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K value for a double shutoff quick disconnect. 1

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SteveFoster

Mechanical
Joined
Jun 28, 2005
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US
I am doing a relatively simple flow analysis on a piping network and I ran into a problem. The piping network was some quick disconnects that are giving me large minor losses. For instance, the manufacturers catalog data shows that at 8gpm the qd has a 5psi(guage) pressure drop. From manometry deltaZ=(deltaP/specific weight) right? Using this simple equation should give the equivalent height(or depth if you prefer) of the fluid corresponding to this pressure change. Also the minor loss of some component could be found by Hm=(K*V^2)/(2*g), where Hm is the head loss. Solving for K gives (2*Hm*g)/V^2. I am after the K value(I am summing all minor losses) but when I substitute in say 8gpm and the 5psi drop, I end up with K being around 45 which seems absurd. I think I am doing this wrong but am not sure. Any ideas/criticism please let me know. Thanks

Pipe ID=0.82 in, g=32.174 ft/s^2, V=Q/A=4.86ft/s,specific weight=43.73 lb/ft^3, rho=1.359 slug/ft^3, Re=360 which is <2300 so laminar, if I forgot anything else just ask me.
 
Steve,

I assume your viscosity is 5.99E+01 cp. Then given the conditions you describe, I estimate the equivalent length of a 5 psi drop is about 17.3 feet of 0.82 pipe or a K value of about 6.4.

Also, are you using psi when you should be using psf?
 
Actually the viscosity is 5.99E+01 cp and I am using psf, I even went back to check in case I thought one thing and wrote another, but it is in psf (5psi *144=720psf). How did you come about with your value for K?
 
If you have a pipe flow calculator, Set Q, D, P1, and P2.... solve for L. That's your equiv L for the given D.
K can be gotten from the relation L/D = K/ft, where ft is the fully turbulent flow friction factor, L is the equiv L and D is the pipe dia(in L units.)
 
Coming out of those awful units:
V=15.9 m/s
[&rho;]=700 kg/m3
velocity head H=[&rho;]V2/2=88000 Pa = 0.88 bar = 13 psi
K=5/13=0.4

prex

Online tools for structural design
 
Steve,

The short answer is that you are spot on with your calcs. It is a long time since I tried to work in pounds force per square foot and slugs per cubic foot! Thank goodness for the SI system. I agree with prex that the old units were awful (NB prex, 4.86 ft/s is 1.48 m/s and not 15.9 m/s).

The reason for your seemingly high K value is the low Reynolds number. Compilations of K values should always include the warning that they apply to turbulent flow only. In turbulent flow the K value of a fitting is more-or-less independent of the Reynolds number, but the K value of a given fitting increases as the Reynolds number decreases below 2300. Hooper introduced the 2-K method and later Darby extended it to the 3-K method in an effort to take changing diameter and Reynolds number into account. These concepts are discussed at
 
Thanks everyone for the help. RGasEng, I was going to solve for L but I don't believe I can use the fully turbulent friction factor because my flow never goes turbulent, or even close. The link katmar posted was very helpful. I know the units are awful but I am using a spreadsheet that was made by another engineer here and he loves english units. I am going to take what ya'll [I'm from Texas;)] have given me and work on this some more. Thanks again for the help.
 
katmar:
Such an astute observation deserves more than one star. Good link. Do you have a copy of the Hooper paper that you can make available?
 
I once heard a lawyer say that the English lawyers are the best in the world. They have to be because the English law is so complicated. Perhaps an engineer who understands formulas expressed in English units is also world class, even if he is a Texan!
 
RGasEng,

I think there would be copyright issues preventing me distributing copies of this article. If you cannot get to a university library yourself I would think that you could request a copy of the article via your local library. The reference was in the link I gave previously
 
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