kp2_responsi Filtration

FILTRATION

Nur istianah,ST,MT,M.Eng – KP2 – THP UB

Constant pressure

( ^{Cs P} ) ^{V A} ( ^{Rm P} )

A V

t

∆ + −

∆

= µα − µ

2 /

t.V ⁄ A

V ⁄ A

Ex1. Constant pressure

Filtration area = 0.01 m² A

Solution density = 1,062 kg/m² ρ

Solution viscosity = 1.6×10^-3 Pa.s µ

Filtration pressure = 200 kPa ∆P

Solid concentration = 3 kg/m³ Cs

Determine specific filter cake resistance and filter medium resistance

resistance

Time (sec)

Time (sec) Volume (cmVolume (cm³³)) 0

0 00

14

14 400400

32

32 800800

55

55 12001200 80

80 16001600 107

107 20002000

Time (sec)

Time (sec) Volume (cmVolume (cm³³)) Volume (mVolume (m³³)) tA/VtA/V V/AV/A 0

0 00 00 00 00

10

10 400400 0.00040.0004 350350 0.040.04 20

20 750750 0.00080.0008 400400 0.080.08 30

30 11401140 0.00120.0012 458.33458.33 0.120.12 40

40 17301730 0.00160.0016 500500 0.160.16 50

50 20102010 0.00200.0020 535535 0.200.20

y = 412,2x - 438

-500 0 500 1000 1500 2000 2500

0 10 20 30 40 50

Series1 Linear (Series1)

kg m

P Slope Cs

/ 10

792 .

9 ) 1175

10 200

( 2

) 3 ( ) 10 6

. 1 (

2 1175

10 3

3

×

=

× =

×

×

∆ =

=

−

α α

µα

P ercept R

y int

^m

= 307 . 67

= µ ∆

m R

R ercept P y

m m

/ 1 10

845 .

3

67 . 10 307

200

) 10 6

. 1 (

67 . 307 int

10 3

3

×

=

× =

×

∆ =

=

−

Ex2. Constant rate

A slurry containing 25.7 kg dry solids/m

³

of filtrate across the filter medium area 2.15 m

²

at a constant rate of 0.00118

m

³

/s. If the pressure drop was observed 4,000 and 8,500 Pa after 150 and 450 seconds of filtration, respectively. The

viscosity of filtrate was 0.001 Pa.s

Determine the specific cake resistance and filter medium

resistance.

The solution

∆ P (Pa)

t (sec) t (sec)

 

 



= 

 

 



= 

t V A

ercept R y

t V A

Slope Cs

µ

m

µα int

2 2

6000 7000 8000 9000

∆∆P (Pa)P (Pa) t (sec)t (sec) 4,000

4,000 150150

8,500

8,500 450450

y = 15x + 1750 R² = 1

0 1000 2000 3000 4000 5000 6000

0 100 200 300 400 500

t (sec)

P (Pa)

V ercept R

y

kg m

t V A

Slope Cs

m

1750

int

/ 10

938 .

1 15 15

. 2

) 00118 .

0 )(

7 . 25 ( ) 001 .

0 (

15

9 2

2 2 2

=

 

 

=

×

=

=

=

 

 



= 

µ α α

µα

m R

R

t ercept A

y

m m

m

/ 1 10

189 .

3 15 1750

. 2

) 00118 .

0 ( )

001 .

0 (

1750 int

×

9

=

=

=

 

 

=

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