INTRODUCTION
Crystal growth fundamentals
Crystals are formed by the process called ‘nucleation’. Nucleation can start either with the solute molecules or with some solid matter which might be an impurity in the solution. The growth normally occurs by aggregation of molecules that are attracted to each other. The number of crystals formed, crystal sizes and shapes generally depend on properties of the solution like, saturation (solute concentration), operating temperature and mechanical disturbances.
In solutions which the solute is near saturation promote fast crystal growth. Supersaturated solutions tend to give crystals which are small in size. If the nucleation is low, such solutions will result in fewer crystals each of larger size. Nucleation is certainly promoted by turbulence and thus mechanical disturbances typically result in smaller crystals.
In general, thermal gradient methods tend to produce high quality crystals. Such methods include slow cooling and zonal heating. The latter employs convection by creating a thermal gradient in the crystal growing vessel. The solution becomes more saturated in the warm part of a vessel and crystal growth occurs when the solution is slowly transferred to a cooler region. [1]
Rest of this report will discuss about how to design a crystallizer for crystallization of aqueous solution of potash under the following operating conditions:
Absolute pressure - 0.8 atm Temperature - 40oC Mean diameter - 2.5 m Length of cylindrical shell - 5 m
1. Material selection
Since an aqueous solution of potash is used as the raw materials for the crystallization process, special attention should paid in order to avoid the corrosion. Therefore it’s better to use Stainless Steel as the fabricating material.
Grade 304 (SA-240) is the most widely used stainless steel which is available in a wider range of forms. It has excellent forming and welding characteristics. Post-weld annealing is not required when welding thin sections. Grade 304 is available in roll formed into a variety of components for applications in the industrial, architectural & transportation fields.
Grade 304L is the low carbon version of 304, does not require post-weld annealing and so is extensively used in heavy gauge components (over about 6mm).
SA-240 also has a excellent corrosion resistance in a wide range of atmospheric environments and many corrosive media. But, it may subject to pitting and crevice corrosion in warm chloride environments, and to stress corrosion cracking above about 60°C. Since the crystallizer is maintained at 40oC this might not be an issue.
Design Pressure & Temperature
i. Design Pressure (PDesign)
Absolute Pressure
The absolute pressure is measured relative to the absolute zero pressure. In other words, relative to the pressure that would occur at absolute vacuum.
Under the given operating conditions, operating pressure inside the crystallizer is 0.8atm (absolute). Therefore, this scenario falls under the category of;
PExternal=PAtmosphere & P Internal<PExternal
Therefore PDesign is given by,
PDesign=PExternal−P Internalabsolute
PExternal=1atm
P Internalabsolute=0.8atm
Therefore; PDesign=(1−0.8 )atm=0.2atm
PDesign=0.2×101325 N /m2=20.265 kN /m2
ii. Design Temperature (T Design)
Since the crystallizer should be operated at 40oC it is required to be heated. Let’s assume, that the vessel is indirectly heated with using a heating coil.
Therefore;
T Design=T Highest temperature of body0 +10oC
Therefore; T Design=50oC
2. Calculation of the wall thickness of the shell economical and safe to PDesign∧T Design
Let’s assume that all the welded joints are butt joints & therefore according to the section II, Part D of ASME, welded joint efficiency (φ) will be 0.7.
Theoretical wall thickness for the cylindrical portion of the vessel can be calculated by;
t actual=20265×2.5
2×0.7×106×106=0.3414 mm
tactual=Pdesign×D
2×ϕ×σdesign
L
h1
h2
Thickness to resist plastic failure;
P=2σφ (t /D o)1
1+1.5U (1−0.2
Do
L)
100( tDo
)
Where, L is the effective length of the vessel.
LEffective=L+ 13h1+
13h2
LEffective=5+ 0.72173
+ 1.2693
=5.6636m
Therefore;
20265=2×106×106×0.7×( t2.5+t
) 1
1+1.5×1.5(1−0.2
2.5+t5.6636
)
100(t
2.5+t)
tT heoretical=4.3657mm
When the actual thickness is calculating, corrosion allowance should be added to the theoretical
thickness. Since SA-240 is used & it is a stainless steel corrosion allowance is not needed.
Therefore; t Actual=tT heoretical=4.3657mm
Critical pressure for elastic failure;
PCritical=K × E×( tDo )
m
Where K & m are constants depends on Do
Leffective ratio.
Do/L(effective) K m0.1 0.185 2.600.2 0.224 2.540.3 0.229 2.470.4 0.246 2.430.6 0.516 2.490.8 0.660 2.481.0 0.879 2.491.5 1.572 2.522.0 2.364 2.543.0 5.144 2.614.0 9.037 2.625.0 10.359 2.58
For this scenario;
Do
Leffective
= 2.55.6636
=0.4414
Assuming linear interpolation is possible K & m were calculated as follows,
Do/L(effective) K m0.4 0.246 2.43
0.4414 0.3019 2.44240.6 0.516 2.49
According to the FIG HA-1 of the page 712 in ASME section II part D, Young’s modulus of SS grade
304 (SA-240) is 193.1GPa.
PCritical=K × E×( tDo )
m
Assuming t<<<Do & thereforeDm≈ Do
PCritical=0.3019×193.1×109×( 4.3657×10−3
2.5 )2.4424
=10709.87Pa
Since PDesign is 20265Pa, PCritical<PDesign therefore, vessel could undergo elastic failure.
Now, let PCritical=PDesign & find the required wall thickness to resist the elastic failure.
20265=0.3019×193.1×109×( t2.5 )
2.4424
=5.6683mm
Therefore, wall thickness of 5.6683mm will resist the vessel for elastic & plastic failure.
Available plate thickness will be 6mm.
Selection of suitable ends & their calculations
Here are some of available heads& closers.
Dished only 80% dished, 10%
knuckle
Flanged only hemispherical
ASME flanged
and dished
Standard flanged
dished
Elliptical Tori-conical
Tori-spherical Conical
When designing a pressure vessel head geometry should be selected based on the design pressure & the fabrication cost. For this vessel, it’s possible to use a flat end. But it will need a thickness which is considerably higher compare to tori-spherical head. Therefore, tori-spherical head is selected. Since this is a crystallizer, bottom end is obviously a conical end.
Tori-spherical head
Here; R - Crown radius,
r – Knuckle radius
D – Outer diameter
t – Thickness of the head
h – Flanged height
According to the ASME code;
R = D
r = 6% Dinner
There are 3 different equations available for calculate h E value.
he=R−√(R−Do
2 )×(R+D o
2−2r )
he=2.5−√ (2.5−1.25 )× (2.5+1.25−(2×0.06×2.5 ))=0.4233
he=Do
2
4 R= 2.52
4×2.5=0.625
he=√ Do×r2
=√ 2.5×0.06×2.52
=0.4332
Minimum value of he is 0.4233
t=D× PDesign×C
2σ Designφ Where; C is the stress concentration factor & given by the following table
t/Do
h E / Do 0.00075 0.0005 0.001 0.002 0.005 0.01 0.02 0.04
0.15 5.34 5.50 5.18 4.55 2.66 2.15 1.95 1.75
0.20 2.55 2.60 2.5 2.3 1.7 1.45 1.37 1.32
0.25 1.48 1.50 1.46 1.38 1.14 1 1 1
0.30 0.98 1.00 0.97 0.92 0.77 0.77 0.77 0.77
0.40 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59
0.50 0.55 0.55 0.55 0.55 0.55 0.55 0.55 0.55
It’s reasonable to assume DOuter = DMean since wall thickness is negligible compare to diameter of the vessel.
Therefore; he
Do
=0.42332.5
=0.1693
Disked section
Knuckle section
Flanged section (cylinder)
R
C
A
O
h
B
a
Assuming linear interpolation is possible C value is calculated for above ratio.
t/Do
h E / Do 0.00075 0.0005 0.001 0.002 0.005 0.01 0.02 0.04
0.15 5.34 5.50 5.18 4.55 2.66 2.15 1.95 1.75
0.1693 4.26 4.38 4.15 - - - - -
0.20 2.55 2.60 2.5 2.3 1.7 1.45 1.37 1.32
When C=4.26;
t=D× PDesign×C
2σ Designφ
t 1=2.5×20265×4.26
2×0.7×106×106=1.45
t 1' =0.00075×Do=0.00075×2.5=1.875
When C=4.38;
t 2=2.5×20265×4.38
2×0.7×106×106=1.5
t 2' =0.0005×D o=0.0005×2.5=1.25
Using above four calculated values, actual C value can be calculated.
Cactual=4.3356
t actual=Cactual×D o=4.3356×2.5=1.48mm
R - Crown radius,
r – Knuckle radius
D – Outer diameter
t – Thickness of the head
h –Height of knuckle section
OC = R = 2.5m
BC = R - r
=2.5−(0.06×2.5 )=2.35m
AB = Din / 2 - r
= 1.25− (0.06×2.5 )=1.1m
AC = (BC2 - AB2) ½
=√2.352−1.12=2.077 m
Sin α = AC/ BC
= 2.077/2.35 = 0.8838
α = 62.10780
h = r ×sin α
=(0.06×2.5)×0.8838
=0.1326m
Dish height = OC –( AC +h)
=2.5-(2.077+0.1326)
=0.2904m
Since thickness of the wall < 25mm;
Dblank = Douter + Douter/42 + 2/3 rknuckle + 2H flanged
Where,
h flanged=23
[ hflanged+hdished+hknuckle ]
8t
8tc
tc
h flanged=23
[ hflanged+0.2904+0.1326 ]
h flanged=0.846 m
Therefore; Dblank
= 2.5 + (2.5/42) + 2/3 × 0.15 + 2×0.846 m
Dblank = 4.3515m
Conical bottom
t – Vessel thickness of the cylindrical portion
tr – Reinforce thickness
α= 600
Pσφ
value for this bottom is 20265
106×106×0.7=0.273×10−3
Pσφ
1×10−3 2×10−3 3×10−3 4 ×10−3 5×10−3 6×10−3
∆ 13o 18o 22o 25o 28o 31o
Assuming extrapolation is possible for the above set of data; corresponding ∆ value can be calculated as 11.11o. Therefore the maximum value of α without reinforcing is 11.11o. But, it’s not practical to use such a small value since it will drastically increase height of the vessel. Let’s assume α= 600.
Then tc is given by;
t c=20265×2.5
2× cos (60)×(106×106×0.7−0.6×20265)=0.683mm
Available plate thickness will be 1mm
Reinforce area is given by;
A= 20265106×106×0.7
×2.52× tan(60)
8×(1−11.11
60 )=301.1mm2
t theoretical=Pdesign×D
2cos α (σ designϕ−0 . 6Pdesign ) .
A= pσϕ
×Di
2 tan α
8 (1− Δα )
A is also given by;
A=2 [ (8 t ×t r )+(8 t c×t r)]
A=16 tr (5.6683+0.683)
Therefore; 301.1=16 t r(5.6683+0.683)
t r=2.963mm
Available plate thickness will be 3mm
Fabrication procedure
Stainless steel is available in several forms & dimensions at market such as Plates, Sheets, Bars and
Forgings. Since this vessel is having moderate diameter it’s suitable to use stainless steel plates for the
fabrication procedure.
Shell fabrication
Cutting the plate to obtain required diameter and height by using an Oxy-acetylene flame or
a laser beam. Then the quality of the edge can be mirror smooth, and a precision of around
0.1mm can be obtained.
Crimping – this was done prior to rolling process to enhance plate rolling roundness and
efficiency. Crimping sets the correct radius on the ends of the plate and eliminates the
waste of excess material
Heating - The plate is then heated and moved to the rolling mill
Rolling- The rollers work the plate to the proper radius. Then ends of the plate meet at the
proper diameter
Welding-After the plate is formed into cylinders; many welding processes are used to
fabricate the rest of the vessel. Here we have selected that the shell should be welded by a
single grove butt welding.
Heat Treating
Testing- then magnetic particle testing was done.
Head and End Fabrication
For both Head and bottom 10mm thick torrispherical ends has been selected
1. Select the 10 mm thickness carbon steel blank
2. It is subjected on pressing.
3. Then it is subjected to spinning using a die.
Shell and ends are welded using single V grove butt joint
Technical drawings of the designed vessel including welding symbols
REFERENCES
http://www.math.wsu.edu/faculty/tsat/files/Potash.pdf
http://www.engineeringtoolbox.com/pressure-d_587.html
http://www.niroinc.com/evaporators_crystallizers/crystallizer_applications.asp
http://www.whiting-equip.com/media/swenson_crystallization.pdf
http://www.matweb.com/search/DataSheet.aspx?MatGUID=25bd2cee70ac40fdaae0acbf5b69dafe
http://www.matweb.com/search/DataSheet.aspx?MatGUID=21eca9c274a2473a8c3587d57d924b52
http://www.walkerep.com/products/products--capabilities/processor-vessels/whey-crystallizer.aspx
http://www.google.cm/imgres?q=crystallizer&hl=fr&sa=X&biw=1366&bih=667&tbm=isch&prmd=imvnsfd&tbnid=j1QepKv5xl39VM:&imgrefurl=http://www.mpi-magdeburg.mpg.de/research/projects/1032/1044&docid=yAPDitdchjtb7M&imgurl=http://www.mpi-magdeburg.mpg.de/research/projects/1032/1044/granulation.png&w=400&h=364&ei=jQdvT-nFE5DxrQfJodygDg&zoom=1&iact=hc&vpx=860&vpy=195&dur=905&hovh=214&hovw=235&tx=125&ty=141&sig=114604428543611562465&page=3&tbnh=143&tbnw=158&start=47&ndsp=26&ved=1t:429,r:11,s:47
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