Question

3. Water is pumped at an initial pressure of 60.0 psi in the basement to a vertical height of 21.0 m on the eighth floor of an apartment building. The pipe is 8.5 inches in diameter in the basement, but in the eighth floor apartments, the pipes are only 0.75 inches in diameter. a) What is the pressure of the water in the pipes on the eighth floor if all the spigots are closed? b) What is the velocity of the water in the basement pipe and what is the velocity in an eighth floor pipe il one spigot is open? c) What is the velocity of the water in the basement pipe if 10 eighth floor spigots are open and water is flowing from all?

Answer 1

Initial pressure Pi = 60psi = 413.685 kPa height = 21m. using Bernaulli's principle we get Pi + 2 put + lgh, = P2 + 1 9 v 2 ² + Sgh- hlo o denotes basement © denotes height h=21m. P, = 413.685 kPa P = density - 1000 kg/m3. g= 9.8 m/s2 .: 413.685 kPa. + 1x 1000 x v? P2 + 1x lovo vez? + 10oo x 9.0 x 21. 415770 -22 - V2² - v, 2. we get Balancing continuity equation Aivi = A₂ V2. I divi = Tid V2

V= dz? V₂ 415770-26- looo di = 8.5 in 0.216 m g de = 0.75 in = 0.02 m.ro ff all spigots closed ::U, = V₂ = 0 a) :: 415770-2P2 - ocrona 1000 P2 = 207.895 kPa If I spigot is opened b) 416270 -62 - V - 20.4 mle. Vi= 0.175 m/s.

If 10 spigots are opened. A, V = 10 A, V2. 10 time as crossectional area becomes .: Vol = 10 da V2. :. 415.770 d, 2 . V, - 21.31 mis . V = 1.83 m/s.