# Toys, toys, toys... and more

### 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:

## Comments

1234Good catch, don't want to claim too much excess, too quickly!

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

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.

@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 )

www.overunity.com/4083/peter-davey-heater/

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

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