My Math Forum Dimensional analysis (Weir Eq.)

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 January 30th, 2012, 09:19 PM #1 Newbie   Joined: Nov 2011 Posts: 17 Thanks: 0 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.
 January 30th, 2012, 09:51 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Dimensional analysis (Weir Eq.) I know that 2.54 cm = 1 in exactly, thus: $1\text{ ft}\cdot\frac{12\text{ in}}{1\text{ ft}}\cdot\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{1\text{ m}}{100\text{ cm}}$ thus: $1\text{ ft}=\frac{12\cdot2.54}{100}\:\text{m}=0.3048\text{ m}$ For your regular contracted weir, the discharge coefficient must have units of $\frac{\sqrt{\text{ft}}}{\text{s}}$ and so our conversion factor is: $\sqrt{\frac{0.3048\text{ m}}{1\text{ ft}}}\approx0.5520869496736904\sqrt{\text{\frac{m} {ft}}}$ and we find: $3.33\cdot0.5520869496736904\approx1.83844954241338 9$
 January 30th, 2012, 10:15 PM #3 Newbie   Joined: Nov 2011 Posts: 17 Thanks: 0 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.
 January 30th, 2012, 10:15 PM #4 Newbie   Joined: Nov 2011 Posts: 17 Thanks: 0 Re: Dimensional analysis (Weir Eq.) Forgot pic woops
 February 6th, 2012, 06:56 PM #5 Newbie   Joined: Nov 2011 Posts: 17 Thanks: 0 Re: Dimensional analysis (Weir Eq.) Any help with my last question?
 February 6th, 2012, 08:05 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Dimensional analysis (Weir Eq.) We would need: $\frac{(0.119^{0.75}(30.4^{1.75}(30.4(0.01)^{ 0.25}}{(2.54 )^{1.25}}\approx242.014627124" />
February 6th, 2012, 08:13 PM   #7
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Re: Dimensional analysis (Weir Eq.)

Quote:
 Originally Posted by MarkFL We would need: $\frac{(0.119^{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)
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