GLOSSARY

Page tree

Versions Compared

Old Version 32

changes.mady.by.user Arthur Aslanyan (Nafta College)

Saved on

compared with

New Version 33

changes.mady.by.user Arthur Aslanyan (Nafta College)

Saved on

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

...

LaTeX Math Inline
body\rho

fluid  density

LaTeX Math Inline
bodyd

orifice diameter

LaTeX Math Inline
bodyD

pipe diameter 

LaTeX Math Inline
body\beta = \frac{d}{D}

orifice narrowing ratio

LaTeX Math Inline
bodyC_d

discharge coefficient 

LaTeX Math Inline
body\epsilon

expansion factor

.

Expand
titleDerivation


Panel
borderColorwheat
bgColorivory
borderWidth7

Assume steady-state, incompressible, inviscid, laminar flow in a horizontal pipe (no change in elevation) with negligible frictional losses with incharge fluid velocity 

LaTeX Math Inline
bodyv_{in}
 and discharge velocity 
LaTeX Math Inline
bodyv_{out}
 at the orifice exit.

The mass conservation (equivalent to continuity equation):

LaTeX Math Block
anchorBXEPW
alignmentleft
\rho \, q = \rho \, v_{in} \, A_{in} = \rho \, v_{out} \, A_{out}  \Leftrightarrow v_{in} = \frac{q}{A_{in}}, \, v_{in} = \frac{q}{A_{out}}

where 

LaTeX Math Block
anchorBXEPW
alignmentleft
A_{in} = \frac{\pi \, D^2}{4} \, A_{out} = \frac{\pi \, d^2}{4}



 Bernoulli's equation reduces to an equation relating the conservation of energy between two points on the same streamline:

LaTeX Math Block
anchorBXEPW
alignmentleft
p_{in} + \frac{1}{2} \rho v^2_{in} = p_{out} + \frac{1}{2} \rho v^2_{out}


LaTeX Math Block
anchorBXEPW
alignmentleft
\Delta p = p_{in} - p_{out} =  \frac{1}{2} \rho v^2_{out} - \frac{1}{2} \rho \, v^2_{in} = \frac{1}{2} \rho \,  v^2_{out} \cdot \left[ 1 - \frac{v^2_{in}}{v^2_{out}} \right]


LaTeX Math Block
anchorBXEPW
alignmentleft
\Delta p = \frac{1}{2} \rho \, \left( \frac{q}{A_{in}} \right)^2 \cdot \left[ 1 - \frac{A^2_{out}}{A^2_{in}} \right] = \frac{\rho \, q^2}{2 \, A^2_{in}}  \cdot \left[ 1 - \frac{d^4}{D^4} \right] = \frac{\rho \, q^2}{2 \, A^2_{in}}  \cdot \left[ 1 - \beta^4 \right] = \frac{\rho \, q^2}{2 \, (\frac{\pi \, d^2}{4})^2}  \cdot \left[ 1 - \beta^4 \right] = \frac{8 \, \rho \, q^2}{\pi^2 \, d^4  }  \cdot \left[ 1 - \beta^4 \right]



...