geankoplis filtration

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FILTRASI pemisahan opadat-cair dengan cara penyaringan menggunakan filter skala besar dibutuhkan cara perhitungan pada proses filtrasi untuk design(pera *Berapakah berapakah luas filter yang dibutuhkan untuk menyaring filtrat slurr *evaluasi proses: berapa lamakah proses filtrasi harus dihentikan untuk pember Laju alir V2-V1 = dV t2-t1 dT untuk melakukan perhitungan filtrasi data" yang dibutuhkan: V: laju alir filtrat miu viskositas Cs konsentrasi slurry A luas filter (-dP) pressure drop alfa tahanan cake Rm tahanan filter

Transcript of geankoplis filtration

Page 1: geankoplis filtration

FILTRASIpemisahan opadat-cair dengan cara penyaringan menggunakan filterskala besar dibutuhkan cara perhitungan pada proses filtrasi untuk design(perancangan filter) dan evaluasi pada saat pengoperasian peralatan

*Berapakah berapakah luas filter yang dibutuhkan untuk menyaring filtrat slurry tahu sebesar 1000L/hari?*evaluasi proses: berapa lamakah proses filtrasi harus dihentikan untuk pembersihan cake?

Laju alir: V2-V1 = dVt2-t1 dT

untuk melakukan perhitungan filtrasi data" yang dibutuhkan:V: laju alir filtratmiu viskositasCs konsentrasi slurryA luas filter(-dP) pressure dropalfa tahanan cakeRm tahanan filter

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skala besar dibutuhkan cara perhitungan pada proses filtrasi untuk design(perancangan filter) dan evaluasi pada saat pengoperasian peralatan

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diket:t V t v t v

4.4 0.000498 34.7 0.002498 73.6 0.0040049.5 0.001 46.1 0.003002 89.4 0.004502

16.3 0.001501 59 0.003506 107.3 0.00500924.6 0.002

t/v V8835.341 0.000498

9500 0.00110859.43 0.001501

12300 0.00213891.11 0.00249815356.43 0.00300216828.29 0.00350618381.62 0.00400419857.84 0.00450221421.44 0.005009

Kp =

Kp =6000000 =

α =

B =6783 =

Rm =

t/v =trial A =t/v =t/V =

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v =kapasitas diganti 500 L t/v =

t =

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slop Kp/2 3000000B= 6783

(-dP)= 338000T= 298.2 K

A= 0.04390.00089*α*23.47/(0.0439)2(338000) Cs= 23.47 kg/m3186334015712.601 m/kg 0.0008937 Pa.s

t= 86400 sv= 1000 L

0.00089*Rm/0.0439(338000)

112618899630.749

864000.260412197836126

Kp + B t1 t285256.5293597173 + 1143.471 86399.99999 86400

Ԉ*α*Cs/A2(-dP) N/m2

µ*α*Cs/A2(-dP) m2

µ=

µ*Rm/A(-dP)

m-1

0 0.001 0.002 0.003 0.004 0.005 0.0060

5000

10000

15000

20000

25000

f(x) = 2884955.53883983 x + 6783.75290220386R² = 0.996513687338502

Grafik t/V Vs V

Column ALinear (Column A)

V

t/V

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500 L 0.5 m31500000 + 6783

753391.5

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Kp 6000000

7060 lbr/ft2

0.473 ft21.465 lbm/ft3

1 m3

dt9.97308E-06

0 0.001 0.002 0.003 0.004 0.005 0.0060

5000

10000

15000

20000

25000

f(x) = 2884955.53883983 x + 6783.75290220386R² = 0.996513687338502

Grafik t/V Vs V

Column ALinear (Column A)

V

t/V

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diket: (-dP) 194400 kN/m2T 298.2 K

A= 0.0439 0.473 ft2Cs= 23.47 kg/m3 1.465 lbm/ft3

0.000894 Pa.sv= 1000 L 1 m3

Vx10^3 t t/V0.0005 6.3 12600

0.001 14.0 140000.0015 24.2 16133.33

0.002 37.0 185000.0025 51.7 20680

0.003 69.0 230000.0035 88.8 25371.43

0.004 110.0 275000.0045 134.0 29777.78

0.005 160.0 32000

intercept: B 9795 s/m3Kp/2 4000000 s/m6kp 8000000 s/m6

Kp= µ*α*Cs B= µ*RmA2(-dP) A(-dP)

α= Kp*(A2(-dP)) Rm= B*(A*(-dP))µ*Cs µ*Rm

α= 142892830984 m/kg Rm= 9.35E+10 m-1

m2

µ=

0 0.001 0.002 0.003 0.004 0.005 0.0060

5000

10000

15000

20000

25000

30000

35000

f(x) = 4421964.4059644 x + 9795.85185185186R² = 0.99860123177719

grafik t/V Vs V

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0 0.001 0.002 0.003 0.004 0.005 0.0060

5000

10000

15000

20000

25000

30000

35000

f(x) = 4421964.4059644 x + 9795.85185185186R² = 0.99860123177719

grafik t/V Vs V

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diket: d 200 µm 0.0002 m ρa interpolasiT 294.3 KP 101.32 kPaρoil 900 kg/m3 ρa 1.202267 kg/m3h 0.457 m µ 1.83E-05 Pa.sg 9.8066 m/s^2

ditanya: Vt?

Vt trial= 5.506E-07Nre= 7.242E-06 3313876.9012 µCd= 3313877 3.765982E-06Vthitung= 0.000768 0.0955513812vtrial-vhit -0.000768 1.67571E-10

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x-x1 = y-y1x2-x1 y2-y1294.3-283.11 = y-1.246311-283.11 1.137-1.246

11.19 = y-1.24627.89 -0.109

(y-1.246)*27.89 = -1.2197127.89y-34.75094 = -1.21971y = 1.2022671208318

x-x1 = y-y1x2-x1 y2-y1294.3-283.11 = y-1.78*10^-5311-283.11 (1.90-1.78)^10-5

11.19 = y-1.78*10^-527.89 1.20E-06

(y-1.78*10^-5)*27.89 = 1.34E-05 -34.75094 -34.8599427.89y-0.000496 = 1.34E-05 27.89 -1.24990827.89y = 5.10E-04

= 1.83E-05-33.53123-1.202267

0.000496

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Misalkan dari rentang D1-D2. bagaimana profil kecepatan settling pada rentang D terseb (misal ex 14.3-3 hal 927 geancop)Silika

Vt Dp Trial Vt Nre Cd Vt hitung0.0000127 0.024827 3.15E-07 7.63E+07 6.001845E-05

0.00002032 0.076075 1.54E-06 1.56E+07 0.00016809840.00002794 0.155497 4.33E-06 5.54E+06 0.0003304510.00003556 0.262998 9.33E-06 2.57E+06 0.00054696220.00004318 0.399539 1.72E-05 1.39E+06 0.0008186187

Dp' 0.0000508 0.563534 2.86E-05 8.40E+05 0.0011437816Diket :

ρ silika = 2.65 g/cm^3 = 0.00265 g/mm^3ρ galena = 7.5 g/cm^3 = 0.0075 g/mm^3μ air = 1.00E-03 kg/m.s = 0.0010005 g/mm.sρ air = 998 kg/m^3 = 0.000998 g/mm^3 g 9.8066 m/s^2= 9806.6 mm/s^2

Silika Murni : Dp gal =(ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

0.00001 0.000015 0.00002 0.000025 0.00003 0.000035 0.00004 0.000045 0.00005 0.0000550

0.000250.0005

0.000750.001

0.001250.0015

0.001750.002

0.002250.0025

0.002750.003

0.003250.0035

0.003750.004

0.004250.0045

0.00475

Grafik Vt Vs Dp

silikagalena

Dp

Vt

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5,21 * 10^ = (ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

Dp gal =(ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

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(misal ex 14.3-3 hal 927 geancop)

Vt trial-Vt hitung Dp 1 = 1.27E-02 mm = 1.27E-05 m0.024767 Dp 5= 5.08E-02 mm = 5.08E-05 m0.075907 interval partikel = 0.000007620.1551670.262451

0.398720.56239

Cd Silika 1 2 3 4Cd 1 7.63E+07 1.635E-07 0.119813 1.363E-08Cd2 1.56E+07 8.018E-07 0.131804 4.861E-08Cd3 5.54E+06 2.254E-06 0.140235 1.111E-07Cd4 2.57E+06 4.851E-06 0.146837 2.052E-07Cd5 1.39E+06 8.949E-06 0.152332 3.349E-07Cd6 8.40E+05 1.485E-05 0.157031 5.022E-07

GalenaDp Trial Vt Nre Cd Vt hitung Vt trial-Vt hitung

1.27E-05 0.096874203 1.23E-06 1.96E+07 0.000235 0.096639 apabila dalam tabung diberi aliran air.2.032E-05 0.299193208 6.06E-06 3.96E+06 0.000661 0.298532 maka partikel dg Vt < Vt air maka partikel tersebut akan naik.2.794E-05 0.610965981 1.70E-05 1.41E+06 0.001299 0.6096663.556E-05 1.034358806 3.67E-05 6.54E+05 0.002152 1.0322074.318E-05 1.568964698 6.76E-05 3.55E+05 0.003218 1.565746

5.08E-05 2.212671205 1.12E-04 2.14E+05 0.004496 2.208175

Cd Galena 1 2 3 4Cd 1 1.96E+07 6.382E-07 0.130011 4.05E-08Cd2 3.96E+06 3.153E-06 0.143091 1.454E-07Cd3 1.41E+06 8.854E-06 0.152235 3.321E-07Cd4 6.54E+05 1.908E-05 0.159411 6.137E-07Cd5 3.55E+05 3.514E-05 0.165361 1E-06Cd6 2.14E+05 5.83E-05 0.170461 1.5E-06

partikel silika yang punya Dp1-Dp2 akan lolos.

Galena Murni :Dp gal =(ρ silika - ρ ) ^0,5

0.00001 0.000015 0.00002 0.000025 0.00003 0.000035 0.00004 0.000045 0.00005 0.0000550

0.000250.0005

0.000750.001

0.001250.0015

0.001750.002

0.002250.0025

0.002750.003

0.003250.0035

0.003750.004

0.004250.0045

0.00475

Grafik Vt Vs Dp

silikagalena

Dp

Vt

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Dp sil ( ρ gal - ρ )

Dp gal = (ρ silika - ρ ) ^0,52,5 * 10^5 ( ρ gal - ρ )

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apabila dalam tabung diberi aliran air.maka partikel dg Vt < Vt air maka partikel tersebut akan naik.

TUGAS MEMBAGI KE FRAKSI2

14

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Misalkan dari rentang D1-D2. bagaimana profil kecepatan settling pada rentang D ter (misal ex 14.3-3 hal 927 geancop)Silika

Vt Dp Trial Vt Nre Cd Vt hitung Vt trial-Vt hitung7.50E-05 0.024827 1.86E-06 1.29E+07 0.000354 0.0244730.00019 0.076075 1.44E-05 1.66E+06 0.001572 0.074503

0.000305 0.155497 4.73E-05 5.07E+05 0.003607 0.151890.00042 0.262998 1.10E-04 2.18E+05 0.00646 0.256538

0.000535 0.399539 2.13E-04 1.13E+05 0.010143 0.389396Dp' 0.00065 0.563534 3.65E-04 6.57E+04 0.014635 0.548899

Diket :ρ silika = 2.65 g/cm^3 = 0.00265 g/mm^3ρ galena = 7.5 g/cm^3 = 0.0075 g/mm^3μ air = 1.00E-03 kg/m.s = 0.001001 g/mm.sρ air = 998 kg/m^3 = 0.000998 g/mm^3 g 9.8066 m/s^2= 9806.6 mm/s^2

0.00E+00 1.00E-04 2.00E-04 3.00E-04 4.00E-04 5.00E-04 6.00E-04 7.00E-040

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

0.055

0.06

0.065Grafik Vt Vs Cd

galenasilika

Dp

Vt

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Silika Murni :

5,21 * 10^ = (ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

Dp gal =(ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

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(misal ex 14.3-3 hal 927 geancop)

Vt trial-Vt hitung Dp 1 = 7.50E-05 m = 0.075 mmDp 5= 6.50E-04 m = 0.65 mminterval partikel = 0.000115

Cd Silika 1 2 3 4Cd 1 1.29E+07 9.658E-07 0.133284 5.642E-08Cd2 1.66E+06 7.497E-06 0.150723 2.907E-07Cd3 5.07E+05 2.46E-05 0.16186 7.521E-07Cd4 2.18E+05 5.73E-05 0.170283 1.479E-06Cd5 1.13E+05 0.000111 0.177163 2.508E-06Cd6 6.57E+04 0.00019 0.182982 3.859E-06

GalenaDp Trial Vt Nre Cd Vt hitung Vt trial-Vt hitung

7.50E-05 0.096874 7.25E-06 3.31E+06 0.001389 0.0954850.00019 0.299193 5.67E-05 4.23E+05 0.006184 0.293009

0.000305 0.610966 1.86E-04 1.29E+05 0.014186 0.596780.00042 1.034359 4.33E-04 5.54E+04 0.025417 1.008942

0.000535 1.568965 8.37E-04 2.87E+04 0.039875 1.529090.00065 2.212671 1.43E-03 1.67E+04 0.057532 2.15514

Cd Galena 1 2 3 4Cd 1 3.31E+06 3.769E-06 0.144629 1.677E-07Cd2 4.23E+05 2.949E-05 0.16363 8.693E-07Cd3 1.29E+05 9.666E-05 0.175711 2.247E-06Cd4 5.54E+04 0.000225 0.184865 4.424E-06Cd5 2.87E+04 0.000435 0.192317 7.492E-06Cd6 1.67E+04 0.000746 0.198632 1.153E-05

0.00E+00 1.00E-04 2.00E-04 3.00E-04 4.00E-04 5.00E-04 6.00E-04 7.00E-040

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

0.055

0.06

0.065Grafik Vt Vs Cd

galenasilika

Dp

Vt

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Galena Murni :Dp gal =(ρ silika - ρ ) ^0,5Dp sil ( ρ gal - ρ )

Dp gal = (ρ silika - ρ ) ^0,52,5 * 10^5 ( ρ gal - ρ )

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apabila dalam tabung diberi aliran air.maka partikel dg Vt < Vt air maka partikel tersebut akan naik.