Doing (at least trying to) some math here....

**I'm NOT sure I got all this under control, I might be waaaay off and wrong in a lot of the following, but hey, at least I'm trying **If I set the process chamber to 500 x 500 mm, the height of the vapor to 200 mm (volume is 0.05 m3)

The temperature is set to 230 deg C, and pressure of 1 atm, I get 1.211 mol required using:

https://www.chemicool.com/idealgas.htmlAccording to

https://www.solvay.com/en/brands/galden-pfpe , the molecular weight of LS/HS/HT-230 is 1020 amu (LS/HS/HT-200 is 870 amu)

So 1 gram of HT-230 is 1/1020 mol (1/870 mol for HT-200).

So to fill the 500 x 500 mm chamber with a 200 mm high "cloud of vapor", I need 1.211 mol * 1020 => 1235 gram of HT-230

If using HT-230 instead (870 amu), I will need 1.211 mol * 870 => 1054 gram of HT-200

Density at 25 deg C of HT-230 is 1.82 g/cm3, so 1235 gram is 679 mL.

Density at 25 deg C of HT-200 is 1.79 g/cm3, so 1054 gram is 589 mL.

Just for fun, compared to water: molecular weight of water is 18.02 amu. 1.211 mol of water weights 21.6 gram

So 21.6 gram of water can fill the chamber with vapor, Galden HT-230 requires 1235 gram/679 mL for the same !

**Getting fluid up to boiling temperature:**Specific heat @ 25 deg C for Galden is 0.23 Cal/g deg C => 0.96 J/g*deg C.

To raise 1253 gram Galden from 20 deg C to 230 deg C requires: 1253 * 210 * 0.96 => 254 kJ.

So if 253 kJ is required, I can bring the Galden to boiling point (from 20 to 230 deg C) in 1 minute using 253 kJ/60 seconds => 4.2 kW

Specific heat for water is 4.186 J/g*deg C.

To raise 1000 gram of water from 20 deg C to 100 deg C requires: 1000 * 80 * 4.186 => 335 kJ

So, the water will boil in 60 seconds if I inject: 335 kJ/60 seconds => 5.58 kW.

**Vaporizing the fluid completely:**Heat of vaporization for Galden when already is at boiling point, for both HT-200, HT-230, LS-200 and LS-230 is 63 J/g (the different types are almost all the same value).

So to vaporize 1235 gram of (already boiling) HT-230 I need 1235 * 63 => 78 kJ.

If I would need to vaporize all the Galden in 1 minute I would be required to supply 78 kJ/60 sec => 1.3 KW of heat during that 1 minute (sounds like an awful small amount ??).

(for HT-200, this would be somewhat smaller, the amount of Galden is smaller, and it would need to be raised 30 deg less than the HT-230)

Heat of vaporization of water already at boling point:2260 J/g

To vaporize 1000 gram of water (already boiling) I need 1000 * 2260 = 2260 kJ

Vaporizing in 60 seconds requires: 2260 kJ/60 sec => 37 kW for 60 seconds

All of this assumes ideal conditions, no loss to the surroundings etc.

Can anyone perhaps confirm the calculations, at first I thought it is a large amount of Galden needed to fill the chamber, I know it is quite large (500 x 500 x 200 mm), but still?

Also, I don't quite understand that the heat of vaporization is 63 J/g for Galden and 2260 J/g for water, is that really true ?

On top of that amount I will need to add the minimum level as Kane pointed out (also a 200 mm high vapor cloud is very much, could probably be much smaller).

EDIT:

If chamber is 500 x 500 with only 130 mm cloud height:

Volume = 0.5*0.5*0.13 = 0.0325 m3

At 230 deg C, 1 atm, 0.0325 m3, 0.787 mol is required.

0.787 mol * 1020 => 803 gram => 441 mL

Raising 803 gram from 20 to 230 deg C in 60 seconds: 803 * 210 * 0.96 => 162 kJ = 162 kJ/60 seconds => 2.7 kW

Vaporizing it in 60 seconds when at boiling point: 803 gram * 63 J/g => 50.6 kJ => 50.6 kJ/60 seconds => 840 Watt