Posted on 05/04/2006 7:17:20 PM PDT by Robert A Cook PE
My dad (long time civil structural engineer) sent the following question this evening to me:
Tomorrow morning, I need the drag of one square foot of plate in 6 mph water. Not edge effects, downstream turbulence, corner losses, etc. Just how much force will I have to provide to hold a square foot of plate still into a 6 mph current? No, not in channel. In open water.
I can't find it ANYWHERE! Durn! Got a structure in a pickle in water, and I really need that drag force to figure piling forces. My Olde Books are EMPTY on the subject.
Anybody got a suggestion of how much it will take? Please?
Ramon in a mess, Durn!
D=(0.5)(rho)(v^2)(A)(Cd)
Assume a common Cd of 1.17 for a regular, square, flat plate.
rho=density of water=62.427lb/cu.ft.
v=flow velocity=8.8ft/sec
A=area of plate=1sq.ft.
D=2830 lb-ft/sec^2
-- OR --
Assume a common Cd of 2.1 for a smooth brick.
rho=density of water=62.427lb/cu.ft.
v=flow velocity=8.8ft/sec
A=area of plate=1sq.ft.
D=5080 lb-ft/sec^2
Does that get into the ballpark?
On a related - there is enough energy in the Gulf Stream to meet the new electricity needs of the entire eastern seaboard for years to come. www.energy.gatech.edu/presentations/mhoover.pdf
Pressure times area = force.
Well, I'd just hook a heavy duty scale, like those used to weigh Tarpon's to a cable attached to the plate at the four corners. Toss 'er in, record the result :)
8<)
Passed along your figures.
Cd = 1.16 (Mech Engr Handbook square plate)
V 8.8 ft/sec
Density = 62.4 lbs/ft^3
g = 32.2 ft/sec^2
F = lbs
V^2= (ft/sec)^2
D = lbs/ft^3
g = ft/sec^2
A = ft^2
F = 174.081272 lbs
It will also depend upon the temperature and make up of the water, all of which influence viscosity - which influences the flow coefficient)
So did I. It all looks chinese to me. Must be rewarding to crunch numbers. I envy them. I must remember from here on out...curiosity killed the cat.
Yep -I hate fluid dynamics. My brain is full. Blow the dang thing up, and the problem goes away. Something about del/phi del/q...this is why one becomes a mechanical engineer. Kill the target.
Cd= 2
density = 1000kg/m3
A= 4ft2 = 0.372m2
v= 6mph = 2.68m/s
drag = 1000 * 0.375 * 2.68 *2.68
drag = 2693N = 606 lbs. (0.225 lb/N)
This is for the shape with a rectangular xsection with the face perp to the flow. If it's rotated to place an edge to the flow, the Cd drops to 1.6. For the triangular columns(piers), the Cd is slightly higher with the blunt face in the flow (Cd=2.2). If the 120o angle apex faces the flow, the Cd drops to 1.7. I think you need to keep the Cd up at ~2.
Forgot to ping you to the last one. The Reynolds number for your 100ft river 20ft deep is 20K, ~10K at 5ft, so those Cd numbers are good and in a flat range.
You guys are giving me a headache.
Headache accepted.
It's an engineering thing, and comes with the territory of analysis, calculation, critique, and determined, focused, explicitly-detailed-point-of-view of seemingly nit-picking technical jargon.
On the other hand, if you don't do things that way, the building (or brdige pier) falls down on those people who don't examine things that way.
Concur with your comment about a Cd of 2.0 .
Civil Engineers ping
I'm filling in for Fierce Allegiance, let me know if you would like on or off the list.
Interesting problem.
I do estimating, so I'd say $37,000 and i'll have your answer to 4 decimal places. Precise, not necessarily accurate, though.
!
Contestant: "Alex, I'll try "More Engineering Answers" for $1000.
Reply: And, the answer is, "More Dollars."
Contestant: "Now, what is the question?"
Naaah, it's 42.
Cheers!
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