hydrosoul

Hello,

I am trying to understand how many BTU are necessary to evaporate water from clay at a certain humidity. Lets say 10% or 15%.
I understand that it is not the same energy necessary to evaporate pure water. Does some know something about this?

Can anyone help me?

Many thanks,
Jorge from Chile
Guest

About 25 years ago I built a monitor to measure the temperature profile of an industrial clay drying oven

The process was:-

1/ Squeeze as much water out of the slurry - effectively the slurry was contained in canvas bags,
2/ Extrude the clay into 'noodles' about 1 cm in diameter.
3/ The noodles go onto an open mesh conveyor belt and are fed into an oven - from memory the oven was about 20m long with 4 loop-backs within - total path length (say) 80m. The noodles spent about 20 minutes in the oven and came out white and crumbly. From memory the maximum temperature inside a noodle was about 60C. The airflow in the oven was considerable - air temperature and humidity levels not known (forgotten).

Clearly the theoretical minimum energy required and the optimum working conditions for an industrial process are going to be very different.

-C2.
Confused2
Sorry - not logged in ^^
Bivalves
QUOTE (Guest+Feb 24 2010, 06:55 PM)
About 25 years ago I built a monitor

Does the gauge needle still rip into the extreme pompous imbecile zone, when you approach it?

Bivalves
QUOTE (hydrosoul+Feb 24 2010, 03:42 PM)
Hello,

I am trying to understand how many BTU are necessary to evaporate water from clay at a certain humidity. Lets say 10% or 15%.
I understand that it is not the same energy necessary to evaporate pure water. Does some know something about this?

Can anyone help me?

Many thanks,
Jorge from Chile

Pointless question - firstly would it be Illite, kaolinite, smectite etc .... clays vary in minerals/water content/particle distributions of the very low µm order. Therefore, calculation of desiccating the stuff in BTU's is nigh on impossible, unless you know its exact composition.

ps;- only an insane idiot (Confused2) would attempt to give you an answer.