{"id":12220,"date":"2010-10-25T09:15:18","date_gmt":"2010-10-25T15:15:18","guid":{"rendered":"http:\/\/rankexploits.com\/musings\/?p=12220"},"modified":"2010-10-26T12:28:28","modified_gmt":"2010-10-26T18:28:28","slug":"gadunkan-relates-to-discussion-of-condensation-in-gcms","status":"publish","type":"post","link":"https:\/\/rankexploits.com\/musings\/2010\/gadunkan-relates-to-discussion-of-condensation-in-gcms\/","title":{"rendered":"Gedanken: Relates to Discussion of Condensation in GCMs"},"content":{"rendered":"

Some of my readers, JeffId, Gavin and others are discussing the possibility that GCM’s do not properly account for condensation on local pressure. I am agnostic on this (I’m trying to think about it.) However, plowing through the comments at The Air Vent<\/a>, I came across an argument between a number of people (Carrick, SteveF and others). I think the question that triggered the argument is the following one:<\/p>\n

94. Gavin said<\/p>\n

Consider a thought experiment. Take a closed container filled with super-saturated air and place it on a scale. Will you be able to detect the moment of condensation by monitoring it\u00e2\u20ac\u2122s weight (i.e. the pressure on the scale)?<\/p><\/blockquote>\n

I bet you all think you know the answer to this question. Right? I’m going to answer this question in agonizing nit-picky detail. <\/p>\n

But before I do, I’m going to show discuss a gedanken problem that contains some of the physics we need to understand before we can fully answer Gavin’s question. This will serve as a preliminary to the full answer to Gavin’s question. The gedanken will also address a debate going on in comments on that thread. By showing this problem we will be able to go forward and figure out if the nit-picky details discussed here could ever have any practical<\/i> importance in any application of interest (and more particularly GCMs.)<\/p>\n

The Thought Experiment:
\nPreliminaries:<\/strong>
\nSuppose we had a very tall massless container full of some viscous fluid, with density \u00cf\u0081f<\/sub> and weight Wf<\/sub> which for the purpose of this discussion will be 1N. The container placed on the read out on the scale will balance the weight of the fluid, W=1N.<\/p>\n

\"\"<\/a>Next, suspend a small diameter, heavy sphere with weight Wp<\/sub> from the ceiling in the room; we select this sphere so that its density is much larger than the fluid \u00cf\u0081f<\/sub>< < \u00cf\u0081p<\/sub>. After immersing the sphere we wait for everything to come to static equilibrium, then read weight registered on the scale. <\/p>\n

Because the weight supported by the scale is the sum of the fluid and the weight of water displaced by the sphere, the scale will now read W = 1N + Fb<\/sub> where Fb<\/sub> is the weight of buoyancy force, and is equal to the weight of displaced fluid. To facilitate discussion, I’ll break this out into a numbered equation:<\/p>\n

(1)<\/eqnumber>Fb<\/sub>~ \u00cf\u0081f<\/sub> g Vp<\/sub> <\/equation>
\nwhere g is the acceleration due to gravity and Vp<\/sub> is the volume of the sphere (or particle).<\/p>\n

Note in the figure to the right, the scale reads W~1N. This is because given the assumption \u00cf\u0081f<\/sub>< < \u00cf\u0081p<\/sub>, the weight of water displaced by the sphere is very small relative to the weight of the solid sphere and my illustration neglects that small weight. <\/p>\n

Begin experiment: Break support<\/strong>
\nAt a time we will call t=0, we will break the support and simultaneously read the weight on the scale. <\/p>\n

\"\"<\/a>First question:<\/em> What should the scale read? The answer is very, very<\/em> close to 1N+Fb<\/sub>. That’s the same weight we read just prior to cutting the support.<\/p>\n

Why?<\/p>\n

The sphere is motionless.<\/em> In this circumstance, the upward force exerted by the fluid on the sphere is approximately equal to that due to buoyancy and equal to the weight of displaced fluid
\n(1)<\/eqnumber>Ffp<\/sub>~ Fb<\/sub> <\/equation><\/p>\n

The particle exerts an equal and opposite force on the fluid. So, for all practical purposes, at this point, the pressure field in the fluid obeys the laws of hydrostatics. The pressure at the bottom of the cylinder must result in a force that balances the the sum of the weight of the fluid only and the weight of fluid displaced by the particle. It need not balance the weight of the particle. (Note: For those who are aware of funky forces like added mass or the Basset force, I’m both neglecting and not discussing them. However I am including ~ in place of = for that reason.) <\/p>\n

Next question:<\/em> What happens to the sphere? The answer is simple: It starts to accelerate downward. This happens because while the force of gravity on the particle is Fg<\/sub> = \u00cf\u0081p<\/sub> g Vp<\/sub>, and acts downward.<\/p>\n

So, the net downward force on the particle is:<\/p>\n

(2)<\/eqnumber>Fnet<\/sub>~ ( \u00cf\u0081f<\/sub>-\u00cf\u0081g<\/sub> ) g Vp<\/sub> <\/equation><\/p>\n

So, at exactly t=0, when the velocity of the particle is exactly 0, the particle will accelerate at <\/p>\n

(3)<\/eqnumber>Mp<\/sub>dU\/dt= Fnet<\/sub> <\/equation>
\nwhere Mp<\/sub> is the mass of the sphere and U is the vertical velocity with positive values pointing down. <\/p>\n

Owing to this acceleration, the velocity of the particle will increase; so at 0< t, the 0< U, with positive values of U corresponding to the downward direction. <\/p>\n

What does the scale register at 0<t?<\/strong>
\nTo figure out what the scale registers after the sphere is released, we must discuss the viscous<\/i> force acting on the particle. For the purpose of further discussion, we will assume that other than in the region near the surface of the particle the fluid velocity is zero, and neglect any acceleration of the bulk fluid when discussing the weight registering on the scale. The magnitude of the fluid acceleration will be small if \u00cf\u0081f<\/sub>< <\u00cf\u0081p<\/sub> as previously assumed. <\/p>\n

Force at time 0< t.<\/em>
\nOnce the sphere’s velocity U is not zero, the force exerted by the fluid on the particle will include both the buoyancy force acting upward and the viscous force, in this case also acting upward. To avoid specifying the constitutive relation for the viscous force, we will represent this as Fv<\/sub> and observe that it is a function of the velocity of the particle relative to the velocity at some point far from the surface of the particle. Given previous assumptions, the viscous force acting on the particle will be a function of particle velocity only, Fv<\/sub>(U). <\/p>\n

\"\"<\/a>Because this viscous force is exerted on the fluid by the particle, an equal and opposite force is exerted by the particle on the fluid. If the bulk of the fluid does not accelerate, the weight exerted by the bottom of the container on the fluid will be equal to the sum of the mass of fluid, the mass of fluid displaced by the sphere and force exerted by the particle on the fluid. So, the scale now registers a weight W of :
\n(4)<\/eqnumber>W=1N+Fb<\/sub>+ Fv<\/sub><\/equation><\/p>\n

This is larger than the previous read out of W=1N+Fb<\/sub>. How much larger will depend on the magnitude of Fv<\/sub>, which will depend on the velocity of the sphere.<\/p>\n

Final state:<\/strong>
\nIf the container is sufficiently tall (i.e. infinitely), the particle will eventually accelerate to its terminal velocity. At that point, the viscous force acting on the particle will exactly balance the weight of the particle. So <\/p>\n

(5)<\/eqnumber> Fv<\/sub>+ Fb<\/sub>= Wp<\/sub>=2N<\/equation>.<\/p>\n

At this point, the weight registered by the scale will be:
\n(4)<\/eqnumber>W=1N+ 2N<\/equation><\/p>\n

That is: Once the particles have achieved their terminal velocity, the weight on the scale will equal the sum of the weight of the fluid and the weight of the solids. Of course, that will also be the weight when the sphere comes to rest at the bottom of the container. <\/p>\n

Observation<\/strong>
\nFor this gedanken experiment, the weight registered by the scale would vary over time, increasing from approximately 1N when the velocity of the sphere is motionless to 3N when the sphere has either reached its terminal velocity or come to rest at the bottom of the container. <\/p>\n

This may seem perplexing to those whose eyes are focused on the outside<\/i> of the container which is motionless giving the false impression the container and its contents are in static equilibrium. If that were so, the weight registered on the scale jump to 3N the exact instant the cable supporting the 2N sphere was broken. <\/p>\n

However, contrary to one’s initial impression, the contents of the container are not<\/i> in static equilibrium. The heavy sphere is accelerating downward. Because the sphere is denser than fluid, this means the center of mass of the contents of the container are accelerating downward. This can only occur if the weight exerted on the container by the scale is less than the weight of the container’s contents.<\/p>\n

This analysis will represent the second<\/i> part of an analysis in my next more complicated, gedanken experiment which may require magic to occur at time t=0. The result will have some consequences if one wishes to provide a full, thorough, complete, true to the tiniest nit answer to Gavin’s question. We’ll see that the answer to his question is: Under circumstances (possibly those that involve magic at time t=0), it might<\/i> be hypothetically possible to detect condensation by monitoring weight with a scale. <\/p>\n

Will the result of this or the second analysis have any practical<\/em> consequences for climate models, or the issue of properly modeling condensation? I have no idea. It may very well have no real consequences. But, recognizing why the answer to Gavin’s question is not simply, “Flat out no, not under any circumstances, no”, will help people do scaling analyses to either show the effect doesn’t matter or show that it does. <\/p>\n","protected":false},"excerpt":{"rendered":"

Some of my readers, JeffId, Gavin and others are discussing the possibility that GCM’s do not properly account for condensation on local pressure. I am agnostic on this (I’m trying to think about it.) However, plowing through the comments at The Air Vent, I came across an argument between a number of people (Carrick, SteveF … Continue reading Gedanken: Relates to Discussion of Condensation in GCMs<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17],"tags":[454,369,370],"class_list":["post-12220","post","type-post","status-publish","format-standard","hentry","category-gcms","tag-gcms","tag-condensation","tag-tav"],"_links":{"self":[{"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/posts\/12220","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/comments?post=12220"}],"version-history":[{"count":0,"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/posts\/12220\/revisions"}],"wp:attachment":[{"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/media?parent=12220"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/categories?post=12220"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rankexploits.com\/musings\/wp-json\/wp\/v2\/tags?post=12220"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}