Question

2-in.-diameter hose 2 ft (3) FIG. P3 Water is siphoned from the tank shown in Fig. P3. Determine (a) the flowrate from the tank, and (b) the pressure at point (1), (2) and (3). (c) The current maximum height of the hose is 2 ft above the water surface. What can the maximum height be without having any cavitation. Take the water temperature to be 60°F.

Answer 1

@ At roof, Pr= 1,767 kPa = 0.2563 psi Now, d=2in = 0.1667 ft - Applying Bernouli Principle at exit of pipe a top of - water level in tank. 3 2x32:2 . Ve = 13.9 ft/s flow have (Ⓡ) > VA "h* 6.167) 3 13.9 1. To= 0.3033 4/ / An @

② Prosure of points . . . 8* 32:2* 1.94 499.2 bbl ft2 TP,= 3.467 psi New Applying Bernout principle at point o r o 2 a I - y U₂2 +2, 24 {2,=722 - Zv,zo } 3.467x144 6204 . + _+ (13-9)? Loon + $ 8 = 12.4 232.2 1 P2 = 312.038 lb/ft2 P₂ = 2.167 psi

Applying Bernout frinciple at point ® a top senjace of Walen A. 2 o foto 62.4 2x32.2 to= en + 43.92 +0 - 0+0+0 - TB = -1.3 psi - ® het maximum height be 'n'. Applying Bernout principe at top of sophon & exit of water. (6.25637147)x145 +(37h) sh=30:3 = 0+(13.902 249.81 + 0 62.4 2x987 The 303 feet 7 An. ©

values of pressure were calculated as gauge pressure, for absolute pressure add atmospheric pressure i.e. 14.7 psi to the answers. tha ks