Add to collection(s) Add to saved
  1. No category
Uploaded by lara.dempsey13

Particle Summary Sheet

advertisement 
Particle Characterisation
Equivalent Sphere Diameter (ESD)
6
π‘₯
ESD, rearrange equation so x is subject,
V = volume of non-sphere i.e for cuboid:
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’:
1
Volume ESD: π‘₯ = (
)
πœ‹
Project Area Diameter: Different value for each
π΄π‘œπ‘›π‘’ π‘“π‘Žπ‘π‘’×4
different face of the shape – π‘₯ = √
π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’
=
Particle Size Distribution
𝑛𝑖 π‘π‘œ. π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’π‘  π‘’π‘›π‘‘π‘’π‘Ÿπ‘ π‘–π‘§π‘’
𝑓𝑖 = =
𝑁
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘œ. 𝑖𝑛 π‘ π‘Žπ‘šπ‘π‘™π‘’
𝐽
√π‘₯𝑓
−
10
√π‘₯𝑖
1
Body-Centred Cubic =
Face-Centred Cubic =
πœ‹π‘Ÿ 3 =
6π‘Ÿ 3
4
2×3πœ‹π‘Ÿ 3
3
4
π‘Ÿ)
√3
4 3
4× πœ‹π‘Ÿ
3
3
(
(2√2π‘Ÿ)
=
=
πœ‹
6
πœ‹√3
8
= 52.4%
= 68.0%
16 3
πœ‹π‘Ÿ
3
3 3
(2√2) π‘Ÿ
= 74%
Crystal Mass Balance
Mixture of 50/50 wt% of 100kg feed, (1) calculate kg of
water needed to dissolve to form a saturated sln at
20˚C , (2) If cooled to 10˚C, what is the product
obtained?
π‘šπ‘Žπ‘ π‘  π‘œπ‘“π‘π‘Ž2 𝑆𝑂4 𝑖𝑛 π‘π‘Ž2 𝑆𝑂4 . 𝐻2 𝑂
π‘€π‘Š(π‘π‘Ž2 𝑆𝑂4 )
=
π‘€π‘Š(π‘π‘Ž2 𝑆𝑂4 . 10𝐻2 𝑂)
π‘š(π‘π‘Ž2 𝑆𝑂4 )
Weight % of π‘π‘Ž2 𝑆𝑂4 =
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘₯ =
π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘Žπ‘ π‘ (100+π‘₯)
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘Žπ‘‘π‘‘π‘’π‘‘
Phase diagram to find wt% by drawing from 50% across
to 20˚C line, down at phase line change
(2) Find wt% at T=10˚C
π‘šπ‘–π‘› = 100 + x (mass mixture + mass 𝐻2 𝑂 added)
Component mass balance:
π‘š1 = π‘š2 + π‘š3
π‘‹π‘π‘Ž2 𝑆𝑂4 (100 + π‘₯) = π‘₯ (
βˆ†π»π‘ ,π‘‘π‘Ÿπ‘’π‘’ =
π‘€π‘Š(π‘Žπ‘£π‘”)
π‘€π‘Š(𝐡𝐴)𝑋𝐡𝐴
𝑃𝑖𝑛𝑙𝑒𝑑
π‘ƒπ‘œπ‘’π‘‘π‘™π‘’π‘‘
𝑃𝑖𝑛𝑙𝑒𝑑
= 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 π‘žπ‘’π‘’π‘ π‘‘π‘–π‘œπ‘›,
11.956−(
Magma leaves at 20˚C w/crystals of dihydrate
(π΅π‘ŽπΆπ‘™2 . 𝐻2 𝑂):
a.
Determine kg/hr of crystals in magma product
Given that solubility 𝑐 = 0.3𝑇 + 30
So solubility @ 100˚C = 60 g/100 g of 𝐻2 𝑂
π‘π΅π‘ŽπΆπ‘™2
Frac. of π΅π‘ŽπΆπ‘™2 𝑖𝑛 𝐻2 𝑂 =
=π‘₯
Frac π΅π‘ŽπΆπ‘™2 𝑖𝑛 𝑒π‘₯𝑖𝑑 =
36+100
= 26.5%
Now, crystal in exit stream = π‘₯1
π‘€π‘Š(π‘π‘Ž2 𝑆𝑂4 )
)
π‘€π‘Š(π‘π‘Ž2 𝑆𝑂4 . 10𝐻2 𝑂)
+ π‘‹π‘π‘Ž2 𝑆𝑂4,π‘œπ‘’π‘‘(100 − π‘₯1 )
Now solve for π‘₯1
Energy Balance on a Vacuum Crystalliser
Solution to example on lecture slide 111.
Determine the temp of the feed solution by the
enthalpy/conc. chart
π‘€π‘Š(π΅π‘ŽπΆπ‘™ )
Component balance π΅π‘ŽπΆπ‘™2 : πΉπ΅π‘ŽπΆπ‘™2 = π‘₯1 (π‘€π‘Š(π΅π‘ŽπΆπ‘™ .2𝐻2 𝑂) +
2
2
(π‘“π‘Ÿπ‘Žπ‘ π΅π‘ŽπΆπ‘™2 𝑖𝑛 π‘ π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ 𝑠𝑙𝑛)(π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑓𝑒𝑒𝑑 − π‘₯1 ) →sub in
values and solve for π‘₯1
Crystals in exit Saturated solution = 10000-π‘₯1
b.
Predominant crystal size (mm)
𝑉𝑀𝐿
𝜏=
= π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘›π‘‘ π‘‘π‘–π‘šπ‘’ (𝑝𝑔. 128)
𝑄𝑀𝐿
π‘˜π‘”
π‘šπ‘Žπ‘ π‘  π‘“π‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’( 𝑠 )
π‘˜π‘”
𝑑𝑒𝑛𝑠𝑖𝑑𝑦 ( 3 )
π‘š
Find 𝑄𝑀𝐿,π‘ π‘œπ‘™π‘–π‘‘ π‘Žπ‘›π‘‘ 𝑄𝑀𝐿,π‘™π‘–π‘ž , add the values together and
π‘š3
divide by 60 to get
β„Žπ‘Ÿ
Sub back into eqn for 𝜏
Now, 𝐿𝑃𝐷 = 3𝐺𝜏, 𝐺 = π‘π‘Ÿπ‘¦π‘ π‘‘π‘Žπ‘™ π‘”π‘Ÿπ‘œπ‘€π‘‘β„Ž π‘Ÿπ‘Žπ‘‘π‘’
c.
Mass frac. crystals in specific size range
𝑧2 𝑧3
π‘‹π‘š = 1 − (1 + 𝑧 + + ) 𝑒 −𝑧
2
6
𝐿
π‘€β„Žπ‘’π‘Ÿπ‘’, 𝑍 =
𝐺𝜏
Need to find π‘‹π‘š,π‘šπ‘Žπ‘₯ π‘Žπ‘›π‘‘ π‘‹π‘š,π‘šπ‘–π‘›
Hence, mass fraction = π‘‹π‘šπ‘Žπ‘₯ = π‘‹π‘šπ‘–π‘›
Falling Film Crystalliser (pg. 135-140)
Feed into falling film crystalliser is 60wt% naphtha. and
40wt% C6H6 at sat. conditions. Inner tube D = 10cm
a.If coolant enters the top @ 10˚C, determine time for
crystal layer thickness to reach 2cm at the top
For a cylindrical wall:
π‘Ÿπ‘–2 −π‘Ÿπ‘ 2
4
−
π‘Ÿπ‘ 2
2
π‘Ÿ
ln ( 𝑖 ) =
π‘Ÿπ‘ 
π‘˜π‘ (π‘‡π‘š −𝑇𝑐 )𝑑
βˆ†π»π‘“ πœŒπ‘
π‘Ÿπ‘  =? = π‘Ÿπ‘–π‘›π‘›π‘’π‘Ÿ − 2
Finding π‘‡π‘š : via a solid-liquid phase diagram, draw a
vertical line from 60wt% up to line separating liquid &
solid phase, draw horizontal line across to temp (π‘‡π‘š )
Now rearrange for t:
π‘Ÿπ‘–2 − π‘Ÿπ‘ 2 π‘Ÿπ‘ 2
π‘Ÿ
− ln ( 𝑖 )
4
2
π‘Ÿπ‘ 
𝑑=
π‘˜ (𝑇 − 𝑇𝑐 )
( 𝑐 π‘š
)
βˆ†π»π‘“ πœŒπ‘
Same process in a planar wall crystalliser with surface
spacing of 10cm, t = ?:
For a planar wall:
4409
)
π‘‡π‘œπ‘’π‘‘π‘™π‘’π‘‘ [=]π‘‘π‘œπ‘Ÿπ‘Ÿ
π‘ƒπ‘œπ‘’π‘‘π‘™π‘’π‘‘ = 10
Sub back into 𝑋𝐡𝐴 = π‘šπ‘œπ‘™π‘’π‘  π‘π‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘‘
Now we can sub back into βˆ†π»π‘ ,π‘‘π‘Ÿπ‘’π‘’ eqn
Then sub βˆ†π»π‘ ,π‘‘π‘Ÿπ‘’π‘’ into:
πœŒπ‘ βˆ†π»π‘ ,π‘‘π‘Ÿπ‘’π‘’ π‘Ÿπ‘ 2
π‘Ÿπ‘ 
1
[ ln ( ) − (π‘Ÿπ‘ 2 − π‘Ÿ02 )]
𝑑=
π‘Ÿ0
4
π‘˜π‘ (𝑇𝑔 − 𝑇𝑐 ) 2
b. No. tubes needed: 𝑁 =
π‘“π‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’ (
π‘˜π‘”
)×π‘‘π‘–π‘šπ‘’(β„Žπ‘Ÿ)
β„Žπ‘Ÿ
π‘šπ‘Žπ‘ π‘  (π‘˜π‘”)
π‘šπ‘Žπ‘ π‘  = πœ‹(π‘Ÿπ‘ 2 − π‘Ÿ02 ) × πΏ × πœŒπ‘ [=] g
Fluid Flow in Porous Media
Powder contained in a vessel to form cylindrical plug
0.8cm in diameter and 3cm long. πœŒπ‘π‘œπ‘€π‘‘π‘’π‘Ÿ=2.5 g/cm3
and 2.2 g powder was used:
a. Find porosity inside plug of powder:
πœ€=
Exit stream solubility @ 20˚C = c=0.3*20+30=36
36
× πΆπ‘ βˆ†π‘‡ + βˆ†π»π‘ 
𝑋𝐡𝐴 =
MSMPR Model οƒ  Application
Aq. Feed of 10000 g/hr, π΅π‘ŽπΆ 𝑙2 @ 100˚C enters a crystalliser
Volume Occupancy of Crystal Structures
πœ‹π‘Ÿ 3
π‘Ÿ0 = 𝑝𝑖𝑝𝑒 π‘œπ‘›π‘™π‘¦, π‘Ÿπ‘  = 𝑝𝑖𝑝𝑒 + π‘‘β„Žπ‘–π‘π‘˜π‘›π‘’π‘ π‘ 
βˆ†π»π‘  𝑔𝑖𝑣𝑒𝑛 = heat of fusion but still need to find “true
1.
Supersat. ratio increases as the rate of nucleation
increases logarithmically for all aq. Solutions
π‘Šπ‘–π‘‘β„Žπ‘‘π‘Ÿπ‘Žπ‘€π‘› π‘Ÿπ‘Žπ‘‘π‘’ = 𝑄𝑀𝐿 =
4
𝑇𝑐 = 𝑖𝑛𝑙𝑒𝑑 𝑓𝑙𝑒𝑖𝑑 π‘‘π‘’π‘šπ‘, 𝑇𝑔 = π‘œπ‘’π‘‘π‘™π‘’π‘‘ π‘”π‘Žπ‘  π‘‘π‘’π‘šπ‘,
Hence, π΅π‘ŽπΆπ‘™2 in feed = x% × 10000π‘˜π‘”/β„Žπ‘Ÿ
Size Distribution After Crushing
Values of breakage dist. 𝑏(𝑖, 𝑗)
Specific rates of breakage 𝑆𝑗
Finding 𝑏(𝑖, 𝑗):
1. πΉπ‘œπ‘Ÿ π‘’π‘Žπ‘β„Ž π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™ π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘Žπ‘› 𝑆𝑗 π‘£π‘Žπ‘™π‘’π‘’:
𝑑𝑦
2. Feed is 𝑑𝑦
→ 1 = 0 − 𝑠1 𝑦1 (𝑦1 = π‘“π‘Ÿπ‘Žπ‘. π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’π‘ )
𝑑𝑑
𝑑𝑑
𝑑𝑦2
= 𝑏(2,1)𝑠1 𝑦1 − 𝑠2 𝑦2
𝑑𝑑
𝑑𝑦3
= [𝑏(3,1)𝑠1 𝑦1 + 𝑏(3,2)𝑠2 𝑦2 ] − 𝑠3 𝑦3
𝑑𝑑
3. Particle dist. Against Interval Graph
3
Simple Cubic = (2π‘Ÿ)
=
3
a. Find time required to reach max thickness of BA of 1.25cm
MW Average = 0.008(π‘€π‘Š(𝐡𝐴)) + 0.992(π‘€π‘Š(𝑁2 ))
100𝑔 π‘œπ‘“ 𝐻2 0+π‘π΅π‘ŽπΆπ‘™2
)
Desublimation Design (pg.141-143)
A Desub unit is sized for recovery of 200 kg/hr of BA
from a gas stream of 0.8mol%BA & 99.2mol% 𝑁2 .
heat of fusion”→ βˆ†π»π‘ ,π‘‘π‘Ÿπ‘’π‘’
3
−16πœ‹π‘£π‘ 2 πœŽπ‘ ,𝐿
π‘π‘Ž
𝐡˚ = 𝐴𝑒π‘₯𝑝 [
]
𝑐 2
3𝑣 2 (𝑅𝑇)3 [ln ( )]
𝑐𝑠
𝑛𝑒𝑐𝑙𝑒𝑖
𝐴 = 1030
π‘π‘š3 𝑠
πœŽπ‘ ,𝐿 = π‘–π‘›π‘‘π‘’π‘Ÿπ‘“π‘Žπ‘π‘–π‘Žπ‘™ π‘‘π‘’π‘›π‘ π‘–π‘œπ‘›
π‘π‘Ž = 6.02 × 1023
𝑐
𝑆 = = π‘ π‘’π‘π‘’π‘Ÿπ‘ π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘–π‘œ
𝑐𝑠
𝐴𝑝
1.Rittinger’s: m = -2, βˆ†πΈ = 𝛾 ( 2) βˆ†π΄ (π‘š2 )
π‘š
1
1
𝐸 = π‘˜π‘… ( − )
π‘₯𝑓 π‘₯𝑖
π‘₯
2.Kick’s: 𝐸 = π‘˜π‘˜ ln( 𝑖 )
𝜌
Primary Nucleation (pg.115)
Primary Homogenous Nucleation
2
Crushing and Mixing (pg.43-45)
𝑣𝑅𝑇𝐷𝑃
Make sure variables in same units
𝑣𝑠 = π‘π‘Ÿπ‘¦π‘ π‘‘π‘Žπ‘™ π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘£π‘œπ‘™. , πœŽπ‘ ,𝐿 =
π‘–π‘›π‘‘π‘’π‘Ÿπ‘“π‘Žπ‘π‘–π‘Žπ‘™ π‘‘π‘’π‘›π‘ π‘–π‘œπ‘›, 𝑣 = π‘›π‘œ. π‘–π‘œπ‘›π‘ 
𝑀𝑀
Finding 𝑣𝑠 =
πœ‹ 3 (6𝑉𝑝)3
π‘šπ‘’π‘Žπ‘› π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’ 𝑠𝑖𝑧𝑒 𝑏𝑦 π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯Μ…0 = ∑ π‘₯̅𝑖 𝑓𝑖
4𝑣𝑠 πœŽπ‘ ,𝐿
2.
3.
πœ‹
1
π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘β„Žπ‘’π‘Ÿπ‘’ π‘œπ‘“ π‘’π‘žπ‘’π‘Žπ‘™ π‘£π‘œπ‘™.
𝑐
𝑐𝑠
ln ( ) = ln(𝑆) =
=
π‘₯𝑓
10
𝐽
(π‘Ÿπ‘– − π‘Ÿπ‘  )2 βˆ†π»π‘“ ( ) 𝜌
𝑔
𝑑=[
]
2π‘˜π‘ (π‘‡π‘š − 𝑇𝑐 )
1.
4.
3.Bond’s: 𝐸 = 𝑀𝑖 (
𝑄𝑖𝑛,(𝑙)
→intersect the new βˆ†H
π‘ π‘œπ‘™π‘–π‘‘
w/composition on diagram for new T
Composition =
Sieve Diameter: Second smallest dimension of the
shape
Wadell’s Sphericity = πœ‘
So, π‘Ÿπ‘– now is 10cm, π‘Ÿπ‘  = 10 − 2 = 0.8π‘š
Sub new info into:
𝑄𝐹2
π‘†π΄π‘π‘’π‘π‘œπ‘–π‘‘
Surface area ESD: π‘₯ = √
2π‘˜π‘ (π‘‡π‘š − 𝑇𝑐 )𝑑
(π‘Ÿπ‘– − π‘Ÿπ‘  ) = √
βˆ†π»π‘“ πœŒπ‘
Effect of Crystal Size on Solubility pg.113
6π‘‰π‘π‘’π‘π‘œπ‘–π‘‘ 3
πœ‹
𝑄𝑖𝑛 , 𝑄𝐹1,(𝑐) , 𝑄𝐹2,(𝑙) , π‘„π‘œπ‘’π‘‘,π‘ π‘œπ‘™π‘–π‘‘ = 𝑔𝑖𝑣𝑒𝑛
For cone creation stream: 𝑄𝑖𝑛 −
(π»π‘’π‘Žπ‘‘ πΉπ‘™π‘œπ‘€ 𝐹𝑒𝑒𝑑 1 × (𝐻𝐹1 )
𝑄
New enthalpy = 𝐹1
𝑉𝑣
𝑉𝐡
𝑑 2
𝑉𝐡 = π‘π‘’π‘™π‘˜ π‘£π‘œπ‘™. = πœ‹ ( ) 𝐿
2
𝑉𝑉 = π‘£π‘œπ‘–π‘‘ π‘ π‘π‘Žπ‘π‘’ = 𝑉𝐡 − π‘‰π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’π‘ 
π‘šπ‘Žπ‘ π‘ 
π‘‰π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘’π‘  =
𝑑𝑒𝑛𝑠𝑖𝑑𝑦
b. Pressure Drop Across Plug (Pa)
1. βˆ†π‘ƒ = πœŒπ‘”β„Ž
𝜌
πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ
*Rearrange for 𝜌, where specific g = 13.6kg/m3
Sub back into eqn 1 where h = 60mm/1000 and density
units match
c. Superficial Gas Velocity: (m/s)
𝑄
π‘ˆ0 = , π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑄[=] π‘š3/𝑠
𝐴
𝐴 = πœ‹π‘Ÿ 2 [=]π‘š2
d. Specific Surface Area per Unit Vol. of Powder
βˆ†π‘ƒ
32−πœ‡
Using Kozeny-Carman eqn: 1.
= 2 𝑒
𝐿
𝑑
βˆ†π‘ƒ
πΎπœ‡
2.
=
×𝑒
2
πœ€
𝐿
[
]
1 − πœ€π‘†π‘‰
π‘ˆ0
𝑒 = π‘–π‘›π‘‘π‘’π‘Ÿπ‘ π‘‘π‘–π‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =
πœ€
Using (2):
πœ€
[=] π‘š−1
𝑆𝑉 =
√πΎπœ‡π‘ˆ0 𝐿 × (1 − πœ€)
βˆ†π‘ƒπœ€
6
e. Sauter Mean Diameter: (um) 𝑆𝑀𝐷 = [=] um
𝐻𝑔: 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔 =
f. Modified Re =
πœŒπ‘Žπ‘–π‘Ÿ π‘ˆ0
𝑆𝑉
(1−πœ€)𝑆𝑉 πœ‡
g. Testing Validity: Since Re < 2, Kozeny Carman eqn is
valid for this scenario
Further Fluid Flow in Porous Media
A cylindrical ion exchange bed composed of spherical
particles is packed in a bed to be used to deionise liq
a. Determine Pressure Drop Using Kozeny-Carman eqn:
(Pa)
βˆ†π‘ƒ
πΎπœ‡
=
2×𝑒
πœ€
𝐿
[
]
1 − πœ€π‘†π‘‰
π‘ˆ0
𝑒 = π‘–π‘›π‘‘π‘’π‘Ÿπ‘ π‘‘π‘–π‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =
πœ€
𝑄
π‘ˆ0 = , 𝐴 = πœ‹π‘Ÿ 2
𝐴
πœ€ = 𝑏𝑒𝑑 π‘£π‘œπ‘–π‘‘π‘Žπ‘”π‘’, π‘₯𝑆𝑉 = π‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’ π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ
𝑆𝑉 =
6
[=]π‘š−1
π‘₯𝑆𝑉
*Can now sub into K-C eqn for βˆ†P [=] Pa
b. Modified Reynolds Number:
πœŒπ‘ˆ0
𝑅𝑒𝑙 =
(pg. 131)
(1−πœ€)𝑆𝑉 πœ‡
π‘ˆ
c. Interstitial Velocity = 0 [=] m/s
πœ€
d. Shear Stress on Ion Exchange Beds:
5
Using Carman Correlation: 𝑓 =
+
𝑅𝑒𝑙
0.4
𝑅𝑒𝑙0.1
Kozeny-Carman eqn for laminar flow:
𝑅
𝑓 = 2 → π‘ π‘œπ‘™π‘£π‘’ π‘“π‘œπ‘Ÿ 𝑅 (π‘ β„Žπ‘’π‘Žπ‘Ÿ π‘ π‘‘π‘Ÿπ‘’π‘ π‘ )
πœŒπ‘’
e. Dynamic Pressure Drop (kPa):
Using:
Where: 𝑓 =
βˆ†π‘ƒ
𝐿
𝑅
=(
πœŒπ‘’2
𝑅
πœŒπ‘’2
)×(
𝑆𝑉 (1−πœ€)πœŒπ‘ˆ02
πœ€3
)
→ rearrange and solve for βˆ†P
Filtration of Liquids – Deep Bed Filtration
A 3-layer filter is used to clarify an effluent containing
60mg/L (ppm) of solids:
Order: π΄π‘›π‘‘β„Žπ‘Ÿπ‘Žπ‘π‘–π‘‘π‘’ → π‘†π‘Žπ‘›π‘‘ → π΄π‘™π‘’π‘šπ‘–π‘›π‘Ž
𝑁 = 𝑁0 𝑒 −πœ†0𝐿
𝑁0 = 𝑖𝑛𝑙𝑒𝑑 π‘π‘œπ‘›π‘, πœ†0 = π‘“π‘–π‘™π‘‘π‘Ÿ. π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘, 𝐿 = π‘‘π‘’π‘π‘‘β„Ž(π‘š)
d. Total filtration pressure after a certain time
Just sub time into initial equation in part(a)
′
e. New Effective Medium Resistance: π‘…π‘š
:
Substitute all values into equation in (c)
f. Total time required to filter remaining slurry:
πœ‡π›Όπ‘
πœ‡π‘…π‘š
𝑑 = 2 𝑉2 +
𝑉 [=] 𝑠
2𝐴 βˆ†π‘ƒ
π΄βˆ†π‘ƒ
g. Total Filtration Time:
π‘‘π‘‘π‘œπ‘‘ = π‘‘π‘π‘Žπ‘Ÿπ‘‘(𝑏) + π‘‘π‘π‘Žπ‘Ÿπ‘‘(𝑓)
Sedimentation and Thickener
Terminal Settling Velocity (pg. 196)
a. Determine the value for 𝑄𝐻
4 (πœŒπ‘  − πœŒπ‘™ )π‘”πœ‡
π‘š
] [=]
𝑄𝐻3 = [
3
πœŒπ‘™2
𝑠
b. Using Heywood Tables (pg.197)
For Particle Diameter, x=1um (= 1/106 m)
log(𝑃𝐻 × π‘₯) = 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ π‘π‘œπ‘™π‘’π‘šπ‘›
π‘ˆπ‘‘
log ( ) = π‘œπ‘›π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’π‘  𝑖𝑛 π‘‘π‘Žπ‘π‘™π‘’
𝑄𝐻
c. Stokes’ are higher than Heywood as former accounts
only for viscous drag for higher x, form drag is larger
which isn’t accounted for, as well as turbulence and
eddies in fluid
d. Find max particle size at which Stokes’ applies:
Re < 0.2 𝑅𝑒 =
Cake Filtration
A filter of 0.1m2 area at a constant pressure of
68.5x104Pa produce the following results: πœ‡π‘“π‘–π‘™π‘‘π‘Ÿπ‘Žπ‘‘π‘’ =
Rearrange for x: π‘₯ 3 =
0.52
b. Dry Solids per Unit Vol. of Filtrate: (pg. 168)
π‘ πœŒ
(𝑝𝑔. 179) [=]π‘˜π‘”/π‘š3
𝑐𝑓 =
1 − π‘šπ‘… 𝑠
π‘Šβ„Žπ‘’π‘Ÿπ‘’, 𝑠 = π‘ π‘™π‘’π‘Ÿπ‘Ÿπ‘¦ π‘π‘œπ‘›π‘. = 0.03
c. Specific Resistance (𝛼)of the Cake and Medium:
For a graph of t/V (s/m3) against V (m3),
y=71660x+546.83, which is comparable to:
𝑑
πœ‡π›Όπ‘
πœ‡π‘…π‘š
1.
=
𝑉+
𝑉 2βˆ†π‘ƒπ΄2
βˆ†π‘ƒπ΄
π‘…π‘’π‘Žπ‘Ÿπ‘Ÿπ‘Žπ‘›π‘”π‘’ π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘ π‘‘π‘œ π‘šπ‘Žπ‘˜π‘’ 𝛼 𝑠𝑒𝑏𝑗𝑒𝑐𝑑:
πœ‡π›Όπ‘
= 71660, 𝛼[=]π‘š−1
2βˆ†π‘ƒπ΄2
2
d. On a 10 m filter, what would the filtrate vol. be
after 2 hours:
𝑑 = 2 β„Žπ‘œπ‘’π‘Ÿπ‘  = 2 × 3600 = 7200 𝑠
First solve for π‘…π‘š :
πœ‡π‘…
546.83 = π‘š, rearrange for Rm
βˆ†π‘ƒπ΄
Put all info found into eqn (1) and solve for V(m3)
e.If the solid density is 2500 kg/m3, what would the
cake thickness be in (d)? (L[=]m):
𝑐𝑉 1 π‘šπ‘… − 1
( +
)
𝐿=
𝐴 πœŒπ‘ 
πœŒπ‘™
Modes of Cake Filtration
The following empirical equation for pressure was
observe: βˆ†P=168(Pa/s)t(s)+6670(Pa)
a. βˆ†π‘ƒ = βˆ†π‘ƒπ‘π‘Žπ‘˜π‘’ + βˆ†π‘ƒπ‘šπ‘’π‘‘π‘–π‘’π‘š
βˆ†π‘ƒπ‘π‘Žπ‘˜π‘’ , π‘£π‘Žπ‘Ÿπ‘–π‘’π‘  π‘€π‘–π‘‘β„Ž π‘‘π‘–π‘šπ‘’ 𝑏𝑒𝑑 βˆ†π‘ƒπ‘šπ‘’π‘‘π‘–π‘’π‘š π‘‘π‘œπ‘’π‘  π‘›π‘œπ‘‘
b. Time to achieve filtr. & constant filtr. rate
πΉπ‘–π‘™π‘‘π‘Ÿπ‘Žπ‘‘π‘’ πΉπ‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’ =
𝑑𝑉
𝑑𝑑
π‘ƒπ‘Ž
πœ‡π›Όπ‘ 𝑉
βˆ†π‘ƒπ‘π‘Žπ‘˜π‘’ (168
𝑑) = 2 𝑉 ( )
𝑠
𝐴
𝑑
Find 𝐾𝑐 π‘‘π‘œ 𝑠𝑒𝑏 π‘–π‘›π‘‘π‘œ 𝛼
3
πœ€
𝐾𝑐 =
, 𝐾 = 5 (π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘)
𝐾(1 − πœ€)2 𝑆𝑉2
Find 𝛼 to sub back into βˆ†π‘ƒπ‘π‘Žπ‘˜π‘’
1
𝛼=
𝐾𝑐 (1 − πœ€)πœŒπ‘ 
𝑆𝑒𝑏 π‘–π‘›π‘‘π‘œ βˆ†π‘ƒπ‘π‘Žπ‘˜π‘’ , π‘ π‘œπ‘™π‘£π‘’ π‘“π‘œπ‘Ÿ 𝑑 (𝑠)
π‘…π‘š : βˆ†π‘ƒπ‘šπ‘’π‘‘π‘–π‘’π‘š
𝑉
′
π‘…π‘š
= π‘…π‘š + 𝑐𝛼
𝐴
πœ‡π‘… 𝑉
= 6670 = π‘š
𝐴
π‘₯π‘ˆπœŒ
πœ‡
π‘Žπ‘›π‘‘ π‘ˆ =
π‘₯ 2 (πœŒπ‘ −πœŒπ‘™ )𝑔
18πœ‡
(0.2πœ‡)(18πœ‡)
(πœŒπ‘  −πœŒπ‘™ )π‘”πœŒπ‘™
[=] π‘š
Particle Sedimentation
a. Max particle d that will be in discharge(um):
π‘₯ 2 (πœŒπ‘  − πœŒπ‘™ )𝑔
1. π‘ˆπ‘‘ =
18πœ‡
𝐻
π‘š
𝑑𝑠 = 𝑠𝑒𝑑𝑑𝑙𝑖𝑛𝑔 π‘‘π‘–π‘šπ‘’ = (𝐻 = β„Žπ‘’π‘–π‘”β„Žπ‘‘)[=]
π‘ˆπ‘‘
𝑠
Rearrange for π‘ˆπ‘‘ and sub into first eqn solve for x
b. Conc. solids below the size calc. in (a)
Find approx. how much settled using trend line,
subtract from 100%, multiply by original conc.
𝑑
=
πœ‹π‘Ÿ 2 𝑙
πœ‹π‘Ÿ02 𝑙
=
π‘Ÿ2
π‘Ÿ02
=
π‘‰π‘œπ‘£π‘’π‘Ÿπ‘“π‘™π‘œπ‘€
𝑉𝑓𝑒𝑒𝑑
π‘Ÿ0 = π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  π‘“π‘Ÿπ‘œπ‘š π‘π‘’π‘›π‘‘π‘’π‘Ÿ π‘‘π‘œ π‘œπ‘’π‘‘π‘’π‘Ÿ 𝑒𝑑𝑔𝑒
π‘Ÿ = π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ 
Inlet Velocity and hence, tangential velocity at 𝑅0 :
π΄π‘π‘œπ‘›π‘’ = πœ‹π‘ŸπΏ, π΄π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 2πœ‹π‘Ÿπ‘™, π‘ π‘’π‘š = π΄π‘‘π‘œπ‘‘π‘Žπ‘™
π‘‚π‘£π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ π‘…π‘Žπ‘‘π‘’[=] π‘š3 /𝑠
π‘š3
π‘“π‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’ ( )
𝑠
Hence, overflow velocity =
π‘Žπ‘Ÿπ‘’π‘Ž (π‘š2 )
𝑑𝑓𝑒𝑒𝑑 𝑖𝑛𝑙𝑒𝑑 2
) [=] π‘š2
𝐴 = πœ‹(
2
Put feed flowrate into m3/s
𝑓𝑒𝑒𝑑 π‘“π‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’
Thus, feed velocity =
𝐴𝑖𝑛𝑙𝑒𝑑
Using conserv. of angular momentum, tangential
velocity at eqm orbit radius: (m/s)
𝑒1 π‘Ÿ1 = 𝑒2 π‘Ÿ2 (π‘Žπ‘‘ π‘π‘œπ‘›π‘ π‘’π‘Ÿπ‘£π‘’π‘‘ π‘Žπ‘›π‘”π‘’π‘Žπ‘™π‘Ÿ π‘šπ‘œπ‘šπ‘’π‘›π‘‘π‘’π‘š)
Separation or cut size (π‘₯50 ) of Hydrocyclone:
π‘₯50 = √
(𝜌
18πœ‡π‘ˆπΎ
2
, π‘ˆπΎ = π‘œπ‘£π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦, π‘Ÿ = π‘Ÿ0
π‘‡π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ = π‘ˆ0 = π‘€π‘Ÿ
π‘ˆ0 2
π‘ˆ02
𝑀2 = ( ) =
π‘Ÿ
π‘Ÿ×π‘Ÿ
π‘ˆ02
𝑀 2π‘Ÿ =
π‘Ÿ
Particle Reynolds number at LZVV:
π‘₯50 π‘ˆπœŒ
𝑅𝑒𝑙 =
πœ‡
Hydrocyclone pressure drop:
π‘₯50 2 (πœŒπ‘  − 𝜌)πΏβˆ†π‘ƒ
3.5 =
πœ‡πœŒπ‘„
Collection Efficiency of a Hydrocyclone:
π‘’π‘›π‘‘π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ π‘šπ‘Žπ‘ π‘ 
a. Total Collection Efficiency (TCE)=
π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘Žπ‘ π‘ 
b. Size dist. In overflow =
π‘šπ‘Žπ‘ π‘ π‘‘π‘œπ‘‘ ∗𝑓𝑒𝑒𝑑 𝑠𝑖𝑧𝑒 𝑑𝑖𝑠𝑑−π‘šπ‘Žπ‘ π‘ π‘’π‘›π‘‘π‘’π‘Ÿπ‘“π‘™π‘œπ‘€∗π‘’π‘›π‘‘π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ 𝑠𝑖𝑧𝑒 𝑑𝑖𝑠𝑑.
Settling Basin Design (pg. 200)
a. Terminal settling velocity (π‘ˆπ‘‘ ) and particle Reynolds
number:
First find π‘ˆπ‘‘ using eqn 1 in part (a) then sub into: 𝑅𝑒 =
π‘₯π‘ˆπœŒ
, if Re<0.2 then Stokes’ is appropriate
πœ‡
b. Critical particle residence time vertically (𝑑𝑉 ) &
critical particle resistance time horizontally (𝑑𝐻 ):
𝐻
𝐻
𝑑𝑠 =
→ 𝑑𝑉 = (𝑠)
π‘ˆπ‘‘
π‘ˆπ‘‘
𝑄
π»πΏπ‘Š
= 𝐿𝑀 → π‘‘β„Ž =
(𝑠)
π‘ˆπ‘‘
𝑄
c. Minimum settling area req. to remove all particles of
diameter given and bigger ones in m2
𝑄
𝐴 = [=] π‘š2
π‘ˆπ‘‘
Continuous Thickener Design
Determine Settling Velocity & Batch Flux:
π‘‘β„Ž
π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =
→ π‘‘π‘Ÿπ‘Žπ‘€ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘  π‘‘π‘œ π‘π‘’π‘Ÿπ‘£π‘’
𝑑𝑑
Determining heights from some arbitrary conc.:
𝐢0 𝐻0 = 𝐢1 𝐻1
Abbreviated flux = settling velocity*solid conc.
Critical flux = tangent to deepest point curve
𝐹𝐢0
Minimum thickener area = 𝑇𝐢
→ π‘‡πΆπ‘ˆ = π‘π‘Ÿπ‘–π‘‘π‘–π‘π‘Žπ‘™ 𝑓𝑙𝑒π‘₯
π‘ˆ
Underflow Rate: 𝐹𝐢0 = π‘ŒπΆπ‘ˆ [=]π‘š3 /β„Žπ‘Ÿ, Y=underflow
withdrawal rate
Overflow Rate: Feed Rate = Underflow+Overflow
Thickener Operation
𝐹𝐢
Underflow conc.→ 𝑇𝐢𝑒 = 0
𝐴
Material balance of solids: 𝐢0 𝐻0 π΄πœŒπ‘  =mass, g
𝐹𝐢
Material balance of thickener feed: π‘‡πΆπ‘ˆ = 0
𝐴
Non-Newtonian Flow
π‘š
Finding Shear Stress: 𝑅 = 𝐾𝛾 , 𝛾 = π‘ β„Žπ‘’π‘’π‘Ÿ π‘Ÿπ‘Žπ‘‘π‘’, π‘š =
π‘“π‘™π‘œπ‘€ 𝑖𝑛𝑑𝑒π‘₯ π‘Žπ‘›π‘‘ 𝐾 = π‘π‘œπ‘›π‘ π‘–π‘ π‘‘π‘’π‘›π‘π‘¦ π‘π‘œπ‘’π‘“π‘“.
Example finding m, 1. 7 = 𝐾7.2π‘š , 2. 9.1 = 𝐾16π‘š
π‘š=
9.1
)
7
16
log ( )
7.2
log (
Suspension in Laminar Flow:
1
c. Medium Resistance (m-1):
𝑉𝑖𝑛𝑠𝑖𝑑𝑒
π‘‰β„Žπ‘¦π‘‘π‘Ÿπ‘œπ‘π‘¦π‘π‘™π‘œπ‘›π‘’
𝑠 −𝜌)π‘Ÿπ‘€
Inlet conc. sand = outlet conc. previous bed
0.0015π‘ƒπ‘Ž. 𝑠, slurry conc. = 3% w/w, cake conc. = 52%
1000π‘˜π‘”
w/w, πœŒπ‘™π‘–π‘ž =
.
π‘š3
a. Moisture Ratio: (pg, 168)
π‘šπ‘Žπ‘ π‘ (𝑀𝑒𝑑 π‘π‘Žπ‘˜π‘’)
π‘šπ‘… =
π‘šπ‘Žπ‘ π‘ (π‘‘π‘Ÿπ‘¦ π‘π‘Žπ‘˜π‘’)
Cake conc. = 52%w/w, assume basis of 1 for wet cake
1
so that π‘šπ‘… =
= 1.92
8πœŒπ‘š π‘ˆ2−π‘š 𝑑 π‘š
𝐾
Hydrocyclone Design: pg.236 for image
Finding equilibrium radius: πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
𝑅𝑒𝑙 =
π‘šπœ‹π‘Ÿ 3 π‘Ÿβˆ†π‘ƒ π‘š
[
]
3π‘š + 1 2𝐿𝐾
𝑄
π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ = π‘ˆ =
𝐴
πΉπ‘™π‘œπ‘€π‘Ÿπ‘Žπ‘‘π‘’, 𝑄 =
π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘Žπ‘ π‘ −π‘’π‘›π‘‘π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ π‘šπ‘Žπ‘ π‘ 
π‘’π‘›π‘‘π‘’π‘Ÿπ‘“π‘™π‘œπ‘€ π‘šπ‘Žπ‘ π‘ 
c. Grad Efficiency =
π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘Žπ‘ π‘ 
Powder Flow and Storage – Hopper
Jenike Shear Cell:
𝛿 = π‘–π‘›π‘‘π‘’π‘Ÿπ‘›π‘Žπ‘™ π‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›, 𝐻(πœƒπ») = 2.2,
πœƒπ» (π‘œπ‘Ÿ πœƒ) = β„Žπ‘œπ‘π‘π‘’π‘Ÿ β„Žπ‘Žπ‘™π‘“ π‘Žπ‘›π‘”π‘™π‘’,
πœ™π‘€ = π‘Žπ‘›π‘”π‘™π‘’ π‘€π‘Žπ‘™π‘™ π‘“π‘Ÿπ‘–π‘. , πœŒπ‘ = π‘π‘œπ‘€π‘‘π‘’π‘Ÿ π‘π‘’π‘™π‘˜ 𝑑𝑒𝑛𝑠𝑖𝑑𝑦
𝐡 = β„Žπ‘œπ‘π‘π‘’π‘Ÿ π‘œπ‘π‘’π‘›π‘–π‘›π‘”
Finding Flow Factor (𝑓𝑓):
1. For πœ™π‘€ 𝑣𝑠 πœƒ graph on pg. 320, draw lines for each axis
value.
2. At the intersection point, take value from the ff curve
closest to this point
Finding HFF Line:
1. Find the gradient of the HFF line, this line will
intersect with specific points on the PFF graph
1
𝐻𝐹𝐹 π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘ =
𝑓𝑓
2. Find eqn of line using 𝑦1 and π‘₯1 , π‘₯2 to find 𝑦2
3. The 3 intersection points (𝑓𝑐1 , 𝑓𝑐2 , 𝑓𝑐3 ) at each time
interval allow us to calculate 𝐡 = π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ π‘œπ‘π‘’π‘›π‘–π‘›π‘” 𝑑
𝑓 𝐻(πœƒ )
4. 𝐡 = 𝑐 𝐻 → 𝑓𝑖𝑛𝑑 𝐡 π‘“π‘œπ‘Ÿ π‘’π‘Žπ‘β„Ž 𝑓𝑐
𝜌 𝑔
𝑏
5. If actual B = 1m, find 𝑓𝑐 value associated with this
and draw a horizontal line from y-axis on PFF graph to
HFF line, approximate time at which this occurs
Hopper Design:
1. 𝐻(πœƒπ» ) graph (pg. 310) → draw vertical line from
given πœƒπ» angle, intersect with correct line i.e. cylindrical
hopper = circular, draw across to y-axis for 𝐻(πœƒπ» ) value
2. Need to plot normal stress (𝜎) vs shear stress (kPa)
to find wall frictions of each time interval (πœ™π‘€ )
3. For each time interval: πœ™π‘€ = tan−1( π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘)
4. Once we have πœ™π‘€ , find ff values from pg. 320 graph
(draw vertical line for πœƒπ» )
5. Now plot 𝜎π‘₯ against 𝑓𝑐 to add HFF lines to find
intersection points (for two diff. time periods there will
be two HFF lines), solve for y2 and draw lines
6. π‘“π‘π‘Ÿπ‘–π‘‘ values are found by horizontally drawing line to
y-axis from intersection of HFF lines with 𝑓𝑐 .
7. Find 2 diameter openings for each year (B values)
Download
advertisement