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Topic: Fill Time Calculation (Read 5756 times) 

Icarus
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Fill Time Calculation
« on: Apr 15^{th}, 2003, 10:17am » 
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I have just finished a caculation in the course of my job that was fairly involved, and while I believe I have it correct, I would really appreciate it if someone could confirm my results. The point of the calculation is this: If an object (say, purely hypothetically, an airplane) that has holes in it is set into water, how long would it take for the interior and exterior water levels to equal? I have made three simplifying assumptions: 1) A single hole of area A at the lowest point of the object. 2) The depth of the hole under the exterior water level at time t varies linearly with the volume of water inside the object (this would be true if the exterior "walls" of the object were vertical). 3) The volume of water inside the object is directly proportional to the depth of the hole under the interior water level (this would be true if the interior "walls" of the object were vertical). The equations I get are dV/dt = A(2P/r)^{1/2} for flow rate through an aperature of area A at pressure P, where r is the density of water. (V=volume passing through, t = time) dV/dt = A(2g(d_{e}d_{i}))^{1/2} for the situation described, where d_{e} is the depth of the hole under the exterior water plane at time t, and d_{i} is the depth of the hole under the interior water plane at time t. (This holds without my assumptions.) If the assumptions are made, then (dV/dt)^{2} = (2gA^{2}D_{0}/V_{F})(V_{F} – V) where D_{0} = d_{e} at time 0. And V_{F} is the total volume of water in the object at the finish.(Those are numbers dependent on the object, which I can take as being given here.) Finally, the fill time T is given by T = V_{F}(1 + 3^{1/2})/A(2gD_{0})^{1/2} It is this last equation in particular that I would like some verification on, to insure I didn't make a calculation error somewhere.

« Last Edit: Apr 15^{th}, 2003, 12:19pm by Icarus » 
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Icarus
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Re: Fill Time Calculation
« Reply #1 on: Apr 15^{th}, 2003, 12:19pm » 
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Okay, a fair bit of that was wrong! So I am starting over. I still am examing the fill rate for an object set in water. The question is, though: How much time does it take for a given amount of water to flow in? More directly for me, the problem is how long until the max depth of the object reaches a certain value? (After that depth, everyone still in the object is pretty much doomed.) The assumptions I am making are: (1) a single hole (aperature) of fixed area A located at the lowest point on the object. (2) If you express the difference between the external and internal water levels at time t as D(t), and the Volume of water inside the object as V(t), then D can be expressed as a function of V (this is a matter of physics, not an assumption). I am assuming that D is essentially linear in V (secondorder and higher terms are much smaller than the firstorder term). This is slightly weaker than the original assumption. The first two equations are the same, except the second is expressed in terms of D as: dV/dt = A(2gD)^{1/2} The final solution I get is: T = (2/g)^{1/2} (V_{T}/A)( (D_{T}^{1/2}  D_{0}^{1/2}) / (D_{T}  D_{0}) ) Can anyone verify?

« Last Edit: Apr 15^{th}, 2003, 12:19pm by Icarus » 
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aero_guy
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Re: Fill Time Calculation
« Reply #2 on: Apr 15^{th}, 2003, 12:33pm » 
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OK, from a first look it seems that you are trying to say, If a plane crashed in the ocean how long would it take to sink?" You make the assumption of a single hole. Without going into the equations it seems, at least at this first crack, that you are making the further assumptions that: 1)There are no losses in travelling through the constriced opening. You will need to consider these later on, or else just say that your answer is conservative in that you will get a faster sinking time than will actually occur. 2)There is a large opening on top to allow for air to escape. If this was an originally pressurized aircraft than this will not hold unless there are structural failures in the top of the aircraft as well. If the openings are small then you need to consider this as part of your analysis. If you consider the roof to have flown off you do not. 3)You assume that there is no pressure gradient from the top of the opening to the bottom. This is probably an OK assumption if the plane is large and the opening is small, but for say a vertical crack it is not. 4)It seems that you have ignored the sinking of the aircraft itself in your second post, which will change the relative depth, D(t). I am going to take a further look at things, but that is what I see from the start.


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aero_guy
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Re: Fill Time Calculation
« Reply #3 on: Apr 15^{th}, 2003, 2:08pm » 
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I can confirm your answer iff time T is the time when D(T)=0, otherwise I do not agree. You can make it more generic by saying: T = (2/g)^{1/2} (a/A)(D_{T}^{1/2}  D_{0}^{1/2}) where a is the linear conversion between volume and height.

« Last Edit: Apr 15^{th}, 2003, 2:19pm by aero_guy » 
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aero_guy
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Re: Fill Time Calculation
« Reply #4 on: Apr 15^{th}, 2003, 2:30pm » 
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Interesting note: If the volume and D are linearly realted, then because of bouyancy effects D WILL NOT CHANGE. As water floods in, the plane will sink at a rate that keeps D constant. This makes me think the original analysis is quite flawed, and we get a linear increase in the depth of the water with: DV/Dt=A(2*g*W/(rho*a))^{1/2} where rho is the density of water, W is the weight of the aircraft, and a is the relation between volume and depth (the area of the base of a rectangularly prismatic aircraft).


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Icarus
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Re: Fill Time Calculation
« Reply #5 on: Apr 15^{th}, 2003, 4:59pm » 
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The situation involves ditching in water. It is assumed the aircraft (I avoid saying plane here, because that word keeps appearing in its geometric interpretation) ditches without damage. There is no trapped air, because the door has to be open for the passengers to get out. The depth that is the limiting time is when the water level outside reaches the door bottom and can start flowing in (the escape hatch is higher, but we are required to provide escape from both exits). You definitely have a misunderstanding yourself if you get D constant when linearly dependent on internal water volume. One case where this linear dependence holds exactly is when the object has constant internal and constant external crosssectional area. Consider a cylinder closed on bottom except for an opening of area A, and assume that the density is sufficiently small for the cylinder itself to be bouyant. By your claim, the cylinder would be floating in the water with the internal water level below the external water level. But experience shows that water will flow into the cylinder until they equal out. As for being conservative, that is very much what I want to be. There are far too many factors going on here to have any hope of successfully analyzing completely the situation. In particular, instead of 1 opening at the lowest portion of the aircraft, I have a number of openings located in numerous places. The lowest point of the aircraft changes as it settles. How it changes depends very much on what interior space is unoccupied, how freely the water can spread to fill that space, and  particularly for the small jets that I am working with  exactly how the aircraft is loaded. Given how much is involved, a good analysis by the fastest supercomputers might possibly manage to finish it by the time that particular model of aircraft is retired. And it almost certainly would be cheaper to simply dump a finished aircraft in a pond and see how long it takes to sink. So conservative is the way I go. It is far better to underestimate the sink time, than it would be to overestimate it. In the first case the luckless passengers will look back in disgust, wondering what the hurry to get out was about. In the latter, they might not be wondering anything for very long. I intentionally expressed "a" as (V_{T} / (D_{T}  D_{0})). The three constants here are ones that I have some hope of obtaining values for. The only way of obtaining a value for "a" is to calculate it from this expression. It is just as general as with an "a", since in both cases it applies to a linear relationship between D and V. And when such a relationship exists, the expression for "a" holds. D(T) = D_{T}. So if it is 0, the formula can be reduced to T = (2/gD_{0})^{1/2}V_{T}/A Concerning your final equation: The pressure P is the water pressure. It is not a function of the weight of the object  only of the depth of the opening  more particularly, of the difference in depths on opposite sides of the opening. Since your calculation provided a false result (the constant D value), I cannot reject my own work on its account. However, I do very much appreciate your interest. It is always helpful to talk things out with someone else. Just posting this in the first place gave me a different perspective which allowed me to restate the problem, and obtain an easier answer.


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aero_guy
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Re: Fill Time Calculation
« Reply #6 on: Apr 15^{th}, 2003, 11:12pm » 
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Actually, your analysis of the cylinder is wrong. You say that experience shows that the water will rush in when that is not the case. I have two forms of proof: 1) Experimental. Take an empty plastic soda bottle and cut a small hole in the bottom. Remove the top and add a handful of bolts for weight. Drop the apparatus into a sink of water. You will see that the distance between the surface of the water in the sink and in the bottle barely changes (due to changes in cross section of bottle) while bottle itself slowly sinks. This works best with a small hole, it gives you time to see the effect. 2) Mathematical. For a cylinder of base a and weight W filled with water of height s from the base where distance between the height of the inner and outer water is D we have an bouyancy equation of: rho*a*s + W = rho*a*(D + s) weight of internal water + weight of bottle = weight of water displaced (assuming negligible wall thickness) Thus we have: D=W/(rho*a) Therefore D is a constant even though s is going to increase linearly. The level of the internal water below the external remains constant while the internal water level rises. This makes the further assumption that the sinking process is slow enough that the bouyancy issue is quasistatic, which is true if there is any possibility of the people getting out alive. As the solution with constant crosssection comes out so much easier than you would expect, it should not be too difficult to expand the results to more complex cross sections. Tell me how it goes.


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aero_guy
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Re: Fill Time Calculation
« Reply #7 on: Apr 16^{th}, 2003, 12:13am » 
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As to your other points, I was wrong about your equation for a, it is right, and I was wrong about the shape of the holes, it doesn't matter so long as they are all initially submerged. Lets solve the whole problem. Assumption 1) the aircraft is a jet liner where bouyancy without any internal water causes the wing to be touching the water, and when the water reaches the top of the wings it has reached the levels of the exits. Thus, we can assume the cross section of the submerging portion of the wings (the part at water level),a, remains constant. Assumption 2) all holes into the inerior are initially submerged. Assumption 3) all the water flows into connected portions of the aircraft which we will model as a thin walled cylinder sitting horizontal in the water with radius r and length l. Assumption 3) there are no viscous losses as the water flows through the holes. Assumption 4) the aircraft does not tilt while sinking. Thus we have a total area of all holes as A. If we say that the bottom of the wings are d below the center of the cylinder we can say that: rho*V + W = rho*a*x + rho*l*(r^{2}*acos((dx)/r)(dx)*sqrt(r^{2}(dx)^{2}) Where V is the volume of water in the aircraft and x is the height of the external water above the base of the wing. When x=h, the thickness of the wing, you are out of luck. We still have the equation: dV/dt=A*sqrt(2*g*D) now we need to relate D, V, and x V=l*(r^{2}*acos((dx+D)/r)(dx+D)*sqrt(r^{2}(dx+D)^{2}) a variable change is in order of dx=q, so our equations are: l*d/dt(r^{2}*acos((q+D)/r)(q+D)*sqrt(r^{2}(q+D)^{2})=A*sqrt(2*g*D) and l*(r^{2}*acos((q+D)/r)(q+D)*sqrt(r^{2}(q+D)^{2}) + W/rho = a*(dq) + l*(r^{2}*acos(q/r)q*sqrt(r^{2}q^{2}) At this point I say a quick numerical integration is in order.

« Last Edit: Apr 16^{th}, 2003, 12:19am by aero_guy » 
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Icarus
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Re: Fill Time Calculation
« Reply #8 on: Apr 16^{th}, 2003, 3:52pm » 
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Concerning the cylinder  I stand corrected: If you can find a material with positive weight, but NO volume to build your cylinder from, then indeed the difference in water level is constant. However for real world situations, it is not. If you will look at my comments on experimental evidence, you will note that I said "the cylinder itself is bouyant". I.e. the particular case I was refering to has a cylinder that floats. By my experience, if you drop such a vessel in the water as I described, it will quickly sink to its "empty" bouyancy level, then slowly fill with water and sinking further, until it reaches it "full" bouyancy level. By your claim, the difference in water levels would be the same the entire time. This demands that when it reaches equilibrium, no more water will flow into the cylinder. But in truth water will still enter until the interior water level reaches the same level as outside. Your experiment works only because you have chosen a situation with a very high effective density for the walls. I don't have time right now to look over the rest, but I will get back later.


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Icarus
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Re: Fill Time Calculation
« Reply #9 on: Apr 16^{th}, 2003, 8:00pm » 
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Quote:Assumption 1) the aircraft is a jet liner where bouyancy without any internal water causes the wing to be touching the water, and when the water reaches the top of the wings it has reached the levels of the exits. Thus, we can assume the cross section of the submerging portion of the wings (the part at water level),a, remains constant. 
 That is somebody else. Note what I said about SMALL jets. My brother works for them. He says they don't have any problems certifying for ditching in water. The wings on my planes are nearly submerged from the start. And the wings are not flat  they have dihedral tilt. Quote:Assumption 2) all holes into the inerior are initially submerged. 
 Working on a "by hole" basis is not possible at this time. Most of my floatation modeling is already done. All I need here is to find as high as reasonable a "minimum" time to reach certain points in the sinking process. The best approach with the resources available right now is to assume a single hole at the lowest portion of the plane, whose area is sufficient to cover the affects of all the actual holes. Conservative estimation is all that is available to me at present. Quote:Assumption 3) all the water flows into connected portions of the aircraft which we will model as a thin walled cylinder sitting horizontal in the water with radius r and length l. 
 Actually I am pretty far removed from this situation. A thin walled cylinder is no more adequate a description than vertical walls. Actually it is less  I believe that the vertical walls situation sinks faster, and again, if I am going to error, it better be on the side of underestimating the times. Quote:Assumption 3) there are no viscous losses as the water flows through the holes. Assumption 4) the aircraft does not tilt while sinking. 
 I did some research on the web today. Flow through a nozzle is calcuated by the equation I came up with: dV/dt = A(2P/r)^{1/2}, except that A is multiplied by a "geometry factor" <1 to account for edge affects. Since A is only an approximation, its value covers any such affects. The aircraft tilts while sinking. I already know what angle it is going to assume at important points on the way. Quote:rho*V + W = rho*a*x + rho*l*(r2*acos((dx)/r)(dx)*sqrt(r2(dx)2) 
 There is no bouyant volume in your model. This is far from the case for an aircraft. For starters, the fuel tanks (the majority of the volume of the wings) are sealed. There are many bottles, a not insignificant amount of structure and systems volumes, though this is actually smaller than it seems like it ought to be. The pressure vessel is for the most part water proof. (The main time I am calculating is how long until water starts poring into the door, so for my calculations, it hasn't started.) The relationship I gave: D = D_{0} + D_{1}V is more justifiable than this, since I can match the two constants with measurable values, so it more accurately represents my plane instead of an abstract model. It might be better yet to add in the V^{2} term, which I suspect to be the last term to make significant contributions, but this introduces a constant that I don't have a good way of obtaining a value for. (Getting the value for D_{1}, "a" in your earlier post, is rather difficult as it is). So I am assuming a value of 0 as a "conservative estimate".


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aero_guy
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Re: Fill Time Calculation
« Reply #10 on: Apr 17^{th}, 2003, 7:03am » 
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I see your points, and the model was quite an abstraction, though it does take into account bouyancy effects. The wing are should have included the area of all sinking nonfillable parts of the aircraft. It sounds like you have a hell of a problem on your hands. Classic engineering: There is no real answer, you need to find a way to get a 'good' answer.


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Icarus
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Re: Fill Time Calculation
« Reply #11 on: Apr 17^{th}, 2003, 7:18pm » 
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It's a common situation. Fortunately all we have to show is that there is a reasonable amount of time for everyone to get off before the plane goes under. This means that as long as we are underestimating sink time, we can be pretty far off and still not have a problem (provided of course that we are able to find enough time). If actual numbers were necessary, we would be spending the money to drop test articles in ponds. Thanks for the discussion.


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