Let's solve the problem step by step: --- ## **Given Data** - **Mass flow rate (\(\dot{m}\))** = 40 kg/s - **Density (\(\rho\))** = 100 kg/m³ - **Diameter (\(D\))** = .1 m - **Elevation difference (\(\Delta z\))** = .5 m - **Atmospheric pressure at exit (\(P_{atm}\))** - **Momentum-flux correction factor (\(\beta\))** = 1.03 - **Neglect weight of elbow and water in it** --- ## **Step 1: Find Velocity at Inlet and Outlet** Both inlet and outlet have the same diameter (.1 m): - Area, \(A = \frac{\pi D^2}{4} = \frac{\pi (.1)^2}{4} = .00785 \, \text{m}^2 \) - Volumetric flow rate, \(Q = \frac{\dot{m}}{\rho} = \frac{40}{100} = .04 \, \text{m}^3/\text{s}\) - Velocity, \(V = \frac{Q}{A} = \frac{.04}{.00785} \approx 5.10 \, \text{m/s}\) --- ## **Step 2: Apply Energy Equation (Bernoulli) for Pressure at Inlet** \[ \frac{P_1}{\rho g} + \frac{\alpha_1 V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{\alpha_2 V_2^2}{2g} + z_2 \] Given: - \(P_2 = P_{atm}\) (atmospheric at exit) - \(z_2 - z_1 = .5\) m (exit is higher) - \(\alpha_1 = \alpha_2 = 1.03\) - \(V_1 = V_2\) - Ignore losses So, \[ \frac{P_1}{\rho g} + \frac{1.03 V^2}{2g} = \frac{P_{atm}}{\rho g} + \frac{1.03 V^2}{2g} + .5 \] \[ \frac{P_1}{\rho g} = \frac{P_{atm}}{\rho g} + .5 \] \[ P_1 = P_{atm} + \rho g (.5) \] **Gage pressure** (relative to atmosphere): \[ P_{gage,1} = P_1 - P_{atm} = \rho g (.5) = 100 \times 9.81 \times .5 = 4905 \, \text{Pa} \] --- ### **(a) Gage Pressure at Inlet** \[ \boxed{P_{gage,1} = 4905 \, \text{Pa}} \] --- ## **Step 3: Find Anchoring Force (Control Volume Momentum Analysis)** ### **Momentum Equation (Control Volume, Steady Flow)** \[ \sum F_{ext,x} = \dot{m} (V_{2x} - V_{1x}) \] \[ \sum F_{ext,y} = \dot{m} (V_{2y} - V_{1y}) \] Let’s define: - \(x\)-axis: original pipe (horizontal) - \(y\)-axis: vertical (upward exit) ### **Forces Acting:** - Pressure force at inlet: \(F_{P1}\) (in \(+x\)) - Pressure force at exit: \(F_{P2}\) (in \(+y\)) - Force by the elbow anchor: \(F_{anchor}\) (opposes resultant) - Ignore weight #### **Pressure Forces:** - At inlet: \(P_1 \cdot A\), in \(+x\) direction - At exit: \(P_2 \cdot A\), in \(+y\) direction (\(P_2 = P_{atm}\), so cancels with atmospheric pressure outside) #### **Net Pressure Forces (relative to atmospheric):** - At inlet: \(P_{gage,1} \cdot A\), in \(+x\) - At exit: (since \(P_2 = P_{atm}\)) #### **Momentum Flux:** - \(V_1 = 5.10~\text{m/s}\) in \(+x\) - \(V_2 = 5.10~\text{m/s}\) in \(+y\) - Correction factor \(\beta = 1.03\) \[ \dot{m} = 40~\text{kg/s} \] #### **x-direction:** \[ \sum F_x = P_{gage,1} A - F_{anchor,x} = \beta \dot{m} ( - V_1) \] \[ P_{gage,1} A - F_{anchor,x} = -\beta \dot{m} V_1 \] \[ F_{anchor,x} = P_{gage,1} A + \beta \dot{m} V_1 \] #### **y-direction:** \[ \sum F_y = - F_{anchor,y} = \beta \dot{m} (V_2 - ) \] \[ -F_{anchor,y} = \beta \dot{m} V_2 \] \[ F_{anchor,y} = -\beta \dot{m} V_2 \] (Negative sign: the anchor must pull downward to counteract upward flow.) --- ### **Area Calculation** \[ A = .00785 \, \text{m}^2 \] --- ### **Plug in Values:** #### **x-direction:** \[ F_{anchor,x} = (4905) \times (.00785) + 1.03 \times 40 \times 5.10 \] \[ = 38.5 + 210.12 \approx 248.6\, \text{N} \] #### **y-direction:** \[ F_{anchor,y} = -1.03 \times 40 \times 5.10 = -210.12 \, \text{N} \] --- ### **Magnitude of the Anchoring Force** \[ F_{anchor} = \sqrt{F_{anchor,x}^2 + F_{anchor,y}^2} \] \[ = \sqrt{(248.6)^2 + (210.12)^2} \] \[ = \sqrt{61803 + 44150} \] \[ = \sqrt{105953} \approx 325.5\, \text{N} \] --- ## **Final Answers** ### **(a) Gage pressure at the inlet:** \[ \boxed{P_{gage,1} = 4905~\text{Pa}} \] ### **(b) Anchoring force needed:** \[ \boxed{F_{anchor} = 325.5~\text{N}} \] (directed downstream and downward) --- If you have any questions on the steps, let me know!