impact of sphere on water

[tbwttihs]

Does anyone know how I can calculate the impact force/energy of a sphere falling onto a fluid?

I know the geometry and mass of the ball, the height it has fallen (so can calculate a velocity at impact), and i know the properties of the fluid its being dropped onto.

the real life situation is a frac ball being dropped down drill pipe when the mud level in the drill pipe is about 150m below drill floor. I am trying to calculate, knowing the impact strength in ft-lb/in of the ball, whether this would be enough to fracture the ball.

 

[eromlignod]

The ball will have a kinetic energy of

KE = 0.5 * m * v^2

If it's moving pretty fast into a confined liquid, it's almost like hitting concrete, so you might be able to neglect any complex fluid calculations if you're just looking for the worst case. The impact strength is just how much energy can be imparted to the material in an impact, per unit of thickness, before it breaks.

Is it a hollow ball?

Don
Kansas City

 

[tbwttihs]

Thanks for the input Don.

Yes, I suppose to go down the road of complex fluid calcs may be overthinking things a bit.

The ball is solid. It is 2-1/4" dia. made of phenolic plastic with a metalic filler. specific gravity of 3.4, so it weighs around 0.3 kg. I don't know the exact co-ef. of drag for it, so terminal velocity when it hit the mud could have been anywhere between 50 - 150 m/sec according to my rough calcs.

So if:

E = 1/2 m v^2 = .5 x .3 x 50^2 = 375 J
thickness = ball dia. = 0.057 m
impact energy = aprox 6580 J/m

The limit of the material is 0.5 ft-lb/in
1 ft-lb/in = 53.4 J/m, so 0.5 ft-lb/in = 26.7 J/m.

So we've exceeded the impact strength by a factor of about 250. This seems excessive (but agrees with what I'm trying to prove, so I'll take it).

Any more thoughts?

 

[dvd]

Terminal velocity of the ball falling through air inside of the pipe comes to mind. Also what is the inside diameter of the pipe? You may never achieve a very high velocity.

 

[dvd]

p.s. Since terminal velocity of the sphere is a function of the fluid density it would be important to use the right air density. Is the air inside of the drill pipe at atmospheric pressure, or something higher?

 

[rb1957]

craig, i think your calc is way too conservative ... you're dealing with the transfer of energy from the ball to the fluid, the ball may well be approximated as rigid, but how will the fluid flow ? is it constrained within the pipe ? (btw, i agree with dvd's comment above, about the air trapped inside the pipe (under the ball) increasing the drag of the ball ... try Hoerner "Fluid Dynamic Drag" for a solution for a ball falling down a pipe). btw, i've never seen the strength of a material dimensioned your way ... ft.lbs/in is (dimensionally) equivalent to lbf. if you could you might try non-linear FE.

 

[zekeman]

Your 50 m/sec is way too high. That is Indianapolis raceway speed.
Do the terminal velocity in free air to get a much smaller conservative term velocity;moreover, being inside a pipe would be much less.

 

[tbwttihs]

The terminal velocity did seem a bit high to me, but hey... it came from NASA.

http://www.grc.nasa.gov/

http://WWW/K-12/airplane/termv.html

on the calculator there i used mass = 0.332 kg, CSA = 0.0026, altitude = 114m, drag coeff. = 0.5 (just looked this up as a typical [high] value for a sphere)

I suppose they're used to objects travelling a bit faster than small plastic balls in drillpipe though, so not necessarily to be trusted....

The drillpipe ID is about 4-3/4 inches.

 

[rb1957]

one mistake i can is (i think) that the mass of the ball isn't 0.3kg (that'd be its weight) but 0.03kgm.

v=sqrt(2gh) is the maximum possible velocity (no drag)
=47 m/sec

including drag ...

mgh = 0.5mv^2+(Cd*0.5rhov^2*CSA)*v

2gh = v^2(1+Cd*rho*CSA*v/m)

if v = 47m/sec the () term is 1+0.5*1.225*0.0026*47/0.03) = 3.5

correct v ... v = sqrt(2*9.8*114/3.5) = 47/sqrt(3.5) = 25m/sec

correct () ... = 2.33
correct v ... = 31m/sec
correct () ... = 2.64
correct v ... = 29 m/sec ...

converging on 30 m/sec

there looks as tho' there's plenty of room in the pipe for the ball (2.25" dia in 4.75" dia pipe) ... still how can the fluid react to the impact ?

 

[amorrison]

See thread404-96282

 

[eromlignod]

It's not going to behave like free fall. The faster it falls, the harder it is for the air to squeeze by it as it passes through the pipe. It's like an elevator falling down a shaft: pressure builds up under the ball and needs to get around it as it falls.

This whole thing is starting to appear to me like one of those cases where you could waste a colossal amount of time (which is money) on assumptions and complex calculations and still not get it right. Can't you just drop one of these balls down the chute and see if it breaks? How much does the ball cost?

Don
Kansas City

 

[KENAT]

Didn't Barnes Wallis spend a lot of time on this kind of thing.

Don't know if there'd be any publications, also it wouldn't address the fact it's in a tube.

KENAT, probably the least qualified checker you'll ever meet...

 

[rb1957]

an elevator is a much closer fit in its shaft than this ball (the tube dia is more than twice the ball dia.).

barnes wallis also spun his bomb (to get it to skip).

 

[40818]

Still reckon the test example would be a good option, you could spend a long time dissapearing up your own arse trying to derive that theoretical answer to this problem, or just chuck it down the pipe!!
One question though, why are you doing it??

 

[KENAT]

Yeah he span it and he had horizontal velocity but it was the closest thing I could think of.

I remember in the film "Dambusters" that his early prototypes would break up on impact so I thought there may be some related info.

KENAT, probably the least qualified checker you'll ever meet...

 

[40818]

A 5 minute google comes up with a greate deal of semi-related information about this topic , without answering it (in a very limited look).
I would like the OP to give the reaason for dropping the ball down the hole and more details on the whole thing.
We could speculate about it, but at the end of the day, all we know is that somebody wants to drop a ball down a hole in the mud.
If the sphere has to be intact upon impact then you can design it to be by picking the worst case if needs be, but without more specifics, then you end up arguing about semantics.

 

[amorrison]

The analysis here will be very similar to that used for a rotameter.

http://archives.sensorsmag.com/articles/1002/14/main.shtml

 

[zekeman]

rb157
quote "mgh = 0.5mv^2+(Cd*0.5rhov^2*CSA)*v"

Where did you get that one?

zeke

 

[tbwttihs]

rb1957 - the mass is 0.3 kg (i.e. the product of it's density and volume). It's weight would be that value multiplied by gravitational acceleration, no?

to explain a bit more history, this ball is used in a tool which is run down into the well. fluid is initially circulated through the tool to wash out the landing area, then when the tool is landed, the ball is dropped down the hole, which blocks a port and allows pressure to be built up. this pressure then shears a pin which opens a port and allows pressure to energise the tool. it sounds a bit daft but it's a field proven tool which has worked great for years and similar designs are used across the industry.

ordinarily the ball would be dropped in and the pipe would be full of fluid right up to the drill floor. in this instance due to the additives which had recently been put in the drilling mud, the fluid level was about 115m below the drill floor when the ball was dropped in.

when the ball went in, they waited the 10 minutes or so for it to land (1500m down to the seabed) then pressured the tool but couldnt build pressure to function the tool. when it was recovered to surface, the ball had fragmented.

essentially what i'm trying to prove is that what caused the ball to fracture was the impact on the fluid after a 115m free-fall, rather than any malfunction inside the tool.

 

[rb1957]

craig83,
"... so it weighs around 0.3 kg", your post 12th (@ 11.07)
i assumed you meant kgf

dia = 2.25", volume = 5.95in3 = 98cc
a liter of water weighs 1kgf, so your ball weighs about 0.3kgf, and its mass is about 0.03kgm.

zekeman,
potential energy = kinetic energy + energy lost overcoming drag (=DV)