MAE 3420 HW 10
Nathan Jones
Distance Traveled by Droplets/Aerosol Released During Human Expiratory Events
(1) Estimate the Reynolds number of jet coming out during breathing, coughing, and
sneezing. Which of these flows are laminar and which of them are turbulent? (10 points)
For all calculations, I used values at 25 °C:
• 𝜇𝑎𝑖𝑟 = 1.849 × 10−5 m2 /s
kg
• 𝜌𝑎𝑖𝑟 = 1.184 m3
Since Re is having its dynamic viscosity defined by the subscript ‘p’ (meaning particle, not air,
since ‘air’ is already defined as a subscript), I assume that the kinematic viscosity should also for
water (𝜈𝑝 = 0.891 × 10−6 m2 /s). I wished the variables were more explicitly defined so I did
not have to make assumptions.
Assuming that breathing has the smallest particle size of 0.5 μm:
𝑅𝑒𝑏𝑟𝑒𝑎𝑡ℎ
𝑢𝐷 (1.5 m/s)(0.5 × 10−6 m)
=
=
= 0.84175
𝜈𝑝
0.891 × 10−6 m2 /s
Assuming that coughing has the median particle size of 8 μm:
• Average speed is (1.5 + 30)/2 = 15.75 m/s
𝑅𝑒𝑐𝑜𝑢𝑔ℎ =
𝑢𝐷 (15.75 m/s)(8 × 10−6 m)
=
= 141.414
𝜈𝑝
0.891 × 10−6 m2 /s
Assuming that sneezing has the largest particle size of 256 μm:
• Average speed is (20 + 50)/2 = 37.5 m/s
𝑅𝑒𝑠𝑛𝑒𝑒𝑧𝑒 =
𝑢𝐷 (37.5 m/s)(256 × 10−6 m)
=
= 10056
𝜈𝑝
0.891 × 10−6 m2 /s
Considering that flow is laminar when 𝑅𝑒 < 2300:
• Breathing and coughing are laminar.
• Sneezing is turbulent.
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(2) Calculate the particle Reynolds number for the given aerosol sizes and expiration
velocities. What is the maximum velocity of release for each aerosol/droplet size, that allows
the aerosol to be in the creeping (Stokes) flow regime? Recall that for practical purposes,
creeping flow is at 𝑹𝒆𝒑 ≈ 𝟏. (10 points)
𝑅𝑒𝑝 =
𝜈𝑝
𝑢𝐷
≈1→𝑢=
𝜈𝑝
𝐷
For given particle size:
• 𝐷 = 0.5 μm: 𝑢 = 1.782 m/s
• 𝐷 = 1 μm: 𝑢 = 0.891 m/s
• 𝐷 = 2 μm: 𝑢 =0.4455 m/s
• 𝐷 = 4 μm: 𝑢 = 0.2228 m/s
• 𝐷 = 8 μm: 𝑢 = 0.1114 m/s
• 𝐷 = 16 μm: 𝑢 = 0.0557 m/s
• 𝐷 = 32 μm: 𝑢 = 0.0278 m/s
• 𝐷 = 64 μm: 𝑢 = 0.0139 m/s
• 𝐷 = 128 μm: 𝑢 = 0.0070 m/s
• 𝐷 = 256 μm: 𝑢 = 0.0035 m/s
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(3) What is the distance traveled by aerosols of different sizes before hitting the floor, when
they are released during different expiratory events from a height of an average human
being (you can use your own height for the calculations)? Also, calculate the time that
aerosols of different sizes take to reach the floor. (60 points)
The aerosols once released will travel in the horizontal and the vertical direction. If we
assume that the direction of the expiration is parallel to the floor, then the horizontal
motion of the aerosols will be propelled by an initial velocity it attains from the jet. The
aerosols will then be slowed down due to drag on them. Drag on the aerosol will depend on
the regime they are in. If they are in the Stokes/Creeping flow regime, the horizontal drag
on the aerosol will be defined by 𝑭𝒙 = 𝟑𝝅𝝁𝒂𝒊𝒓 𝑫𝒖. Where D is diameter of the aerosol, u is
velocity of the aerosol, and 𝑭𝒙 is the drag force on the aerosol. In the vertical direction, the
aerosol will accelerate towards the floor due to gravity. Whereas buoyancy and drag in the
vertical direction will slow down the vertical motion of the aerosols. Though, in case the
direction of expiration is not parallel to the floor, then the motion of the aerosols will be
altered according to the direction of their initial velocity.
On a given particle for parallel expiration
∑ 𝐹𝑥 = −3𝜋𝜇𝑎𝑖𝑟 𝐷𝑢𝑥
∑ 𝐹𝑦 = −3𝜋𝜇𝑎𝑖𝑟 𝐷𝑢𝑦 + 𝜌𝑎𝑖𝑟
𝜋 3
𝜋
𝐷 𝑔 − 𝜌𝑝 𝐷3 𝑔
6
6
From the sum of the forces, the acceleration in x and y can be determined by dividing by the
mass of the particle:
𝜋
𝑚𝑝 = 𝜌𝑝 𝐷3
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∑ 𝐹𝑥 → 𝑎𝑥 =
∑ 𝐹𝑦 → 𝑎𝑦 =
−3𝜋𝜇𝑎𝑖𝑟 𝐷𝑢𝑥 −18𝜇𝑎𝑖𝑟 𝑢𝑥
=
𝜋
𝜌𝑝 𝐷2
𝜌𝑝 6 𝐷3
1
𝜋 3
𝜋 3
𝜋 3 (−3𝜋𝜇𝑎𝑖𝑟 𝐷𝑢𝑦 + 𝜌𝑎𝑖𝑟 6 𝐷 𝑔 − 𝜌𝑝 6 𝐷 𝑔)
𝜌𝑝 6 𝐷
Used values:
• Average height of a man: 5 foot 9 inches or 1.7526 m (from CDC)
• 𝑔 = 9.81 m/𝑠 2 (in negative y direction)
NOTE: When attempting to plot the sizes of below 64 μm, the small sizes resulted in overflow
errors and caused that portion to not function (Specifically, the mass of the particles were very
small, causing their respective accelerations to be incredibly large). Because of this, only sizes 64
μm and above were plotted.
3
Time, distance (for 1.5 m/s, coughing):
• 𝐷 = 64 μm: 14.547 s, 0.0185 m
• 𝐷 = 128 μm: 3.684 s, 0.0738 m
• 𝐷 = 256 μm: 1.106 s, 0.2943 m
4
Time, distance (for 15.75 m/s, coughing):
• 𝐷 = 64 μm: 14.547 s, 0.1938 m
• 𝐷 = 128 μm: 3.684 s, 0.7753 m
• 𝐷 = 256 μm: 1.106 s, 3.0901 m
5
Time, distance (for 37.5 m/s, coughing):
• 𝐷 = 64 μm: 14.547 s, 0.4615 m
• 𝐷 = 128 μm: 3.684 s, 1.8460 m
• 𝐷 = 256 μm: 1.106 s, 7.357 m
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(4) Determine the angle with respect to the horizontal at which if the expiration occurs, will
result in the aerosols/droplets traveling the longest distance? (20 points)
NOTE: When attempting to plot the sizes of 64 μm, the small sizes resulted in overflow errors
and caused that portion to not function (Specifically, the mass of the particles were very small,
causing their respective accelerations to be incredibly large). Because of this, only sizes 64 μm
and above were plotted.
By iterating through angles from 0 to 90 degrees, this angle can be found:
• For 64 μm:
o 1.5 m/s: 0 degrees (0.0185 m)
o 15.75 m/s: 0 degrees (0.1938 m)
o 37.5 m/s: 0 degrees (0.4615 m)
• For 128 μm:
o 1.5 m/s: 0 degrees (0.07384 m)
o 15.75 m/s: 0 degrees (0.7753 m)
o 37.5 m/s: 0 degrees (1.8460 m)
• For 256 μm:
o 1.5 m/s: 0 degrees (0.2943 m)
o 15.75 m/s: 1.41 degrees (3.0912 m)
o 37.5 m/s: 2.03 degrees (7.3661 m)
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(5) If the moving aerosols are not in the Stokes/Creeping flow regime, then the drag on the
𝟏
aerosol is defined by 𝑭𝒙 = 𝟐 𝑪𝑫 𝝆𝒂𝒊𝒓 𝑨𝒖𝟐 , where 𝑪𝑫 is the drag coefficient; 𝑨 = 𝝅𝑫𝟐 /𝟒, the
𝟐𝟒
frontal area of the aerosol; u is the velocity of the particle. 𝑪𝑫 is estimated as 𝑪𝑫 = 𝑹𝒆 +
𝒑
𝟏. 𝟓 where the particle Reynolds number 𝑹𝒆𝒑 =
𝒖𝑫
𝝂
=
𝝆𝒑 𝒖𝑫
𝝁
. In the Stokes/Creeping flow
regime, 𝑹𝒆𝒑 < 𝟏, and the second term in the 𝑪𝑫 relationship becomes negligible compared
to 𝟐𝟒/𝑹𝒆𝒑. This makes the drag on the aerosol similar to the drag calculated using the
formula in the previous section. Ideally, when the transport of the aerosol particles is
calculated, it is more accurate to use the drag formula that works across different regimes,
that are Stokes and laminar/turbulent. How do your answers for part 3 and part 4 change
if you use the more accurate formula for calculating drag. (100 points)
On a given particle for parallel expiration
1
∑ 𝐹𝑥 = ± 𝐶𝐷 𝜌𝑎𝑖𝑟 𝐴𝑢𝑥 2
2
1
𝜋
𝜋
∑ 𝐹𝑦 = ± 𝐶𝐷 𝜌𝑎𝑖𝑟 𝐴𝑢𝑦 2 + 𝜌𝑎𝑖𝑟 𝐷3 𝑔 − 𝜌𝑝 𝐷3 𝑔
2
6
6
•
The sign on the drag force (±) should be opposite of the sign of the given velocity
From the sum of the forces, the acceleration in x and y can be determined by dividing by the
mass of the particle:
𝜋
𝑚𝑝 = 𝜌𝑝 𝐷3
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1
± 2 𝐶𝐷 𝜌𝑎𝑖𝑟 𝐴𝑢𝑥 2 ±3𝐶𝐷 𝜌𝑎𝑖𝑟 𝐴𝑢𝑥 2
∑ 𝐹𝑥 → 𝑎𝑥 =
=
𝜋
𝜌𝑝 𝜋𝐷2
𝜌𝑝 6 𝐷3
∑ 𝐹𝑦 → 𝑎𝑦 =
1
1
𝜋 3
𝜋 3
2
𝜋 3 (± 2 𝐶𝐷 𝜌𝑎𝑖𝑟 𝐴𝑢𝑦 + 𝜌𝑎𝑖𝑟 6 𝐷 𝑔 − 𝜌𝑝 6 𝐷 𝑔)
𝜌𝑝 6 𝐷
Used values:
• Average height of a man: 5 foot 9 inches or 1.7526 m (from CDC)
• 𝑔 = 9.81 m/𝑠 2 (in negative y direction)
NOTE: When attempting to plot the sizes of below 16 μm, the small sizes resulted in overflow
errors and caused that portion to not function (Specifically, the mass of the particles were very
small, causing their respective accelerations to be incredibly large). Because of this, only sizes 64
μm and above were plotted.
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Time, distance (for 1.5 m/s, breathing):
• 𝐷 = 16 μm: 15.0055 s, 0.0212 m
• 𝐷 = 32 μm: 5.6506 s, 0.0756 m
• 𝐷 = 64 μm: 3.0423 s, 0.1747 m
• 𝐷 = 128 μm: 1.9782 s, 0.3075 m
• 𝐷 = 256 μm: 1.4001 s, 0.4639 m
9
Time, distance (for 15.75 m/s, coughing):
• 𝐷 = 16 μm: 15.0055 s, 0.0446 m
• 𝐷 = 32 μm: 5.6506 s, 0.1271 m
• 𝐷 = 64 μm: 3.0423 s, 0.2825m
• 𝐷 = 128 μm: 1.9782 s, 0.5261 m
• 𝐷 = 256 μm: 1.4001 s, 0.8965 m
10
Time, distance (for 37.5 m/s, sneezing):
• 𝐷 = 16 μm: 15.0055 s, 0.0547 m
• 𝐷 = 32 μm: 5.6506 s, 0.1476 m
• 𝐷 = 64 μm: 3.0423 s, 0.3238 m
• 𝐷 = 128 μm: 1.9782 s, 0.6090 m
• 𝐷 = 256 μm: 1.4001 s, 1.0619 m
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Finding new angles:
NOTE: When attempting to plot the sizes of below 16 μm, the small sizes resulted in overflow
errors and caused that portion to not function (Specifically, the mass of the particles were very
small, causing their respective accelerations to be incredibly large). Because of this, only sizes 64
μm and above were plotted.
By iterating through angles from 0 to 90 degrees, this angle can be found:
• For 16 μm:
o 1.5 m/s: 0 degrees (0.0212 m)
o 15.75 m/s: 14.97 degrees (0.0449 m)
o 37.5 m/s: 12.74 degrees (0.0554 m)
• For 32 μm:
o 1.5 m/s: 0 degrees (0.0756 m)
o 15.75 m/s: 0 degrees (0.1271 m)
o 37.5 m/s: 0 degrees (0.1476 m)
• For 64 μm:
o 1.5 m/s: 0 degrees (0.1747m)
o 15.75 m/s: 0 degrees (0.2825m)
o 37.5 m/s: 0 degrees (0.3238 m)
• For 128 μm:
o 1.5 m/s: 1.26 degrees (0.3075 m)
o 15.75 m/s: 12.44 degrees (0.5315 m)
o 37.5 m/s: 12.52 degrees (0.6185 m)
• For 256 μm:
o 1.5 m/s: 5.64 degrees (0.4647 m)
o 15.75 m/s: 16.75 degrees (0.9278 m)
o 37.5 m/s: 15.97 degrees (1.108 m)
Compared to previous answers:
• The graphs exhibited a steeper descent for the particles, meaning that drag less of
an effect on the particles, speeding up their acceleration.
• The angle which produces the maximum distance for a given size and speed also
increased.
Here are some examples of the plots generated during the solving of angles:
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