Re: [PSUBS-MAILIST] fairings drag vrs entrained mass

[EM <emuller@naic.edu>]

Fairing is something I'm curious about too.. 
It seems to me that fairing defines a comoving volume that you have to
cart along, which really WILL affect the acceleration.. unless you allow
the water to flow THROUGH the fairing, which will just double your drag 
(since you now have TWO surfaces exposed to the flow).

So you loose a lot with fairing. In one case you carry around a lot of
water, and your acceleration dies, in the other you just up your drag
(though hopefully the whole point of installing the fairing was to
re-divert flow away from lossy turbulent eddies etc.).

To be honest, the only use I can imagine for fairing is to re-divert the
flow away from potentially turbulent corners, and at the speeds
we're talking about.. is that often a big concern?

I think somone should make a flow-box.


EM.



On Mon, 28 Nov 2005, rick miller wrote:

> 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 ----- 
>   From: Paul Kreemer 
>   To: personal_submersibles@psubs.org 
>   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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