1. The problem statement, all variables and given/known data A 1 cm radius pipe connects a 3 m tall water tower to a faucet. (a) What is the gauge pressure at the faucet when the faucet is closed (b)12 What is the gauge pressure at the faucet if the faucet is opene d so that there is a flow of 0.2 Liter/s? Assume that the water flow is laminar. 2. Relevant equations P_1 + .5d(v_1)^2 + d(y_1) = P_2 + .5d(v_2)^2 + d(y_2) d = density , P = pressure ,y = height, v = velocity (A_1)(v_1) = (A_2)(v_2) A = area v = velocity 3. The attempt at a solution for part a) if the faucet is closed, gauge pressure at the faucet would simply be P = dgy = (1000 kg/m^2)(9.8 m/s^2)(3 m) for part b) i must use bernoulli's equation. looking at the picture i drew I can say P_2 is 0, v_2 is 0 (because it is negligible), y_2 = 3 m , and y_1 = 0 to find v_1 : use Av where v = l/t Av = A(l/t) = Vol/t (which they give us as .2 liter/s = 2E-4 m^3/s) A = pi r^2 = pi (.01 m) ^2 = 3.14 E-4 Av = Vol/t (3.14E-4)v = 2E-4 v = 2/3.14 = .637 = v_1 using bernoullis: P_1 + .5d(v_1)^2 + d(y_1) = P_2 + .5d(v_2)^2 + d(y_2) P_1 + .5d(.637)^2 + d(0) = 0 + .5d(0)^2 + d(3) P_1 + .5(1000)(.637)^2 = 3(1000) P_1 = 3000 - .5(1000)(.637)^2 im not as worried about the actual numbers but is my method correct?