TUGAS UNIT OPERASI DAN PROSES

UNIT FILTRASI

Disusun Oleh:

Nama : Febry Dahyani

NPM : 0906636825

Jurusan : Teknik Lingkungan

TUGAS UNIT OPERASI DAN PROSES

UNIT FILTRASI

Disusun Oleh:

Nama : Febry Dahyani

NPM : 0906636825

Jurusan : 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 = 2,65

Shape factor (ϕ) = 0,75

Porosity (ɛ) = 0.41

Filtration rate = 2,25 gpm/ft2

Operating temperature = 500F (100C)

Table 1. 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

10.1. Va = (2,25 gal/min) (ft3/7,48gal) (min/60 sec) = 0,005013 ft/sec

v = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec

NRE = ϕ d Va / v ; CD = 24/ NRE

Table 2. Stratified Data From Sieve Analysis

Sieve Size Weight Retained d (ft) NRE CD CD x/d (ft-1)14-20 0,44 0,003283 0,876 27,38 3720-28 14,33 0,002333 0,623 38,54 236728-32 43,22 0,001779 0,475 50,54 1227832-35 27,07 0,001500 0,400 59,94 1081635-42 9,76 0,001258 0,336 71,47 554542-48 4,22 0,001058 0,282 84,98 338948-60 0,54 0,000888 0,237 101,24 61660-65 0,29 0,000746 0,199 120,51 46865-100 0,13 0,000583 0,156 154,21 344Σ CD x/d 35860 ft-1

Rose equation:

ℎ = 1,067 = 1,0670,75 2,532,17 0,0050130,41 35860 = 3,53 ft

10.3. To be backwashed in ft

The backwashed velocity to expand the bed requires the settling velocity of the largestparticles. The settling velocity, VS, is given by

= 43 ( − 1) /The drag coeffficient, Cd, for the transition range that applies to this problem is given by

= 24 + 3 + 0,34The NRE value is

=For each sieve size, convert d(ft) to d(cm) , d(cm) = d(ft) (30,48 cm/ft). From figure belowthis 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

SieveSize

WeightRetained

d (ft) NRE CDVs

(ft/sec)Vb

(ft/sec)ɛe 1 −

14-20 0,44 0,003283 80,2 0,974 0,488 0,00884 0,414 0,00820-28 14,33 0,002333 40,7 1,400 0,343 0,00621 0,414 0,24428-32 43,22 0,001779 27,9 1,767 0,267 0,00483 0,414 0,73732-35 27,07 0,001500 18,3 2,351 0,213 0,00385 0,414 0,46235-42 9,76 0,001258 11,0 3,432 0,161 0,00291 0,414 0,16642-48 4,22 0,001058 7,4 4,694 0,126 0,00229 0,414 0,07248-60 0,54 0,000888 4,6 6,894 0,095 0,00173 0,414 0,00960-65 0,29 0,000746 2,6 11,418 0,068 0,00123 0,414 0,00565-100 0,13 0,000583 1,5 18,497 0,047 0,00085 0,414 0,002Σ 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 3,9656820,006214 0,372861 2,7890040,00483 0,289821 2,1678630,003845 0,230702 1,7256540,002914 0,17485 1,3078810,002285 0,137115 1,0256180,001728 0,10365 0,7753050,00123 0,073823 0,5521930,000855 0,051274 0,383533

To determine head loss at the beginning of the backwash, 1-ɛ and D are substitued for 1-ɛe

and De

hL = (Ss – 1) (1-ɛ) D = (2,65-1) (1-0,41) (2,5) = 2,434 ft (c)

De = (1-ɛ) D Σ = (1–0,41)(2,5) (1,706) = 2,52 ft (d)The backwash rate in this problem is the minimum required to expand the bed. Actually, 15to 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 largestparticles. The settling velocity, VS, is given by

= 43 ( − 1) /The drag coeffficient, Cd, for the transition range that applies to this problem is given by

= 24 + 3 + 0,34The NRE value is

=For each sieve size, convert d(ft) to d(cm) , d(cm) = d(ft) (30,48 cm/ft). From figure belowthis 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

SieveSize

WeightRetained

d (m) NRE CDVs

(m/s)Vb

(m/s)ɛe 1 −

14-20 0,44 0,0010006 24,4 1,929 0,106 0,0019 0,414 0,00820-28 14,33 0,0007111 12,4 3,126 0,070 0,0013 0,414 0,24428-32 43,22 0,0005422 8,5 4,187 0,053 0,0010 0,414 0,73732-35 27,07 0,0004572 5,6 5,907 0,041 0,0007 0,414 0,46235-42 9,76 0,0003843 3,4 9,137 0,030 0,0005 0,414 0,16642-48 4,22 0,0003225 2,3 13,002 0,023 0,0004 0,414 0,07248-60 0,54 0,0002707 1,4 19,797 0,017 0,0003 0,414 0,00960-65 0,29 0,0002274 0,8 33,950 0,012 0,0002 0,414 0,00565-100 0,13 0,0001777 0,5 56,340 0,008 0,0001 0,414 0,002Σ 1,706

Table 6. Backwash Flow in gpm/ft2 Require to Expand The Bed

Vb (m/s) Rate (lt/s-m2)0,003467 3,4671050,002296 2,2958420,001732 1,7322350,001339 1,3391420,000987 0,987210,000758 0,75810,000563 0,5628720,000394 0,3939470,00027 0,270333

To determine head los at the beginning of the backwash, 1-ɛ and D are substitued for 1-ɛe andDe

hL = (Ss – 1) (1-ɛ) D = (2,65-1) (1-0,41) (0,76) = 0,74 m (c)

De = (1-ɛ) D Σ = (1–0,41)(0,76) (1,706) = 0,765 m (d)The backwash rate in this problem is the minimum required to expand the bed. Actually, 10,2to 13,6 l/s -m2 is used to expand a bed for proper agitation cleansing.

10.5. Trimedia Filter

Shape factor (ϕ) = 0,9

Filtration Rate = 5 gpm/ft2

Water temperature = 500F (100C)

- Anthracite coal layer

Specific Gravity = 1,2

Porosity (ɛ) = 0,4

d = 1,5 mm = 0,005 ft

D = 18 in = 1,5 ft

Va = (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec

v = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec

= = 0,9 0,005 .0,01111,4084 10 = 3,55Because NRE >1, the drag equation is

= 24 + 3 + 0,34 = 243,55 + 33,55 + 0,34 = 8,7

ℎ = 1,607 1 = 1,607 .8,7 .1,5 . (0,0111)0,9 .32,17 . (0,4) . 0,005 = 0,00260,0037 = 0,703ft- Sand layer

Specific Gravity = 2,65

Porosity (ɛ) = 0,45

d = 0,8 mm = 0,0026 ft

D = 9 in = 0,75 ft

Va = (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec

v = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec

= = 0,9 0,0026 .0,01111,4084 10 = 1,84Because NRE >1, the drag equation is

= 24 + 3 + 0,34 = 241,84 + 3√1,84 + 0,34 = 15,6ℎ = 1,607 1 = 1,607 .15,6 .0,75 . (0,0111)0,9 .32,17 . (0,45) . 0,0026 = 0,00230,0031 = 0,751ft

- Garnet layer

Specific Gravity = 1,2

Porosity (ɛ) = 0,5

d = 0,3 mm = 0,001 ft

D = 3 in = 0,25 ft

Va = (5 gal/min) (ft3/7,48gal) (min/60 sec) = 0,0111 ft/sec

v = 1,3101 centitokes at 500F = (1,3101) (1,075 x 10-5) = 1,4084 x 10-6 ft2/sec

= = 0,9 0,001 .0,01111,4084 10 = 7,09Because NRE >1, the drag equation is

= 24 + 3 + 0,34 = 247,09 + 3√7,09 + 0,34 = 4,85ℎ = 1,607 1 = 1,607 .4,85 .0,25 . (0,0111)0,9 .32,17 . (0,5) . 0,001 = 0,000240,0018 = 0,133ftTotal Head Loss = hL1 + hL2 + hL3 = 0,703 + 0,751 + 0,133 = 1,587 ft