Computational Fluid Dynamics · Chapter 5 of 10 · Advanced
Pressure-Velocity Coupling: SIMPLE
There is no separate equation for pressure. CFD must invent one, deducing the pressure that forces the velocity field to conserve mass. The SIMPLE algorithm is how almost every code does it.
Readiness check
This chapter solves the hardest part of incompressible CFD. Tick only what you can do closed-notes.
- Recall incompressible continuity div(u) = 0.
- State that pressure drives flow through its gradient.
- Recall the momentum coefficient aP.
- Add signed mass flows across a cell's faces.
- Follow an iterative guess-and-correct loop.
The core idea
Incompressible flow has no pressure equation, only the constraint that mass is conserved. SIMPLE guesses a pressure, finds the velocity it drives, then corrects the pressure to remove the mass imbalance.
u′ = d(p′P − p′E), d = A/aPp′ source = mass imbalancep = p* + p′, u = u* + u′The momentum equations give velocity from a pressure field, but continuity gives no equation for pressure itself. SIMPLE breaks this deadlock iteratively. It guesses a pressure p*, solves momentum for a velocity u* that generally violates continuity, then solves a pressure-correction equation whose source is exactly the leftover mass imbalance in each cell. The correction p′ adjusts both the pressure and the velocity, through u′ = d(p′P − p′E) with d = A/aP, until the imbalance vanishes. A staggered grid, with velocities stored at cell faces and pressure at centres, prevents a checkerboard pressure field that a collocated arrangement would allow.
The skills, taught in order
Five skills explain the pressure problem, the staggered grid, the correction equation, the SIMPLE loop, and under-relaxation.
5.1 The pressure problem
In incompressible flow the momentum equations give velocity from a known pressure, but there is no equation that gives pressure directly. Pressure is whatever it must be so that the velocity field satisfies continuity. CFD has to construct this link.
5.2 The staggered grid
Storing velocity and pressure at the same nodes lets a wavy checkerboard pressure look uniform to the momentum equation. The staggered grid fixes this by placing velocities at cell faces, exactly where pressure differences drive them, so every velocity feels its adjacent pressures.
5.3 The pressure-correction equation
From a guessed pressure p*, the momentum solution u* leaves a mass imbalance in each cell. A pressure correction p′ is introduced; its equation has the form of a diffusion equation whose source is that mass imbalance, and the velocity correction is u′ = d(p′P − p′E) with d = A/aP.
5.4 The SIMPLE algorithm
The Semi-Implicit Method for Pressure-Linked Equations cycles: guess p*, solve momentum for u*, solve the p′ equation, correct p and u, then solve the other transport equations, and repeat until the mass imbalance and residuals are small.
| Step | Action |
|---|---|
| 1 | guess the pressure field p* |
| 2 | solve momentum for u*, v* |
| 3 | solve the pressure-correction p′ equation |
| 4 | correct p = p* + p′ and the velocities |
| 5 | solve other equations, then iterate |
5.5 Under-relaxation
The corrections can overshoot, so each update is damped by an under-relaxation factor (typically 0.7 for momentum, 0.3 for pressure). It slows each step but keeps the nonlinear iteration stable, the practical art of getting SIMPLE to converge.
Engineering connection: SIMPLE and its variants (SIMPLEC, PISO) are the pressure-velocity solvers in virtually every commercial incompressible CFD code.
Worked example 1: the velocity correction
A staggered momentum cell has a central coefficient aP = 100 kg/s and a face area A = 0.01 m². Find the d coefficient, and the velocity correction when the neighbouring pressure corrections differ by p′P − p′E = 200 Pa.
- ProblemFind d and the velocity correction for the face in Figure 1.
- Given / findaP = 100 kg/s, A = 0.01 m², p′P − p′E = 200 Pa. Find d and u′.
- AssumptionsSIMPLE velocity correction; the neighbour correction terms are dropped, as SIMPLE does.
- ModelThe d coefficient is the face area over the momentum coefficient; the correction is d times the pressure-correction difference.
- Equationsd = A/aPu′ = d(p′P − p′E)
- Solved = 0.01/100 = 1×10⁻⁴. u′ = 1×10⁻⁴ × 200 = 0.02 m/s.
- CheckThe units work: (m²/(kg/s)) × Pa = (m²·s/kg)(N/m²) = m/s. A small correction like 0.02 m/s is what each SIMPLE iteration applies as it nudges the field toward continuity.
- ConclusionThe d coefficient links a pressure correction to a velocity correction. Larger d, from a smaller momentum coefficient, means a given pressure correction moves the velocity more.
Worked example 2: the mass imbalance source
A two-dimensional cell has face mass flows after the momentum step: in through the west 0.05 kg/s and the south 0.04 kg/s, out through the east 0.06 kg/s and the north 0.02 kg/s. Find the net mass imbalance that drives the pressure-correction equation.
- ProblemFind the net mass imbalance for the cell in Figure 2.
- Given / findIn: 0.05 (west) and 0.04 (south); out: 0.06 (east) and 0.02 (north), all kg/s. Find the imbalance b′.
- AssumptionsSteady incompressible flow; the imbalance is what continuity has not yet enforced.
- ModelThe imbalance is the sum of inflows minus the sum of outflows across the faces.
- Equationsb′ = (ṁw + ṁs) − (ṁe + ṁn)
- Solveb′ = (0.05 + 0.04) − (0.06 + 0.02) = 0.09 − 0.08 = 0.01 kg/s.
- CheckA nonzero imbalance means continuity is not yet satisfied, so the field is not converged. This 0.01 kg/s is exactly the source the pressure-correction equation uses to adjust the pressures.
- ConclusionThe mass imbalance is both the convergence monitor and the engine of the pressure correction. SIMPLE iterates until it falls below tolerance everywhere.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Collocated without a remedy | Checkerboard pressure field | "Where are velocity and pressure stored?" | Use a staggered grid or Rhie-Chow interpolation. |
| Expecting a pressure equation | Looking for an explicit pressure PDE | "What equation gives pressure?" | Pressure comes from enforcing continuity, via the p′ equation. |
| No under-relaxation | SIMPLE diverges or oscillates | "Are the corrections damped?" | Apply under-relaxation, around 0.7 for momentum and 0.3 for pressure. |
| Reading an unconverged field | Mass imbalance still large | "Has b′ fallen below tolerance?" | Iterate until the imbalance and residuals are small. |
Practice ladder
A momentum cell has aP = 50 kg/s and A = 0.02 m². Find the d coefficient.
Show answer
d = A/aP = 0.02/50 = 4×10⁻⁴.
For the Worked Example 1 cell, what velocity correction results from a pressure-correction difference of 50 Pa?
Show answer
u′ = d × 50 = 1×10⁻⁴ × 50 = 0.005 m/s. The correction scales linearly with the pressure-correction difference.
A cell has inflows 0.12 (west) and 0.08 (south) and outflows 0.15 (east) and 0.04 (north) kg/s. Find the imbalance and say whether it is converged.
Show answer
b′ = (0.12 + 0.08) − (0.15 + 0.04) = 0.20 − 0.19 = 0.01 kg/s. Nonzero, so not yet converged; the p′ equation will reduce it.
Explain to a colleague why an incompressible CFD run needs a pressure-correction scheme at all, in terms of the missing pressure equation.
What good work looks like
A clear statement that momentum gives velocity from pressure but continuity gives no pressure equation, so pressure must be found by forcing mass conservation iteratively.
Working with AI, and proving it yourself
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Portfolio task
Describe the SIMPLE loop for a real incompressible case: the staggered or remedied grid, the pressure-correction source, the velocity correction, and the under-relaxation you would use.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Why is pressure hard in incompressible CFD?
There is no explicit pressure equation; pressure is whatever makes the velocity conserve mass.
2. What does the staggered grid prevent?
A checkerboard pressure field, by coupling face velocities to adjacent pressures.
3. What is the source of the p′ equation?
The mass imbalance left in each cell after the momentum step.
4. Write the velocity correction.
u′ = d(p′P − p′E) with d = A/aP.
5. Why under-relax?
To damp overshooting corrections and keep the nonlinear iteration stable.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Versteeg and Malalasekera, An Introduction to Computational Fluid Dynamics, Chapter 6 (Solution Algorithms for Pressure-Velocity Coupling) |
| Cross-reference | Patankar, Ch. 6 · Ferziger and Peric, Ch. 7 |
| Core topics | 5.1 The pressure problem · 5.2 Staggered grid · 5.3 Pressure-correction equation · 5.4 SIMPLE · 5.5 Under-relaxation |
| Engineering connection | SIMPLE and its variants run inside almost every incompressible CFD code. |
| Read next | Chapter 6: Solving the Discretized Equations. |