# Tank Fluid Mechanics Problem: Step-by-Step Solution Let's organize the information and solve each part step by step. --- ## **Given Data** - **Tank cross-sectional area:** \( 120\,\text{cm} \times 180\,\text{cm} = 21600\,\text{cm}^2 = 2.16\,\text{m}^2 \) - **Tank height:** \( 150\,\text{cm} = 1.5\,\text{m} \) - **Water density:** \( \rho_w = 100\,\text{kg/m}^3 \) - **Oil specific gravity:** \( SG_o = .85 \implies \rho_o = .85 \times 100 = 850\,\text{kg/m}^3 \) - **Pressure of air-vapor mixture at top:** \( P_{\text{air}} = 100\,\text{kPa} \) - **Pressure gauge reading at lid:** \( -7.35\,\text{psi} \) (negative means below atmospheric) - **Gauge pressure conversion:** \( 1\,\text{psi} = 6.895\,\text{kPa} \implies -7.35\,\text{psi} = -50.7\,\text{kPa} \) - **Fluid arrangement (from bottom up):** 1. **Water:** bottom half of tank 2. **Oil:** 2/3 of the remaining upper space 3. **Air-vapor:** remainder at the very top --- ## **1. Fluid Heights** - **Total height:** \( 1.5\,\text{m} \) - **Height of water:** \( h_w = .5 \times 1.5 = .75\,\text{m} \) - **Remaining space:** \( 1.5 - .75 = .75\,\text{m} \) - **Height of oil:** \( h_o = \frac{2}{3} \times .75 = .5\,\text{m} \) - **Height of air-vapor:** \( h_a = .75 - .5 = .25\,\text{m} \) --- ## **2. Atmospheric Pressure Reference** - **Gauge pressure at lid:** \( -50.7\,\text{kPa} \) (relative to atmospheric) - **Absolute pressure at lid:** \( P_{\text{atm}} + (-50.7\,\text{kPa}) \) - Assume \( P_{\text{atm}} = 101.3\,\text{kPa} \) (standard) - So, **absolute pressure at lid:** \( 101.3 - 50.7 = 50.6\,\text{kPa} \) --- ## **A) Force on the Side Walls Due to Water** ### **Step 1: Hydrostatic Pressure on Side Walls** - Pressure at depth \( h \): \( p = \rho_w g h \) - The **average pressure** on the side wall (from \( h= \) to \( h=.75\,\text{m} \)): - Varies linearly from to \( p_{\text{max}} = \rho_w g h_w \) - **Average pressure:** \( p_{\text{avg}} = \frac{1}{2} \rho_w g h_w \) ### **Step 2: Area of Side Walls in Contact with Water** - **Perimeter** of tank: \( 2 \times (1.2 + 1.8) = 6\,\text{m} \) - **Height in contact with water:** \( .75\,\text{m} \) - **Total side wall area:** \( A_{\text{side,water}} = 6 \times .75 = 4.5\,\text{m}^2 \) ### **Step 3: Total Force** \[ F_{\text{water,side}} = p_{\text{avg}} \times A_{\text{side,water}} = \left( \frac{1}{2} \rho_w g h_w \right) \times A_{\text{side,water}} \] Plug in values: - \( \rho_w = 100\,\text{kg/m}^3 \) - \( g = 9.81\,\text{m/s}^2 \) - \( h_w = .75\,\text{m} \) - \( A_{\text{side,water}} = 4.5\,\text{m}^2 \) \[ F_{\text{water,side}} = .5 \times 100 \times 9.81 \times .75 \times 4.5 \] \[ = .5 \times 100 \times 9.81 \times 3.375 \] \[ = .5 \times 100 \times 33.09375 \] \[ = .5 \times 33,093.75 \] \[ = 16,546.875\ \text{N} \] **Final Answer (A):** > The force on the side walls due to water is **16,547 N** (rounded to the nearest newton). --- ## **B) Gauge Pressure at the Bottom of the Tank** ### **Step 1: Gauges Pressure at Bottom** \[ p_{\text{gauge, bottom}} = p_{\text{gauge, lid}} + \rho_a g h_a + \rho_o g h_o + \rho_w g h_w \] But the air-vapor mixture exerts negligible hydrostatic pressure compared to liquids (its density is very small), so: \[ p_{\text{gauge, bottom}} = p_{\text{gauge, lid}} + \rho_o g h_o + \rho_w g h_w \] Plug in values: - \( p_{\text{gauge, lid}} = -50,700\,\text{Pa} \) - \( \rho_o = 850\,\text{kg/m}^3 \), \( h_o = .5\,\text{m} \) - \( \rho_w = 100\,\text{kg/m}^3 \), \( h_w = .75\,\text{m} \) - \( g = 9.81\,\text{m/s}^2 \) \[ p_{\text{gauge, bottom}} = -50,700 + (850 \times 9.81 \times .5) + (100 \times 9.81 \times .75) \] \[ = -50,700 + (4,174.25) + (7,357.5) \] \[ = -50,700 + 11,531.75 \] \[ = -39,168.25\ \text{Pa} \] **Final Answer (B):** > Gauge pressure at the bottom of the tank is **\(-39,168\) Pa** (rounded to the nearest pascal), or **\(-39.2\,\text{kPa}\)**. --- ## **C) Absolute Pressure at the Oil-Water Interface** - The oil-water interface is at a depth measured from the top: - From top: \( h_a + h_o = .25 + .5 = .75\,\text{m} \) - From bottom: \( h_w = .75\,\text{m} \) - **Absolute pressure at lid:** \( 50,600\,\text{Pa} \) - **Pressure increases by oil above interface:** \[ p_{\text{abs, int}} = p_{\text{abs, lid}} + \rho_o g h_o \] \[ = 50,600 + 850 \times 9.81 \times .5 \] \[ = 50,600 + 4,174.25 \] \[ = 54,774.25\,\text{Pa} \] **Final Answer (C):** > Absolute pressure at the oil-water interface is **54,774 Pa** or **54.8 kPa** (rounded). --- ## **D) Absolute Pressure Difference Across the Oil Layer** - Pressure at the bottom of the oil: \( p_{\text{abs, int}} = 54,774\,\text{Pa} \) - Pressure at top of the oil (just below air-vapor): \( p_{\text{abs, lid}} = 50,600\,\text{Pa} \) - \(\Delta p = p_{\text{abs, int}} - p_{\text{abs, lid}} = 4,174\,\text{Pa}\) **Final Answer (D):** > Absolute pressure difference across the oil layer is **4,174 Pa** or **4.17 kPa**. --- ## **E) Force on the Bottom Due to the Oil** - **Force = pressure due to oil at interface \(\times\) area** - Pressure at oil-water interface: \( \rho_o g h_o \) - Area: \( 2.16\,\text{m}^2 \) \[ F_{\text{oil, bottom}} = (\rho_o g h_o) \times A = (850 \times 9.81 \times .5) \times 2.16 \] \[ = 4,174.25 \times 2.16 \] \[ = 9,017.4\,\text{N} \] **Final Answer (E):** > The force on the bottom of the tank due to the oil is **9,017 N** (rounded). --- # **Summary Table** | Part | Quantity | Value | |------|-------------------------------------------------|------------------------------| | A | Force on side walls due to water | \(16,547\,\text{N}\) | | B | Gauge pressure at tank bottom | \(-39,168\,\text{Pa}\) | | C | Absolute pressure at oil-water interface | \(54,774\,\text{Pa}\) | | D | Absolute pressure difference across oil layer | \(4,174\,\text{Pa}\) | | E | Force on bottom due to oil | \(9,017\,\text{N}\) | --- **If you need explanations for why each step was done a certain way, or more on hydrostatics, just ask!**