Sublimation Science HOME

Intermediate Sublimation in a Lyophilizer

Saturation

This page is for individuals who already have an understanding of phase diagrams and how or why water ice will sublimate in a vacuum.  This document will focus on the saturation relationship between temperature and pressure.  Saturation is actually a term from refrigeration.  It refers to the temperature of a liquid (for refrigeration) at which that liquid is in equilibrium with its own partial pressure in the gas phase.   One might suggest that we couldn't have made that concept much more difficult. So let's try again. If you know what dew point is, saturation is dew point.  As it applies to water vapor, the saturation point is the temperature at which water vapor in the air becomes saturated and condensation begins. If you have Javascript enabled, there is a widget here that will let you look at temperature and dew point, which is saturation point.

To understand the widget, slowly raise the Temperature slider until the temperature is about 25°C.   Next raise the Dew Point slider as high as it will go. The Dew Point temperature and the air temperature will match. Notice that the relative humidity will be 100%. That defines "saturation" for your selected temperature.   However,  of more interest to lyophilization is the "saturation pressure" listed for your selected temperature.   In this case, the saturation pressure is the partial pressure of water in the air.  No more water can be put into the air.   It won't hold any more.   It is saturated.   Now of course, there doesn't have to be that much water in the air.   The relative humidity might not be 100% and in that case, the dew point temperature will be lower.   Slide the dew point slider down a little and you will see both the dew point temperature and the relative humidity decrease.

In order to input an exact temperature and obtain a matching saturation pressure, put a temperature value into the top box of the 4 boxes just below the button in the middle of the page.  For example, type in 37.60809°C. Click the "button" and you will see that the saturation pressure for that temperature is exactly 50 torr (not to be confused with mTorr).   The mathematical relationship that correlates saturation temperature and pressure was derived by Rudolf Clausius and Émile Clapeyron.  Although usually presented in tables, it can be expressed as shown below with suitable constants.  R is the molar gas constant and is 8.314 J/mol•K.   Temperature must be put in as degrees Kelvin [°C+ 273.15°K].  The formala returns pressure in Pascals.  Use this widget to convert.  Despite the large number of decimal places provided, the formula is approximate due to numerical fitting. It is, however, good enough for use in its range, which is NOT the range typically needed in lyophilization.


Clearly, the partial pressure of the water vapor plus the partial pressure of the remaining air is equal to the total "air" pressure, as measured with a barometer.  So at 37.60809°C and 100% relative humidity, and when the barometric pressure is exactly 1 atmosphere [760 torr], then the partial pressure of the remaining air is 760 - 50 = 710 torr. Looked at as molecules of water and molecules of dry air, the molecules of water constitute 7% of the total. (Arithmetic here does not work on your calculator until you convert everthing to SI units. Torr is not an SI unit of pressure.)



Partial Pressure of Water Vapor Over Ice

Moving along to something of slightly more interest, one can fit the same Clausius equation above to the careful measurements of the partial pressure of water vapor over ice at various temperatures. Indeed, those would be equilibrium temperatures. That is, the temperature and partial pressure of water vapor measured exactly at the ice/gas interface. Besides having an equation, most lyophilization sites should provide a table for this purpose and I hope to explain why one might ever refer to the table.

The driving force for sublimation during primary drying is the disruption of the saturation equilibrium as a consequence of using vacuum to remove water vapor from the inside of the lyophilization chamber. At the ice/gas interface, equilibrium will be constantly re-established by water molecules coming off of the ice interface and moving into the gas phase.

Vial Pic

For sublimation in primary to proceed expeditiously but without any melting (or 'partial melting', 'softening', 'entering the glass transition area', etc.) it is necessary to control the temperature of the ice interface. Several factors effect that interface temperature.
•  Chamber Vacuum - through the Clausius-Clapeyron relationship
•  Shelf Temperature - but hardly direct
•  Resistance to gas diffusion out of the vial

In the figure to the right is shown the Clausius relationship for vapor pressure over ice. The enthalpy of sublimation has a different value than the enthalpy of vaporization shown earlier. Also, the enthalpy value shown is for a temperature of ‑40°C. Indeed the enthalpy values are a little bit temperature dependent, even though I have used it as if it were a constant.   If you look in the table, you will see that while it changes with temperature, it doesn't change much.  If you calculate with this equation, your units will be in torr, since 'A' is in torr.

The Problem

Now let's look at something interesting about this curve (or the table - since they represent the same data).   In a typical lyophilization cycle, we would put the chamber vacuum at 100 mtorr (0.1 torr) and we might measure the ice temperature with thermocouples and it might be about ‑18°C and relatively flat for a time in primary. But according to the table, 100 mtorr isn't close to an equilibrium pressure with ‑18°C !  Indeed, 100 mtorr would be in equilibrium with ‑39.7°C and ‑18°C is in equilibrium with 940 mtorr - rather a long way off.  This could all be happening while the shelf temperature was close to 0°C. The explanation is that dry layer and vial stopper resistance are acting to bottle up the subliming water vapor and causing the pressure at the ice interface to be ~940 mtorr. Since we know that it is critical to keep the interface temperature below the collapse temperature, other than 'trial and error' how can we pick the proper chamber vacuum and shelf temperature?

Solution Outline

It has been obvious from the beginning of lyophilization that the only controls available to the machine were temperatures of the shelves and condensers, depth of the vacuum, and possibly exchange rate of inert gasses input to control pressure. Further, I have seen arguments, as recently as this decade, saying that condenser temperature and capacity were sublimation driving forces. They are not! The only purpose of a condenser is to protect the vacuum pumps from getting water in them. Invent a vacuum pump that can drink water and go to the levels needed and we can remove the condenser.

It is better understood that the ice temperature must be critically controlled. From the discussion here you realize that pressure at the ice interface and temperature at the ice interface are essentially one interconvertible item. You also realize that chamber pressure is not very similar to ice interface pressure. So let's develop a relationship between shelf temperature, ice temperature, ice interface pressure and chamber pressure. That's a tall order, but something we can do. After going through this exercise you will understand how one selects a shelf temperature and chamber pressure as well as appreciatate why 'trial and error' has prevailed.

Ohm's Law

In an electrical system we accept the E = I x R. That is, Voltage pressure is the product of current flow and resistance to flow. In a hydrodynamic system it is the same. We will use that as a starting point to develop a dynamic relation. Let's start by observing that flow is the change of mass per unit time.

Next, if we write flow as pressure divided by resistance and continue with the dynamic differential we get the following.

It would have been correct to have written dP/R instead of ΔP/R, but ΔP can be separated into Pi and Pc where the subscripts stand for pressure of the ice and pressure of the chamber respectively. Pressure of the chamber can be held constant, but both resistance and pressure of the ice are changing throughout the lyophilization. A definite criticism of the method we are about to show is that we will assume and average resistance and therefore hold R constant. We will also assume an constant ice temperature. In fact, these are not terrible assumptions since we have observed on many occassions that the thermocouple measuring ice temperature flattens appreciably throughtout the primary sublimation. It is only as the thermocouple tip begins coming out of the ice and going through the ice interface that apparant temperature rises.

In lyophilization, dMass/dt is a rate term and has preferred units of gm/(cm2•hr). While I would like to say that the term on the right above 'obviously' has those units, it obviously doesn't. If pressure is in torr and flow is in gm/hr, then the units of Resistance will be torr•hr/gm, and pressure/resistance will be gm/hr. So we need to introduce an area term to normalize the sublimation rate to the surface area of the ice.

One of the problems with the method being presented here is the need to choose a value for resistance. In a still more advanced presentation, it is possible to discuss methods for measuring the resistance but for the purposes here, we are going to be satisfied with selecting one based on experience. Let's say that the range of resistance is from 1 to 24 when expressed in the units shown. But in fact, the range is more commonly from 1 to 10 and still more often between 1 and 5. Since the resistance of pure water might be expected to be zero (a rather difficult concept in light of the equation above!), the lowest value of 1 can be attributed to the glass walls and stoppers. Sublimation is inherently a surface phenomenon and thus it must be and is rate limited by the available surface. Ultimately, the rate of water molecules escaping the surface is limited by a pre-exponential factor, which may be thought of as a 'flap' frequency for the vibration of molecular water. It is approximately 1013/sec.  As a pratical matter, the lowest resistance will come from the shortest and least tight cakes, where 'least tight' partially translates into least amount of solids by percent of solution.   This cake porosity can be influenced by the freezing step through freeze rate or annealing and, as well, the choice of excipients. Some salts and sugars just have a tighter pore structure and thus a higher resistance to gas permeation (flow).

Newton's Law of Cooling

Newton’s Law of Cooling says that the rate at which an object gains or losses heat is proportional to the difference between its temperature and the ambient temperature.   The temperature inside a building (across a wall) will lag behind the temperature that is directly in the heat source (sun).  For the purposes here, think of the wall as being the walls and bottom of the vial and think of the heat source as being the shelves. 'k' is an "overall heat transfer coefficient". 'Ambient' is the shelf temperature (Ts) and 'Temp' is the temperature at the bottom of the ice (Tb), since it resides immediately on the other side of the wall.   We could argue that the 'wall', being the entire glass vial, supports 'Temp' being everywhere along the ice-Vial contact area.   In the interest of getting through this, let's not.   Also, whether the equation is written as the "ambient - Temp", or the "Temp - ambient" is solely a matter of which is cooling and which is heating.

A refinement of Newton's Law of Cooling says that the energy transfer rate (dQ/dt) is proportional to the heat transfer coefficient (h), the surface area for heat transfer (A) and the temperature difference across the wall.  Here Avrefers to the area of the vial bottom and Kv refers to a vial heat transfer coefficient.

We are less interested in the temperature of the vial bottom and more interested in the temperature at the top of the ice, so we will introduce a simple equality and add it to the equation. ΔT is just the temperature difference between the top and bottom of the vial and Ti is the temperature of the ice interface. Remember that Ti is what we are searching for.



Thermodynamics of Open Systems

To make this as simple as possible, we are saying that there is another way to express the change in heat of the ice. Since the only thing happening is the loss of mass as a consequence of sublimation, it must be that the change in heat over time is equal to the change in mass times the sublimation enthalpy. We divide by the molecular weight of water because the enthalpy value is given in Joules/mole.



Clausius Clapeyron

The Clausius Equation is mentioned again just for completion. See the discussion at the start of this article if it isn't clear.  The equation is written in function notation Pi(Ti) to emphasize that pressure is a function of ice interface temperature and a reminder that ice interface temperature is what we seek.



Assembly

Algebra Happens!   Collect together the various equations and begin rearranging them in order to isolate Ti.   The resultant equation cannot be solved algebraically but can be solved numerically.   In this discussion the variable 'A' has been used for three different items. Ap represents the area of the product inside the vial.   Thus it is the vial inner diameter.   Av represents the area of the vial bottom and is thus he vial outer diameter.   Finally, 'A' with no subscript represents a constant on the front of the Clausius equation and has units of pressure.   When doing algebra, these are three different items.



More algebra. Also note again that Pi(Ti) is function notation. We have only solved the above equation for Pi.



Finally, substitute in the Clausius equation to obtain the end result.  Ti appears on both sides of the equation and there is no algebraic way to isolate it.   Still, it is a single equation in one unknown, namely Ti.



Back to Saturation

The original problem was how to pick a shelf temperature and vacuum setting that would result in some desired ice interface temperature.  Clearly, one way is to use the equation above with trial and error picks for all the variables.   It is a whole lot faster to do trial and error with an equation than it is on a lyophilizer.   With real numbers inserted, one can easily solve for Ti with a numerical method.   Indeed, this site has a tool for doing just that and it helps you to select values for resistance and Kv.