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Created by
user-9c36d, last updated on Mar 04, 2008
1 minute read
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\large
$$
\left( {\Delta h_
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A_
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\rho g} \right)L_
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= \left( {\rho g\Delta h_
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A_
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} \right)L_
Unknown macro: {valve;lever;arm}
$$
where
Unknown macro: {latex}
$\Delta h_
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$
is the change in depth of the liquid level in the constant head tank and
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$A_
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$
is the cross sectional area of the cylindrical float. Thus
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$\Delta h_
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A_
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$
is the submerged volume of the float that when multiplied by the density,
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$\rho$
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 resisting moment provided by the pressure of water acting over the