How to simulate wellhead compression: what the reservoir gives and what it costs to move

A gas well is not a feed whose flow you type in. The rate is an OUTCOME of how hard the facility pulls on the reservoir, and this example starts there. The **well** runs the Vogel inflow-performance relationship: at a 250 bar reservoir pressure and a 5e-5 mol/s/Pa productivity index its absolute open flow is `J x p_r / 1.8` = **694 mol/s**, and the fraction of that you actually get follows `1 - 0.2(p_wf/p_r) - 0.8(p_wf/p_r)^2`. Holding 180 bar at the bottomhole delivers **306 mol/s, 44% of open flow**. Squeeze to 150 bar and it rises to 411; back off to 240 and it collapses to 49. The curve is deliberately not a straight line — that curvature is why compression pays for itself, and why the last increment of drawdown buys less than the first. **Compression is where the stage count earns its keep.** Taking that gas from 30 to 150 bar in ONE stage lands the discharge at **508 K (235 C)** — past what reciprocating machine valves and lube oil tolerate, before any efficiency argument. Split it into three with intercooling to 313 K and the discharge is **362 K**; four stages give 349 K. The temperature, not the power, is what sets the stage count on a real machine. **Read the reported duty carefully.** `duty` on this unit is the NET of the compression work and the intercooler heat removed, summed into one number — it goes from +2094 kW at one stage to -384 kW at three, not because compression got cheaper but because there is now intercooling to subtract. It is not shaft power, and this example does not present it as such.

Frequently asked questions

What does the Wellhead compression: what the reservoir gives and what it costs to move model simulate?

A gas well is not a feed whose flow you type in. The rate is an OUTCOME of how hard the facility pulls on the reservoir, and this example starts there. The **well** runs the Vogel inflow-performance relationship: at a 250 bar reservoir pressure and a 5e-5 mol/s/Pa productivity index its absolute open flow is `J x p_r / 1.8` = **694 mol/s**, and the fraction of that you actually get follows `1 - 0.2(p_wf/p_r) - 0.8(p_wf/p_r)^2`. Holding 180 bar at the bottomhole delivers **306 mol/s, 44% of open flow**. Squeeze to 150 bar and it rises to 411; back off to 240 and it collapses to 49. The curve is deliberately not a straight line — that curvature is why compression pays for itself, and why the last increment of drawdown buys less than the first. **Compression is where the stage count earns its keep.** Taking that gas from 30 to 150 bar in ONE stage lands the discharge at **508 K (235 C)** — past what reciprocating machine valves and lube oil tolerate, before any efficiency argument. Split it into three with intercooling to 313 K and the discharge is **362 K**; four stages give 349 K. The temperature, not the power, is what sets the stage count on a real machine. **Read the reported duty carefully.** `duty` on this unit is the NET of the compression work and the intercooler heat removed, summed into one number — it goes from +2094 kW at one stage to -384 kW at three, not because compression got cheaper but because there is now intercooling to subtract. It is not shaft power, and this example does not present it as such.

Which thermodynamic method does it use?

The PENG-ROBINSON property package, over methane, ethane, propane, co2 — already selected. You can switch the method on the canvas before running.

Which unit operations are in the flowsheet?

It chains WELL, COMP. Every block is a real, solvable unit operation you can reconfigure, add to, or remove.

Do I need to install software or buy a license?

No. Wellhead compression: what the reservoir gives and what it costs to move runs entirely in your browser on MaximaLabs — free, no install, no license. Open the model to load it live and run the deterministic solver.