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 Carroll

Wes 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

D

dxfdvv

dPdyg

2

22

21

21

1 22

Pvgyh

Pvgy 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 0.949

C2 0.025

C3 0.002

N2 0.016

CO2 0.007

C4 0.0003

iC4 0.0003

C5 0.0001

iC5 0.0001

O2 0.0002

Equation Example

Modified-Colebrook

2/12

105.2

2122

21

Re

825.27.3

log5972.26

06843.0

avgavg

st

avgavg

avg zTdL

fDD

PQ

zT

PHHdPP

Modified-Colebrook Results

125 175 225 275 3250

200

400

600

800

1000

1200

1400

1600

1800

NPS = 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)

Pre

ssu

re D

rop (

psi

a)

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)

Per

cen

t 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 – 2,240

Colebrook 2.4 – 10% 538 – 2,240

Modified-Colebrook

1.0 – 8.8% 224 – 1,971

AGA 0.2 – 15% 45 – 3,360

IGT 7.6 – 17% 1,702 – 3,808

Mueller 13 – 20% 2,912 – 4,480

Mathematical Model

General 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-II

NPS = 16; Analytical

NPS =18; Pro-II

NPS = 18; Analytical

NPS = 20; Pro-II

NPS = 20; Analytical

NPS = 22; Pro-II

NPS = 22; Analytical

Flow Rate (MMSCFD)

Pre

ssu

re D

rop (

psi

a)

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)P

erce

nt

Err

or

Equation Name

Range of Error

Cost of Error ($ of fuel

cost/compressor/yr)

Mathematical Model

0 – 0.9% 0 – 200

THE MATHEMATICAL MODEL VS. J-CURVE ANALYSIS

Landon Carroll

Wes Hudkins

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)

TA

C p

er

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)

TA

C p

er

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)

TA

C p

er

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)

TA

C p

er

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)

TA

C p

er M

CF

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 Seg...

Flow Rate (MMSCFD)

TA

C p

er M

CF

Order 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)

TA

C p

er M

CF

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)

TA

C p

er

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 Required

For 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 $ 66

2 850 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 Network

Pipe 1 Pipe 2

Pipe Diameter (in)

22 22

Compressor Work (hp)

10,740 0

Pressure Drop (psi)

1,830 1,490

TAC Model $ 0.596

TAC J-Curves $ 0.616

THE MATHEMATICAL MODEL

Landon Carroll

Wes Hudkins

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

Model

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

Economics

Project 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)

Hydraulics

Gas 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 Equation

Pressures

Model Logic - Output

Physical

Pipe LocationsPipe DiametersDemand at Each PeriodFlowratesInlet and Outlet PressuresCompressor LocationsCompressor Capacities

Economics

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

CASE STUDYLandon Carroll

Wes Hudkins

Case Study - Given

Fairfield

Supply P (kPa)

3548.7

Supply T (°R)

529.67

MinOP (kPa)

10050.5

MaxOP (kPa)

4200

Elevation (km)

0.185928

Mavis Mayberry Split Beaumont Travis

Initial Demand (Mcmd)

283.17 566.34 0 2831.7 1699

Price ($/m3) 0.32 0.33 0 0.3 0.3

Elevation (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 - Results

Non-Graphical Results

Pipe 1 ID (in.) 22

Pipe 2 ID (in.) 22

Pipe 3 ID (in.) 22

Pipe 4 ID (in.) 18

Pipe 5 ID (in.) 12

NPV ($) 4,392,078,000

NPTAC ($) 243,706,100

Pipe Cost ($) 185,720,700

Supplier Compressor Capacity (hp)

22,929.16

Consumer1 Compressor Capacity (hp)

13,365.09

Consumer2 Compressor Capacity (hp)

13,293.76

Consumer3 Compressor Capacity (hp)

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

illion

FCIinit=$303,036,750

CONCLUSIONSLandon Carroll

Wes 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.