Below some responses...
Not possible to show the steam at the outlet because that would require cutting the copper piping (and then having to reconnect), and second I don't want to dump a bunch of steam into my closet (my wife would kill me)!Crawdaddy wrote: A second video of the steam right at the point of generation would be awesome! (I know it's too much work for bruce tho).
Incidentally, these steam showers seem like they might cook you if you aren't careful1
As far as cooking myself, well, that's the whole point of a steam shower! There is a diverter that connects on the end of the pipe outlet so steam isn't just shooting straight out like I shown in the beginning of the video. It is directed at the floor. It's pretty similar to any steam shower in the world. The link to the exact model is in the original post.
Uh... The steam is used for the steam shower, hence the glass enclosed structure I'm standing in with a bucket.sparkyy0007 wrote: Looking at your video again at :20, the steam is completely invisible for the first 3 or so inches. What is this steam generator used for?
I connected the 11' rubber heater hose (5/8" ID) to the steam outlet at the wall so I could show what a 10kW off the shelf steam generator would produce (after all the losses in the copper tubing and the heater hose) just for comparison purposes. The heater hose is just for the test, normally there is a diverter that connects right at the wall to direct the steam down at the floor. The generator (located in the closet) is connected to the shower via ~25 feet of 1/2" insulated copper tubing.
The weight measurements were made on a cheap scale (you know, one of those scales used to check your weight?). The exact measurements weresparkyy0007 wrote: Your steam is very dry, too dry (there's a little experimental error there)
Total energy transferred to cold water:
Q= (energy in pail at end) - (energy in pail at start)
Q=Cp(end mass*end temp - start mass* start temp)
=4.186KJ/Kg*k(13.8*81 - 12.3 * 21.3)
=3582.42 KJ
If that energy is transferred by latent heat only, how much excess water will accumulate ?
delH vap= 2260KJ/kg
Kg of steam = 3582.42/2260
= 1.585kg
the del Mass is 13.8-12.3 = 1.5 kg(reported)
So all the latent energy of the added water would be necessary for
3582KJ
You have a few percent error in you measurement of weights, no problem.
Its very dry!
bucket: 2.2 lbs
bucket + water start: 29.4lbs
bucket + water end: 32.6lbs
So really the delta was 1.45kg +/- some error on my cheap scale.
The scale is surely not accurate enough to be drawing conclusions on the dryness of the steam. If it will add additional clarity I can re-run with a better scale to get more exact numbers.
If you're interested I uploaded another little video so you get a better picture of the setup.
<http://www.youtube.com/watch?v=JOd9wc6tCok>
If you have other questions let me know and I'll try to clarify.
Bruce