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January 30th, 2012, 09:19 PM   #1
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Dimensional analysis (Weir Eq.)

So
Code:
 Q=3.33(L-0.2h)(h)^1.5
and I need to convert from field units (ft^3/sec, ft) to SI units (m^3/sec, meters)
I know the answer should be
Code:
Q=1.84(L-0.2h)(h)^1.5
but I am having trouble getting that constant to come out.
Some help would be appreciated, thank you.
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January 30th, 2012, 09:51 PM   #2
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Re: Dimensional analysis (Weir Eq.)

I know that 2.54 cm = 1 in exactly, thus:



thus:



For your regular contracted weir, the discharge coefficient must have units of

and so our conversion factor is:



and we find:

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January 30th, 2012, 10:15 PM   #3
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Re: Dimensional analysis (Weir Eq.)

Ah thanks, that did it.

This problem's a little harder, was wondering how you would tackle it. I found the conversion for each unit from Enlight Engineering units to CGS units:
Code:
psi -> g/cm*s^2                  = 68948
lb/gal -> g/cm*s^3               = 0.1198
in -> cm                         = 2.54 
ft/sec -> cm/s                   = 30.48
ft -> cm                         = 30.48
cp -> g/cm*s                     = 0.01
I ended up getting 3.5*10^-9 but I am not sure about this answer.
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January 30th, 2012, 10:15 PM   #4
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Re: Dimensional analysis (Weir Eq.)

Forgot pic woops
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February 6th, 2012, 06:56 PM   #5
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Re: Dimensional analysis (Weir Eq.)

Any help with my last question?
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February 6th, 2012, 08:05 PM   #6
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Re: Dimensional analysis (Weir Eq.)

We would need:

^{0.75}(30.4^{1.75}(30.4(0.01)^{ 0.25}}{(2.54 )^{1.25}}\approx242.014627124" />
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February 6th, 2012, 08:13 PM   #7
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Re: Dimensional analysis (Weir Eq.)

Quote:
Originally Posted by MarkFL
We would need:

^{0.75}(30.4^{1.75}(30.4(0.01)^{ 0.25}}{(2.54 )^{1.25}}\approx242.014627124" />
Oh okay, I didn't use the exponents when I got my answer. I guess that would explain it.

for the attached image, does this look right?
?Pf = mupLv / (1.2x10^-3)(d0-di)^2 + YbL / (.065)(d0-di)
Attached Images
File Type: jpg 4.jpg (34.4 KB, 550 views)
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