Pipeline Hydraulics Fundamentals

1. Overview & Applications

Pipeline hydraulics governs the flow of natural gas through transmission and distribution systems. Understanding gas flow equations, pressure drop mechanisms, and compression requirements is fundamental to pipeline design, capacity analysis, and system optimization.

Pipeline design

Diameter & MAOP

Select pipe diameter and maximum allowable operating pressure for target capacity.

Capacity analysis

Flow rate calculations

Determine maximum throughput at operating conditions for existing systems.

Compression design

Station spacing & HP

Calculate compression station locations, discharge pressure, and horsepower.

System optimization

Operating efficiency

Optimize pressure profiles, loop sections, and fuel consumption.

Key Concepts

Flow rate (Q): Volumetric throughput at standard conditions, typically MMscfd or m³/day
Pressure drop (ΔP): Reduction in pressure due to friction and elevation, psi or bar
Friction factor (f): Dimensionless coefficient quantifying pipe wall resistance
Transmission factor (F): Efficiency factor accounting for pipe roughness and flow regime
Compressibility (Z): Real gas deviation factor, varies with pressure and temperature
Base conditions: Standard reference state (typically 14.73 psia, 60°F in US)

Why hydraulics matter: A 50 psi pressure drop across 100 miles may seem small, but it represents significant energy loss requiring expensive compression. Accurate hydraulic modeling prevents over-design (excessive capital cost) and under-design (insufficient capacity, inability to meet contract deliverability).

Flow Regime Characteristics

Flow Regime	Reynolds Number	Characteristics	Applicable Equations
Laminar	Re < 2,000	Smooth, layered flow; rare in pipelines	Hagen-Poiseuille
Transitional	2,000 < Re < 4,000	Unstable flow; avoid in design	Colebrook-White
Turbulent	Re > 4,000	Friction depends on roughness and Re	Panhandle A, Modified Colebrook
Fully turbulent	Re > 4,000,000	Friction independent of Re (rough pipe)	Weymouth, Panhandle B, AGA

Transmission pipelines typically operate at Re > 10 million, firmly in the fully turbulent regime. Distribution lines at lower pressures may operate in transitional turbulent flow.

2. Gas Flow Equations

Multiple empirical and semi-theoretical equations have been developed to predict gas flow in pipelines. Each equation has specific applicability ranges and assumptions. Selection depends on pipe diameter, flow regime, and industry standard practice.

General Flow Equation Form

Generalized Gas Flow Equation: All gas flow equations can be expressed in the form: Q = C × E × (Tb/Pb) × √[(P₁² - P₂²) / (G × Tf × Lm × Z)] × D^n Where: Q = Flow rate at base conditions (scf/day) C = Equation-specific constant E = Pipeline efficiency factor (0.92–0.98 typical) Tb = Base temperature (°R, usually 519.67°R = 60°F) Pb = Base pressure (psia, usually 14.73 psia) P₁ = Upstream pressure (psia) P₂ = Downstream pressure (psia) G = Gas specific gravity (dimensionless, air = 1.0) Tf = Average flowing temperature (°R) Lm = Pipeline length (miles) Z = Average compressibility factor (dimensionless) D = Inside diameter (inches) n = Diameter exponent (varies by equation: 2.5 to 2.667) The diameter exponent 'n' distinguishes the major equations.

Weymouth Equation

The Weymouth equation is the most conservative and widely used for large-diameter (≥ 12") high-pressure transmission lines. Recommended by AGA Report 8 for fully turbulent flow.

Weymouth Equation: Q = 433.5 × E × (Tb/Pb) × √[(P₁² - P₂²) / (G × Tf × Lm × Z)] × D^2.667 US customary units: Q = flow rate (scf/day at 14.73 psia, 60°F) P₁, P₂ = pressure (psia) D = inside diameter (inches) Lm = length (miles) Tf = average temperature (°R) Transmission factor form: Q = 433.5 × E × (Tb/Pb) × [(P₁² - P₂²) / (G × Tf × Lm × Z)]^0.5 × D^2.667 Or using transmission factor F: Q = F × (Tb/Pb) × [(P₁² - P₂²) / (G × Tf × Lm × Z)]^0.5 × D^2.667 Where F = 433.5 × E (transmission factor) Typical applicability: - Large diameter (D ≥ 12 inches) - High Reynolds number (Re > 4 million) - Rough pipe (commercial steel, ε ≈ 700 microinches) - Fully turbulent flow

Panhandle A Equation

Developed for smooth pipe and partially turbulent flow. Less conservative than Weymouth. Suitable for smaller diameter lines and lower Reynolds numbers.

Panhandle A Equation: Q = 435.87 × E × (Tb/Pb) × [(P₁² - P₂²) / (G^0.8539 × Tf × Lm × Z)]^0.5394 × D^2.6182 Key differences from Weymouth: - Gas gravity exponent: G^0.8539 (instead of G^1.0) - Pressure term exponent: 0.5394 (instead of 0.5) - Diameter exponent: 2.6182 (instead of 2.667) These exponents make Panhandle A predict higher flow rates than Weymouth for same conditions. Typical applicability: - Smooth pipe (new, clean steel) - Reynolds number: 5 million to 14 million - Partially turbulent to fully turbulent transition - D = 6 to 24 inches Warning: Panhandle A overestimates capacity for rough pipe.

Panhandle B Equation

Modified version of Panhandle A for fully turbulent flow in intermediate-roughness pipe. More accurate than Panhandle A for aged pipelines.

Panhandle B Equation: Q = 737 × E × (Tb/Pb) × [(P₁² - P₂²) / (G^0.961 × Tf × Lm × Z)]^0.51 × D^2.53 Key characteristics: - Higher leading constant (737 vs 435.87) - Gas gravity exponent: G^0.961 (closer to 1.0) - Pressure term exponent: 0.51 (closer to 0.5) - Diameter exponent: 2.53 (lower than Panhandle A) Results typically fall between Panhandle A and Weymouth. Typical applicability: - Average pipe roughness (ε = 400–600 microinches) - Reynolds number: Re > 10 million - Fully turbulent flow - D = 10 to 36 inches Most accurate for intermediate-age pipelines with moderate roughness.

AGA Fully Turbulent Equation

The American Gas Association fully turbulent equation uses the Colebrook-White friction factor in the fully turbulent limit. Provides theoretical basis for flow calculations.

AGA Fully Turbulent Equation: Q = 38.77 × E × D^2.5 × (Tb/Pb) × √[(P₁² - P₂²) / (f × G × Tf × Lm × Z)] Where friction factor f is calculated from: f = 1 / [2 × log₁₀(3.7 × D/ε)]² Or equivalently: f = 0.25 / [log₁₀(ε / (3.7 × D))]² Where: ε = absolute pipe roughness (inches) D = inside diameter (inches) Common roughness values: - New steel pipe: ε = 0.0007" (700 microinches) - Commercial steel (avg): ε = 0.0018" (1800 microinches) - Aged/corroded pipe: ε = 0.002 to 0.003" This equation allows custom roughness input and is preferred for detailed design. Advantage: Theoretically rigorous, based on Darcy-Weisbach equation. Disadvantage: Requires roughness estimate, which varies with pipe age and condition.

Equation Comparison

Equation	Diameter Exponent	Applicability	Conservatism	Industry Use
Weymouth	2.667	D ≥ 12", rough pipe, Re > 4M	Most conservative	AGA standard, transmission lines
Panhandle A	2.618	D = 6–24", smooth pipe, Re 5–14M	Least conservative	Smooth new pipe, gathering
Panhandle B	2.530	D = 10–36", avg roughness, Re > 10M	Moderate	General transmission, aged pipe
AGA Fully Turbulent	2.500	All D, custom roughness	Adjustable	Detailed design, capacity studies

Equation selection guideline: Use Weymouth for MAOP design (conservative, safe). Use Panhandle B or AGA for capacity analysis of existing lines (more accurate). Avoid Panhandle A for rough or aged pipe (over-predicts capacity). When in doubt, Weymouth is the safest choice and most widely accepted by regulators.

Example Calculation: Compare Equations

Calculate flow rate for a 20-inch ID pipeline, 100 miles long, with inlet pressure 1000 psia, outlet pressure 800 psia. Gas: SG = 0.6, Tf = 60°F, Z = 0.92, E = 0.95.

Given: D = 20 inches Lm = 100 miles P₁ = 1000 psia P₂ = 800 psia G = 0.6 Tf = 60 + 459.67 = 519.67 °R Z = 0.92 E = 0.95 Tb/Pb = 519.67/14.73 = 35.28 Pressure term: P₁² - P₂² = 1,000,000 - 640,000 = 360,000 psi² Weymouth: Q = 433.5 × 0.95 × 35.28 × √[360,000 / (0.6 × 519.67 × 100 × 0.92)] × 20^2.667 Q = 14,528 × √[360,000 / 28,724] × 5,278 Q = 14,528 × 3.541 × 5,278 Q = 271.8 MMscfd Panhandle A: Q = 435.87 × 0.95 × 35.28 × [360,000 / (0.6^0.8539 × 519.67 × 100 × 0.92)]^0.5394 × 20^2.6182 Q = 14,598 × [360,000 / 31,285]^0.5394 × 5,100 Q = 14,598 × 3.847 × 5,100 Q = 286.3 MMscfd (+5.3% vs Weymouth) Panhandle B: Q = 737 × 0.95 × 35.28 × [360,000 / (0.6^0.961 × 519.67 × 100 × 0.92)]^0.51 × 20^2.53 Q = 24,690 × [360,000 / 29,562]^0.51 × 4,162 Q = 24,690 × 3.706 × 4,162 Q = 278.1 MMscfd (+2.3% vs Weymouth) Summary: Weymouth: 271.8 MMscfd (most conservative) Panhandle B: 278.1 MMscfd (+2.3%) Panhandle A: 286.3 MMscfd (+5.3%, least conservative) For design: use Weymouth (271.8 MMscfd) For capacity analysis: use Panhandle B or AGA with known roughness

3. Pressure Drop Calculations

Pressure drop in gas pipelines results from friction losses and elevation changes. Accurate pressure drop prediction is essential for determining required inlet pressure, compression station locations, and pipeline operating costs.

Friction Factor

The Darcy-Weisbach friction factor quantifies resistance to flow. For fully turbulent flow, friction factor depends only on relative roughness (ε/D), not Reynolds number.

Darcy-Weisbach Friction Factor: For fully turbulent flow (Re > 4,000,000): f = 0.25 / [log₁₀(ε / (3.7 × D))]² Or equivalently (Colebrook-White in fully turbulent limit): 1/√f = -2 × log₁₀(ε / (3.7 × D)) Where: f = Darcy friction factor (dimensionless) ε = absolute roughness (inches or mm) D = inside diameter (inches or mm, same units as ε) Relative roughness: ε/D = absolute roughness / diameter Example: D = 20 inches, ε = 0.0018 inches (1800 microinches, commercial steel) ε/D = 0.0018 / 20 = 0.00009 f = 0.25 / [log₁₀(0.00009 / 3.7)]² = 0.25 / [log₁₀(0.0000243)]² f = 0.25 / [-4.614]² = 0.25 / 21.29 = 0.0117

Transmission Factor

The transmission factor F combines efficiency and friction effects into a single empirical parameter. Widely used in flow equation applications.

Transmission Factor: F = 2 / √f Where f is the Moody friction factor. Typical values: - New smooth pipe: F = 22–23 (f = 0.008) - Average pipe condition: F = 18–20 (f = 0.01–0.0123) - Rough/corroded pipe: F = 15–17 (f = 0.014–0.018) - Very rough pipe: F = 13–14 (f = 0.020–0.024) Relationship to efficiency E: E = F / (theoretical maximum F) For Weymouth equation: F_Weymouth = 433.5 × E If E = 0.95 (typical): F = 433.5 × 0.95 = 411.8 The transmission factor accounts for: - Pipe internal roughness - Bends and fittings - Valves and restrictions - Flow disturbances

Typical Friction Factors and Roughness

Pipe Condition	Absolute Roughness ε	Friction Factor f (D=20")	Transmission Factor F
New, smooth steel	0.0007" (700 μin)	0.0098	20.2
Commercial steel (new)	0.0018" (1800 μin)	0.0117	18.5
Average service	0.002" (2000 μin)	0.0123	18.0
Aged/light corrosion	0.003" (3000 μin)	0.0145	16.6
Heavily corroded	0.005" (5000 μin)	0.0178	15.0

Elevation Effects

Elevation change affects pressure drop through hydrostatic head. Uphill flow requires additional pressure; downhill flow reduces required inlet pressure.

Elevation Correction: ΔP_elevation = 0.0375 × ρ_avg × ΔZ Where: ΔP_elevation = pressure change due to elevation (psi) ρ_avg = average gas density (lb/ft³) ΔZ = elevation change (feet, positive for uphill) Gas density: ρ_avg = (P_avg × MW) / (Z_avg × R × T_avg) ρ_avg = (P_avg × G × 28.97) / (Z_avg × 10.73 × T_avg) For significant elevation changes: P₂² = P₁² - [(ΔP_friction)² + ΔP_elevation²]^0.5 Or use segment calculation for complex terrain. Example: P_avg = 900 psia, T_avg = 60°F (519.67°R), G = 0.6, Z = 0.92 ρ_avg = (900 × 0.6 × 28.97) / (0.92 × 10.73 × 519.67) = 3.05 lb/ft³ Elevation increase: ΔZ = +1000 feet ΔP_elevation = 0.0375 × 3.05 × 1000 = 114 psi For uphill flow, required inlet pressure increases by ~114 psi to overcome gravity. For downhill flow (-1000 ft), inlet pressure can be 114 psi lower.

Pressure Drop Calculation Methods

Method 1: Average Pressure Method

Simplest method, suitable for ΔP < 10% of P₁. Uses average pressure and temperature.

Average Pressure Method: P_avg = (P₁ + P₂) / 2 T_avg = (T₁ + T₂) / 2 Z_avg = Z(P_avg, T_avg) Calculate Q using P_avg in flow equation, or solve for P₂. Limitation: Inaccurate for large pressure drops (ΔP > 100 psi).

Method 2: Pressure Squared Method

Most common method in industry. Implicit in all major flow equations (Weymouth, Panhandle).

Pressure Squared Method: Flow equations naturally use (P₁² - P₂²) term. For average properties: P_avg = √[(P₁² + P₂²) / 2] (root-mean-square average) Or: P_avg = (P₁ + P₂) / 2 (arithmetic average, common simplification) Calculate Z at P_avg. This method is built into Weymouth, Panhandle A/B equations.

Method 3: Segmented Calculation

Most accurate for long pipelines, complex terrain, or variable pipe diameter. Divide pipeline into segments and calculate sequentially.

Segmented Method: 1. Divide pipeline into n segments (10–50 segments typical) 2. For each segment i: - Calculate P_avg,i = (P_i + P_{i+1}) / 2 - Calculate Z_i = Z(P_avg,i, T_avg,i) - Apply elevation correction if needed - Calculate P_{i+1} using flow equation 3. Outlet pressure = P_n Advantages: - Handles variable elevation accurately - Accounts for changing Z along pipeline - Allows variable diameter (taper or loops) - Required for pipes longer than 50 miles with elevation changes Computational note: Requires iterative solution if flow rate is given.

Example: Pressure Drop Calculation

Calculate pressure drop for 100 MMscfd in a 16-inch ID pipeline, 75 miles long. Inlet P₁ = 1200 psia, T = 70°F, G = 0.65, Z = 0.88, E = 0.92.

Given: Q = 100 MMscfd = 100 × 10⁶ scfd D = 16 inches Lm = 75 miles P₁ = 1200 psia Tf = 70 + 459.67 = 529.67 °R G = 0.65 Z = 0.88 E = 0.92 Tb/Pb = 519.67/14.73 = 35.28 Step 1: Rearrange Weymouth for P₂² Q = 433.5 × E × (Tb/Pb) × √[(P₁² - P₂²) / (G × Tf × Lm × Z)] × D^2.667 Solve for P₂²: P₁² - P₂² = [Q / (433.5 × E × (Tb/Pb) × D^2.667)]² × (G × Tf × Lm × Z) Step 2: Calculate terms D^2.667 = 16^2.667 = 3,249 Flow coefficient: 433.5 × 0.92 × 35.28 × 3,249 = 45,742,000 Normalized flow: Q_norm = 100,000,000 / 45,742,000 = 2.186 Resistance term: G × Tf × Lm × Z = 0.65 × 529.67 × 75 × 0.88 = 22,728 Step 3: Calculate pressure drop P₁² - P₂² = (2.186)² × 22,728 = 108,603 psi² P₂² = P₁² - 108,603 P₂² = 1,440,000 - 108,603 = 1,331,397 psi² P₂ = √1,331,397 = 1154 psia Result: Outlet pressure: P₂ = 1154 psia Pressure drop: ΔP = 1200 - 1154 = 46 psi Pressure drop per mile: 46 psi / 75 mi = 0.61 psi/mi Pressure drop per 100 miles: 61 psi/100 mi (typical range)

4. Compression Stations

Compression stations restore pressure lost to friction and elevation, enabling long-distance gas transmission. Proper compression design balances capital cost (number and size of stations) against operating cost (fuel consumption).

Compression Station Spacing

Station spacing depends on allowable pressure drop, pipeline diameter, flow rate, and MAOP. Optimal spacing minimizes total lifecycle cost.

Maximum Spacing Calculation: Given MAOP (Maximum Allowable Operating Pressure) and minimum delivery pressure P_min: Maximum spacing between stations: Lm_max = (P_discharge² - P_min²) / [(Q / (C × E × D^n))² × (G × Tf × Z)] Where: P_discharge = Compressor station discharge pressure (psia) ≤ MAOP (usually MAOP × 0.95 for safety margin) P_min = Minimum suction pressure for next station or delivery (psia) Typically 600–800 psia for transmission Practical considerations: - Avoid suction pressure below 400 psia (compressor efficiency drops) - Discharge pressure limited by MAOP and pipe class - Add 10% margin for flow variation and elevation Example spacing for common pipeline: MAOP = 1440 psia (80% SMYS, 0.500" WT, X70) P_discharge = 1440 × 0.95 = 1368 psia P_min = 700 psia (typical) D = 24 inches Q = 400 MMscfd Lm_max = 100–150 miles (typical result)

Compression Ratio

Compression ratio is the outlet pressure divided by inlet pressure. Affects compressor efficiency, fuel consumption, and discharge temperature.

Compression Ratio: r_c = P_discharge / P_suction Typical ranges: - Single-stage centrifugal: r_c = 1.2 to 1.8 (max 2.0) - Multi-stage centrifugal: r_c = 1.05 to 1.5 per stage - Reciprocating: r_c = 2.0 to 4.0 per stage Guidelines: - Lower ratios → better efficiency, lower discharge temperature - Higher ratios → fewer stations, higher capital cost per station - Optimal: r_c = 1.3–1.5 for centrifugal compressors Discharge temperature (adiabatic): T_discharge = T_suction × r_c^[(k-1)/k] Where k = Cp/Cv ≈ 1.27 for natural gas Example: T_suction = 80°F (539.67°R) r_c = 1.5 T_discharge = 539.67 × 1.5^[(1.27-1)/1.27] = 539.67 × 1.5^0.2126 T_discharge = 539.67 × 1.093 = 590°R = 130°F Cooling often required if T_discharge > 140°F to prevent thermal stress.

Compression Horsepower

Compressor horsepower depends on flow rate, compression ratio, inlet conditions, and compressor efficiency. Accurately predicting horsepower is critical for sizing prime movers and estimating fuel costs.

Adiabatic Compression Horsepower: BHP = (Q_inlet × P_suction × Z_avg) / (229.1 × η_c) × [(k/(k-1)) × (r_c^[(k-1)/k] - 1)] Where: BHP = Brake horsepower required (hp) Q_inlet = Inlet volumetric flow rate (MMscfd at actual P, T) P_suction = Suction pressure (psia) Z_avg = Average compressibility factor ≈ (Z_suction + Z_discharge)/2 η_c = Compressor adiabatic efficiency (0.75–0.85 typical) k = Specific heat ratio Cp/Cv (≈ 1.27 for natural gas) r_c = Compression ratio P_discharge / P_suction Convert standard flow to inlet volumetric flow: Q_inlet = Q_std × (P_base/P_suction) × (T_suction/T_base) × (Z_suction/Z_base) Simplified form (for quick estimates): BHP ≈ Q_std × [(P_discharge - P_suction) / P_suction] / 16.5 Where Q_std is in MMscfd. Example: Q_std = 400 MMscfd P_suction = 700 psia P_discharge = 1350 psia r_c = 1350/700 = 1.929 T_suction = 80°F = 539.67°R Z_avg = 0.88 η_c = 0.80 k = 1.27 BHP = (400 × 700 × 0.88) / (229.1 × 0.80) × [(1.27/0.27) × (1.929^0.2126 - 1)] BHP = (246,400) / (183.3) × [4.704 × (1.187 - 1)] BHP = 1,344 × 4.704 × 0.187 BHP = 1,181 hp Typical prime mover: 1,250 hp or 1,500 hp gas turbine or engine.

Fuel Consumption

Station fuel consumption is a major operating expense. Fuel gas is consumed by prime movers (gas turbines or engines) driving compressors.

Fuel Gas Consumption: Fuel consumption (scfd) = BHP × BSFC × 24 Where: BHP = Brake horsepower from calculation above BSFC = Brake specific fuel consumption Gas turbine: 9,000–11,000 Btu/hp·hr (7–8 scf/hp·hr for 1000 Btu/scf gas) Gas engine: 7,000–8,500 Btu/hp·hr (5.5–6.5 scf/hp·hr) 24 = hours per day For gas turbine (BSFC = 10,000 Btu/hp·hr): Fuel (scfd) = 1,181 hp × 10,000 Btu/hp·hr × 24 hr/day / 1000 Btu/scf Fuel (scfd) = 1,181 × 10 × 24 = 283,440 scfd Fuel (MMscfd) = 0.283 MMscfd As percentage of throughput: Fuel % = 0.283 / 400 × 100% = 0.071% (typical for single station) For multi-station pipeline: If 5 stations @ 1,181 hp each: Total fuel = 5 × 0.283 = 1.42 MMscfd Fuel % = 1.42 / 400 = 0.35% of throughput Fuel cost (at $3.00/MMBtu): Annual cost per station = 0.283 MMscfd × 365 days × $3.00/MMBtu × 1000 Annual cost = $309,900 per station

Compression Station Design Parameters

Parameter	Typical Range	Design Consideration
Suction pressure	600–900 psia	Avoid < 400 psia (efficiency loss)
Discharge pressure	1200–1440 psia	Limited by MAOP (class, WT, grade)
Compression ratio (centrifugal)	1.3–1.8	Optimal 1.4–1.5 for efficiency
Station spacing	50–150 miles	Balance capital vs operating cost
Compressor efficiency	75–85%	Modern centrifugal 80–83%
Discharge temperature	100–140°F	Aftercooling if > 140°F
Fuel consumption	0.3–0.5% throughput	Major operating expense

Station Configuration Options

Series configuration: Multiple compressor units in series increase compression ratio. Used for high ΔP requirements.
Parallel configuration: Multiple units in parallel increase total flow capacity. Provides redundancy and flexibility.
Series-parallel: Combination provides both high pressure ratio and high capacity with redundancy.
Spare capacity: Typical design: N+1 (one spare) or 3×50% (any two provide full capacity).

Compression optimization: Adding more stations with lower compression ratios improves fuel efficiency but increases capital cost. Optimal design minimizes net present value of capital plus 20-year operating costs. Rule of thumb: station spacing of 75–100 miles with r_c = 1.4–1.5 balances these factors for most transmission pipelines.

5. Practical Applications

Pipeline Design Example

Design a pipeline to transport 500 MMscfd natural gas over 200 miles. Determine required diameter, MAOP, and number of compression stations.

Design Criteria: Q = 500 MMscfd (design flow) Lm = 200 miles G = 0.60 Tf = 60°F = 519.67°R Z = 0.90 (assume average) E = 0.95 (efficiency) Step 1: Select trial diameter Rule of thumb: D (inches) ≈ 3.5 × Q^0.5 (for Q in MMscfd) D_trial = 3.5 × √500 = 3.5 × 22.4 = 78 inches Too large. Try D = 30 inches (common large pipeline size). Step 2: Determine MAOP based on pipe class Assume X70 steel (SMYS = 70,000 psi), 0.500" wall thickness Design factor F = 0.72 (ASME B31.8 Class 1 location) MAOP = (2 × SMYS × WT × F × E_weld × T) / OD Where: SMYS = 70,000 psi WT = 0.500 inches F = 0.72 E_weld = 1.0 (seamless or 100% radiography) T = 1.0 (temperature derating factor) OD = 30 inches MAOP = (2 × 70,000 × 0.500 × 0.72 × 1.0 × 1.0) / 30 MAOP = 50,400 / 30 = 1,680 psia Round to: MAOP = 1440 psia (standard operating limit) Step 3: Calculate required inlet pressure (no compression) Using Weymouth equation, solve for P₁ given P₂ (delivery pressure): P₂ = 1000 psia (delivery requirement) P₁² = P₂² + [Q / (433.5 × E × (Tb/Pb) × D^2.667)]² × (G × Tf × Lm × Z) D^2.667 = 30^2.667 = 13,572 Flow term = 500,000,000 / (433.5 × 0.95 × 35.28 × 13,572) = 2.686 Resistance = 0.60 × 519.67 × 200 × 0.90 = 56,184 P₁² = 1,000,000 + (2.686)² × 56,184 P₁² = 1,000,000 + 405,488 = 1,405,488 P₁ = 1,185 psia (within MAOP, OK) Result without compression: 30-inch pipeline can deliver 500 MMscfd over 200 miles with inlet P₁ = 1185 psia (no stations). Step 4: Optimize with compression Alternative: Use compression to maintain higher average pressure (reduce diameter). Try D = 24 inches: D^2.667 = 24^2.667 = 7,003 Without compression over 200 miles: P₁² = 1,000,000 + [500M / (433.5 × 0.95 × 35.28 × 7,003)]² × 56,184 P₁² = 1,000,000 + (5.203)² × 56,184 = 1,000,000 + 1,521,175 P₁ = 1,588 psia (exceeds MAOP = 1440 psia, NOT OK) Add 1 compressor station at midpoint (100 miles): Segment 1: P₁ = 1400 psia → P₂ = ? (100 miles) P₂² = 1,960,000 - [5.203² × (56,184/2)] P₂² = 1,960,000 - 760,588 = 1,199,412 P₂ = 1,095 psia (station suction) Compress to P₃ = 1400 psia (r_c = 1.28) Segment 2: P₃ = 1400 psia → P₄ = ? (100 miles) P₄² = 1,960,000 - 760,588 = 1,199,412 P₄ = 1,095 psia (exceeds delivery requirement of 1000 psia, OK) Final design: 24-inch pipeline, MAOP 1440 psia, 1 compressor station at mile 100 Capacity: 500 MMscfd with P₁ = 1400 psia, P_delivery = 1095 psia Compression ratio: 1.28, BHP ≈ 1,850 hp per station Capital cost comparison: - 30" no compression: Higher pipe cost, $0 compression cost - 24" with 1 station: Lower pipe cost, ~$15M station cost Typically 24" with compression is more economical.

Capacity Analysis of Existing Pipeline

Existing pipeline: 20-inch ID, 150 miles, MAOP 1200 psia. Current flow 200 MMscfd. Can capacity be increased to 300 MMscfd?

Given: D = 20 inches (ID) Lm = 150 miles MAOP = 1200 psia P_delivery = 800 psia (contract requirement) Current Q = 200 MMscfd Step 1: Calculate maximum capacity at MAOP Using Weymouth equation: Q_max = 433.5 × 0.92 × 35.28 × √[(1200² - 800²) / (0.6 × 520 × 150 × 0.90)] × 20^2.667 P₁² - P₂² = 1,440,000 - 640,000 = 800,000 psi² Denominator = 0.6 × 520 × 150 × 0.90 = 42,120 Q_max = 14,064 × √[800,000 / 42,120] × 5,278 Q_max = 14,064 × 4.36 × 5,278 Q_max = 323.5 MMscfd Result: Maximum capacity = 323.5 MMscfd (exceeds target of 300 MMscfd, OK) Step 2: Calculate required inlet pressure for 300 MMscfd P₁² = P₂² + [300M / (433.5 × 0.92 × 35.28 × 5,278)]² × 42,120 P₁² = 640,000 + (3.918)² × 42,120 P₁² = 640,000 + 646,285 = 1,286,285 P₁ = 1,134 psia (within MAOP, OK) Conclusion: Existing 20-inch pipeline can accommodate 300 MMscfd increase by raising inlet pressure to 1,134 psia. No diameter increase or compression required. Verification check: - Operating at 94.5% of MAOP (1134/1200) - Pressure drop: 334 psi over 150 miles = 2.23 psi/mile - Within typical range for transmission pipelines - Recommendation: Proceed with uprate

Loop Line Design

Loop lines are parallel pipe sections installed to increase capacity without increasing operating pressure. Common for system expansions.

Loop Line Capacity Increase: For two parallel pipes of equal length: Q_total = √(Q₁² + Q₂²) If D₁ = D₂ (same diameter): Q_total = √2 × Q₁ = 1.414 × Q₁ If D_loop < D_main: Q_loop = Q_main × (D_loop/D_main)^2.667 (Weymouth exponent) Example: Existing 24" line: Q = 400 MMscfd Add 20" parallel loop for 50 miles (of 100 mile total length) Capacity increase from loop: Q_loop = 400 × (20/24)^2.667 = 400 × (0.833)^2.667 = 400 × 0.719 = 288 MMscfd But loop is only 50 miles (half the length): Effective increase ≈ 50% of 288 = 144 MMscfd additional capacity New total capacity ≈ 400 + 144 = 544 MMscfd Accurate calculation requires segmented analysis: - Segment 1 (0-50 mi): Parallel 24" + 20" pipes - Segment 2 (50-100 mi): Single 24" pipe Loop lines provide capacity increase at fraction of full replacement cost.

Erosional Velocity Limit

High gas velocity can cause erosion and premature failure. API RP 14E provides velocity limits.

API RP 14E Erosional Velocity: V_max = C / √ρ Where: V_max = Maximum safe velocity (ft/s) C = Empirical constant C = 100 (continuous service, non-corrosive) C = 125 (intermittent service, clean gas) C = 150 (temporary operation) ρ = Gas density at operating conditions (lb/ft³) Gas density: ρ = (P × MW) / (Z × R × T) = (P × G × 28.97) / (Z × 10.73 × T) Check actual velocity: V_actual = Q_actual / A Where: Q_actual = volumetric flow at operating P, T (ft³/s) A = pipe cross-sectional area (ft²) Example: Pipeline: D = 24" (ID), P = 1000 psia, T = 80°F, Q = 500 MMscfd, G = 0.6, Z = 0.9 Step 1: Calculate density T = 80 + 459.67 = 539.67°R ρ = (1000 × 0.6 × 28.97) / (0.9 × 10.73 × 539.67) = 3.35 lb/ft³ Step 2: Calculate erosional velocity limit V_max = 100 / √3.35 = 100 / 1.83 = 54.6 ft/s Step 3: Calculate actual velocity Convert Q to actual conditions: Q_actual = Q_std × (P_base/P) × (T/T_base) × (Z/Z_base) Q_actual = 500 × 10⁶ × (14.73/1000) × (539.67/519.67) × (0.9/1.0) scfd Q_actual = 500 × 10⁶ × 0.01473 × 1.0385 × 0.9 = 6,903,000 scfd Q_actual = 6,903,000 / 86,400 = 79.9 ft³/s Area: A = π × (24/12)² / 4 = 3.14 ft² V_actual = 79.9 / 3.14 = 25.4 ft/s Step 4: Check ratio V_actual / V_max = 25.4 / 54.6 = 0.465 (46.5% of limit, OK) Erosional velocity is rarely limiting in large transmission pipelines but can be critical in smaller gathering lines or high-flow scenarios.

Optimization: Pipe Diameter Selection

Select optimal diameter by minimizing total cost (capital + operating) over project life.

Economic Optimization: Total cost = Capital cost + Present value of operating costs Capital cost: - Pipeline material and installation: ∝ D × Lm - Compression stations: ∝ BHP × (number of stations) Operating cost (annual): - Fuel consumption: ∝ BHP × stations - Maintenance and O&M Trade-offs: - Larger diameter → higher capital cost, lower pressure drop, fewer stations - Smaller diameter → lower capital cost, higher pressure drop, more stations, higher fuel cost Optimal diameter typically results in: - Pressure drop: 100–150 psi per 100 miles - Velocity: 20–40 ft/s at average conditions - Station spacing: 75–125 miles Sensitivity analysis required for: - Fuel cost ($/MMBtu) - Interest rate / discount rate - Expected throughput growth - Project lifetime (typically 30–40 years) Rule of thumb: If fuel cost doubles, optimal diameter increases ~1–2 sizes.

Common Design Errors and Pitfalls

Using gauge pressure instead of absolute: Flow equations require absolute pressure. Always add atmospheric (14.7 psia).
Incorrect base conditions: Verify contract base (14.73 psia and 60°F is common, but 15.025 psia and 60°F also used).
Ignoring elevation: 1000 ft elevation change ≈ 40–50 psi pressure effect. Critical in mountainous terrain.
Assuming constant Z-factor: Z varies with P and T. Use average or segment calculation for long lines.
Wrong flow equation for application: Panhandle A overestimates capacity for rough pipe. Use Weymouth for design conservatism.
Neglecting pipeline efficiency: E = 1.0 is theoretical maximum. Use E = 0.92–0.95 to account for bends, fittings, age.
Ignoring future corrosion: Internal roughness increases over time. Design with aged roughness assumption.
Insufficient MAOP margin: Operating at 100% MAOP leaves no flexibility. Design for 95% MAOP max continuous operation.

Industry Standards and References

AGA Report No. 8: Compressibility Factors of Natural Gas and Other Related Hydrocarbon Gases
ASME B31.8: Gas Transmission and Distribution Piping Systems (design pressure, safety factors)
API RP 14E: Recommended Practice for Design and Installation of Offshore Production Platform Piping Systems (erosional velocity)
GPA 2145: Physical Constants of Paraffin Hydrocarbons and Other Components of Natural Gas
GPSA: Section 12 (Gas Transmission), Section 13 (Compressors)
49 CFR Part 192: Transportation of Natural and Other Gas by Pipeline: Minimum Federal Safety Standards

Best practice: Use conservative assumptions for design (Weymouth equation, E = 0.92, aged pipe roughness, 95% MAOP limit). Use realistic assumptions for capacity analysis (Panhandle B or AGA, measured roughness, actual operating pressure). Document all assumptions clearly. Independent verification recommended for major projects.

Pipeline Hydraulics Calculations

Fully turbulent

Weymouth for NPS ≥ 12"

50-200 psi / 100 mi

Contents

Use this guide when you need to: