Sinking Boat: Calculations & Results

[haruspex]

water weight..m= Vρ = A*h_i *ρ
F_B = V * ϱ * g = A* h₀* ϱ * g

M*g + m*g =A ( h0)*ρ * g
A*c*ρ_boat *g+ A*h_i*ρ_water*g = A *h_o *ρ*g...

c*ρ_boat = ρ_water (h₀-h_i)
density of water and boat is constant...

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.
In terms of these variables, what is part a asking you to prove?

 

[charlie05]

M*g + A*h_i*ρ*g = A*h₀*ρ*g
M = A *ρ*(h₀-h_i)
M,A and ρ is constant...(h₀-h_i) is constant too?

 

[BvU]

Dear Charles,

Your exercise is very helpful in part a) by positioning you on the track

(h0-hi) 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,
properly embellished with symbols for $h_*$ (I leave that to you)

Here is my artistic rendering, and as you see, I already cashed in on the one item that you are asked to prove:

Note that the darker blue only distinguihes water inside from water outside: its $\rho$ is the same (hint hint)

 

[charlie05]

thank you for the sketch, I can now imagine it, it is clear...
h_0utside-h_inside = 1,5m constant, in all cases

 

[BvU]

That's how I suggestively have drawn it -- can you explain why it is so ?

 

[charlie05]

because there is still a buoyant force an empty ship?
amount of displaced water is about (A * 1.5) m³ greater than the volume of water into ship

Are posts 50,52, 54 right or not?

 

[mfb]

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?

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.

amount of displaced water is about (A * 1.5) m³ greater than the volume of water into ship

Correct.

 

[haruspex]

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.

That is true also for a tank emptying into open air, no? I am concerned about a different issue.
The water that has passed through the hole is slowed by the surrounding water. That implies it is flowing against a pressure gradient. Thus the pressure on the immediate downstream side of the hole is less than the ambient pressure at that level inside the boat.
Note that this makes it different from the emptying tank case.

 

[mfb]

That is true also for a tank emptying into open air

Right.

That implies it is flowing against a pressure gradient.

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]

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

We're waltzing around each other.
At the level of the hole, there is an ambient pressure p_u on the upstream side and p_d on the downstream side, determined by the water depths. Along the stream line between the two, the pressure varies. Water accelerates into the hole, so the pressure declines there, but decelerates on the other side, implying a pressure increase. The situation is analogous to a ball rolling into a dip and up the other side.
Say the pressure minimum along the path is p_m. The velocity at the hole itself is determined by p_u-p_m>p_u-p_d.

 

[mfb]

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.

 

[haruspex]

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.

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 5.24

 

[mfb]

I cannot access the previous page in the preview, which makes the formula unclear.

 

[haruspex]

I cannot access the previous page in the preview, which makes the formula unclear.

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]

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.

 

[haruspex]

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.

Ok.