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