CHE441 Lecture: Tank & Vessel

Introduction of Vessel Applications of Vessels:  (1) as a liquid surge drum for liquid storage  (2) as a vapor/liquid separator to separate vapor/liquid mixture to vapor and liquid streams  (3) as a liquid/liquid separator to separate liquid/liquid mixture to light liquid and heavy liquid streams action  (4) as a distillation column with installed packings or trays for fractionation  (5) as a reactor with installed catalyst beds for chemical reactions  (6) as a filter with installed filter media for filtration  (7) as a dryer with installed drying media to dry a vapor or liquid stream. 2 Yu, Process Design for Chemical Engineers

Vertical Drum Design  Must specify (1) liquid flow rate and (2) surge time (EX gpm and 10 minutes)   Optimum length (L)/diameter(D) = 3, but the range 2 to 5 is common. 3

Example: 1000 gpm, Minimum surge time = 10 min (between LLL and NLL)  Volume between LLL and HLL = (20 min)(1000 gpm)(1 ft 3 /7.48 gal) = 2673 ft 3  Assume total volume = (1.5)(2673) = 4011 ft 3  Since (L/D) = 3, or πD 2 (3D)/4 = 4011, or D = 12 ft  Let L 1 = height between LLL and HLL, then L 1 = 23.7 ft  The disengagement height is H D = 0.75D = (0.75)(12) = 9 ft  Total height = 6/ / = 35.2  36 ft  Check L/D = 36/12 = 3, design ok Note: the answer is not unique. 4

Vessel Liquid Surge Requirements Flow Rate BasisDestination and Condition Minimum Surge Time Minutes, NLL to LLL Accumulators Overhead Product Rate Reflux rate To storage2 To process directly10 To column5 Column Bottom Product Rate To storage, no process heat exchanger 2 To storage, with process heat exchanger 5 To process directly5 To kettle reboiler2 To recycle, no process heat exchanger 2 To recycle, with process heat exchanger 5 To cracking furnaces5 To reactors or licensed systems 10 5 Flow Rate BasisDestination and Condition Minimum Surge Time Minutes, NLL to LLL Feed Tanks or Vessels Feed Rate To fired heater/column10 To catalytic reactor10 To other process units10 Steam Drums for Unfired Generators Boiler Feed Water Rate Process Steam Drum HP Steam Drum Larger of 2 minutes or 1/3 volume of steam drum and boiler piping Deaerated Water Storage Boiler Feed Water Rate To steam generation 5 KO Drums Condensate Rate To pump To other process systems To sewer, disposal, storage Compressor KO Drums Condensate Rate Shutdown trip From HLL alarm to trip 2

Horizontal Drum Design 6 = R 2 cos -1 (1 – H/R) – (R – H) where R = D/2 H = height of liquid A sector – A triangle A segment =

Consider the same example 7

Knock-Out Drums  Knock-out drums are used for removing mist or liquid droplets in a gas stream.  The knock-out drum diameter is based on gas velocity and size of droplets to be removed.  The size of a knock-out drum must also satisfied the surge time for liquid.  For laminar flow Re < 1  Vertical tank, V actual = 0.75 V. Horizontal tank, V actual = 1.25 V.  Diameter: 8

Example  Gas at the rate of 3000 cfm and liquid at 25 cfm enter a drum in which entrainment is to be removed. Holdup of liquid in the drum is 10 min. The properties are those of air and water at atmospheric conditions. Find the size of the drum needed to remove droplets greater than 200  m diameter.  Note: In such a problem, both the liquid and gas flow rates should be given. So are the holdup time and diameter of droplets that should be knocked out. 9

Solution – Vertical tank  V = 3.87 ft/s (using Stokes equation)  V actual = 0.75V = (0.75)(3.87) = 2.9 ft/s  Based on gas flow rate, D d = 4.7 ft  5.0 ft  Now check if this also satisfies the surge time requirement, if not, you may increase D d  Area between HLL and LLL is known (250 ft 3 ), then L 1 = 12.7 ft.  L = L x 5 = 19 ft  Check L/D = 19/5 = 3.8, design is ok. 10

Solution - Horizontal tank  V actual = 1.25V = (1.25)(3.87) = 4.84 ft/s  Let  = the fraction of the cross-sectional area occupied by vapor 11 h/D  D(ft)L(ft)L/D

Pressure Vessel  Most reactors, separation columns, flash drums, heat exchangers, surge tanks, and other vessels in a chemical plant will need to be designed as pressure vessels.  A specialized subject in mechanical engineering. However, the chemical engineer will be responsible for developing and specifying the basic design information for a particular vessel, and needs to have a general appreciation of pressure vessel design.  Work effectively with the specialist designer  Mechanical constraints can cause significant cost thresholds in design  ASME code (The American Society of Mechanical Engineers) 12

Pressure Vessel Design Factors  Design pressure  10% margin between the normal operating pressure and the design pressure  Design temperature  50 F margin between the normal operating T and the design T  Materials  Carbon steel, low-alloy steel, stainless steel 304, 316, Nickel alloy, Monel alloy, Chromium, etc. on process stream corrosive issue.  Maximum Allowable Stress  Depends on the design temperature and the material of construction.  Welded Joint Efficiency  The strength of a welded joint will depend on the type of joint and the quality of the welding.  Others  Corrosion Allowance (1/8 inch), loads from earthquake/wind, minimal thickness, thickness level, etc. 13

ASME Wall Thickness  tP = wall thickness in inches to withstand the internal pressure  Pd = internal design gauge pressure in psig  Di = inside shell diameter in inches,  S = maximum allowable stress (psi)  E = fractional weld efficiency carbon steel <= 1.25”, E=0.85 carbon steel >1,25”, E=1.0 14

Minimal Wall Thickness & Level Vessel inside Diameter (ft)Min. wall thickness (inch) <=4¼ 4-65/ / / ½ 15 1/16 in. increments for 3/16 to 1/2 in. inclusive 1/8 in. increments for 5/8 to 2 in. inclusive 1/4 in. increments for 2-1/4 to 3 in. inclusive Corrosion Allowance (1/8 inch)

Example  An adiabatic reactor consists of a cylindrical vessel with elliptical heads, with an inside diameter of 6.5 ft (78 in.) and a tangent-to-tangent length of 40 ft (480 in.). Gas enters the reactor at a pressure of 420 psia and 650 F. Exit conditions are 410 psia and 850 F. The vessel will be oriented in a horizontal position. The vessel contains no internals and the gas is non-corrosive. Estimate the vessel thickness in inches. 16

Solution  Have 10% margin on design pressure: Pd = 110%*420 – 14.7 = psig  Design temperature = = 700 F  Select carbon steel, S = psi from table  Assume thickness > 1.25”, E = 1.0  tP = 447.3*78/(2*11500*1-1.2*447.3) = inch.  Which is greater than the assumed 1.25in. And also greater than the minimum value of 3/8 in. required for rigidity. Because the orientation of the vessel is horizontal, the vessel is not subject to wind load or earthquake considerations. Adding a corrosion allowance of 1/8 in. gives a total thickness of in. Specify the thickness to 1.75 inch. 17

Vessel Weight and Cost Estimation 18