Question

Two containers on the same elevation contain water and oil (sp. gr. = 0.88). They are connected by a U-tube manometer with the manometric liquid having a specific gravity of 1.25. If the manometric liquid in the limb connecting the water is 2 m higher than the other; then find $P_A - P_B$

Answer 1

$1.9 \rho_w g$ or $18620 \, \text{Pa}$

Answer 2

We apply the principle of hydrostatic pressure. Let's choose a reference level at the interface between the manometric liquid and the oil.

Pressure from the right side (oil container): $P_{right} = P_B + \rho_{oil} \times g \times 5$

Pressure from the left side (water container), considering the manometric liquid is 2 m higher: $P_{left} = P_A + \rho_{manometer} \times g \times 2$

Equating pressures: $P_B + \rho_{oil} \times g \times 5 = P_A + \rho_{manometer} \times g \times 2$

Using specific gravities ($S_{oil} = 0.88$, $S_{manometer} = 1.25$): $\rho_{oil} = 0.88 \rho_w$ $\rho_{manometer} = 1.25 \rho_w$

Substituting: $P_B + (0.88 \rho_w) \times g \times 5 = P_A + (1.25 \rho_w) \times g \times 2$ $P_B + 4.4 \rho_w g = P_A + 2.5 \rho_w g$

Rearranging for $P_A - P_B$: $P_A - P_B = 4.4 \rho_w g - 2.5 \rho_w g$ $P_A - P_B = 1.9 \rho_w g$

Using $\rho_w = 1000 \, \text{kg/m}^3$ and $g = 9.8 \, \text{m/s}^2$: $P_A - P_B = 1.9 \times 1000 \times 9.8 = 18620 \, \text{Pa}$