K-Factor of Valve

When the valve is turned, does it increase or decrease vapor into a liquid stream or liquid into a vapor stream, i.e. two or three port (or more)?

To the best of my knowledge it's a resistance factor.

http://www.pipeflowcalculations.com/resistancecoefficientK/index.htm


Cv and Kv are conductance based.... analogous to mho the reciprocal of resistance in ohms.

See if you can get your hands on a copy of Crane's Tech Pub 410 "Flow of Fluids ...". There you will find K = (891*d^4) / Cv^2.

Crane gives the equation Cv = 29.9 * d^2 / sqrt K

The conversion is in error as the sqaure of 29.9 is 894.01

probably matters not a jot... but.

K= 894.01 * d^4 / Cv^2

Regards Woody

K factor is dimensioneless. Under the Darcy-Weibasch equation it is equivalent to a turbulent friction factor times the equivalent length divided by the diameter. This is a consideration taken when calculating the minor losses in series or parallel pipe systems.

K = (Le/D)ft

You can find a well developed explanation here:

http://www.cheresources.com/eqlength.shtml

Regards.

K factor is dimensioneless. Under the Darcy-Weibasch equation it is equivalent to a turbulent friction factor times the equivalent length divided by the diameter. This is a consideration taken when calculating the minor losses in series or parallel pipe systems.

K = (Le/D)ft

You can find a well developed explanation here:

http://www.cheresources.com/eqlength.shtml

I copied the above reply, because I hadn´t logged in. To explain this better, when the valve is nearly closed obviously it will be harder for the fluid to cross the valve against being fully opened. This is the reason why at 10% the loss is higher then at 100% opened. If you have a fluid flow of 135gpm (0.3ft3/s) you will have a loss at 10% open of 0.14ft while at 100% opened it will be 0.006ft. This is why its considered a minor loss since pipe is the major factor in losses in any system that is well designed of course.