Briggs Plume Rise (Neutral and Unstable)

Briggs' two-thirds law is the result of asking how far a buoyant jet climbs while the atmosphere entrains air into it at a rate proportional to its own rise velocity. The answer is that the rise grows as the two-thirds power of downwind distance and falls inversely with wind speed: $\Delta h = 1.6 F^{1/3} x^{2/3}/u$. Take a plume with $F = 64$ m⁴/s³, at 1000 m downwind in a 5 m/s wind. The cube root of 64 is 4, the two-thirds power of 1000 is 100, so $\Delta h = 1.6 \times 4 \times 100/5 = 128$ m. The 1.6 is not derived, it is fitted, and it comes from Briggs' own 1975 review of plume photographs and lidar traverses at power stations.

Read the exponents and the design advice falls out. Rise depends on the cube root of the buoyancy flux, so quadrupling the heat release buys only 59 percent more rise, which is why nobody solves a dispersion problem by burning more fuel. Rise depends inversely on wind speed, so a plume that climbs 128 m at 5 m/s climbs only 64 m at 10 m/s. That inverse dependence fights with the direct dilution the same wind provides, and the fight is what produces a critical wind speed at which ground-level concentrations are worst, usually somewhere in the 3 to 8 m/s range for a tall buoyant source.

The honesty this equation needs is about its range. It describes TRANSITIONAL rise, the climb the plume is still doing, and it has no idea that the plume eventually levels off. Briggs puts final rise at a downwind distance of $3.5x^*$, with $x^* = 14F^{5/8}$ below 55 m⁴/s³ and $34F^{2/5}$ above it. For the example above, $F = 64$ gives $x^* = 34 \times 64^{0.4} = 34 \times 5.28 = 179$ m, so final rise happens at about 628 m, and the true rise is $1.6 \times 4 \times 628^{2/3}/5 = 1.6 \times 4 \times 73.3/5 = 94$ m. Feeding 1000 m into the equation returned 128 m, an over-prediction of 36 percent, and the error grows without limit the further out you go. Compute $3.5x^*$ first, and if your distance of interest is beyond it, use the rise at $3.5x^*$ and stop.

Two more boundaries. This form is for neutral and unstable air only. In stable air the plume runs out of buoyancy against the ambient stratification and levels off at $\Delta h = 2.6(F/us)^{1/3}$, where s is the stability parameter $g(\partial \theta/\partial z)/T_a$, and in stable calm it is $5F^{1/4}/s^{3/8}$. The equation also divides by wind speed, so it blows up at calm, which is not a prediction that plumes rise infinitely high on still nights but a statement that the model has left its domain. Finally, use the wind speed at STACK TOP, not the anemometer reading at 10 m, or the rise will be over-predicted by whatever the wind profile happens to be worth.