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Created by
user-9c36d, last updated on Mar 04, 2008
1 minute read
Unknown macro: {latex}
\large
$$
\left( {\Delta h_
Unknown macro: {FCM}
A_
Unknown macro: {float}
\rho g} \right)L_
Unknown macro: {float;lever;arm}
= \left( {\rho g\Delta h_
Unknown macro: {stock}
A_
Unknown macro: {orifice}
} \right)L_
Unknown macro: {valve;lever;arm}
$$
where
Unknown macro: {latex}
$\Delta h_
Unknown macro: {FCM}
$
is the change in depth of the liquid level in the constant head tank and
Unknown macro: {latex}
$A_
Unknown macro: {float}
$
is the cross sectional area of the cylindrical float. Thus
Unknown macro: {latex}
$\Delta h_
Unknown macro: {FCM}
A_
Unknown macro: {float}
$
is the submerged volume of the float that when multiplied by the density,
Unknown macro: {latex}
$\rho$
and by acceleration due to gravity is equal to the total buoyant force acting on the float. The lever arm for the float has a length
Unknown macro: {latex}
$L_
Unknown macro: {float;lever;arm}
$
. The moment acting to open the valve is provided by the pressure of liquid from the stock tank,
Unknown macro: {latex}
$\rho g\Delta h_
Unknown macro: {stock}
$
, acting over the area of the valve opening
Unknown macro: {latex}
$A_
Unknown macro: {orifice}
$
. The lever arm for the opening moment is
Unknown macro: {latex}
$A_
Unknown macro: {orifice}
$
Unknown macro: {latex}
The derivative of the function $f
$ at the point $x_0$ is
\begin
Unknown macro: {equation}
f'(x_0) =
\lim_
Unknown macro: {x rightarrow x_0}
\frac
Unknown macro: {f(x) - f(x_0)}
Unknown macro: {x - x_0}
\end