Sizing a Solar Array To Boil Water

[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: