Theory of the propeller

a propeller has for object of accelerating the speed of a fluid passing through .
the energy needed is equal to   
E =
  M  x ( V2 - V1)*2
M= mass of water and is a function of the propeller diameter  .
V2 and V1 are speed after and before  .

The propulsion force is directly  f(E) .

The propeller main advantage is that all water movements are happening outside the ship and in a continuous way.

The main factor in the ship speed limit is friction of the keel with the surrounding water .
The resistance or friction depends on many factors like shape of the nose ,quality of keel surface ,
water temperature ,speed etc etc .

Propellers of big sizes are real manufacturing challenges .
the biggest propellers today are 6 meters in diameter casted in one piece and weights approx 130 tons (5 blades).
they must be perfectly balanced to avoid vibrations due to the high rotation speed (100 RPM) .
the main shaft is about 80 meters long for about 0.5 meter in diameter .the energies involved can reach 20 to 30 MW .
a propeller of 6 meters for a ship at 25 knots can process approx 340 CBM of water /second .

we can simplify the propulsion force by saying  Pf  = M expelled /second - friction forces .
it is very close to a rocket reaction where the power = M V
the difference being that in a ship the mass expelled is water and a ship has unlimited amount of water available .
Renewable energies             Wave energy                              Wind energy                                              Solar energy

If we want to build a ship that does not use a propeller ,the main challenge will be to process enormous amount of water inside the ship and to
have an engine that can process water with the same or higher efficiency than a propeller .
Just the collection of water poses a challenge because the water has a kinetic energy relative to the moving ship  and any attempt to collect
and store water will simply produce a force that will slow down the ship .The stored water has to lose its energy first.
It seems that the only way will be to transform the water kinetic energy into potential energy .
Water flowing at 25 knots (46 kph) has energy of : P= MV . ==> MgH .
1 kg at 46 Kph = 1kg at 12.7 m/s  = 1kg at 1.3 meter high - frictions
1 kg at 60 Kph =  1 kg at 16.6 m/s  = 1 kg at 1.70 meter high - friction .

To expel water at high speed in the cheapest way is to use compressed air .Water being incompressible ,all pressure forces will be
transmitted .
Super cavitation

Supercavitation is the use of
cavitation effects to create a large
bubble of gas inside a liquid, allowing an
object to travel at great speed through
the liquid by being wholly enveloped by
the bubble. The cavity (i.e., the bubble)
reduces the drag on the object and this
makes supercavitation an attractive
technology; drag is normally about
1,000 times greater in water than in air.

A cavitation forms behind an object
passed by a rapidly streaming liquid.

In 1977, Soviet engineers developed the
VA-111 Shkval ("Squall") torpedo. This
can travel at 230 mph (100 m/s)
underwater, compared to the top speed
of about 80 mph (35 m/s) for
conventional aquatic craft, but it is
reportedly not steerable. Even faster
speeds of about 310 mph (ca. 140 m/s)
and higher have also been rumored.
News of the device reached the West in
the 1990s.
source wikipedia
Japanese in 1999 have experimented
cavitation (they call it micro bubbles) on
ship successfully reducing friction by 20
% for speed of 7m/s .
Generating gas pressure by using hydrogen / oxygen combustion  .(hydrogen at 250 bars and oxygen at 1 bar )

Universal law : PV =  n R T     for perfect gas (air is not a perfect gas ,but a mix of perfect gas ) .
P= pressure (atm)   V = volume(liter)  T = temperature (Kelvin)  R = constant (about 8,3145 joules/mol/kelvin)  n=number of mol .
For K = 273.15  and  P = 1 atm  n= 1 mol   V= 22.4 liters .
Specific heat of air  : 1000 joules /m3 or kg /degree           specific heat for water 4.18 kj /kg/degree

for an explosion chamber of 1 CBM .at 1 atm 1 CBM of air contains 20% in oxygen .= 200 liters = 8.92 mol = 160.5 grs of oxygen .
For a perfect combustion it takes 17.84 mol of hydrogen ( 2H2 + O2 = 2 H2O)   
1mol of H2 burning in the air releases 286 KJ .  17.4 mol of hydrogen will release : 4 976 KJ .

Hydrogen at 250 bars contains 1 mol/0.0896 litres . it takes then 1.56 liters(at 250 bars) for 17.4 mol .
By expending 1.56 liters of H2 into the explosion chamber we have a new pressure of 1.39 bars .
For 1kg of air and an energy increase of  4 976.KJ(after combustion/explosion) then the new temp = approx  4500 degrees .
This temperature is extremely high ,but it has been reported by car manufacturer having developed hydrogen internal combustion engines
much higher temperature than with gasoline or natural gas .(only testing will verify this point , is it 4200 or 4800 degrees ?) .
Hydrogen has about 3 times more energetic value than gasoline or diesel .
Remark: hydrogen has a much higher propagation rate allowing faster mixing with oxygen and a much faster combustion speed.

Since V and n and R are constant then P1/T1=P2/T2  or P1 T2 = P2 T1      1.39x 4500 = P2 x 273              P2=
23 bars approx .
This calculation is just an approx ,because we are not dealing with perfect gas and there is water steam created by the combustion .
For every mol of hydrogen we can produce 57 liters of air at 23 bars at 4500 degrees .
In order to reduce the temperature the way would be to increase the normal pressure in the combustion chamber without changing the
quantity of hydrogen  (or to decrease the injection of hydrogen ).

if we add water in the combustion chamber before explosion /combustion then
the number of gas mol will increase because water will be transform into steam .
the addition of water will in the same time cool down the combustion chamber and absorb calories .
if the chamber is not cooled down the quantity of air admissible  will be lessen .

1 kg of air + 313 grs of H2O (result of combustion) + 1kg of water ==>  
it takes 2550 J to vaporize 1 gr of water   . Latent heat of vaporization : 2250 J/g   . Specific heat of steam :2J/gr/degree.
Then temperature in chamber = approx 500 degrees Celsius
for 1313 grs of water = 72.9 mol
for 1000 grs of air (O2 = 32gr   N2 = 28grs) = approx 34 mol of mix .
1.39 x 800 x (72.9 + 34) = P2 x 300 x 34             P2 =
11.65 bars

for 1 kg of air + 313 grs + 500 grs of water then temp = 1050 degrees Celsius .
for 813 grs of water = 45 mol  
1.39 x 1350 x ( 34 + 45 ) = P2 x 300 x 34            P2=
14.5 bars

for 1kg of air + 313 grs + 100 grs of water then temp = 2000 degrees Celsius
for 413 grs of water = 23 mol
1.39 x 2300 x ( 23+34) = P2 x 300 x 34              P2=
17.8 bars

in the previous calculation i have found 4500 degrees and 23 bars by neglecting the H2O produced
by including it we found approx
2500 degrees and most probably around 20 bars .Sounds more reasonable .

If water is added it must be fresh water ,if not there will be  depot of salt and other and the combustion chamber
will have to be cleaned very often.
When the gas pressure will be applied to sea water ,the action has to be very fast ,because steam will condense and
temperature will drop fast .

It seems then possible to build a ship using hydrogen as energy source without propeller and without a conventional engine .
Except the valves ,it wont be too many moving parts and by using cavitation the drag will be reduced significantly .
It is of course just an academic thought at the moment and testing have to be carried out .
The engine will be close to a jet engine (like in jet ski ) but without mechanical parts (more like a water rocket).

We can still of course use more conventional systems :
internal combustion engine directly connected to the shaft with a propeller.
Steam engine where hydrogen/oxygen will vaporize water and then steam will power a turbine linked to a propeller .

Existing propulsion performances .
A conventional diesel engine = 70 %
A standard shaft                      = 97 %
A good propeller                     = 60 %

we have then a ratio of
40 % efficiency   thrusting power /energy used

Real example . Oil tanker  .
Length  :280 m
dead weight : 150 000 tons
draft : 16.5 meters
power :  15 300 kW
speed : 15 knots = 28 kmh
fuel used : 268 euro /ton     21 000 tons over 280 days .
Consumption : 111 kg of fuel /km = approx 3 330 MJ /km = 28 MJ/s = 28 000 KW .
approx 1500 to 2000k Newton force of thrust  150  to 200 tons of thrust ????? to verify
= 38 US$ /km in fuel .
Pressure = Force / Surface .  1 bar = 100 000 Newton / M2 . or 10 Newton/ cm2

for  1  500 000 Newton / 1 M2  = 15.00 bars
1  500 000 Newton / 4 M2 =    3.75 bars .

Speed of a water jet :  V*2 = 2gH
Force of water jet on a surface . F = M ( V-v )    V=water jet speed m/s   v=speed of the surface m/s   F in Newtons .
or also  F = p Q (V-v)     p=specific mass of the water =1 kg /liter   Q = flow in M3/s
mechanical power of a water jet .  P = 0.5 Q p V*2    =  0.5 S p V*3    S = area of the jet in M2 .  P  in watts .