# Water Flow in a Horizontal Garden Hose Let's break down the problem into three parts, step by step. --- ## Given Data - **Mass flow rate,** \(\dot{m} = 30~\text{kg/min}\) - **Inlet velocity,** \(V_1 = 1.5~\text{m/s}\) - **Outlet velocity,** \(V_2 = 14.5~\text{m/s}\) - **Momentum-flux correction factor,** \(\beta_1 = \beta_2 = 1.04\) - **Density of water,** \(\rho \approx 100~\text{kg/m}^3\) - **Hose is horizontal** (ignore gravity) - **Ignore weight of hose and water** - **Atmospheric pressure at outlet** (since the outlet is open to the atmosphere) --- ## (a) Calculate the Inlet Area **Step 1: Convert mass flow rate to SI units** \[ \dot{m} = 30~\text{kg/min} = \frac{30}{60}~\text{kg/s} = .5~\text{kg/s} \] **Step 2: Find volumetric flow rate** \[ Q = \frac{\dot{m}}{\rho} = \frac{.5~\text{kg/s}}{100~\text{kg/m}^3} = .0005~\text{m}^3/\text{s} \] **Step 3: Use continuity to find area at inlet** \[ Q = A_1 V_1 \implies A_1 = \frac{Q}{V_1} \] \[ A_1 = \frac{.0005~\text{m}^3/\text{s}}{1.5~\text{m/s}} = .000333~\text{m}^2 \] **Final Answer (a):** \[ \boxed{A_1 = 3.33 \times 10^{-4}~\text{m}^2} \] --- ## (b) Calculate the Gauge Pressure at the Hose Inlet ### Step 1: Apply the steady-flow energy equation (Bernoulli's equation with momentum correction) Since the hose is, and neglecting gravity and friction losses: \[ \frac{P_1}{\rho} + \beta1 \frac{V_1^2}{2} = \frac{P_2}{\rho} + \beta_2 \frac{V_2^2}{2} \] At the outlet, \(P_2\) is atmospheric, so gauge pressure is zero (\(P_2 = ~\text{gauge}\)). ### Step 2: Solve for \(P_1\) (gauge pressure) \[ P_1 = \rho \left[ \beta_2 \frac{V_2^2}{2} - \beta_1 \frac{V_1^2}{2} \right] \] Insert values: \[ P_1 = 100 \left[ 1.04 \cdot \frac{(14.5)^2}{2} - 1.04 \cdot \frac{(1.5)^2}{2} \right] \] Calculate each term: - \( (14.5)^2 = 210.25 \) - \( (1.5)^2 = 2.25 \) - So, \[ P_1 = 100 \left[ 1.04 \cdot \frac{210.25 - 2.25}{2} \right] \] \[ = 100 \left[ 1.04 \cdot \frac{208}{2} \right] = 100 \left[ 1.04 \cdot 104 \right] = 100 \times 108.16 = 108160~\text{Pa} \] **Final Answer (b):** \[ \boxed{P_1 = 1.08 \times 10^5~\text{Pa (gauge)}} \] --- ## (c) Calculate the Anchoring Force (Magnitude and Direction) ### Step 1: Use the Linear Momentum Equation The force required to hold the hose is equal to the net change in momentum flux plus the pressure forces: \[ F_x = \left[ \beta_2 \dot{m} V_2 - \beta_1 \dot{m} V_1 \right] + (P_2 A_2 - P_1 A_1) \] Since the outlet is open to the atmosphere, \(P_2 = \), and \(A_2 = Q / V_2\). #### Calculate \(A_2\): \[ A_2 = \frac{Q}{V_2} = \frac{.0005}{14.5} = 3.45 \times 10^{-5}~\text{m}^2 \] ### Step 2: Compute momentum terms \[ \beta_2 \dot{m} V_2 = 1.04 \times .5 \times 14.5 = 7.54~\text{N} \] \[ \beta_1 \dot{m} V_1 = 1.04 \times .5 \times 1.5 = .78~\text{N} \] \[ \text{Net momentum flux:}~7.54 - .78 = 6.76~\text{N} \] ### Step 3: Compute pressure force term \[ P_2 A_2 = \] \[ P_1 A_1 = 108160 \times .000333 = 36.03~\text{N} \] So, the pressure force acts **backward** (opposite to flow). ### Step 4: Total anchoring force \[ F_x = (\text{momentum flux}) - (\text{pressure force}) = 6.76 - 36.03 = -29.27~\text{N} \] The negative sign indicates the force must be applied **opposite to the direction of flow** (i.e., to the left if water is flowing to the right). **Final Answer (c):** \[ \boxed{ \begin{aligned} &\text{Magnitude:}~29.3~\text{N} \\ &\text{Direction: Opposite to the direction of water flow} \end{aligned} } \] --- ## **Summary Table** | Part | Quantity | Value | |------|------------------------|-------------------------------------| | (a) | Inlet area \(A_1\) | \(3.33 \times 10^{-4}\) m² | | (b) | Inlet gauge pressure | \(1.08 \times 10^5\) Pa | | (c) | Anchoring force | 29.3 N, opposite to water flow |