Novel fast and heavy ship, Corrected on 5.10.2020

Military value of 360 km/h heavy ship of my design – corrected on 5.10.2020
Author: David Judbarovski, systems engineering, principle inventor, retired
judbarovski@gmail.com

Note: My sea transport can be used for military quick response, and as a civil cargo ship too. At first, I disclose my idea quite popularly.

Abstract
Being compared with typical destroyer (Type 055, Japan) = USD, 60,000,000,00; 13,000 ton (dwt, deadweight); 112,000 kW; 56 km/h,
my ship is better than the said destroyer in 4.5 times by CAPEX (capital expenditures)  and 3 times by OPEX (operational expenditures), and can substitute 3 destroyers, by much cheaper way

Very shorty,
Two key elements of my ship are
(1) Slightly lower than sea level, the ship’s nose is tilted back and like a wedge it bends the upstream water upward in the air, and it substitutes the enormous water resistance by the air resistance being in three orders of magnitude less one. Nevertheless, the drag (frontal)
force would be quite big. Such effects are inherent for very big velocity, because the water can be considered as incompressible, but flexible matter in that case.

It is new kind of sea ship, breaking all stereotypes.
(2) to add a thin air cocoon created by trickles of pressed air through and around a perforated sectioned second wall of the ship. It reduces the power losses by the water viscosity practically to zero.

All those can be proved quite simply by simplified modeling in laboratory.

The frontal power losses W (kW) = S (m2) * H (m) * V (m/s) + 1.3 (kg/m3) * l (m)* B (m) * 0.001 * V^3 / 2 (see Note 2)
S is the ship’s cross-section submerged, and H and l are height and  length of the ship’s nose correspondingly, and V is its velocity.
While for conventional ships it is W = k * S * V^3 / 2

Let a ship of my design be
H = 6 m, B – 10 m, S = 60 m2, l = 10 m, 6500 ton; and V = 360 km/h – 100 m/s

So for the said ship of my design :
W = 6 * 60 * 100 = 36,000 kW (for front water weight lifting) + 1.3 * 10 * 10 * 0.001 * 100^3 / 2 = 65,000 kW (aerodynamics resistance) + ~10,000 kW (for the abovementioned cocoon at trickles velocity being 2.0 m/s)
So W ~= 100,000 kW totally.


If compared with the said destroyer:
(6500 /13,000) * (360/56) = 3.2. my transport instead of the three ones,
and by cost  3 * 60 million) /40 million (military prices ordinary are overstated) =  4.5 times by CAPEX,
while 3 * 112,000 /111,000 = times by OPEX

Note 2.

Power for front water weight lifting in the air is:
W1 = Q * H = ( l * B * H / T ) * H = B * H^2  * V = S * H * V; Q.E.D!
Here: W1 (kW) – power; Q = G / T (m3/s}- water rate; G = H * B * l; H (m) – ship’s draft =
nose height submerged; B (m) – ship’s nose beam submerged; S (m2) = B
* H; l (m)  nose length submerged; V (m/s) – velocity = V (km/h) /3.6; T (sec)= l / V
So
W1 = 0.28 * S * H * V (km/h), hence the power is proportional to a speed !!!.
While for conventional ships it is W = a * S * V^3 / 2

Aerodynamics share of frontal power doesn’t needs a special consideration, as power for cocoon too, taking in mind classical formulas for aerodynamic resistance, and for flow rate through small holes, and gases compression power losses.


 

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