Gas and droplets in the gravity field
Anastassia Makarieva
scientists which I would like to discuss with physicists and would be
most appreciative of any feedback.
Consider gas -- e.g. water vapor -- in the gravity field. As is well-
known, in hydrostatic equilibrium gas pressure at any point is equal
to the weight of gas column above that point. Now suppose we have
condensed some part of the gas. The total mass of substance in the
column has not changed, but it is now present in the form of gas and
liquid droplets.
The statement under discussion is that until the droplets fall out (so
that the mass of the column changes), the gas pressure at the surface
remains unchanged.
The answer I argue for is that the characteristic time scale of the
gas pressure adjustment is different from the time scale of droplet
movement and will take place independently of the latter. Static
equilibrium of gas and droplets cannot exist.
Regards,
Anastassia
Ian Parker
fall at a terminal velocity that depends on its size. At terminal
velocity the momentum of the Gas/Liquid combination does not change
and the pressure supports gravity operating on both the gas and the
droplets.
This is a theoretical model. In practice in drizzle water falls out of
a uniform atmosphere at terminal velocity. Terminal velocities are
given here.
http://www.madsci.org/posts/archives/2000-07/962626446.Ph.r.html
However rain does not fall from an atmosphere in equilibrium. Gliders
and birds use thermal currents and can therefore fly without expending
energy. Heavy rain in fact falls from up drafts, the most extreme
being thunderstorms where up currents can be as high as 100km/h.
People have baled out of aircraft and the up draft has taken them and
their parachute into the clouds. In fact large water droplets are held
be up currents, which is how we get heavy rain. You might say "Ah well
5m/sec, rain clouds are only about 1km up, it can only rain for
200sec". Wrong, the water is held by up currents.
You are right in a theoretical equilibrium case. However it does not
rain in equilibrium conditions and there will be changes in pressure
due to the dynamic motion of air.
- Ian Parker
Anastassia Makarieva
Many thanks. Of course, water droplets are held by updrafts. And
updrafts are produced by the non-equilibrium vertical pressure
distribution -- there must be an upward pressure-gradient force to
accelerate air. Condensation rarifies the air, creates a pressure
imbalance and thus supports the updrafts. In their turn, the updrafts
bring more vapor up and sustain condensation. Recently we have offered
a dynamic explanation for hurricane and tornado energetics based on
this positive feedback, http://www.bioticregulation.ru/pubs/absn.php?ref=pla09b&lang=en
But here my point is about static equilibrium in a model static
atmosphere. The statement I oppose to is that "until all the drops
fall out, air pressure at the surface will not change". In reality
though the air will start expanding upwards independently of whether
the drops are falling out or not. There can be so tiny drops, for
example, that they will not be falling out at all at first, but will
be carried upwards by the upward air flow initiated by condensation.
In other words, relaxation of the gas pressure field will start
immediately upon condensation, the air will not "wait" until the drops
fall out.
Uncle Al
The total mass in the dropletted column cannot be the same. Dense
material is volumetrically replacing air.
100 tonnes of lead pellets are dropped from a plane at 30,000 feet
directly above you. The cloud falls vertically. 30 seconds later,
has your air pressure increased?
The fun is that a cloud is floating not falling at terminal velocity.
You are a scuba diver at 100 feet depth. You swim from sky above you
to under an ocean liner. Are you crushed?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
Richard D. Saam
Dimensionally quantifying this a bit.
What is the fluid shear (dv/dx) required to suspend particles
in that fluid having height(h) and horizontal area x^2.
In fluid mechanics and waste water treatment facilities in particular
dv/dx = sqrt(mixing power(P)/(fluid viscosity(mu)*mixing volume(V))
= sqrt(mixing power(P)/(fluid viscosity(mu)*mixing volume(V)
= sqrt(P/(mu*V))
also the particle diameter(D) settling velocity(v)
in that volume(V) where V = x^2 * h
is the well known Stokes' laminar relationship
v = g * (delta rho) * D^2 / (18*mu)
g = gravity acceleration
delta rho = density difference between particle and fluid.
let v = average velocity over height h such that
dv/dx = dv/dh = v/h
Do the math manipulation then the power(P)
required per height distance(h)
P/V = g^2 (delta rho)^2 * D^4 / (18^2 * mu * x^2)
or
P/h = g^2 (delta rho)^2 * D^4 / (18^2 * mu)
or in terms of suspended particle diameter
D^4 = P/h * 18^2 * mu / (g^2 (delta rho)^2)
In conclusion,
a shear (dv/dx = dv/dh) orthokinetic condition exists
for gas and droplets equilibrium (suspension).
It is obvious that increased atmospheric energy dissipation(P/h)
will effect atmospheric particle suspension.
But the problem is a bit more complicated
for fluid particles (rain drops).
Under dv/dx , dv/dh shear condition,
what is the size of the rain drops(D).
Dimensionally it can be shown that:
D = sigma * C / (mu * dv/dx)
where:
sigma = interfacial gas-liquid tension
between gas and fluid (73 dyne/cm for air-water)
C = fluid volumetric concentration (volume fluid/volume gas)
mu = supporting fluid viscosity as above.
Things get more complicated in turbulent conditions
but the basic principles remain.
Many years ago, I wrote several thousand lines of pascal computer code
that dimensionally handled all of this for any supporting fluid (gas or
liquid) and suspended fluid (gas or liquid) particle
in laminar and turbulent conditions.
Richard D. Saam
Chalky
wrote:
>
> Many thanks. Of course, water droplets are held by updrafts. And
> updrafts are produced by the non-equilibrium vertical pressure
> distribution -- there must be an upward pressure-gradient force to
> accelerate air. Condensation rarifies the air, creates a pressure
> imbalance and thus supports the updrafts. In their turn, the updrafts
> bring more vapor up and sustain condensation. Recently we have offered
> a dynamic explanation for hurricane and tornado energetics based on
> this positive feedback,http://www.bioticregulation.ru/pubs/absn.php?ref=pla09b〈=en
>
> But here my point is about static equilibrium in a model static
> atmosphere. The statement I oppose to is that "until all the drops
> fall out, air pressure at the surface will not change". In reality
> though the air will start expanding upwards independently of whether
> the drops are falling out or not. There can be so tiny drops, for
> example, that they will not be falling out at all at first, but will
> be carried upwards by the upward air flow initiated by condensation.
> In other words, relaxation of the gas pressure field will start
> immediately upon condensation, the air will not "wait" until the drops
> fall out.
Seems to me you are 'almost' right. Merely by the act of condensing
into droplets, the constituent water molecules will cease to be active
participants in the ideal gas law, so pressure will drop.
However, I suspect that the observable consequences will not be
immediate, but will propagate at the speed of sound.
Anastassia Makarieva
on that.
Chalky
> Seems to me you are 'almost' right. Merely by the act of condensing
> into droplets, the constituent water molecules will cease to be active
> participants in the ideal gas law, so pressure will drop.
This is exactly my point.
> However, I suspect that the observable consequences will not be
> immediate, but will propagate at the speed of sound.
There is another velocity scale in this problem. Let air density be
rho and condensation-related pressure drop be dp [dimension 1 N/m^2 =
J/m^3]. Then applying Bernoulli's equation we observe that maximum
vertical velocity V that can be formed is found from the relationship
dp = rho V^2/2, which is V = sqrt(2 dp/rho). A unit air volume
accelerated over distance h having non-equilibrium pressure difference
dp (and non-equilibrium pressure gradient dp/h -- it has the dimension
of force per unit volume) will have velocity V. (Sound waves do not
transfer mass.)
So I totally agree the effect will not be immediate. Rather, its time
scale will be completely different from that of the droplets falling
out.
For typical atmospheric values, where water vapor partial pressure is
about 3x10^3 Pa, we obtain vertical velocity about 60 m/s, of the type
observable in compact circulation patterns like tornadoes.
Richard D. Saam
>let v = average velocity over height h such that
>dv/dx = dv/dh = v/h
For most atmospheric circulation patterns horizontal dimension x >> h.
For this reason the condensation-induced pressure disequilibrium is
relaxed much more rapidly along the vertical dimension, making air
pressure at the surface smaller by dp in the area of condensation.
This will cause air to accelerate along the surface (i.e., in
horizontal dimension) towards the condensation area, so a typical V
velocity will that of horizontal wind, i.e. we have a hurricane.
Uncle Al
>The fun is that a cloud is floating not falling at terminal velocity.
Droplets of the cloud are falling at terminal velocity. But as there
is a supporting air updraft with the same velocity, the net outcome is
that the droplets and the cloud remain where they are.
Anastassia Makarieva
on that.
Chalky
> Seems to me you are 'almost' right. Merely by the act of condensing
> into droplets, the constituent water molecules will cease to be active
> participants in the ideal gas law, so pressure will drop.
This is exactly my point.
> However, I suspect that the observable consequences will not be
> immediate, but will propagate at the speed of sound.
There is another velocity scale in this problem. Let air density be
Richard D. Saam
> Richard D. Saam
>> let v = average velocity over height h such that
>> dv/dx = dv/dh = v/h
>
> For most atmospheric circulation patterns horizontal dimension x >> h.
> For this reason the condensation-induced pressure disequilibrium is
> relaxed much more rapidly along the vertical dimension, making air
> pressure at the surface smaller by dp in the area of condensation.
> This will cause air to accelerate along the surface (i.e., in
> horizontal dimension) towards the condensation area, so a typical V
> velocity will that of horizontal wind, i.e. we have a hurricane.
>
Yes, but at the < cm level
dv/dx = dv/dh = v/h
not related to gross weather hurricane dimensions (100 of km)
and the density of that < cm volume
is of the droplet air mixture.
Condensation evaporation only changes the local density
by changing the droplet contribution to density
with pressure change (rho (delta v)^2) of the < cm volume
related to the velocity change over that volume (energy dissipation)
dv/dx = dv/dh = v/h
to maintain suspension.
By such a method, large drops and hail stones are formed
until dv/dx = dv/dh = v/h can not hold them and they drop to the earth.
Densities of particle fluid mixtures are real.
Don't sail a ship over a destabilized subsurface natural gas hydrate field.
Richard D. Saam