# Step-by-Step Solution --- ## Given Data | Symbol | Meaning | Value | | :------------------------- | :-------------------------------------- | :------------------------------------------------- | | \(\dot{m}\) | mass flow rate | \(.25\ \{kg/s) | | \(\rho\) | density of water | \(100\ \text{kg/m}^3\) | | \(P_1\) | pressure before pump | \(110\ \text{kPa} = 110{,}000\ \text{Pa}\) | | \(P_2\) | pressure at faucet exit | \(100\ \text{kPa} = 100{,}000\ \text{Pa}\) | | \(z_2 - z_1\) | height difference | \(10\ \text{m}\) | | \(D_2\) | faucet diameter | \(2\ \text{cm} = .02\ \text{m}\) | | \(g\) | gravitational acceleration | \(9.8\ \text{m/s}^2\) | | Flow type | Steady, incompressible, adiabatic | - | --- ## (a) Find the outlet velocity \(v_2\) ### Step 1: Cross-sectional area at faucet \[ A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (.02)^2}{4} = \frac{\pi \times .0004}{4} \approx 3.142 \times 10^{-4}\ \text{m}^2 \] ### Step 2: Velocity at faucet \[ v_2 = \frac{\dot{m}}{\rho A_2} = \frac{.25}{100 \times 3.142 \times 10^{-4}} \approx .796\ \text{m/s} \] ### **Final answer:** \[ \boxed{ v_2 \approx .80\ \text{m/s} } \] --- ## (b) Find the shaft power supplied by the pump ### Step 1: Apply Bernoulli's equation with pump work \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 + \frac{W_{\text{pump}}}{g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \] Rearranged for \(W_{\text{pump}}\): \[ W_{\text{pump}} = \left( \frac{P_2 - P_1}{\rho} \right) + \frac{v_2^2 - v_1^2}{2} + g (z_2 - z_1) \] Assuming \(v_1 \approx \): \[ W_{\text{pump}} = \frac{100{,}000 - 110{,}000}{100} + \frac{(.796)^2 - ^2}{2} + 9.8 \times 10 \] Calculate each term: - Pressure difference term: \[ \frac{-10{,}000}{100} = -10\ \text{J/kg} \] - Velocity term: \[ \frac{(.796)^2}{2} = \frac{.634}{2} = .317\ \text{J/kg} \] - Height difference term: \[ 9.8 \times 10 = 98\ \text{J/kg} \] ### Step 2: Total work per unit mass \[ W_{\text{pump}} = -10 + .317 + 98 = 88.317\ \text{J/kg} \] ### Step 3: Power supplied by the pump \[ \dot{W} = \dot{m} \times W_{\text{pump}} = .25 \times 88.317 \approx 22.1\ \text{W} \] ### **Final answer:** \[ \boxed{ \dot{W}_{\text{pump}} \approx 22.1\ \text{W} } \] --- ## **Summary of Results** | Quantity | Symbol | Result | | :--------------------------- | :------------------------------ | :---------------- | | Outlet velocity at faucet | \(v_2\) | **.80 m/s** | | Pump shaft power | \(\dot{W}_{\text{pump}}\) | **22.1 W** |

# Step-by-Step Solution --- ## Given Data | Symbol | Meaning | Value | | :------------------------- | :-------------------------------------- | :------------------------------------------------- | | \(\dot{m}\) | mass flow rate | \(0.25\ \text{kg/s}\) | | \(\rho\) | density of water | \(1000\ \text{kg/m}^3\) | | \(P_1\) | pressure before pump | \(110\ \text{kPa} = 110{,}000\ \text{Pa}\) | | \(P_2\) | pressure at faucet exit | \(100\ \text{kPa} = 100{,}000\ \text{Pa}\) | | \(z_2 - z_1\) | height difference | \(10\ \text{m}\) | | \(D_2\) | faucet diameter | \(2\ \text{cm} = 0.02\ \text{m}\) | | \(g\) | gravitational acceleration | \(9.8\ \text{m/s}^2\) | | Flow type | Steady, incompressible, adiabatic | - | --- ## (a) Find the outlet velocity \(v_2\) ### Step 1: Cross-sectional area at faucet \[ A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.02)^2}{4} = \frac{\pi \times 0.0004}{4} \approx 3.142 \times 10^{-4}\ \text{m}^2 \] ### Step 2: Velocity at faucet \[ v_2 = \frac{\dot{m}}{\rho A_2} = \frac{0.25}{1000 \times 3.142 \times 10^{-4}} \approx 0.796\ \text{m/s} \] ### **Final answer:** \[ \boxed{v_2 \approx 0.80\ \text{m/s}} \] --- ## (b) Find the shaft power supplied by the pump ### Step 1: Apply Bernoulli's equation with pump work \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 + \frac{W_{\text{pump}}}{g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \] Rearranged for \(W_{\text{pump}}\): \[ W_{\text{pump}} = \left( \frac{P_2 - P_1}{\rho} \right) + \frac{v_2^2 - v_1^2}{2} + g (z_2 - z_1) \] Assuming \(v_1 \approx 0\): \[ W_{\text{pump}} = \frac{100{,}000 - 110{,}000}{1000} + \frac{(0.796)^2}{2} + 9.8 \times 10 \] ### Step 2: Calculate each term - Pressure difference term: \[ \frac{-10{,}000}{1000} = -10\ \text{J/kg} \] - Velocity term: \[ \frac{(0.796)^2}{2} = \frac{0.634}{2} = 0.317\ \text{J/kg} \] - Height difference term: \[ 9.8 \times 10 = 98\ \text{J/kg} \] ### Step 3: Total work per unit mass \[ W_{\text{pump}} = -10 + 0.317 + 98 = 88.317\ \text{J/kg} \] ### Step 4: Power supplied by the pump \[ \dot{W} = \dot{m} \times W_{\text{pump}} = 0.25 \times 88.317 \approx 22.1\ \text{W} \] ### **Final answer:** \[ \boxed{\dot{W}_{\text{pump}} \approx 22.1\ \text{W}} \] --- ## **Summary of Results** | Quantity | Symbol | Result | | :--------------------------- | :------------------------------ | :---------------- | | Outlet velocity at faucet | \(v_2\) | **0.80 m/s** | | Pump shaft power | \(\dot{W}_{\text{pump}}\) | **22.1 W** |

# Given Data - **Mass flow rate** \(\dot{m} = 0.25\ \text{kg/s}\) - **Density of water** \(\rho = 1000\ \text{kg/m}^3\) - **Pressure before pump** \(P_1 = 110\ \text{kPa} = 110{,}000\ \text{Pa}\) - **Pressure at faucet exit** \(P_2 = 100\ \text{kPa} = 100{,}000\ \text{Pa}\) - **Height difference** \(z_2 - z_1 = 10\ \text{m}\) - **Faucet diameter** \(D_2 = 2\ \text{cm} = 0.02\ \text{m}\) - **Gravitational acceleration** \(g = 9.8\ \text{m/s}^2\) - **Flow type**: Steady, incompressible, adiabatic --- # (a) Find the Outlet Velocity \(v_2\) ## Step 1: Cross-Sectional Area at Faucet \[ A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.02)^2}{4} \approx 3.142 \times 10^{-4}\ \text{m}^2 \] ## Step 2: Velocity at Faucet \[ v_2 = \frac{\dot{m}}{\rho A_2} = \frac{0.25}{1000 \times 3.142 \times 10^{-4}} \approx 0.796\ \text{m/s} \] ### **Final Answer:** \[ \boxed{v_2 \approx 0.80\ \text{m/s}} \] --- # (b) Find the Shaft Power Supplied by the Pump ## Step 1: Apply Bernoulli's Equation with Pump Work \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 + \frac{W_{\text{pump}}}{g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \] Rearranged for \(W_{\text{pump}}\): \[ W_{\text{pump}} = \left( \frac{P_2 - P_1}{\rho} \right) + \frac{v_2^2 - v_1^2}{2} + g (z_2 - z_1) \] Assuming \(v_1 \approx 0\): \[ W_{\text{pump}} = \frac{100{,}000 - 110{,}000}{1000} + \frac{(0.796)^2}{2} + 9.8 \times 10 \] ## Step 2: Calculate Each Term - **Pressure Difference Term**: \[ \frac{-10{,}000}{1000} = -10\ \text{J/kg} \] - **Velocity Term**: \[ \frac{(0.796)^2}{2} = \frac{0.634}{2} \approx 0.317\ \text{J/kg} \] - **Height Difference Term**: \[ 9.8 \times 10 = 98\ \text{J/kg} \] ## Step 3: Total Work per Unit Mass \[ W_{\text{pump}} = -10 + 0.317 + 98 = 88.317\ \text{J/kg} \] ## Step 4: Power Supplied by the Pump \[ \dot{W} = \dot{m} \times W_{\text{pump}} = 0.25 \times 88.317 \approx 22.1\ \text{W} \] ### **Final Answer:** \[ \boxed{\dot{W}_{\text{pump}} \approx 22.1\ \text{W}} \] --- # Summary of Results - **Outlet velocity at faucet** \(v_2 \approx 0.80\ \text{m/s}\) - **Pump shaft power** \(\dot{W}_{\text{pump}} \approx 22.1\ \text{W}\)