A building is \( 30 \) m high and \( 20 \) m in projected width. The good engineering practice stack height is
Answer: A
Correct answer: 60 m
The lesser of the two dimensions is the width, 20 m:
\[ H_{GEP} = H_b + 1.5L = 30 + 1.5 \times 20 = 60\ \text{m} \]The trap is to use the larger dimension, or the height, without checking which is smaller.
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A stack of \( 2.0 \) m diameter passes \( 47 \) m\(^3\)/s. The exit velocity is
Answer: B
Correct answer: 15.0 m/s
\[ A = \frac{\pi d^2}{4} = \frac{\pi (2.0)^2}{4} = 3.14\ \text{m}^2 \]\[ v = \frac{Q}{A} = \frac{47}{3.14} = 15.0\ \text{m/s} \]
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Wind at stack height is \( 9 \) m/s. The minimum exit velocity to avoid stack tip downwash is
Answer: C
Correct answer: 13.5 m/s
The usual rule is an exit velocity of at least 1.5 times the wind speed at stack height:
\[ v_{min} = 1.5 \times 9 = 13.5\ \text{m/s} \]Below that the plume is liable to be drawn into the stack’s own wake.
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Halving the stack diameter at constant volumetric flow changes the exit velocity by a factor of
Answer: D
Velocity is flow divided by area and area goes as the square of diameter, so halving the diameter quadruples the velocity. The pressure drop rises steeply as well, which is what sets the practical limit on narrowing it.
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A physical stack of \( 80 \) m has a plume rise of \( 55 \) m. Compared with a stack of the same physical height and no rise, the maximum ground level concentration falls by a factor of about
Answer: A
Correct answer: 2.86
Effective height rises from 80 m to \(80 + 55 = 135\) m, and the maximum ground-level concentration goes as \(1/H^2\):
\[ \frac{C_1}{C_2} = \left(\frac{135}{80}\right)^2 = (1.6875)^2 = 2.85 \]So the concentration falls to about a third - the whole value of plume rise in one number.
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Holland's formula gives momentum rise proportional to \( v_s d / u \). Doubling exit velocity at constant diameter and wind speed changes that term by a factor of
Answer: B
Correct answer: 2
The momentum term is proportional to \(v_s d / u\), and it is linear in exit velocity. With \(d\) and \(u\) unchanged, doubling \(v_s\) doubles the term.
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A stack emits \( 60 \) g/s and produces a maximum ground level concentration of \( 180 \) micrograms per cubic metre. Halving the emission gives
Answer: C
Correct answer: 90 micrograms per cubic metre
Ground-level concentration is directly proportional to the emission rate, everything else being unchanged - the dispersion is the same whatever the plume carries. So halving \(Q\) halves the concentration:
\[ 180 \times \tfrac{1}{2} = 90\ \mu\text{g/m}^3 \]
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Wind at \( 10 \) m is \( 5.0 \) m/s and the power law exponent is \( 0.15 \). The wind at \( 80 \) m is about
Answer: D
Correct answer: 6.7 m/s
\[ u_2 = u_1\left(\frac{z_2}{z_1}\right)^{p} = 5.0 \times \left(\frac{80}{10}\right)^{0.15} = 5.0 \times 8^{0.15} \]\[ 8^{0.15} = e^{0.15 \ln 8} = e^{0.312} = 1.37 \]\[ u_2 = 6.8 \approx 6.7\ \text{m/s} \]Note how modest the increase is: eight times the height gives only about a third more wind.
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A flue gas at \( 145 \) degrees is cooled by a wet scrubber to \( 55 \) degrees against an acid dew point of \( 120 \) degrees. The gas is now
Answer: A
Correct answer: Below the dew point; acid will condense
At 55°C the gas is well below its acid dew point of 120°C, so sulphuric acid will condense on the stack lining and corrode it. This is exactly why a wet scrubber is followed either by reheat - raising the gas back above the dew point - or by an acid-resistant lining. Reheat costs fuel but also restores some buoyant plume rise.
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An effective stack height of \( 120 \) m is needed. If plume rise is \( 40 \) m, the physical stack must be
Answer: B
Correct answer: 80 m
Effective height is physical height plus plume rise:
\[ H_e = h_s + \Delta h \quad\Rightarrow\quad h_s = 120 - 40 = 80\ \text{m} \]The 40 m of rise is worth 40 m of steelwork, which is why raising exit velocity or gas temperature is often cheaper than building higher.
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