MECHANICAL DESIGN FOR SIEVE PLATE COLUMN
Inner diameter of column = 1.5 m = 1500 mm 2
2
Design pressure = 1200 kN/m = 1.2 N/mm o
o
Design temperature = 150 C = 302 F Material of construction used is Stainless steel 18 Cr, 8 Ni Grade 304
Determine the maximum allowable stress, S, based on the design temperature given To find the maximum allowable stress, S, we need to refer to table below.
Table 1 Taken from Gavin Towler, Ray Sinnot, Chemical Engineering Design Principles, Practice and Economics of Plant and Process Design, © 2008, Table 13.2
o
The maximum temperature for Stainless steel 18 Cr, 8 Ni Grade 304 is 1500 F, therefore for this problem
the
o
design
temperature
do
not
exceed
the
given
maximum
temperature.
o
Assume 302 F ≈ 300 F. 2
S = 15.0 ksi = 103.422 N/mm
To calculate the minimum thickness required for a cylindrical shell, refer equation given by the ASME BPV Code (Sec. VIII D.1 Part UG-27)
Therefore,
Take into consideration of 2 mm of corrosion allowance. Therefore the final thickness is,
Column thickness design The height specification of column is 37 m, which is less than 15 m. By right, we have to consider the wind load and dead weight load of column. The approach we take in designing our column is that along the way from the base to the top of the column, thickness must be thicker at the bottom than that at the top. Therefore, the column is divided to 5-equal section and we increase it by 2 mm on each section. This is done to prevent buckling.
Figure 1 Column thickness that has been distributed according to respective section
Now that the column has undergo those thickness change, we can now calculate the mean thickness of the column.
Dead weight of vessel From equation 13.74, total weight of shell, excluding internal fittings for a steel vessel is given by,
Where C w = a factor to account for the weight of nozzles, etc Hv = length of the cylindrical section, m Dm = mean diameter of vessel = = 1.5 + (15*10^-3) = 1.515 m Therefore,
Weight of sieve trays and plates : Assumption has to be made, that the diameter of sieve plate is equal to the inside diameter of column. Therefore, area of 1-tray,
Therefore, weight of plate including water
For 50 sieve plates,
Weight of insulation : 3
Insulations used is mineral wool. For mineral wool, the density is 130 kg/m Approximate volume of insulation are,
Weight of insulation,
Double that value to allow for fittings
Total weight : Shell
239.67
Plates and contents
106.2
Insulation
22.2
Total
368.07
Wind Loading Our design required the column height to be 37 m. Therefore, we have to consider for wind loading because only structures less than 30 m can we neglect the wind loading Dynamic wind pressure: 2
Pw = 0.05Uw (for smooth column) 2
Pw =0.07Uw (for column with ladder etc) For this case, where we have access ladder and so on, we can assume the wind speed as 160 kph, that is 2
equivalent to a wind pressure of 1280 N/m
2
Pw =0.07Uw
Pw =0.07(160) = 1792 N/mm
2
Mean diameter, including insulation: = D i + 2(t s +t ins ) x 10 = 1.5 + 2(11+50) x 10
-3
-3
= 2.72 m Loading (per linear meter), F w : W= 1792 N/m 2 x 2.72 m = 4874.24 N/m Bending moment at bottom tangent line:
2
M x = X Hv
M x =
X (37)
2
M x =3342509.3 Nm
Analysis of Stresses At bottom tangent line, Pressure stresses : Longitudinal stress,
σL =
2
(N/mm )
σL =
σL = 30 N/mm
Circumferential stress,
σh =
2
(N/mm )
2
2
(N/mm )
σh =
2
(N/mm )
σh = 60 N/mm
2
Dead weight stress: σw = σw =
2
(N/mm )
2
(N/mm )
2
σw = 5.142 N/mm (compressive) Bending stresses : Do = Di + 2 X t = 1500 mm + 2 (15 mm) = 1530 mm Iv = Iv =
4
4
(Do + Di )
4
4
(1530 + 1500 )
11
Iv = 5.17 X 10 mm
σb = ±
( +t)
σb = ±
(
σb = ± 4.95 N/mm2
+ 15 mm)
The resultant longitudinal stress is σz = σL + σw ± σb σw is compressive therefore it is negative 2
σz (upwind) = 30 + (-5.142) + 4.95 = 29.808 N/mm
2
σz (downwind) = 30 + (-5.142) – 4.95 = 19.908 N/mm
2
2
σz = 19.908 N/mm
σz = 29.808 N/mm
σh = 60 N/mm
2
2
σh = 60 N/mm
Figure 3 Up-wind
Figure 2 Down-wind
Differences in Principal stress for: Upwind,
= 60 – 29.808 2
= 30.192 N/mm Downwind,
= 60 – 19.908 = 40.092 N/mm
2
2
2
Since both upwind (30.192 N/mm ) and downwind stress (40.092 N/mm ) are much more less than the 2
maximum allowable stress, S = 103.422 N/mm this design with metal thickness of 15 mm is okay and satisfactory.
MECHANICAL DESIGN FOR SKIRT SUPPORT o
A straight type skirt support is selected, where Ө = 90 . Material of construction used is plain carbon 2
2
steel maximum allowable design stress, S =89 N/mm and Young’s modulus, E =200,000N/mm at ambient temperature. The welding efficiency , E =0.85.
Figure 4 Taken from Gavin Towler, Ray Sinnot, Chemical Engineering Design Principles, Practice and Economics of Plant and Process Design, © 2008, Figure 13.21
Maximum dead-weight load on the skirt will occur when the vessel is full of water. Since hydrocarbon materials has density that is lower than water, then we use density of water to complete our calculations. Approximate weight
2
= (π/4 X 1.5 X 37)(1000)(9.81) = 641420.9 N = 641.4 kN
Weight of vessel from the previous column design
= 368 kN
Total weight of vessel and vessel full of water
= 641.4 + 368 = 1009.4 kN
Wind loading from previous column design
=4874.24 N/m =4.9 kN/m
Bending moment at base of skirt
2
= (Fw)(x )
2
= (4.9)(39.5 )
= 3822.6 kNm
Bending stress in the skirt: σbs = σbs =
σbs = 89 N/mm2 Dead-weight stress in the skirt : σws =
σws(test) =
= 11.133 N/mm
2
σws(operating) =
= 4.059 N/mm
2
Dead-weight stress in “test” condition is with column full of water, while “ operating” is just the column’s weight. σws(compressive) = 89-11.133 σws(compressive) = 77.867 N/mm
2
σws(tensile) = 89-4.059 σws(tensile) = 84.941 N/mm
2
For the skirt design to work, 2 conditions must be met. Those conditions are σs(tensile) < Ss.E.sinӨ σs(compressive)<< 0.125.E.(t/D s)sinӨ Substitute the value σs(tensile) < (89)(0.85)(1) 2
77.867 N/mm >75.65 N/mm
2
σs(compressive)<< 0.125.200000.(19/1500)(1) 2
84.941 N/mm <<316.67 N/mm
2
The skirt design obeys and satisfy the second rule which stated that the σs(tensile) must be much more less than the product of 0.125.E.(t/D s)sinӨ. But for the first, condition, some problems has been detected because the calculated σws(compressive) have higher values that the product of S s.E.sinӨ. So, as an act to counter this, we can use different t sk assumption, if possible higher than 19 mm so that when the calculation for σbs it will become lower than the original value. σbs =
σws(compressive)= σbs - σws(test)
DIMENSIONED SKETCH OF DESIGNED COLUMN
