Let's break this problem down step by step: --- ## **Given Data** - **Fluid:** 10°C liquid water (density, ρ ≈ 100 kg/m³) - **A₁ =** .8 m² (Inlet area) - **A₂ =** .2 m² (Outlet area) - **V₁ =** 10 m/s (Inlet velocity) - **p₁ =** p_atm + 20 kPa (Inlet pressure) - **p₂ =** p_atm (Outlet pressure) - **Outlet angle:** 45° - **Neglect gravity** --- ## **Step 1: Find V₂ using Continuity Equation** \[ Q = A_1 V_1 = A_2 V_2 \implies V_2 = \frac{A_1}{A_2} V_1 \] \[ V_2 = \frac{.8}{.2} \times 10 = 4 \times 10 = 40\, \text{m/s} \] --- ## **Step 2: Find Mass Flow Rate (\(\dot{m}\))** \[ \dot{m} = \rho Q = \rho A_1 V_1 = 100 \times .8 \times 10 = 800\, \text{kg/s} \] --- ## **Step 3: Apply Linear Momentum Equation** The control volume is the nozzle. The forces needed to hold it stationary are: \[ \vec{F} = \sum (\text{Momentum out}) - \sum (\text{Momentum in}) + \sum(\text{Pressure forces}) \] ### **Component Form:** #### **X-direction:** \[ F_x = (\dot{m} V_{2x}) - (\dot{m} V_{1x}) + (p_1 A_1 - p_2 A_2 \cos 45^\circ) \] #### **Y-direction:** \[ F_y = (\dot{m} V_{2y}) - (\dot{m} V_{1y}) - (p_2 A_2 \sin 45^\circ) \] --- ## **Step 4: Calculate Velocity Components** - At inlet (1): \(V_{1x} = V_1 = 10\, \text{m/s}\), \(V_{1y} = \) - At outlet (2): \(V_{2x} = V_2 \cos 45^\circ = 40 \times \frac{1}{\sqrt{2}} = 28.28\, \text{m/s}\) - \(V_{2y} = V_2 \sin 45^\circ = 40 \times \frac{1}{\sqrt{2}} = 28.28\, \text{m/s}\) --- ## **Step 5: Pressure Forces** - At inlet: \(p_1 A_1\) - At outlet, only atmospheric acts, but since both outside and at outlet is \(p_{atm}\), the net force due to atmospheric cancels except for the extra \(20\, \text{kPa}\) at the inlet. \[ p_1 - p_{atm} = 20,000\, \text{Pa} \] --- ## **Step 6: Plug in the Numbers** ### **X-direction:** \[ F_x = (\dot{m} V_{2x}) - (\dot{m} V_{1x}) + [p_1 A_1 - p_2 A_2 \cos 45^\circ] \] Since \(p_2 = p_{atm}\) and the outside is also at \(p_{atm}\), the \(p_2\) force cancels with the atmospheric pressure on the nozzle, so only the gauge pressure at inlet matters. \[ F_x = (800 \times 28.28) - (800 \times 10) + [20,000 \times .8] \] \[ = (226,240) - (80,000) + 16,000 \] \[ = 146,240 + 16,000 = 162,240\, \text{N} \] \[ F_x = 162.24\, \text{kN} \] --- ### **Y-direction:** \[ F_y = (\dot{m} V_{2y}) - (\dot{m} V_{1y}) - (p_2 A_2 \sin 45^\circ) \] But again, \(p_2 = p_{atm}\), and atmospheric cancels, so just use the momentum: \[ F_y = 800 \times 28.28 - \] \[ = 226,240\, \text{N} \] \[ F_y = 226.24\, \text{kN} \] --- ## **Final Answers** \[ \boxed{ F_x = 162.24\, \text{kN} \\ F_y = 226.24\, \text{kN} } \] --- ### **Summary Table** | Direction | Force (kN) | |-----------|------------| | \(F_x\) | 162.24 | | \(F_y\) | 226.24 | --- **Let me know if you want a breakdown of the pressure force terms or other clarifications!**