{latex}
\large
$$
\left( {\Delta h_{FCM} A_{float} \rho g} \right)L_{float\;lever\;arm} = \left( {\rho g\Delta h_{stock} A_{orifice} } \right)L_{valve\;lever\;arm}
$$
{latex}
where {latex}$\Delta h_{FCM}${latex} is the change in depth of the liquid level in the constant head tank and {latex}$A_{float}${latex} is the cross sectional area of the cylindrical float. Thus {latex}$\Delta h_{FCM} A_{float}${latex} is the submerged volume of the float that when multiplied by the density, {latex}$\rho${latex} 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 {latex}$L_{float\;lever\;arm}${latex}. The resisting moment provided by the pressure of water acting over the |