ 57%. Add some wiring and connection losses and you have ~=50% panel to output.
[Actual battery: 12 Volt. Type unspecified but wording used suggests lead acid.]
A lead acid battery is assumed only for getting a feel for charge/discharge efficiencies. There are many other factors in battery choice but the most significant one is liable to be operating temperature range. In subzero conditions none of the "traditional" batteries do really well.
Overall, if lowest lifetime cost is wanted plus operation in a wide range of temperatures, efficient use of the PV panel then the battery technology of choice is Lithium Ferro Phosphate (LiFePO4). About the only factor which may cause it not to be chosen is initial cost. The mass and volumetric energy densities are lower than for LiIon and for top NimH batteries, but this is unimportant in this role.
Gaisma Chicago
Related material:
"Assumptions"
A user question shows a misunderstanding of the engineering concept and implication of "assumptions".
A student of the art said ...
You make unnecessary assumptions about the system ...
This is extremely important.
An "assumption" is not a restriction per se but the assignment of an initial value to an equation set.
To "make an assumption" is NOT to set a value in stone but just the opposite - it is to say "this is the value we are using but you may wish to vary it depending on what parameters are considered important" etc. The initial values assigned to "assumptions" should not be random but would be expected to be the best available engineering guesstimates based on known data and conditions.
If you can assign a value to something AND if varying that value will affect the result then it is not "unnecessary". If you leave out something which can affect a result to "make things simpler" you risk making them, as Einstein warned "simpler than they can be". It may be that a variable has potential effect but that the solution is insensitive enough that it can be left as a constant or implied in other calculations. Here the volume of the bin may be decided to be unimportant NOT because it does not affect the end result but because all concerned have a general feel for the range of sizes that a rubbish bin is liable to take. My power & energy estimates carries an implicit "assumption" that eg we were not dealing with a 14 cubic metre dumpster.
By identifying factors that affect the result and assigning explicit values you make your answer usefully flexible and allow its limitations to be determined. By leaving possible factors unstated you deem them unimportant. If you find that you have included factors that the result is insensitive to they are simply assigned as constants.
Lets examine the "assumptions in my answer and see which ones are "unnecessary".
Depends on how much sun it gets,
how big the panels are and
what technology they use and
energy required for trash compaction.
4 points. All are key. Substantially vary any one and the result varies accordingly. Next ...
For a 20 W monocrystalline silicon PV panel and
lead acid battery
you'd probably get 10-15 compactions/day at a bad location (sun wise) in Chicago in midwinter on an average December or January day. For all except 2.5 months of the year you'd get 2+ x that.
That sounds useful.
All the above is a statement - its based on the following calculations. A 20 Watt panel is the sort of size seen on road signs and similar in typical city use. Lead acid is the battery technology of choice for industrial use. It's not the best by most measures, but it has low capital cost and some other advantages and it's liable to be what they use in the bins now.
The table below is from the wonderful www.gaisma.com site - provides solar and insolation & wind & more information from a vast range of sites worldwide.
The "assumption" here is that hard data on available solar energy will be "useful".
The meaning of the data table is explained. Degree of and language is targeted at typical site users.
...
Lets assume a 20 W panel for now
That's based on a paragraph of explanation .
A 20 W panel in December gives an average of 30 W.h /day and in June - July it gives 120 Wh/day. By the time that gets stored and then used to power the compactor you'll get maybe 50%.
Based on known real world performance.
In winter 30 W.h x 50% = 15 W.h. Assume the compactor uses a 1/4 HP or ~= 200 Watt motor and that a compaction cycle takes 10 seconds (both of which I'd think should be very very adequate for a single garbage bin.).
More assumptions. Stated so users can change them. Based on (my) real world experience, but clearly stated so that anyone change them.
200W x 10s x 1/3600 s/hr = 0.555 Wh/compaction - say 0.5 Wh.
So with 15 Wh available you get 15/.5 = 30 compactions/day.
Actual calculations so users can see how assumptions are used.
BUT that's with a fully illuminated panel that gets what Chicago Winter sun it can when its available. At reduced light levels you get less.
I'll put compactions/day in [[square brackets]] in the following assuming 30 on a good winters day with panel pointed at sun.
That's all fact based.
A bright clouded sky when you can't see the sun location but it almost hurts your eyes to look at can approach 0.5 suns (50,000 lux)[[15]]. A good bright clouded sky, sun not obvious and not dazzling bright can be 20% of a sun = 20,000 lux[[6]]. Dimly overcast and in deep shaded skyscraper valleys etc can go from 10% / 10,000 lux on down [[<= 3]].
Facts from experience.
I think the 200 Watts/10 seconds per compaction is probably rather higher than needed. 200 W = 20kg.metre/second. At say 50% electrical to mechanical that's say 10 kg force over 10 metres or 100kg over 1 metre with 10 seconds of operation. You'd have to have some rather stroppy rubbish and a big bin to need this -
Assumption modification. Clearly stated. Clearly reasoned.
so you may be able to get say 3 to 5 x as many compactions/day as above.
ie 10 - 15 compactions on an average winter day in a rather unfavourable location.
Reassess based on the above.
The above was based on a 20 Watt panel. Resize as required.
...
I said 50% panel to output via storage.
Battery has current storage efficiency - say 85% for lead acid, and
and voltage conversion efficiency = Vbattery out / Vpanel_rated.
* Using a lead acid battery (most usual) a panel rated at 18V (usually) delivers output from the battery at ~~ 12V so that's 2/3 efficient to start and current charge efficiency of LA is good but not 00% so say 2/3 x 85% =~ 57%. Add some wiring and connection losses and you have ~=50% panel to output.
Figures based on experience.
Hmmm.
We seem to be at the end.
I don't see anything you'd want to risk leaving out.
I don't see any fudging, miracles, etc.
Uncertainties? Sure. But stated.
How many watts?
How long a compaction cycle ?
How many cycles a day? ...
If you can offer constructive way to improve on that I'd be pleased to see them.
Assume the power input to the bulb is 10 Watts.
Assume for now 100% efficiency from battery output to bulb input.
Efficiency of energy storage by the battery of energy supplied to it will vary with battery chemistry and how well the charger is designed. Best case using a Lithium battery of some sort, over 90% efficiency may be achievable. Lower or much lower efficiency is often achieved in practice.
Efficiency of energy provided at the battery terminals compared with energy out of the PV panel will depend on the interface design and will also vary with battery state of charge.
Power output from the panel at any moment (Wp) and compared to the maximum power the panel can make under ideal conditions (Wmpp) will vary with insolation level (sunshine level), panel conditions, atmospheric conditions and more.
SO overall, a say 100 Watt panel will produce 100 Watts in full sunshine when new and will produce the equivalent of 2 or 3 hours of equivalent sunshine in most continental US locations in winter and 5 to 6 hours of equivalent full sunshine.
ie you get 200 to 700 Watt-hours per day depending on season.
With the very best interface equipment (MPPT, intelligent battery sizing to minimise resistive losses, ... you may get 95% + of this energy at the battery terminals and, as above, 90%+ of this actually stored into the battery.
So PV Watts rating x 0.95 x 0.9 x hours_equivalent_per_day = Watt-hours available. Say 85%. Using 80% would be safer and still very optimistic in many cases.
At the start I assumed 100% battery out to bulb in power.
Regardless of load type (which is usually LED in this context), if you want constant brightness as battery varies or constant "bulb" input there will be some conversion losses. 90% from battery to bulb or LED would usually be excellent.
So overall PV "nameplate rating" watt-hours to 'bulb' input watt-hours is at best about 75%. Usually less.
When the sun is providing energy, some gains can be had by running the bulb from the panel without battery storage. This gain is useful but still a small part of the total energy needed via the battery. I'll ignore it in the following and it can be factored in later if needed.
From the above:
Watt hours available = (Panel Watts rated) x 75% x Sunshine hours.
Watt hours wanted = Load_Watts x 24.
Rearranging the above -
Panel Watts needed = Load Watts x 24 / (0.75 x Sunshine hours )
= Load_Watts x 32 / Sunshine_Hours
So eg 10 Watt load in winter with 2 hours/day sunshine hours /day (= equivalent full sunshine).
Panel Watts needed = 10 x 32 / 2 = 160 Watts !!!
10 Watt load in Summer with 6 sunshine hours/day.
Panel watts needed = 10 x 32 / 6 = 53 Watts.
In practice higher Watts will be needed.
Averge sunshine hours per day can be found at the wonderful Gaisma site here - this example is for Houston
Top line is insoltaion in kWh/m^2/day = sunshine hours/day = hours of equivalent full sunshine. I = January, II = February etc.
2.34 hours/day in January.
5.98 hours/day in July
These are means for many years and any year and any day in the montyh may vary widely from this. That's weather for you :-)
More later ...
Best Answer
If they wrote the firmware right, a reading in kilowatts (kW) is the output of the solar array at that moment. The total "work", taken over time, will be in kilowatt-hours (kWh). The time period is arbitrary; it may be taken over the course of a day, or a month, or it may be a lifetime total.
If you want to work it out yourself, by recording the output each second, you still have to assume that the output over each second was constant. But that will normally be close enough. One second's worth of power, by the way, gives you watt-seconds, a unit not normally interesting unless you are looking at something like an impulse from a capacitor. 3600 consecutive 1-second readings will give you the output that was generated during that hour, and it will be in kWh. Remember, it's a cumulative total, just like filling a swimming pool.