Maximum deepness of atomic submarines: classical equations

As supplement to material №102 about spherical elements of U-boat, in this article we will appreciate:

* How much times grows work deepness, if we use spheres instead of cylinders?
* What value can reach work deepness of submarines during nearest decades?
* Is it possible to create atomic submarines with work deepness 4600 – 5500 meters, which allows to reach the bottom of all oceans, correspondingly, on 60% and on 90% of ocean’s surface area?

Also as we remember, usage of supercritical LWRs requires some more then 225 atmospheres of pressure, which corresponds to depth 2200 meters. So as external tubes are recognized as most sensitive component of submarine, usage of such LWRs automatically provides deepness 2200 meters by factor of external tube system.
Ability to reach bottom of the ocean in any chosen place, gives an opportunity to do new hiding tactics, especially if submarine is atomic and equipped with nuclear warheads in torpedoes. Even if not, anti-submarine operations will require the third dimension (altitude) because work depth will become many times bigger then effective radius of unguided sea weapons.

Lets imagine, that robust housing of submarine is made using cylinder with length ‘L‘, radius ‘R‘ and half-spheres on end-faces, radius ‘R‘. Wall thickness everywhere is ‘d‘.
So volume of replaced water for this form will be V = cylinder + two halfspheres =(Pi*L*R^2) + (2*(1/2)*(4Pi/3)*R^3) = (Pi*R^2)*(L+(4R/3)) ;
Lets mark the value X = (length/diameter) = (L+ 2R)/(2R).
We can see that 1<X<infinity, and that L = 2R(X-1) ;
Volume of the replaced water is V = (Pi*R^2)*(2RX – 2R + (4R/3)) = (2Pi/3)*(R^3)*(3X-1) ;

Lets suppose, that thickness of shield ‘d’ is d<<R, density of steel is ‘p‘, in this case mass of shield is:
M = V(cylinder)*p + V(sphere)*p = 2Pi*R*L*p*d + 4Pi(R^2)*dp = {L=2R(X-1)} =
= 2PiR*p*d*(2R(X-1)) + 4Pi(R^2)*p*d = 4Pi*(R^2)*X*p*d ;

Force which is rising the shield must be bigger then its mass multiplied to ‘g’. Also reserve of swimability ‘Y‘ is usually taken 30% of replacement, so Y=0.3:
maximum replaced mass of water is
m = (2Pi/3)*(R^3)*(p_water)*(3X-1)*(1-Y) ;
and equilibrium equation will be:
(2Pi/3)*(R^3)*(p_water)*(3X-1)*(1-Y) >= 4Pi*(R^2)*X*p*d
from which we can find swimability criterium for thickness of submarine wall and its radius:
(d/R) <= (3X-1)*(1-Y)*(p_water)/(6*X*p_steel) with conditions 1<=X<=infinity;  0<Y<1.

Lets suppose, that ‘d’ is everywhere equal: at cylinder and spherical parts.
In this case we use correlation for maximum internal pressure of cylinder shield: (d/R) = (P/sigma) where sigma is maximum (‘sigma 0.2‘) fluidity strength of chosen shield material, it is measured in Pascales and can be found in literature. This formulae is case of internal pressure, for external pressure case it is some optimistic, and we will use it for appreciation.
At the same time, hydrostatic pressure is P= (p_water)*g*h; Value ‘h‘ means maximum depth of submarine.
So h = P/(p_water*g) = sigma*(d/R)/(p_water*g).
Density of water dissapears and we get formulae:
h <= (3X-1)*(1-Y)*(sigma_steel/p_steel)/(6*X*g) ;

Analyzing this correlation, we note that if X=1, formulae not gives the right value for spherical case, and reason is understandable:
is the beginning we supposed, that form of shield has cylinder component.
If we will suppose that only spheres and no cylinder elements, we can take correlation (d/R) = (P/(2*sigma_of_steel)) and depth, i.e. ‘h‘ value, will be bigger.
Certainly, in this case light shield can be a cylinder which is necessary for high speed, and robust housing can be chain of 6 or 9 spheres inside light shield.
But their intersection radius, if thickness ‘d‘ is constant everywhere, must be made not bigger then (R/2) in other case we get equivalent of cylindrical tunnel between spheres and must use formulae (d/R) = (P/sigma) without multiplier ‘2’. Such mathematics appears due to the fact, that if there is a cylinder with external or internal pressure, applied strength in parallel cross section to its axis is 2 times bigger then in perpendicular cross section, and also perpendicular cross section strength is equal to strength if the case of sphere.

Making few obvious appreciations, we can see numerical values of achievable deepness of submarines.
Lets write previous equation as:
h <= (3X-1)*(1-Y)*Z*(sigma_steel/p_steel)/(6*X*g) ;
here ‘Z‘ is percent of shield’s mass in all replacement value. We can suppose, that:
submarine with ballistic missiles has robust housing which is 20% of all mass;
fast submarine with cruise missiles has 40% of all mass in the shield;
deepwater submarine has 60% of all mass in the shield;
and batiscaphe can have shield’s share up to 80% of all rising force.

Lets suggest, that shield’s material is steel with sigma=70 (kg/mm^2) = 700 MPa, density 7800 (kg/m^3), swimability reserve Y=0.3, X=9, in this case
maximum depth:
h = (3*9 -1)*(1-0.3)*Z*(700E6/7800)/(6*9*9.8) = 3086*Z meters;
If 20% of submarine’s mass is in robust housing, Z=0.2 and achievable deepness is 617 meters.
If safety coefficient is 1.5 then we get  operational depth 400 meters, which is easily achievable using modern technologies.

Certainly, submarine must be fast i.e. with powerful engine, with weapons, equipment and stocks for operators, but we can note that if shield’s mass share is 60%, then easilly achievable deepness for usual steel and cylindrical form is 1200 meters.

Now we compare with required 4600 – 5500 meters and remember, that we have two reserves: spherical form and non-steel materials with better (strength/density) coefficient. Such known materials are, for example, aviation aluminum alloys and titanium non-magnet alloys.
Titanium alloys have density 4500 kg/m^3 and sigma up to 1000 MPa. Calculation gives:
h = (3*9-1)*(1-0.3)*Z*(1E9/4500)/(6*9*9.8) = 7642*Z meters.
If Z=0.6 then h=4585 meters.

Also we can use reserve, which appears if we choose for the form of submarine a chain of spheres inside cylinder light shield:

IN WORK

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