EH2208O
DESIGN PROJECT II (MARCH 2017) PRODUCTION OF 20,000 METRIC TONNES OF POLYTETRAFLUOROETHYLENEPER POLYTETRAFLUOROETH YLENEPER YEAR
CHAPTER 3: MULTICOMPONENT DISTILLATION COLUMN (T-102) STUDENT’S NAME:
NURBADAYU BINTI BASIRON 2013493592
SUPERVISOR: MS CHRISTINA VARGIS
FACULTY OF CHEMICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA SHAH ALAM
CHAPTER 3
EQUIPMENT DESIGN
DESIGNED DESIGNED BY: NURBADAYU BINTI B ASIRON ASIRON
3.6.1
MULTICOMPONENT DISTILLATION DISTILLAT ION COLUMN (T-102)
3.6.1.1 INTRODUCTION
This distillation column is installed to separate between the product which are TFE and HCL, and the waste which is the water before it goes to the treatment plant. The components are separate based on their boiling points. The higher the boiling point is less volatile the component, therefore it being discharge at the bottom outlet as the liquid. Since the water is less volatile component, it will be removing at the bottom product as the nearly pure product. The top component is recycling back to the process to avoid any waste of the raw materials.
℃
This column operates at pressure and temperature,43.23
℃
and 450 kPa at top and 550
℃
℃
kPa, 155.5 at bottom part with boiling point of TFE is about -40.9 , HCL at -85.05 and
℃
water at 100 . A design consists of chemical and mechanical evaluation to calculate for total stages, height, thickness, support, as well as the insulation. A key in distillation d istillation process is relative volatility, α.
431
585 kg/h 0.9755 TFE 0.0245 HCL
778 kg/h 0.2486 Water 0.7329 TFE 0.0184 HCL
193 kg/h 1.00 Water
Figure 3.1: Flow diagram of T-102
Table 3.1: Inlet and o outlet utlet compositions Components
Feed
Mole
Top
Mole
Bottom
Mole
(kmol/hr)
fraction,xf
(kmol/hr)
fraction,xd
(kmol,hr)
fraction,xb
TFE
5.71
0.34
5.71
0.94
Water
10.74
0.64
-
-
10.74
1.00
HCL
0.39
0.02
0.39
0.06
-
-
Total
16.84
1.00
6.1 6 .1
1.00
10.74
1.00
-
Table 3.2: Properties inlet and ou tlet stream Parameter
Feed column
Top (distillate) column
Bottom column
Temperature ( 0K)
323
233
428
Pressure (Kpa)
2000
460
540
432
3.6.1.2 CHEMICAL CHEMICAL DESIGN OF T-102
Step Step 1: Calculation of b ubble point, T bp and dew point T dp
For multi component mixture, trial and error procedure is required to calculate bubble and dew point by using the Antoine Equation. By using Antoine Equation to find the vapor pressure (P ) with related at equ ilibrium, Constant K, °
iK PP
Ln P mm HHgg A TKB + C Table 3.3: Antoine constant Components
A
B
C
TFE
15.5602
1704.80
-41.30
HCL
16.5040
1714.25
-14.45
Water
18.3036
3816.44
-46.13
Bubble poin t (Feed (Feed column)
:Σ
=Σ
= 1
Bubble point is calculated by using Microsoft Excel Goal Seek function. The goal is set so that the summation of yi is equal to one by changing the guess temperature. Different temperature will yield different Ki value. After calculating, bubb le point for feed is 39 10K.
Temperature : 391 0K Components
PO (kpa)
Ki
xi
Yi=kixi
TFE
5871.50404
2.93575202
0.34
0.99815569
HCL
20809.0583
10.4045291
0.64
6.65889866
Water
188.710109
0.09435505
0.02
0.0018871 0.99950786
433
Dew Dew po int (Top column )
yx 1 :∑ i=∑
Dew point is calculated by using Microsoft Excel Goal Seek function. The goal is set so that the summation of xi is equal to one by changing the guess temperature. Different temperature will yield different Ki value. After calculating, dew point for top column is 2690K.
Temperature : 2690K Components
PO (kpa)
Ki
yi
Xi=yi/ki
TFE
424.4487
0.943219
0.94
0.985985
HCL
2318.352
5.151893
0.06
0.013587
Water
0.426361
0.000947
-
0.999572
Bubble point (bottom column)
:Σ
=Σ
= 1
Bubble point is calculated by using Microsoft Excel Goal Seek function. The goal is set so that the summation of y i is equal to one by changing the guess temperature. Different temperature will yield different Ki value. After calculating, calculating, bubble point for bottom column column is 4220 K
Temperature : 422 0K Components
PO (kpa)
TFE
8710.34011
HCL Water
Ki
xi
Yi=kixi
15.836982
0.01
0.15836982
29356.0183
53.3745788
-
467.298909
0.84963438
0.99
0.84113804 0.99950786
434
Step 2: Determination of the Relative volatilit y
The relative volatility of two components can be expressed as the ratio of their K va lue: αi= Ki/K j where, Ki= light components K j= heavy components
ℎ ℎ
In this case, at distillate, the light component is HCL and the heavy one is TFE. Thus,
5.0.1951893 43219 5.462031 For bottom part,only volatility of water is taken into calculation because the HCl contain in bottom part is being considered as trace component. Thus,
0.849634
Average relative volatility is calculated by using equation:
√
= 2.154235
Step 3: Minimum n umber of stages for to tal reflux
For number of stages, the value of α i of the light component is used in Fenske’s Equation:
] l o g[ log, 0. 0 66. 1 0 0. 9 910. 7 3 l o g[ ] 0. 9 46. 1 0 0. 0 110. 7 3 log2.154235
Substituting into equation above,
= 2.4022 theoretical stages = 3 stages.
435
Step 4: Calculation o f feed conditi on, q, at feed stream
The phase of the feed mixture is classified by the feed quality, q value. Equation below is used to calculate the q value.
Where,
,,+,
q = Feed quality Cp = Heat capacity at feed, kJ/kmol·°C Tbubble = Bubble point temperature at feed, °C T = Temperature at feed, °C Hv = Latent heat of vaporization at feed, kJ/kmol Table 3.4: Latent heat of vaporization and heat capacity Component
X I,F
Latent heat of vaporization,hv(kJ/kmol)
Heat Capacit y
(kJ/kmol.
TFE
0.34
20211
57.30
HCL
0.02
16366
75.3747
Water
0.64
40641
75.3027
Cp = (0.32 x 57.30) + (0.02 x 75.3757) + ( 0.64 x 75.3027) CP = 68.037 kJ/kmol.0K HV= (0.32 x 20211 ) + (0.02 x 16366) + (0.64 x 40641) HV = 32805.08 kJ/kmol
6323.0829+32805.08 q 68.037 391.532805. q = 1.1415
Since q is large than 1, feed condition is in sub-cooled liquid.
436
Step 5: Estimation of Minimum reflux r atio
The minimum reflux ration is that reflux ratio which will require an infinite number of trays for given separation of the key components. For multi component distillation, two pinch point or zones of constant composition occur: one in the section above the feed plate and another below the feed tray. Underwood’s method is being used to determine the minimum reflux ratio:
Where,
1∑ +1∑
the relative volatility of component i with respect to some reference component,
usually the heavy key Rm = the minimum reflux ratio XiF = concentration of component i in the tops at minimum reflux
Table 3.5: Relative volatility is calculate using water as referen ce Component x If Relative volatility TFE
0.34
31.11388
HCL
0.02
110.27
Water
0.64
1
1∑ 0. 3 4 110. 2 7 0. 0 2 1 0. 6 4 11.1415 31.31.11388 + + 11388 110.27 1
By using goal seek analysis in Microsoft Excel, the va lue of is determined to be 2.2116. Table 3.6: Relative volatility is calculate using TFE as reference Component
x iD
TFE HCL Water
0.94 0.06 -
Relative volatility 1 5.462 0.001
437
+1∑ 1 0.2116 94 + 5.5.44622. 62 0.2116 06 R +1 12. Rm + 1 = -0.675 Rm= 0.325
The optimum reflux ratio is 1.5 of R m. Thus, R= 0.325 x 1.5 = 0.4875 Next, Erbar-Maddox correlation is used to determine the number of stages
+1 0.0.325+1 325 0.245 +1 0.0.4875+1 4875 0.328 From the chart, Nm/N is 0.47. From step 3, N m is 3, N = 3/ 0.35 = 8.57 = 9 stages
438
Figure 3.2: Erbar-Maddox correlation
Step 6: Determination of column efficiency using O’conell’s correlation A quick estimation of the overall column efficiency can be obtained from the correlation by O’Connel (1946).
0=
Where,
51 − 32.5 log(
)
= Molar average viscosity, mNs/m 2
αavg = Average volatility of light key 439
From ASPEN HYSYS, molar average viscosity of the mixture is 0.9233 kg/m∙s. Taking average volatility from previous step, the ov erall column efficiency is then calculated,
0=
51 − 32.5 log(0.9233 × 2.154235)
0 =
41% efficiency.
Step 7: Determination of th e actual number of stages
Actual stages, Nact are stages required for the column after factoring the efficiency of the stages. The stages can be calculated by using equation below:
= 21.95 ≈ 22
0.941 21.95
Step 8: Estimation of f eed-plate location
Kirkbride has devised an approximate method to estimate the number of theoretical stages above and below the feed which can be used to estimate the feed-stage location. This empirical relation is as follows:
0.206log[,, ,, ]
Where, Ne = Number of stages above the feed, including condenser Ns = Number of stages below the feed, including reboiler B = Molar flow of bottom outlet, kmol/h D = Molar flow of distillate outlet, kmol/h xf,HK = Mole fraction of the heavy key in the feed x,f,LK = Mole fraction of the light key in the feed xd,HK = Mole fraction of the heavy key in the distillate xb,LK = Mole fraction of the light key in the bottom 440
0.206log[0.0.6042 10.5.7713 0.0.9031 ] 0.359 Also, Ne + Ns = 0.359NS + NS =N, N = 22 Stages. Solving, Ns = 16and Ne = 6. This means there are 16 stages below the feed including reboiler
Step 8: Pressur e Drop
Pressure at top column: 460 kPa Pressure at bottom column: 540 kPa Thus the pressure drop is,
Δ = 540
− 460
= 80
Step 9: Determination of relative molecular mass
Relative molecular mass (RMM) is the average molecular mass of the composition of component in distillate and bottom product in the distillation column.
= Σxi
Where, xi = Component composition MW = Molecular weight xf = composition of component at feed
xd = composition of component at d istillate xb = composition of component at bottom
Table 3.7: Relative molecular weight of feed, distillate and bottom Component MW(kg/kmol) x f RMM Xd RMM Xb
RMM
TFE
86.468
0.37
29.399
0.94
80.415
-
-
HCL
36.461
0.61
23.335
0.06
2.5527
0.01
0.36461
Water
18.015
0.02
0.3603
-
-
0.99
17.835
441
∑xIRMM
53.093
82.9677
18.19961
RMM at feed = 53.093 kg/kmol RMM at distillate = 82.9677 kg/kmol RMM at bottom = 18.19961 kg/kmol
Step 10: Calculation the density of mixtu re The density of the mixture can be calculated by using the following equations:
∶, ,=∑ Liquid :
Table 3.8: Density of pure component Component
Density (kg/m 3)
TFE
1409.172
HCL
1192.98
Water
1000
Liquid density: Distillate,D = (0.93 X 1409.172) + (0.07 X 1192.98) =1394 kg/m3 Bottom,B = (0.01 X 1192.98) + (0.99 X 1000) = 1002 kg/m3
Vapor densit y: Distillate,D =
.. .
= 19.52 kg/m 3
Bottom,B =
.. . .
= 2.81 kg/m 3
Table 3.9: Density of component at distillate and bottom 442
Variable
Distillate
Bottom
Liquid densit y ( kg/m 3)
1394
1002
Vapor d ensity ( k g/m 3)
19.52
2.81
Step 11: Determination o f vapor and liqu id flo w rate at distillate and bottom
The value of vapor and liquid flow rate at distillate and bottom is taken from ASPEN HYSYS and tabulated in Table 3.10.
Table 3.10: Liquid and vapor flow rate at bottom and distillate Location
Liquid flow rate (kmol/h)
Vapor flow rate (kmol/h)
Distillate
165.2
749.8
Bottom
257.8
64.34
Step 12: Maximum vapor volumetric flo w rate
Maximum volumetric flow rate,Q = Where,
Vm = vapor molar flowrate (kmol/h) RM= relative molecular mass
= vapor density
QTop =
.. . 0.885/ .. . 0.1157/
QBottom =
Step 13: Calculation of flooding velocity
The diameter of column can be determined by using equation below: Q=AUf Rearranging,A=
Where flooding velocity, Uf is 443
− To find the value of ki, the liquid-vapor flow factor has to be determined first. The liquidvapor flow factor, FLV can be determined by below equation:
= ( )() .
Where, LW = Liquid flow rate, kmol/h VW = Vapor flow rate, kmol/h pv = Vapor density, kg/m 3 pL = Liquid density, kg/m 3
At the top column,
,= ... .
= 0.026
At the bottom column,
,= .. . .
= 0.212
444
Figure 3.3: Flooding velocity,sieve plate
Assume 0.5 m of tray spacing, from Figure 3.5, value of Ki can be determined. Ki top = 0.1 Ki bottom = 0.7
Calculating the flooding velocity,
, − 5 2 , 0.1 139419. 19.52 0.84 / , − 445
10022. 8 11 , 0.07 2.811 1.32 / For design, the optimum value is 80 -85% of flooding, 91% of downspout area, 95% for foaming at maximum flow rate ( Sinnott,2005 ). Thus,
Top = 0.84 x 0.85 x 0.91 x 0.95 = 0.617 m/s Bottom = 1.32 x 0.85x 0.91 x 0.95 = 0.97 m/s
Step 14: Calculati on of Net area requir ed
Net required, An =
ℎ,2005
Atop= 0.885/0.617= 1.434 m2 Abottom = 0.1157/0.95 = 0.122 m2
Step 15: Calculation of co lumn diameter
4 41.434 1.35
40.122 0.4 Step 16: Calculation o f colu mn height
Column height, Hc is the product of number of trays and tray spacing.
=
= 22 × 0.5
×
= 11
446
According to Geankoplis, 2003, the rule of thumb in designing the column height is to add 1.2 m at top of column to minimize entrainment and extra spacing for man hole and add 1.8m at the bottom of reboiler. Thus,
= 11 + 1.2 + 1.8 = 14
Step 17: Liquid flow pattern
Maximum volumetric vapor flow rate, Ql,max can be calculated using equation below:
Maximum liquid flow rate,Ql,max = Where,
Lm = liquid molar flowrate (kmol/h) RM= relative molecular mass
= liquid density
QL,top =
. . 0.00273/ . . 0.0013/
QL,bottom =
Liquid flow pattern to be used for this column is single-pass plate.
Step 18: Provis ion al Plate Design In designing plate, column area, down comer area, active area, holes area, holes size and weir weight are needed to be determine first. The column area, Ac is calculated by using the largest diameter, Dc equal to 1.35 m.
1. 3 5 4 1.43
The down comer area, Ad is assumed to be 12% of the total column area.
= 0.12 x 1.43 =0.17 m2
The net area, An is then calculated, =
−
= 1.43-0.17 =1.26 m2
447
Follows by calculation of active area,
=
−2
= 1.43- 2x0.17 = 1.09 m 2
The holes area, Ah, is calculated by taking 7% of total act ive area, Aa, h =
0.07 × 1.09= 0.0763 m2
A chart on the relation between down comer area and weir length from Sinnott, 2005, is used.
100% 0.1.1473 100%11.88%
Figure 3.4: Relation between downcomer area and weir length From Figure 3.6,
= 0.76 = 0.76x 1.43 =1.0868 m
Table 3.11: Summary of provisional plate de sign Column diameter, Dc
1.35 m
Column area,Ac
1.43 m2 448
Downcomer area, Ad
0.17 m2
Net area, An
1.26 m2
Active area,Aa
1.09 m2
Hole area,Ah
0.0763 m2
Weir length,lw
1.0868 m
Step 19: Check for w eeping
From the book of Chemical Engineering Design Volume 6, the su ggested dimension for weir and plate are; Weir height, hw = 45 mm Hole diameter = 5 mm Plate thickness = 3 mm
LW 3600
. .
= 1.303 kg/s
Minimum liquid rate at 70 % turn-down ratio = 0.7 x 1.303 = 0.9121 kg/s
Step 20: Weir liqui d crest
Francis weir formula is used to estimate the height of the liquid crest over the weir. For a segmental down comer, the formula can b e written as,
ℎ 750 / Where, how = Weir crest, mm liquid Lw = Liquid flowrate, kg/s lw = Weir length, m
449
1.3030868) 8.453 ℎ 750(10021. At minimum rate, clear liquid depth:
h + h = 8.453 + 45 =53.45
From Figure 3.7, the weep point correlation, K2 is found, K2 = 30.1
Figure 3.5: Weep point correlation
Step 21: Weep Point
The vapor velocity at weep point is the minimum value for a stable operation. Minimum vapor velocity through the holes based on the holes area, Uh is calculated by: 450
4 ] [ 0.9025. .
Where,
Uh = Minimum vapor velocity through the hole dh = Hole diameter, mm K2 = Constant depending on the depth of clear liquid on the plate
4 5] 30.10.2.98025. 1. 7.00 / The minimum vapor velocity through the hole will be compared against the operating vapor velocity. Vapor velocity is the ratio of minimum vapor volume tric rate over the holes area.
,, , 0.1157 , 0.0.01763157 1.516 /
The vapor volumetric rate was calculated in step 1 5, Q =
m3/s
The vapor velocity is above the minimum va por velocity which will cause weeping. Thus, no weeping will occur.
Step 22: Plate pressure dro p
To calculate pressure drop, orifice coefficient has to be known first. Orifice coefficient, Co can be determine through the relationship between percent perforated area with orifice in Figure 3.8
ℎ ℎ 0.0.000305 0.6 ( ) 100 0.1.076309 1007 ,, , 0.0.0876384 11.58/
From the figure 3.8, Co = 0.71
451
Figure 3.6: Orrifice coefficient and per cent perforated area correlation
According to Sinnott, 2005, pressure drop through the holes can be predicted from a modification of equation for flow through an orifice.
Where,
ℎ 51( )
Uh = Vapor velocity throughout holes, m/s pv = Vapor density pl = Liquid density 452
Co = Orifice coefficient The orifice coefficient is a function of the plate thickness, hole diameter and the hole of perforated area ratio.
ℎ 51(11.0.7518) (19.139452)190
Calculate residual head Residual head can be calculated using Hunt et al (1955) equation:
12. 5 10 12. 5 10 ℎ 1394 8.97 Total pressure drop
,ℎ ℎ+ℎ+ℎ+ℎ ℎ 190+53.45+8.97252.42
Step 23: Down comer liq uid back up
,ℎ ℎ 10451035 , ℎ 0.035 1.08680.038 Since Aap is less than Ad, Aap is used to calculated head loss in down comer, hdc
ℎ 166 1.303038)5.68 ℎ 166(10020. ,ℎ ℎ+ ℎ+ℎ+ ℎ ℎ 53.45+ 69.45+ 5.68128.58 0.12858
According to Thomas and Shah, 1964, for safe design,the liquid backup in downcomer, hb, should not exceed half the plate spacing plus weir length to avoiding flooding.
ℎ < 12 + ℎ 0.128 < 12 0.5+0.045
453
0.128 <0.2725 In this case , plate spacing is acceptable.
Furthermore, according to Kister, 1992, he
recommends that the height in down comer should not greater than 80% of the tray spacing.
ℎ0.5 100%<80% 0.0.1528 100% 25.6% , ℎ >3 0.170.1.01868251002 19.59
Thus, the plate spacing is acceptable for b oth recommendations.
is greater than 3s as recommended by Sinnott, 2005.
Step 24: Entrainment check Entrainment can be estimated from the correlation given by Fair (1961). Figure 3.9 gives the fractional entrainment ψ (kg/kg gross liquid flow) as a function of liquid -vapor factor, FLV with the percentage approach to flooding as a parameter. The percentage flooding is calculated as,
100% 0.1.121576 0.0918 / 0.0.01918157 100%79.3% From Figure 3.9 by using the value of F LV = 0.212 and flooding percentage = 79.3%, the value of ψ is 0.012. The value is below than 0.1, thus the column diameter proposed earlier is acceptable (Sinnott, 2005).
454
Figure 3.7: Fractional entrainment and liquid-v apor flow factor correlation
455
Step 25: Perforated area
Figure 3.8: relaxation between angle subtended by chord, chord height and chord length
1.1.038685 0.81 456
From the Figure 3.10,
0.21 108 Mean length of unperforated edge strips, La Assume 0.05 m width of unperforated strips around plate edge, ws
180 1.350.05 108180 2.38 Area of unperforated edge strip, As
2.38 0.050.119
Approximation of mean length of calming zone
ℎ , + 1.0868+0.051.136 , 2 2 1.136 0.050.1136 Total area available for perforation, Ap
The ratio of Ah / Ap is,
+ 1.090.119+0.11360.8574 0.0.08763574 0.088 457
From Figure 3.11, the value of lP /dh is 3.1 It is satisfactory as it is within 2.5 to 4 ( Sinnott, 2005)
Figure 3.9: Ah / Ap and lp /dh correlation
0. 0 05 ℎ, 4 4 1.963 10− ℎ 1.90.630763 10− 3886.9 ≈3887 ℎ
458
3.6.2 SUMMARY OF CHEMICAL DESIGN FOR T-102
Table 3.12: Chemical design sheet for distillation column (T-102) Specification
Value
Unit
Design Data Total feed inlet
778
Kg/h
Distillate
585
Kg/h
Bottom
193
Kg/h
Key components Heavy component
Water
Light component
TFE,HCL
Bubble point (feed)
391.56
K
Dew point ( top)
269
K
Bubble point ( bottom)
422
K
Number of tray
22
Stages
Feed point location
6nd
Stages
Column efficiency
41
%
Minimum reflux ratio
0.325
Column height
14
Column diameter
1.35
Column area
1.43
Liquid flow pattern
Single pass
m m m2
Plate Desig n Downcomer area
0.17
m2
Net area
1.26
m2
Active area
1.09
m2
Hole area
0.0763
m2
Area of one hole
1.963x10-5
m2
Number of hole per plate
3887
holes
Plate spacing
0.5
m
Plate thickness
3
mm
Weir length Weir height
45
mm 459
Hole diameter
5
mm
Weir length
1.0868
m
Weir liq uid crest Minimum vapor velocity Weep point
1.516
m/s
7
m/s
Plate pressure dr op Dry pressure drop
190
mm
Residual head
8.97
mm
Downcomer design Downcomer pressure loss
35
mm
Head loss in downcomer
5.68
mm
Backup in downcomer
0.128
m
Residence time
19.59
s
Perforated area
0.8574
m2
2.38
m
0.119
m2
1.136
m
Mean length, unperforated edge strips Area of unperforated edge strips Mean length, calming zone
460
3.6.3
MECHANICAL DESIGN OF DISTILLATION COLUMN (T-102)
Step 1: Design temp erature
The strength of metals decreases with increasing of temperature. Thus, it is vital to identify the design temperature before design stress be evaluated. The operating temperature for T102 is shown in table below
Table 3.13: Operating temperature of T-102
℃
Column
Temperature ( )
Distillate
-43
Bottom
155
Therefore, 155°C is decided to be the design temperature as it is the highest. 10% allowance in the wall temperature Design temperature,
℃
= 155 x 1.1 = 171
℃
Step 2: Design pressure
Design for pressure is taken at 50% above the operating pressure. It includes 10% pressure at which the relief device is set and 40% excess to avoid crack of the column during pressure make up above 25% of normal working pressure.
Table 3.14: Operating temperature of T-102 Column
Pressure ( kpa)
Distillate
450
Bottom
550
Therefore, 550 kPa is taken as the operating pressure as it is the highest. Designing at 10% above the operating pressure, Design pressure, PD = 550 kPa x 1.1 = 605 kPa = 0.605 N/mm2
461
Step 3: Material of cons tructi on
Selection of suitable materials must take into account in order to make the compatibility with the process environment of the design T-102 and also the suitability of materials for fabrication.T-102 has 3 components, TFE, Water and HCL and the nature of these components have to be taken into account when selecting the material of construction. For chemical plant, the materials are chosen based on the ability of the material to resist corrosion and also based on the economical factor. It is preferable to choose the material that has lowest cost over the working life of plant, allowing for maintenance and replacement. By reviewing and considering the factors, the most suitable and economical material of construction for these components is HASTELLOY® B-2 alloy is a nickel-base wrought alloy with excellent resistance to hydrochloric acid at all concentrations and temperatures. It also withstands hydrogen chloride, sulfuric, acetic and phosphoric acids. The alloy has excellent resistance to pitting, to stress corrosion cracking and to knife line and heat-affected zone attack. It resists the formation of grain-boundary carbide precipitates in the weld heat affected zone, thus making it suitable for most chemical process applications in the a s welded condition.
Step 4: Design s tress, S
It is necessary to determine the maximum allowable stress that can be accepted or withstand by the material without failure under operating condition. To calculate the maximum allowable stress, the material design stress factor is being used. By referring to Figure 1.1,
Material of construction : HASTELLOY,B-2 alloy Design temperature : 170.5
℃
Tensile strength : 849.4 N/mm 2 Design stress,S : 350.25 N/mm2
462
Figure 3.10: HASTELLOY® B-2 alloy (Source: http://specialmetals.ir/images/technical_info/nickel%20base/hastelloy%20B2.pdf )
Step 5: Welded Joint Efficiency Joint efficiency selected is 1.0 because it can allow the use of all material. If the joint factor is lower, the vessel will be thicker and heavier. The use of lower joint factors in design, though saving costs on radiography, will result in a thicker, heavier, vessel, and the designe r must balance any cost savings on inspection and fabrication against the increased cost of materials. Wield joint factor, J = 1.0
Table 3.15: Maximum allowable joint efficiency. ( Ref : Coulson Richardson, Chemical Engineering Design,4th edition) Type of joint
Degree of radiography 100 percent
spot
none
Double-welded butt or equivalent
1.0
0.85
0.7
Single-weld butt joint with bonding strip
0.9
0.80
0.65
463
Step 6: Corrosi on Allow ance
The “corrosion allowance” is the additional thickness of metal added to allow for material lost by corrosion and erosion, or scaling. For light corrosion is expected, a corrosion allowance of 3.0 mm should be used.
Step 7: Determination of vessel thick ness
Minimum thickness of vessel is required to resist the internal pressure. A much thicker wall will be needed at the column base to withstand the wind and dead weigh load. Hence, the minimum thickness of column is calculated based on the equation below.
Minimum vessel thickness, t = Where,
+ −.
Pi = Design pressure, N/mm 2 Di = Column diameter, mm S = Design stress, N/mm2 E = Joint factor (1 for double-welded joint) c = Corrosion allowance (3 mm )
Minimum vessel thickness, t =
. .−.. +34.17
Step 8: Determination of h ead and c losures
1. Hemispherical head design
4 + 40. 605135040.605 +33.58 4350.0.2510. 0.8851 + 0.
2. Torispherical head design
464
Where, RC= Crown radius = D i
6 051350 350.0.82850. 510.10.605 +35.06 3. Ellipsoidal head design
2 + 20. 605135020.605 +34.17 2350.0.2510.
Table 3.16: Minimum thickness for hemispherical, torispherical and ellipsoidal head Type of head Minimum
thickness,t
Hemispherical
Torispherical
Ellipsoidal
3.58
5.06
4.17
(mm)
From table 3.17, the minimum thickness for hemispherical, torispherical, and ellipsoidal head is 3.58 mm, 5.06 mm and 4.17 mm respectively. In this case, ellipsoidal head is chosen as the thickness is the closest to the vessel’s wall thickness. Ellipsoidal head also usually proves to be the most economical closure to use whereas the cost of forming hemispherical head will be higher than that for a shallow ellipsoidal head even though hemispherical head is best known for the strongest shape.
Step 9: Determination o f weight loads
a) Dead weight of vessel, Wv The major sources of the dead weight loads are, 1. The vessel shell 2. The vessel fittings; manhole, nozzles 3. Internal fitting; plates 4. External fitting 5. Auxiliary equipment which is not self-supp orted 6. Insulation 7. The weight of liquid to fill the vessel 465
Dead weight of vessel can be calculated by using Equation below:
= 240
m(
+ 0.8
)
Where, Wv = Total weight of the shell, excluding internal fittings Cw = Factor to account for the weight of nozzles, manways, internal supports (1.5 for distillation column) Dm = Mean diameter of vessel (D m = Di + t x 10 -3), m Hv = Height, m t = Wall thickness, mm
Recall minimum wall thickness is 4.17
mm, divide the column into five sections, with
thickness increasing by 2 mm per section. Try 4.17 mm, 6.17 mm, 8.17 mm, 10.17 mm and 12.17 mm. Thus, rough estimation of the weight of this vessel by using the average thickness is 8.17 mm.
Dm = 1.35 + (8.17 x 10-3) = 1.36 m Hv = 14 m Wv = 240(1.5)(1.36)[14+(0.8 x 1.36)] 8.17 Wv = 60.35 kN
b) Weight of plate, Wp
Weight of steel contacting plate including typical liquid loading may be estimated by factor of 1.2 kN/m 2 area given by Nelson (1963).
4 4 1.35 1.43 ℎ 1.2 1.2 1.431.716 ℎ , 1.716 . 1.7162237.752 466
c)
Weight of insulation, Wi
The distillation column should be well insulated to prevent loss of heat and to protect against burns. On columns located outdoors, because of size and safety, the insulation bracket will shield the column from increased heat loss due to wind currents. According to Kvaalen et al.,(1990), two to three inches of fiberglass insulation is good. Fiberglass insulation can withstand temperature range from -30°C to 540°C. The temperature fit in the design temperature, 170.5°C. In this design, the fiberglass insulation is assumed to be 2 inch which is approximately 50mm.
− , 1.36 14 50 10 3 ℎ ,
Fiberglass density, = 100 kg/m3
Thickness of insulation, ti = 50 mm
Vi =
Wi = 3 x 100 x 9.81 = 2943 N
This weight is double to allow for f itting, thus, Wi = 2943 x 2 =5886 N = 5.886 kN
Total weight: Wt = Wv + Wp + Wi = 60.35 + Wt = 103.988 kN.
37.752
+ 5.886
Step 10: Wind l oad
Wind loading is significant due to the column’s height and its open space installation. Its value depends on the dynamic wind pressure and column area. A column must be designed to withstand the highest wind speed that is likely to be considered at the site during the life of the plant. A wind speed of 160 km/h can be used for the preliminary design studies, equivalent to a wind pressure of 1280 N/m (Sinnott, 2005).
Where,
, , + 2+
t1 = Wall thickness, m 467
ti = Insulation thickness, m Dc = Internal diameter, m
1.35+2
0.00817 + 0.05) = 1.47 m
The loading unit per length of the column, Fw = Pw x Deff Fw = 1280 x 1.47 = 1881.6 N/m The bending moment,M at the bottom b ase:
14 2 1881.6 2 184397
Step 11: Analysis of st resses
The resultant stresses from all loads should be determined to ensure that the maximum allowable stress intensity is not exceeding at any po int (Sinnott, 2005).
Pressure stress:
, 4
Where,
P = Operating pressure, N/mm 2 D = Column diameter T = Column thickness, m
Dead weight stress:
Where,
0.5458.1350 22. 7 2 / 17 , 2 0.5258.1350 17 45.44 / +
Wt = Total weight of the column, N Dc = Column diameter, mm t = Column thickness, mm 468
Bending stress:
103. 9 88 10 1350+8.178.17 2.98 / + 2 1350+2 8.171366.3 64 64 1366.3 13508.01810 ± (2 +) 184397 10 ± 8.018 10 (1366.2 3 +8.17)±15.89 /
Principle stress: The resulted longitudinal stress is,
Since
+ ± ,22.722.25+15.8936.36 / ,22.722.2515.894.58 /
is compressive stress, therefore it becomes negative sign, therefore,
The greatest difference between the principle stresses will be on the downwind side,
, 45.444.5840.86 /
The value obtained, 40.86 N/mm 2 is well below the maximum allowable design stress which is 350.25 N/mm2.
Step 12: Check elastic stability
A column design must be checked to ensure that the maximum value of the resultant axial stress does not exceed the critical value at which b uckling will occur.
, 2 10 ()
469
8.17 )121 / 2 10 (1350 + 2.25+15.1418.14 /
The maximum compressive stress will occur when the vessel is not under pressure,
The design is satisfactory as the value is below the critical buckling stress.
Step 13: Design for vessel supp ort
The method used to support a vessel depends on size, shape and weight of the vessel, the design temperature and pressure, the vessel location and arrangeme nt, and the internal and external fittings and attachments. The supports must be designed to carry the weight of the vessel and contents, and any superimposed loads, such as wind loads. There are three types of support which are; 1. Skirt supports; used for tall and v ertical columns 2. Brackets, or lugs; used for all types of vessel 3. Saddle support; used for horizontal vessel.
Step 14: Skirt s upport d esign
Skirt support is recommended for vertical vessels as it does not impose concentrated loads on the vessel shell. In other word, it is particularly suitable for use will tall columns subject to wind loading. Thus, skirt support design will be used to suppo rt the bottom of reactor column and it is illustrated in Figure 3.12.
470
Figure 3.11: Skirt support design
Type of support
: Straight cylindrical skirt
Material of construction
: Carbon steel
Design stress, S
: 89 N/mm2
Young modulus
: 20000 N/mm2
Angle, θs
: 90°
Approximate weight,
4 1.35 14 1002 9.81196.98
From previous calculation, weight of vessel = 103.98 kN New net total weight,W total, Wtotal = 103.98 + 196.98 = 300 kN
Bending moment at base skirt, Ms By taking skirt support equal to 3, the equation become,
+ 2 14+3 1881.6 2 271.89 471
The resultant stresses in the skirt will be,
(
)=
−
(
)=
+
Where,
σbs = Bending stress at the skirt σws = Dead weight stress at the skirt Bending stress at the skirt is given by,
4+
Where,
Ds = Inside diameter of the skirt, at the base ts = skirt thickness
4 271.8910001000 1350+8. 178.171350 23.95 / , , + 300 1000 , 1350+8. 8. 6 0 / 178.17
Dead weight stress,
For testing, the weight is equal to 300 kN
For testing, the weight is equal to 103.98 kN
103.98 1000 , 1350+8. 2. 9 8 / 178.17
Thus, the resulting stress in the skirt is,
23.95 23.95 (
(
)=
−
,
– 2.98 = 20.97 N/mm2
)=
(
)=
(
)=
+
,
+ 8.6 = 32.55 N/mm 2.
The criteria for designing are,
(
,
)<
472
(
,
) < 0.125
Where,
= Young modulus
Ss = Design stress of skirt
For σs (maximum, tensile), by taking joint factor, E equal to 0.85, 20.97 < 89 × 0.85 × sin (90) 20.97 < 75.65
.
For (
,
32.55 < 0.125 x 200000 x 32.55 < 151.3
),
x sin (90)
Since both of the criteria are satisfied, the design is acceptable.
Step 15: Base ring and anchor bolt s
The loads carried by the skirt are transmitted to the foundation slab by the skirt base ring (bearing plate). A variety of base ring designs are used with skirt supports. The preliminary design of base ring is based on Scheiman’s short cut method. Scheiman gives the following guide rules which can be used for the selection of the anchor bolts:
Table 3.17: Design base ring Approximate pitch circle
Dappro(b) = Di + 2(t +t insulation) Dappro(b) = 1350 + 2(8.17 + 50) = 1466.34 mm
Circumference of bolt circle
= 1466.34
= 4606 mm Recommended spacing between bolts
600mm
Minimum number of bolts required, Nb
4606/600 = 7.67 Closest = 8
Bending moment at skirt base, Ms Total weight of vessel
271890 103980 N
473
125 N/mm2
Bolts design stress, S b
The bolt area is given by Sinnott,2005,
1 (4 )
Where,
Ab = Area of one bolt at the root of the thread, mm 2 Nb = Number of bolts Sb = Maximum allowable bolt stress, N/mm 2; typically 125 N/mm 2 Ms = Bending moment at the base W = Weight of the vessel Db = Bolt circle diameter
8 1125 (4 1.271890 4634 103980)639
From Table 3.18, it can be estimated that the conventional bolt size is M36 bolts.
Table 3.18: Anchor bolt chair design
474
Figure 3.12: Anchor bolt illustration
Table 3.19: Bolt dimension for T-102 Item
A
B
C
D
E
F
G
Dimension(mm)
57
102
76
16
32
42
42
Bolt diameter, Db
4 6394 28.5
475
Figure 3.13: Base ring illustration
The base ring must be sufficiently wide to distribute the load to the foundation. According to Sinnott,2005, total compressive load on the base ring per unit length can be estimated by,
4 + 2
Where,
Fb = the compressive load on the base ring, N/m Ds = skirt diameter,m
1 03980 4271890 + 1.35 21.350 228.46 /
Minimum width of base ring, L b
Where,
101
Lb = base ring width,mm
476
Fc = maximum allowable bearing pressure on the concrete foundation pad, which will depend on the mix, used and will typically range from 305 to 7N/mm2 (200 to 1000 psi). Taking the average, 156 will be used.
228.15646 101 0.00146
Actual width required, Lb = Lr + ts + 50 Lb = 102 + 8.17 + 50 =160.17 mm
Actual bearing pressure on concrete foundation,
4 6 , 228. 1. 4 3 / 160.17
Actual minimum base thickness,
, 3
Where, Lr = The distance from the edge of the skirt to the outer edge of the ring,mm To = Base ring thickness,mm f c = Actual bearing pressure on base, N/mm2 f r = Allowable design stress in the ring material, typically 140 N/mm 2
102 31.14043 17.86 Step 16: Design o f pip ing
There are three nozzles in the distillation column, which are feed inlet, distillate outlet and bottom outlet. By assuming that the flow of the pipe is turbulent flow, the optimum duct diameter is,
Where,
293.−.
G = Mass flowrate, kg/s 477
= Density of components, kg/m 3
The nozzle thickness, t p, can be calculate from equation:
20+
Where, Ps = Operating pressure, N/mm2
= Design stress, N/mm2
Table 3.20: Flowrate and density at 3 locations Feed
Distillate
Bottom
Mass flowrate
0.216
0.1625
0.0536
Density (kg/m3)
1143
1394
1002
Feed stream
G = 0.216 kg/s
= 1143 kg/m 3 = 350.25 N/mm 2
Ps = 2 N/mm 2
, 2930.216.1143−. 9.6 20350.29.265+2 0.00274 , + 0.00274+33.00274 Distillate stream
G = 0.1625 kg/s
= 1394 kg/m 3 = 350.25 N/mm 2
Ps = 0.46 N/mm 2 478
, 2930.1625.1394−. 7.68 0.467.25+0. 68 46 0.00044 20350. , + 0.00044+33.00044 Bottom stream
G = 0.0536 kg/s
= 1002 kg/m 3 = 350.25 N/mm 2
Ps = 0.54 N/mm 2
, 2930.0536.1002−. 4.82 0.544.25+0. 82 54 0.00037 20350. , + 0.00037+33.00037 From the calculation, the optimum diameter is around 9-10 mm. From Table 3.21, the most suitable flange to be used is the nominal size 10.
479
Figure 3.14: Standard flange design
Table 3.21: Typical standard flange size and dimension
480
3.6.4
SUMMARY OF MECHANICAL DESIGN FOR T-102
Table 3.22: Mechanical design sheet for distillation column (T-102) SPECIFICATION
VALUE
UNIT
Design Operation Type of construction material
HASTELLOY, B-2 alloy
Type of insulation
fiberglass
Design pressure
0.605
Design temperature
171
Design stress
350.25
N/mm2
℃
N/mm2
Design o f head and closure Type
Ellipsoidal
Joint factor
1
Minimum thickness
4.17
mm
Corrosion allowance
3
mm
Column weight Column weight of vessel
60.35
kN
Weight of plate
37.752
kN
Weight of insulation
5.886
kN
Total weight
103.988
kN
Bending moment
184397
Nm
Design of suppo rt type
skirt
Skirt thickness
8.6
mm
Skirt height
3
m
Design of b ase ring and anchor bol ts Pitch circle diameter
1466.34
mm
Area of bolts
639
mm2
Bolts size
M36
Minimum base ring thickness
17.86
mm
481
