# Step-by-Step Solution to Manometer Problem We will calculate the gauge pressure indicated by the pressure gauge, given a multi-fluid manometer. Here’s a generic approach since the diagram isn’t visible, but all necessary parameters are provided. ## Step 1: **Identify Known Data** - Densities: - Liquid 1: \( \rho_1 = 780\, \text{kg/m}^3 \) - Liquid 2: \( \rho_2 = 123\, \text{kg/m}^3 \) - Liquid 3: \( \rho_3 = 105\, \text{kg/m}^3 \) - Semicircular section radius: \( r = 22\, \text{mm} = .022\, \text{m} \) - Air above beaker at atmospheric pressure (\( P_{\text{atm}} \)) - Manometer contains air and three liquids (order assumed based on typical manometer setups) - **Neglect surface tension and height variations across diameter** ## Step 2: **Understand Pressure Relationships** - The manometer **compares the pressure at the gauge** (\( P_{\text{gauge}} \)) **to atmospheric pressure**. - The pressure changes only due to the **vertical height differences** in fluids of different densities. - **Semicircular section**: The **vertical height change** is equal to its diameter: \( h_{\text{semi}} = 2r = .044\, \text{m} \). ## Step 3: **Write the Pressure Equation** Let’s start at the atmospheric end (open to air) and move through the manometer to the gauge, summing pressure changes. Let’s assume the following liquid order from atmospheric side to gauge: 1. Down through Liquid 1: \( h_1 \) 2. Up through semicircular section (Liquid 2): \( h_{\text{semi}} \) 3. Down through Liquid 3: \( h_3 \) 4. Up or down through air (negligible density, ignore contribution) Let the heights of each liquid column be \( h_1, h_{\text{semi}}, h_3 \). In the absence of a diagram, we’ll use \( h_1 \), \( h_2 \), \( h_3 \) as the heights for each respective liquid. The **gauge pressure** is: \[ P_{\text{gauge}} - P_{\text{atm}} = \sum (\pm) \rho_i g h_i \] The sign is positive when moving **downward** through a column (pressure increases), and negative when moving **upward**. ### Typical Manometer Path Suppose starting at atmospheric side: 1. **Down \( h_1 \) in Liquid 1:** \( +\rho_1 g h_1 \) 2. **Up \( h_2 \) in Liquid 2 (semicircle):** \( -\rho_2 g h_2 \) 3. **Down \( h_3 \) in Liquid 3:** \( +\rho_3 g h_3 \) So, \[ P_{\text{gauge}} - P_{\text{atm}} = \rho_1 g h_1 - \rho_2 g h_2 + \rho_3 g h_3 \] ## Step 4: **Plug in Heights** If the semicircular section is the only vertical height given (\( h_2 = .044\, \text{m} \)), and if the other heights (\( h_1, h_3 \)) are not specified or cancel (as in a U-tube with equal levels), the **only net pressure difference** comes from the semicircular section. ## Step 5: **Calculate Gauge Pressure** \[ P_{\text{gauge}} - P_{\text{atm}} = -\rho_2 g h_2 \] Because the semicircular section is traversed **upwards** (from bottom of the semicircle to the top), the pressure decreases, so the sign is negative. Plug values: - \( \rho_2 = 123\, \text{kg/m}^3 \) - \( g = 9.81\, \text{m/s}^2 \) - \( h_2 = .044\, \text{m} \) \[ P_{\text{gauge}} - P_{\text{atm}} = - (123)(9.81)(.044) \] \[ P_{\text{gauge}} - P_{\text{atm}} = - (123 \times .044) \times 9.81 \] \[ = - (54.12) \times 9.81 \] \[ = -531\, \text{Pa} \] ## Step 6: **Final Answer** **Gauge pressure reading:** \[ \boxed{-531\, \text{Pa}} \] ## Step 7: **Summary and Assumptions** - Only the **semicircular section** contributes to the net gauge pressure (as per given data and typical setups). - **Negative sign** means the gauge reads **531 Pa below atmospheric pressure** (i.e., a partial vacuum). --- **If other heights (\( h_1, h_3 \)) are specified or not equal, add their contributions with the appropriate densities and signs as shown in Step 3.** --- ## Diagram (Conceptual) ``` [Atmospheric] ──(down h₁, ρ₁)──┐ | (up semicircle, h₂, ρ₂) | [Gauge] ──(down h₃, ρ₃)──┘ ``` *(Diagram shows path from atmospheric side through liquids 1, 2, and 3 to the gauge)* --- **If you provide the actual heights for liquids 1 and 3, substitute them in the formula:** \[ P_{\text{gauge}} - P_{\text{atm}} = \rho_1 g h_1 - \rho_2 g h_2 + \rho_3 g h_3 \]