Donald Smith Devices too good to be true

[JohnStone]

Your looking at a light bulb immersed in water , during a time to heat ...power used test!!

A side walk Calorimetry test, to see if the fellows that are lighting bulbs can use this method with a Killowatt meter [like I used in the link above]
1 Liter of water heated until I used .01 KWH {10 watts} took 14 minutes temp rise 5.2 C

Hi all,
please correct my calculation if not correct!

INPUT ENERGY:
given: 1 KWh => 3600KJ
current measurement: 0.01 kWh => 36 KJ

OUTPUT ENERGY:
given: Thermal capacity of water: 4,18 KJ / kg.°C
current measurement: 14min. / 5.2°C corresponds 22.28 °C within 1h
calculation: total energy input to 1 kg of water = 1 (kg)22.29 (°C) *4.18 (KJ/kg.°C) = 93.15 KJ

COP:
calculation: 93.13/36=2.59 CONGRATULATIONS

[Slovenia]

Cool Thread

Cool Thread

Great things going on here. Thanks for sharing!!

[Slovenia]

Zilano Thanks!!

Zilano,

Thanks very much for sharing your secrets with us. You have great knowledge indeed.

Best Regards,
Slovenia

its in auto transformer mode! and resonance can be achieved

ref : kapanadze secret! see image!

[sinergicus]

About coils tuning

Hi guys;I found an interesting article about how to tune primary and secondary of your coils precisely.

The old fashion when we moving one coil in another are not precise ;this is because between primary and secondary may appear multiple secondary resonance spots ( but not the maximum we need) and we may think we achieved the best resonance point..

Read more about this here: Jamie Oliver's Tesla Coil Demonstration Page

[Ares123]

Hi all,
please correct my calculation if not correct!

INPUT ENERGY:
given: 1 KWh => 3600KJ
current measurement: 0.01 kWh => 36 KJ

OUTPUT ENERGY:
given: Thermal capacity of water: 4,18 KJ / kg.°C
current measurement: 14min. / 5.2°C corresponds 22.28 °C within 1h
calculation: total energy input to 1 kg of water = 1 (kg)22.29 (°C) *4.18 (KJ/kg.°C) = 93.15 KJ

COP:
calculation: 93.13/36=2.59 CONGRATULATIONS

given: 1 KWh => 3600KJ
current measurement: 0.01 kWh => 36 KJ (cca15min) so * 4 = 144[/B]

COP = 93.13/144= 0,647

[JohnStone]

given: 1 KWh => 3600KJ
current measurement: 0.01 kWh => 36 KJ (cca15min) so * 4 = 144[/B]

COP = 93.13/144= 0,647

Ohhhh. Sorry. I forgot the factor 4.
You are right
sad John

[a.king21]

Open Sourcing

Zilano,
Any chance of you joining in the party and open sourcing like Wesley's team?
Think of the good Karma it'll bring!

Meanwhile
HH

[quantumuppercut]

Your looking at a light bulb immersed in water , during a time to heat ...power used test!!

A side walk Calorimetry test, to see if the fellows that are lighting bulbs can use this method with a Killowatt meter [like I used in the link above]
1 Liter of water heated until I used .01 KWH {10 watts} took 14 minutes temp rise 5.2 C

Energy require to raise water is heat capacity x change in temperature.

4.18KJ/C x 5.2C = 21.736 KJ

Power = 21.736 KJ / 14 min = 25.9 watts

So one needs at least 25.9 watts to change water 5.2C in 14 min.
Why RAM has .01 KWH ?

[RAMSET]

Quantum
That is a good question!

Perhaps we need to look a little closer at this??

I am Quite busy right now ...........

But I will run this test again with whatever recommendations or adjustments necessary for more proper evaluation .

Thanks for you time[calculations] and your comment.


Chet

[a.king21]

Is It Ou?

Hi,

experimenting with resonance my lights finally have 100% brightness

Does anyone have some idea to harvest useful power from this setup? I would like to charge 24V battery bank

All suggestions welcome

Kind rgds D

Hi I'm interested in your setup. Does it output more than input? There are several ways to harvest power. One way is the Bedini way where he treats batteries as a capacitor, but they need charging with cold electricity and conditioning to get the effect. You could pulse DC them, but watch out for arcing. Then you would get a grid tie inverter and output to grid. This has the effect of legally turning your meter backwards, so the grid acts as a perfect storage medium for your generated power. It is vital to know if you are OU or there is no point to the exercise.
Additionally you could close the loop using a capacitor bank to drive your nst. If it's OU why not clone it and use more than one for different uses. We really need more details than a schematic to comment properly.

[Duncan]

Playing with your numbers!

Energy require to raise water is heat capacity x change in temperature.

4.18KJ/C x 5.2C = 21.736 KJ

Power = 21.736 KJ / 14 min = 25.9 watts

So one needs at least 25.9 watts to change water 5.2C in 14 min.
Why RAM has .01 KWH ?

So there's a few liberties I've taken here for instance one Litre of water = 1KG and so it has from 1901 – 1964 Litre - Wikipedia, the free encyclopedia however I don't think we need bother about that gnats ball in this situation, also I guess “pure water” and such a point on the globe blah blah blah..
also not of to much interest in the rough indication required here.
Yahoo answers says....
Watt is a unit of measurement of electrical power where as Joule is a unit of measurement mechanical (heat) energy. However, the equivalent can be stated as 1Joule/sec = 1 Watt.

As you can see I have taken the liberty of dividing both sides of this equation by seconds and given myself the privilege of the Joule and the watt/second I doubt either actually exists as an “official unit” But it would seem to serve here so each can see what the other is demonstrating ,,,

Ramset has These figures 0.01Kwh and 14 minuets and 5.2 deg
Ramset is ultimately wanting to get to calorific exchange and the exchange point is basically

1watt/sec = 1 Joule so first we must operate on 0.01Kwh
0.01Kw/h x 60 = 0.6 Kw/min
0.6Kw/min x 60 = 36 Kw/sec
36Kw/sec x 1000 = 36000 Watt/sec
And so the energy input required to raise this litre of water 5.2 deg = 36000watt/sec or 36000Joules

Using quantumuppercuts figures for perfect energy exchange

4.18KJ/C x 5.2C = 21.736 KJ

Power = 21.736 KJ / 14 min = 25.9 watts (quantum has rounded up here from 25.876 … I've gone one decimal place up)

So with perfect energy transfer it needs at least 25.9 watts to change water 5.2C in 14 min.
Why RAM has .01 KWH ?

(Working backwards just to demonstrate)
25.88 Watts x 14min = 362.3 watt/min 362.6 x 60 = 21739 watt/sec or Joules

Assuming no external ambient advantage C.O.P = efficiency =

Perfect Energy Transfer as calculated by quantumuppercuts
The actual measurement taken from Ramsets experiment

= 21739 x100% = 60.3% efficient which is pretty good for a light bulb thrown in a jug!
36000

I suspect if there's any teacher's reading this I'd get detention for “slack work” but I dare say the general Idea of what Ramset is suggesting is clear ?

[RAMSET]

And Ramset Is very happy that others understand ....
And can do the Math..
I just collect Data and take recommendations on Test Procedure!!

Here's another example From Daemonbart with his new Ikea Boiler

Test of new setup!
Hi,

50 liter of water rise 30 deg celsius in one h, = 1500 X 4,18 = 6270 kJ
6270 / 3600 = 1,74 kWh

Input energy 360W = 360 Wh = 0.360 kwH

COP = 1,74 / 0,360 = 4,83

True COP = 4,83 X 1,2 = 5,8 (I have about 20% loss from test tank)

This setup based on IKEA thermos.

Kind rgds D
__________

Thanks
Chet

[quantumuppercut]

So there's a few liberties I've taken here for instance one Litre of water = 1KG and so it has from 1901 – 1964 Litre - Wikipedia, the free encyclopedia however I don't think we need bother about that gnats ball in this situation, also I guess “pure water” and such a point on the globe blah blah blah..
also not of to much interest in the rough indication required here.
Yahoo answers says....
Watt is a unit of measurement of electrical power where as Joule is a unit of measurement mechanical (heat) energy. However, the equivalent can be stated as 1Joule/sec = 1 Watt.

As you can see I have taken the liberty of dividing both sides of this equation by seconds and given myself the privilege of the Joule and the watt/second I doubt either actually exists as an “official unit” But it would seem to serve here so each can see what the other is demonstrating ,,,

Ramset has These figures 0.01Kwh and 14 minuets and 5.2 deg
Ramset is ultimately wanting to get to calorific exchange and the exchange point is basically

1watt/sec = 1 Joule so first we must operate on 0.01Kwh
0.01Kw/h x 60 = 0.6 Kw/min
0.6Kw/min x 60 = 36 Kw/sec
36Kw/sec x 1000 = 36000 Watt/sec
And so the energy input required to raise this litre of water 5.2 deg = 36000watt/sec or 36000Joules

Using quantumuppercuts figures for perfect energy exchange

4.18KJ/C x 5.2C = 21.736 KJ

Power = 21.736 KJ / 14 min = 25.9 watts (quantum has rounded up here from 25.876 … I've gone one decimal place up)

So with perfect energy transfer it needs at least 25.9 watts to change water 5.2C in 14 min.
Why RAM has .01 KWH ?

(Working backwards just to demonstrate)
25.88 Watts x 14min = 362.3 watt/min 362.6 x 60 = 21739 watt/sec or Joules

Assuming no external ambient advantage C.O.P = efficiency =

Perfect Energy Transfer as calculated by quantumuppercuts
The actual measurement taken from Ramsets experiment

= 21739 x100% = 60.3% efficient which is pretty good for a light bulb thrown in a jug!
36000

I suspect if there's any teacher's reading this I'd get detention for “slack work” but I dare say the general Idea of what Ramset is suggesting is clear ?

Hope we're not diverging too much from Don's device. Z, you're interesting.

Yes, specific heat capacity change with temperature too, but as long as we keep the temperature within certain range, the error is minimal. It is also important to have temperature to spread equally or at least the temp probe indicate a fair temp between the high and low. These factors are controlled experiment. However, with COP this high, the error is a small percentage.

Joule = Watt x second

When the meter reads .01 KWH , I don't exactly sure what it is reading since I'm not familiar with those meter. I'm thinking it means "if you operate at this rate, each hour you will consume 36 KJ." Which is 10 watts in average.

When you calculate 21739 KJ/ 36000 KJ x 100% = 60.3% you assuming the whole process operate in 1 hour. It only operates 14 min. Hope I'm clear.

[RAMSET]

percentage of Killowatt hour .01

Quantum
I am discussing this here also as a possible test procedure for these devices!!

Thanks
Chet

[daemonbart]

OU?

Hi,

thanks for support!

Well, power input is 60mA X 220V = 13,8W (ohms law not the right way here, I know..) and output "light power" about 60W

I will connect more F-tubes and see where it ends, for now I know that one or two F-tubes does not affect input (13,8W).

I used 3 ferrite cores, like putting hockey pucks on top of each other, primary coil in the middle. Windings are outside of rings only, each winding not toching the other. I will draw some pic later

Kind rgds D

Hi I'm interested in your setup. Does it output more than input? There are several ways to harvest power. One way is the Bedini way where he treats batteries as a capacitor, but they need charging with cold electricity and conditioning to get the effect. You could pulse DC them, but watch out for arcing. Then you would get a grid tie inverter and output to grid. This has the effect of legally turning your meter backwards, so the grid acts as a perfect storage medium for your generated power. It is vital to know if you are OU or there is no point to the exercise.
Additionally you could close the loop using a capacitor bank to drive your nst. If it's OU why not clone it and use more than one for different uses. We really need more details than a schematic to comment properly.