Let's solve each question step by step. --- ## **Q1: Shear Stress in a Newtonian Fluid** **Given:** - Specific gravity, \( S = .90 \) - Kinematic viscosity, \( \nu = 3 \times 10^{-5} \) m²/s - Velocity profile: \[ u = \frac{4y}{5h}U - \frac{2}{5}\left(\frac{y}{h}\right)^2 U - \frac{1}{5}U \] - \( y \): distance from the fixed surface, \( h \): total height, \( U \): some velocity #### **Step 1: Find Dynamic Viscosity (\( \mu \))** \[ S = \frac{\rho}{\rho_{\text{water}}} \] \[ \rho = S \cdot \rho_{\text{water}} = .9 \times 100 = 900 \ \text{kg/m}^3 \] \[ \mu = \nu \cdot \rho = 3 \times 10^{-5} \times 900 = .027 \ \text{kg/(m·s)} \] #### **Step 2: Shear Stress Formula** Shear stress at a surface: \[ \tau = \mu \left. \frac{du}{dy} \right|_{y=} \] #### **Step 3: Differentiate the Velocity Profile** \[ u = \frac{4y}{5h}U - \frac{2}{5}\left(\frac{y}{h}\right)^2 U - \frac{1}{5}U \] Differentiate with respect to \( y \): \[ \frac{du}{dy} = \frac{4U}{5h} - \frac{2}{5}U \cdot 2\frac{y}{h^2} \] \[ \frac{du}{dy} = \frac{4U}{5h} - \frac{4U y}{5h^2} \] At \( y = \): \[ \left. \frac{du}{dy} \right|_{y=} = \frac{4U}{5h} \] #### **Step 4: Calculate Shear Stress** \[ \tau = \mu \cdot \frac{4U}{5h} \] \[ \tau = .027 \cdot \frac{4U}{5h} \] \[ \tau = \frac{.108U}{5h} \] \[ \tau = \frac{.0216U}{h} \] --- **Final Answer for Q1:** \[ \boxed{\tau = \frac{.0216U}{h}} \] where \( U \) is in m/s and \( h \) is in m. --- ## **Q2: Pressure in the Tank Using Manometer** **Given:** - \( h_1 = .1 \) m (water) - \( h_2 = .2 \) m (oil) - \( h_3 = .35 \) m (mercury) - Densities: - Water: \( 100 \) kg/m³ - Oil: \( 850 \) kg/m³ - Mercury: \( 13,600 \) kg/m³ - Atmospheric pressure = \( 85.6 \) kPa - Manometer as shown (left to right: water \(\rightarrow\) oil \(\rightarrow\) mercury) - Find absolute pressure in the tank #### **Step 1: Pressure at Each Point** Let's start from the open end (atmospheric) and move through the manometer: Let the tank pressure be \( P_{\text{tank}} \). 1. **Atmospheric side:** \( P_ = 85.6 \) kPa \( = 85,600 \) Pa 2. **Move down \( h_3 \) (mercury):** Pressure increases \( +\rho_{\text{Hg}} g h_3 \) 3. **Move up \( h_2 \) (oil):** Pressure decreases \( -\rho_{\text{oil}} g h_2 \) 4. **Move up \( h_1 \) (water):** Pressure decreases \( -\rho_{\text{water}} g h_1 \) 5. **Arrive at tank:** \( P_{\text{tank}} \) So, \[ P_{\text{tank}} = P_ + \rho_{\text{Hg}} g h_3 - \rho_{\text{oil}} g h_2 - \rho_{\text{water}} g h_1 \] #### **Step 2: Plug in Values** - \( g = 9.81 \) m/s² \[ P_{\text{tank}} = 85,600 + (13,600 \times 9.81 \times .35) - (850 \times 9.81 \times .2) - (100 \times 9.81 \times .1) \] Calculate each term: - Mercury: \( 13,600 \times 9.81 \times .35 = 46,734.6 \) - Oil: \( 850 \times 9.81 \times .2 = 1,668.3 \) - Water: \( 100 \times 9.81 \times .1 = 981 \) Now substitute: \[ P_{\text{tank}} = 85,600 + 46,734.6 - 1,668.3 - 981 \] \[ P_{\text{tank}} = 85,600 + 46,734.6 - 2,649.3 \] \[ P_{\text{tank}} = 85,600 + 44,085.3 \] \[ P_{\text{tank}} = 129,685.3 \ \text{Pa} \] --- **Final Answer for Q2:** \[ \boxed{P_{\text{tank}} = 129,685 \ \text{Pa} \ (\text{or} \ 129.7 \ \text{kPa, 3 sig. figs})} \] --- ### **Summary** - Q1: \( \boxed{\tau = \frac{.0216U}{h}} \) - Q2: \( \boxed{P_{\text{tank}} = 129,685 \ \text{Pa}} \) Let me know if you need more details or explanations!