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© Institute for International Research, Inc. 2006. All rights reserved. Module 5B: Cycle Optimization Continued.

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Presentation on theme: "© Institute for International Research, Inc. 2006. All rights reserved. Module 5B: Cycle Optimization Continued."— Presentation transcript:

1 © Institute for International Research, Inc. 2006. All rights reserved. Module 5B: Cycle Optimization Continued

2 2 international Module 5B Purpose and Objectives  Module Purpose:  Process optimization requires understanding the process. The student will continue with the examination of variables involved with sublimation.  Module Objectives: After this module, you will be able to  Understand how one selects a Shelf Temperature & Chamber Pressure  Appreciate why trial and error has prevailed.

3 3 international Ohm’s Law

4 4 international Resistance Range is from 1 to 24 torr·cm 2 ·hr / gm. (Usually 1 to 5)  Common Sense may help you to select a value. Actual measurement is possible and quite involved.  Short cakes have lower resistance.  Loose and porous cakes have lower resistance  To calculate a shelf temperature, you must select a value.

5 5 international 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. Ambient Temp

6 6 international Newton’s Law of Cooling In Lyophilization, the “wall” is the “vial” and we would like to include the “ice”. The “Ambient” temperature is the shelf temperature The “Temp” is the temperature at the top of the ice.

7 7 international ΔT Over Length of Ice This value is the change in temperature from the bottom to the top of the ice. It is less than 2 deg C. per cm. In this experiment, it was clear by direct thermocouple measurement that the temperature changed by 10 degrees over a linear distance of ~6 cm. Literature values used by others are often 0.1 to 0.5 degrees per cm.

8 8 international From Newton to Lyophilization k, is an overall heat transfer coefficient = Kv, times the vial bottom area = Av. The constant temperature source, “Ambient” temperature, is the lyophilizer shelf, Ts. Tb is the temperature on the inside bottom of the vial. However, the “Temp” of interest is at the top of the ice.

9 9 international From Newton to Lyophilization  Unit Analysis: Heat Transfer Coefficient

10 10 international From Newton to Lyophilization Ts is the Shelf Temperature Tb is the Ice Bottom Temperature  T is the temperature difference per linear distance across water ice. Ti is the temperature at the top of the ice, also called Temp of the Interface. The bottom is warmer than the top (interface).

11 11 international Thermodynamics of Open Systems Often, M (0.018 kg/mol) is omitted, but since I use ΔH in units of J/mol, one needs the conversion factor. Mass can be in either moles or grams. Unit Analysis

12 12 international Clausius Clapeyron Equation Enough Already !

13 13 international Assembly Newton’s Law of Cooling applied to sublimation. Thermodynamics of Open Systems Ohm’s Law applied to Sublimation

14 14 international Solve for P i Note: P i (T i ) is function notation, not multiplication. Finally, Add the Clausius Eq. in place of P i (T i ) to get one equation in one unknown. The unknown is T i.

15 15 international Solution by Newton’s Method

16 16 international The Function & Its Derivative

17 17 international Another Way to Look at Rate Since Rate is measurable (and fairly linear) and since Pi can be calculated and Pc is measured, the Rp can be calculated from a singe run.

18 18 international Rate vs Pressure vs Temperatures

19 19 international Exercise Module 5b  Write a Function that when given Ti (interface temperature) and Pc (chamber pressure) will return Ts (shelf temperature).


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