Worked Example: Gas Hydrate Prediction & Inhibitor Sizing Step-by-Step

In this worked example, we perform a complete, step-by-step flow assurance calculation to predict hydrate formation conditions and size chemical inhibitor injection rates for an offshore subsea gas tieback.

Problem Statement

A subsea natural gas well is tied back to a host platform via a 35 km flowline: * Operating Pressure ($P$): $85 \text{ barg}$ ($86.01 \text{ bar(a)}$) * Minimum Seabed Temperature ($T_{op}$): $4.0^\circ\text{C}$ ($39.2^\circ\text{F}$) * Gas Flow Rate ($Q$): $3,000,000 \text{ Sm}^3/\text{day}$ ($\approx 106 \text{ MMSCFD}$) * Gas Specific Gravity ($G$): $0.65$ (relative to air) * Inlet Wellhead Temperature: $45^\circ\text{C}$ (water saturated) * Required Safety Margin: $3.0^\circ\text{C}$

Step 1: Predict Hydrate Formation Temperature ($T_{hyd}$)

Convert pipeline pressure to kilopascals: $$P_{kPa} = 86.01 \times 100 = 8,601 \text{ kPa} \implies \log_{10}(P_{kPa}) = 3.9345$$

Apply the Motiee correlation: $$T_{hyd} = -238.29 + 78.997(3.9345) - 5.352(3.9345)^2 + [34.904 + 0.5527(3.9345)](0.65)$$ $$T_{hyd} = -238.29 + 310.81 - 82.85 + (37.078)(0.65) = 13.78^\circ\text{C}$$

Conclusion: At $85 \text{ barg}$, hydrates will crystallize at any temperature below $13.8^\circ\text{C}$. Because the seabed is at $4.0^\circ\text{C}$, severe hydrate plugging will occur without inhibition!

Step 2: Determine Required Temperature Depression ($\Delta T$)

To ensure the flowline operates safely outside the hydrate envelope: $$T_{target} = T_{op} - \Delta T_{safe} = 4.0^\circ\text{C} - 3.0^\circ\text{C} = 1.0^\circ\text{C}$$ $$\Delta T = T_{hyd} - T_{target} = 13.8^\circ\text{C} - 1.0^\circ\text{C} = 12.8^\circ\text{C}$$

Step 3: Solve Inhibitor Concentration ($W$) via Hammerschmidt

Using the inverted Hammerschmidt equation ($W = \frac{100 M \Delta T}{K_H + M \Delta T}$):

For Methanol ($M = 32.04, K_H = 1297$):

$$W_{MeOH} = \frac{100 \times 32.04 \times 12.8}{1297 + (32.04 \times 12.8)} = \frac{41,011}{1,297 + 410.1} = 24.0 \text{ wt}\%$$

For MEG ($M = 62.07, K_H = 1500$):

$$W_{MEG} = \frac{100 \times 62.07 \times 12.8}{1500 + (62.07 \times 12.8)} = \frac{79,450}{1,500 + 794.5} = 34.6 \text{ wt}\%$$

Step 4: Moisture Condensation & Water Rate

Using the Bukacek correlation: * Water content at wellhead ($45^\circ\text{C}, 85\text{ bar}$): $\approx 1.25 \text{ g/Sm}^3$ * Water content at seabed ($4^\circ\text{C}, 85\text{ bar}$): $\approx 0.18 \text{ g/Sm}^3$ * Water condensed out: $\Delta W = 1.25 - 0.18 = 1.07 \text{ g/Sm}^3$

Daily free liquid water rate: $$\dot{m}_{water} = 1.07 \text{ g/Sm}^3 \times 3,000,000 \text{ Sm}^3/\text{day} = 3,210 \text{ kg/day}$$

Step 5: Chemical Dosage & Dosing Pump Rate

Liquid Phase Dosage:

$$\dot{m}_{MeOH, liq} = \dot{m}_{water} \left( \frac{W}{100 - W} \right) = 3,210 \left( \frac{24.0}{76.0} \right) = 1,014 \text{ kg/day}$$

Methanol Vaporization Loss Allowance:

At $85\text{ barg}$ ($\approx 1247\text{ psia}$), vapor loss $\approx 10.5\text{ lb/MMSCF} = 4.76\text{ kg/MMSCF}$. $$\dot{m}_{vap} = 106 \text{ MMSCFD} \times 4.76 \text{ kg/MMSCF} = 505 \text{ kg/day}$$

Total Required Methanol Injection:

$$\dot{m}_{total} = 1,014 + 505 = 1,519 \text{ kg/day} \quad (63.3 \text{ kg/h})$$

Volumetric injection rate ($\rho_{MeOH} = 792 \text{ kg/m}^3$): $$\dot{V} = \frac{63.3}{792} \times 1000 = 79.9 \text{ L/hr} \quad (0.352 \text{ US GPM})$$