Sizing a Solar Array To Boil Water

[mcharbs55]

Hello,

I am having a hard time deciding the size of a solar array to boil water for a school project. Before putting any thought into the subject I simply thought that I would just take the wattage of the panels and add them up since they are added in both series or parallel and then select an element close to that wattage. Apparently this is not the case.

I am trying to do my calculations for a regular household element that is rated for 1500W at 120vac with panel specifications as 31.5V and 8.89A. My calculations go as follows:
With 5 panels connected in series=157.5V, 8.89A
Element Resistance: 120^2/1500= 9.6ohms
Element Power: 157.5^2/9.6= 2,583.98W
Element Current Draw: 2,583.98/9.6= sqrt269.16= 16.41A *This value cannot be supplied*

How would I go about calculating the voltage drop caused by the current draw being higher than the panels are capable of handling?
regular vdrop calculation would be: IR= (8.89)(9.6)=85.34V
if the power is recalculated with the new voltage of 72.16V I get a wattage of 542.4W. This seems way too low...also, performing this vdrop calculation does not seem to make sense in this scenario. If it is correct, then why are panels that are rated to supply 1400W providing a value so small? Would rearranging the panels make any difference (1 in parallel and 4 in series)?

Thanks for any help you can provide.

 

[Khashishi]

There's no need for photovoltaics if you are boiling water. You can use a parabolic reflector to heat much more efficiently.

 

[russ_watters]

I am trying to do my calculations for a regular household element that is rated for 1500W at 120vac with panel specifications as 31.5V and 8.89A. My calculations go as follows:
With 5 panels connected in series=157.5V, 8.89A
... Would rearranging the panels make any difference (1 in parallel and 4 in series)?

You can't connect an AC load to a DC source directly. The panels should be wired in parallel to a controller/inverter to change the current to AC.

What level of schooling is this and what level of supervision are you getting? This can be dangerous for someone lacking the required knowledge.

 

[Khashishi]

Are you talking about a store-bought heater? If it's a simple resistive heating element, then it can be made to work with AC or DC at various voltages, but a store-bought heater will probably only work with 120VAC.

 

[sophiecentaur]

If this is a school project then there is absolutely no need to be trying to mimic a domestic installation. It is potentially very dangerous and needlessly expensive. PV panels are available in all maximum power ratings from 1W to 1kW and a small version would be a far better project.
There is a range of low cost panels will provide you with 12V DC at a few tens of Watts. That voltage would operate many readily available pieces of DC equipment, including lamps and radios.
But it is very inefficient to use PV as a source of heating only. Direct heating with a parabolic style reflector is far better value and a lot cheaper.
I used an old 0.9m microwave dish with cooking foil stuck to it to boil a few hundred cc of water in a black can in a very few minutes. It needed full sunlight, of course. But so would your PV version. The focus was not at all critical because it was better to heat most of the can's side surface with the defocussed beam. A perfectly adequate reflector shape could be made with paper mache´ built up on an exercise ball or any other handy shaped object.
I have to point out the danger of direct sunlight, focussed onto a small area and the risk of fire and blinding. A diffuse reflector is easier to make and much safer but would produce the same effect without a dangerously bright focussed image of the Sun. You should not consider using more than, say 1m² of area, which would give just a safe few hundred Watts of heating. Your teacher should be aware of this in any case (I would hope)
Edit: If you really want to go for the PV solution, you need to use an appropriate load. If you try to load a PV panel with a lower resistance than corresponds to its rated power, the power output drops drastically (the Volts drop like a stone) You should ask teachers for specific help with this.

 

[NTL2009]

If this is a school project then there is absolutely no need to be trying to mimic a domestic installation. It is potentially very dangerous and needlessly expensive. PV panels are available in all maximum power ratings from 1W to 1kW and a small version would be a far better project.
There is a range of low cost panels will provide you with 12V DC at a few tens of Watts. That voltage would operate many readily available pieces of DC equipment, including lamps and radios. ... .

Good advice - if this is all about demonstrating principles, keep it scaled down. As others mentioned, the OP is proposing dangerous voltages.

Go with a 12V panel, and if you still want to boil water, boil a single cup with one of these (~ 140 watts):

https://www.amazon.com/dp/B001AQZI64/?tag=pfamazon01-20

 

[sophiecentaur]

Go with a 12V panel, and if you still want to boil water, boil a single cup with one of these (~ 140 watts):

That could be a fun project - if you have the money for 0.5m2 with of panel area (Around £200). But what would you do with the panel afterwards, to justify that cost? You would need a battery and controller if you wanted it for other uses. I have such a system which lights my shed. It has actually worked for over a year (all seasons) to power internal lights and security lights. It uses only a 40W panel.
A much smaller solar reflector than 0.5m² would do the same job as 140W worth of panels and the parts would be very low cost.
I could suggest that boiling water would not be the best way to demonstrate a PV based system; it't not the tool for the job. People can be very impressed by moving parts, light and sound, powered from a solar panel. Going for more than a low power panel would increase impact because 'everyone' knows about 10W panels charging phone batteries.

 

[OmCheeto]

... Would rearranging the panels make any difference (1 in parallel and 4 in series)?

Thanks for any help you can provide.

Interesting question.

My guess is no.

ps. I'm not willing to do the experiment, as I only have 3 of 4 of my original solar panels left.
pps. Here's a graph of me trying to analyze my PVs last summer:

 

[gleem]

Typically solar panels are rated for max power, open circuit (max) voltage (I=0) and short circuit (max) current(V=0).

Below is a typical characteristic curve for a solar cell for different light intensities with different loads showing the resulting currents and voltages. The max power occurs near the top of the knee around 0.8 V_max. You can see that the power produced is highly dependent on the choice of load resistance.

 

[davenn]

Hello,

I am having a hard time deciding the size of a solar array to boil water for a school project. Before putting any thought into the subject I simply thought that I would just take the wattage of the panels and add them up since they are added in both series or parallel and then select an element close to that wattage. Apparently this is not the case.

I am trying to do my calculations for a regular household element that is rated for 1500W at 120vac with panel specifications as 31.5V and 8.89A. My calculations go as follows:
With 5 panels connected in series=157.5V, 8.89A
Element Resistance: 120^2/1500= 9.6ohms
Element Power: 157.5^2/9.6= 2,583.98W
Element Current Draw: 2,583.98/9.6= sqrt269.16= 16.41A *This value cannot be supplied*

How would I go about calculating the voltage drop caused by the current draw being higher than the panels are capable of handling?
regular vdrop calculation would be: IR= (8.89)(9.6)=85.34V
if the power is recalculated with the new voltage of 72.16V I get a wattage of 542.4W. This seems way too low...also, performing this vdrop calculation does not seem to make sense in this scenario. If it is correct, then why are panels that are rated to supply 1400W providing a value so small? Would rearranging the panels make any difference (1 in parallel and 4 in series)?

Thanks for any help you can provide.

you are approaching this the wrong way

you only need around 12VDC ... there are 12V water heating elements available
they are used for water heating to run off a car battery used for camping

depending on the wattage of the panel(s) and the wattage of the 12V element ... you may only need 1 panel or maybe a couple in parallel

as others have said, keep the project small scaled to keep it safe

You can't connect an AC load to a DC source directly.

most of them are plain resistive element ... it will work on AC or DC Dave

 

[sophiecentaur]

you only need around 12VDC ... there are 12V water heating elements available

Unfortunately, the OP ( I now realize) seems already to have 30V panels (?) and they would be difficult to butcher to work at lower volts.
But power resistors are not that expensive and could be used as heating elements. Perhaps two 12V heaters could be connected in series. For Auto use, any heater would be capable of running on 15V. If you chose to use a Maximum Power Point Tracking (MPPT) Solar Charge Controller you could optimise the available heat.

 

[davenn]

Unfortunately, the OP ( I now realize) seems already to have 30V panels (?) and they would be difficult to butcher to work at lower volts.

Doh ... yeah you are correct, I did read that but overlooked it in my reply...
don't know what immersion heater elements are available around that voltage ... google hahaor put 2 or 3 12V elements in series, problem solved

actually probably only need 2 elements as panels usually specify open circuit voltages
so it will drop under load

 

[NTL2009]

Unfortunately, the OP ( I now realize) seems already to have 30V panels (?) and they would be difficult to butcher to work at lower volts.
But power resistors are not that expensive and could be used as heating elements. Perhaps two 12V heaters could be connected in series. For Auto use, any heater would be capable of running on 15V. If you chose to use a Maximum Power Point Tracking (MPPT) Solar Charge Controller you could optimise the available heat.

It's not clear to me if this is still in the planning or hypothetical stage, or OP actually has these panels and is trying to configure them.

Hopefully, OP comes back to clarify. IMO, this whole thread is rather muddy. Is this a learning exercise, or an application, or both? OP should have learned it takes a lot of PV panels to boil a significant amount of water!

 

[OmCheeto]

It's not clear to me if this is still in the planning or hypothetical stage, or OP actually has these panels and is trying to configure them.

Hopefully, OP comes back to clarify. IMO, this whole thread is rather muddy. Is this a learning exercise, or an application, or both? OP should have learned it takes a lot of PV panels to boil a significant amount of water!

"still in the design phase" according to "PV panels to boil water 1.0" thread.

 

[davenn]

OP should have learned it takes a lot of PV panels to boil a significant amount of water!

good grief ... I doubt he's trying to heat a swimming pool ! ... it's a school project, so you could safely assume a litre or 3 of water

my suggestion above will do that very easily with just one of the panels that he stated
for an element rated at 24V up to ~ 200W

ebay and other places are full of examples

Dave

 

[sophiecentaur]

my suggestion above will do that very easily with just one of the panels that he stated
for an element rated at 24V up to ~ 200W

. . . and that's quite enough if you want to avoid hot wires[(use really fat connecting wires - 10mmSUP]2[/SUP] ) and too much hot water. If this is for a demo, I would think regularly boiling small amounts of water would have more impact, rather than many litres at a time. A glass beaker with Good thermal insulation (with a hole in the insulation for people to see the bubbles) would add to the impact. Audience concentration span is a factor here.

 

[NTL2009]

good grief ... I doubt he's trying to heat a swimming pool ! ... it's a school project, so you could safely assume a litre or 3 of water

my suggestion above will do that very easily with just one of the panels that he stated
for an element rated at 24V up to ~ 200W

ebay and other places are full of examples

Dave

I agree, but the OP seemed "hot" (pun intended) on using a 1500W element to its full rating, and that takes a lot of panels Also, those elements need to be fully submerged when used near their rated power or they overheat and burn out quickly (seconds?), that adds to the set up complexity, along with safety issues of 120 V and water.

Also agree with sophiecentaur, small volumes boiled quickly would make a better presentation. Heck, I'd go all the way down to a 12V 30W soldering iron, and boil a shot glass sized (should be an SI unit!) amount of water. That should be cheap and small enough to make the point. And a small panel can be easily handled, people could easily experiment with changing the angle, shading a bit of it, etc.

 

[sophiecentaur]

I still think that PV for heating is not a good message to be giving. Anyone who is serious about a domestic PV supply in the absence of the Feed In Tariff that totally distorts things, should plan a dual system. Solar thermal for heating and PV for when you actually need Electricity.
If the OP has the toys already, there is no argument that he should use them. But why use them for an inappropriate purpose such as heating? If this project will be presented to 'the public' then he should be prepared for someone to bring this up and to have a good answer ready to an embarrassing question and have demos of other more impressive uses for the power from his PV. 140W can do some pretty impressive mechanical tasks.

 

[OmCheeto]

It seems to me that we've exhausted the ohmic logic here.

Solar panels are not ohmic devices. They are ≈constant current devices on one end of the curve, and ≈constant voltage devices at the other end of the curve.

I really liked mcharbs55's question; "Is there a way to hook these things up, like resistors, or batteries?" [paraphrased]

The above schematic is grossly simplified, in the context of this question, of course.

The question is, "Does the single panel on the left, limit the current capability of the two panels on the right?"

I spent about an hour yesterday trying to find an answer, and could not find one.
Hence, my "No" guess.

ps. I have no idea how solar panels work, other than; "quite well".
pps. If the panel on the left is NOT capable of carrying 17.8 amps, then the problem becomes very trivial.

Spoiler: and the answer is...

8

 

[sophiecentaur]

ps. I have no idea how solar panels work, other than; "quite well".

That should be carved in stone, somewhere. I love it.

 

[OmCheeto]

That should be carved in stone, somewhere. I love it.

Ha! I was recalling when I once had a recommendation for a youngster.
I assumed I would never be able to find the post. but*...

ronaldor9;

"One day i would like to work on improving and designing Photovoltaic solar panels. I was wondering which degree in university is generally took, if one wants to research in solar panels. I for sure know that mechanical engineers research with thermal solar panels, but what about voltaics?

thanks"​

...
Om;

"Since PV panels are quantum devices, I would recommend going into the field of quantum mechanics if you want to improve their basic design. But all of the fields will contribute to the ultimate systems, so it's up to you.

It may impractical to create a 100% efficient PV panel, so I'm sure it was a ME that pointed out that the thermal waste byproduct of current PV panels can be put to good use.

And any half-witted EE would be able tell you that PV's in series vs parallel will cut your transmission losses by the square."
"​

hmmm...
8 years to the day: May 6, 2009 [ref]
hmmmm...

(counts fingers to make sure 17-9 = 8)
-----------------------------
*pfoogleing: When did Om use the terms 'engineer' and 'quantum' in a single post? Bingo!

 

[davenn]

I agree, but the OP seemed "hot" (pun intended) on using a 1500W element to its full rating, and that takes a lot of panels Also, those elements need to be fully submerged when used near their rated power or they overheat and burn out quickly (seconds?), that adds to the set up complexity, along with safety issues of 120 V and water.

he also wanted to use multiple high power panels ... it was obvious that he didn't realize that there were lower power/voltage heater elements available

 

[NTL2009]

I still think that PV for heating is not a good message to be giving.

... But why use them for an inappropriate purpose such as heating? If this project will be presented to 'the public' then he should be prepared for someone to bring this up and to have a good answer ready to an embarrassing question ...

Well, it depends upon what the goal is.

If your goal is to show that PV is not the best way to heat water, this could make quite an impression!

And I do feel that the OP really hasn't outlined his/her goal, just threw out a bunch of different scenarios. The first step in solving a problem is defining the problem.

That should be carved in stone, somewhere. I love it.

 

[sophiecentaur]

Well, it depends upon what the goal is.

If your goal is to show that PV is not the best way to heat water, this could make quite an impression!

And I do feel that the OP really hasn't outlined his/her goal, just threw out a bunch of different scenarios. The first step in solving a problem is defining the problem.

Yes you are right but there is such a lot of misinformation about 'Energy'. In the UK, I don't think anyone uses large scale off-grid energy and the Feed In Tariff affects choices. But PV as a heat source (not just for boiling a kettle of water) is a poor solution. I had a neighbour (in the urban centre of Brighton) who has built an Eco-house, doing it properly, (zero footprint) and has used dual source solar energy - and that decision included the FIT factor.
Imo, even opportunity should be taken (especially by a teacher) to educate the public. PV is marginal enough value in the short term, as it is.
(loved the stone)

 

[gleem]

If your goal is to show that PV is not the best way to heat water, this could make quite an impression!

And I do feel that the OP really hasn't outlined his/her goal, just threw out a bunch of different scenarios. The first step in solving a problem is defining the problem.

We haven't heard from the OP regarding suggestions. The characteristic curve of the chosen solar panel must be consulted from which you choose the appropriate resistance for maximum power. Trying to fit the panels to the element is not a good approach. If the resistance is too high you kill the output.

We do not know what kind of project this is, a DIY hot water heater or a science project. One might possibly consider an alternative method of heating the water using a passive heating system and a solar cell driven pump to circulate the water which will use much less energy . Solar cells are only about 20% efficient.

 

[sophiecentaur]

a passive heating system and a solar cell driven pump

That's very elegant and I have seen it somewhere before. No battery needed for the pump because you only need it to run when the sun is up.

 

[OmCheeto]

We haven't heard from the OP regarding suggestions. The characteristic curve of the chosen solar panel must be consulted from which you choose the appropriate resistance for maximum power. Trying to fit the panels to the element is not a good approach. If the resistance is too high you kill the output.

We do not know what kind of project this is, a DIY hot water heater or a science project. One might possibly consider an alternative method of heating the water using a passive heating system and a solar cell driven pump to circulate the water which will use much less energy . Solar cells are only about 20% efficient.

bolding mine

hmmmm...
(pfoogle pfoogle pfoogle)
ah ha!

Let's see... 24 watts of electrical input over 3.5 hours yields 84 watt hours consumed.
Energy gained by the system was 2300 watt hours.
2300 - 84 = 2216 net watt hours
system efficiency: η = Pout / Pin
= 2216/84 = 2338% efficiency

Ha! Take that wikipaedia!

Efficiencies may not exceed 100%

Dullards have apparently never heard of the Kobayashi Maru. When in doubt, cheat.

[ref]

Can't remember what the total cost of the system was.

1. 50 watt solar panel: inherited from my dead dad. (I think he probably paid $500 for it. Early adopter. Scaled cost today would be around $40)
2. 100' of black irrigation tube: ≈$15
3. bilge pump: I have 5 very old wooden boats... They are ubiquitous in my house. (I don't have a garage, nor shed, so shovels, rakes, ladders, and solar panels occupy my most all of the rooms in my house.)

answers:

$15 for me
$70 from scratch​

ps. The sun is in and out today. I may do the experiment I was referring to earlier.

 

[OmCheeto]

...
I may do the experiment I was referring to earlier.

hmmmm...
Looks like I guessed correctly.

Interesting how Kyocera's "optimum current" matches the power and voltage with current intersection point of my graphs so well.
Guessing by "optimum", they meant maximum.

 

[gleem]

Guessing by "optimum", they meant maximum

Actually the optimum voltage is that voltage that gives the maximum power output for the maximum sunlight. The max voltage occurs at zero current which is about 25% higher.

 

[OmCheeto]

Actually the optimum voltage is that voltage that gives the maximum power output for the maximum sunlight. The max voltage occurs at zero current which is about 25% higher.

Believe it or not, I was lying in bed this morning worrying about my comment, hoping I had edited it out, before I pushed the "post reply" button.

But anyways, just scratching my head, I can't figure out what the minimum number of panels is to boil the water.
One moment...

In series, the answer is 6 panels. (189 VDC @ 8.89 amps)
In series-parallel, the answer is 8 panels. (126 VDC @ 13.13 amps)
With a power optimizer circuitry, the answer is 5.4 panels, which is effectively 6.

So, I guess I'm done experimenting, and have learned something.

ps. Good lord I'm slow. I just realized why with the 3 panels in parallel, the voltage and power graphs didn't droop. There's a 9 amp limit!
hmmmm... Maybe I will do more experimenting. I'll bet I can make my resistors REALLY smoke.

pps. Regarding a comment in mcharbs55's "Solar panels & coffee pots part 1" thread:

One thing to keep in mind - manufacturers often specify wattage rating at operating temperature. Heater element wire has a positive temperature coefficient (often abbreviated as 'tempco') which means resistance increases as temperature increases, so don't be alarmed if an ohmmeter measurement at room temperature yields a value somewhat lower than the calculated 9.6 ohms.

I have a Kill-A-Watt meter, and measured the power consumption of my coffee pot. I also measured the resistance of the heating element. I did not notice an appreciable deviation in wattage from what I calculated it should be from the "no load" resistance. My guess is that the heating element temperature did not go much over the boiling point of water.

hmmmmm... This gives me an idea for experiment #4.

[edited for typo: it ≠ in]

 

[gleem]

It series, the answer is 6 panels. (189 VDC @ 8.89 amps)
In series-parallel, the answer is 8 panels. (126 VDC @ 13.13 amps)
With a power optimizer circuitry, the answer is 5.4 panels, which is effectively 6.

I hope you realize that you cannot use the same resistance heater for both the series and parallel setups and get the same power transfer. And how did you get 13.13A for the parallel-series setup?

 

[OmCheeto]

I hope you realize that you cannot use the same resistance heater for both the series and parallel setups and get the same power transfer.

I'm still kind of working on the OP's problem, so the resistance is fixed.

And how did you get 13.13A for the parallel-series setup?

I just kept adding sets until I reached or exceeded the requested power level.

 

[gleem]

Ideally you need the specific characteristic curve ( the current- voltage relationship) for his SP to determine the actual current and voltage for the 9.6Ω for max power. If you are constrained to a certain resistance heater you should determine the voltage to deliver the needed power with the max current available 8.89A or determine the current for the optimum voltage or close to it. So we have a max current of 8.89 A and a voltage of 31.5 V for max power and a 9.6Ω resistor.with a power rating of 1500W. If we can deliver 8.89A we need168.7V which we get by connecting 5.3 panels in series. If we connect 5.3 panels in parallel we get the same power but at a lower safer voltage. So while not optimal we settle for 5 panels.

 

[OmCheeto]

Ideally you need the specific characteristic curve ( the current- voltage relationship) for his SP to determine the actual current and voltage for the 9.6Ω for max power. If you are constrained to a certain resistance heater you should determine the voltage to deliver the needed power with the max current available 8.89A or determine the current for the optimum voltage or close to it. So we have a max current of 8.89 A and a voltage of 31.5 V for max power and a 9.6Ω resistor.with a power rating of 1500W. If we can deliver 8.89A we need168.7V which we get by connecting 5.3 panels in series.

That's kind of what I inferred above:

In series, the answer is 6 panels. (189 VDC @ 8.89 amps)
In series-parallel, the answer is 8 panels. (126 VDC @ 13.13 amps)
With a power optimizer circuitry, the answer is 5.4 panels, which is effectively 6.

If we connect 5.3 panels in parallel we get the same power but at a lower safer voltage. So while not optimal we settle for 5 panels.

Not with a fixed resistance. In that case we would always get the same power output.

ps. Asymptotic was correct. The resistance of my coffee maker element rose from 14.4Ω to 15.0Ω when void of water. Observed wattage dropped from 979 to 938.

pps. On a humorous side note, my coffee maker is so old, it is ungoogleable*. The closest I can come is an ad on Etsy:

Overview

• Vintage item from the 1970s

-------------------------

* Other are welcome to try:

PROCTOR-SILEX
2209 SULPHUR SPRING ROAD
BALTIMORE,MD.21227
120 VOLTS A.C.ONLY
1000 WATTS 60 HZ
MADE IN U.S.A.
TYPE A06
MODEL A6126
SERIES A 3288​

ppps. I knew it was probably old when I saw "MADE IN U.S.A.", but not that old.

 

[sophiecentaur]

What you need is a switch mode controller - equivalent to a MPPT for charging batteries. That will give you the best Voltage for the available light. I know that there are some good designs of charge control but, whether that goes for resistive loads, I have no idea. Could be an interesting project on its own. Problem is that the VI curve has a fairly sharp knee so the benefit would not be that great - just allowing you to use a slightly lower resistor (= more power at times).

 

[Asymptotic]

That's kind of what I inferred above:Not with a fixed resistance. In that case we would always get the same power output.

ps. Asymptotic was correct. The resistance of my coffee maker element rose from 14.4Ω to 15.0Ω when void of water. Observed wattage dropped from 979 to 938.

pps. On a humorous side note, my coffee maker is so old, it is ungoogleable*. The closest I can come is an ad on Etsy:

Overview

• Vintage item from the 1970s

-------------------------

* Other are welcome to try:

PROCTOR-SILEX
2209 SULPHUR SPRING ROAD
BALTIMORE,MD.21227
120 VOLTS A.C.ONLY
1000 WATTS 60 HZ
MADE IN U.S.A.
TYPE A06
MODEL A6126
SERIES A 3288​

ppps. I knew it was probably old when I saw "MADE IN U.S.A.", but not that old.

That sounds about right. Immersed, element wire temperature won't rise very much above the temperature of the water surrounding the heater sheath, and limits how much the electrical resistance changes. I wonder how much the longevity of your coffee maker is due to wattage rating versus robust construction; lower power rated elements tend to last longer.

Better check my math (gray-haired American engineers often think in BTU), but about 234 watts are required to increase 15°C water in a 10 cup (US measure; 2.5 liter) coffeemaker to 100°C at sea level. Ignoring thermal losses, this takes about 14 minutes at 1000 watts, and 9.3 minutes at 1500W, and may help explain why in this impatient age the more modern kettle was designed at a higher power rating.

 

[OmCheeto]

...

Better check my math (gray-haired American engineers often think in BTU), but about 234 watts are required to increase 15°C water in a 10 cup (US measure; 2.5 liter) coffeemaker to 100°C at sea level. Ignoring thermal losses, this takes about 14 minutes at 1000 watts, and 9.3 minutes at 1500W, and may help explain why in this impatient age the more modern kettle was designed at a higher power rating.

bolding mine

You might want to double check your maths.
Fortunately for me, I used a Pyrex measuring cup to check how much water was in my carafe.

10 of my coffeemaker cups ≈ 1.4 liters​

according to wiki:

10 coffeemaker cups = 1.5 liters

https://en.wikipedia.org/wiki/Cup_(unit)#Metric_cup

A "coffee cup" is 1.5 dl or 150 millilitres or 5.07 US customary fluid ounces, and is occasionally used in recipes. It is also used in the US to specify coffeemaker sizes (what can be referred to as a Tasse à café). A "12-cup" US coffeemaker makes 57.6 US customary fluid ounces of coffee, or 6.8 metric cups of coffee. In older recipes cup may mean "coffee cup".​

It took my coffee maker ≈9 minutes to brew 1.35 liters of water.
Which from my calculations, indicates an output of 858 watts, which is 90% of the measured wattage of ≈958 watts.
Tc = 291 K (measured)
Th = 373 K (theoretical. I measured outlet water temperature at only 361 K, but decided that didn't make sense, if my assumption on how drip coffee makers work is correct.)

I wonder how much the longevity of your coffee maker is due to wattage rating versus robust construction; lower power rated elements tend to last longer.

My guess is construction. I just dug my mothers other older coffee percolator out from the back of my top shelf kitchen cabinets. From the date of when the company went out of business, it's ≥ 52 years old.

And you are probably right about people getting more impatient. This one is rated at 600 watts.
The power cord has become dissociated, but I'm sure it still works. Measured resistance: 26 Ω

ps. Still wondering why all these non-electronic coffee makers of mine say; "A.C. ONLY". Did people have 120 VDC available back in the 50's, 60's, and 70's? There were no solar panels back then. hmmmm...

---------------------
UNIVERSAL
MADE BY
LANDERS, FRARY & CLARK
NEW BRITAIN, CONN., U.S.A.
NO. C4580
VOLTS 110-120
WATTS 600
UL®
A.C. ONLY
DO NOT IMMERSE IN WATER

 

[Asymptotic]

bolding mine

You might want to double check your maths.
Fortunately for me, I used a Pyrex measuring cup to check how much water was in my carafe.

10 of my coffeemaker cups ≈ 1.4 liters​

according to wiki:

10 coffeemaker cups = 1.5 liters

https://en.wikipedia.org/wiki/Cup_(unit)#Metric_cup

A "coffee cup" is 1.5 dl or 150 millilitres or 5.07 US customary fluid ounces, and is occasionally used in recipes. It is also used in the US to specify coffeemaker sizes (what can be referred to as a Tasse à café). A "12-cup" US coffeemaker makes 57.6 US customary fluid ounces of coffee, or 6.8 metric cups of coffee. In older recipes cup may mean "coffee cup".​

It took my coffee maker ≈9 minutes to brew 1.35 liters of water.
Which from my calculations, indicates an output of 858 watts, which is 90% of the measured wattage of ≈958 watts.
Tc = 291 K (measured)
Th = 373 K (theoretical. I measured outlet water temperature at only 361 K, but decided that didn't make sense, if my assumption on how drip coffee makers work is correct.)
My guess is construction. I just dug my mothers other older coffee percolator out from the back of my top shelf kitchen cabinets. From the date of when the company went out of business, it's ≥ 52 years old.

And you are probably right about people getting more impatient. This one is rated at 600 watts.
The power cord has become dissociated, but I'm sure it still works. Measured resistance: 26 Ω

ps. Still wondering why all these non-electronic coffee makers of mine say; "A.C. ONLY". Did people have 120 VDC available back in the 50's, 60's, and 70's? There were no solar panels back then. hmmmm...

---------------------
UNIVERSAL
MADE BY
LANDERS, FRARY & CLARK
NEW BRITAIN, CONN., U.S.A.
NO. C4580
VOLTS 110-120
WATTS 600
UL®
A.C. ONLY
DO NOT IMMERSE IN WATER

Yep, I blew it ... figured a US cup as 8 fluid ounces (didn't realize there was such a thing as a "coffee cup" measure; on the plus side, I've learned something new ;)

In New York City, DC utility voltage held out until 2007 when ConEd shut down the last run, and I believe San Francisco and Chicago both still have small segments of their original DC distributions in operation!.

The AC only ratings are probably dictated by thermostat switch contact ratings. A typical rating is 250V/10A, 125V/15A for AC, but unrated for DC (or, if it is, it would be limited to 30 volts at the 125 VAC current rating). The arc that occurs when the contacts open extinguishes when AC crosses zero volts, but a DC arc sustains until their separation distance becomes great enough. Not only does this generate a lot more heat, but in a physically small switch like a thermostat the distance between contacts may never open up enough to fully extinguish the arc, and they'll melt in short order.

 

[OmCheeto]

...The arc that occurs when the contacts open extinguishes when AC crosses zero volts, but a DC arc sustains until their separation distance becomes great enough. Not only does this generate a lot more heat, but in a physically small switch like a thermostat the distance between contacts may never open up enough to fully extinguish the arc, and they'll melt in short order.

10/10 star explanation!

 

[sophiecentaur]

The thermostat argument is highly relevant in a real life situation but this demo would only need a thermal fuse and a current fuse. No one would use DC to supply their domestic sockets. But neither would anyone use naked PV as the only source for their domestic electricity supply; some storage (grid or batteries) would be essential. The thread has had the PF treatment where no one has fixed the rules of engagement. Good fun, of course but what is the consensus as to what the OP should actually be doing on his / her demo?

 

[NTL2009]

... The thread has had the PF treatment where no one has fixed the rules of engagement. Good fun, of course but what is the consensus as to what the OP should actually be doing on his / her demo?

The OP should define the goal of the demo.

We don't have that (at least not that I see), so that is why we have all these drifting discussions (which have been interesting and informative, especially the AC/DC relay switching in post #38 from Asymptotic ), I never comprehended that relay issue as fully as that succinct post explained.

 

[OmCheeto]

...
No one would use DC to supply their domestic sockets.

Maybe not exclusively, but I would like having multiple types of power available in my house.
But, I'm an ex-submariner, and am used to the concept.

But neither would anyone use naked PV as the only source for their domestic electricity supply

Doesn't mean I didn't try that yesterday. I discovered this morning why that experiment failed: Over supply voltage cutoff protection: 15.5 ± 0.5 vdc.
My panels supply ≈20 vdc at no load, so my 12vdc-120vac inverter would not operate.

; some storage (grid or batteries) would be essential.

My deep cycle battery in my living room seems to be adding mass every year. I was not willing to move it.

The thread has had the PF treatment where no one has fixed the rules of engagement. Good fun, of course but what is the consensus as to what the OP should actually be doing on his / her demo?

Consensus? I think we've answered all of the important "why is that" questions. So, I'm just having fun figuring out how coffee pots, solar panels, and dc-ac inverters work.

Yesterday I determined that my solar panels have indeed lost 0.4% of their capacity per year.
This morning I measured the capacity of the heating element system from a coffee pot that one of my neighbors gave me last year. The coffee pot didn't work, even though it was manufactured ≥ 2009. Being that the "electronics" portion of the pot was toasted, from the discoloration of the printed circuit board, I decided to keep the heating element, and recycle the rest.

Anyways, the capacity is 10 ml, and the element has a resistance of 14.8 Ω. (same as my current drip coffee maker)
I calculated that one of mcharbs55's solar panels can boil water in the tube in 33.3 seconds, which is about 10 times longer than for my 1970's coffee maker(3.6 seconds @ 120 VAC).
This took a bit of finagling, as plugging in the max wattage, produced an excessive voltage:
280 watts @ 9.6 Ω yields 5.4 amps(ok, 8.89 amp limit)
but 51.8 vdc(not ok, 31.5 vcd limit)

Which is, I believe, what mcharbs55 was asking about, in the original post.

So, another answer is: It only takes one mcharbs55 panel to boil water. It will just take longer.

----------------------
ps. Other things I found out:
Where my missing funnel was. It was next to moms old coffee pot.
I have about 5000 coffee mugs, of which 2 were purchased by me, around 1982.
how to spell finagling

 

[Asymptotic]

The thermostat argument is highly relevant in a real life situation but this demo would only need a thermal fuse and a current fuse. No one would use DC to supply their domestic sockets. But neither would anyone use naked PV as the only source for their domestic electricity supply; some storage (grid or batteries) would be essential. The thread has had the PF treatment where no one has fixed the rules of engagement. Good fun, of course but what is the consensus as to what the OP should actually be doing on his / her demo?

Point taken.

Several other pieces of information are required to answer the OP's question, provided an element rating of 1500W/120V, and (for instance) using OmCheeto's Kyocera panel measurement data. At zero thermal losses, 1.35 liters, and a 65°F (291.5°K) start temperature, the answer could be only 2 panels (103 watts), but would take 75 minutes to hit 100°C. Four panels (413W) is 19 minutes, 6 panels (930W) is 8 minutes, and 8 panels (1654W) is 5 minutes to reach 100°C.

However, thermal efficiency isn't anywhere close to 100%, and it is necessary to consider the characteristics of the vessel the heater is in. For example, a reflective stainless steel pot radiates less energy than a black enameled pot of the same surface area, and while it's a sure bet water will come to a boil using 8 panels, there is a distinct possibility 2 panels wouldn't be enough to overcome losses, and the water would never reach 100°C.

If the 1500W/120V isn't a precondition for the demo, then I like the idea of using a small quantity of water, and safer, lower voltage element, then figuring for an appropriate solar panel.

 

[Asymptotic]

...
This morning I measured the capacity of the heating element system from a coffee pot that one of my neighbors gave me last year. The coffee pot didn't work, even though it was manufactured ≥ 2009. Being that the "electronics" portion of the pot was toasted, from the discoloration of the printed circuit board, I decided to keep the heating element, and recycle the rest.

View attachment 203318

Anyways, the capacity is 10 ml, and the element has a resistance of 14.8 Ω. (same as my current drip coffee maker)
I calculated that one of mcharbs55's solar panels can boil water in the tube in 33.3 seconds, which is about 10 times longer than for my 1970's coffee maker(3.6 seconds @ 120 VAC).
This took a bit of finagling, as plugging in the max wattage, produced an excessive voltage:
280 watts @ 9.6 Ω yields 5.4 amps(ok, 8.89 amp limit)
but 51.8 vdc(not ok, 31.5 vcd limit)

Which is, I believe, what mcharbs55 was asking about, in the original post.

So, another answer is: It only takes one mcharbs55 panel to boil water. It will just take longer.

----------------------
ps. Other things I found out:
Where my missing funnel was. It was next to moms old coffee pot.
I have about 5000 coffee mugs, of which 2 were purchased by me, around 1982.
how to spell finagling

I never gave it any thought because both electric tea kettles I've had (neither of which lasted more than 6 months) used immersion heaters, but if the heater mcharbs55 has is like the one you've pictured here, then one of his panels would do the trick, and seeing hot water and steam spitting out of delivery end of the pipe would enhance the demonstration.

 

[Asymptotic]

The OP should define the goal of the demo.

We don't have that (at least not that I see), so that is why we have all these drifting discussions (which have been interesting and informative, especially the AC/DC relay switching in post #38 from Asymptotic ), I never comprehended that relay issue as fully as that succinct post explained.

Thanks! I was thinking about it some more, and the 'AC arc extinguishing at zero crossing' is only mostly true, and relies on contact distance increasing quickly enough to emerge from the ionized region. This isn't guaranteed after hundreds of thousands of cycles in a sluggish spring return switch in the process of losing it's "springiness":

 

[sophiecentaur]

Thanks! I was thinking about it some more, and the 'AC arc extinguishing at zero crossing' is only mostly true, and relies on contact distance increasing quickly enough to emerge from the ionized region. This isn't guaranteed after hundreds of thousands of cycles in a sluggish spring return switch in the process of losing it's "springiness":

Solid state circuit breakers are available, cheap and amazingly robust. Let's face it, if the DC is to be converted to AC, it will already involve such technology.

 

[OmCheeto]

Point taken.

Several other pieces of information are required to answer the OP's question

Several? Try about a bazillion!
After experimenting for the last two days, I can understand why few lay people post the data from their experiments.
Good god.

2017.05.09 ≈10-11 am

Om puts his #4 panel out and notices the no-load voltage is dropping and records:

vdc time frame
20.6 initially
19.9 after a couple of minutes...​

conclusion: "I don't have that many multimeters!"​

Clear, but somewhat hazy sky...

Om gets out his 4x4 foot aluminum reflector and takes measurements:

Reflector     Ω  Volts   Amps
no         21.7  18.32  0.785
yes        21.7  19.14  0.820
no            ∞  19.52  0
yes           ∞  20.0   0

conclusion: atmospheric moisture power attenuation = 8.4%​

, provided an element rating of 1500W/120V, and (for instance) using OmCheeto's Kyocera panel measurement data. At zero thermal losses, 1.35 liters, and a 65°F (291.5°K) start temperature, the answer could be only 2 panels (103 watts), but would take 75 minutes to hit 100°C. Four panels (413W) is 19 minutes, 6 panels (930W) is 8 minutes, and 8 panels (1654W) is 5 minutes to reach 100°C.

However, thermal efficiency isn't anywhere close to 100%, and it is necessary to consider the characteristics of the vessel the heater is in. For example, a reflective stainless steel pot radiates less energy than a black enameled pot of the same surface area, and while it's a sure bet water will come to a boil using 8 panels, there is a distinct possibility 2 panels wouldn't be enough to overcome losses, and the water would never reach 100°C.

If the 1500W/120V isn't a precondition for the demo, then I like the idea of using a small quantity of water, and safer, lower voltage element, then figuring for an appropriate solar panel.

I did those experiments yesterday, and it was quite fun. My local weather forecaster said it was going to rain today, so I did just what you described.

Unfortunately, my numbers for experiment #2 were off by an order of magnitude, and I haven't resolved them yet. But I have a very good idea why:

mass of water to be boiled: 0.01 kg (used in calculation)
mass of the mostly aluminum boiler device: 0.12 kg (which I did not factor into the above calculation)
test device, which is part of the experiment, is 12 times more massive than the stuff being tested (could be a problem)​

I may have to take a nap before analyzing the data though., as Al has a specific heat capacity that is 4.65 less than that of water

Experiment #1 was quite easy: No matter how long I waited, my single solar panel would not heat the water past a certain temperature.

If anyone would like to figure out my thermal losses, whilst I take my nap, here are the equations from the data I collected yesterday. 3 x 50 watt 12 vdc panels wired in series:

temp climb
K = -0.0006 * sec^2 + 0.4865 * sec + 295.77​

temp decay

K = 661 * sec^(-0.11)​

No load voltage measured was 55.1 vdc.

ps. Obligatory picture, of how science is not always pretty: