a) Briefly explain what would happen to solution droplets formed on this 10-16 g ammonium sulphate nucleus in an environment with constant saturation ratio=1.0025. Separately consider 3 droplets with initial radius of i) r = 0.05μm, ii) r = 0.2 µm and iii) r = 1µm. b) The growth of a small solute droplet in an environment at an ambient saturation ratio, S = eles(T) where e is ambient vapour pressure far away from the droplet, is controlled (approximately) by, r dr/dt = [(S-1) - a/r + b/r³]{1, (7.18 from Rogers and Yau) where ₁ (>0) is a growth parameter dependent on T and p and molecular diffusion properties of air. Explain the relationship between this growth equation and the Kohler curve shown above. c) Suppose that the ambient saturation ratio is held at 0.995 and there are two solution droplets containing 10-16g of ammonium sulphate. Droplet A has an initial radius of 0.02 um while droplet B has a radius of 0.2 μm. What happens to each of them. Without attempting detailed calculations speculate on which would you expect to achieve an equilibrium fastest.
a) Briefly explain what would happen to solution droplets formed on this 10-16 g ammonium sulphate nucleus in an environment with constant saturation ratio=1.0025. Separately consider 3 droplets with initial radius of i) r = 0.05μm, ii) r = 0.2 µm and iii) r = 1µm. b) The growth of a small solute droplet in an environment at an ambient saturation ratio, S = eles(T) where e is ambient vapour pressure far away from the droplet, is controlled (approximately) by, r dr/dt = [(S-1) - a/r + b/r³]{1, (7.18 from Rogers and Yau) where ₁ (>0) is a growth parameter dependent on T and p and molecular diffusion properties of air. Explain the relationship between this growth equation and the Kohler curve shown above. c) Suppose that the ambient saturation ratio is held at 0.995 and there are two solution droplets containing 10-16g of ammonium sulphate. Droplet A has an initial radius of 0.02 um while droplet B has a radius of 0.2 μm. What happens to each of them. Without attempting detailed calculations speculate on which would you expect to achieve an equilibrium fastest.
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
Related questions
Question
a=1.5x10^-18, b = 7.51x10^-18
![For warm clouds (T> 0°C) Kohler curves help explain aspects of the diffusional growth of cloud
droplets on cloud condensation nuclei. These are based on the S(r,M) equation,
S(r,ms) = es'(r,M)/ es() = (1 - b/r³) exp(a/r), (Rogers and Yau, p 87)
where es' (r,M) is the saturation vapour pressure on the surface of a spherical droplet of radius r,
containing mass, M of solute and es() is the saturation vapour pressure over a plane surface (r = ∞)
of pure water. It will vary with temperature. Also a = 20/(PLRT) and b = 3im,M/(4TTPLMs) where
symbols have their usual meanings.
Fig 6.2 from Rogers and Yau is an example.
Saturation Ratio
1-02
1.01
$50
1-00
0-99-
exp (a/r)
0-1 r
I
Droplet Radius, m
FIG. 6.2. Equilibrium saturation ratio of a solution droplet formed on an
ammonium sulfate condensation nucleus of mass 10-16 g.
a) Briefly explain what would happen to solution droplets formed on this 10-16 g ammonium sulphate
nucleus in an environment with constant saturation ratio=1.0025. Separately consider 3 droplets with
initial radius of i) r = 0.05μm, ii) r = 0.2 µm and iii) r = 1µm.
b) The growth of a small solute droplet in an environment at an ambient saturation ratio, S = e/es(T)
where e is ambient vapour pressure far away from the droplet, is controlled (approximately) by,
r dr/dt = [(S-1) - a/r + b/r³] {1, (7.18 from Rogers and Yau)
where ₁ (>0) is a growth parameter dependent on T and p and molecular diffusion properties of air.
Explain the relationship between this growth equation and the Kohler curve shown above.
c) Suppose that the ambient saturation ratio is held at 0.995 and there are two solution droplets
containing 10-16g of ammonium sulphate. Droplet A has an initial radius of 0.02 μm while droplet B
has a radius of 0.2 μm. What happens to each of them. Without attempting detailed calculations
speculate on which would you expect to achieve an equilibrium fastest.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd75659fd-39c5-4de8-858f-2f21c841e55f%2Fbba6ccc1-ad2c-4554-9d35-e4797a71ed45%2Fkeageut_processed.png&w=3840&q=75)
Transcribed Image Text:For warm clouds (T> 0°C) Kohler curves help explain aspects of the diffusional growth of cloud
droplets on cloud condensation nuclei. These are based on the S(r,M) equation,
S(r,ms) = es'(r,M)/ es() = (1 - b/r³) exp(a/r), (Rogers and Yau, p 87)
where es' (r,M) is the saturation vapour pressure on the surface of a spherical droplet of radius r,
containing mass, M of solute and es() is the saturation vapour pressure over a plane surface (r = ∞)
of pure water. It will vary with temperature. Also a = 20/(PLRT) and b = 3im,M/(4TTPLMs) where
symbols have their usual meanings.
Fig 6.2 from Rogers and Yau is an example.
Saturation Ratio
1-02
1.01
$50
1-00
0-99-
exp (a/r)
0-1 r
I
Droplet Radius, m
FIG. 6.2. Equilibrium saturation ratio of a solution droplet formed on an
ammonium sulfate condensation nucleus of mass 10-16 g.
a) Briefly explain what would happen to solution droplets formed on this 10-16 g ammonium sulphate
nucleus in an environment with constant saturation ratio=1.0025. Separately consider 3 droplets with
initial radius of i) r = 0.05μm, ii) r = 0.2 µm and iii) r = 1µm.
b) The growth of a small solute droplet in an environment at an ambient saturation ratio, S = e/es(T)
where e is ambient vapour pressure far away from the droplet, is controlled (approximately) by,
r dr/dt = [(S-1) - a/r + b/r³] {1, (7.18 from Rogers and Yau)
where ₁ (>0) is a growth parameter dependent on T and p and molecular diffusion properties of air.
Explain the relationship between this growth equation and the Kohler curve shown above.
c) Suppose that the ambient saturation ratio is held at 0.995 and there are two solution droplets
containing 10-16g of ammonium sulphate. Droplet A has an initial radius of 0.02 μm while droplet B
has a radius of 0.2 μm. What happens to each of them. Without attempting detailed calculations
speculate on which would you expect to achieve an equilibrium fastest.
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