It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

Propulsion systems for SuperSonic speeds movement underwater, how?

page: 2
0
<< 1   >>

log in

join
share:

posted on Apr, 10 2006 @ 09:02 AM
link   
you posted:

"The speed at which the cavities form depends on a number of things, but are mostly related to pressure. The maths can get quite complex. If i get chance over the next few days i will sort some infomation out about it."

Good point ..thanks, As I recall pressure also changes the speed at which one can run quiet at higher speeds due to this cavitating effect. I am surmising by this that one does not run cavitation speeds at great depths. It would be much easier to achieve this type of cavitation at shallow depths. Obviously running quiet is not a issue in this type of propulsion.

I belive the pressure of which you speak is the drop in pressure due to a given shape nosecone at which the bubble/cavitation forms at a given depth.

Thanks for your post,
Orangetom




posted on Apr, 10 2006 @ 09:42 AM
link   
cavitation occurs when the local pressure at a point in the fluid falls below the local vapour pressure (depends on material properties such as temp and density) of the fluid.

The local pressure is the product of the static pressure at that point (atmos pressure + rho x g x d, where rho is density of seawater ~1025kg/m^3, g is accelartion due to gravity and d = the depth) + the dynamic pressure due to the motion of the fluid at that point (faster flow equals lower pressure from bernoulli's eqn).

If the combination of static and dynamic pressure drop below the vapour pressure the fluid can not hold itself together as a fluid in that area and turns to vapour, forming a cavity (hence cavitation), when the local pressure rises back above the vapour pressure the cavitys collapse (more correctly they implode) back into fluid (thats what makes the noise and damages things like propellers).

Looking at this principle, the deeper into a fluid a caviator is th higher the static pressure will be, this in turn means that to form a cavity either the vapour pressure is raised (easy in a test-tank and you can apply a vacuum, not so easy in the ocean) or the effect of dynamic pressure have to be increased (higher forward speed or larger discontinuties).

If you imagine a flat circular disc travelling through the water perpendicular to the flow. With the plate moving slowly through the water there is some diturbance behind it, as the speed increases, the pressure behind the plate drops till a cavity is formed behind the disc. As speed increases the cavity will stretch out behind the disc getting longer as the speed increases. If the projectile is in this cavity, then it will be supercavitating.



posted on Apr, 10 2006 @ 10:18 AM
link   
As promised a basic mathematical description of the shape of a cavity behind a circular disc cavitator. This method can easily be put into software to look at such effects of pressure, speed depth etc.


This is a very simple and is based upon:

OPTIMUM DESIGN OD A SUPERCAVITATING TORPEDO CONSIDERING OVERALL SIZE, SHAPE, AND STRUCTURAL CONFIGURATION

Edward Alyanak, Ramana Grandhi, Ravi Penmetsa

Department of Mechanical and Materials Engineering, Wright State University, Dayton, Ohio, USA

Found in International journal of solids and structures Vol 43 (2006)

NOTE: This is there work so please include the reference if you recite this anywhere.


“Cavitation is described by the cavitation number (supercavitation is characterised by very low cavitation numbers), it is a non-dimensional quantity that represents the extent of cavitation.

It is given by:
sigma (cav no) = (P-Pcavity)/(0.5*rho*V^2)

P is the pressure outside the cavity
Pcavity represents the pressure inside the cavity
Rho is density of seawater
V = torpedo speed

For supercavitation to occur sigma must approach zero, thus Pcavity approaches P. Therefore if a torpedo is at a given depth the pressure P is equivalent to the depth pressure P=rho*g*h (density * gravity * depth). As a result of this the depth pressure must be considered for supercavitating torpedo structuaral survivability. Even at high dynamic pressures this argument still holds. For example, at a velocity of 120m/s, Pcavity is within 98% of P for a cavitation number of 0.01 and a depth of 600m.

The flow can be characterised by a torpedo speed V, the cavitation number, sigma, and the caviator diameter d. The drag coefficient can be determined for a flat disc cavitator by the relation

CD=0.815(1+sigma) where the drag coefficient is defined by
CD= (D/(0.5*rho*V^2*A))

Where D is the drag force and A is the cavitator area, in this case it is (pi*d^2)/4

Form these equations the drag on the torpedo can be defined as
D=(pi/8)*CD*rho*(Vd)^2

The maximum cavity diameter dm can be defined by sigma

Dm = d(CD/(sigma-0.132*sigma^(8/7)))^0.5

From which cavity length Lc can be determined

Lc = dm ( (1.067*sigma^-0.658) – (0.52*sigma^0.465) )

The cavity shape can be determined using the above equations using the logvinovich principle for stationary cavities.

S-So/ Sk-So = t/tk* (2-(t/tk))

Where S is the cavity cross-sectional area at time t, So is the initatil cavity area, Sk is the maximum cavity area and tk is the time taken for the area to grow from So to Sk. These quantise are defined as

So = (pi/4) *d^2

Sk = (pi/4)*dm^2

Assuming the x axis runs down the longitudinal axis of the cavity and is zero at the cavitator then

T =x/V

Tk =Lc/2V

Yielding

t/tk = 2x/Lc


This allows us to write the cavity radius Rc as a function of distance x from the cavitator


Rc(x) = ( 1/4 (2x/Lc(2- 2x/Lc)(dm^2-d^2) +d^2))^1/2


Hope you can understand all this.

Paperplane.



posted on Apr, 10 2006 @ 10:57 AM
link   

Originally posted by iori_komei
Ok, I was thinking about how slow submarines go, compared to Jet's, and I got to thinking, how could a device be made to allow submarines and torpedo's to go at SuperSonic speeds underwater.

Hydrodynamics isn't that much different than aerodynamics. Water obviously has a higher density, of course. But the problems are basically the same. I think the trick would be to:
1) figure a way to move either huge amounts of water (or air?) through the jet
2) get super-tough materials along leading edges that would cut through with minimum disruption
3) find a very slick surface to minimize drag and create laminar flow.

The first one would probably be the toughest. Getting a lot of water to flow at supersonic speeds would require an awful lot of power.

Hey, how about using high-powered lasers or masers along the leading edge of the ship to boil the water? It would decrease the density, and possibly provide air which could be fed into a modified air jet engine for propulsion? Maybe that would work.

Look out, little fishies!





[edit on 10-4-2006 by Enkidu]



posted on Apr, 13 2006 @ 04:01 AM
link   
a company called pursuit dynamics tried to do something akin to an underwater jet engine a few years ago. It somehow used steam to accelerate the water. Not sure how far they got with it (must be 5 years ago now). It had no moving parts below the waterline!



new topics

top topics
 
0
<< 1   >>

log in

join