pipeline network design presentation 2

PIPELINE NETWORK DESIGN

Landon Carroll & Wes Hudkins

Overview Goals Background Information Conventional Pipeline

Optimization Analysis Mathematical Model

Analysis Expansion Conventional Comparison Application

Conclusion and Recommendations

Goal Create a program that will design an

optimal pipeline network, which is faster and more accurate than conventional design methods

Natural Gas Industry The US consumes 1.5 to

2.5 million cubic feet (MMscf) per month

97% of this gas is piped from the well all the way to your furnace

Large upside due to clean Natural Gas power plants and Compressed Natural Gas CNG automobiles

Natural Gas Price Breakdown

*Standard Heating value Gas of 1000 Btu/scf. Thus, $12/Mscf = $12/MMBtu

Pipeline Optimization Basics

Pipeline Optimization Methods Hydraulic Analysis

Conventional Various Equations

derived from The General Flow Equation

New Method General Equation

combines constants into two parameters, A and B

Economic Analysis Conventional

J-Curves New Method

Mathematical Programming using a General Algebraic Modeling System (GAMS) interface

NATURAL GAS HYDRAULICS

Landon CarrollWes Hudkins

Natural Gas Hydraulics 101 Steady State Mechanical Energy Balance on Pipe:

(PE) + (ΔP) + (KE) + (Friction Loss) = 0

In most liquids, density is constant:

Natural Gas:

Therefore, Integration is slightly more difficult

Use average z, T, and P to simplify integration:

02

2

v

DdxfdvvdPdyg

2

22

21

21

1 22PvgyhPvgy L

Natural Gas Hydraulics 101 KE: Negligible ∆P:

PE:

Friction Loss: ,

Therefore, Combine, solve for Q:

General Flow Equation

Conventional Hydraulic Equations are derived from this equation; just insert different values for the friction factor, f

Conventional Hydraulic Equations

1. Colebrook-White2. Modified Colebrook-

White3. AGA4. Panhandle5. Weymouth6. IGT7. Spitzglass8. Mueller9. Fritzsche

Equation Accuracy Analysis Theoretical Pipe

Set the Temperature, Inlet Pressure and Natural Gas Flow Rate

Solve ∆P with Equation for various diameters and elevation changes

Simulate Pipe: Pro/II Set same conditions

Compare Results

Natural Gas Composition Used

Natural Gas Component

Mole Fraction

C1 0.949C2 0.025C3 0.002N2 0.016

CO2 0.007C4 0.0003iC4 0.0003C5 0.0001iC5 0.0001O2 0.0002

Equation Example Modified-

Colebrook

2/12

105.2

2122

21

Re825.2

7.3log5972.26

06843.0

avgavg

st

avgavg

avg zTdL

fDD

PQzT

PHHdPP

Modified-Colebrook Results

125 175 225 275 3250

200

400

600

800

1000

1200

1400

1600

1800NPS = 16; Pro-IINPS = 16; Ana-lyticalNPS =18; Pro-IINPS = 18; Ana-lyticalNPS = 20; Pro-IINPS = 20; Ana-lyticalNPS = 22; Pro-IINPS = 22; Ana-lytical

Flowrate (MMSCFD)

Pres

sure

Dro

p (p

sia)

150 170 190 210 230 250 270 290 310 330 3500

5

10

15

20

25

30

35

40

45

50

NPS = 16

NPS =18

NPS =20

NPS = 22

Flowrate (MMSCFD)

Perc

ent E

rror

Costly Error! One Pipeline

Flowing 200 (MMscfd) Operating 350 days/year Averaging $8 per Mcf EIA States 3-5% of gas flow is

used for compressor fuel 1% of hydraulic error is $224

wasted Natural Gas per compressor per year!

Range of Error

Equation Name

Range of Error

Cost of Error ($ of fuel

cost/compressor/yr)Panhandle 3.5 – 10% 784 – 2,240Colebrook 2.4 – 10% 538 – 2,240 Modified-Colebrook

1.0 – 8.8% 224 – 1,971

AGA 0.2 – 15% 45 – 3,360IGT 7.6 – 17% 1,702 – 3,808

Mueller 13 – 20% 2,912 – 4,480

Mathematical ModelGeneral Flow Equation:

Where, Equation becomes:

Rearrange:Where,

Thus:

Mathematical Model Analysis

50 100 150 200 250 300 350 4000

200

400

600

800

1000

1200

1400

1600

1800NPS = 16; Pro-IINPS = 16; AnalyticalNPS =18; Pro-IINPS = 18; AnalyticalNPS = 20; Pro-IINPS = 20; AnalyticalNPS = 22; Pro-IINPS = 22; Analytical

Flow Rate (MMSCFD)

Pres

sure

Dro

p (p

sia)

50 100 150 200 250 300 350 4000

10

20

30

40

50

60

70

80

90

100

NPS = 16

NPS =18

NPS =20

NPS = 22

Flow Rate (MMSCFD)Pe

rcen

t Err

or

Equation Name

Range of Error

Cost of Error ($ of fuel

cost/compressor/yr)Mathematical

Model0 – 0.9% 0 – 200

THE MATHEMATICAL MODEL VS. J-CURVE ANALYSIS


J-Curve - Simulator Trials Simulations are used to generate

diameter/flowrate/pressure drop correlations for the J-curves Three pressure parameters (P3) were selected discretely–

750, 800, and 850 psig. Both segments will have distinct optimums.

P1 = 800 psig

P2 P5 = 800 psig

L = 60 mi

Q = 100 – 500 MMSCFD

L = 60 mi

P3 P4

Q = 50 MMSCFD

J-Curve - Procedure Simulations are run to generate pressure drop at

a given flowrate and diameter Cost calculations are completed for these

pressure drops which relate to compressor and operating costs

Plot cost vs. flowrate Repeat at various diameters and/or pressures The lowest cost at the desired flowrate ‘wins’

100 150 200 250 300 350 400 450 500$0.00$0.20$0.40$0.60$0.80$1.00$1.20$1.40$1.60

NPS = 16NPS = 18NPS = 20

Flow Rate (MMSCFD)

TAC

per

MC

F

100 150 200 250 300 350 400 450 500$0.00$0.20$0.40$0.60$0.80$1.00$1.20$1.40$1.60

$0.34

0.356

Segment 1P = 850

NPS = 16NPS = 18NPS = 20NPS = 22

Flow Rate (MMSCFD)

TAC

per

MC

F

100 150 200 250 300 350 400 450 500$0.00$0.20$0.40$0.60$0.80$1.00$1.20$1.40$1.60

0.353

Segment 1P = 800

NPS = 16NPS = 18NPS = 20NPS = 22

Flow Rate (MMSCFD)

TAC

per

MC

F

100 150 200 250 300 350 400 450 500$0.00$0.20$0.40$0.60$0.80$1.00$1.20$1.40$1.60

0.3287

Segment 1P = 750

NPS = 16NPS = 18NPS = 20NPS = 22

Flow Rate (MMSCFD)

TAC

per

MC

FJ-Curve – Segment 1 Optimum

The lowest TAC at Q=300 is achieved with NPS = 18 for all three pressures

P = 750 gives the lowest overall TAC for NPS = 18

Why so many decimal places? At high flowrates, these fractions of cents per MCF can become millions of dollars.

J-Curve - Segment 2 Optimum

Since P = 750 is the optimum pressure parameter for Segment 1, we then determine the optimum diameter for Segment 2 at P = 750

The optimum diameter is then NPS = 18 Then, optimize the system starting with segment 2

100 150 200 250 300 350 400 450 500$0.00

$0.50

$1.00

$1.50

$2.00

$2.50

$3.00

0.3026

0.3063

Segment 2P = 750 NPS = 16

NPS = 18NPS = 20NPS = 22

Flow Rate (MMSCFD)

TAC

per

MC

F

100 150 200 250 300 350 400 450 500$0.40$0.50$0.60$0.70$0.80$0.90$1.00$1.10$1.20

0.907

0.7340.652 0.616 0.607 0.613 0.629 0.652

Optimizing Segment 2 First; P=850

NPS = 18 Segment 1 & NPS = 18 Segm...

Flow Rate (MMSCFD)

TAC

per

MC

FOrder of Optimization

100 150 200 250 300 350 400 450 500$0.4

$0.6

$0.8

$1.0

$1.2

$1.4

1.299

0.908

0.7400.663 0.631 0.626 0.635 0.654 0.679

Optimizing Segment 1 First; P = 750 NPS = 18 Seg-ment 1 & NPS = 18 Segment 2

Flow Rate (MMSCFD)

TAC

per

MC

F

Optimizing segment 2 first results in the

optimum design

Overall Optimum & Relevance of Optimum

100 200 300 400 500$0.5

$0.6

$0.7

$0.8

$0.9

$1.0

$1.1

$1.2

$1.3

$1.4

$1.5

0.6164

0.6174

Overall OptimumP=850

NPS = 18 Segment 1 & NPS = 16 Segment 2

NPS = 18 Segment 1 & NPS = 18 Segment 2

Flow Rate (MMSCFD)

TAC

per

MC

FThe optimum pressure is 850 psig, and the optimum pipe sizes are 18 inches in both segments.

Shown: Optimization of Segment 1 at Segment 2’s optimum pressure.

# J-Curves RequiredFor un-branched pipeline networks such as this one, the number of J-Curves required for optimization is:

As the number of pipes in a pipelines network increases, the number of J-Curves required for optimization increases exponentially.

# pipes

# discrete pressures

# orders

# diameters

Economic Optimums

Segment

Optimum

Pressure

Optimum

Diameters

TAC per MCF

Total Annual Cost

(millions)

1 750 18 & 18 $ 0.631 $ 662 850 18 & 18 $ 0.616 $ 65

Both* 850 18 & 18 $ 0.616 $ 65

Two-Segment Network

Optimizing Segment 1 first gave the incorrect solution. All possible combinations must be analyzed to find an overall

optimum. In order to analyze both segments at once, 48 J-curves must be

analyzed for even this simple two pipe network!

Mathematical Model Results

The mathematical model reached an optimum of $2,000 per MCF less than the J-curve method. Why? The J-curve method ignores volume buildup, time value of money, inflation, and many cost variations over time.

Remember, this required 48 J-curves and 432 simulations with the conventional method and the results are not even accurate!

Nonlinear Model – 2 Pipe NetworkPipe 1 Pipe 2

Pipe Diameter (in)

22 22

Compressor Work (hp)

10,740 0

Pressure Drop (psi)

1,830 1,490

TAC Model $ 0.596TAC J-Curves $ 0.616

THE MATHEMATICAL MODEL


Model Expansion Willbros, Inc.

Friday, February 20th, 2008

Diameter Coating cost Transportation

cost Quadruple

random length joints

Dr. Bagajewicz Installation cost Pipe maintenance

cost Compressor

maintenance cost

Model Logic Linear Model

Generates discrete pressures Minimizes net present total annual cost Gives optimum diameters, compressor locations,

compressor installation time, and compressor size Nonlinear model

These optimums are then input into the nonlinear model

Minimizes net present total annual cost

Model Logic - Input

ModelDiameter OptionsSupplier TemperaturesSupplier PressuresConsumer Demands (V/t)Demand Increase (%/yr)Min/Max Operating PressureCompressor Location OptionsElevationsPipe ConnectionsDistances

EconomicsProject LifetimeOperating Cost ($/P*t)Maintenance Cost ($/hp,%TAC)Operating Hours (hr/yr)Interest RateConsumer Price ($/V)Steel Cost(d) ($/L)Coating Cost(d) ($/L)Transportation Cost(d) ($/L)Installation Cost(d) ($/L)

HydraulicsGas DensityCompressor EfficiencyCompressibility FactorCompressibility RatioHeat Capacity

Model Logic – Economic Calculations

Objective Function: Net Present Total Annual Cost

Total Annual Cost

Compressor Cost

Pipe Cost

Maintenance Cost

Operating Cost

TAC(t)

Pipe Cost

Compressor Cost Maintenance Cost

Operating Cost

Capacities and Works come from hydraulic calculations.

Model Logic – Linear Hydraulic Calculations

Capacities to Compressor Cost and Maintenance Cost Equations

Works To Operating Cost EquationCapacity Limits

Maximum Capacity

Pressure Work

Total Demand

Compressor Work

Hydraulic Equation Part A

Hydraulic Equation Part B

Discrete Pressures

DPDZ

Discrete Pressures

Pressures

Pressure Works

Total Demand

Max Comp Capacity

Model Logic – Nonlinear Hydraulic Calculations

Works to Operating Cost Equation

To Compressor Cost Equation and Maintenance Equation

Capacity Limits

Compressor Work

Hydraulic EquationPressures

Model Logic - Output

PhysicalPipe LocationsPipe DiametersDemand at Each PeriodFlowratesInlet and Outlet PressuresCompressor LocationsCompressor Capacities

EconomicsNet Present ValueNet Present Total Annual CostTotal Annual Cost at Each PeriodFixed Capital InvestmentRevenueOperating CostPipe CostCompressor CostMaintenance CostPenalties

CASE STUDYLandon CarrollWes Hudkins

Case Study - Given

FairfieldSupply P (kPa)

3548.7

Supply T (°R)

529.67

MinOP (kPa)

10050.5

MaxOP (kPa)

4200

Elevation (km)

0.185928

Mavis Mayberry Split Beaumont TravisInitial Demand (Mcmd)

283.17 566.34 0 2831.7 1699

Price ($/m3) 0.32 0.33 0 0.3 0.3Elevation (km) 0.56376 0.54864 0.2286 0.10668 0.12816

• 10% Annual Demand Increase• Season Demand Variation• 8 Year Project Lifetime

Case Study - J-Curves

# simulations per curve

# diameters # discrete pressures

# pipes # possible compressor

location configurations

# possible orders of

optimization

Optimization of this case study using J-curves would require 293,932,800 simulations!

If a person were to run this many simulations 24/7 at 5 minutes per simulation, it would take 2796 years!

If this person only worked the standard 40 hours per week, it would take 11,776 years!

In order to accomplish the design in 6 months, it would require 23,552 employees!

At minimum wage, that’s $153,088,000!

Case Study - ResultsNon-Graphical Results

Pipe 1 ID (in.) 22Pipe 2 ID (in.) 22Pipe 3 ID (in.) 22Pipe 4 ID (in.) 18Pipe 5 ID (in.) 12NPV ($) 4,392,078,00

0NPTAC ($) 243,706,100Pipe Cost ($) 185,720,700Supplier Compressor Capacity (hp)

22,929.16

Consumer1 Compressor Capacity (hp)

13,365.09


13,293.76


8,439.168

This took 1 person about 1 hour!

0 2 4 6 8 10 12 14 16012345678 Year Consumer Demand

Consumer1 DemandConsumer2 DemandConsumer4 DemandConsumer5 Demand

Time periods (6 months)

MM

scm

d

0 2 4 6 8 10 12 14 160

20

40

60

80

1008 Year Economics

TACFCIOperating CostCompressor Cost

Time Periods (6 Months)

$ m

illio

n

FCIinit=$303,036,750

CONCLUSIONSLandon CarrollWes Hudkins

Recommendations Expand model to

incorporate more pipeline details (i.e. thickness, friction due to fittings, heat transfer)

Make more user friendly

GAMS coupled with GAMS data exchanger (GDX) to create user interface

Uncertainty added to model

Conclusion Conventional hydraulic equations

inaccurate

J-Curve analysis inaccurate and time consuming. Does not allow for complex networks.

Mathematical model produces accurate results and when coupled with GAMS saves time and money

Special Thanks Willbros, Inc. – industry feedback and

input Debora Faria – original program author Chase Waite – last year’s group member Vi Pham – teaching assistant Mark Bothamley – industrial feedback

and input Miguel Bagajewicz - professor

Any Questions

Please see us at our poster with questions.

pipeline network design presentation 2

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