PRESSURE_VESSELS final & Introduction 12

RahulRakhade 21 views 45 slides Sep 25, 2024
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About This Presentation

Statistical analysis is the process of collecting, reviewing, and interpreting data to discover patterns, trends, relationships, or insights. It is a fundamental tool in research, business, and data science for making informed decisions based on data.

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PRESSURE VESSELS

PRESSURE VESSELS Cylinder: Cylinder is a mechanical device used for storing, receiving and processing the fluid . Cylinder may be Pressure vessel Storage vessel Pipe Engine cylinder Cylindrical Pressure Vessel Thin Cylinder :- If The ratio of D i / t > 15, …….Boiler shells, pipes, tubes etc. Thick cylinder :- If the ratio D i / t ≤ 15, ……… high pressure pipes, gun barrels etc.

Stresses in Thin Cylinder Circumferential or Hoop Stress , σ t Longitudinal Stress, σ l Radial Stress, σ r Radial stress is equal to internal pressure at inner surface and zero at outer surface. Forces due to internal pressure are balanced by shear stresses in wall.

Thickness of Thin Cylinder It is observed that circumferential stress is twice the longitudinal stress By Considering longitudinal and circumferential joints equation is modified as, η c = Efficiency of Circumferential Joint η l = Efficiency of Longitudinal Joint

Stresses in Thin Spherical Vessels Circumferential or Hoop Stress : Thickness of Pressure Vessel The Capacity of the Spherical Vessel,

Thick Cylinders Considering Equilibrium of vertical forces acting on half portion of ring, It is assumed that longitudinal stress is constant over the cylinder wall thickness. The longitudinal strain Ɛ l ……..a Neglect,

Thick Cylinders The modulus of elasticity E, Poisson's ratio μ , longitudinal strain are constant. Therefore right hand side is constant and is denoted by (-2C 1 ). Subtract equation ( a) from ( b) From equation (c) and (d), ………………………….b We know that……… …………………….c ………………………d By Integrating, or …………e Multiplying both sides by r,

Substitute the above value in equation ( b) The C 1 and C 2 can find by using following boundary condition, At, ; ; We get, …………….f

Thick Cylinders By substituting C 1 and C 2 , in equation (e) and (f) The negative sign in the expression for σ r denotes compressive stress,

Thick Cylinders Putting Boundary following Condition, At inner surface,

Thick Cylinders The longitudinal stress, by considering forces in axial direction, At the outer surface,

Thick Cylinders Sr. NO. Principal Stress Subjected to Internal Pressure (Pi) Subjected to External Pressure (Po) 1. Circumferential Stress ( σ max) 2. Radial Stress ( σ min) 3. Longitudinal Stress

Autofrettage or Pre- Stressing of Thick Cylinders Compound Cylinders Wire Wound Cylinders Tape Wound Cylinders

Compound Cylinders

Stress Distribution in Compound Cylinder

Wire Wound Cylinders Tape Wound Cylinders

Residual Stresses Due to Shrink Fit Sr. No. Parameter Residual Stress at Inner Surface Residual Stress at Outer Surface Jacket (Outer Cylinder) 1. 2. (Inner Cylinder) 1. 2.

1. Increase in Inner Diameter of Jacket (Outer Cylinder) 2. Decrease in Outer Diameter of Inner Cylinder 3. Total deformation (Neglecting the negative sign) : The negative sign indicate contraction.

Unfired Pressure Vessel Class 1 Class 2 Max. shell thickness up to 38 mm Class 3 Max. shell thickness up to 16 mm Classification Pressure vessel are container or pipe line for storing, receiving or carrying the fluids under a pressure. Unfired pressure vessel are: storage vessel, reaction vessel, heat exchanger, evaporator Other: Steam boiler, nuclear pressure vessel

Types of Welded Joints in Used Pressure Vessel Weld Joint Efficiency = Ratio of the strength of the weld joint to the strength of the plates

Categories of Welded Joints in Unfired Pressure Vessel Category A Category B Category C Category D

Materials Used for Unfired Pressure Vessel Cast Irons Plain Carbon Steels Alloy Steels Aluminum Steels Copper and Copper Alloys Nickel and Nickel Alloys

Selection of Design Parameters for Unfired Pressure Vessels Maximum Working Pressure ( P max ) Design Pressure (P i ) P i = 1.05 P max Hydrostatic Test Pressure ( P b ) P b = 1.3 P i Allowable Stress, Corrosion Allowance (CA) As per IS code As per ASME code As per DIN code

Design of Unfired Pressure Vessel 1. Pressure Vessel Shell Where, Considering Equilibrium of vertical forces acting on half portion of the cylinder, ….a As, Adding Corrosion Allowance, Mean Diameter, Put in above equation (a), We get,

Spherical Pressure Vessel Shell

Pressure Vessel Under Combined Loading Other loads on Pressure Vessel: Wt. of the Vessel with its content Wind load Load due to offset piping's Stresses induced in the cylindrical vessel due to Combined Loading 1. Stress in circumferential or Tangential Direction ( σ t ) 2. Stress in longitudinal or Axial Direction Due to Internal Pressure Where, D = Mean Diameter, mm

Stress in Longitudinal direction due to weight of the vessel its content in vertical pressure vessels Bending Stress due to total wt. Pressure Vessel Under Combined Loading Where, D = Mean diameter, mm

Bending moment acting on horizontal pressure vessel shell

Bending moment acting on Vertical pressure vessel shell P 1 = Wind Pressure up to shell Ht. 20 m = 4.623 x 10 -8 .V w 2 N/mm 2 P 2 = Wind Pressure after shell Ht. 20 m = 4.623 x 10 -8 .V w 2 N/mm 2 Where, V w 2 = Wind Speed in km/hr M = Maximum of M w and M s Total Stress in longitudinal direction = Force due to wind load acting on lower part of shell Force due to wind load acting on upper part of shell

3. Torsional Shear Stress due to offset piping : 4. Resultant Stress : Or We can write, According to distortion energy theory,

Design of End Closures Flat Head Formed head Plain Formed Head Torispherical Dished Head Semi-Elliptical Dished Head Hemispherical Head Conical Head

Plain Formed Head ii. Torispherical Dished Head S f = 3t or 20 mm whichever is larger S f = 3t or 20 mm whichever is larger (Straight Flange Length) Knuckle Radius Crown Radius (L)

Semi-Elliptical Dished Head Hemispherical Head S f = 3t or 20 mm whichever is larger

Conical Head S f = 3 t or 20 mm whichever is larger

Design of nozzles and openings in unfired pressure vessels In the portion of the shell as excess thickness In the portion of the nozzle outside the vessel as excess thickness In the portion of the nozzle inside the vessel as excess thickness In the reinforcement pad or compensation ring Area for the Area Method of Compensation The area of the material removed is compensated by providing additional area of material

Loss of Area in the opening, The inner diameter of the nozzle in corroded condition is, The required thickness of the cylindrical shell is , ƞ = Weld joint eff. in nozzle The limiting dimension X parallel to the wall of the shell is given by, or ......Whichever is Maximum The limiting Dimensions h 1 and h 1 are given by, h 1 = 2.5 (t – CA) h 1 = 2.5 ( t n – CA) Whichever is smaller Where, t = Total thickness of the wall of cylindrical shell t n = Total thickness of nozzle wall t r = Required thickness of cy. shell

Estimation of Compensation : The total A 1 of excess thickness in the vessel shell, which is available for reinforcement (A 1 ) : The total A 2 of excess thickness in the nozzle wall is given by, The thickness required for the nozzle wall to be able to withstand the pressure, The area A 3 of the inside extension of the nozzle is given by, The total area available for reinforcement is ( A 1 + A 2 + A 3 ) The When above condition is not satisfied , then area required for compensation is ( A 4 ):

Estimation of Compensation Dimensions of reinforcing pad : d po = Outer diameter of the reinforcing pad, mm d pi = Inner diameter of the reinforcing pad, mm t p = Thickness of the reinforcing pad, mm Where,

A pressure vessel consists of cylindrical shell with an inner diameter of 1500 mm and thickness of 25 mm. It is provided with a nozzle of inner diameter 250 mm and thickness 15 mm. The yield strength of the material for the shell and nozzle 220 N/mm 2 and design pressure is 2.5 Mpa . The extension of the nozzle inside the vessel is 15 mm. The corrosion allowance is 2.5 mm. while the welding joint efficiency is 0.85. Neglecting the area of welds determine whether or not a reinforcing pad is required for the opening, if so, determine the dimensions of pad made from a plate 15 mm thickness. Given: For Cylinder Shell Di = 1500 mm, t = 25 mm For Nozzle : di = 250 mm, tn = 15 mm, h 2 = 15 mm, Pi = 2.5 Mpa , Syt = 250 N/mm 2 , CA = 2.5 mm; η = 0.85 For pad, t = 15 mm Loss of Area in the opening,

The limiting dimension X ( higher of the following two values ) parallel to the wall of the shell is given by, or X = 255 mm Selected X = 157.5 mm or The limiting Dimensions h 1 and h 1 are given by (Lower of following two values is selected), h 1 = 2.5 (t – CA ) = 56.25 mm h 1 = 2.5 ( t n – CA ) = 31.25 mm Therefore, h 1 = 31.25 mm h 2 = 15 mm

The area available for reinforcement: The total A 1 of excess thickness in the vessel shell, which is available for reinforcement (A 1 ) : The total A 2 of excess thickness in the nozzle wall is given by, The thickness required for the nozzle wall to be able to withstand the pressure, The area A 3 of the inside extension of the nozzle is given by, The total area available for reinforcement is ( A 1 + A 2 + A 3 )

The When above condition is not satisfied , then area required for compensation is ( A 4 ): Dimensions of reinforcing pad : d po = Outer diameter of the reinforcing pad = 355 mm d pi = Inner diameter of the reinforcing pad = 280 mm t p = Thickness of the reinforcing pad = 15 mm

Methods of Reinforcement Balanced Reinforcement Unbalanced Reinforcement

Supports for Vertical Pressure Vessels Bracket or Lug Support Skirt Support

Supports for Horizontal Pressure Vessels Saddle Support Plate Type Ring Type
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