Nodal Analysis: IPR Meets VLP

Pick a node - usually the bottomhole - and plot what the reservoir can deliver against what the tubing can lift. Where the two curves cross, supply meets demand, and that crossing is the one rate and pressure at which the well actually produces.

Petroleum EngineeringProduction EngineeringFree preview
⏱️ About 20 min
Nodal Analysis: IPR Meets VLP — illustration
Illustrative image (AI-generated).

Plot what the reservoir can deliver against what the tubing can lift - where the two curves cross is the one rate and pressure at which the well actually produces.

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The big idea: Nodal analysis picks a node (usually the bottomhole) and plots two curves against rate - inflow (the IPR, reservoir delivering fluid to the node) and outflow (the VLP, tubing carrying fluid from the node to surface); their intersection is the operating point, the unique rate and bottomhole flowing pressure at which supply equals demand, and anything that shifts either curve moves the operating point.
🎯 By the end, you'll be able to
  • Define nodal analysis as choosing a node (usually the bottomhole) and plotting inflow and outflow curves against rate
  • Identify the inflow curve as the IPR (reservoir delivering fluid to the node) and the outflow curve as the VLP (tubing carrying fluid from the node to surface)
  • Explain that the intersection of the two curves is the operating point - the unique rate and bottomhole flowing pressure at which supply equals demand
  • Solve for the operating rate and pressure when inflow equals outflow, and predict how a shifted curve moves the operating point

Pick a Node, Plot Two Curves

A producing well is a chain of components - reservoir, near-wellbore zone, tubing, wellhead, surface choke - and each one relates pressure to rate in its own way. Nodal analysis simplifies that chain by picking a single point, the node - almost always the bottomhole - and looking at what flows into it and what flows out of it. From below, the inflow curve is the reservoir's IPR: the rate the formation can deliver to the node as the bottomhole pressure is drawn down. From above, the outflow curve is the tubing's VLP: the bottomhole pressure needed to carry that rate from the node up to surface. Plotted together against rate, the two curves describe the whole producing system on one graph - supply from below, demand from above.

The Operating Point: Where Supply Meets Demand

The reservoir will only deliver what the tubing can lift, and the tubing will only lift what the reservoir supplies - so the well must settle where the two agree. That place is the operating point: the unique rate and bottomhole flowing pressure at which inflow equals outflow, the intersection of the IPR and VLP curves. At any other rate the curves disagree - supply would exceed demand, or demand exceed supply - and the well cannot stay there. The power of nodal analysis is that anything that shifts either curve moves the operating point: a bigger tubing lowers the VLP, a lower wellhead pressure lowers the VLP, a higher productivity index raises the IPR, and adding artificial lift reshapes the outflow curve entirely. To predict what a change does to production, shift the curve and read the new intersection.

📝 Worked example: A well has a straight-line IPR $p_{wf} = \bar p - q/J$ with average reservoir pressure $\bar p = 3{,}000$ psi and productivity index $J = 1.0$ STB/d/psi (so $p_{wf} = 3000 - q$), and a simple tubing VLP $p_{wf} = p_{wh,\text{head}} + c\,q^2$ with a head term $p_{wh,\text{head}} = 1{,}000$ psi and coefficient $c = 0.001$ psi/(STB/d)$^2$ (so $p_{wf} = 1000 + 0.001\,q^2$). Find the operating rate and the operating bottomhole flowing pressure.
  1. At the operating point inflow equals outflow: $3000 - q = 1000 + 0.001\,q^2$.
  2. Rearrange: $0.001\,q^2 + q - 2000 = 0$, or multiplying by 1,000, $q^2 + 1000\,q - 2{,}000{,}000 = 0$.
  3. Solve the quadratic: $q = \dfrac{-1000 + \sqrt{1000^2 + 4(2{,}000{,}000)}}{2} = \dfrac{-1000 + \sqrt{9{,}000{,}000}}{2} = \dfrac{-1000 + 3000}{2}$.
  4. $q = 2000/2 = 1{,}000$ STB/d.
  5. Operating pressure from the IPR: $p_{wf} = 3000 - 1000 = 2{,}000$ psi (check with the VLP: $1000 + 0.001 \times 1000^2 = 1000 + 1000 = 2{,}000$ psi).
✓ The well settles at an operating rate of $q = 1{,}000$ STB/d and an operating bottomhole flowing pressure of $p_{wf} = 2{,}000$ psi - the one point where the reservoir's inflow and the tubing's outflow agree.
✨ Drag the Curves Yourself

The whole point of nodal analysis is seeing how a shifted curve moves the operating point - and the best way to feel that is to move the curves yourself. This module's interactive IPR/VLP nodal-analysis plot lets you slide the reservoir pressure and productivity index (which reshape the inflow IPR) and the tubing head and friction (which reshape the outflow VLP), and watch the operating point slide along to a new rate and pressure. Try raising the productivity index $J$ (lifts the IPR) and watch the operating rate climb; try lowering the tubing head or friction, as gas lift would, and watch it climb further. The worked example above is exactly the kind of intersection the tool computes - and displays - in real time.

🎮 Nodal Analysis: IPR ∩ VLP Operating Point LIVE
Predict first: At the default settings the inflow and outflow cross at 1,000 STB/d and 2,000 psi - the worked example. Now raise J or lower the VLP head/friction: which way does the operating point move?
Illustrative nodal-analysis model. IPR: p_wf = p_bar - q/J (straight line). VLP: p_wf = head + c*q^2 (a simple rising friction curve; a real multiphase VLP is J-shaped). At the defaults (p_bar = 3000 psi, J = 1.0, head = 1000 psi, c = 0.001) the operating point is q = 1000 STB/d, p_wf = 2000 psi, matching the worked example.

Check your understanding

1. With the straight-line IPR p_wf = 3000 - q and the tubing VLP p_wf = 1000 + 0.001 q^2, what is the operating rate at the intersection?
Setting inflow equal to outflow, $3000 - q = 1000 + 0.001\,q^2$, gives $q^2 + 1000\,q - 2{,}000{,}000 = 0$, whose positive root is $q = \dfrac{-1000 + 3000}{2} = 1{,}000$ STB/d.
2. At that operating rate of 1,000 STB/d, what is the operating bottomhole flowing pressure?
From the IPR, $p_{wf} = 3000 - 1000 = 2{,}000$ psi; the VLP agrees, $1000 + 0.001 \times 1000^2 = 2{,}000$ psi. The operating point is 1,000 STB/d at 2,000 psi.
✅ Key takeaways
  • Nodal analysis picks a node (usually the bottomhole) and plots inflow (the IPR, reservoir delivering to the node) and outflow (the VLP, tubing lifting from the node) against rate.
  • The intersection of the two curves is the operating point - the unique rate and bottomhole flowing pressure at which supply equals demand; anything that shifts either curve moves it.
  • For the worked well (IPR p_wf = 3000 - q, VLP p_wf = 1000 + 0.001 q^2), the operating point is q = 1,000 STB/d at p_wf = 2,000 psi.
➡️ When the operating point sits too low - the well cannot flow on its own, or flows too slowly - the engineer's lever is artificial lift. The next lesson starts with the most common continuous method: gas lift.