Solar Powered Heater - Is it possible?

[physicso]

Hi..I joined this forum a couple of days ago and I find that this is the best place for newb like me to share and seek some ideas.

I have a question to ask and I don't know if this had been asked before. Would it be possible to heat water to its boiling point by solar?

The idea is to collect the energy from sun by photovoltaic panel and store it in a 12V 7.2A-hour rechargeable battery and from that battery, I want to heat water to it's boiling point.

I have made a few calculations, correct me if I'm wrong.
Water = 350ml = 0.35kg
Water Specific Heat = 4186 joule kg-1 C-1
T_f - T_i = (100 - 30)C = 70C
*assuming Ti is 30 degree Celsius.

Q = 0.35*4186*70 = 102557 J = 102.6kJ

let say the heating period is around 3 hours (I don't mind about the time taken, as for now, the main objective is to get water boils.)

P = Q/t = 102.6kJ/(3x60x60) = 9.49 Watt ~= 10 Watt.

Power supplied from battery

1) for first hour
P = VI = 12V*7.2A-hour*(1/1hour) = 86.4 Watt

2) for second hour
P = VI = 12V*7.2A-hour*(1/2hour) = 43.2 Watt

3) for third hour
P = VI = 12V*7.2A-hour*(1/3hour) = 28.8 Watt

to support my calculation, I try to figure out the length of the heating element needed.

P = I²*R
R = P/I² where P = 10 watt, I = (7.2A-hour)*(1/3hour) = 2.4A
R = 1.736 ohm

material used is nichrome with resistivity 1.5 x 10^-6 ohm-m.

R = resistivity*(Length/Cross Area)
Length/Cross Area = R/resistivity = 1.736/(1.5 x 10^-6) = 1,157,407 m/m²
Length = Cross Area * 1,157,407 m/m² where Cross Area = 1000 circular mils = 0.506707479 mm³
Length = 0.586 m *approx. half a meter

*it seems to me that the power supplied by 12V 7.2A-hour battery is more than enough to boil 350ml water but I never build it yet because I have to finish the paperwork/calculations first before start assemble. My teacher told me that it is not possible to boil water using battery. He told me something about start up energy/power/heat that requires a lot of power. Anyone can explain this to me?

Questions.
1. Is the calculation right?
2. Any other idea to heat water with minimum power?

Thank You. Really appreciate your help.

 

[chaoseverlasting]

You don't need to do any of that. One way is to use a parabolic reflector to focus solar energy. Another is to use a glass top. What this does is that it prevents the IR radiation from escaping the container and furthur increases the internal temperature. You could build a fairly cheap solar heater that way. The parabolic reflector is the biggest issue in such a construction.

 

[vk6kro]

Have a look at this item on Ebay:
http://cgi.ebay.com.au/Kettle-Car-Pot-Hot-Water-Heater-12V-Heated-Travel-/250722084351?pt=UK_CarsParts_Vehicles_CarParts_SM&hash=item3a603365ff

It is a heater for use in a car and it uses 8 amps at 12 volts. (I'm not suggesting that you get one, just as an example.)

# Voltage: DC12V
# Current: 8A
# Power: 96W
# Capacity: 200-650 ML
# Boiled time: 15-30 mintues

So, it looks like it would boil 350 ml of water in about 20 minutes.

Your battery may be able to supply 8 amps, and if it did this for 20 minutes, this would be 2.6 amp-hours. ( 8 amps * 1/3 hour ).
So, your battery would not be anywhere near flat.

 

[physicso]

@chaoseverlasting how does a glass top work? I googled glass top but nothing came close to this topic. Does it work like a green house effect? I have thought about the parabolic reflector before and it works,effectively. Thanks.

@vk6kro If a product specs state that '96W',what does it mean? Is it the power required to start up the device or the operational device power? I usually get confused on this.

 

[sophiecentaur]

@physicso: I don't see where you get different times for the three successive hours in your op. Remember, the answers to Energy calculations should be in Joules (or Watt seconds) rather than in Watts.

@chaoseverlasting: Direct heating has to be the best (and cheapest) solution if you want to use the boilingwater at the time. To store the energy for later (much later) use, the battery would probably be a better solution. Even though the conversion efficiency of PV cells and a battery is not good, once you have the energy stored, you could well lose less energy than from even a well lagged container of hot water )at 100 C, if you have wait a long time.
PV cells are v. expensive and a reflector could be very cheap! - but I guess this is more of a thought experiment. . . . ?

 

[physicso]

@sophiecentaur I made assumption that the operation is around 3 hours which equals to 3hrs x 60 mins/hrs x 60 secs/mins = 10800s. From that assumption, I can calculate the power for the operation. Power (Watts) = Energy (Joules)/time (s).

 

[vk6kro]

@vk6kro If a product specs state that '96W',what does it mean? Is it the power required to start up the device or the operational device power? I usually get confused on this.

The device uses 8 amps at 12 volts.

So, it uses 96 watts all the time it is being used. Power = Voltage * current.

Not all this power will be used to heat water, though.
Some will heat the plastic of the appliance and some will be lost due to convection etc. This water heater seems quite efficient, though.
We have a kettle for boiling water and it uses 2400 watts. This one only uses 96 watts and still boils the water in 20 minutes.

 

[physicso]

@vk6kro Is it true that the power required for a device to start up before its in operational state is aorund 2 times the operational power? I might get wrong about this.

 

[vk6kro]

There is some truth in that.

The heating element will have less resistance when it is cold than when it is hot.
So, for a few seconds, the element may draw more current than it will when it warms up.

This extra power may be twice the normal power or it could be something else.
A water heater element will not get as hot as the same element is air because the water conducts heat away from the element.

 

[sophiecentaur]

@sophiecentaur I made assumption that the operation is around 3 hours which equals to 3hrs x 60 mins/hrs x 60 secs/mins = 10800s. From that assumption, I can calculate the power for the operation. Power (Watts) = Energy (Joules)/time (s).

But why use different operating times for each of the three hours? I don't understand your reasoning because you would surely run the heater continuously.
Would it not be easier, after you have your required energy (106.2kJ) to say that the required power is that value divided by the time (10800s), say what that represents in current, for the 12V you will use and then the Ah battery capacity needs to be that value times 3hours?

The resistance of nichrome doesn't change much between 30C and 100C, does it? (0.00017 per Ohm per degree, I think) The battery volts will change more than that whilst it is being drained - discharging and getting a bit warm.

What does the "startup power" refer to? Could it be a confusion with tungsten lightbulbs, which take an initial value of current which is about ten times their running value for a few ms? Their temperature range is a bit higher than 30 to 100C, though. Or it could refer to electric motors with a high stall (startup) current.

Yes this is very feasible and could give you a small supply of boiling water whenever you want it. You will also, of course, need to calculate the area / spec of PV panel required. Even a simple 20W panel will cost a fair bit and the ratio of charge to discharge time will significantly exceed the ratio of estimated charge /discharge currents. It may be cloudy, early morning or late afternoon when it is charging.