Question

Consider water flowing to three faucets in a building through the very simple plumbing network shown below. Every section of pipe in the figure is 25 ft long. Ali of the piping is nominal half-inch type L copper piping. The inlet pressure to the main pipe is constant at 50 psi and the pressure of the water leaving each faucet is 0 psi. The water has a temperature of 140°F. For the purposes of this analysis assume that there are no pressure losses in the fittings (elbows, tees, etc.) and that the only pressure losses are given by the following equation D 2 where f= 0.02, L is the length of pipe, D is the pipe diameter, Vis the water velocity, and ρ is the water density. What is the water flow rate in gallons per minute from each faucet A, B, and C? 5 ft 25 ft 25 ft 25 ft

Answer 1

f = 0.02 L = 25� T = 140 degrees F (60 degrees C)o pi = 0.9832 gm/cc 62.4511/ft^3 delta p = (f.L.v^2/2D) () implies delta p = f (v^2) (50 - 0)/(1/144) = 0.02 times (25 times 2) times (v)^2 times 62.45/2 times (0.5/12) v = 3.1 ft/sec QA = pi/4 (0.5/12)^2 times 3.1 = 4.227 times 10^-3 ft^3/sec (or) Q_A = 1.9 gallons/min (50 - 0) times 144 = 0.02 times (25 times 3) times v^2 times 62.45/2 times (0.5/12) v = 2.53 ft/sec Q_B = pi/4 (0.5/12)^2 times 2.53 = 3.45 times 10^-3 ft^3/Sec = 1.548 gal/min (50 - 0) times 144 = 0.02 times (25 times 4) times v^2 times 62.45/2 times (0.5/12) v = 2.19 ft/sec QC = pi/4 (0.5/12)^2 times 2.19 = 3 times 10^-3 ft^3/sec = 1.346 gal/min