Average Velocity of Sublimation

From a sublimation rate, one can calculate the average velocity of water vapor exiting a vial. This is not the same as molecular velocity, but is analogous to "wind speed". It is easy to obtain a sublimation rate in gm / (m^{2} x s). That value can be directly measured by opening the lyophilizer __before sublimation is complete__ and measuring the weight remaining in a vial or collection of vials if an average is desired. Because sublimation is a zero order reaction, a one point (in time) measurement is sufficient to obtain the sublimation rate.

The expected value will be on the order of 0.02 to 0.08 gm/(cm^{2} x hr). Alternatively, the average sublimation rate can be measured from a graph of product ice temperature versus time. This assumes that the thermocouple temperature is coming from the bottom of the vial. The amount sublimed in grams is approximately the total water in the vial.

Assuming an ideal gas approximation for the amount of water vapor in any element of volume above the lyophile (cake), one can know how much water vapour can be in that volume element.

<= Weight (in grams) of water vapor at the chamber pressure and the saturation temperature in the vial above the lyophile for an arbitrary and unspecified volume area.

From the sublimation rate and the same arbitrary volume element, one can obtain a mass exit rate in (gm x m) / sec.

<= Mass exit rate in (gm x m) / sec for some arbitrary volume.

If the Mass exit rate is divided by the water vapor mass, one is left with just the exit rate in meters per second without ever having to specify the arbitrary volume!

<= Calculation of Exit Velocity from the Sublimation Rate and known chamber pressure. The temperature to use is the "saturation" temperature for the defined chamber pressure because the sublimation interface is at the saturation temperature.

Example:

<= 150 mTorr

The saturation temperature can either be looked up, or calculated from the Classius Clapeyron equation. I will calculate it.

<= Degrees C