A house with its own well has a pump in the basement with an output pipe of inner radius \(6.3 \mathrm{mm}\). Assume that the pump can maintain a gauge pressure of \(410 \mathrm{kPa}\) in the output pipe. A shower head on the second floor (6.7 \(\mathrm{m}\) above the pump's output pipe) has 36 holes, each of radius \(0.33 \mathrm{mm} .\) The shower is on "full blast" and no other faucet in the house is open. (a) Ignoring viscosity, with what speed does water leave the shower head? (b) With what speed does water move through the output pipe of the pump?

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 \(0.58\mathrm{m/s}\).