Calculate Time & Force for Steel/Bakelite Ball Through Sea Water

  • Thread starter Thread starter eightace148
  • Start date Start date
  • Tags Tags
    Calculation
AI Thread Summary
To calculate the time for a 1.0" diameter stainless steel ball with a density of 8.0 g/cm³ to travel through 2000m of seawater (density 1026 kg/m³), one can use the concept of terminal velocity, which is reached quickly in this scenario. The formula for time can be simplified to time = 2000m/terminal_velocity, assuming constant seawater density and smooth flow. For the Bakelite ball with a density of 1.25 g/cm³, similar calculations apply. Impact force is more complex, as it depends on how quickly the ball decelerates upon hitting the seabed, which is influenced by the stiffness of the surface. Proper equations must be used, as terminal velocity differs in liquids compared to gases, and the impact scenario involves a hard surface.
eightace148
Messages
7
Reaction score
0
hello everyone

Iam wonderind if somone could help me?

I need to find out how long it would take for a Stainless steel ball 1.0" Dia With a density of 8.0g/cm qubed through 2000m of sea water with a density of 1026kg/m qubed and what impact force it would genorate.

again this calc would need to be done with a Bakelite ball with a densaty of 1.25g/cm qubed.

If somone could point me in the right direction it would be grately appreciated.
 
Physics news on Phys.org
The terminal velocity in a fluid is fairly simple (http://en.wikipedia.org/wiki/Terminal_velocity#Terminal_velocity_in_the_presence_of_buoyancy_force)
Note: Cd for a sphere is about 0.47
Assuming that the density of seawater doesn't change over 2000m and that the flow is smooth.

Is you assume it reaches terminal velocity pretty quickly compared to the 2000m total then you can just assume that it takes time = 2000m/terminal_velocity.

Impact force is a little more difficult - the FORCE doesn't depend on the speed it hits with but the rate at which it slows down as it hits. So it is totally dependant on the stiffness of the sea floor - this is why an object falling from the same height onto say concrete would break but would survive landing in mud.
 
I assume the ball has accererated to it's teminal velocity. Be carefull what equation you use. Terminal velocity is different for objects falling in liquids comopared to gasses.
 
The ball will be coming to rest on a steel section of tooling so pritty hard stuff.

Thanks for the imput folks.
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
comparing a flat solar panel of area 2π r² and a hemisphere of the same area, the hemispherical solar panel would only occupy the area π r² of while the flat panel would occupy an entire 2π r² of land. wouldn't the hemispherical version have the same area of panel exposed to the sun, occupy less land space and can therefore increase the number of panels one land can have fitted? this would increase the power output proportionally as well. when I searched it up I wasn't satisfied with...
Back
Top