Sinking Boat: Calculations & Results

[charlie05]

AHA :-) :-) :-)

now I need to calculate b / velocity
0,5 ρ v² = hρg ?

 

[haruspex]

AHA :-) :-) :-)

now I need to calculate b / velocity
0,5 ρ v² = hρg ?

Where h is...?

 

[charlie05]

that is the question :-)... column of water above the hole. ? h=1,5 m ?

 

[haruspex]

that is the question :-)... column of water above the hole. ? h=1,5 m ?

What force pushes the water in through the hole? What height is responsible for that?

 

[charlie05]

hydrostatic pressure - is generated by weight of water.

 

[haruspex]

hydrostatic pressure - is generated by weight of water.

Right, but there is water on both sides of the hole, so two hydrostatic pressures.

 

[charlie05]

act against each other? pressure of column 5m is reduced by the pressure of column 1,5m?

 

[haruspex]

act against each other? pressure of column 5m is reduced by the pressure of column 1,5m?

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]

so in fact the hydrostatic pressure makes only 2 m column?

 

[mfb]

Where do the 2 meters come from? It is wrong, and I don't see any way to get that with the given 5m and 3.5m.

Did you draw a sketch?

 

[charlie05]

p_htotal = h₁ρg - h₂ρg = ( 5-3,5) ρg

 

[haruspex]

p_htotal = h₁ρg - h₂ρg = ( 5-3,5) ρg

Right.

 

[charlie05]

aha :-) so:

p_htotal = 1/2 ρv²...v = √ 2g(h₁-h₂) = √ 2g*(5-3,5) = 5,425 m/s

 

[charlie05]

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

 

[haruspex]

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

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?

 

[charlie05]

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 ?

 

[haruspex]

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 ?

I thought you had solved a) in order to write your post #29.
Yes, it is to do with Archimedes' principle, as I wrote earlier.
Try to answer my post #19.

 

[charlie05]

a/ Show that difference between the levels inside and outside the ship is not changed.

without hole...outside 1,5m, inside 0
Water flows in1 m...outside 2,5m...inside 1m
ship sinking...outside 5 m..inside 3,5m

...level difference is always 1,5m?

can I prove it generally?
Archimedes' principle without hole...M*g = A ( h₀)*ρ * g
Archimedes' principle with hole...M*g + mg = F_B = ?

 

[haruspex]

Archimedes' principle without hole...M*g = A ( h0)*ρ * g

In post #19 I wrote h_o, not h₀, as being the height of water outside the boat relative to the hole. Your equation is correct, but h_o is a variable. Ah_o is the volume displaced at any time, not just at the start.

Archimedes' principle with hole...M*g + mg = F_B = ?

Write m in terms of h_i, A and ρ. For F_B, use the volume displacement mentioned above.

 

[charlie05]

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...

 

[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.