HYDRAULICS
Pipeline transportation study and application of fluid mechanics principles to the flow of liquids or gases through pipes. It encompasses the analysis of factors such as pressure, flow rate, velocity, friction, and fluid properties to ensure efficient and safe transportation of substances through pipelines.
This dataset and calculations are for educational purposes only.
Modeled and Analyzed by the author Roderick Paulino
problem definition
A civil and geodetic engineer surveyed the land between the pump station and the storage tank to determine the optimal route for the pipeline. Soil testing was conducted to identify suitable locations for pipe supports that could provide adequate foundation strength. If the soil at any chosen pipe support location is unsuitable, piling is considered as an alternative to support the foundation. The provided sketch illustrates the proposed layout. Design the tank (30m-diameter, 20m-height), pump, pipe size, and bends necessary to transport crude oil from the pump station to the uphill storage tank at atmospheric pressure near the harbor. Check if the terrain can withstand the pipe and the tank to be erected. Q. E. D.
The Factor of Safety is too low for this setup and we need to increase it more than the current value of 0.865. The solution has converged after 100 iterations identifying the critical slip plane.
Slice 2 - Morgenstern-Price Method
Factor of Safety 0.865
Column Base Material Upper Layer
Phi Angle 20 °
C (Strength) 5 kPa
Pore Water Pressure -71.357 kPa
Pore Water Force -136.29 kN
Pore Air Pressure 0 kPa
Pore Air Force 0 kN
Phi B Angle 0 °
Slice Width 1.1152 m
Mid-Height 2.5844 m
Base Length 1.9099 m
Inclination -54.274 °
Anisotropic Strength Mod. 1
Applied Lambda 0.44463
Weight (incl. Vert. Seismic) 43.232 kN
Base Normal Force 35.649 kN
Base Normal Stress 18.665 kPa
Base Shear Res. Force 22.525 kN
Base Shear Res. Stress 11.794 kPa
Base Shear Mob. Force 26.038 kN
Base Shear Mob. Stress 13.633 kPa
Left Side Normal Force -2.9678 kN
Left Side Shear Force -0.15671 kN
Right Side Normal Force 10.808 kN
Right Side Shear Force 1.1333 kN
Horizontal Seismic Force 0 kN
X-Column Force 0 kN
Y-Column Force 0 kN
X-Column Force Location 0 m
Y-Column Force Location 0 m
Point Load 0 kN
Pullout Force 0 kN
Shear Force 0 kN
Surcharge Load 0 kN
Polygon Closure 0.64848 kN
Top Left Coordinate 10.884806, 18 m
Top Right Coordinate 12, 18 m
Bottom Left Coordinate 10.884806, 16.190826 m
Bottom Right Coordinate 12, 14.640329 m
The proposed solution is to excavate and adjust the slope of the hill to achieve a more favorable grade. While a sensitivity analysis could be used, in this case, the entire design is redrawn and recalculated to reach a factor of safety of 1.330, improving from the current safety factor of 0.865 to stabilize the hill terrain. The green color indicates with bounder white line is the critical slip while red color is where the factor of safety was calculated.
Slope Stability Material: Mohr-Coulomb
Slope Stability Model: Morgentern-Price
tank settlement
Assuming the proposed revision is approved by the regulatory authorities, we must determine the settlement of the tank to be constructed by assessing soil displacement due to the tank’s full capacity weight. These findings will inform the tank's foundation design, aimed at minimizing vertical displacement. To achieve this, the tank will be positioned 25 meters away from the hill's curve as shown below. Q. E. D.
The graph above was generated based on soil displacement values relative to the height of the tank to be constructed, showing that displacement values tend to increase as the tank height Y(m) increases.
barrier WALLS
The objective of this study is to evaluate the localized instability of rocky elements above the hill subjected to seismic activity and water pressures within intersurface fractures. The analysis will determine sliding and overturning safety factors, which will be used to assess the stability of the rock mass. If necessary, the findings will inform the design of stabilization measures, including the use of active or passive anchors and nails. After studying of the rock formation, there is a possibility that a boulder could reach the pump house. So, a barrier is needed to be designed based on the predicted projectile of incoming boulders as follows.
This is the designed barrier (3.75 meters) to stop the boulder and can absorbed the load based on the predicted final energy of the boulder at impact. Note the reaction forces of the boulder depends on the different surfaces along the slope of the hill and needs to be determined in the actual site geographical survey.
RETAINING WALLS
Embankment protection, erosion control, reinforced steep slopes, retaining walls. The drawings below are automatically generated based on prompts through GUI using different kind of geographical models.
Computed Retaining Walls and resultant reinforcements based on soil investigation and loads using a double pile row. Q.E.D.
Thermodynamic properties
Before we can transport the crude oil, we need to characterize the thermodynamic properties of the crude oil using discrete hypothetical components assuming that the crude oil data for light ends and atmospheric TBP distillation are available. After blending using Peng-Robinson Property Package Model, the following charts are now available for sizing a pump in a suction lift pump system.
This boiling temperature will be used to check for cavitation of the crude oil in the pump later for transient analysis. Note that at this blend, our critical pressure is at approximately 524.32 kPa when it starts to vaporize from the pure liquid volume state.
Since the computed bubble point pressure 57.94 kPa is lower than the vapor pressure in a mixture, it becomes the limiting factor for pump design. If the pressure in the suction line drops below the bubble point pressure, vapor bubbles can form even before the liquid reaches the vapor pressure of any individual component.
In our mixtures, the bubble point pressure is the pressure at which vapor begins to form, meaning that it indicates the point where cavitation is likely to start. Because the bubble point pressure can be significantly lower (10 times lower than the vapor pressure), it is more conservative to use in pump design to ensure that vapor formation (and cavitation) is avoided.
pump sizing
We established a baseline for the volumetric flow, beginning at the supply manifold of an oil tank with a pressure of 1000 kPag. The system delivers to a storage tank located at an elevation of 200 meters, connected by 1500 meters of piping. The flow rate required is 128 m³/h, with a flow velocity of 1.102 m/s through an 8-inch Schedule 40 stainless steel pipe. The pump selected for this configuration operates with an efficiency of 62.12% and requires 88.729 kW of power from a demo catalogue. This can be used to find the suitable pump using manufacturer's catalog.
Performance Curves for Pumps in series configuration.
Performance curve for Pumps in parallel configuration.
With this information, we can compute the ideal power of pump to move this mixture per meter length of pipe using 8-in in diameter using this python script.
# Given values
density = 866.2 # kg/m^3
# Update viscosity value in cP to Pa·s viscosity_cp = 8.038 # viscosity in centipoise (cP) viscosity_pa_s = viscosity_cp * 1e-3 # convert to Pa·s
viscosity = 8.038e-3 # Pa·s
Q_m3_per_hr = 128 # m^3/hr
diameter = 0.2 # m (200mm)
g = 9.81 # m/s^2
hf_per_meter = 1 # frictional loss per meter of pipe
# Step 1: Convert volumetric flow rate from m³/hr to m³/s
Q_m3_per_s = Q_m3_per_hr / 3600 # m^3/s
# Step 2: Calculate the flow velocity (v)
area = math.pi * (diameter / 2) ** 2 # Cross-sectional area of the pipe
velocity = Q_m3_per_s / area # m/s
# Step 3: Calculate the Reynolds number (Re)
Re = (density * velocity * diameter) / viscosity
# Step 4: Estimate the Darcy friction factor (f) using Blasius equation (valid for turbulent flow)
f = 0.079 * Re ** (-0.25)
# Step 5: Calculate frictional head loss (hf) per meter of pipe
hf = f * (velocity ** 2) / (2 * g)
# Step 6: Calculate the power (P) required per meter of pipe
power_per_meter = density * g * Q_m3_per_s * hf # Power in watts
# Convert power from watts to kilowatts
power_per_meter_kW = power_per_meter / 1000 # kW
Power required per meter of pipe: 0.000125 kW or 0.125W
pipe sizing
Here are the baseline pipe sizes based on our stream using Stainless Steel 40S. Note of the velocity should not exceed 9 ft/s as the criteria for sizing to prevent erosion as per API RP 14E specifications. However, your Design Criteria from the client should be followed and adjustment is needed to determine the pipe sizes calculations accordingly. Computed pressure drop for suction is 115.1 kPa while 379.0 kPa for discharge using Tulsa Unified Model (2-Phase) as a baseline when selecting pump specifications.
pipe sizing VARIATIONS
Here is the sample of computed pressure for suction line 50m in length from oil tanker manifold to the inlet of the pump assuming adiabatic condition using 8" of pipe where mixture velocity was reduced from the baseline calculation of 4"-dia using larger pipe diameter. Note that this can vary based on available data from manufacturer's catalog for proper selection of pump.
ADDITIONAL sizing COMPUTATIONS
Here is an example of pumping a SAE SW-40 synthetic oil to simulate pipeline transportation from pressurized storage tank to atmospheric tank using stainless steel pipe S40.
STORAGE TANK DESIGN
For the storage tank, we will design a supported cone roof type storage tank with a diameter of 30m and height of 20m per API 650.
API-620/650/653 General Tank Data:
------------------------------------------------------------------------------
API Design Code .................................... 650
Design Method (V, O, or A) ......................... One Foot
Design Temperature ................................. 25.000 C
Design Pressure at Top ............................. 0.50000 KPa
Shell Material ..................................... A-573,70
Internal Pressure Combination Factor ...........[Fp] 0.30000
Shell Design Stress .....................[Sd or Sts] 0.19305E+06 KPa
Shell Hydro Test Stress ........................[St] 0.20684E+06 KPa
Tank Nominal Diameter ...........................[D] 30.000 m.
Tank Shell Height .............................[HTK] 20.000 m.
Design Liquid Level .............................[H] 18.000 m.
Liquid Specific Gravity .........................[G] 1.2000
Weight of Attachments/Structures.................... 0.00000 N.
Distance down to Top Wind Girder ................... 0.00000 m.
Joint Efficiency (App A or 653) .................[E] 1.0000
Wind Velocity ...................................... 30.000 M./sec.
Insulation Thickness ............................... 0.00000 mm.
Insulation Density ................................. 0.00000 kg./cu.cm
Include Annular Base Plate Details ................. Yes
Include Wind Moment in Appendix F_4_2 Calculations . Yes
Minimum Yield Strength of Bottom Plate: ............ 0.26199E+06 KPa
Number of Shell Courses ............................ 7
Note that the plate thickness here are not standard. The plate thickness should be changed of standard plate thickness in the market based on the MTC EN 10204.
Final calculated thickness of the plates for fabrication as designed.
Note that anchors are not needed for this tank size .
CALCULATIONS
Determine Maximum allowed Fluid Heights:
For Course # 1, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.6693 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 62.026 ft. [18.905 m.]
For Course # 1, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.6693 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 79.462 ft. [24.220 m.]
For Course # 2, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.5288 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 49.214 ft. [15.000 m.]
For Course # 2, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.5288 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 62.989 ft. [19.199 m.]
For Course # 3, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.4208 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 39.371 ft. [12.000 m.]
For Course # 3, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.4208 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 50.334 ft. [15.342 m.]
For Course # 4, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.3150 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 29.718 ft. [9.058 m.]
For Course # 4, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.3150 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 37.923 ft. [11.559 m.]
For Course # 5, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 22.538 ft. [6.870 m.]
For Course # 5, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 28.692 ft. [8.745 m.]
For Course # 6, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 22.538 ft. [6.870 m.]
For Course # 6, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 28.692 ft. [8.745 m.]
For Course # 7, Operating Case
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )28000.00/(2.6 * 98.4 * 1.200 ) + 12.000/12
= 22.538 ft. [6.870 m.]
For Course # 7, Hydro Case, not corroded
Allowed fluid height above this course based on the 1ft method
= (Td - ca)Sd/(2.6 * D * G ) + X/12
= (0.2362 - 0.0000 )30000.00/(2.6 * 98.4 * 1.000 ) + 12.000/12
= 28.692 ft. [8.745 m.]
Appendix M Yield Stress reduction factor, M.3.6 [Mfact]:
= Yield Reduction Factor * Fy / Fyamb
= 1.000 * 42000.0/30000.0
= 1.000 Must be less than or equal to 1.0
Note:
Computing the roof weight not including the corrosion allowance
per user request.
Roof Analysis:
Total Design External Pressure Loading per V.7 [Pr]:
= max[ DL + (Lr or S) + Fpe*Pe, DL + Pe + 0.4(Lr or S)]
= max[19.294 + max[26.232, 0.000 ] + 0.400 * 0.000,
19.294 + 0.000 + 0.4 * max[26.232, 0.000 ]]
= max[ 45.526, 29.787 ]
= 45.526 psf [2179.786 N/M2]
Appendix M Yield Stress reduction factor, M.3.6 [Mfact]:
= Yield Reduction Factor * Fy / Fyamb
= 1.000 * 38000.0/30000.0
= 1.000 Must be less than or equal to 1.0
Appendix M Yield Stress reduction factor, M.3.6 [Mfact]:
= Yield Reduction Factor * Fy / Fyamb
= 1.000 * 38000.0/30000.0
= 1.000 Must be less than or equal to 1.0
Appendix M Yield Stress reduction factor, M.3.6 [Mfact]:
= Yield Reduction Factor * Fy / Fyamb
= 1.000 * 38000.0/30000.0
= 1.000 Must be less than or equal to 1.0
Appendix M Yield Stress reduction factor, M.3.6 [Mfact]:
= Yield Reduction Factor * Fy / Fyamb
= 1.000 * 38000.0/30000.0
= 1.000 Must be less than or equal to 1.0
Wind Load Calculations:
Wind velocity factor 5.2.1 [Vfact]:
= ( Wind Velocity / 120 )^2
= ( 67.1/120 )^2
= 0.313
Vertical projected wind area of tank [VArea]:
= Tank Height * Insulated Tank Diameter
= 65.62 * 98.43
= 6458.346 ft^2 [600.000 m.^2]
Velocity pressure per 5.9.7.1 [p]:
= 0.00256 * Kz * Kzt * kd * V^2 * I * G + 5.0
= 0.00256 * 1.040 * 1.000 * 0.950 * 67.108^2 * 1.000 * 0.850 + 5.0
= 14.682 psf [702.976 N/M2]
Wind moment on shell [Moment]:
= Wind Pressure * Wind Area * Tank Height/2
= 14.682 * 6458.3 * 65.62/2
= 0.31109E+07 ft.lb. [0.42177E+07 N.m.]
Vertical projected wind area of Cone roof:
= Roof Height * Roof triangular base
= 5.17 * 49.21
= 254.550 ft^2 [23.648 m.^2]
Area Exposed to Wind:
= Wind area of tank + Wind area of roof
= 6458.346 + 254.550
= 6712.896 ft.^2 [623.648 m.^2]
Wind moment at the base including the cone roof [Total Moment]:
= Moment + ( Wind P * Roof Area( Tank Height + Roof Height/3))
= 3110941 + ( 14.682 * 254.5 ( 65.6 + 5.17/3))
= 0.33626E+07 ft.lb. [0.45589E+07 N.m.]
Wind shear force acting on the tank:
= Wind Pressure * (tank area + roof area)
= 14.682 * 6712.9
= 0.98559E+05 lbf. [0.43839E+06 N.]
Moment about shell-to-bottom joint [MDL]:
= ( Shell Weight + Attachement Weight + Framing Weight * % )D/2
= ( 318648.5 + 0.0 + 28725.2 * 0.25 )98.4/2
= 16040426.0 ft.lb [21746968.0 N.m.]
Resisting Weight of Fluid [WL]:
= min( 5.6 * tb * sqrt( Fby * H * min( 0.7, G )), 0.45 * H * D )
= min(5.6*0.3937*sqrt(38000.0*59.06*min(0.7,1.200)),0.45*59.1*98.43)
= min( 2763.3, 2615.6 )
= 2615.6 lbf/ft [38.2 N./mm.]
Moment from Fluid Weight [MF]:
= WL * ( PI * D ) * ( D / 2 )
= 2615.6 * ( 3.1416 * 98.43 ) * ( 98.43/2 )
= 39802356.0 ft.lb [53962444.0 N.m.]
Moment from Design Internal Pressure [MPI]:
= Design Pressure * pi( Diameter in inches / 2 )^2 * ( D / 2 )
= 0.0725 * 3.1416 ( 1181.10/2 )^2 * ( 98.43/2 )
= 3910254.2 ft.lb [5301366.5 N.m.]
Design Pressure per F.4.1 [P]:
= (0.962 * A * Fy * Mfact * tan(theta))/(D^2) + 0.245 * DLR/D^2
= (0.962*12.734*42000.0*1.000*tan(6.00))/(98.4^2)+0.245*176331/98.4^2
= 10.042 in. of water [255.056 mm. of water]
= 0.363 psi [2.500 KPa]
Note: Design Pressure was less than maximum computed per F.4.1
Moment about shell-to-bottom joint [MWS]:
= Pws * D * H^2/2
= 18 * Vfact * (D + Insulation) * H^2/2
= 18 * 0.3127 (98.4 + 0.0 ) * 65.6^2/2
= 1192795.5 ft.lb [1617144.4 N.m.]
Moment about shell-to-bottom joint [MW]:
= MWS + Wind Uplift * VArea * D / 2
= 1192795 + 0.0652 * 1095632 * 98.4/2
= 4705868.0 ft.lb [6380027.5 N.m.]
Moment about shell-to-bottom joint [MDLR]:
= ( Roof Weight )D/2
= ( 147606.1 )98.4/2
= 7264080.5 ft.lb [9848350.0 N.m.]
Determine required thickness per the selected Method:
Shell Course Thickness per the One Foot Method:
Required Thickness for Course # 1
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 59.06 - 1 )1.200/28000.0
= 0.637 in. [16.172 mm.]
Minimum thickness per 5.6.1.1 : = 0.637 in. [16.172 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.63671 in. [16.17250 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 59.06 - 1 )/30000.0
= 0.49522 in. [12.57861 mm.]
Minimum thickness per 5.6.1.1 : = 0.495 in. [12.579 mm.]
Final Selected Thickness including other Code allowances:
= 0.49522 in. [12.57861 mm.]
Required Thickness for Course # 2
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 49.21 - 1 )1.200/28000.0
= 0.529 in. [13.431 mm.]
Minimum thickness per 5.6.1.1 : = 0.529 in. [13.431 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.52877 in. [13.43065 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 49.21 - 1 )/30000.0
= 0.41126 in. [10.44606 mm.]
Minimum thickness per 5.6.1.1 : = 0.411 in. [10.446 mm.]
Final Selected Thickness including other Code allowances:
= 0.41126 in. [10.44606 mm.]
Required Thickness for Course # 3
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 39.37 - 1 )1.200/28000.0
= 0.421 in. [10.689 mm.]
Minimum thickness per 5.6.1.1 : = 0.421 in. [10.689 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.42082 in. [10.68881 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 39.37 - 1 )/30000.0
= 0.32730 in. [8.31352 mm.]
Minimum thickness per 5.6.1.1 : = 0.327 in. [8.314 mm.]
Final Selected Thickness including other Code allowances:
= 0.32730 in. [8.31352 mm.]
Required Thickness for Course # 4
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 29.53 - 1 )1.200/28000.0
= 0.313 in. [7.947 mm.]
Minimum thickness per 5.6.1.1 : = 0.313 in. [7.947 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.31287 in. [7.94696 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 29.53 - 1 )/30000.0
= 0.24335 in. [6.18097 mm.]
Minimum thickness per 5.6.1.1 : = 0.243 in. [6.181 mm.]
Final Selected Thickness including other Code allowances:
= 0.24335 in. [6.18097 mm.]
Required Thickness for Course # 5
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 19.69 - 1 )1.200/28000.0
= 0.205 in. [5.205 mm.]
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.23622 in. [6.00000 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 19.69 - 1 )/30000.0
= 0.15939 in. [4.04842 mm.]
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including other Code allowances:
= 0.23622 in. [6.00000 mm.]
Required Thickness for Course # 6
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 2.6D( H - 1 )G/Sd
= 2.6 * 98.425 ( 9.84 - 1 )1.200/28000.0
= 0.097 in. [2.463 mm.]
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.23622 in. [6.00000 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 2.6D( H - 1 )/St
= 2.6 * 98.425 ( 9.84 - 1 )/30000.0
= 0.07543 in. [1.91588 mm.]
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including other Code allowances:
= 0.23622 in. [6.00000 mm.]
Required Thickness for Course # 7
Required thickness based on the 1ft method per 5.6.3.2, Operating [td]:
= 0.000 in.
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including corrosion and other allowances:
= 0.23622 in. [6.00000 mm.]
Required thickness per 5.6.3.2, Hydrotest [tt]:
= 0.00000 in.
Minimum thickness per 5.6.1.1 : = 0.236 in. [6.000 mm.]
Final Selected Thickness including other Code allowances:
= 0.23622 in. [6.00000 mm.]
Storage tank with spiral ladder with platforms showing the computed girders, rafters, girder rings, columns for supported cone roof.
After construction, the final tank settlement needs to be re-assessed per API 653 specifications. The above chart is the final tank displacement curve passing for this code.
3d model
Accurately modeling the pipeline to make sure that all alignments are precise for fabrication.
The figures above are 3D model to accurately measure the fabricated elbow to be machined. This pipeline segment is a rolling offset starting from the pump discharge line going uphill at different rise of elevation. To make the precision for curve angle we created axis at different curvature to fit the elbow and the pipes making sure of the grading elevation per topographical survey.
Checking the dihedral quality of each pipe joints at each bend to make sure that no overlapping component or segments are present even at its curvature.
stress analysis
Although we have captured the fluid dynamics of our crude oil, we are yet to investigate the proper sizing of pipe based on the static stress of our chosen pipe thickness. Total deformation of pipe using ASTM 106 Grade B and Stainless Steel material for comparison. The left end node is fixed while right end node is free at Y, 5mm for Z and X axis. However note that the X axis has a total deformation of 6.574mm. At this point of design, we need to identify the stress points to be supported.
Above is the total Deformation for ASTM 106 for reference in using the Stainless Steel (207 Mpa compressive and tensile Yield Strength). STATIC.
Above is the total Deformation for Stainless Steel. Also inset is the factor of safety at the pump discharge line with a total of 182.34 kN forces acting axially on pipe. Appropriate line stop support is needed . However, this load is just for the load of the pipe and does not include the dynamics of the fluid which needs to be added in the solution.
To add the hydrostatic pressure of the crude oil (add the fluid weight), there is a slight decrease of the safety factor but not significant to affect the integrity of the pipe material for this 18 meters length in total.
PIPE SUPPORTs
To create a dynamic analysis for pipe support design, we need to find the forces acting on the pipe due to fluid dynamics and vise versa. This will facilitate the two-way interaction between the pipe structure and the fluid. Below is the illustration diagram of probable location of pipe supports based on the total stress above from modal analysis as predicted (sensitivity analysis not shown here) instead of monte-carlo algorithm.
In this study, we will analyze the effects of fluid dynamics on the pipeline and its supports to assess the adequacy of restraints in managing potential displacements or movements. We will also investigate how stress propagates through the system during pipeline operation to ensure proper selection of pipe supports. Additionally, the weight of the fluid due to gravity has been included in our fluid dynamics model to mimic the actual field conditions. The illustration above includes the piping after the isolation valve where we assumed that the valves are properly restrained in the pump side.
Based on the deformation, we can generalized that the piping system is structurally sound to support the weight, fluid and forces acting on it. The restraints were able to distribute uniformly the safety factor which increases the integrity of the pipe.
Looking at the pipe support 1, the stresses induced by the piping system are minimal based on the predicted support location of the pipe.
In this deformation contour, we stabilized the movement of the pipe by analyzing the deformation in all directions. As shown in the deformation graph (inset), we can visually confirm that the movement has been successfully restrained, resulting in reduced stress on the pipe.
fluid dynamics
Showing quantitative post processing results with a computed pressure drop of 129.4205 Pa at the outlet. This represents the pressure associated with the motion of a fluid. It is a measure of the kinetic energy per unit volume of a fluid. This dynamic pressure helps determine the force exerted by the flow on surfaces. Theoretically, the dynamic pressure is approximately 121.03 Pa or 0.121 kPa after 1 meter of pipe length which is within the design criteria of selected pipe material.Q.E.D.
