### The parenteral Society Leak Rate

```Leak Rate Problem
Using a specification from the Parenteral Society, the leak rate for a new, clean empty
freeze dryer should be less than 0.02mBar-L/sec. a) If one had a leak of that
magnitude, and if it came from multiple sources, where each source was a round hole
with diameter 0.2 microns, then how many such holes would be present? b) If the leak
rate were changed to X mBar-L/sec, then how many such holes would be present?
What follows will be the derivation of an equation that will allow us to solve the
problem above as well as many similar problems. The equation that will be derived is
2 ⋅ π ⋅R ⋅T ⋅ M
where RL is the "Leak Rate" in mbar-Liters/sec or PV / time.
P ⋅R ⋅ T
M is molecular weight of air, P is pressure outside the LYO (or more technically, it is ∆P
between the inside and outside of the Lyo), R is the Molar Gas constant,
joule
8.314472 ⋅
and T is room temperature in degrees K. [DO ALL CALCULATIONS
mole ⋅K
IN SI UNITS ALL THE TIME AND THEN CONVERT THE FINAL ANSWER TO
WHATEVER UNITS YOU WANT.]
Area = RL ⋅
I use MathCad. I can multiply ( 2ft ⋅ 3mm ⋅0.05furlong) = 18.395 L and MathCad assures
that my answer and units are correct. If you aren't using MathCad, Be Careful.
Kinetic theory defines ν as the number of
collisions
2
that will occur between a gas some
m ⋅sec
area of a wall.
n' ⋅ ν m
ν=
EQ #1. n' is defined as the number of molecules per unit volume.
4
molecules
n' =
and ν is the mean velocity of gas molecules from kinetic theory
3
m
m
νm =
8 ⋅R ⋅T
π ⋅M
Thus we began with 1 equation and 2 definitions of terms.
n' ⋅ν m
EQ #1 Let ν = the number of collisions per square meter in a sec.
4
moles
molecules
n' =
Definition n' is a concentration 'like' term (conc. is
)
3
3
m
m
8 ⋅R ⋅T
J
kg
νm =
M := 0.029
Defiition Note: R := 8.314472
= air molecular
mol ⋅K
mol
π ⋅M
wt.
and T := 298.15K = room temperature.
ν=
Since we know that the collisions are done with air, we can estimate the mass, w,
of the collisions.
w=
2
ν ⋅M
EQ#2 w = total mass of collisions per m in a second.
Na
M = mass of each air molecule
23
−1
Na = Avogodro's number Na := 6.0221415 ⋅10 ⋅mole
as a check on what has been done so far, we can look at a units only calculation:
 collisions  ⋅.029 kg


2
mol
m
⋅
s
kg


w=
=
23
−1
2
6.0221415 ⋅10 ⋅mole
m ⋅s
Substituting ν from EQ#1 into EQ#2 and solving for the number of molecules/m3
n' ⋅ν m
ν ⋅M
ν=
EQ#1 w =
EQ#2
4
Na
n' ⋅ν m
w=
n' =
4
Na
⋅M
4 ⋅ w ⋅ Na
ν m ⋅M
<= This is the substitution
EQ#3 <= Here it is solved for the concentration 'like' term.
molecules
3
n'
m
mol
=
=
Now verify the units for n', the concentration 'like' term:
molecules
3
Na
m
mole
note that n'/Na is an actual concentration term and can be substituted into the gas law.
Gas Law: P =
n
n
n'
is the same as
. Both are
V
V
Na
concentration.
P=
n'
4 ⋅w
⋅R ⋅ T =
⋅R ⋅ T
Na
ν m ⋅M
EQ#4 Notice that for n' we substituted in
EQ#3 and cancelled out Na
P=
4w
ν m ⋅M
⋅R ⋅ T
EQ#4
Now substitute in an expression for the mean velocity of molecules, ν m
P=
4 ⋅w ⋅R ⋅ T
8 ⋅ R ⋅T
π ⋅M
and simplify to get
P = w⋅
⋅M
2 ⋅π ⋅ R ⋅T
M
EQ#5
'w' is the mass of molecules/sec striking some area and is easily obtained. From the
problem, we know that 0.02 mbar-liter /s of molecules are striking an unknown area.
And we can safely assume that they are air molecules with M = 29 g/mol. The mass
of the molecules was given by the problem, where the Leak Rate was defined as
mbar-Liter/sec.
Writing a differential form of the gas law,
d 
 P  ⋅V
 dt 
d 
d
n=
we should note that  P  ⋅V is in fact our leak rate. We should call it
R ⋅T
dt
 dt 
d 
RL =  P  ⋅V
 dt 
d
P ⋅V ⋅M
M
dt
kg
d
=w=
but notice that n ⋅
Not obvious? w has units of
2
R ⋅T ⋅Area
dt m2
m ⋅s
So substitute in RL and we get
RL ⋅M
w=
which can now be put back into EQ#5 to get
R ⋅ T ⋅Area
RL
2 ⋅ π ⋅R ⋅T
P=
⋅M ⋅
which simplifies to
R ⋅T ⋅Area
M
Area = RL ⋅
2 ⋅π ⋅ R ⋅T ⋅M
P ⋅R ⋅T
Now let's put in some numbers so that we can compute. mbar := .001bar
RL := 0.02 ⋅
mbar ⋅liter
s
R := 8.314472
J
mol ⋅K
5
<= definition
Leak Rate
Molar Gas Constant
P := 1.013 × 10 Pa − 13.332Pa
Pressure differential between atmospheric and Lyo
T := 298.16K
M := 29
gm
mol
Area := RL ⋅
Molecular weight of Air
2 ⋅ π ⋅R ⋅T ⋅ M
P ⋅R ⋅ T
dia = 14.681 µm
− 10
Area = 1.693 × 10
m
2
dia := 2 ⋅
Area
π
<= diameter if it were a single hole
The original question was how many 0.2µm holes must exist to obtain the observed leak
rate?
Area of a .2µm hole is
Area
= 5389
BugArea
2
 .2µm 
BugArea := 
 ⋅π
 2 
2
BugArea = 0.031 µm
That many pores of 0.2µm diameter can exist with the Society Leak
Rate.
```