Hello good people,
I would like to present you free DIY for copper magnetic heater, We have integrated at our home and its working very well and now we want to share with all of you :)
So please if you have any tip or question please post it here or contact us direct on https://emolio.com/Magnetic-Heater
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Emolio.com
What is its performance? How long does it take to heat up its water through, say, 60 degrees centigrade?
here are the data:
from 22 till 95 Celsius of 18 liter oil goes in 17 minutes and I did it with s2 shell oil and it used 0.8kWh
Quote from: dakanadaka on July 12, 2015, 10:26:35 AM
here are the data:
from 22 till 95 Celsius of 18 liter oil goes in 17 minutes and I did it with s2 shell oil and it used 0.8kWh
0.830kg/liter * 1.8kJ/kg * 18liters * 73C = 1.96MJ = 0.55kWh
0.55kWh/0.8kWh = 68% efficiency
Hmm. Let's see....
0.8 kwh = 800 watt-hours
800 watt-hours x 60 minutes/hour x 60 seconds/minute = 2,880,000 watt-seconds or Joules of energy.
Using rough figures for mineral oil: Specific heat 1.67 J/gram-degree, density 0.85 gm/ml
So 18 liters of oil would weigh about 18000 ml x 0.85 gm/ml = 15,300 gm
To raise this amount of oil by (95-22) = 73 degrees should then take about
15,300 gm x 73 degrees x 1.67 J/gm-degree = 1,865,223 Joules of energy
And 17 minutes = 1020 seconds, so the power needed, if all power goes into the oil, is about 1,865,223 Joules / 1020 seconds = 1829 Watts.
1829 Watts applied for 17 minutes / 60 minutes/hour = 0.28 hour is therefore about 1.829 kW x 0.28 hour = 0.512 kWh of energy.
So, if all applied power goes into heating the oil, it would take about 0.512 kWh. But it actually took 0.8 kWh. So the efficiency is about 0.512/0.8 = 0.64 or 64 percent.
Will someone please check my math?
ETA: I see MarkE has already done the same problem, using slightly different values for specific heat and density.
Quote from: MarkE on July 12, 2015, 01:02:38 PM
0.830kg/liter * 1.8kJ/kg * 18liters * 73C = 1.96MJ = 0.55kWh
0.55kWh/0.8kWh = 68% efficiency
here is my math with given data:
this is input:
time | temp | kwh |
19:26 | 15 | 4108.3 |
19:31 | 88.5 | 4108.8 |
output:
Oil coeff | ΔT | ΔkW | Δtime |
1.8 kJ/kg.C | 73.5 C (88.5 - 15) | 0.5 kWh (4108.8 -4108.3) | 5 min |
and now Qinput = Qoutput
this is output:
18 liters water x 0.858 kg/liter(massa of oil) = 15.4 massa kg x 1.8 kJ/kg.C x 73.5 C = 2043kJ
input of electricity:
0.5 kWh
and now we have input of 0.5kWh and output of 2043kJ
Quote from: dakanadaka on July 12, 2015, 01:42:27 PM
here is my math with given data:
this is input:
time | temp | kwh |
19:26 | 15 | 4108.3 |
19:31 | 88.5 | 4108.8 |
output:
Oil coeff | ΔT | ΔkW | Δtime |
1.8 kJ/kg.C | 73.5 C (88.5 - 15) | 0.5 kWh (4108.8 -4108.3) | 5 min |
and now Qinput = Qoutput
this is output:
18 liters water x 0.858 kg/liter(massa of oil) = 15.4 massa kg x 1.8 kJ/kg.C x 73.5 C = 2043kJ
input of electricity:
0.5 kWh
and now we have input of 0.5kWh and output of 2043kJ
hi dakanadaka,
It's interesting using N,S,N,S poles of neo magnets to heat up metal similar to induction heater except that the magnetic field is set by the strong magnets which the field seems to last like forever.
I think i have seen the original video somewhere in youtube.I do believe this is a good way to heat up water and then playing with rpm in order to maintain to a preset heat.
I do wonder if you put magnet on both sides it would create even more change in magnetic flux which would heat up water via the metal in shorter amount of time.
This technique seems to be more efficient than a induction heater.
I do wonder if the typical induction heater can be replaced by a flat disc motor(with step up gear) which is rotating those neo magnets(which cover the surface area of disc) under a rather thin metal pan. ;)
Here are the data you listed at first:
Quote from: dakanadaka on July 12, 2015, 10:26:35 AM
here are the data:
from 22 till 95 Celsius of 18 liter oil goes in 17 minutes and I did it with s2 shell oil and it used 0.8kWh
Quotethis is input:
timetempkwh19:26154108.319:3188.54108.8
That makes no sense.
Quote18 liters water x 0.858 kg/liter(massa of oil) = 15.4 massa kg x 1.8 kJ/kg.C x 73.5 C = 2043kJ
That is approximately correct.
QuoteOil coeffΔTΔkWΔtime1.8 kJ/kg.C73.5 C (88.5 - 15)0.5 kWh (4108.8 -4108.3)5 min
Where did the "5 min" come from? At first you said 17 minutes.
And you really should try to express your math in some kind of standard form so that it is clear what you mean.
Quoteinput of
electricity:
0.5 kWh
But you said at first that the input was 0.8 kWh.
Both MarkE and I used the numbers you cited originally and we came up with very close agreement in our results. Now you seem to have changed the numbers to something rather implausible. You went from 17 minutes down to 5 minutes. You changed the input from 0.8 kWh to 0.5 kWh.
Maybe you should make a video showing your device operating and your measurements.
Yes I started with lathe and placing magnts north, south, north .... and did heat the copper pipe an I saw that on youtube video last year and then toaght lets try to make real thing
with a lot of work and most difficult part was the copper part, because it needs to be welded and can hold 4bar.
But formula to see efficiency was always the problem for me... My math was input is output, but I see here two people already made almost the same value and that is around 68%...
Thats why I will give you the video where it says:
17 liter oil s2 shell
start point:
10:18 temperature 17.3 C with starting 4308.9 kWh
end point:
10:35 temperature 95.4 C with end of 4309.7kWh
and that is:
in 18 minutes from 17.3 C to 95.4 C used 0.8kWh
and here is video:
https://www.youtube.com/watch?v=KWzhLtqdKVE
the water if I want to see how much oil is in there then there is 18 liter, but in the copper part there are three other walls inside so it can be less then 18 liter thats why I would say 17 liter
and in the attachment I will place drawing of copper part
Quote from: dakanadaka on July 12, 2015, 01:42:27 PM
here is my math with given data:
this is input:
time | temp | kwh |
19:26 | 15 | 4108.3 |
19:31 | 88.5 | 4108.8 |
output:
Oil coeff | ΔT | ΔkW | Δtime |
1.8 kJ/kg.C | 73.5 C (88.5 - 15) | 0.5 kWh (4108.8 -4108.3) | 5 min |
and now Qinput = Qoutput
this is output:
18 liters water x 0.858 kg/liter(massa of oil) = 15.4 massa kg x 1.8 kJ/kg.C x 73.5 C = 2043kJ
input of electricity:
0.5 kWh
and now we have input of 0.5kWh and output of 2043kJ
Before you reported your input electrical energy at 0.8kWh. Now that we all agree that 18 liters of oil heated by 73C is around 0.50 - 0.55kWh you report that your input electricity is 0.5kWh. How do you explain the change in your reported electrical input?
Submersible resistance heating elements are 98%+ efficient and very low cost. Your device is an eddy current brake. It's efficiency heating the thermal transfer oil is hindered by the efficiency of the motor, and the heat conduction and radiation of the copper block. Even just putting thermal insulation around the expensive copper block would help. But the whole thing is economically ridiculous due to the cost and losses of the motor.
If you want a non-contact system: use ingot iron for the thermal transfer block inside insulation surrounded by induction coils. If you design the induction coils properly, then you can get 95% - 99% plus power transfer into eddy currents in the iron. Induction heater design is old, very well-understood art.
Quote from: dakanadaka on July 12, 2015, 02:49:17 PM
the water if I want to see how much oil is in there then there is 18 liter, but in the copper part there are three other walls inside so it can be less then 18 liter thats why I would say 17 liter
and in the attachment I will place drawing of copper part
Fill it and then drain it into a measuring container.
Quote from: magpwr on July 12, 2015, 02:12:09 PM
hi dakanadaka,
It's interesting using N,S,N,S poles of neo magnets to heat up metal similar to induction heater except that the magnetic field is set by the strong magnets which the field seems to last like forever.
I think i have seen the original video somewhere in youtube.I do believe this is a good way to heat up water and then playing with rpm in order to maintain to a preset heat.
I do wonder if you put magnet on both sides it would create even more change in magnetic flux which would heat up water via the metal in shorter amount of time.
This technique seems to be more efficient than a induction heater.
I do wonder if the typical induction heater can be replaced by a flat disc motor(with step up gear) which is rotating those neo magnets(which cover the surface area of disc) under a rather thin metal pan. ;)
But it is an induction heater. The changing magnetic field comes from the moving magnets. It is otherwise known as an eddy current brake. Bury heating elements in a heat exchange block that is well insulated and you'll spend a lot less money and get much better results.
Quote from: MarkE on July 12, 2015, 05:15:50 PM
But it is an induction heater. The changing magnetic field comes from the moving magnets. It is otherwise known as an eddy current brake. Bury heating elements in a heat exchange block that is well insulated and you'll spend a lot less money and get much better results.
hi MarkE,
There is
no "heat exchange or transfer" involved in this type this induction heater concept regardless if it's is using magnets or pancake coil with litz wire since the metal pan(thin or thick) itself is getting hot.
It's the primary reason why induction heater is more efficient than a typical heating element design since
heat exchange part is eliminated.
Quote from: magpwr on July 13, 2015, 02:23:18 AM
hi MarkE,
There is no "heat exchange or transfer" involved in this type this induction heater concept regardless if it's is using magnets or pancake coil with litz wire since the metal pan(thin or thick) itself is getting hot.
The copper structure most definitely performs as a heat exchanger to the oil thermal transfer fluid. Eddy currents heat the copper block and the block rejects heat to the oil running through the inside of the block and air surrounding the block.
Quote
It's the primary reason why induction heater is more efficient than a typical heating element design since heat exchange part is eliminated.
You are wrong on both accounts. Properly wired and located, resistance heating elements are 98%+ efficient. Induction heating is useful in many circumstances where the thing that we want to heat is something other than fluid captured within the heater, such as: water in a pot, or some piece of metal that we are working. The copper block that is your heating element is also your heat exchanger to the oil heat exchange fluid.
If you were to throw away the motor and magnet disc, and instead mount one or more resistance heaters in the copper block, and then insulate the copper block, your power efficiency would improve substantially, and your costs would drop dramatically.
Resistance heating elements are beauties in simplicity and low cost. They create however a lot of problems, due to water residue minerals attracted to the surface, and the internal temperature os such element is very high and once the mineral deposit on the external wall is built this temperature even rise and is the reason of malfunction.
Now , induction heaters can solve much of those problems.
offtopic:
I have a few questions about resistive heaters used for AC power. MarkE,can I ask you ?
Ok Tnx for all math and reply... Today I hope I will take the water out of the copper box to see how much exacly oil does it contain and then I could then conclude exacly how efficient this heater is and then try to make more efficiency :)
This project cost me a lot to research it and do it on my own and the thing was there are people on the youtube that are showing this kind of heaters, but no one showed exacly how can it be done with all math on efficiency etc.
I saw one patent on it on air heater and there is one german guy who is searching for investitors, but no one who gave everything to the public and then we all could profit and help each other to make it even better, so that is what I want to achieve to share and to help each other to build more efficiency.
Im not saying that this path could lead to nothing then research that we couldnt achieve nothing more then 67% of efficiency, but then other people dont have to go so far as I did with financiel costs :)
So one more question that could help me on this trip:
Do you have any tip on how to inprove efficiency of this kind of heating system or any research already done on this?
I saw electrical existence heater and yes they are saying 98%, and I also build the friction heater and didnt go much ferder on it, but I miss that guy on youtube oliepigg username he just dissappierd with his friction heater :(
Quote from: forest on July 13, 2015, 04:21:43 AM
Resistance heating elements are beauties in simplicity and low cost. They create however a lot of problems, due to water residue minerals attracted to the surface, and the internal temperature os such element is very high and once the mineral deposit on the external wall is built this temperature even rise and is the reason of malfunction.
The mineralization issue occurs in
water heaters with an insufficient sacrificial anode. We are talking about an oil heater. Mineralization is not an issue.
The internal temperature of a resistive heater element can be whatever one chooses from just above room temperature to well over 1000C. Hotter transfers more thermal energy for a given size heating element.
Quote
Now , induction heaters can solve much of those problems.
Induction doesn't solve a problem that doesn't exist: mineralization in an oil heater, or higher working temperature than is desired in the heating elements.
Quote
offtopic:
I have a few questions about resistive heaters used for AC power. MarkE,can I ask you ?
Do you mean resistance heaters powered by AC?
Quote from: dakanadaka on July 13, 2015, 05:09:36 AM
Ok Tnx for all math and reply... Today I hope I will take the water out of the copper box to see how much exacly oil does it contain and then I could then conclude exacly how efficient this heater is and then try to make more efficiency :)
This project cost me a lot to research it and do it on my own and the thing was there are people on the youtube that are showing this kind of heaters, but no one showed exacly how can it be done with all math on efficiency etc.
I saw one patent on it on air heater and there is one german guy who is searching for investitors, but no one who gave everything to the public and then we all could profit and help each other to make it even better, so that is what I want to achieve to share and to help each other to build more efficiency.
Im not saying that this path could lead to nothing then research that we couldnt achieve nothing more then 67% of efficiency, but then other people dont have to go so far as I did with financiel costs :)
So one more question that could help me on this trip:
Do you have any tip on how to inprove efficiency of this kind of heating system or any research already done on this?
I saw electrical existence heater and yes they are saying 98%, and I also build the friction heater and didnt go much ferder on it, but I miss that guy on youtube oliepigg username he just dissappierd with his friction heater :(
Hydrodynamics have built a mechanically driven cavitation heater for years. At one time they thought it was overunity because they did their calculations based on the wrong feed water temperature. If you want to heat efficiently, then no matter what you use as the heat source, you need to insulate that and your heated output from the local environment.
Quote from: MarkE on July 13, 2015, 06:44:21 AM
Hydrodynamics have built a mechanically driven cavitation heater for years. At one time they thought it was overunity because they did their calculations based on the wrong feed water temperature. If you want to heat efficiently, then no matter what you use as the heat source, you need to insulate that and your heated output from the local environment.
Do you mean these guys: https://m.youtube.com/watch?v=yh_-DUKQ4Uw ?
And what about this solution:
https://m.youtube.com/watch?v=A5llLsfZYXw does anyone knows how to build this one?
As long as I saw the frequency is something that could bring overunity... As start I found very interesting the chladni plate and cymatics and vortex coil all working with frequency so maybe the rotation frequency of motor could be solution to gain overunity or better results, but of course then we are going into the beter motor design to use less elecrticity.
I think we all saw the qeg, maybe that could be solution? But again then it is not making better this kind of heater... But only combination of using engine....
Quote from: dakanadaka on July 13, 2015, 08:01:03 AM
Do you mean these guys: https://m.youtube.com/watch?v=yh_-DUKQ4Uw ?
And what about this solution:
https://m.youtube.com/watch?v=A5llLsfZYXw does anyone knows how to build this one?
Yes, Griggs is the guy behind hydrodynamics now proven under unity water heater. Griggs and many observers fooled themselves by plugging the wrong numbers into their equations. Hall Puthoff of Earth Tech International figrued out the mistake over ten years ago and published a paper on it.
Quote from: dakanadaka on July 13, 2015, 08:09:29 AM
As long as I saw the frequency is something that could bring overunity... As start I found very interesting the chladni plate and cymatics and vortex coil all working with frequency so maybe the rotation frequency of motor could be solution to gain overunity or better results, but of course then we are going into the beter motor design to use less elecrticity.
I think we all saw the qeg, maybe that could be solution? But again then it is not making better this kind of heater... But only combination of using engine....
I am sorry but that is all so much nonsense.
MarkE
I have resistive heater 2000W for 230AC 50hz current. Can I power it by pure DC if I know the resistance of heater ? Any problems ? Do you know if somebody tested the solution of such heater immersed in water to eliminate the sludge of minerals on heater surface ? Like for example high frequency impulses send sometime to the heater maybe ? If I can power it by pure DC then if I connect negative of DC to the stainless steel container of heater and the negative electrode and all to the grounded rod and I will be checking the current to the heaters the it would be reletively safe ? What do you think ?
Quote from: forest on July 14, 2015, 03:41:26 AM
MarkE
I have resistive heater 2000W for 230AC 50hz current. Can I power it by pure DC if I know the resistance of heater ?
You can power it with 230VDC. That would be 0.71X the voltage you would have if you applied 230AC full-wave rectified into a capacitor. You could get close by half-wave rectifying 230VAC.
QuoteAny problems ? Do you know if somebody tested the solution of such heater immersed in water to eliminate the sludge of minerals on heater surface ? Like for example high frequency impulses send sometime to the heater maybe ? If I can power it by pure DC then if I connect negative of DC to the stainless steel container of heater and the negative electrode and all to the grounded rod and I will be checking the current to the heaters the it would be reletively safe ? What do you think ?
DO NOT CONNECT THE HEATER CASE TO ANY POTENTIAL OTHER THAN EARTH. That includes leads that are intended to be at earth potential but may become disconnected from earth due to corrosion, human error, etc. If you want to connect one side of the heater element to earth do that with a separate dedicated lead at the power supply output to earth. But absolutely maintain the insulation between the heater element connections and the local safety ground bonding of the element case. This may not seem intuitive, but it is essential. You want to guard with your life (in a very real sense) against any possibility of a wiring failure electrifying the water in the heater. If you electrify the water in your heater, then you can get electrocuted at any of your faucets or in the bath or shower.
As much as I would love to have more efficiency these are the formules that MarkE has given:
Quote0.830kg/liter * 1.8kJ/kg * 18liters * 73C = 1.96MJ = 0.55kWh
0.55kWh/0.8kWh = 68% efficiency
Teselkoala did the next:
Quote0.8 kwh = 800 watt-hours
800 watt-hours x 60 minutes/hour x 60 seconds/minute = 2,880,000 watt-seconds or Joules of energy.
Using rough figures for mineral oil: Specific heat 1.67 J/gram-degree, density 0.85 gm/ml
So 18 liters of oil would weigh about 18000 ml x 0.85 gm/ml = 15,300 gm
To raise this amount of oil by (95-22) = 73 degrees should then take about
15,300 gm x 73 degrees x 1.67 J/gm-degree = 1,865,223 Joules of energy
And 17 minutes = 1020 seconds, so the power needed, if all power goes into the oil, is about 1,865,223 Joules / 1020 seconds = 1829 Watts.
1829 Watts applied for 17 minutes / 60 minutes/hour = 0.28 hour is therefore about 1.829 kW x 0.28 hour = 0.512 kWh of energy.
So, if all applied power goes into heating the oil, it would take about 0.512 kWh. But it actually took 0.8 kWh. So the efficiency is about 0.512/0.8 = 0.64 or 64 percent.
And if this is true then we have around 64 or 67 % of the efficiency.......
the live video of how much it used to reach 95 Celsius you can see it at: https://www.youtube.com/watch?v=KWzhLtqdKVE
So if anyone can say that this efficiency percentage is not true please replay and if you think about building one:
QuoteDO NOT begin before you see our experience that we shared on https://emolio.com/Magnetic-Heater , so you can see what challenges you will meet!
as for now this post is then closed :)
Quote from: MarkE on July 14, 2015, 06:16:40 PM
You can power it with 230VDC. That would be 0.71X the voltage you would have if you applied 230AC full-wave rectified into a capacitor. You could get close by half-wave rectifying 230VAC.DO NOT CONNECT THE HEATER CASE TO ANY POTENTIAL OTHER THAN EARTH. That includes leads that are intended to be at earth potential but may become disconnected from earth due to corrosion, human error, etc. If you want to connect one side of the heater element to earth do that with a separate dedicated lead at the power supply output to earth. But absolutely maintain the insulation between the heater element connections and the local safety ground bonding of the element case. This may not seem intuitive, but it is essential. You want to guard with your life (in a very real sense) against any possibility of a wiring failure electrifying the water in the heater. If you electrify the water in your heater, then you can get electrocuted at any of your faucets or in the bath or shower.
MarkE ,Thank You! AS it become much offtopic here I sent you PW message, I appreciate your response.