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### Transcript of Tugas Unit Operasi Dan Proses - Unit Filtrasi

TUGAS UNIT OPERASI DAN PROSES UNIT FILTRASI

Disusun Oleh: Nama NPM Jurusan : Febry Dahyani : 0906636825 : Teknik Lingkungan

For Problem 10.1 ; 10.3. ; and 10.4. A rapid sand filter has a sand bed 30 in. in depth. Pertinent data are: Specific grafity of the sand Shape factor () Porosity () = 2,65 = 0,75 = 0.41 Table 1. Sieve AnalysisSieve Size 14-20 20-28 28-32 32-35 35-42 42-48 48-60 60-65 65-100 Weight Retained 0,44 14,33 43,22 27,07 9,76 4,22 0,54 0,29 0,13

Filtration rate Operating temperature

= 2,25 gpm/ft2 = 500F (100C)

10.1.

Va v NRE

= (2,25 gal/min) (ft3/7,48gal) (min/60 sec) = 0,005013 ft/sec = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec = d Va / v ; CD = 24/ NRE

Table 2. Stratified Data From Sieve Analysis Sieve Size Weight Retained 14-20 0,44 20-28 14,33 28-32 43,22 32-35 27,07 35-42 9,76 42-48 4,22 48-60 0,54 60-65 0,29 65-100 0,13 Rose equation: = 1,067 = d (ft) 0,003283 0,002333 0,001779 0,001500 0,001258 0,001058 0,000888 0,000746 0,000583 NRE 0,876 0,623 0,475 0,400 0,336 0,282 0,237 0,199 0,156 CD CD x/d (ft-1) 27,38 37 38,54 2367 50,54 12278 59,94 10816 71,47 5545 84,98 3389 101,24 616 120,51 468 154,21 344 35860 ft-1 CD x/d

1,067 2,5 0,005013 35860 = 3,53 ft 0,75 32,17 0,41

10.3. To be backwashed in ft The backwashed velocity to expand the bed requires the settling velocity of the largest particles. The settling velocity, VS, is given by 4 = 3 = ( 1)/

The drag coeffficient, Cd, for the transition range that applies to this problem is given by + + 0,34

24

3

The NRE value is

For each sieve size, convert d(ft) to d(cm) , d(cm) = d(ft) (30,48 cm/ft). From figure below this sentence, a particle d(cm) in diameter and having gravity of 2,65 has settling velocity, VS. The viscosity, v, is 1,3101 centitokes = 1,3101 x 10-2 cm2/sec.

=

Table 3. Reduced Data From Sieve Analysis Sieve Size 14-20 20-28 28-32 32-35 35-42 42-48 48-60 60-65 65-100 Weight Retained 0,44 14,33 43,22 27,07 9,76 4,22 0,54 0,29 0,13 d (ft) 0,003283 0,002333 0,001779 0,001500 0,001258 0,001058 0,000888 0,000746 0,000583 NRE 80,2 40,7 27,9 18,3 11,0 7,4 4,6 2,6 1,5 CD 0,974 1,400 1,767 2,351 3,432 4,694 6,894 11,418 18,497 Vs (ft/sec) 0,488 0,343 0,267 0,213 0,161 0,126 0,095 0,068 0,047 Vb (ft/sec) 0,00884 0,00621 0,00483 0,00385 0,00291 0,00229 0,00173 0,00123 0,00085 e 0,414 0,414 0,414 0,414 0,414 0,414 0,414 0,414 0,414 1 0,008 0,244 0,737 0,462 0,166 0,072 0,009 0,005 0,002 1,706

Table 4. Backwash Flow in gpm/ft2 Require to Expand The Bed Vb (ft/sec) Vb (ft/min) Rate (gpm/ft2) 0,008836 0,530171 3,965682 0,006214 0,372861 2,789004 0,00483 0,289821 2,167863 0,003845 0,230702 1,725654 0,002914 0,17485 1,307881 0,002285 0,137115 1,025618 0,001728 0,10365 0,775305 0,00123 0,073823 0,552193 0,000855 0,051274 0,383533 To determine head loss at the beginning of the backwash, 1- and D are substitued for 1-e and De De = (1-) D hL = (Ss 1) (1-) D = (2,65-1) (1-0,41) (2,5) = 2,434 ft (c) = (10,41)(2,5) (1,706) = 2,52 ft (d)

The backwash rate in this problem is the minimum required to expand the bed. Actually, 15 to 20 gpm/ft2 is used to expand a bed for proper agitation cleansing.

10.4. To be backwashed in SI The backwashed velocity to expand the bed requires the settling velocity of the largest particles. The settling velocity, VS, is given by 4 = 3 = ( 1)/

The drag coeffficient, Cd, for the transition range that applies to this problem is given by + + 0,34

24

3

The NRE value is

For each sieve size, convert d(ft) to d(cm) , d(cm) = d(ft) (30,48 cm/ft). From figure below this sentence, a particle d(cm) in diameter and having gravity of 2,65 has settling velocity, VS. The viscosity, v, is 1,3101 centitokes = 1,3101 x 10-2 cm2/sec.

=

Table 5. Reduced Data From Sieve Analysis Sieve Size 14-20 20-28 28-32 32-35 35-42 42-48 48-60 60-65 65-100 Weight Retained 0,44 14,33 43,22 27,07 9,76 4,22 0,54 0,29 0,13 d (m) 0,0010006 0,0007111 0,0005422 0,0004572 0,0003843 0,0003225 0,0002707 0,0002274 0,0001777 NRE 24,4 12,4 8,5 5,6 3,4 2,3 1,4 0,8 0,5 CD 1,929 3,126 4,187 5,907 9,137 13,002 19,797 33,950 56,340 Vs (m/s) 0,106 0,070 0,053 0,041 0,030 0,023 0,017 0,012 0,008 Vb (m/s) 0,0019 0,0013 0,0010 0,0007 0,0005 0,0004 0,0003 0,0002 0,0001 e 0,414 0,414 0,414 0,414 0,414 0,414 0,414 0,414 0,414 1 0,008 0,244 0,737 0,462 0,166 0,072 0,009 0,005 0,002 1,706

Table 6. Backwash Flow in gpm/ft2 Require to Expand The Bed Vb (m/s) Rate (lt/s-m2) 0,003467 3,467105 0,002296 2,295842 0,001732 1,732235 0,001339 1,339142 0,000987 0,98721 0,000758 0,7581 0,000563 0,562872 0,000394 0,393947 0,00027 0,270333 To determine head los at the beginning of the backwash, 1- and D are substitued for 1-e and De De = (1-) D hL = (Ss 1) (1-) D = (2,65-1) (1-0,41) (0,76) = 0,74 m (c) = (10,41)(0,76) (1,706) = 0,765 m (d)

The backwash rate in this problem is the minimum required to expand the bed. Actually, 10,2 to 13,6 l/s -m2 is used to expand a bed for proper agitation cleansing.

10.5. Trimedia Filter Shape factor () Filtration Rate Water temperature - Anthracite coal layer Specific Gravity Porosity () Va v = = 1,2 = 0,4 d D = 1,5 mm = 0,005 ft = 18 in = 1,5 ft = 0,9 = 5 gpm/ft2 = 500F (100C)

= (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec = 0,9 3 0,005 .0,0111 = 3,55 1,4084 10

Because NRE >1, the drag equation is = + + 0,34 =

24

24 3 + + 0,34 = 8,7 3,55 3,55

- Sand layer Specific Gravity Porosity () Va v =

=

1,607

1

=

1,607 .8,7 .1,5 . (0,0111) 0,0026 = = 0,703ft 0,9 .32,17 . (0,4) . 0,005 0,0037 d D = 0,8 mm = 0,0026 ft = 9 in = 0,75 ft

= 2,65 = 0,45

= (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec = 0,9 3 0,0026 .0,0111 = 1,84 1,4084 10 1

Because NRE >1, the drag equation is = + + 0,34 = =

=

1,607

24

1,607 .15,6 .0,75 . (0,0111) 0,0023 = = 0,751ft 0,9 .32,17 . (0,45) . 0,0026 0,0031

24 3 + + 0,34 = 15,6 1,84 1,84

- Garnet layer Specific Gravity Porosity () Va v = = 1,2 = 0,5 d D = 0,3 mm = 0,001 ft = 3 in = 0,25 ft

= (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec = 0,9 3 0,001 .0,0111 = 7,09 1,4084 10 1

Because NRE >1, the drag equation is = + + 0,34 = =

Total Head Loss = hL1 + hL2 + hL3 = 0,703 + 0,751 + 0,133 = 1,587 ft

=

1,607

24

1,607 .4,85 .0,25 . (0,0111) 0,00024 = = 0,133ft 0,9 .32,17 . (0,5) . 0,001 0,0018

24 3 + + 0,34 = 4,85 7,09 7,09