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thanks paul
i guess i got stupid and forgot
basic physics. once a body is in motion it stays in motion, unless acted upon by
external forces. So while the fairing will allow for more speed the weight
penalty is only a factor durring acelleration.
how much reduction in the cd can
be reasonably obtained using a boat tail
fairing? 5%?
rick m
----- Original Message -----
Sent: Monday, November 28, 2005 11:32
AM
Subject: Re: [PSUBS-MAILIST] fairings
drag vrs entrained mass
Rick, some numbers to discuss. I kept it in US standard
units which isn't as nice for the math but the end results are easier for us
to look at. :-) There are a lot of numbers here and there's
probably a mistake somewhere. But I assumed only:
A = 9
ft^2 (frontal area)
F = 60 lb (max propulsion thrust from 0
to 3.5 kts)
The drag varies a lot depending on the shape of the
submerged body and I did the numbers only with the drags of airship and
compact car shapes. But there's a web page below that lists some more drag
coefficients.
It seems to me that while underwater your drag would
mainly be a concern when dealing with a current and in prolonging your running
time - letting you cover more ground on a battery charge. With a more
streamlined shape you could run at a lower power setting.
If you're
interested in operating in currents you'd also want to look at drag when the
current hits you from the side or stern. Right? But it doesn't
take that long to accelerate your 6000lb so I don't think that's much of a
concern. I guess you would hit
something a bit harder with the extra mass...
take
care-
Paul
==================================
Acceleration:
F= m * a
m
= 6000lb/32.2 = 186 slugs (mass value - not weight)
v = 3.5 kts =
5.8 ft/s
v = 1/2 * a * t^2
a = F / m
v = 1/2 * F / m *
t^2
5.8 ft/s = 1/2 * 60 / 186 slug * t^2
t = ~6 seconds to
accelerate 6K lb to 3.5 kts, ignoring the hydro drag
This is the
simpler analysis that doesn't consider the hydrodynamic drag of the sub. Sorry
but I'm not up for digging out this equation right now. Looking at the
estimated drags below, I'd just do a WAG that the hull drag would average
15lbs for the Ford Escort shape over the 0 to 2.4kt range and come up
with:
4.37 ft/s = 1/2 * (60-15) / 186 slug * t^2
t = 6 sec to accelerate to
2.4kts
This might be a
reasonable estimate. I'm guessing that you don't care too much about the
acceleration times so I didn't spend much time on this.
:-)
==================================
Top speed:
Here's the drag
force at 3.5kts, first with a very streamlined shape, equivalent
to an
airship:
F = 1/2 * density * V^2 * A * Cd
density = 1.94
slugs/ft^3
F= 1/2 * density * 33.6 * 9 * 0.02 = 5.9 lb drag at
3.5kts
That's not much drag, huh? But if you have a more
practical shape, like that of a Ford Escort, you'd have 104 lb of drag at
3.5kts. Oops, that's more drag than we have thrust!
Now
I'd look at solving for top speed with a Ford Escort shape and 60lb
thrust:
60 lb thrust = 1/2 * density * V^2 * 9 * 0.36
V = 4.37
ft/s = 2.6 kts top
speed
=================================================
Drag
coefficients:
Here's a nice table of drag
coefficients. You've got to love a list that includes both tractor
trailers and aircraft wings. These are for higher speeds
than what you're running, but probably close enough. I think the main
time you really see a difference is with bluff bodies like a sphere or a
pickup truck where you get different vortices at different Reynolds
numbers.
http://aerodyn.org/Drag/tables.html
Cd
= 0.02 : airship
Cd = 0.36 : Ford Escort
Cd = 0.5
: pickup truck
Cd = 1.2 : flat plate,
normal to flow
On 11/27/05, rick
miller <rickm@pegasuscontrols.com > wrote:
hi guys
Rethinking fairings and have
a question. how much saving in hydrodynamic drag do i have to save to
justify the the entrained mass that i will carry in traped water? especially
with a speed of only 3.5 kts. diplacement about 6000 lbs
rick
m
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