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Right, but there is water on both sides of the hole, so two hydrostatic pressures.charlie05 said:hydrostatic pressure - is generated by weight of water.
Right, but there is water on both sides of the hole, so two hydrostatic pressures.charlie05 said:hydrostatic pressure - is generated by weight of water.
The only time there is a 5m height of water is when the boat is about to sink, the water outside being 5m above the hole. What column is 1.5m at that time? Did you mean 3.5m?charlie05 said:act against each other? pressure of column 5m is reduced by the pressure of column 1,5m?
Right.charlie05 said:phtotal = h1ρg - h2ρg = ( 5-3,5) ρg
Looks right.charlie05 said:hole area...S = π * 0,012 = 0,000314 m2
in 1 s...V1 = v*S = 5,425*0,000314 = 0,0017 m3
T = V /V1 = 1750/0,0017 = 1 029 412 s = 285,95 hour
I thought you had solved a) in order to write your post #29.charlie05 said:Thanks very much - can we please have a look at the task a/ ?
I think the proff is based on the the definitions boundancy force ?
In post #19 I wrote ho, not h0, as being the height of water outside the boat relative to the hole. Your equation is correct, but ho is a variable. Aho is the volume displaced at any time, not just at the start.charlie05 said:Archimedes' principle without hole...M*g = A ( h0)*ρ * g
Write m in terms of hi, A and ρ. For FB, use the volume displacement mentioned above.charlie05 said:Archimedes' principle with hole...M*g + mg = FB = ?
That's good, except there is no point in trying to express the boat's mass in terms of a density. Just leave that as M.charlie05 said:water weight..m= Vρ = A*hi *ρ
FB = V * ϱ * g = A* h0* ϱ * g
M*g + m*g =A ( h0)*ρ * g
A*c*ρboat *g+ A*hi*ρwater*g = A *ho *ρ*g...
c*ρboat = ρwater (h0-hi)
density of water and boat is constant...
It should dawn on you why that is so. And for that, what you really want is a graphic screenplay of this Titanic disaster,charlie05 said:(h0-hi) is constant
That velocity should be an upper estimate, with the actual velocity a bit lower. We have to add a prefactor - but that is beyond the scope of this problem.haruspex said:Looks right.
I am a little uncertain about the applicability of v^2=2gΔh here. The water that flows in then has to be arrested by the water already present. That suggests the pressure in that stream increases a little after entering the boat, which in turn suggests the water will flow in faster than that equation says. It is different from the case of water pouring out of a tank into the open air. However, I am unable to find any discussion of this on the net.
@mfb, any comments?
Correct.charlie05 said:amount of displaced water is about (A * 1.5) m3 greater than the volume of water into ship
That is true also for a tank emptying into open air, no? I am concerned about a different issue.mfb said:That velocity should be an upper estimate, with the actual velocity a bit lower. We have to add a prefactor - but that is beyond the scope of this problem.
Right.haruspex said:That is true also for a tank emptying into open air
That is countered by the water being slowed. The water cannot get more energy than necessary to flow into the boat at that speed, no matter how exactly it is slowed down.haruspex said:That implies it is flowing against a pressure gradient.
We're waltzing around each other.mfb said:That is countered by the water being slowed. The water cannot get more energy than necessary to flow into the boat at that speed, no matter how exactly it is slowed down
I had trouble finding anything on this that is not behind a paywall. Finally found https://books.google.com.au/books?i...zAG#v=onepage&q=flow submerged sluice&f=false. See eqn 5.24mfb said:I don't see how a pressure notably below ambient downstream pressure could be maintained in the stream. Such a pressure difference sucks in water from the outside. That is happening, of course, but I would expect that to form a somewhat stable vortex without large pressure differences.
It has a diagram with H1 as the depth in front of the sluice, H2 as the general depth downstream of the sluice, and H3 as the depth just after the sluice. The surface water is lower just after the sluice than further downstream, so H3 < H2, as per the formula. The height difference used for the flow rate is H1-H3.mfb said:I cannot access the previous page in the preview, which makes the formula unclear.
Ok.mfb said:Interesting, opening the PDF again I can now see the previous page, but not the page I saw before.
It discusses a notable effect as long as "a" (the height of the opening) is relevant compared to H2. For a negligible hole size, and assuming we can adapt this to the vertical case, all the a/H terms vanish.