Model Formula for Fluid Contamination Levels

1.Attenuation of Fluid Contaminants by Cyclic Multipass Filtration

Let us consider "To" particles as initial number of contaminants in a fluid container staged
for contaminant attenuation by cyclic filtration.

The reduction of particles can be described in the formula shown below,

Tn/To = (1-f*E)n...equation 1
Where "n" is the No. of Total Volume Circulations or Cycles.
For example, given a 50-l fluid volume that is circulated at a 200-l rate will have a n=4.

Example 1
Let's assume an initial fluid contamination of 5-micron particles of 100,000 staged for a filtration efficiency of 90% at a circulation or cyclic rate of 0.7. Let us calculate the reduction of the number
of particles per no. of total volume circulations or cycles.

Using the above equation,Tn/To = (1-f*E)n

One circulation will yield a reduction of particles as shown,
from T1/100,000=(1-0.9 x 0.7)1 will result to T1=37,000

The reduction of particles until 10 volume cycles is shown at the table below;

No. of Volume Cycles	Attenuation of No. of Fluid Particles
0	100000
1	37000
2	13690
3	5065
4	1874
5	693
6	257
7	95
8	35
9	13
10	5

Increasing the no. of total volume cycles will dramatically reduce the no. of fluid particles
from an initial contamination level.

2. Cases for Intermittent Contaminant Introduction during Washing Process

For cases wherein contaminants are introduced constantly at intermittent periods into a given fluid, and assuming the introduced no. of contaminants as constant, the no. of particles at a certain interval can be described by the formula given below,

TE+TW = TW(To/Tn)/[(To/Tn)-1] ... equation 2
where: TE : no. of particles settling within the fluid
TW : no. of particles introduced from the washed item

Example 2
Let's assume the no. of 5-micron particles from the washed item at 100,000 at a volume cycle of 1
washed once in a cyclic filtration process.
Filtration Efficiency: 90%
Circulation or Cyclic Rate : 70%

From equation 1:
Tn/To = (1-f*E)n
Tn/To = (1 - 0.9 x 0.7)1
= 0.37

From equation 2:
TE+TW = TW(To/Tn)/[(To/Tn)-1]
TE+TW = 100000(100000/37000)/[(100000/37000)-1]
TE+TW = 100000 x 2.703/(2.703-1)
TE+TW = 158719.9
Therefore, as the washed item is introduced once, TE+TW = 158,720

Further computation of the above equation at a series of washing is shown at the table below;

No. of washed itemIntroduction	After Introduction	After 1 Volume Cycles	No. of washed itemIntroduction	After Introduction	After 1 Volume Cycles
0	0	0	15	158730.11	58730.14
1	100000.00	37000.00	16	158730.14	58730.15
2	137000.00	50690.00	17	158730.15	58730.16
3	150690.00	55755.30	18	158730.16	58730.16
4	155755.30	57629.46	19	158730.16	58730.16
5	157629.46	58322.90	20	158730.16	58730.16
…			…
10	158722.53	58727.33	25	158730.16	58730.16

It can be observed that the numerical values are nearly identical at intervals.
For the given case above wherein the no. of introduced contaminants are assumed constant at given intervals, the number of particles at every volume cycle nearly settles at a certain level.

Example 2. Case 1

Let's have the same assumption as the above example only having no. of volume cycles at 2.
From equation 1:
Tn/To = (1-f*E)n
Tn/To = (1-0.9 x 0.7)2
= 0.1369

From equation 2:
E+TW = TW(To/Tn)/[(To/Tn)-1]
TE+TW = 100000(100000/13690)/[(100000/13690)-1]
TE+TW = 100000 x 7.305/(7.305-1)
TE+TW = 115860.4

Therefore, as the washed item is introduced once, TE+TW = 158,720

Further computation of the above equation at a series of washing is shown at the table below;

No. of washed itemIntroduction	After Introduction	After 2 Volume Cycles	No. of washed itemIntroduction	After Introduction	After 2 Volume Cycles
0	0	0	8	115874.84	15874.85
1	100000.00	13700.00	9	115874.85	15874.85
2	113700.00	15576.90	10	115874.85	15874.86
3	115576.90	15834.04	11	115874.86	15874.86
4	115834.04	15869.26	12	115874.86	15874.86
5	115869.26	15874.09	…
6	115874.09	15874.75	…
7	115874.75	15874.84	20	115874.86	15874.86

Example 2. Case 2

Let's have the same assumption as the above example only having no. of volume cycles at 3.
From equation 1:
Tn/To = (1-f*E)n
Tn/To = (1-0.9 x 0.7)3
= 0.051

From equation 2:
E+TW = TW(To/Tn)/[(To/Tn)-1]
TE+TW = 100000(100000/5065)/[(100000/5065)-1]
TE+TW = 100000 x 19.74/(19.74-1)
TE+TW = 105336.2

Therefore, as the washed item is introduced once, TE+TW = 105,336

Further computation of the above equation at a series of washing is shown at the table below;

No. of washed itemIntroduction	After Introduction	After 3 Volume Cycles	No. of washed itemIntroduction	After Introduction	After 3 Volume Cycles
0	0	0	6	105374.08	5374.08
1	100000.00	5100.00	7	105374.08	5374.08
2	105100.00	5360.10	8	105374.08	5374.08
3	105360.10	5373.37	9	105374.08	5374.08
4	105373.37	5374.04	10	105374.08	5374.08
5	105374.04	5374.08

In conclusion, for an intermittent introduction of washed items with the given above conditions for a 5-micron particle size, at a no. of volume cycle = 1, will be reduced at 58,720; at a no.
of volume cycles = 2, will be reduced at 15,860, and further at 5 volume cycles, will yield to
5,336.
The value of the contaminant level in the system and the gap between the rise and fall of contaminant level at a certain time interval can be decreased further by increasing the number of volume circulations or cycles.

Related Sites

Yamashin Filter Mfg, Corp.

Hydraulic Filter Group

http://www.yamashin-filter.co.jp