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Re: [PSUBS-MAILIST] Electric Propulsion
Hoerner's Fluid Dynamic Drag, Chapter 6 has what you need - specifically
the formula in the bottom left column of page 6-16. For a body with an
l/d of 5, the friction coefficient at high Reynolds numbers seems to be
about .003, so the overall drag coefficient based on frontal area will be
CsubDo = .09 + .003(15+3*sqrt(.2)) = .139
For a body 50 feet long with a maximum diameter of 10 feet, this gives a
drag of 1.376 x 10^5 newtons at 30 knots, or about 30,000 lbf.
Power required is 2,150 hp. This would have to be divided by the
propulsive efficiency, which will be pretty high if the propeller is
centrally mounted on the tail of the body, to give net power required.
Best,
Marc de Piolenc
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On 9/30/2010 11:45 PM, ShellyDalg@aol.com wrote:
In a message dated 9/29/2010 11:21:45 P.M. Eastern Daylight Time,
irox@ix.netcom.com writes:
The idea of 25-30Kts is rather infeasible with battery power
alone (I will be happy to be proven wrong, of course).
Although reaching a top speed of 30 knots while submerged isn't feasible
without nuclear power, it would be a fun exercise to calculate.
The Albacore hull would be the choice for it's overall drag numbers, but
just how much horsepower is required to push one at 30 Knots ?
Jon said 50 feet long ? That's a good number. Anybody know the formulas
to calc this ?
I remember the S101 and how much space was taken up for batteries. There
wasn't much room left for people, for sure.
With the required horsepower known we could figure out how many
batteries it takes to push that hard for how long. Then get a percentage
of displacement needed to supply that amount of power.
My guess is it's over 90%. You're really looking at building a 50 foot
torpedo if 30 knots is the goal.
Might be fun to figure out though.
Frank D.
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