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Toys, toys, toys... and more

on 16 July 2013.

Video preview of some toys on display at ICCF

As part of our preparations for ICCF18, next week, our US lab is scrambling to get a number of example apparatus together to help show people who might be following the experiments what they actually look like. Pictures are nice, videos are even better, but seeing it in person, touching it, and asking questions right there is way better. We will be bringing a whole table full of examples including some new, never revealed designs. Here is a preview video.

Donate your old toys

Do you have lab equipment sitting in your store room or collecting dust on a shelf? Your unused equipment might be just what MFMP needs to try a new test or help set up a new collaborating lab. Read this:

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+1 #35 Ron B 2013-07-25 04:41

Speaking of toys, the data viewer is not giving all the data fields for older records of experiments. I wanted to look at the runs from Jan/Feb 2013 on the macor and mica cells but all the fields are not available. I wanted to review this data as a result of gaining more information from the recent ICCF. Is there a chance that this can be fixed?

+1 #34 charlie tapp 2013-07-22 16:01

so do i mail it to mfmp now so they can check it out

+1 #33 Malachi Heder 2013-07-22 15:02

@ Jordan

Good catch, don't want to claim too much excess, too quickly!

+2 #32 Jordan D Maiers 2013-07-22 14:59

@Malachi
I think you might have shifted the decimal place in your calculation. I'm pretty sure the calculation itself is right, but when I do:

4.1855J/gK*70.87g*75K

I get: 22,248.2J = 22.25kJ

Total energy 159kJ+22.25kJ = =181.25 kJ, COP 181.25kJ/120kJ = 1.51

0 #31 Malachi Heder 2013-07-22 14:42

@ Johan

The latent heat in your calculation is to vaporize water that is already at 100C. There is also the energy required to raise water temp to 100C to consider(only calculating for the 2.5 oz, though much of the rest of the water will increase in temp).

70g of water from 25 to 100C using 4.1855 J/gK:

4.1855*70.8738*(374-299)= 222.5 kJ

So, I believe it would be a total of 382 kJ and a COP of 3.19.

That makes it a little more interesting, I think :) Someone should make sure I'm doing the correct calculations, this is how I remember it from my heat transfer classes.

0 #30 Johan Eriksson 2013-07-21 21:25

@Charlie
@Ryan

2.5 oz = 70g

2272 kj/kg for phase exchange (water to steam).

0.07 kg =>159 kJ

5 min = 300 s

120V * 3.34A = 400W ( 400 J/S )

400 * 300 = 120000 J = 120kJ

So this means that if your measurements are accurate you have a "efficiency" of 159/120 = 133%

I would say that the measuement doesn't give any clear indication that there is any anomaly here..

What you should do is to swap the amp meter with an "plug in" power meter. (Measuring amps doesn't work if the load is anything else that a old style lightbulb / resisitive load, you could easily read values that is 50% off if you have a non resistive/non-l inear load.)

Also boil more water and take weight measurements at more times, for example, measure temp and weight every 5 mins and run for an hour.
And then turn of the power and continue to measure temp & weight until you get roughly to room temperature. (To estimate heat loss due to radiation & convection)

Uneven water temperature is not an issue here, since we are only interested in the parts that boils of. However we want to measure weights when we have reached a steady-state on the temperature.

EDIT: didn't account for altitude, however this is irrelevant unless the energy input is properly measured. Clamp meter doesn't work here unless it is verified that the load is linear & purely resistive. ( Highly unlikely here )

0 #29 Ryan Hunt 2013-07-19 18:09

@ Charlie - Thanks for doing the test. Sorry, I don't have time to run the calculations at the moment - we are busy packing up for a road trip. Can anyone else run the rough calculation of power in vs heat of evaporation?

0 #28 charlie tapp 2013-07-19 17:51

@ron b 1700 feet ajo arizona

0 #27 James Bryant 2013-07-19 13:27

In regards Peter Davey's device, this Overunity forum discussion may be helpful:
overunity.com/.../...

0 #26 Robert Greenyer 2013-07-19 11:17

@charlie tapp

If you can sign up to a google account - you get google drive. Upload to there and then share the link to the asset.

+1 #25 Ron B 2013-07-19 06:00

Charlie, What is the altitude at your house?

This is needed to allow an accurate calculation of the energy output.

+1 #24 charlie tapp 2013-07-18 23:14

Ok made a video to long to email , tried to put on YouTube it said two hours to go never loaded. When the other half gets home she will figure it out. But in the mean time I did what Ryan suggested. Here goes (1lb) 16 oz of water warmed up. Ran for 5 min. Recorded amps every minute because they drop as it goes. Average 3.34 amps . Useing old postage scale not digital but pretty close, boiled off 2.5 oz in the 5 min temp are hard to figure it boils from the top so the bottom is cooler needs a stirrer on it, note oz. is weight not liquid measurement. Is this good or should I stop talking about it? Oh yea 119.9 volts

+1 #23 Robert Greenyer 2013-07-18 22:40

@charlie tapp

Excellent - sounds exciting...

+1 #22 charlie tapp 2013-07-18 18:27

Ill send it right now in operation then ill do your tests. And post results.

+1 #21 Ryan Hunt 2013-07-18 15:17

@ Charlie - Because we are going to be gone at ICCF next week, maybe it would speed things along if we could ask you to do a few simple tests for us first before you pack it up.
First, A short video of it working would be great.
Second, measure the volume or weight of the water you are heating and boiling.
Third, put a thermometer in the water.
Do you have a plug in power meter?

Or this procedure:
- Weigh the empty cup , then add water
-Get the water boiling, take the device out. Weigh the cup with the water
- then put the device in and boil till almost half the water is gone. Note how much time that took. (or video it)
- Then weigh the cup + the water again.
From the amount of water evaporated we can figure out how much heat was created. If you can find a plug in power meter, then we can know how much electricity was put in.

If you can do any part of the above, it will help us greatly. Thanks!

+1 #20 charlie tapp 2013-07-18 15:04

@robert greenyer luck i guess i saw it and started researching the net, not alot on it so looked it up on you tube a couple people had some that worked but i didnt believe it becaause of the ac power input. i had an old box of doorknobs and decided to build one and it worked. i havent tried 220 volts ac yet but 120 works pretty good mabee you guys can tell me if it has cop or not.going to send it today.

+1 #19 Robert Greenyer 2013-07-18 11:24

@charlie tapp

Will be great to test this device. How did you come by the necessary geometry and materials as it says it took Davey 25 years to get the first one working.

+1 #18 charlie tapp 2013-07-18 00:08

@ron b that link had nothing on it in the way of pics weird, also weird no one has heard of this. I just stumbled on it did not believe it because it uses ac power . I always mess with Meyers type cells so I new that ac would not work so I built one because I am stubborn well you know what I was wrong that dam thing scared me when it first went into the water instant boil. They say it is like cop 200????? Mabee mfmp can tell me.

+1 #17 charlie tapp 2013-07-17 23:58

Look up Peter davey I will try to send pic to you in am . Very simple device . he was a saxophone player and called it a sonic boiler it warmed his coffee. Magee I can send video when my son gets home.

+1 #16 Ron B 2013-07-17 23:31

Davey Device.. possibly this is what he's talking about:

multimedia.stuff.co.nz/.../...

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