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 find the speed of the water jet emerging from the nozzle, we need to apply Bernoulli's principle, which relates the speed of fluid flow to pressure and height in a fluid system. Using Bernoulli’s equation, the total mechanical energy per unit volume of the fluid remains constant. We consider two points: the pump outlet and the nozzle. 1. At the pump outlet: - Pressure = \( P_{\text{pump}}^{(\text{gauge})} = 464.707 \, \text{kPa} \) - Height = 0 m - Velocity = \( v_1 = 0 \) (assuming it starts from rest initially due to the pressure conversion) 2. At the nozzle: - Pressure = \( P_{\text{nozzle}} = 0 \) (atmospheric pressure) - Height = \( h = 14 \, \text{m} \) - Velocity = \( v_2 \) (unknown, what we want to find) Now, we can rewrite Bernoulli’s equation in terms of these points: \[ P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \] Putting in the values, we have: \[ 464707 \, \text{Pa} + 1000 \cdot 9.8 \cdot 0 = 0 + 1000 \cdot 9.8 \cdot 14 + \frac{1}{2} \cdot 1000 \cdot v_2^2 \] This simplifies to: \[ 464707 = 0 + 137200 + 500 v_2^2 \] Now, rearranging gives us: \[ 464707 - 137200 = 500 v_2^2 \] \[ 327507 = 500 v_2^2 \] Dividing by 500: \[ v_2^2 = \frac{327507}{500} = 655.014 \] Finally, take the square root to find \( v_2 \): \[ v_2 = \sqrt{655.014} \approx 25.62 \, \text{m/s} \] Therefore, the speed of the water jet emerging from the nozzle is approximately \( 25.62 \, \text{m/s} \).