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Author Topic: Finally magnetic heater DIY for FREE  (Read 12945 times)

Offline dakanadaka

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Finally magnetic heater DIY for FREE
« on: July 12, 2015, 03:10:28 PM »
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

Kind regards
Emolio.com

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Offline Paul-R

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Re: Finally magnetic heater DIY for FREE
« Reply #1 on: July 12, 2015, 04:15:00 PM »
What is its performance? How long does it take to heat up its water through, say, 60 degrees centigrade?

Offline dakanadaka

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Re: Finally magnetic heater DIY for FREE
« Reply #2 on: July 12, 2015, 04:26:35 PM »
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

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Re: Finally magnetic heater DIY for FREE
« Reply #2 on: July 12, 2015, 04:26:35 PM »
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Offline MarkE

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Re: Finally magnetic heater DIY for FREE
« Reply #3 on: July 12, 2015, 07:02:38 PM »
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

Offline TinselKoala

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Re: Finally magnetic heater DIY for FREE
« Reply #4 on: July 12, 2015, 07:21:51 PM »
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.





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Re: Finally magnetic heater DIY for FREE
« Reply #4 on: July 12, 2015, 07:21:51 PM »
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Offline dakanadaka

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Re: Finally magnetic heater DIY for FREE
« Reply #5 on: July 12, 2015, 07:42:27 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:
timetempkwh19:26154108.319:3188.54108.8
output:
Oil coeffΔTΔkWΔtime1.8 kJ/kg.C73.5 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

Offline magpwr

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Re: Finally magnetic heater DIY for FREE
« Reply #6 on: July 12, 2015, 08:12:09 PM »
here is my math with given data:
this is input:
timetempkwh19:26154108.319:3188.54108.8
output:
Oil coeffΔTΔkWΔtime1.8 kJ/kg.C73.5 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
. ;)






 

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Re: Finally magnetic heater DIY for FREE
« Reply #6 on: July 12, 2015, 08:12:09 PM »
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Offline TinselKoala

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Re: Finally magnetic heater DIY for FREE
« Reply #7 on: July 12, 2015, 08:20:27 PM »
Here are the data you listed at first:

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
this is input:
timetempkwh19:26154108.319:3188.54108.8

That makes no sense.

Quote
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

That is approximately correct.
Quote
Oil 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.

Quote
input 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.










Offline dakanadaka

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Re: Finally magnetic heater DIY for FREE
« Reply #8 on: July 12, 2015, 08:36:16 PM »
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

Free Energy | searching for free energy and discussing free energy

Re: Finally magnetic heater DIY for FREE
« Reply #8 on: July 12, 2015, 08:36:16 PM »
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Offline dakanadaka

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Re: Finally magnetic heater DIY for FREE
« Reply #9 on: July 12, 2015, 08:49:17 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

Offline MarkE

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Re: Finally magnetic heater DIY for FREE
« Reply #10 on: July 12, 2015, 11:08:17 PM »
here is my math with given data:
this is input:
timetempkwh19:26154108.319:3188.54108.8
output:
Oil coeffΔTΔkWΔtime1.8 kJ/kg.C73.5 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.

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Re: Finally magnetic heater DIY for FREE
« Reply #10 on: July 12, 2015, 11:08:17 PM »
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Offline MarkE

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Re: Finally magnetic heater DIY for FREE
« Reply #11 on: July 12, 2015, 11:10:58 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.

Offline MarkE

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Re: Finally magnetic heater DIY for FREE
« Reply #12 on: July 12, 2015, 11:15:50 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.


Offline magpwr

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Re: Finally magnetic heater DIY for FREE
« Reply #13 on: July 13, 2015, 08:23:18 AM »
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.

Offline MarkE

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Re: Finally magnetic heater DIY for FREE
« Reply #14 on: July 13, 2015, 09:38:29 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.

 

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