Finally magnetic heater DIY for FREE

[dakanadaka]

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

[magpwr]

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.

[TinselKoala]

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

this is input:
timetempkwh19:26154108.319:3188.54108.8

That makes no sense.

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.

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.

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.

[dakanadaka]

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

[dakanadaka]

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