Calculation of Total Pressure Required For Transport
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2016© UPES
Ques.1.
 A gas pipeline, NPS 16 with 0.250 in. wall thickness, 50 mi long,
transports natural gas (specific gravity=0.6 and viscosity =0.000008
lb/ft-s) at a flow rate of 100 MMSCFD at an inlet temperature of
60°F.
 Assuming isothermal flow, calculate the inlet pressure required if the
required delivery pressure at the pipeline terminus is 870 psig.
 The base pressure and base temperature are 14.7 psia and 60°F,
respectively.
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 Use the Colebrook equation with pipe roughness of 0.0007 in.
 Case A—Consider horizontal pipeline with no elevation.
 Case B—Consider elevation changes as follows: inlet elevation of 100
ft and elevation at delivery point of 450 ft, with elevation at the
midpoint of 250 ft. For easy calculation, assume a compressibility
factor of 0.8666 throughout for this case.
 As an initial approximation use average pressure as 110 % times
the output pressure in psia.
 Use Generalized flow equations and Colebrook White friction factor.
 Solve until the inlet pressure is within 
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2psia.
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Sol.
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 To calculate the compressibility factor Z , the average pressure is required.
Since the inlet pressure is unknown, we will calculate an approximate value
of Z using a value of 110% of the delivery pressure for the average pressure.
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 Solving Gives: Inlet pressure as 1008.34 psia or 993.64 psig.
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Pipe Analysis
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BARLOW’S EQUATION
 When a circular pipe is subject to internal pressure, the pipe
material at any point has two stress components that act at right
angles to each other.
1. Circumferential stress (Hoop stress) : acts in
circumferential direction.
2.Axial stress: acts in a direction parallel to the pipe axis.
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 Hoop Stress:
P*D
Sh 
2t
 Sh=Hoop or circumferential Stress in Pipeline, psi
 P=Internal Pressure, psi.
 D=pipe outside diameter, in.
 t= Pipe Wall Thickness
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 Axial Stress
P*D
Sa 
4t
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Calculation of Internal Design Pressure
 The following form of Barlow’s equation is used in design
pipelines for gas transportation systems.
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PMAOP ,i
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2  ti  S  E  F  T

D
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1.Specified Minimum yield stress
 It is an indication of the minimum stress a pipe may experience
that causes plastic (permanent) deformation.
 Steel pipes are generally used in gas pipeline systems. They
generally conform to API 5LX specifications.
 These are manufactured in grades ranging from X42 to X90
with SMYS
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Joint Factor
 Pipelines generally use bends for changing the direction of gas
flow.
 These bends are welded in pipeline, giving joints.
 Joint efficiency ‘E’ in a pipeline represents a generic level of
confidence in the overall strength of the weld seam considering
the methods that were used to produce the seam.
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Calculation of Seam Joint Factor
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Calculation of Design Factor (F)
 Design factor of a pipeline is obtained by making the use of
class location.
 The class location unit (CLU) is defined as an area that extends
220 yards (201 meters) on either side of the center line of a 1-mi
(1.6km / 5280 ft.) section of pipe, as indicated in Figure.
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CLASS LOCATION
(The following definitions of class 1 through
class 4 are taken from DOT 49 CFR, Part 192).
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 Class 1
Offshore gas pipelines are Class 1 locations. For onshore
pipelines, any class location unit that has 10 or fewer buildings
intended for human occupancy is termed Class 1.
 Class 2
This is any class location unit that has more than 10 but fewer
than 46 buildings intended for human occupancy.
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 Class 3
This is any class location unit that has 46 or more buildings
intended for human occupancy
 Class 4
This is any class location unit where buildings with four or
more stories above ground exist.
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Design Factor
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Temperature Deration Factors
 When pipelines are to be operated at temperatures greater than 121°C, the
maximum allowable operating pressure should be reduced by applying a
thermal derating factor. This is done to make pipeline to work under safe
operations.
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 Example
An offshore gas pipeline is constructed of API 5L, X65 steel,
NPS 16, 0.250 in. wall thickness. Calculate the MAOP of this
pipeline for class 1 through class 4 locations. Assume Seamless
Joint.
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MAINLINE VALVES
 Mainline valves are installed in gas pipelines so that portions of
the pipeline can be isolated for testing leakage and maintenance.
 ASME B31.8 recommends the following spacing for mainline
valves.
.
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 It can be seen from the preceding that the valve spacing is
shorter as the pipeline traverses high-population areas.
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HYDROSTATIC TEST PRESSURE
 When a pipeline is designed to operate at a certain MOP, it must
be tested to ensure that it is structurally sound and can withstand
safely the internal pressure before being put into service.
 Generally, gas pipelines are hydro tested with water by filling
the test section of the pipe with water and pumping the pressure
up to a value higher than the MAOP and holding it at this test
pressure for a period of 4 to 8 hours.
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 The magnitude of the test pressure is usually 125% of the max.
operating pressure.
 Thus, a pipeline designed to operate continuously at 1000 psig
will be hydro tested to a minimum pressure of 1250 psig.
 If the pipeline is designed to be below ground, the test pressure
is held constant for a period of 8 hour.
 Above-ground pipelines are tested for a period of 4 hours.
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Ques.
 Consider a pipeline NPS 24, with 0.375 in. wall thickness,
constructed of API 5L X65 pipe. For a cross country pipeline,
using a temperature deration factor of 1.00, and class 1 location,
calculate MAOP, and the hydrostatic test pressure. Assume
seamless joint.
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 So ,
 Hydrostatic pressure = 1.25 *1462.5 =
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1828.125psig
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DETERMINING PIPE TONNAGE
 Frequently in pipeline design, we are interested in knowing the
amount of pipe used so that we can determine the total cost of
pipe.
 A convenient formula for calculating the weight per unit length
of pipe used by pipe vendors is given in Equation:
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w  10.68* t *( D  t )
 w=pipe weight, lb/ft.
 D=pipe outside diameter, in.
 T= pipe wall thickness, in.
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 The constant 10.68 in the above Eqn. includes the density of
steel and, therefore,
the equation is only applicable to steel pipe.
 For other pipe material, we can ratio the densities to obtain the
pipe weight for non steel pipe.
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 In SI units, the equation:
w  0.0246* t *( D  t )
 w=pipe weight, kg/m.
 D=pipe outside diameter, mm.
 T= pipe wall thickness, mm.
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 Example 5
Calculate the total amount of steel pipe in a 10 mi pipeline,
NPS 20, 0.500 in wall thickness. If pipe costs $700 per ton,
determine the total pipeline cost.
1 mile = 5280 ft and 1 Ton= 2000 lb.
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Solution :
 w = 10.68 × 0.500 × (20 – 0.500) = 104.13 lb/ft
 I mile=5280 feet.
 1ton = 2000 lb
 Therefore, the total pipe tonnage in 10 miles of pipe is
Tonnage = 104.13 × 5280 × 10/2000 = 2749.03 tons
 Total pipeline cost = 2749.03 × 700 = $1924322.4
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Example 6
 A 60 km pipeline consists of 20 km of DN 500, 12 mm wall
thickness pipe connected to a 40 km length of DN 400, 10 mm
wall thickness pipe. What are the total metric tons of pipe?
 DN500=500mm pipe; 1 metric ton=1000kg
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 w = 0.0246 × 12 × (500 – 12) = 144.06 kg/m and the weight per meter of DN 400 pipe is:
 w = 0.0246 × 10 × (400 – 10) = 95.94 kg/m
 Therefore, the total pipe weight for 20 km of DN 500 pipe and 40 km of DN 400 pipe is :
Weight = (20 × 144.06)x1000 + (40 × 95.94)X1000 = 6719x1000 Kg
= 6719 ton=6719 metric tons.
Total metric tons = 6719
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 Example 7
 Calculate the MOP for NPS 16 pipeline, 0.250 in wall thickness, constructed
of API 5LX-52 steel.
 What minimum wall thickness is required for an internal working pressure of
1440 psi?
 Use class 2 construction and for an operating temperature below 250°F.
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Ques.
 A natural gas pipeline, 600 km long, is constructed of DN 800
pipe and has a required operating pressure of 9 MPa . Compare
the cost of using X-60 and X-70 steel pipe. The material costs
of the two grades of pipe are as follows:
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 Use a class 1 design factor and temperature deration factor of 1.00
 800DN=800mm
 X60=414MPa
 X70=483MPa.
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Solution.
The wall thickness of pipe required to withstand the operating pressure of 9
M Pa. So, the pipe wall thickness required for X-60 pipe (60,000 psi =414 M
Pa) is
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 Similarly, the pipe wall thickness required for X-70 pipe (70,000 psi = 483
M Pa) is:
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 The pipe weight in kg/m will be calculated . For X-60
pipe,Weight per meter = 0.0246 × 13 × (800 –13) = 251.68 kg/m.
 Therefore, the total cost of 600 km pipeline at $800 per ton of X60 pipe is :
 Total cost = 600 × 251.68 × 800 = $120.81*10^(6)= 120.81
million
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 Similarly, the pipe weight in kg/m for X-70 pipe is :
 Weight per meter = 0.0246 × 11 × (800 – 11) = 213.50 kg/m
 Therefore, the total cost of 600 km pipeline at $900 per ton of X-70 pipe is :
 Total cost = 600 × 213.50 × 900 = $115.29 *10^(6)=115.29 million
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 Therefore, the X-70 pipe will cost less than the X-60 pipe. The difference in
cost is :
 $120.81 – $115.29 = $5.52 million.
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