\( 018 \quad 10.0 \) 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 \[ P_{\text {pump }}^{\text {(gauge })}=P_{\text {pump }}^{(\text {abs })}-P_{\text {atm }} \] \( =67.4 \mathrm{PSI}=464.707 \mathrm{kPa} \). Calculate the speed of the water jet emerg- ing from the nozzle. Assume 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 \( \mathrm{m} / \mathrm{s} \).
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To find the speed of the water jet emerging from the nozzle, we can use Bernoulli's equation: \[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \] where: - \( P_1 \) is the gauge pressure at the pump, - \( v_1 \) is the speed of water at the pump (which we can assume to be negligible), - \( h_1 \) is the height of the pump above a reference level (14 m below the nozzle), - \( P_2 \) is atmospheric pressure (does not affect this calculation as we're using gauge pressure), - \( v_2 \) is the speed of the water coming out of the nozzle, - \( h_2 \) is the height at the nozzle which is 14 m above the pump. Since we can ignore \( v_1 \) and use the height difference, the equation simplifies to: \[ P_{\text{pump}}^{\text{(gauge)}} + \rho g h_1 = \frac{1}{2} \rho v_2^2 + \rho g h_2 \] Plugging in the values: - \( P_{\text{pump}}^{\text{(gauge)}} = 464.707 \, \text{kPa} = 464707 \, \text{Pa} \) - \( \rho = 1000 \, \text{kg/m}^3 \) - \( g = 9.8 \, \text{m/s}^2 \) - \( h_1 = 0 \, \text{m} \) - \( h_2 = 14 \, \text{m} \) Thus, rewriting the equation gives: \[ 464707 + 1000 \cdot 9.8 \cdot 0 = \frac{1}{2} \cdot 1000 \cdot v_2^2 + 1000 \cdot 9.8 \cdot 14 \] Calculating the gravitational potential energy term: \[ 1000 \cdot 9.8 \cdot 14 = 137200 \, \text{Pa} \] Now we can plug this back into the equation: \[ 464707 = \frac{1}{2} \cdot 1000 \cdot v_2^2 + 137200 \] Rearranging gives: \[ 464707 - 137200 = \frac{1}{2} \cdot 1000 \cdot v_2^2 \] Thus: \[ 327507 = \frac{1}{2} \cdot 1000 \cdot v_2^2 \] \[ v_2^2 = \frac{2 \cdot 327507}{1000} = 655.014 \] And finally: \[ v_2 = \sqrt{655.014} \approx 25.6 \, \text{m/s} \] The speed of the water jet emerging from the nozzle is approximately \( 25.6 \, \text{m/s} \).
