Solved Problem on Fluid Mechanics
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A U-shaped tube with open ends contains three immiscible liquids of densities ρ1,
ρ2 and ρ3. If the liquids are in equilibrium, find the
density ρ1 as a function of the densities ρ2 and
ρ3.
Problem data:
- Density of liquid 1: ρ1;
- Density of liquid 2: ρ2;
- Density of liquid 3: ρ3.
We take as reference the lowest interface between two liquids, densities ρ1 and
ρ3 (Figure 1). Points 1 and 2 of the liquid are at the same height, and the
pressures acting on these points are equal. The atmospheric pressure P0, the pressure
of the liquid column of density ρ1, and the pressure of the liquid column of
density ρ2 act on the left-hand branch, and the atmospheric pressure
P0 and the pressure of the liquid column of density act on the right-hand branch
ρ3.
Solution
The pressure due to a column of liquid is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{P=\rho gh}
\end{gather}
\]
\[
\begin{gather}
P_{0}+P_{1}+P_{2}=P_{0}+P_{3}\\[5pt]
P_{0}+\rho_{1}g\frac{1}{8}h+\rho _{2}gh=P_{0}+\rho _{3}g\frac{3}{4}h\\[5pt]
\rho_{1}g\frac{1}{8}h=P_{0}-P_{0}+\rho_{3}g\frac{3}{4}h-\rho_{2}gh\\[5pt]
\rho_{1}\cancel{g}\frac{1}{8}\cancel{h}=\rho_{3}\cancel{g}\frac{3}{4}\cancel{h}-\rho_{2}\cancel{g}\cancel{h}\\[5pt]
\rho_{1}\frac{1}{8}=\rho_{3}\frac{3}{4}-\rho_{2}\\[5pt]
\rho_{1}=8\times\left(\rho_{3}\frac{3}{4}-\rho_{2}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\rho_{1}=8\times\left(0.75\rho_{3}-\rho_{2}\right)}
\end{gather}
\]
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