The various equations for finding required Cv for valve sizing contain what look like dimensional constants that depend on the units chosen. For example:
But, I'm having trouble using this equation in Mathcad, since I don't know what is included in the constant 63.4 and I don't know if Cv should have any units or not. Anybody have any experience or ideas? Please see the worksheet attached. Thanks for any help you can lend.
I would assume that Cv is unitless (described as a "coefficient" and no unit listed).
I can't open your file, but here are two possible ways to approach equations like this:
Thanks Mark. I like your idea of using the "k" factor to deal with the units within the equation. I think it's rather elegant. However, it leaves unanswered the question of what goes into the 63.4 constant. I wish I could see where that came from. Regarding Cv, the discussions of its origins say that it was the gallons per minute of 60 degree F water that would flow through a valve with a 1 psi pressure drop. I think that would mean that Cv would probably have some units, even though, as you point out, some similar things, like discharge coefficients, are dimensionless.
I noticed something else yesterday. In the English unit system, 2g often appears in equations. 2g is 2*32.17405=64.35, which is close to the 63.4 constant in the equation. I wonder if that means anything?
Maybe it's 2g, maybe not. Those kind of equations was established by the 'ancients', or manufacturers. Sometimes carries with the units coefficients, but others also include empirical factors. If you don't reference from where came the equation, have less chance to obtain some help, but those equations sometimes are so obscures that even with the complete book no one can say from where came those coefficients.
Possibly, the density of water in lb/ft^3. I've seen it referenced as 62.4 lb/ft^3.
Alternatively, If you search for a worksheet in this collab, Colebrook-Friction in Pipes that may be of interest.
I hope this helps.
Equations appear to be from Crane's 4-10 Note Book. At the front of the book the variables are defined with units. The equation in the notebook should be used without units. The conversions are included in the equation. I have run into this issue with other empirical function. One way to resolve the units is define the variable in the units you are using, i.e. delta P in pascals. Divide delta by psi for the equation. This gives you the unitless value, but the correct number for the equation. This approach also works for Mathcad 15 Solve function.
One regular method for comversting these empirical equations into a coherent SI equation (in mathcad) is to, for each variable, divide it the variable by the requested units (within the equation), and then multiply the final answer (which will then be numeric) by the stated output result units.
This works for all types of equation that have the conventional style of units and dimensions. It can fail for advaced phyics where they can use plank units and loose all feeling for scale (every constant is unity, and many convetional 'variables' disappear). Though this works fine for regular engineering calculations.
Luc's method gives a nice feel for what is hidden in that final constant.
If the equation's conversion can be taken in stages it is possible in some cases to re-align the equation to being 'science' with understandable material constants being added.
By basic balancing of dimensions, your 63.4 needs to be an area. (There is also probably a unit conversion factor or 5 thrown in):
The bottom equation would report the same flow coefficient regardless of the units used.
The units are area, but A = 63.4*ft^2. The conversion 63.4 is for US units. Adding the units to the 63.4 will allow the use any units.