Michael Fawcett wrote:
On 3/6/07, Deane Yang <deane_yang@yahoo.com> wrote:
Noah Roberts wrote:
Deane Yang wrote:
Can you provide a concrete example? Formula for Fluid Compressibility through a Venturi Tube:
Y = {[kt^(2/k)/(k-1)][(1-b^4)/(1-b^4t^(2/k))][(1-t^((k-1)/k)/(1-t)]}^.5
With k being the SP heat ratio of the fluid passing through the venturi. Static dimensional analysis is impossible here. In fact dimensional analysis at all just isn't appropriate in these odd cases.
So you're saying that the dimension/unit library does *not* need to worry about formulas like this, right? If so, my comments below are off-topic.
What I'm saying is that there are just times when you'll have to break out. You'll have to get the value out of the quantity, use it, and put the result into a quantity with the dimensions it's supposed to have. There's just no way around it that I can see. In other words I don't think there's much use in trying to solve ALL problems that might come up, especially wrt exponents.
But I'm still a skeptic. I don't see why the trick I outlined before can't be played here, too. Tell me what each variable means and what units it is in (if any). Or give me an online reference for this formula.
It appears that he's referring to the equation given here:
Could be. It's the Fluid Compressibility function for a Venturi Tube or Nozzle shaped orifice. I'm not an engineer...I just do what I'm told on that end and ask questions /if I need to/. Y = fluid compressibility factor t = fluid temp (F I believe) b = beta - orifice/pipe size. k = SP Heat
Look for the equation under the heading "Flow of gases through an orifice". Above it and below it are references to the variable units.
The only variable that has units is temperature. The rest are ratios. But in that EQ t^2k is the problem that will work in the end. k is an unknown so you can't statically create a result type for that EQ. You have to take the temp value out and use it as a double and then plug in the compressibility into a dimensional quantity (it has none but it's good to have it in one that's labeled as such). There are other occasions when you might plug that value into a mass flow rate or something. You could say that t^x is a dimensionless quantity and tell people to multiply that by 1 * whatever dimensions they expect. I don't like that idea though because it doesn't really convey the importance that they get it right that breaking out of the static system does. When you have to cast out of the quantity system into doubles you're going to be careful. This problem can be solved with runtime dimensional analysis...unfortunately it doesn't do it right most of the time because we don't care what the true dimension of t^x is...it's empirical and doesn't use dimensional analysis.