dp = 5 μm P₁ = 1.5 g/cm³ FILTER d = 50 cm Q = 50 min dc = 600 μm E = 0.4 σ = 1 removal = 90% How deep should filter be? Ce/Co -(3/1207(1-6)/dc)L In Calle (72an (1-ε\/ad = -(3½2 L = an (1-E)/dc) en Итот NI = . = NG= ท + 76+ NB (de/dc)² мм 600 мм 3/2 ( 5μm 60 = 1.04 × 10-4 (ep-ec) g de² .. 18 мѵ V₁ = (1/2)² V= CM/S V= 0.424 M = 0.01 9/CM.S 9=981 CM152 PL = 1 9 CM³ No = m)² сусо (50 min) (10³ CM³) π (25 CM)² (60 s/min) @ 20ć (assumed) см (1.5%/CM² - 1% CM³) (981 cm/s²) (5 μm) (10-4 cm)]² мм 18. (0.01 %/CM.S) (0.424 (M/S) NG = 1.61×10 RT -3 2/3 no = 0.9 (un derdev) 2 = 0.9 1.38×10-16 9 cm² 52 K 298K им (0.01% (M.S) (600×10" CM) (5x10" (M) (0.424 CM/S). = 4.24×10 -S И тот = 1.75 × 10 L= 88 CM -3 @ 90% remove) 213
dp = 5 μm P₁ = 1.5 g/cm³ FILTER d = 50 cm Q = 50 min dc = 600 μm E = 0.4 σ = 1 removal = 90% How deep should filter be? Ce/Co -(3/1207(1-6)/dc)L In Calle (72an (1-ε\/ad = -(3½2 L = an (1-E)/dc) en Итот NI = . = NG= ท + 76+ NB (de/dc)² мм 600 мм 3/2 ( 5μm 60 = 1.04 × 10-4 (ep-ec) g de² .. 18 мѵ V₁ = (1/2)² V= CM/S V= 0.424 M = 0.01 9/CM.S 9=981 CM152 PL = 1 9 CM³ No = m)² сусо (50 min) (10³ CM³) π (25 CM)² (60 s/min) @ 20ć (assumed) см (1.5%/CM² - 1% CM³) (981 cm/s²) (5 μm) (10-4 cm)]² мм 18. (0.01 %/CM.S) (0.424 (M/S) NG = 1.61×10 RT -3 2/3 no = 0.9 (un derdev) 2 = 0.9 1.38×10-16 9 cm² 52 K 298K им (0.01% (M.S) (600×10" CM) (5x10" (M) (0.424 CM/S). = 4.24×10 -S И тот = 1.75 × 10 L= 88 CM -3 @ 90% remove) 213
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Question
how was L calculated for from calcualtions shown
![dp
= 5 μm
P₁ = 1.5 g/cm³
FILTER
d = 50 cm
Q = 50 min
dc = 600 μm
E = 0.4
σ = 1
removal = 90%
How deep should filter be?
Ce/Co
-(3/1207(1-6)/dc)L
In Calle (72an (1-ε\/ad
=
-(3½2
L = an (1-E)/dc) en
Итот
NI
=
.
=
NG=
ท
+ 76+ NB
(de/dc)²
мм
600 мм
3/2 ( 5μm 60
= 1.04 × 10-4
(ep-ec) g de²
..
18 мѵ
V₁ = (1/2)²
V=
CM/S
V= 0.424
M = 0.01 9/CM.S
9=981 CM152
PL = 1 9 CM³
No =
m)²
сусо
(50 min) (10³ CM³)
π (25 CM)² (60 s/min)
@ 20ć (assumed)
см
(1.5%/CM² - 1% CM³) (981 cm/s²) (5 μm) (10-4 cm)]²
мм
18. (0.01 %/CM.S) (0.424 (M/S)
NG = 1.61×10
RT
-3
2/3
no = 0.9 (un derdev) 2
= 0.9
1.38×10-16 9 cm²
52 K
298K
им
(0.01% (M.S) (600×10" CM) (5x10" (M) (0.424 CM/S).
= 4.24×10
-S
И тот
= 1.75 × 10
L= 88 CM
-3
@ 90% remove)
213](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e532244-1a69-487f-88f4-10ae0a919295%2F85d5c9a0-0c44-466f-8c49-1a51f84677c2%2Fywzbldq_processed.png&w=3840&q=75)
Transcribed Image Text:dp
= 5 μm
P₁ = 1.5 g/cm³
FILTER
d = 50 cm
Q = 50 min
dc = 600 μm
E = 0.4
σ = 1
removal = 90%
How deep should filter be?
Ce/Co
-(3/1207(1-6)/dc)L
In Calle (72an (1-ε\/ad
=
-(3½2
L = an (1-E)/dc) en
Итот
NI
=
.
=
NG=
ท
+ 76+ NB
(de/dc)²
мм
600 мм
3/2 ( 5μm 60
= 1.04 × 10-4
(ep-ec) g de²
..
18 мѵ
V₁ = (1/2)²
V=
CM/S
V= 0.424
M = 0.01 9/CM.S
9=981 CM152
PL = 1 9 CM³
No =
m)²
сусо
(50 min) (10³ CM³)
π (25 CM)² (60 s/min)
@ 20ć (assumed)
см
(1.5%/CM² - 1% CM³) (981 cm/s²) (5 μm) (10-4 cm)]²
мм
18. (0.01 %/CM.S) (0.424 (M/S)
NG = 1.61×10
RT
-3
2/3
no = 0.9 (un derdev) 2
= 0.9
1.38×10-16 9 cm²
52 K
298K
им
(0.01% (M.S) (600×10" CM) (5x10" (M) (0.424 CM/S).
= 4.24×10
-S
И тот
= 1.75 × 10
L= 88 CM
-3
@ 90% remove)
213
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