Horizontal tank example question

[b-rad22]

Hi, 

I have a project that deals with a horizontal cylindrical tank in labview and I am trying to figure out a few things. 

1) The example model in labview uses the equation: dh/dt = (qi - Cv*sqrt(h)) / (2*L*sqrt((D-h)^2)) where: qi is flow in, h is height of fluid level, D is diameter of tank and L is the length. 

Now I understand this is of the form, dh/dt = (qi - k*sqrt(h))/ (A)

Going based off of examples found from places such as example of a conical tank for modelling, 

 I do not understand where the form of 2*L fits into the equation for cross sectional area of the horizontal cylindrical tank. 

Any assistance would be appreciated. 

[camerond]

This question has nothing to do with LabVIEW, you can solve it in any computer language (or manually) you want. For basic math formulas, look in basic math texts.

BTW, your reference link is broken.

Cameron

[Taki1999]

camerond is right.

I like math though.

Don't think of it as a 2*L term

Think of it as an L term multiplied by a segment length in order to get cross-sectional area.

2*(sqrt(D-h)^2) is your segment length.