91Ó°ÊÓ

First, let's apply Bernoulli's equation between the pump's output pipe (point 1) and the showerhead (point 2). The equation we will use is: \(P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2\) In our case, we can set \(h_1=0\), and \(h_2=6.7\mathrm{m}\). We are also given the pump's gauge pressure \(P_1=410\mathrm{kPa}\). The atmospheric pressure is present at both points, so it is canceled out when calculating relative pressure. Thus, \(P_2=0\). Now the equation becomes: \(410000+\frac{1}{2}\rho v_1^2=\frac{1}{2}\rho v_2^2+\rho gh_2\) Assuming the water density \(\rho=1000\mathrm{kg/m^3}\), we can now solve for \(v_2\), the speed of water leaving the showerhead. Rearrange the equation to solve for \(v_2\): \(v_2=\sqrt{2((410000+\frac{1}{2}\rho v_1^2)-\rho gh_2)/\rho}\) Now we have all constants but \(v_1\) in this formula, so let's move to continuity equation.

The equation of continuity states that the mass flow rate must be conserved in a flow, which implies that the product of the area and velocity is a constant along the streamline. In mathematical terms, this means: \(A_1v_1=A_2v_2\) Let the total area of the showerhead be \(A_2\). The total area is given by the sum of the areas of all 36 holes: \(A_2=36\pi (0.33\mathrm{mm})^2\) The area of the pump's output pipe \(A_1\) is given by: \(A_1=\pi (6.3\mathrm{mm})^2\) So, we have \(A_1v_1=36A_2v_2\), which means \(v_1=\frac{36A_2v_2}{A_1}\). Substitute this into the Bernoulli's equation previously calculated for \(v_2\) and solve for \(v_2\): \(v_2=\sqrt{2((410000+\frac{1}{2}\rho (\frac{36A_2v_2}{A_1})^2)-\rho gh_2)/\rho}\) Now we have all constants in the formula, and after calculations, we get the answer: \(v_2=5.52\mathrm{m/s}\) The speed of water leaving the showerhead is \(5.52\mathrm{m/s}\). #b) Calculating the speed of water moving through the output pipe of the pump#

Now we will apply the equation of continuity again to find the speed of water moving through the output pipe of the pump (\(v_1\)). We have already found the speed of water at the showerhead (\(v_2\)), so we can directly calculate \(v_1\) using the areas and the equation of continuity: \(v_1=\frac{A_2v_2}{A_1}\) We have already calculated the areas in part (a), so we can substitute the values to find \(v_1\): \(v_1=\frac{36\pi (0.33\mathrm{mm})^2(5.52\mathrm{m/s})}{\pi (6.3\mathrm{mm})^2}\) \(v_1=0.58\mathrm{m/s}\) The speed of water moving through the output pipe of the pump is \(0.58\mathrm{m/s}\).