Very interesting it’s a vicious cycle to go the speed you
are looking for you will need big HP and for big HP you need lots of fuel
adding more weight and more power is needed to offset the weight, one EMD 16-567
Natural will produce 1800 hp at 900 RPM but weighs 25 ton and requires 9072 cubic
inches of air for each complete rotation and the engine is 12 feet long and 6
feet wide and 8 feet tall, there are a lot of nice 4 stroke engines out there that
are a bit smaller and produce the same hp but are huge money, just for my own
curiosity what are you planning to do with the sub, the German and Russian government
have diesel electric boats for sale that might suite your needs for far
less than you might think, the Canadian government just sold 3 subs for almost
nothing
Brian V. Ryder
brian@subatlantic.com
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From:
owner-personal_submersibles@psubs.org
[mailto:owner-personal_submersibles@psubs.org] On Behalf Of Yohnathan
Aldana
Sent: Saturday, October 16, 2010 10:46 PM
To: personal_submersibles@psubs.org
Subject: RE: [PSUBS-MAILIST] Electric Propulsion
Could
you explain this further? and give the full equation?
john
> Date: Fri, 1 Oct 2010 08:35:50 +0800
> From: piolenc@archivale.com
> To: personal_submersibles@psubs.org
> Subject: 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
>
> Archivale catalog: http://www.archivale.com/catalog
> Polymath weblog: http://www.archivale.com/weblog
> Translation services: http://www.proz.com/profile/639380
> Ducted fans: http://massflow.archivale.com/
>
> 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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