018 10.0 points A fireman standing on a 14 m high ladder operates a water hose with a round nozzle of diameter 2.66 inch. The lower end of the hose \( (14 \mathrm{~m} \) below the nozzle) is connected to the pump outlet of diameter 3.34 inch . The gauge pressure of the water at the pump is \[ \begin{array}{l} P_{\text {pump }}^{(\text {gauge })}=P_{\text {pump }}^{(\text {(abs })}-P_{\text {atm }} \\ =67.4 \mathrm{PSI}=464.707 \mathrm{kPa} \text {. } \\ \text { Calculate the speed of the water jet emerg- }\end{array} \] ing from the nozzle. Assime that water is an incompressible liquid of density \( 1000 \mathrm{~kg} / \mathrm{m}^{3} \) and negligible viscosity. The acceleration of gravity is \( 9.8 \mathrm{~m} / \mathrm{s}^{2} \). Answer in units of \( \mathrm{m} / \mathrm{s} \).

To calculate the speed of the water jet emerging from the nozzle, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a streamline flow. The equation can be expressed as: \[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \] Where: - \( P_1 \) is the pressure at the pump, - \( v_1 \) is the velocity of the water at the pump (which we can assume to be negligible, \( v_1 \approx 0 \)), - \( h_1 \) is the height at the pump (which is \( 0 \) m), - \( P_2 \) is the pressure at the nozzle, - \( v_2 \) is the velocity of the water at the nozzle (which we want to find), - \( h_2 \) is the height at the nozzle (which is \( 14 \) m), - \( \rho \) is the density of water (\( 1000 \, \text{kg/m}^3 \)), - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)). ### Step 1: Convert the gauge pressure to absolute pressure The gauge pressure at the pump is given as \( P_{\text{pump}}^{(\text{gauge})} = 464.707 \, \text{kPa} \). The absolute pressure at the pump is: \[ P_{\text{pump}}^{(\text{abs})} = P_{\text{pump}}^{(\text{gauge})} + P_{\text{atm}} \] Assuming atmospheric pressure \( P_{\text{atm}} \approx 101.325 \, \text{kPa} \): \[ P_{\text{pump}}^{(\text{abs})} = 464.707 \, \text{kPa} + 101.325 \, \text{kPa} = 566.032 \, \text{kPa} \] ### Step 2: Apply Bernoulli's equation Substituting the known values into Bernoulli's equation: \[ 566032 + 0 + 0 = P_2 + \frac{1}{2} \cdot 1000 \cdot v_2^2 + 1000 \cdot 9.8 \cdot 14 \] ### Step 3: Calculate \( P_2 \) The pressure at the nozzle \( P_2 \) can be calculated as: \[ P_2 = P_{\text{atm}} = 101.325 \, \text{kPa} \text{ (since the water exits to the atmosphere)} \] ### Step 4: Substitute and solve for \( v_2 \) Now substituting \( P_2 \) back into the equation: \[ 566032 = 101325 + \frac{1}{2} \cdot 1000 \cdot v_2^2 + 1000 \cdot 9.8 \cdot 14 \] Calculating the potential energy term: \[ 1000 \cdot 9.8 \cdot 14 = 137200 \, \text{Pa} \] Now substituting this value into the equation: \[ 566032 = 101325 + \frac{1}{2} \cdot 1000 \cdot v_2^2 + 137200 \] Combining the terms: \[ 566032 = 238525 + \frac{1}{2} \cdot 1000 \cdot v_2^2 \] Subtracting \( 238525 \) from both sides: \[ 327507 = \frac{1}{2} \cdot 1000 \cdot v_2^2 \] Multiplying both sides by \( 2 \): \[ 655014 = 1000 \cdot v_2^2 \] Dividing by \( 1000 \): \[ v_2^2 = 655.014 \] Taking the square root: \[ v_2 = \sqrt{655.014} \] Now, let's calculate \( v_2 \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{655.014}\) - step1: Rewrite the expression: \(\sqrt{\frac{327507}{500}}\) - step2: Use the properties of radicals: \(\frac{\sqrt{327507}}{\sqrt{500}}\) - step3: Simplify the expression: \(\frac{\sqrt{327507}}{10\sqrt{5}}\) - step4: Multiply by the Conjugate: \(\frac{\sqrt{327507}\times \sqrt{5}}{10\sqrt{5}\times \sqrt{5}}\) - step5: Multiply the numbers: \(\frac{\sqrt{1637535}}{10\sqrt{5}\times \sqrt{5}}\) - step6: Multiply the numbers: \(\frac{\sqrt{1637535}}{50}\) The speed of the water jet emerging from the nozzle is approximately: \[ v_2 \approx 25.59 \, \text{m/s} \] Thus, the final answer is: \[ \text{Speed of the water jet} \approx 25.59 \, \text{m/s} \]