A 45 m stack discharges 140 °C gas at 12 m/s through a 2 m opening into a 15 °C day with a 4 m/s wind. The source emits 2.4 g/s of SO₂, and the screening tables put the spread ratio σz/σy at 0.5 for the stability class in hand. The permit reviewer wants the classic screening answer: how much does the hot plume climb above the stack on its own buoyancy at 600 m downwind, what effective height does that give the source, and what is the worst ground-level concentration anywhere downwind.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
v_s = 12 m/s — Stack exit velocity
d = 2 m — Stack inside diameter
T_s = 140 °C — Stack gas temperature
T_a = 15 °C — Ambient temperature
u = 4 m/s — Wind speed
h_s = 45 m — Physical stack height
Q = 2.4 g/s — SO₂ emission rate
Determine
(a)the buoyancy flux of the hot discharge
(b)the plume rise at 600 m downwind
(c)the effective stack height
(d)the maximum ground-level concentration
Step 1 of 4(a) · solve for Buoyancy flux (m⁴/s³)
Everything Briggs does downstream rides on this one number, and it is bought almost entirely by the temperature difference: 125 degrees of excess heat through a 2 m throat gives F = 35.6 m⁴/s³. The temperatures go in absolute — the 4T_s in the denominator is kelvins, and feeding it Celsius is the classic way to triple a plume rise.
Rise goes as the cube root of F — burn three times hotter to climb 44 % higher — and inversely with wind, because wind bends the plume over before buoyancy is done lifting it. At 600 m out in the 4 m/s wind this plume has climbed 93.6 m, more than twice the stack's own height.
Rearranged for Δh
Δh=u1.6F1/3x2/3
35.604 —carried from step 1
Your values, in your units
Δh=(4m/s)1.6⋅(35.6044m4/s3)1/3⋅(600m)2/3
Answer
Δh=93.611m
Carried onward at full precision, not this rounded figure.
The atmosphere doesn't know where the concrete stops: what disperses is the sum. 45 m of steel plus 93.6 m of buoyancy is an effective source at 138.6 m — the cheap two-thirds of this stack was bought with heat, not height.
Rearranged for H
H=hs+Δh
93.611 mcarried from step 2
Your values, in your units
H=(45m)+(93.6114m)
Answer
H=138.61m
Carried onward at full precision, not this rounded figure.
The closed form answers the reviewer's only real question — not what arrives at one fence, but the worst anywhere. With H in the denominator squared, the answer is 3.66 µg/m³: double the effective height and the worst case falls by four.
Rearranged for C_max
Cmax=eπuH22Q⋅σyσz
138.61 mcarried from step 3
Your values, in your units
Cmax=eπ⋅(4m/s)⋅(138.611m)22⋅(2.4g/s)⋅(0.5)
Answer
Cmax=3.6569e−09kg/m3
Carried onward at full precision, not this rounded figure.
Therefore the discharge carries a buoyancy flux of 35.6 m⁴/s³, climbs 93.6 m above the stack in the wind, disperses from an effective height of 138.6 m — and the worst ground-level concentration anywhere downwind is 3.66 µg/m³, two-thirds of it earned by heat rather than steel.
Why this order
Screening dispersion runs in exactly this order because each number is the only input the next one needs: exit conditions make a buoyancy flux, the flux and the wind make a rise, the rise and the concrete make an effective height, and the height caps the worst case. The physics worth keeping is the pair of sensitivities — rise goes as F^{1/3}, so heat is a blunt instrument, while the worst-case concentration goes as 1/H², so height is a sharp one. That asymmetry is why reheating a scrubbed plume was ever a practice, and why the tall-stack era built what it built.
Two traps. The kelvin one in step (a) is arithmetic and merciless. The subtler one is in step (b): Briggs' neutral form keeps growing with x^{2/3} forever, but real plumes stop rising at final rise — quote the 600 m figure as “rise at 600 m”, not “the rise”, or the effective height inherits an optimism the worst-case number then squares.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
