{‘7{MARIANA:iIIIséiIIIIIE‘IS'JQII'EIEI-{'.2;i;-.g::.1:_;~'.flg 'j.

GAS AND VAPOR. PERMEABILITY OF
THE DOUBLE WALL COMPARED TO
SINGLE WALL PLASTIC PACKAGES

Thesis for the Degree of ‘M. S.
MICHIGAN STATE UNIVERSITY
ISTVAN GYESZLI
1971

.....

 

N
\‘\ '
t"‘l"' I. I
‘ h...‘ I .h' ,

 

IIIIIIIIIIIIIIIIIIIIIIIIIIIIII

3 1293 00085 4251

 

I..__w

 

 

 

 

 

ABSTRACT

GAS AND VAPOR PERMEABILITY OF THE DOUBLE WALL
COMPARED TO SINGLE WALL PLASTIC PACKAGES

By

Istvan Gyeszli

A package should be designed for a certain shelf
life, which is the length of time during which a product
has acceptable quality, when stored under conditions of
its usual channels of distribution. Many factors can
affect the shelf life. If we consider only those factors
which can be determined by the package, two of the most
important are the internal gas and vapor concentrations.

The production researcher should tell the package
designer under what conditions the quality of the product
will not be suitable for consumers. (In this paper we
are considering only the internal gas and water vapor
Concentration as the factors affecting the shelf life.)

If we assume the internal gas and water vapor
concentrations are dependent only on the conditions at

the time of package closure and the permeability of the

Istvan Gyeszli

package, for given initial conditions many packages with
different permeabilities can be designed. Different pack-
ages will result in different shelf lives at different
costs. After consultation with the market expert, the
best one can be chosen.

The cost is one of the most important factors.
This paper introduces a hypothesis and its proofs. Using
the hypothesis we can reduce the package material expense.
The hypothesis is: the internal, partial pressure of a
gas or vapor is smaller in the double wall package as
compared with the single wall package until a certain time
tc’ if the quality and quantity of the material used, and
the temperature, and partial pressure in the packages and
between double walls are the same at time t = 0. The
time tC:

--is dependent on properties (material, surface
area, thicknesses, the space between double walls) of the
packages.

--is directly prOportional to the difference of
the external and internal partial pressure (in packages

and between double walls) at time t = 0.

The above hypothesis is proved by mathematical proof and
also by calculation of an analog computer.

If the needed shelf life is t5 and the limit of
the internal partial pressure for the given gas or vapor

is ps, we can design a single wall package which solves

Istvan Gyeszli

this problem. We assume we did and that package has the
minimum material expense.

Following the above, we design a double wall pack-
age using the same quality and quantity of the package
material and the same sized package. Then internal partial
pressure will be pd at time ts. Using the hypothesis: if

t<t

s c’ then pd < ps. However, the limit for the internal

partial pressure is ps, so pd may increase to p5. But pd
increases if the quantity of the package material decreases.
The size of the package is fixed, so we can reduce the
quantity of package material if we use thinner walls or
if we use a smaller space between the double walls. We
can do those until pd = ps. Then we get the needed shelf

life in spite of the fact we use less package material.

GAS AND VAPOR PERMEABILITY OF THE DOUBLE WALL
COMPARED TO SINGLE WALL PLASTIC PACKAGES

By

Istvan Gyeszli

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

MASTER OF SCIENCE

School of Packaging
1971

ACKNOWLEDGMENT

I would like to extend my appreciation to Dr. James

W. Goff who helped make this thesis possible.

ii

TABLE OF CONTENTS
Page
INTRODUCTION . . . . . . . . . . . . . . . . . . . . 1

DEVELOPMENT OF THE EQUATION OF THE GAS AND VAPOR
PERMEABILITY OF THE PLASTIC PACKAGES . . . . . . 3

PROBLEM I . . . . . . . . . . . . . . . . . . . . . 11
PROBLEM 2 . . . . . . . . . . . . . . . . . . . . . 15
PROBLEM 3 . . . . . . . . . . . . . . . . . . . . . 21
PROBLEM 4 . . . . . . . . . . . . . . . . . . . . . 22
PROBLEM 5 . . . . . . . . . . . . . . . . . . . . . 26
PROBLEM 6 . . . . . . . . . . . . . . . . . . . . . 31
PROBLEM 7 . . . . . . . . . . . . . . . . . . . . . 34
SUMMARY AND CONCLUSIONS . . . . . . . . . . . . . . 36
LIST OF REFERENCES . . . . . . . . . . . . . . . . . 38

iii

Table

LIST OF TABLES

Internal partial pressure of the oxygen
in the single wall package at different
times . . . . . . . . . . .

Internal partial pressure of the oxygen
for the double wall package at different
times, different wall thicknesses and
different ratios of the diameters

Internal partial pressure of the oxygen
in the double wall package using 90%,
93%, 96% of original quantity of the
package material (wall thicknesses
decreased) . . . . . . . . . . . .

Internal partial pressure of the oxygen
in the double wall package using 90%
of original quantity of package material
(ratios of diameters decreased)

Internal partial pressure of the oxygen
in the double wall package at different
initial conditions and at different
times

iv

Page

24

30

33

33

35

LIST OF FIGURES
Figure Page

1 Analog computer circuit to simulate the
internal partial pressure change of the
oxygen of the single wall package as a
function of time . . . . . . . . . . . . . . 25

2 Analog computer circuit to simulate the
internal partial pressure change of the
double wall package as a function of
time . . . . . . . . . . . . . . . . . . . . 29

INTRODUCTION

A package should be designed for a certain shelf
life, which is the length of time during which a product
has acceptable quality, when stored under conditions of
its usual channels of distribution. If we consider only
those factors which can be determined by the package, two
of the most important are the internal gas and vapor con-
centrations. For example, meat changes color, because
of oxygen; apples turn to yellow because of carbon dioxide;
dry food cannot be used over a certain moisture content;
bread is unusable beyond certain moisture levels.

In this work we examine what the differences are
in gas or vapor permeability of single and double wall
plastic packages. We have developed the equation of the
gas and vapor permeability of plastic packages. Using
the developed equation we posed seven problems and we
solved them. Some of them are mathematical proofs, others
are real situations solved by analog computer. To draw
conclusions we find a hypothesis, which can be useful for
package design, because by using the rule we can save
material.

The theory of the gas and vapor permeability of

the plastic materials is very well studied. Some excellent

reference books are available (3, 4, 10) and many articles
were written, summarized by Lebovits (11) until 1966.

Much data on gas and water vapor permeability can be found
in the literature but only for sheet material (2, 6, 9,
12, 13, 16, 17). Some data are about the gas permeability
of the real packages (15). More publications are about
water vapor permeability and the relationship between that
and the shelf life (1, 7, 8).

Loudenslagel and Roe (l4) wrote about simulation
of the gas permeation in plastic containers by analog
computer. Stannett and Szwarc (18) tried to find a simple
relationship for gas permeability between polymer and gas,
which can be used generally, without real tests. They

tried to calculate one permeability constant from others.

DEVELOPMENT OF THE EQUATION OF THE GAS AND VAPOR
PERMEABILITY OF THE PLASTIC PACKAGES

Mass transport through polymeric materials occurs
by activated diffusion. This takes place in three steps.
First the permeant dissolves in the permeable membrane on
the side of its higher concentration. Then it diffuses
through the membrane towards the side of lower concentra-
tion, a process which depends on the formation of "holes"
in the plastic network due to thermal agitation of the
chain segments. Finally the permeant becomes desorbed on
the side of the lower concentration. The equation of

diffusion is given by Fick's first law:‘

8c

dn = D 3? dt (1)
n = number of diffused molecules
D = diffusion constant
%% = concentration gradient, which is assumed as constant
t = time

The number of diffused molecules is equal to the
diffusion constant times the concentration gradient times
time, if the area of the diffusion is unity. The concen-
tration gradient is a constant, if we integrate between 11:

12, and c1 and c2 it gives

dn _ C1'C2
11-12 = L, the thickness of membrane
c1 = concentration of the gas molecules on one
side of the membrane
c2 = concentration of the gas molecules on the

other side of the membrane
If the area of the membrane is equal to A,
dn _ C1’C2
2le—1)——];_A (3)
The permeation is a combined process between diffusion and
solution, thus the solubility coefficient must be con-

sidered, which is given from Henry's law.

c = Sp (4)
C = concentration of the gas on the surface of the
membrane
S = solubility coefficient
p = partial pressure of the gas in the gas space

After above:

 

dn = DS C1’C2 A (5)
3f' __r__

D - S E Pm = permeability constant

P = (number of diffused mols. x thickness )

m area x time x concentration difference

dn Cl-CZ

HIE-Pm—TA (6)

Equation 6 is the equation of the mass permeability
of polymeric materials. If we assume ideal gas laws for
gases and vapors (which is not completely true, but use
of them results in only negligible error) then concentration
of the gases and vapors are equal to their partial pres-
sures,

dn _ p1132
3? - Pm _—I—— A (7)

p1 and p2 are the partial pressures of the gas or
vapor on the two sides of the membrane. In the literature
the permeability constant is given commonly as:

volume of diffusedAgas or vapor (STP) x thickness
area x time x pressure difference

 

 

 

 

_ L
Pm - number of mol x A x t x AP
P = volume of diff mol (STP) x A x E x AP
Em = number of mol = n
fi volume Of_moI STP VISTPI
V = volume of mol (STP)
n = m—pov (8)
o
poV p
~ _ - _ - o
Pm — P RT; - P TR (9)
‘V‘ 0

- P -
O

po = 1 atm
= 0
To 273 K
R = Regnard gas constant

Let's take into consideration a real package. We
want to know the internal gas or vapor concentration as
a function of time (which is equal to partial pressure of

gas or vapor). For example:

pe(t) = the external partial pressure of the gas
or vapor
pi(t) = the internal partial pressure of the gas
or vapor
V = volume of the internal gas or vapor space.

This is a constant.

If dn molecules go through the wall of packages,

the internal partial pressure is changed by dp.

n = %¥ (ll)
dn = §E¥ (12)

From equation 10 and equation 12:

 

d V _ p0 pe(t)-pi(t)

‘HT' RTo L

d _

55- = P J;- p0 r: (pectrpicm , (14)

Equation 14 gives the internal partial pressure of
the gas or vapor as a function of time.

We must consider for what condition the equation is
true for determination of package life. We must also con-
sider the possible change of every component of the equation
as a function of time, temperature and pressure.

There may not be any chemical reactions between the
gases and solids (the material of the package or the
packaged product) or liquids (the packaged product). The
commonly considered gases and vapor relatively inert under
usual conditions and package materials ordinarily are
chemically inactive.

The permeation of gases or vapors are independent,
so the equation is true for.each gas or vapor separately.

We assume the velocity of the gases or vapors is
zero inside and also outside of the package.

The time required to reach steady state is negli-
gible compared to the storage time.

The package is faultless. There is no leakage, no
damage, no mechanical distortion.

The sign of an infinitisimally small change of the
internal pressure (dp) depends on the direction of the
permeation. pi is always the internal partial pressure
of the gas or vapor. The sign of dpi is positive, if
the direction of permeation is from outside to inside of

the package.

The infinitesimally short time change (dt) is
practically zero as compared to the storage time.

The area of permeation (A) is the area that gas or
vapor can go through. The area is constant for all time,
temperature, and pressure.

The thickness of package wall (L) must be the same
everywhere, which for a curved surface is sometimes not
true. To correct for this we might use average thickness.
The change of thickness may be a function of time, for
example degradation. It might be estimated, but usually
this is not necessary. The thickness is independent of
time, temperature and pressure.

The volume of gas space (V) is only the actual gas
or void space. For the package of a very porous material
we might consider the volume of the pores. The volume of
the gas space is independent of time, temperature and

pressure.

run

is the permeability constant.

“Us

ED°S

The diffusion constant is assumed independent of
the concentration of the permeant in the membrane. The
solubility coefficient is also assumed independent of the
pressure of the permeant in the phase in equilibrium with
the membrane. In this case the permeability constant is
independent of the pressure. The temperature dependence

of D and S is expressed by Arrhenius type equations.

D
II

DO exp(-Ed/RT) (15)

CD
II

So exp(-AHS/RT) (16)

['11
II

d activation energy for the diffusion process

HS = the heat consumed on dissolving a mole of
permeant in the membrane
R = Regnard gas constant
P 5 DS = DO exp(-Ed/RT) So exp(-AHS/RT) (17)

F =

0 D050 (18)
Ep = Ed + AHS (19)
P = P0 exp(-Ep/RT) (20)

Ep and P0 can be determined from two permeability
constants, which were measured at different temperatures.
The permeability constant is independent of time and
pressure.

po is 1 atm
To is 273°K

T is the temperature at any time. This may be
Changed by function of time.

pe is the external partial pressure of the gas or
vapor. This is constant or zero (N2, 02’ C02) or function
of time and temperature.

7 pi is the internal partial pressure of the gas

or vapor. This is a function of temperature.

10

_ T
pi ’ pio 77? (21)

T = temperature °K

= internal partial pressure at 273°K

The internal partial pressure of gas or vapor is
changed by time.
Then equation of gas or vapor permeability of the

plastic package is

A pio(t)T(t)
“a? ‘ o T po Vt (Pe(t) ' 273 I (22)

 

 

PROBLEM I

We have to design a package with volume V. The
package has no product and is filled with N2 gas. The
oxygen concentration is zero at time t = 0. At tf time
the internal partial pressure of the oxygen may not be
more than pf in the closed package. (The increase of the
internal partial pressure of the oxygen is caused by
oxygen permeation only.) The temperature is constant
Tf°K. The external gas is air.

Requirement: The material expense should be a
minimum.

Solution: After we have decided the shape of the
package we can calculate the surface area A of that, from
the volume of the package. We can use the equation of
the gas and vapor permeability of the plastic packages,

equation 22.

dP‘ - E p- (t)T(t)
-3% = P0 exp(-§T%fy) IAEl pO V? (pe(t) - 10273 ) (23)

O

 

T is constant and equal to Tf. Using equation 19,

"Un
II

f PO exp(-Ep/RTf) ' (24)

11

12

PT is constant. p0; T0 are constant also. After

 

 

f
above,
dpi A
*3; = k VI (Pe(t)‘Pi(t)) (25)
where
- Tf
k = P - p (26)
Tf TO 0
and
p- (t) T
pi(t) = 1373 f (27)

The external partial pressure of the oxygen is

constant P, because the external gas is air.

dpi - k A (P - (t 28
'3f" VI Pi )) ( )

Let's integrate the equation 28 between 0 and t,

and p(o) and p(t).

pict) = P[1 - expc-k (7% tn 4131(0) exp(-k (7% t) (29)

where pi(o) is the internal partial pressure of the oxygen

at t = 0 time. But pi(o) = 0.

pi(t) = P[1 - exp(-k VE‘tII (30)

Since pi(t) may not be larger than pf at time

t = tf we can use

PiLtf) = pf (31)

13
Using equation 29 and equation 30,
_ A
pf — P[1 - exp(-k VF tf)] (32)

A and V are constants. So from equation 32,

 

 

Pf
A I (33)
V tf
Using equation 26 and equation 33,
p —
VT in (1 - —£) PT
0 P _ f 4
A T p ‘ ‘r‘ (3 )
l f o

If we assume the expense of the process of package
making is independent of the quality of the material used,
then the minimum material expense will be obtained by use
Of the minimum material. The quantity of the material is
equal to surface area times thickness. The surface area is
fixed, so minimum thickness results in minimum quantity
of material.

From the literature we can pick the oxygen permea-
bility constants of different materials. (We can calculate

the PT from two different oxygen permeability constants,

f
which were measured at two different temperatures. Using
a semi-log graph, where the permeability constants are
plotted against reciprocal of absolute temperature, we

can draw a straight line through the two points. We can

read the oxygen permeability constant for Tf°K tempera-

ture.)

14

Using equation 34 we can see the left side of the

equation is constant B, so

"UI

L = Tf (35)
T

From PT and B we can calculate L.
f

N = L x A x U (36)

Where N is the cost of material, U is the unit
price. If we make this calculation for many materials,

we can find the minimum material expense.

Conclusion: If the material, M, has an oxygen
permeability constant Pm at temperature Tf, the wall
thickness should be Lm for a given shelf life problem.

We cannot reduce the material expense, because in equation

34 everything is fixed.

PROBLEM 2

How does the internal partial pressure of the
oxygen change as a function of time for a single wall
package compared with a double wall package? Assume
that the internal volume of the package and quantity and
quality of the package material are the same, the tempera-
ture is constant, and the external gas is air.

a. The internal partial pressure of oxygen is
zero at time t = 0 for both packages and between double
wall.

b. The internal partial pressure of oxygen is
the same for both packages, and is not zero at time t = 0.

Solution: The single wall package has volume V1,

surface area A1 and wall thickness L1. The double wall

package can be considered as two packages. The smaller

1 and wall thickness L3.

The larger package has a volume V2 (which is equal to the

package has a volume V1, area A

full volume minus the volume of the smaller package);
surface area A2 and wall thickness L2. The quantity and

quality of material is the same; so

AlLl = A1L3 + AZLZ . (37)

15

16
Using equation 22 for single wall package,

dp.
~3§ = 90 exp(-§%) TE pO VIE? p(e)(t) - pi(t)) (38)

but pe(t) is constant because the external gas is air.

Because T is constant

dpi = k A1 (P - (t)) (39)
‘3? VIII Pi
where Po (exp - fig) TE po = k (40)

Using equation 25 for double wall package,

dp: A1 *
~33 = k VII; (pm(t) - pi(t)). (41)
where pm(t) is the internal partial pressure of the oxygen
between the double wall.
Some gas molecules will go into the double walls
from outside.

1

dpm _ A2
*3? - k WEIE'LP ‘ PmLt)) (42)

and some gas molecules will leave from between double walls

and they move into the smaller package at the same time.

dp A1 *
W = k V—2L3 (piCt) - pm(t)) ‘ (43)

17

or
dpi _ A1 *
‘3? — ‘k VEE; (PmLt) ‘ PiLt)) (44)

dpm = dpm + dpm (45)
at (It at

Combination of the equation 42, equation 43 and
equation 45,

dpm A2 A

7&- = (75-5-2- (P - pm(t)) - k V2113; (pm(t) - p:(t)) (46)

k is always the same constant, because the same material

and temperature are used.

Let's say:
k VEIE'LPmLt) ' PiLt)) = Z(t) (47)
Z(o) = 0 because of the same initial condition.
*
Z(w) = 0 because after t = w, pm and pi will be equal
to P.
*
Z(t) > 0 pm(t) is always larger than pi(t).

If we assume Z(t) = 0, then equation 46 will be:

+
dpm A2

.3? = k 7—1—2 2 (P - pm(t)) (48)

but, because Z(t) ; O,

18

+
dp dp
m m
‘H—t' 3- 7E (49)

The combination of equation 48 and equation 49,

dpm > k .121. (P - (t)) (50)
—HE-= VZLZ pm

If we integrate the equation 50 between 0 and t,

pm(o) and pm(t).

A A
pm(t) ; P (l-expc-k V§%; t)) + pm(o) exp(—k vgf; t) (51)
if pm(o) = 0 (situation a),
A2
PmLt) ; P[l-exp(-k vgfg't3I (52)

t
For p(o) = 0, pi(o) = 0, pm(o) = 0 (situation a)

we can combine equation 41 and equation 52.

dp: A1 A2 *

“af'é k VIE; [P (l-exp(-k vgr; t)) ‘ PiLt)I (53)
or

dp: A1 * A2

Compare equation 54 with equation 39.

dpi A

—df = VI%I (P ‘ Pi(t))

A A
1 1 . . I
k v—lrg- > k Vl-Ti- because L1 > L3. BUt untll a certaln t1me

19

A A
2 1 . .
- P exp(-k V—L_ t) k VPL—' is larger than the difference
2 2 1 3
which is caused by L1 > L3. It can be seen that the certain
A

- - 2
time tC 15 dependent on the constants. If L3 and k'VZf;

decrease, then tC increases. It can be proven if V2 and A2
are increasing, A2 is decreasing. In this way we need
2

large L3, A2, V2 and L2. Equation 37 gives a relationship

among them for problem 2.

A1L1 = A1L3 + Asz

Since A1 and L1 are fixed, so increase of all
factors (A2, L2, L3) is impossible, so we can suppose we
have to find an optimum value for A2, L2, L3 which gives

the longest tC.

For pi(o) = p:(o) = pm(o) f 0 (situation b) we

can combine equation 41 and equation 51.

*
dpi A A

75.: k VIE} (P - p:(t) - (P - pm(o)) exp(-k (312,—; t)) (55)

Compare equation 55 with equation 39 and it can be
seen if pi(o) = p:(o) = pm(o) increase, the time tc de-
creases and approaches zero, but never can reach that if
pi(o) = p;(o) = pm(o) < P, because we can always find time
when the second part of the right side of equation 55 is

larger than the difference which was caused by L1 > L3.

20

Conclusion: The internal partial pressure of
the oxygen is always smaller in the double wall package
Compared with single wall package until a certain time tC,
if the quality and quantity of the used material are the
same and the temperature and partial pressure in the
packages and between double walls are the same. The time
tC is dependent on the properties of the package (material,
surface area, volume, wall thicknesses, the space between
double walls), and the internal partial pressure of the oxygen
in the packages and between double walls at time t = 0.
The time tC can not be zero ever, if the external partial

pressure of the oxygen is higher than the internal partial

pressure of the oxygen at time t = 0.

PROBLEM 3

This is the same as problem 1, but uses a double

wall package and results of the problem 2. tf < tc.
*
Solution: If tf < tc’ then dpi < dpi (from problem
at at

2). But pi(o) = p:(o), so

*
*
BUt Pi(tf) = Pf, 50 piLtf) < Pf-

The internal partial pressure may not be more than

*
pf, so can be pf. So pi(tf) may increase to pf. But

*
pi(tf) is increasing, then we use less material. We can
a!
use always less and less material until pi(tf) = pf. In
this way we can reduce the quantity of materials. We can

do this two ways, using thinner L2 and L or using smaller

3,
A2, which is given by smaller V2.

Conclusion: If the needed shelf life time tf is
shorter than tc, then by using the double wall package, we

can save package materials.

21

PROBLEM 4

We have a cylindrical package. The height of the
package is equal to the diameter. The package is empty,
and it is filled up with nitrogen gas. The material is
polyethylene. The package is closed and the oxygen con-
centration is zero at time t = 0. The diameter of the
package is 10 cm. The temperature is constant 25°C. The
external gas is air.

The oxygen permeability constant of the poly-

ethylene at 25°C is 0.66 cm3 x mm

 

cm? x month x atm
The wall thickness is 1 mm. What will the internal
partial pressure of the oxygen be at 0.25, 0.5, 1.0 and
1.5 months?
Solution: Using the equation of the gas permea—

bility of the packages, equation 22,

dp

 

' - T A
“3% = P T_ po VI (Pe(t) ‘ PiLt))
o
3
P = 0.66 2 cm X cm
cm x month x atm
T = 298°K
T = 273°K
o

22

23

pe(t) = 0.21. The oxygen concentration is a

constant 21% in the air.
p0 = 1 atm

After the above,

dpl _ A 0
‘3? — 0.734 VI ( .21 - Pi(t))

D = diameter 10 cm
H = weight of package 10 cm
2
A = £211 + DUH
but D = H
A _ 3Dzn
__2_
v = Dzn-H = D30
‘"1"' 4

Using equation 58 and equation 57,

<m>
I
DIG)

‘ 0.6

<m>
I

D = 10 cm, so

After the above,

dpi
at

0.440 (0.21 - pi(t))

(56)

(57)

(58)

(59)

To solve this problem we can make a simple analog

computer circuit in Figure 1.

24

Because of p(o) = 0, the initial condition is 0.

The results are given in Table 1.

Conclusion: The internal partial pressure of the
oxygen reaches 50% of the external partial pressure after

1.5 months.

Table 1. Internal partial pressure of the
oxygen in the single wall package
at different times.

 

 

Time (months)

0.25 0.5 1.0 1.5

 

Internal partial
pressure of the 0.022 0.042 0.076 0.104
oxygen (atm)

 

25

.mefip mo cowuocsm mm mmmxomm Ham: mamcwm may we comxxo on» mo omcmco .
whammohm Hmwuhmm Hangouaw map oumfisefim ou ufisouwo yoHSQEou moamc< .H whamwm

 

 

Epw HN.o u >OOH Or

 

 

 

 

 

 

figcsa-a Alecaa-av- Adena- “scam- finscaa-acs..o

 

 

 

 

 

Sum Hm.o u >ooa
youmfimom muvflm

 

 

PROBLEM 5

The problem is the same as problem 4, but we use
double wall package instead of single. The quantity and
quality of the used material are the same. The partial
pressure of the oxygen is zero between double wall t = 0
time. The conditions are the same. What will be the
internal partial pressure of the oxygen at 0.25, 0.5, 1.0
and 1.5 months?

Solution: The double wall package consists of

two single packages. Symbols are the same as in problem

2.
D2 = XD1 and H2 = XH1
From equation 57,
3Din 3D§n
A = A =
l 2 2 2
3X2Din
But D2 = XDl’ so A2 = ——7———-
D30
v=1
1T
3 3 3 3 3
D20 X D10 D10 D10

26

27

Using equation 41 and equation 46,

3
dpi - A1 *
—af" k VIE; (PmLt) ' PiLt))

dpm — A2 A1 *
—3f - k vgf; (P ' PmLt) ‘ k var; (PmLt) ' PiLt))

k = 0.734 (from problem 4).

The used material should be the same as in problem 4.

AlLl = A1L3 + AZLZ (60)
or
A2
L1 = L3 + KT L2 (61)
L1 = 1 mm (from problem 4)
3Dln 3X2Din
A1=—T Az‘T‘
After the above,
1 = L + XZL (62)
3 2

Because D1 = 10 cm

 

 

 

k A1 = 0.734 - 0.6 = 0.44
51133 L3 I3
k A2 = 0.734 - 0.6 x2 = 0.44 x2
U2E2 L2(x3-1) L2(x3-1)
k A1 = 0.734 - 0.6 = 0.44
U2E3 (x3-1)L3 (xx-1n3

28

*
95? = #05:“ (pm(t) - p:(t)) (63)

dpm 0 44 x2 0 44 *
= - (p _ (t)) - ° ( (t) " -(t)) (64)
7*? {Tl—(353‘?) pm (x3-1)L3 p’“ p1

where P is 0.21 (the partial pressure of the oxygen in the
air).

Using different ratios, X of the diameters, we can
calculate the constants.

We can solve this problem by an analog computer
circuit in Figure 2. Because p:(o) = pm(o) = 0, the initial
conditions are 0.

Choosing X = 1.05, 1.10, 1.2, and using L = 0.3,

2
0.4, 0.5, 0.6 mm after calculation of constants and use of
analog computer circuit, Figure 2, the results are given

in Table 2.

Conclusion: The internal partial pressure of the
oxygen is smaller than in the single wall package (problem
4) until a certain time tc' This way the conclusion of
problem 2 is proved. Using larger ratios X of the diameters
time tC increases. If the ratio X of the diameters is
constant, time tC increases if the ratio of wall thick-

nesses approaches 1.

.osfiu mo

:oflpocsm mm ommxomm Ham: cansow may mo «mango whammonm
Hmfiuuma Hangman“ on» mamasswm ou uflsonflo wouSQEoo monc< .N ouswfim

 

sum H~.o u 39an

 

29

 

vv.o

 

 

noumflwom fil/Mpvfin all!
t
Ema- _\ E a 8.2.98:

 

«I: w-I‘

 

 

 

ahead-

 

wand.

 

 

Sum H~.o u >OOH

 

iH- xv AH- xv~3
a S 59/: 2:... a E is LIEEPe LII

  

anew. a. mug EdvIIIII

we. o ev.o

 

 

 

2:593-

fiH- xv a

Xv¢.o

Table 2.

30

Internal partial pressure of the
oxygen for the double wall package
at different times, different wall
thicknesses and different ratios
of the diameters.

 

 

L

Internal partial pressure of the
oxygen at time (in atm)

 

cm?) (mi) .34: .23. 2.2. man-.3.
.05 0.685 0.300 0.017 0.038 0.074 0.103
.05 0.580 0.400 0.016 0.038 0.076 0.107
.05 0.475 0.500 0.018 0.039 0.075 9.105
.05 0.370 0.600 0.017 0.039 0.076 0.103
.10 0.637 0.300 0.014 0.034 0.072 0.104
.10 0.516 0.400 0.014 0.034 0.073 0.105
.10 0.395 0.500 0.014 0.036 0.073 0.105
.10 0.274 0.600 0.015 0.038 0.075 0.107
.20 0.856 0.100 0.014 0.036 0.072 0.101
.20 0.568 0.300 0.011 0.032 0.069 0.104
.20 0.424 0.400 0.010 0.030 0.068 0.103
.20 0.280 0.500 0.013 0.034 0.073 0.107
.20 0.136 0.600 0.015 0.037 0.075 0.109
.20 0.092 0.700 0.017 0.039 0.076 0.110

 

PROBLEM 6

Everything is the same as in problem 5. How will
the results change:

a. If we use only 90% of the material decreasing

b. If we use only 96%, 93%, 90% of the material

decreasing L2 and L3?

Solution:
a. Using equation 54,
A A
_ 2 2 _ 2
Ll-L3+KIL2,bUtK1-‘Xo

For equation 55,

 

L1 = L3 + x L2
X2 _ L1'L3
- I:2
but
* —
L1 - 0.9 L1
*
§2 _ L1'L3
- I:2
X ‘ L
2

31

32

*
If L2 and L3 are given, we can calculate X and
t
X can be used in equation 63 and equation 64 instead of

X. To do this calculation for

L2 0.400 0.500 0.600

L3 0.424 0.280 0.136

and using an analog computer in Figure 2, the results are

shown in Table 3.

b. Using equation 62,

If we use Y% of the material (Y = 90, 93, 96),

YL1 = YL3 + YXZLZ, but X is constant, so if we multiply

L3 and L2 by Y we will use Y% of the material.

Using YL3 and YL2 in equation 63 and equation 64
instead of L3 and L2, we can get the answer using analog
computer. The results are in Table 4 for X = 1.2 and

L2 = 0.4 and 0.500, L3 = 0.424 and 0.280.

Conclusion: To reduce the quantity of the package
material, time tC decreases. The time tC decreases at a
slower rate if the quantity of package material is reduced
in wall thickness, and tC decreases at a faster rate than
noted above if the quantity of material is reduced by

decreasing the space between the double walls.

33

 

 

 

Table 3. Internal partial pressure of the
oxygen in the double wall package
using 90%, 93%, 96% of original
quantity of the package material
(wall thicknesses dedreased).
Internal partial pressure of the
L L oxygen at time (in atm)
X 3 2
(mm) (mm) 0.25 0.5 1.0 1.5
month month month month
1.2 0.407 0.384 0.013 0.035 0.075 0.109***
1.2 0.395 0.372 0.014 0.036 0.079 0.111**
1.2 0.383 0.360 0.015 0.038 0.082 0.115*
1 2 0.269 0.480 0.014 0.037 0.078 0.115***
1.2 0.262 0.465 0.015 0.039 0.082 0.119**
1.2 0.255 0.450 0.016 0.043 0.087 0.122*

 

* = 90% ** = 93% *** = 96°

0\

 

 

 

Table 4. Internal partial pressure of the
oxygen in the double wall package
using 90% of original quantity of
package material (ratios of
diameters decreased)
Internal partial pressure of the
L L oxygen at time (in atm)
X 3 2
(mm) (mm) 0.25 0.5 1.0 1.5
month month month month
1.104 0.424 0.400 0.022 0.040 0.084' 0.116

1.090 0.280 0.500 0.020 0.045 0.088 0.126

 

PROBLEM 7

How are results of problem 4 and problem 5 changed

if p,(o) = pm(o) = p:(o) P 0?

Solution: Using the analog computer circuits in
Figure 1 and Figure 2, we use initial conditions for the

integrations. The results are in Table 5 for

*
pi(o) = pm(o) = pi(o) = 0.042; 0,064; 0.084; 0.105;

0.147; 0.168 atm. X = 1.2; L2 = 0.4; L3 = 0.424.

Conclusion: Time tC decreases if pi(o) = pm(o)

*
= pi(o) increase.

34

35

Table 5. Internal partial pressure of the
oxygen in the double wall package
at different initial conditions
and at different times.

 

 

Internal partial pressure of
the oxygen at time (in atm)

 

igigigi°n wgiikigethe 0.5 1.0 1.5
month month month
0.042 Single 0.078 0.102 0.127
0.064 " 0 092 0.117 0.137
0.084 " 0.111 0.131 0.148
0.105 " 0.128 0.145 0.158
0.147 " 0.166 0.171 0.179
0.168 " 0.176 0.181 0.189
0 042 Double 0.067 0.100 0.129
0.064 " 0.084 0.113 0.139
0.084 " 0.103 0.128 0.150
0.105 H 0.119 0.143 0.160
0.147 " 0.155 0.169 0.181

0.168 " 0.172 0.183 0.191

 

SUMMARY AND CONCLUSIONS

If we evaluate the solutions and results of the
seven problems, we see the following:

The internal partial pressure of a gas or vapor is
always smaller in the double wall package compared with
single wall package until a certain time tC if the quality
and quantity of the used material are the same, and the
temperature and partial pressure in the packages and
between double walls are the same at time t = 0.

Time tC increases if the space between double walls
increases.

For a given double wall package time tC is longer
as the ratio of the wall thicknesses approaches 1.

The time tC decreases at a slower rate if the
quantity of material is reduced in wall thickness and
tC decreases at a faster rate than noted above if the
quantity of material is reduced by decreasing the space
between the double walls.

The time tC is inversely proportional to the
internal partial pressure at time t = 0.

The time tC can be very small but cannot be zero
ever if the external partial pressure is higher than the

internal partial pressure at time t = 0.

36

37

If the required shelf life for a package (considering
internal partial pressure) is shorter than time tC using
a double wall package material can be saved.

After above we can formulate the hypothesis: The
internal partial pressure of a gas or vapor is always
smaller in the double wall package compared with single wall
package until a certain time tc, if the quality and quantity
of the material used and the temperature are the same and
partial pressure in the packages and between double walls
are the same at time t = 0. The time tC

--is dependent on properties of the packages
(material, volume, surface area, wall thickness, the
space between double walls).

--is directly pr0portiona1 to the difference of
the external and internal partial pressure (in packages

and between the double walls) at time t = 0.

10.

11.

12.

13.

LIST OF REFERENCES

Alvaz, G. E. WVT through plastic containers. Modern
Packagigg, July (1967), p. 123.

 

Anderson, E. Test result from fast WVT unit. Modern
Packaging, Dec. (1967), p. 141.

Barrer. Diffusion in and throu h solids. Cambridge
at the University Press, 19 1..

 

Brown, L. R. Permeability and shed life. Modern
Packaging, Oct. (1964), p. 184.

Crank, J., Park, G. S. Diffusion in polymers. Academic
Press, London and New York, 1968.

 

Gillespie, F. Diffusion of water vapour through a
hydrophilic polymer film. Polymer Sci., Vol. 4,
A-l, No. 4, Apr. (1966), p. 939.

 

Heiss, R. Shelf—life determinations. Modern Packaging,
Aug. (1958), p. 119.

 

Hu, K. H. Estimating shelf life of products affected
by moisture. Packaging Eng., Feb. (1963), p. 62.

 

Huang, R. Y. M., Kanitz, P. J. F. Permeation of gases
through modified polymer films. J. App. Poly. Sci.
13 (1969), p. 669.

 

Jost, W. Diffusion in solids, liguids, gases.
Academic Press Inc. Publishers, New York, 1957.

 

Lebovits, A. Permeability of polymers to gases, vapors
and liquids. Mod. Plastics, March (1966), p. 139.

 

Lindergren, C. R. Polybuthylene; properties of a
packaging material. Polymer Eng. and Sci., Vol.
10, No. 3, May (1970), p. 163.

 

Loudenslagel, K. D., Roe, S. F. Gas transmission in
thick plastic sections. Modern Packgging, Feb.
(1967).

 

38

14.

15.

16.

17.

18.

39

Loudenslagel, K. D., Roe, S. F. Analog simulation of
permeation. Modern Plastics, March (1969).

 

Loudenslagel, K. E., Floate, W. Whole-package trans-
mission test. Modern Packaging, Sept. (1970).

 

Peterson, C. M. Permeability studies on heterogeneous
polymer films. J. of App. Pol. Sci. 12 (1968),
p. 269.

 

Schneider, N. Water vapor permeability of ultra-thin
polyurethane films. J. Macromol Sci. B., Vol. 3,
No. 4, Dec. (1969), p. 761.

 

Stannett, V., Szwarc, M. The permeability of polymer
films to gases-~a simple relationship. J. of
Polymer Sci., Vol. 16, No. 81 (1955), p. 89.

 

‘11111111111111Ifs