Module 5B:
Cycle Optimization Continued
© Institute for International Research, Inc. 2006. All rights reserved.
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.
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Ohm’s Law
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Resistance
• Range is from 1 to 24 torr·cm2·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.
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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
d
Temp
dt
k  ( Ambient  Temp )
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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.
d
Temp
dt
k  ( Ambient  Temp )
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ΔT Over Length of Ice
This value is the change in temperature from the bottom
to the top of the ice.
In this experiment, it
It is less than 2 deg C. per cm.
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.
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From Newton to Lyophilization
d
Temp
dt
k  ( Ambient  Temp )
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.
d
Q
dt
Av  Kv   T s  T b
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From Newton to Lyophilization
 Unit Analysis:
d
Q
dt
Heat Transfer Coefficient
Av  Kv   T s  T b
J
2
m 
2
m sK
 ( K)
J
s
Watts
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From Newton to Lyophilization
d
Q
dt
Av  Kv   T s  T b
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.
d
Q
dt

T i  T
Tb
The bottom is warmer than
the top (interface).
Av  Kv  T s  T i  T

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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
d
Q
dt
Hsub
d
mass 
Mw
dt
J
mol
kg

kg
s
J
s
Watts
mol
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Clausius Clapeyron Equation
 Hsub
P i T i
A e
RT
 i
Enough Already !
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Assembly
d
Q
dt

Av  Kv  T s  T i  T
Hsub
d
mass 
Mw
dt
Pi  Pc
d
mass Ap 
Rp
dt
d
Q
dt
d
Q
dt
Ap   P i  P c  Hsub

Rp
Mw

Newton’s Law of Cooling applied to sublimation.
Thermodynamics of Open Systems
Ohm’s Law applied to Sublimation
d
Q
dt

Av  Kv  T s  T i  T

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Solve for Pi
Note: Pi(Ti) is function notation, not multiplication.
P i T i

 Ap  Hsub  P c  Av  Kv  Rp  Mw  T s  Av  Kv  Rp  Mw  T i  Av  Kv  Rp  Mw  T

Ap  Hsub
 Hsub
P i T i
 Hsub
A e
RT
 i
A e
RT
 i
Finally, Add the Clausius Eq. in place of Pi(Ti)
to get one equation in one unknown. The
unknown is Ti.

 Ap  Hsub  P c  Av  Kv  Rp  Mw  T s  Av  Kv  Rp  Mw  T i  Av  Kv  Rp  Mw  T
Ap  Hsub
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
Solution by Newton’s Method
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The Function & Its Derivative

 Ap  Hsub  P c  Av  Kv  Rp  Mw  T s  Av  Kv  Rp  Mw  T i  Av  Kv  Rp  Mw  T
f  T i
Ap  Hsub
d
f  T i
dT i
n
x0
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Ap  Hsub
 A
Hsub
2
 A e
RT
 i
 Hsub
e
RT
 i
R  Ti
0 1 2 3 4
( 273.15K  25K)
xn 1
x
Av  Kv  Rp 
Mw

 Hsub
xn 
f  xn
d
f  xn
dx
 248.15 


240.047


 236.561 

K
236.115


 236.109 
 236.109 


<= Guess Value to get started
<= The Computer Program
<= The Answer: Ti = Temperature at the interface for
some chosen Ts (shelf temperature) and Pc (chamber
pressure).
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Another Way to Look at Rate

 
Pi Ti  Pc
Rp

Rate T i  P c
P
force
area
P
Rp
N
2
N1
m
Units_of_Rp
kgm
2
s
kg  m
2
s
2
kg  m
2
s
2
 torr m2 hr 


kg


 kgm

 2

s
2

m s 
 m2



kg


m
2
m
kg
2
m s
torr cm hr
gm
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.
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Rate vs Pressure vs Temperatures
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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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