# Mercury Manometer Problem Breakdown ## Problem Statement A mercury manometer is attached to a sealed tank containing air, oil, and water. The following data are provided: - **Elevations:** - Top: 6 m - Next: 5 m - Oil-water interface at 2 m - Water-mercury interface at 0 m - **Fluids:** - Air (pressurized) - Oil (density, \( \rho_{\text{oil}} = 823 \) kg/m³, note the negative sign in "Oils -0.823 Flex" likely means 0.823 of water) - Water (\( \rho_{\text{water}} = 1000 \) kg/m³) - Mercury (\( \rho_{\text{Hg}} = 13,600 \) kg/m³) - **Atmospheric Pressure:** \( P_{\text{atm}} = 101.3 \) kPa - **Manometer Reading:** The deflection in the mercury column is \( y \). - **Goal:** Find the deflection \( y \) for static equilibrium. --- ## Diagram Description ``` |<--- 6 m (top, air) ---| Pressure = ? |<--- 5 m (oil) ---| Oil, ρ = 823 kg/m³ |<--- 2 m (water) ---| Water, ρ = 1000 kg/m³ |<--- 0 m (Hg interface) ---| Mercury, ρ = 13,600 kg/m³, y = ? ``` --- ## Approach 1. **Apply Hydrostatic Pressure Differences:** - Pressure change across each layer: \( \Delta P = \rho g h \) - For the manometer, sum all pressure changes from one atmospheric side to the other, accounting for the direction. 2. **Set Up Pressure Loop:** - Start at atmospheric side, go through each fluid layer to the air space in the tank. --- ## Solution Setup ### 1. Pressure at Air Space Let \( P_{\text{air}} \) be the pressure of the air in the tank (at 6 m elevation). - **Moving down through tank:** - From air to oil: height \( h_1 = 1 \) m (\(6 \to 5\) m) - Oil to water: height \( h_2 = 3 \) m (\(5 \to 2\) m) - Water to Hg: height \( h_3 = 2 \) m (\(2 \to 0\) m) - Down mercury column: height \( y \) - **Manometer U-tube connects to atmosphere at the same elevation as mercury interface** ### 2. Pressure Equation Pressure at the air side (top of tank): \[ P_{\text{air}} \] Pressure at the atmospheric side (bottom of manometer): \[ P_{\text{atm}} \] The pressure at the bottom of the manometer (across the mercury column) must be equal for equilibrium: \[ P_{\text{air}} + \Delta P_{\text{oil}} + \Delta P_{\text{water}} + \Delta P_{\text{Hg}} = P_{\text{atm}} \] Where: - \( \Delta P_{\text{oil}} = \rho_{\text{oil}} g h_{\text{oil}} \) (down) - \( \Delta P_{\text{water}} = \rho_{\text{water}} g h_{\text{water}} \) (down) - \( \Delta P_{\text{Hg}} = \rho_{\text{Hg}} g y \) (down if moving down in manometer, up if moving up) --- ## 3. Manometer Deflection The air pressure pushes down, causing the mercury to rise by \( y \) on one side. The difference in mercury levels between the two sides is \( y \). So for the pressure difference: \[ P_{\text{air}} + (\rho_{\text{oil}} g (5-2)) + (\rho_{\text{water}} g (2-0)) = P_{\text{atm}} + (\rho_{\text{Hg}} g y) \] Solving for \( y \): \[ P_{\text{air}} + \rho_{\text{oil}} g (3) + \rho_{\text{water}} g (2) = P_{\text{atm}} + \rho_{\text{Hg}} g y \] \[ P_{\text{air}} - P_{\text{atm}} = \rho_{\text{Hg}} g y - \rho_{\text{oil}} g (3) - \rho_{\text{water}} g (2) \] If the manometer is open to the atmosphere and the tank is sealed. If the air pressure is not given, but the manometer reading relates the pressure difference, rearrange: \[ y = \frac{P_{\text{air}} - P_{\text{atm}} + \rho_{\text{oil}} g (3) + \rho_{\text{water}} g (2)}{\rho_{\text{Hg}} g} \] --- ## 4. Plug in Values **Given:** - \( \rho_{\text{oil}} = 823 \) kg/m³ - \( h_{\text{oil}} = 3 \) m - \( \rho_{\text{water}} = 1000 \) kg/m³ - \( h_{\text{water}} = 2 \) m - \( \rho_{\text{Hg}} = 13,600 \) kg/m³ - \( g = 9.81 \) m/s² - \( P_{\text{atm}} = 101,300 \) Pa If \( P_{\text{air}} \) is not specified, \( y \) is left in terms of \( P_{\text{air}} \). --- ### **Final Answer (in terms of air pressure):** \[ y = \frac{P_{\text{air}} - 101,300 + (823)(9.81)(3) + (1000)(9.81)(2)}{13,600 \times 9.81} \] --- **If \( P_{\text{air}} \) is known, substitute to find \( y \).** --- ## **Summary** - Use hydrostatic pressure differences through oil and water. - Relate pressure difference to mercury deflection in manometer. - Use the formula above to compute \( y \). --- **[Insert diagram of sealed tank with manometer and labeled elevations/fluids]** *Alt text: A schematic of a tank with air, oil, and water layers connected to a U-tube mercury manometer, showing elevations and indicating the deflection y in the mercury column.*