91Ó°ÊÓ

The Bernoulli's equation for an incompressible fluid can be written 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 \(P_1\) and \(P_2\) are pressures of fluid at points 1 and 2, \(\rho\) is the fluid density, \(v_1\) and \(v_2\) are the fluid speeds at points 1 and 2, \(h_1\) and \(h_2\) are the height of the fluid at points 1 and 2, respectively. Since the problem states that the initial flow speed is negligible, we can assume \(v_1 = 0\). And since its the same fluid and the heights are not given, the term \(\rho g h\) can be dropped (as it would be equal for both the points). The equation will then reduce to: $$ P_1 = P_2 + \frac{1}{2} \rho v_2^2 $$

To find the flow speed at which water begins to boil, we will first calculate the boiling pressure by adding the vapor pressure to the initial pressure (remember that the boiling pressure is the pressure at which the liquid transitions into a vapor). The vapor pressure is given to be \(2.3388 \mathrm{kPa}\), and the initial pressure is \(101.3 \mathrm{kPa}\). Therefore, $$ P_{boil} = P_1 - P_{vapor} = 101.3 - 2.3388 = 98.9612\, \mathrm{kPa} $$

Now that we have the boiling pressure, we can use Bernoulli's equation to calculate the flow speed at the point where cavitation begins. Substitute the boiling pressure for \(P_2\) in the equation from step 1: $$ P_1 = P_{boil} + \frac{1}{2} \rho v_2^2 $$ We are given the initial pressure \(P_1 = 101.3 \mathrm{kPa}\), the density of water \(\rho = 998.2 \mathrm{~kg} / \mathrm{m}^{3}\), so we can solve for the flow speed \(v_2\). Rearrange the equation to solve for \(v_2\): $$ v_2 = \sqrt{\frac{2(P_1 - P_{boil})}{\rho}} $$ Plug in the known values: $$ v_2 = \sqrt{\frac{2(101.3 - 98.9612) \times 10^3}{998.2}} = \sqrt{\frac{4676}{998.2}} $$ Calculate the flow speed: $$ v_2 \approx 3.05 \mathrm{m/s} $$ The water will begin to boil when its flow speed reaches \(3.05 \mathrm{m/s}\) due to cavitation.