Pipeline System Design

Pipeline System Design:Pipeline Network

Mike Yoon, Ph.D.

© 2010 Yoon Consulting2

• System Approach to Design• Pipeline Components• Pipeline Configuration

Key Topics


• There are four aspects of design that are interrelated in the system approach to design:– Hydraulic design– Mechanical design– Geo-technical design– Operations and maintenance design

• Decisions in one area of design directly affect or limit the options in another area.

System Approach to Design


• The hydraulic design is the process of evaluating:– The physical characteristics and quantities of the fluid to be

transported, – The number and location of pump stations,– The pipeline route and topology, – The range of pressure and temperatures, – The environmental conditions along the route

• Several hydraulic designs can be viable for any given design basis and route, but the best design should satisfy early use requirements and future capacity plans for the system.

Hydraulic Design


• For a hydraulic design, a number of mechanical designs can be developed to meet the criteria of the design basis.

• The mechanical design is governed by the codes and standards, focusing on selection of pipe material and the specification of pipe properties, type, size, and power required of pumps and other equipment or ancillary facilities such as heaters, and the support or burial requirements for the pipeline.

• The pipe diameter is selected based on the design flow, with little mechanical design required. However, internal and external pressure, allowable stress, and other considerations affect the final design of the wall thickness for the selected diameter.

Mechanical Design


• The pipeline company is responsible to protect the environment in the vicinity of the pipeline and its business interest from potentially adverse environmental conditions.

• Geo-technical design can affect the cost and safety significantly, if the pipeline route lies in challenging environments.

• Geo-technical design issues to be addressed are:– River crossings– Horizontal directional drilling– Buoyancy control– Geohazard management

Geo-Technical Design


• The operation and maintenance design takes into consideration the day-to-day tasks of operating and maintaining the functional integrity of the system.

• These include the necessary control systems to operate the system within its design parameters safely and continuously.

Operation and Maintenance Design


• System Approach to Design• Pipeline Components

– Line pipe and design pressure– Valves – Joints and Fittings

• Pipeline Configuration

Key Topics


• Specifications for line pipe are given in the following standards:– API 5L, 5LX: Specifications for Line Pipe– ANSI/ASME B36.10M: Welded and Seamless Steel Pipe

• Line pipe is manufactured by several methods, the most common being seamless (SMLS), electric resistance welded (ERW), and submerged arc welded (SAW) in the form of longitudinal and spiral welds.

Line Pipe


• The SMYS of the pipe material means the specified minimum yield strength for steel pipe manufactured in accordance with a listed specification. This is a common term used in the oil and gas industry for steel pipe.

• API 5L specifies various strength grades, ranging from Grade B, rated at 35,000 psi (241 MPa) to Grade X120, rated at 120,000 psi (827 MPa), where the Grade X120 refers to the SMYS in 1000 psi.

SMYS


SMYS vs. Pressure

• The pipe yield strength varies with the pipe grade. Shown in the plot is a pipe stress vs. pressure for a given pipe grade.

• SMYS is the maximum stress, beyond which the pipe is deformed with additional applied pressure.


Pipe Specification and Category

• API 5L: This is the umbrella specification for API 5LX, 5LS, etc. It covers Seamless, Electric resistance welded, Electric flash welded, Submerged arc welded, Furnace lap welded, and Furnace butt welded.

• API 5LX: Seamless, Electric resistance welded, Electric flash welded, Submerged arc welded

• API 5LS: Electric resistance welded, Submerged arc welded


Examples of Allowable Stresses

1.0070,000(482)X70API 5LXElectric flash welded

1.0070,000(482)X70API 5LX,API 5LS

Submerged arc welded

1.0070,000(482)X70API 5LX, API 5LS

Electric resistance0.8028,000(193)Cls 2API 5LFurnace lap0.8025,000(172)Cls 1API 5LFurnace lap0.6025,000(172)A25API 5LFurnace butt 1.0070,000(482)X70API 5LXSeamless

Joint Factor

SMYSPSI(MPA)

GradeSpecsWeld Joint


Usage for Types of Line Pipe

Less expensive than seamless, more than ERW

Less expensive than seamless

More expensive than ERW

Relative cost

All servicesNot offshoreAll servicesService

B to X80B to X70B to X80Grades

NPS 64NPS 26NPS 26Max. Diameter

NPS 16<=2.375”<=2.375”Min. Diameter

SAWERWSeamless


Design Pressure Calculation• Use Barlow’s formula and safety factors to

determine the design pressure for a given pipe grade and pipe wall thickness.Pdesign = (2S*t/Do) * F*J*Twhere S = specified minimum yield strength of pipe, psigt = pipe wall thickness, inchesDo= outside pipe diameter, inchesF = design factorJ = joint factorT = temperature derating factor (applicable only to gas

pipelines)


Design Factor


Location Classification

Not applicableDwelling >= 4 stories4

Not applicableDwelling units >= 46, Institute that are difficult to evacuate

3

Not applicable10 < dwelling units < 46,Building or area occupied by more than 20 people,Industrial installation susceptible to environmentally hazardous conditions

2Not applicableDwelling units <= 101

B31.4Z662Class


Comparison of Design Factors

0.720.800.80Low vapor pressure liquid0.720.64- Stations0.720.50- Railways0.720.64- Roads0.720.64High vapor pressure liquid

B31.4 *Class 2Class 1Application

* ASME B31.4 does not define a design factor, but opts to factor the specified minimum yield strength by 0.72.


Temperature Factor

• Section 401.3.1 of B31.4 states that it is not necessary to vary allowable design stress for metal temperature between -30oC and 120oC.

• For application where ground or air temperature is expected to be extremely low, seasonally or locally, the properties of pipe component materials at low temperature should be considered to verify that the design will be adequate.

• Below -30oC, steel toughness should be taken into consideration, but not strength.


Temperature Derating Factor

1.00<120oC

1.00>-30oC

0.91<200oC

0.93<180oC

0.97<150oC

1.00<120oC

B31.4Z662Temperature


Comparison of Joint Factors

0.600.60Furnace butt welded1.001.00Submerged are welded

1.001.000.80

1.00Electric welded- Resistance welded- Induction/flash welded- Fusion arc welded

1.001.00SeamlessB31.4Z662Class


Hydrotest Pressure

• B31.4 Codes specifies that the magnitude of hydrotestpressure is 125% of the maximum operating pressure.

• Therefore, by hydrotesting the pipe at 1.25 times the MAOP, the pipe is stressed to 90% (72% * 1.25) of the SMYS.

• For example, a pipeline designed to operate continuously at 1,000 psig will be hydrotested to a minimum pressure of 1,250 psig.


Example: Design Pressure• Pipe: 16” OD x 0.250”wall thickness API 5L grade

X56 Electric resistance welded• Location: Class 1, so F = 0.72• Joint factor: E = 1.0

Pmax = (2 * 56,000 * 0.250/16) * 0.72 * 1.0= 1,260 psig

• This pressure is called maximum allowable operating pressure (MAOP), and the internal pressure that will cause the hoop stress to reach the yield stress of 56,000 psig is 1,260/0.72 = 1750.

• Hydrotest pressure = 1.25 * 1,260 = 1,575 psig


Wall Thickness Calculation• Minimum wall thickness, t, is a function of internal pressure,

Pi, pipe outside diameter, D, and the allowable stress, S.t = Pi * D/(2*S)

• Nominal wall thickness, tn, includes an allowance for manufacturing tolerance.tn = t + allowance

• The actual wall thickness will be equal to or greater than the nominal value.

• Wall thickness for calculation of the MAOP excludes additional thickness for corrosion allowance, imposed stresses such as concentrated loads at supports, thermal expansion or contraction, and bending.


Wall Thickness Selection• Specific sections of the system may have different wall

thickness requirements as determined by the internal pressure and other imposed stresses, and making use of the hydraulic gradient.

• For economic benefits, thinner wall may be installed at some distance downstream of pumping stations as the operating pressure there declines.

• However, there are other considerations such as complications in construction of a system with frequent variations in wall thickness.

• Changes in wall thickness should be limited with anticipated growth or expansion of the system capacity.


Buried Pipe

• Pipelines are generally buried, because:– Surface use of pipeline corridor– Protection from intentional or accidental damage– Protection against expansion and contraction from ambient

temperature changes– Minimize variations of ambient temperature and resultant

effects on fluid viscosity– Provide restraint longitudinally along pipeline length


Aboveground Pipe

• Pipelines may be partially or wholly installed aboveground for reasons of economy of construction, maintenance, requiring insulation, or heat tracing.

• Longitudinal restraints are required at certain locations such that expansion or contraction due to temperature or pressure changes is absorbed by axial or longitudinal stress in addition to radial expansion.





Key Topics


Valve Functions

• Mainline isolation valves have three functions:– Sectionalize the pipeline into smaller segments that can be

isolated in order to minimize in the event of a line rupture,– Change flow direction to an interconnected pipeline or to a

delivery facility such as tanks,– Isolate process equipment such as plant for safety,

maintenance, or operating purposes.

• Mainline valves must allow passage of pigs.


Valve Types

• Block valves– Mainline valve

– Side valve to isolate a lateral from the mainline

• Pressure relief valves

• Pressure regulating or control valves

• Surge tanks


Valve Selection

• Valve selection takes into account the following:– Function– Operating characteristics– Location and terrain – Fluid service– Cost – Materials– Space availability– Maintenance– Repair capability


Valve Location

• In general, valves are installed at locations where safety is of primary concern and to be enhanced.

• Valves are installed at the origin and terminal points of a pipeline, branch points to isolate a section and to facilitate hydrostatic testing, i.e., anywhere that the test pressure is differentiated such as sections of higher operating pressure or a change in wall thickness.

• Sectionalizing block valves should be located in easily accessible positions, e.g., aboveground on a buried pipeline.


Block Valve • Block valves are used to isolate sections of mainline

or laterals in the vent of a line break or during maintenance.

• Code requirements for maximum valve spacing are:

12 KmNR15 Km412 KmNR15 Km312 KmNR15 Km212 KmNRNR1B31.4LVPHVPClass

Z662Z662


• ANSI B16.5, B16.9, B16.10, B16.11, B16.25, B16.28: Flanges, Fittings, Valves

• API 6D Pipeline Valves, Gate, Plug, Ball, and Check Valves

• API 600, 602, 603: Valves

Standards for Valves and Fittings


Control Valve

• A basic control valve system would normally consist of the following components:– Control valves – Actuators– Controllers – Sensors

• Control valves adjust processes by alterning flow rate and/or differential pressure which is defined as the difference between the pressure at the valve inlet and the pressure at the valve outlet.

• In general, the greater the differential pressure the greater the flowrate for any given valve orifice size.


Actuator

• The operation of a control valve involves positioning its movable part relative to the stationary seat of the valve. The purpose of the valve actuator is to accurately locate the valve plug in a position dictated by the control signal.

• The actuator accepts a signal from the control system and, in response, moves the valve to a fully-open or fully-closed position, or a more open or a more closed position (depending on whether 'on / off' or 'continuous' control action is used).

• Two popular ways of providing this actuation are pneumatic and electric types.


Pressure Regulating Valve

Flow

Pipe segment to be protected

Pressure regulating valve





Key Topics


• Pressure losses occur through joints and fittings due to energy losses resulting from changes in the magnitude or direction of liquid velocity in the pipeline.

• Pipe joints include welds and flanges, while fittings pipe enlargements and contractions, pipe bends and elbows, and restrictions such as valves.

• In addition to joints and fittings, the following contributes minor pressure losses:– Valves– Orifice meter

Joints and Fittings


Minor Pressure Losses

• In most long-distance pipelines, pressure losses due to valves and bends are comparatively small, so such pressure losses are called minor pressure losses.

• In long transmission lines, the minor losses are so small that they can be ignored without significant errors.

• In short pipelines such as plant piping, the pressure loss due to the above can be substantial.


Calculation of Pressure Losses

• The minor pressure losses can be expressed by a kinetic energy term:ΔP = Kρv2/2

where K = pressure loss coefficient (dimensionless).ΔP = Minor pressure loss

v = velocity of liquid through valve, joint or fitting

• The value of K is determined mainly by the flow geometry or the shape of the device. K values are almost constant at high Reynolds number and compiled in hydraulic data handbooks.


K Values of Valves

0.220.230.270.31Plug

0.300.350.680.77Butterfly

0.040.040.050.05Ball

4.14.45.15.1Globe

0.100.100.120.12Gate

18-24”12-16”8-10”6”Valves


Pressure Loss Coefficients


Example: Minor Pressure Losses

• A tank open to the atmosphere is filled with 40oAPI oil to a height of 10m from the bottom. A tap at the bottom of the tank is opened, and oil flows out from the outlet.

• Determine the outlet pressure.• Assume that the flow is

steady and incompressible.

10mOil

200m

16” OD

1

2


Applicable Equations

• From the momentum conservation in steady state, we have (P1 + ρv1

2/2 + ρgz1) - (P2 + ρv22/2 + ρgz2) = fρv2/2*L + Kρv2

2/2

where f is the Darcy friction factor and L the pipe length, and K = Kentrance + Kexit are the pressure loss coefficients for pipe entrance and exit, respectively.

• P1 is atmospheric pressures, v1 = 0, v2 = v, z1 = 10m, and z2 = 0. Therefore, the equation becomesP2 = ρgz1 - ρv2/2 - fρv2/2*L – (Kentrance + Kexit)ρv2

2/2

where Kentrance = 0.5 and Kexit = 1.0.


Example Data and Friction Factor

• Data:• Pipe = 16” OD x 0.25 wt: inside diameter = 0.394m• Pipe roughness = 0.0018” = 0.0457mm• Crude viscosity = 3 cSt of 40 oAPI crude• Density = 825 kg/m3

• Flow rate = 1,000 m3/hour or velocity = 2.28 m/sec• Friction factor calculation:

• Re = 2.28*0.394/0.000003 = 299,000: partially turbulent• Relative roughness = 0.0000457m/0.394m = 0.000116• f = 0.0156


Example: Friction Pressure Loss • Pf/L= 0.0156*825kg/m3 * (2.28 m/sec)2/(2*0.394m)

= 84.9kg/(m2 sec2) = 84.9 Pa/m = 0.0849 kPa/m• Total pressure drop for 200m long pipe = 0.0849 * 200

= 17.0kPa • Pressure at the tank outlet = ρgz1 = 825*9.8*10/1000 =

80.9kPa • The velocity pressure, ρv2/2 = 825 * (2.28)2/2 =

2,144pa = 2.14kPa• The pressure losses due to entrance and exit are

(Kentrance + Kexit)ρv22/2 = 1.5*2.14 = 3.21

• Pressure at the pipe discharge point = 80.9 – 2.14 -17.0 – 3.21 = 58.6kPa


Example Discussion

• The pressure losses due to pipe entrance and exit can be relatively significant for low pressure system, whereas they are negligibly small for large pressure system such as transmission pipeline.


Enlargement

• Consider liquid flowing through a pipe of diameter D1. If the diameter enlarges to D2, the pressure loss can be calculated as follows:ΔP = Kρ(v1 – v2)2/2 = Kρ(A2/A1 – 1)2 v2

2/2

where v1 and v2 are the velocity of the liquid in D1and D2 pipes and A1 and A2 the areas.

• The value of K depends on the diameter ratio and the different angle due to the enlargement.


Gradual Enlarger - Reducer

D2

D1 D2

Gradual Pipe Enlargement and Reduction


Enlarger Pressure Loss Coefficient


Enlargement Example• Calculate the pressure loss due to a gradual

enlargement in a pipe that flows 1,800m3/hr of diesel from a 8” diameter to a 12” with an angle of 60o. Both sizes are internal diameters.

• Solution: The liquid velocity in the 12” pipe size is v2

= 1.714m/sec, and diameter ratio = 12/8 = 1.5From the diagram, the value of K is 1.2 for area ratio = 2.25 and angle = 60o.Therefore, pressure loss due to gradual enlargement is

ΔP = 1.2 * 850 * (2.25 – 1)2 * 6.8562/2 = 37,456 Pa = 37.5 kPa


Contraction

• For flow through an abrupt contraction, the flow from the larger pipe to a smaller pipe results in the formation of a vena contracta, just after the diameter change.

• At the vena contracta, the flow area reduces to Ac with increased velocity of vc, and subsequently to v2.

• The pressure loss for sudden contraction is:ΔP = ρ(1/Cc – 1)2 v2

2/2

Where Cc is the contraction coefficient


Contraction Coefficients

1.00.8920.8130.7550.712Cc

1.00.90.80.70.6A2/A1

0.6810.6590.6430.6320.624Cc

0.50.40.30.20.1A2/A1


Equivalent Lengths

100 Swing check18Plug3Ball

340Globe8Gate

L/DValves

16 Long-radius elbow (90o)

16Standard (45o)

30Standard (90o)

L/DElbow

• The minor pressure losses can be accounted for by means of an equivalent length method. K is analogous to the term fL/D for a straight length in the Darcy equation.

• It is better to use K, because K is less dependent on the Reynolds number and relative roughness.


Equivalent Length Example

• A piping system of a pump station is 140m of NPS 24 pipe that has two 24” gate valves, three 24” ball valves, on swing check valve, and two 90o standard elbows. Using the equivalent length concept, calculate the total pipe length of the station.

• Convert all valves and fittings in terms of 24” pipe as follows:– Two 24” gate valves = 2* 24”* 8 * 0.0254 = 9.75m– Three 24” ball valves = 3* 24” * 3 * 0.0254 = 5.49m– One 24” swing check valve = 1* 24” * 100 * 0.0254 = 60.96m– Two 90o elbows = 2* 24” * 30 * 0.0254 = 36.58m– Total equivalent length of straight pipe and all fittings = 140m + 112.78m =

252.78m– The pressure drop due to friction is calculated based on 252.78m of pipe.


• System Approach to Design• Mechanical Design• Pipeline Configuration

Key Topics


• Liquid may be delivered off the pipeline (stripping) at intermediate locations, reducing the main line flow rate and theremainder continues to the pipeline terminal. Since the downstream flow is lower, the frictional pressure drop is lower.If the stripping rate is high, then the pipe size of the downstream section can be reduced.

• Liquid may enter the main pipeline at intermediate locations (side-stream injection), adding flow rate to the main line flow and the remainder continues to the pipeline terminal. Since the downstream flow is higher, the frictional pressure drop is higher. If the side stream injection rate is high, then the pipesize of the downstream section should be increased.

• A pump is installed to increase the injection pressure on the injection side, while a control valve is installed to control the delivery pressure.

Injections and Deliveries


• Liquid may be delivered out of the main pipeline into branch lines or injected from branch lines into the main pipeline at intermediate locations.

• The pressure at the flow lifting point should be so high that the pressure at the injection point is higher than the main line pressure. A pump is normally installed at the flow lifting point.

• The main line pressure should be high enough to meet the delivery pressure requirement. To regulate the delivery pressure, a control valve is installed on the branch line or the main line pressure at the junction is locked.

Pipe Branches


Elements of Pipeline System - 1

• Injection station, also known as receipt or inlet station, is where the product is injected into the line. Storage facilities, such as tank terminals, as well as other devices to push the product through the line, like booster pumps are usually located at these locations.

• Delivery station, also known as terminal, is where the product will be distributed to the final consumer. It could be a tank terminal or a connection to other pipelines.

• An intermediate station can be side stream injection or delivery station. These stations allow the pipeline operator to inject or deliver part of the product being transported.


Elements of Pipeline System - 2

• Pump stations are located along the line to move the liquid through the pipeline.

• Block Valve Stations are the first line of protection for pipelines. With these valves the operator can isolated any segment of the line to perform some specific maintenance work or isolate a rupture or leak. Block valve stations are usually located every 20 to 30 miles, depending on the type of pipeline.

• Regulating station is a special type of valve station, where either pressure or flow is controlled. Pressure regulators are usually located at the downhill side of a peak, while flow regulators at delivery stations.


• The following pipe arrangements are made to build a complex pipeline network:– Pipes in series – refer to the connection of pipes of different

diameters in series.– Parallel or looped pipes – increase flow rates and reduce

pressure drop. – Branch or lateral lines – connect to/from other pipelines or

facilities from/to the main line.• A junction is required at the point where more than

two pipes join together. Normally, valves are installed at a junction to control the flow or pressure.

Pipeline Network


• The same flow rate goes through the pipes connected in series.

• The larger the pipe diameter, the slower the velocity, the smaller the friction factor, and the lower the friction pressure loss.

Q = v1A1 = v2A2 = v3A3 and Ptotal = Σ (ΔPi)

• The total pressure loss is obtained by adding the pressure losses of all the pipe segments, which may have different pipe diameters.

Pipes in Series - 1


• Equivalent length is based on the same pressure drop in the equivalent pipe as the original pipe diameter.Le = L2 (D1/D2)5 * (f2/f1)If D1and D2 are similar, the ratio f2/f1 is almost 1.

• A pipe is connected in series where there is a large flow increase or decrease due to side-stream injection or delivery.

• A pipe in series is not practical for petroleum liquid pipelines, because it doesn’t allow easy pigging operations unless a pig trap and launch facility is installed.

Pipes in Series - 2


Series Piping - 1

D1 D2 D3

L1 L2 L3


D1

K1

Qb

P1 P2 P3 P4

D2

K2

Qb

D3

K3 Qb

A system of pipeline with different lengths and diameters connected in series. Note the flow resistance changes with diameter.

Series Piping - 2


• Under steady state conditions, the pipe pressure decreases from pump station to the terminal in the direction of flow. The pipeline segment immediately downstream of a pump station will be subject to higher pressures while the tail end of that segment will be subject to lower pressures.

• If we use the same wall thickness throughout thee pipeline, the downstream portion of the pipeline will be underutilized.

• A more efficient approach would be to reduce the pipe wall thickness as we move away from a pump station toward the suction side of the next pump station or the delivery terminal.

• Wall thickness reductions must be able to handle the higher pressures that result from shut-down of intermediate station and surge pressure caused by pump trip.

Telescoping Pipe Wall Thickness


• Like telescoping pipe wall thickness to compensate for lower pressures, the pipe grade may be changed.

• The higher pressure sections are constructed of higher grade pipe material, whereas the lower pressure sections are constructed ofsomewhat lower grade steel.

• Sometimes a combination of telescoping and grade tapering is used to minimize pipe cost.

• Wall thickness reductions must be able to handle the higher pressures that result from shut-down of intermediate stations and surge pressure caused by pump trip.

Pipe Grade Tapering


• Two pipes are connected in series with a pipe reducer and gate valves as follows:– One 24” gate valve at the beginning of 24” pipeline – 24” pipeline, 0.394” wall thickness, 10,000m long– One 24” to 20” sudden reducer– 20” pipeline, 0.306” wall thickness, 25,400m long– One 20” gate valve at the end of 20” pipeline

• Determine the total equivalent length on 20”diameter pipeline.

Series Piping Example


• The total equivalent length on 24” pipe is the sum of the following:– One gate valve = 8 * (24” – 2*0.394) * 25.4/1,000 = 4.7m– Pipe length = 10,000m– Total equivalent length on 24” = 10,004.7m

• The equivalent length of the 24” pipe = 10,004.7 * (23.212/19.388)5 = 24,609m

• The total equivalent length on 20” pipe is the sum of the following:– One gate valve = 8 * (20” – 2*0.306) * 25.4/1,000 = 3.9m– Pipe length = 25,400m– Total equivalent length on 24” = 25,404m

• The total equivalent length on 20” diameter pipe is 50,013m. The pressure losses due to the gate valves and reducer are insignificant.

Solution: Series Piping Example


• For parallel pipes, the flow through each pipe is different, whereas the pressure drops across the parallel pipes are the same.

• A pipeline is looped to increase throughput or flow rate. Since the frictional pressure drop is lower with a looped pipe, so is pumping requirement.

• Continuity equation yields the total flow as follows:Q = Σ (Qi) and

v1 = Q1/A1, v2 = Q2/A2, v3 = Q3/A3, and

ΔP1 = ΔP2 = ΔP3

Parallel or Looped Pipes - 1


• Assuming the pipe length is the same, the flow through each branch in terms of the inlet flow Q is as follows:Q1 = Q/[1 + (f1/f2)0.5 (D2/D1)2.5]

Q2 = (f1/f2)0.5 (D2/D1)2.5 *Q/[1 + (f1/f2)0.5 (D2/D1)2.5]If the diameter difference is small, the ratio of the friction factor is almost 1.

• The equivalent diameter of these two pipes isDe = D1 (Q/Q1)0.4

• If the two diameters are equal, then Q1 = Q2 = Q/2 and De = 1.3195*D1

Parallel or Looped Pipes - 2


P1 P2

K2

Qb Qb

Qb1 K1

D1

D2

Qb2

A system of pipelines connected in parallel (Looped)

Pressure Drop Can be Alleviated By Looping


P1

L, D1

L - X

P2

XQ1 K1

K’1

D2, K’2, X

Pipeline Segmental Looping

Partial Sections of a Line may be Looped to Achieve Required Pressure Drop


• A pipeline network can be analyzed using the following network theorem:– The algebraic sum of the pressure drops around each loop in

the network is zero.– Continuity equation must be satisfied at each junction or

node. This means that all flows going into a junction must equal to flows leaving the junction.

• Use the total pressure loss equations in terms of the flow rate, including the static pressure due to elevation changes.

• A set of non-linear equations is arranged in a sparse matrix to take advantage of many zero vales in off-diagonal elements.

Pipeline Network Analysis

Pipeline System Design

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