Let's break down and solve each part step by step: --- ## **8(a) Mass of Air Entered the Tank** ### **Given:** - Initial tank volume, \( V = .75 \, m^3 \) - Initial air density, \( \rho_1 = 1.18 \, kg/m^3 \) - Final air density, \( \rho_2 = 4.95 \, kg/m^3 \) - The tank is rigid (volume does not change). ### **Required:** Determine the mass of air that has entered the tank. --- ### **Step 1: Calculate Initial Mass of Air** \[ m_1 = \rho_1 \cdot V \] \[ m_1 = 1.18 \times .75 = .885 \, kg \] --- ### **Step 2: Calculate Final Mass of Air** \[ m_2 = \rho_2 \cdot V \] \[ m_2 = 4.95 \times .75 = 3.7125 \, kg \] --- ### **Step 3: Mass of Air Entered** \[ \Delta m = m_2 - m_1 \] \[ \Delta m = 3.7125 - .885 = 2.8275 \, kg \] --- ### **Final Answer for 8(a):** \[ \boxed{2.83 \, \text{kg}} \] of air has entered the tank (rounded to two decimal places). --- ## **8(b) Force on the Elbow Flanges** ### **Given:** - Water flow rate, \( Q = .16 \, m^3/s \) - Area, \( A = .015 \, m^2 \) - \( D = 30 \, cm = .3 \, m \) - \( d = 10 \, cm = .1 \, m \) - \( h = 50 \, cm = .5 \, m \) - Volume = \( .03 \, m^3 \) - Angle = \( 60^\circ \) - Ignore weight of elbow material, water in elbow, and friction. ### **Required:** Determine the force acting on the flanges of the elbow. --- ### **Step 1: Find Water Velocity** \[ Q = A \cdot v \implies v = \frac{Q}{A} \] \[ v = \frac{.16}{.015} = 10.67 \, m/s \] --- ### **Step 2: Calculate the Force Using Momentum Equation** The force exerted by the fluid on the elbow is based on the change in momentum (linear momentum equation): \[ \vec{F} = \dot{m} (\vec{v}_{out} - \vec{v}_{in}) \] Where: - \(\dot{m} = \rho Q\) - Water density \(\rho \approx 100 \, kg/m^3\) #### **Inlet and Outlet Velocities:** Let’s assume: - Inlet is horizontal (x-direction) - Outlet is downward at \(60^\circ\) to the horizontal #### **Velocity Components:** - Inlet: \( \vec{v}_{in} = v \hat{i} \) - Outlet: \( \vec{v}_{out} = v (\cos 60^\circ \hat{i} - \sin 60^\circ \hat{j}) \) \[ \cos 60^\circ = .5,\ \sin 60^\circ = .866 \] So: - \( \vec{v}_{out} = v (.5 \hat{i} - .866 \hat{j}) \) #### **Change in Velocity:** \[ \vec{v}_{out} - \vec{v}_{in} = v (.5 \hat{i} - .866 \hat{j}) - v \hat{i} = v(-.5 \hat{i} - .866 \hat{j}) \] #### **Calculate Mass Flow Rate:** \[ \dot{m} = \rho Q = 100 \times .16 = 160 \, kg/s \] --- ### **Step 3: Calculate Force Components** \[ \vec{F} = \dot{m} \times [(-.5 v) \hat{i} + (-.866 v) \hat{j}] \] Plug in values: - \( v = 10.67 \, m/s \) - \( \dot{m} = 160 \, kg/s \) \[ F_x = 160 \times (-.5) \times 10.67 = -853.6 \, N \] \[ F_y = 160 \times (-.866) \times 10.67 = -1,478.5 \, N \] --- ### **Step 4: Magnitude of Total Force** \[ F = \sqrt{F_x^2 + F_y^2} \] \[ F = \sqrt{(-853.6)^2 + (-1,478.5)^2} \] \[ F = \sqrt{728,636.96 + 2,186,003.25} = \sqrt{2,914,640.21} \] \[ F \approx 1,707 \, N \] --- ### **Final Answer for 8(b):** - **Horizontal force (\(x\)):** \(\boxed{-854 \, N}\) (to the left) - **Vertical force (\(y\)):** \(\boxed{-1,479 \, N}\) (downward) - **Magnitude:** \(\boxed{1,707 \, N}\) --- ### **Summary Table** | Part | Step | Expression | Value/Answer | |---|---|---|---| | 8(a) | Mass entered | \(m_2 - m_1\) | \(2.83 \, kg\) | | 8(b) | Force (\(x\)) | \(-.5 \dot{m} v\) | \(-854 \, N\) | | 8(b) | Force (\(y\)) | \(-.866 \dot{m} v\) | \(-1,479 \, N\) | | 8(b) | Magnitude | \( \sqrt{F_x^2 + F_y^2} \) | \(1,707 \, N\) | --- Let me know if you'd like a detailed diagram or further clarification on any step!