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Iain,
"Can anyone help with this problem. What is an achievable maximum
speed for a small 2 person 1atm sub? About 4m long, 1.5m diameter, hemisphere
front and convergent tail cone."
Below are a set of simple equations that can be entered into Excel to
give you drag and power for flow past a submerged object.
There are two basic ways to calculate drag on a submerged body.
The first is based on the projected cross sectional area and a drag
coefficient. The second is based on the wetted surface area and a
resistance coefficient due to friction.
Drag Based on Projected Area:
Using the first of these techniques, the overall drag D can be expressed
as:
D=Cd*Ap*Rho*v^2/2
where, Cd is the overall drag coefficient, Ap is total cross sectional area normal to the flow field, Rho is the water density and v is the velocity of the body. Use any consistent set of units. The drag on the submersible can be expressed in terms of the contribution of parts as, D=D1 + D2 + . . .Dn
Where for example D1 might be the hull, D2 the external battery pods
etc. For each part that contributes to drag, a separate drag coefficient
and cross sectional area would be used.
The drag coefficient for each part can be determined from a wind or water tunnel experiment or by towing the boat at a known speed, measureing load and solving for Cd. Since most psubbers do not have access to a test facility, an estimate of Cd is normally used. To bracket the value of Cd, for turbulent flow, a Cd for a flat circular disk pushed normally to the flow would be 1.2 while for a streamlined airship hull, the value would be 0.04. The value for ?K boats? is probably on the order of 1. As an approximation, you can assume the drag coefficient is a constant but in reality, it is velocity dependent. More accurately it is dependent on the dimensionless Reynolds Number which is calculated as Re=(Rho*L*v)/mu
where Rho is the water density, "L" is a characteristic length, v is the velocity and mu is the dynamic viscosity. For parts such as the hull and battery pods ?L? is the maximum diameter of the projected area, for wings, it is the thickness of the projected area. Drag Based on Surface Area: Using the second of these techniques, the overall drag D can be expressed
as
D=C*S*Rho*v^2/2 where C is the resistance coefficient due to friction, Rho is again the water density, S is the reference surface area or total wetted surface area and v is the velocity. As with the frontal area technique, C is Reynolds Number dependent. Again, this equation would be used for each part of the boat subjected to frictional drag. For streamlined shapes such as an airplane or an Albacore shaped sub hull,
the technique that is based on surface area is normally used while for bluff
bodies, the technique that uses cross sectional area is more often used.
There are a number of empirical surface resistance (skin friction)
equations that can be used. One example for turbulent flow, is
C=0.075/(Log(Rn)-2)^2
Where the Reynolds Number, Re as before for each part can be expressed as Re=(Rho*L*v)/mu
For this technique, the characteristic length ?L? is normally taken to be
the length of part parallel to flow. As an example, for the pressure hull,
it would be the axial length of the pressure hull.
Power:
After the drag has been calculated from either technique, then the power
required is then calculated as
P=D*v For your example using the drag based on projected area (1.5m
diameter), a speed of 5 mph, (4.34 knots), the power required for a range
of Cd?s is
Cd Power 1.2 8.1 hp 0.8 5.4 hp 0.1 0.7 hp If your shape was very streamlined (NASA low drag profile or Albacore class
hull shape), the second technique would calculate the power required at 5 mph
(4.34 knots) to be about 0.4 hp assuming there are no appendages such as battery
pods.
Hope this helps.
Cliff
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